Newsgroups: sci.math From: hoey@zogwarg.etl.army.mil (Dan Hoey) Date: 4 Jan 93 16:49:23 GMT Subject: Re: Lunde's problem (least number with n divisors) shal...@graceland.uwaterloo.ca (Jeffrey Shallit) writes: >Basically, you're looking for the smallest integer M with n positive >*divisors* (although most mathematicians would include 1 and M in the >enumeration). >Beiler also sketches an algorithm that will find such numbers. To >the best of my knowledge, it is not known how to produce the prime >factorization of the solution in polynomial time (in log n, of course). If it were, we would have a polynomial-time factoring algorithm, since one plus the exponent of 2 is a factor of n, and a proper factor if n is composite. If the prime factorization of n were given as input, would there then be a polynomial-time algorithm for this function? Dan --- Newsgroups: rec.games.abstract From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Wed, 6 Jan 1993 23:43:48 GMT Subject: Sprouts and Nim (was Re: defects in abstract games) kwh...@arthur.uchicago.edu (Kevin Whyte) writes: > ... The games which can be reduced to Nim > (Sprouts can't, and hasn't been solved ....) are exactly ^^^^^ > those where > 1) The game always terminates, even it the players don't > alternate moves. > 2) From every position both players have exactly the same > legal moves, and exactly the same winning positions, etc.... Sprouts has these features, and _can_ be reduced to Nim, at least with in normal play (last player wins). Misere Sprouts, where the last player loses, has positions that aren't encountered in Nim, but they can be dealt with using the extended Sprague-Grundy theory you can find in On_Numbers_And_Games or Winning_Ways. I suspect that ``can't'' is a misprint for ``can''. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: comp.programming From: hoey@zogwarg.etl.army.mil (Dan Hoey) Date: 7 Jan 93 17:01:06 GMT Subject: Re: Locating duplicates k...@cs.cornell.edu (David Karr) writes: >A shortcoming of course is that the running-time analysis is valid >only if you can repeatedly allocate large amounts of zeroed-out memory >in a small (essentially constant) time. You can actually skip the zeroing out at a cost of a constant factor at access time. It's well-known; email me if you're interested. There's a known linear-time solution for the element uniqueness problem on the real-ram-with-floor model (none of this bogus 32-bit integer restriction) that uses a kind of hash table in this way. It was in JACM in the seventies, if I recall. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: comp.theory From: hoey@zogwarg.etl.army.mil (Dan Hoey) Date: 13 Jan 93 17:49:38 GMT Subject: Re: solution or information needed an3842@anon.penet.fi writes: >Does anybody have the solutions to the exercises in the classic >computer theory book? "Introduction to automata theory, languages, and >computation" by Hopcroft and Ullman in 1979. If you have any, please >post your reply or send it to me by e-mail. Thank you. >Enrico I wonder if this was posted anonymously because ``Enrico'' doesn't want his professor to know he's doing his homework by netnews. We've a lot of plagiarism attempts before in this newsgroup, and it's somewhat annoying to see them start to be relatively untraceable. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.games.abstract From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Wed, 13 Jan 1993 23:38:55 GMT Subject: Re: TwiXt rules db...@csugrad.cs.vt.edu (David Bush) writes: > ... there was an error in an earlier posting regarding what > constitutes a drawn position. The error was later corrected. You > can indeed link up in tight situations, as long as there is no link > which actually crosses the path of your intended link. Not always, as I mentioned during the discussion of this back in November. The position - - - - O - O - - - - - - - - - - O - O - - - - - - - - O - O - - - - - - - - - - O - O - - - - - X - X O X O X - X - X X - X - X O X O X - X - - X - X O X O X - X - X X - X - X O X O X - X - - - - - O - O - - - - - - - - - - O - O - - - - - - - - O - O - - - - - - - - - - O - O - - - - is a forced draw, even though there are no links, nor were there links in any ancestor position. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: misc.misc, news.admin.policy Followup-To: misc.misc From: hoey@zogwarg.etl.army.mil (Dan Hoey) Date: 14 Jan 93 05:27:42 GMT Subject: Posting peoples addresses (was Re: Language Censors) dshol...@csl36h.csl.ncsu.edu (Doug Holtsinger) writes: > shede...@gw.home.vix.com (Anne P. Mitchell, J.D.) writes: > > Ahem...Ms. Foo's email address is: > > f...@bar.edu > steven....@bar.edu Professor > Arlene....@bar.edu Department Head... [except that I have substituted foo, bar, baz, and zog for the posted information.] Possibly Ms. Mitchell is as naive as she says she is, but Mr. Holtzinger should have been paying attention to the followups, which cover a few of the reasons why I say: *** DON'T POST PEOPLE'S NET ADDRESSES WITHOUT THEIR PERMISSION. *** EVEN IF YOU FOUND IT IN A PUBLIC DIRECTORY. It's just possible that Mitchell or Holtzinger received permission from the people whose addresses they posted, in which case I have no objection (though in that case mentioning that it was by permission would have been helpful). But they show no indication of having asked permission. Mitchell in particular misses the point with pinpoint accuracy when she writes: > "Finding" her address was only a matter of looking it up on a public > access netfind .. it's a listed 'address'. It was NO different than > looking someone's phone number up in the phone book..if it's there, > it's public information. I would *never* *ever* reveal, by post or > *otherwise*, a non-public way to contact someone. There is a difference between the publicness of a telephone book or a netfind and the publicness of posting the information. Looking up the information in netfind and corresponding with the person is fine, that's what netfind is for. Publishing the information in another medium, say by posting it, makes it MORE public. That is an invasion of their privacy if you do it without their permission. The presence of the information in a public directory does not imply such permission. Now there are times when such invasion of privacy might be justified, but I don't believe the fact that their names were mentioned in a controversial newspaper story [which was reposted to the net] constitutes such justification. In fact, it is hard to imagine that anyone could post those addresses in a followup to the news story without considering the possibility that the addresses would be used to send email that is annoying or even abusive. In such circumstances, posting the information without permission would be shamefully reckless. So, Ms. Mitchell, you said in your post that you corresponded with the person ``to ask her if the Wall Street Journal's account was factually accurate.'' Did you think to mention your plan to post her Internet address to the world? Have you ever noticed that sometimes the people who appear in newspaper stories have their telephone numbers listed in the telephone book? Have you ever wondered why the newspaper doesn't print the phone numbers, just in case a few hundred readers want to check on the details of the story? Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math Followup-To: rec.collecting From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 26 Jan 93 23:41:16 GMT Subject: Re: Help me find a "Math class is tough" Barbie rrw...@canrem.com (Roy Wood), perhaps slyly, writes: > Now that Mattel has pulled the "Math class is tough" Barbie, it's > pretty much impossible to get one, but I'd still like to. I don't > suppose that anyone out there has one that they'd be willing to part > with? Or perhaps you know of somewhere that I can get one? Since collectors will pay thousands of dollars for such a doll, (not too surprising, since fewer than 1.5% of the $25 dolls would say any given phrase) it's pretty unlikely you'll find it cheap. A recent article in rec.collecting has several addresses of publications devoted to Barbie collecting, for the benefit of anyone who really, really wants such an artifact (or who has one and really, really wants thousands of dollars.) Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: comp.programming From: hoey@zogwarg.etl.army.mil (Dan Hoey) Date: 27 Jan 93 19:01:08 GMT Subject: Re: Virus Programs 4 Sale (Virus, Trojans, etc) a...@access.digex.com (Albatross) writes: > VIRUS'es FOR SALE >Cost: $20.00 per disk >Contents: 10 Viri per disk > I will hold NO responsiblity for such actions, > if incidents are incurred. Be sure to mention that to the judge. It always helps to inject some lighthearted courtroom humor to set the tone for your sentencing. sm...@ctron.com (Larry Smith) writes: >On the other hand, the original poster's suggestion that it be >used to test virus detection software or procedures is a valid one. >It is also easy to imagine someone with a neat new idea for zapping >compuviruses and would like a few real-life examples to see if it >works before taking it to market. And I'd much rather he test his >new software with _known_ viruses, rather than writing some to see >if it works, if you take my drift... It is possible to manufacture files that appear to be infected but are not infectious, and programs that appear to act as viruses but do not spread. These are quite sufficient to test your software. On the other hand, testing your virus software by trying to infect your disk with real viruses is like testing your vaccination by buying a rabid dog to bite you. It is so ridiculous that no one would take seriously such a pretext from a mad dog vendor. d...@cs.arizona.edu (Dave Schaumann) writes: >kramer...@snake.cs.uidaho.edu (Brian Kramer) writes: >>There is no difference between one piece of software and another if >>they are made of the same binary bits. >Could you explain what you mean by this? He means the usual thing: ``My contraband is legal because it is made of the same stuff as this obviously legal thing.'' Like ``My heroin is legal because it's made of the same carbon, nitrogen, and oxygen as my petunias'' or ``My atom bomb is legal because it's made of the same protons, neutrons, and electrons as my toaster.'' Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: alt.folklore.urban, news.admin.misc From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Thu, 28 Jan 1993 02:55:58 GMT Subject: Old news (was Re: Mnemonics...) bwil...@austin.ibm.com (Brad Wilson) writes: +Now, I have a question ... [details deleted] sys...@codewks.nacjack.gen.nz (Wayne McDougall) writes: +That doesn't seem likely, since .... [details deleted] twc...@tennyson.lbl.gov (Terry Chan) writes: +Uh, Wayne, much as we appreciate your thoughts, but you know, Brad +posted this in December, and two famous people I know (well, maybe +1.5), Larry Doering and Derek Tearne posted replies... dated 11 +December 1992 and 13 December 1992, respectively. Are you having +like, a really slow feed down there? ... It's worse than that. Wayne posted his reply on 18 December 1992, and it appeared on alt.folkore.urban that weekend. But someone changed the date to 27 January 1993 and reinjected it into the news stream. They did that to nine other messages of his, too. At least on AFU Wayne seems to be the only victim. In case you're interested, I include the the message-ids and dates at the end of this message. The prime suspects are copper!mercury.cair.du.edu!mnemosyne.cs.du.edu, since they all appear in the duplicates and don't appear in the originals. ObUL: Resubmitting old news with new dates is happening to the entire contents of several newsgroups, all the time. But the delay is several years, so we don't recognize the duplicates, we just feel deja vu. Sometimes we are flami^Whaving discussions with people who have left the net entirely, or even died. Of course, with the rate of increase in Usenet volume, a prior year's traffic is only a small fraction of this year's. Dan Hoey Hoey@AIC.NRL.Navy.Mil ================================================================ <7oa1VB4w1...@codewks.nacjack.gen.nz>: 27 Jan 93 02:39:39 GMT / Fri, 18 Dec 92 13:44:41 NZST : 27 Jan 93 02:40:08 GMT / Fri, 18 Dec 92 14:51:36 NZST : 27 Jan 93 02:40:14 GMT / Fri, 18 Dec 92 14:57:22 NZST : 27 Jan 93 02:39:42 GMT / Fri, 18 Dec 92 13:59:34 NZST : 27 Jan 93 02:39:34 GMT / Fri, 18 Dec 92 13:43:12 NZST : 27 Jan 93 02:39:44 GMT / Fri, 18 Dec 92 14:24:08 NZST : 27 Jan 93 02:39:52 GMT / Fri, 18 Dec 92 14:36:24 NZST : 27 Jan 93 02:39:59 GMT / Fri, 18 Dec 92 14:37:57 NZST : 27 Jan 93 02:40:06 GMT / Fri, 18 Dec 92 14:47:31 NZST : 27 Jan 93 02:40:18 GMT / 19 Dec 92 00:32:19 GMT --- Newsgroups: comp.programming From: hoey@zogwarg.etl.army.mil (Dan Hoey) Date: 28 Jan 93 19:52:08 GMT Subject: Multiple precision and symbolic math software (Re: Questions) tmjo...@eos.ncsu.edu (Tommie Miles Jones) writes: > Is there something like Maple that I can get source code to. Or > possibly a book on some of maple's algorithms like how does it take > derivatives in the function form? For multiple-precision stuff, there's a list in netlib, under "c/free-c". For information on accessing netlib: mail net...@ornl.gov mail net...@uunet.uu.net Subject:(hit return) Body-of-letter: send index Offers: Software thru email For general symbolic mathematics, there's a newsgroup called sci.math.symbolic, and a list of software maintained by deso...@math.berkeley.edu (Paulo de Souza). His list refers to files in the directory pub/Symbolic_Math in the anonymous FTP of "math.berkeley.edu". There's a bibliography there, too. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: alt.music.hardcore, alt.cyberpunk, alt.thrash, alt.vampyres, alt.fandom.misc, rec.arts.sf.misc Followup-To: alt.zines, rec.mag From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Thu, 28 Jan 1993 22:13:53 GMT Subject: Zines (was Re: Latest F5 info) aa...@cleveland.Freenet.Edu (Jennifer Engel) writes: > re: zines are only "punk" if they aren't sold for money, which is > when they become [miniature] magazines Sorry, Jennifer, you're misquoting Johan Schimanski's message, though you're doing it so mysteriously that already kindly old Bob Conrad (Ick! Shun!) has taken issue with you over it as your own assertion. The reason it's so mysterious is that Johan posted his message only to alt.zines, though he set the followup-to (most of) the newsgroups where the F5 announcement started this topic (plus rec.arts.sf.misc, who are probably even more mystified than most of us (but IAWOL for them)). What joha...@hedda.uio.no (Johan Schimanski) actually said was: > Well, all of the punkzines I have seen (and liked) have been SOLD > for MONEY. Whereas the sf/fannish zine scene has been around since > 1930 (already reaching a large production rate by the 40s), > developing a culture which set the criteria for much of the rest of > zinedom, and mostly subsisted on distribution WITHOUT money. I think > this is a challenge to the status of punkzines. Are they really > fanzines, or just amateur magazines? So the question is not what is "punk" but what is a "fanzine" (and what is "amateur" (not to mention "semi-pro")). Jennifer continues: > so how do you expect to produce one? doing mine myself, and having > access to free copying, the postage is enough to warrant asking for > money. The way it (mostly) worked is that you wouldn't send your zine free to just anyone who wanted one, but only to people who sent you their zine, or sent art for your zine, or wrote an article or letter of comment. This cuts down the subscription list to something you can maybe afford, especially since you no longer need to pay for reading material, and all your other hobbies disappear. (And most of them didn't have free copying. They bought their own mimeographs (and jelly and hecto and ditto and carbon paper and even linotypes)). But nowadays, when a fanzine can get nominated for the Hugo after it has stopped accepting zines in trade, I don't think this economic purity issue has much bite. > everyone should stop worrying about what "punk" is and trying to > define it and just live their life. Once again, no one's wondering what's punk. They are wondering what's zines and what is a ``little bastard faction'' as Cometbus so coyly put it. My own belief is that you can recognize a little bastard faction by its members. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@zogwarg.etl.army.mil (Dan Hoey) Date: 3 Feb 93 00:14:50 GMT Subject: Re: an algorithm for determining the shortest Roman numeral string? d...@scott.skidmore.edu (douglas cho) writes: >I'm wondering if there is an efficient algorithm for converting a >decimal number into a Roman numeral while retaining the shortest form >possible. While a series of div and mod commands will give out a Roman >numeral equivalent, trying to shorten the string seems somewhat tricky. >For example, how would one convert 99=LXXXXVIIII=IC ? For that matter, shouldn't 8 be IIX rather than VIII? The hard part is figuring out whether IXC should mean 89 or 91. bleve...@magnus.acs.ohio-state.edu (Bryan D Levenson) writes: >When I was working w/ Roman numerals I always found the rules to include the >convention that a numeral could only be placed before the next two higher >numerals. (e.g. I can only be placed before V and X, V before X and L, etc.) >So, by this you shouldn't use IC. Well, did you get these `rules' from anyone who is empowered to make rules about Roman numerals? If you know of any authoritative standards for Roman numerals I'd like to know about it. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- From: hoey@zogwarg.etl.army.mil (Dan Hoey) Newsgroups: rec.puzzles Subject: Re: word used as its own opposite Date: 3 Feb 93 00:40:13 GMT RVESTERM@vma.cc.nd.edu (Bob Vesterman) wrote: > there was a post here a week or so ago about the lack of english words > which were their own opposite. anyway, i just thought of this: don't > many people use the phrase "i could care less" to mean "i couldn't care > less"? When I say ``I could care less,'' it is short for the phrase ``I could care less, but I'd have to try.'' I think that's a fairly wry way of expressing apathy so great that it need not compete for the extremes of apathy. I express that phrase in its contracted form to indicate that apathy prevents me from finishing it. Careful listeners can hear the ellipsis. If they care. I am told this phrase originated in the first half of this century with the meaning I ascribe to it. The fact that many people seem to have forgotten its origins doesn't excuse such silly presumptiousness as djweisbe@unix.amherst.edu (David Weisberger) writing: > Yes, but people are, well, wrong. They don't know whether they want to > say "I couldn't care less," or "I could care." If it was worth the effort, I would insult him back for it. --- Newsgroups: rec.puzzles Followup-To: alt.peeves From: hoey@zogwarg.etl.army.mil (Dan Hoey) Date: 3 Feb 93 13:55:13 GMT Subject: Re: word used as its own opposite RVEST...@vma.cc.nd.edu (Bob Vesterman) writes: >"inflammable" means "very flammable". >"flammable" does not mean "very flammable". Not according to Webster's 3d New International Dictionary, The Random House Dictionary (2d Ed), The American Heritage Dictionary (1st Ed), or Webster's II New Riverside Dictionary, all of which list ``flammable'' as a synonym of ``inflammable.'' The last goes on to explain: ``_Inflammable_'' is derived ultimately from the Latin prefix _in-_, ``in,'' and the noun _flamma_, ``flame.'' There is another prefix _in-_... that means ``not....'' In order to eliminate possibly dangerous confusion about the combustibility of various materials, safety officials in the 20th century have adopted the term _flammable_, which had a brief life in the early 19th century, to mean ``able to burn.'' If you wish to demonstrate a difference between the meanings of the words you might cite some authority, lest we think it's just a peculiarity of your dialect. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: comp.admin.policy, comp.security.misc Followup-To: comp.admin.policy From: hoey@zogwarg.etl.army.mil (Dan Hoey) Date: 5 Feb 93 19:18:52 GMT Subject: Re: Logging anon. ftp usage m...@kraken.ucsd.edu (Mark Anderson) writes: >And I want people to fix their software, so that if I tell it >my email address is "m...@cs.ucsd.edu", it doesn't throw a fit.... Of course, if it got that address out of a PASS command, there's a nonnegligible chance it's actually the user's real password. People are creatures of habit, and occasionally type the wrong thing, especially if their user agent elicited it by printing ``Password: ''. This is why I avoid keeping logs of the responses (or fingering them, for that matter). I'm concerned that the logs themselves may constitute a security risk. If anyone actually has seen real passwords show up in ftp logs, I'd like to hear my suspicions corroborated (by email, unless there's something of really general interest). Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: comp.mail.misc From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 9 Feb 93 14:07:47 GMT Subject: Re: Bill Clinton's E-Mail Address gc...@horus.cem.msu.EDU (Greg Cook|xxxxxx) writes: >t...@Veritas.COM (Ronald S. Karr): >>s...@csi.compuserve.com (Sam Neely) writes: >>>Try 75300.3...@compuserve.com instead. (Notice number of zeros) >> Is this address intended for public use, then? Is there much point in >> sending a message, for example, supporting (or decrying) a gas tax? >Good luck getting mail to this address! Compuserve only allows 100 messages >in the mail spool at a time, and this address is always full!!!!!! I received the following message from the INFO-NETS mailing list. It seems use of the e-mail address is not yet advised. : From: Robert L Krawitz : Subject: [Tim...@Think.COM: Don't e-mail White House yet] : Date: Mon, 8 Feb 93 19:43:41 EST : : As this subject has come up on info-nets, I thought I'd send out this : update. : : January 31, 1993 : : Important Information RE: E-Mail to the White House : : Yesterday, I saw several postings related to the E-mail address : for the White House. Along with a good number of others, I worked : throughout the campaign as part of a network of E-mail volunteers : for the Clinton campaign, so I can pass along some important : information about that E-mail account. The account is actually the : personal compuserve account of Jock Gill. Jock worked hard (along : with a handful of programming volunteers, BBS operators, listserver : maintainers, and computer sophisticates at places such as Marist : College, MIT, San Francisco, Chicago, and elsewhere) during the : campaign to put together an E-mail system for national campaigning. : The system was later expanded to accommodate all three major : Presidential campaigns. It was an innovative, highly successful : effort and it played a huge role in getting campaign position : statements out to a wide public. Things posted from that address : found their way into the virtual reality as the messages got passed : along many networks from their original posting. Several weeks : before the Inauguration of President Clinton, Jeff Eller was : appointed by the President-Elect to have overall charge of : establishing something which has never existed--an interactive : public access E-mail system into the White House and into other : offices of the administration. Jock Gill was then hired by the : administration to work under Jeff Eller. Currently, Jock Gill is : working in an office located in the Old Executive Office Building : across the street from the White House. At this point, he is : working alone, without a staff. His current assignment is to use : the E-mail system (as during the campaign) to issue official copies : of White House statements, the texts of press briefings and press : conferences, copies of Executive Orders and Presidential Memos, and : the like to the virtual world of E-mail. Since the compuserve box : is a regular personal mail box, it gets filled quickly, especially : given the high volume of mail now beginning to arrive with the broad : dissemination of his address. Those of you who have sent E-mail : to that address may well have received an error message stating : that the box is full. That's another way of saying it has been : overwhelmed. Jock has asked those of us who have been part of the : volunteer E-mail team to help him out while he works to get a good : interactive system up and running. Basically, he has asked that : everyone cooperate and not begin sending a barrage of E-mail to that : compuserve address. The White House itself employs a large staff to : handle snail mail. Actually, at this point in the development of : the White House E-mail system, you will probably get your message : through to the administration quicker through ordinary snail mail : and telephone. Later, once the administration's E-mail team : develops the system they want and need, E-mail contacts should : became the easier route. All things in their time. Once the E-mail : address was circulated together with the heading the "White House", : everyone understandably believed a real system was up and running. : Not quite yet. : : SUGGESTION: Use the compuserve address you have judiciously, : reserving it for absolutely vital contacts. Until such time that : a real public access White house E-mail system is operational, : consider relying on the traditional means of contacting the : administration. Given what they had to start with from the previous : administration (scratch), I have every reason to expect that Jeff : Eller and Jock Gill will work well--and as quickly as possible--to : get an interactive system up and running. But it will take time : and patience. We can all help them achieve that effort best if we : refrain from acting as if that non-existent system were already in : place. PLEASE HELP RELAY THIS CONTEXT AND SUGGESTION TO OTHER : NETWORKS AND INDIVIDUALS. Thanks. : : Snail Mail Address and Phone Numbers -- White House : : White House Numbers: : The President (202) 456-1414 : White House Comment Line (202) 456-1111 : (To register your opinion on an issue) : When bill signed or vetoed (202) 456-2226 : : Vice President (202) 456-2326 : (202) 456-7125 : : Mailing Address: : : The White House : 1600 Pennsylvania Avenue, N.W. : Washington DC 20500 : : ------ : Jon Darling : PITT/Johnstown -- January 31, 1993 Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: news.admin.policy From: hoey@zogwarg.etl.army.mil (Dan Hoey) Date: 9 Feb 93 14:55:10 GMT Subject: Re: Anonymous postings to non-personals newsgroups Having seen several attempts by students in introductory computer science courses to get their homework done by the readers of comp.theory, I was somewhat dismayed to see the anonymous posting service used for the purpose. So far there has been only one such use, though that seems to be the only use of the service on comp.theory so far. While there has never been any real security against anonymous or forged postings on Usenet, the process has until now been sufficiently inconvenient, error-prone, and undocumented to limit its use by persons who have not learned the culture of the net. On the other hand, a recent use of the anonymous posting service on sci.math seemed seemed to be a student asking help on a homework problem. It has now been attributed to a teacher, asking for an explanation of a dubious answer in his teaching guide. He says his news posting is broken, so he is using the anonymous service as a mail-to-news gateway. Gateways from mail to news have provided a fairly easy method of forgery and anonymity in the past, though they are somewhat less prone to abuse for several reasons. First, their use does not provide anonymity by default, nor does it promise anonymity. Such gateways may in fact keep enough records to trace back to the poster in most cases. Second, the gateways do not provide a covert address for private, anonymous communication with the poster. Thus the cheating student must either provide a private address or make do with that information that gets posted in the newsgroup. The inherently greater risk of detection would be likely to reduce such abuse. For these reasons, I think such gateways are much less harmful than the anonymous posting service. It is unfortunate that no well-known mail-to-news gateways seem to exist outside of the anonymous posting service. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math, sci.physics, comp.theory.cell-automata Followup-To: comp.theory.cell-automata From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Wed, 10 Feb 1993 23:37:24 GMT Subject: Big life and blt life (Re: This Week's Finds...) When b...@guitar.ucr.edu (John Baez) writes: > Sorry, I was speaking as a statistical mechanic just then. What I > meant was that a random initial configuration will *almost* always > settle down into a boring configuration that is static except for > small regions executing periodic motions (in practice I have never > seen anything except blinkers). Indeed, I think that in the > infinite-volume limit my "almost always" could be made precise in > the sense of probability theory, namely, "with probability 1". > One has to have seen a number of large-screen displays of Life to > realize just how depressingly dull the average final state is! > (With the program I have now, I can get one cell per pixel, for a > total of about 1000x1000. I boggle at the spectacle of a physicist talking about a 1000x1000 array as ``large.'' In contrast to John's conjecture, I think there are some conjectures that for _large_ starting arrays most of the starting positions never settle down. But I don't know if anyone has the computrons to get a handle on typical 1e9 x 1e9 arrays (and I'm not sure they are large enough to exhibit macroscopic effects). step...@dogmatix.inmos.co.uk (Stephen Collyer) writes > Image [sic! joke?] your lattice as a 2D image containing only 2 > values, 1 and 0 which correspond to a cell being occupied or empty. > Now perform a 2D discrete convolution of the image with the > following 3 by 3 kernel: > 1 2 4 > 128 256 8 > 64 32 16 > ...this gives a unique result for each of the 512 different 3 by 3 > image squares (remember each cell can only have 2 values).... feed > the output from a 3 by 3 convolver into a look-up-table.... But that's profligate for Life. If you use the kernel 1 1 1 1 7 1 1 1 1 you only need a 16-element table, versus 512 for your kernel. In fact, I've seen something much like this method where the (binary) addition is converted to Boolean operations, and implemented in bitblts. I found the algorithm, credited to Byte magazine, on Lisp machines. It used about 55 bitblts, but I found out how to cut the number of bitblts in half by doing horizontal convolution followed by vertical convolution. > (Does anyone know where I can get a copy of xlife from) I don't know if it's the most recent version, but /contrib/xlife.tar.Z is on export.mit.edu. r...@ark.abg.sub.org (Ralf Stephan) mentions the puffer train as a configuration that keeps producing new matter, but that is a relatively stable configuration, since the matter turns out to be very regular and static. When I was talking about positions that don't settle down, I was talking about truly chaotic positions. Perhaps the best definition is positions whose population increases without bound, but you can't prove it. This stuff is the topic of comp.theory.cell-automata, which I recommend to people interested in talking about it. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Thu, 25 Feb 1993 00:48:22 GMT Subject: Re: "Card-shuffling" algorithms m...@caesun6.epg.harris.com (Matt Mahoney) writes: >The code (deleted) contains an error in that it does not reorder the >shuffled deck between iterations. I did each of the 27 possible >permutations by hand and got the following results: >012 = 4 >021 = 5 >102 = 5 >120 = 6 >201 = 3 >210 = 4 >total = 27 The 6 should be a 5, and the 3 a 4. dhep...@cup.hp.com (Dan Hepner) writes: > Thanks, and apologies. After fixing the code, here's the new results: > result 012 = 147911 > result 021 = 184700 > result 102 = 185064 > result 120 = 185953 > result 201 = 148152 > result 210 = 148220 ... ... Believable approximations to the actual probabilities, .148 and .185. > So it seems to be "bad" by having the winners overrepresented by > over 10%, while the losers are underrepresented by a corresponding > amount. 11 1/9%, to be precise. To see that the situation does not even out for large decks, it's not hard to see that, the preponderance of odd permutations in odd decks and even permutations in even decks increases for larger decks. The probability of the preponderant parity approaches (1+e^-2)/2, or about 56.8%. So we can be sure some permutations will get over- or under- represented by 13.5% for large decks. At 52 cards, it's 13.0%. Now for a three-card deck the actual bias (5:4) for the permutations exceeds the exceeds lower bound (14:13) implied by the parity bias. I wonder if the actual bias approaches the parity bias for large decks? Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math Followup-To: comp.misc From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 2 Mar 93 00:54:42 GMT Subject: Number theory questions ``(John Forrest)'' writes: > Here's a few problems that I've been unsuccessful in solving. I'd > appreciate any help. Thanks in advance. .. > I don't have an e-mail account so post these answers if you are able to > solve them. > John and ``(William F. Miller)'' writes: > Here's another number theory question. .. > Bill Here's a question. If ``John'' comes from UCLA and ``Bill'' comes from Univ. of Oregon like their Organization: lines say, then why were their messages posted from Trinity College in Hartford? What's all the interest in questions that would be a simple exercise in any Introductory Number Theory course? Why don't they have email addresses? Followups to comp.misc, where they are discussing a student who got caught asking comp.lang.c to do his homework for him. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math Followup-To: comp.misc From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 3 Mar 93 22:45:28 GMT Subject: Re: Number theory questions When I posed the question: >If ``John'' comes from UCLA and ``Bill'' comes from Univ. of Oregon >like their Organization: lines say, then why were their messages >posted from Trinity College in Hartford? What's all the interest in >questions that would be a simple exercise in any Introductory Number >Theory course? Why don't they have email addresses? eapu...@orion.oac.uci.edu (Lee Furnival) deletes some stuff and writes: > Well that is mighty perceptive of you, but who cares? As I mentioned in the stuff that Lee deleted, > Followups to comp.misc, where they are discussing a student who got > caught asking comp.lang.c to do his homework for him. People ask questions to which they want the answers; I tend to provide answers when I can. However, I deeply resent my help being taken advantage of by lazy college students who are looking for someone else to do their exercises. I am not the only one here with that outlook, so when I notice an apparent attempt to take advantage of us I tend to let the newsgroup know. It might save someone from being taken in by the ruse. In the case of ``John Forrest'' and ``William F. Miller'' it is not clear whether this is a case of homework fraud; I can only surmise it from the character of the questions and the peculiar anonymity of the messages. Perhaps they wish to explain the situtation; I cannot ask them more directly, as they have not provided email addresses. As for Tal Kubo's curiosity about how I find the time for about how I find the time for ``so much netcop detective work,'' I take it from the time I would rather spend answering questions from people who are interested in learning something. It's an unfortunate loss, but you can't get rid of cockroaches by feeding them. Tal also wonders how I was appointed our fearless protector from these homework hacks. I suppose I was appointed in the same way he was appointed to protect us from people who wish to expose them. I regret that this posting lacks any information relevant to mathematics. On the other hand, at least I haven't provided any mathematical misinformation. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Thu, 11 Mar 1993 21:57:28 GMT Subject: Computer dating (was Re: Computer logic) emc...@bnr.ca (hume smith) asks on what day of the week the 13th is most likely to fall. ab...@Freenet.carleton.ca (Guy Cousineau) answers: >In the last 400 years, which should be a reasonably full cycle, >Christmas has fallen on a particular day of the week: I should note that Guy's figures for the days of Christmas are correct for the Gregorian calendar, and he then calculates the probable day of December 13th correctly. Unfortunately, December 13th is not typical of thirteenths, so he answers the question wrong. But while the last 400 years were a full cycle in countries that were controlled by Pope Gregory in the 16th century, England and her colonies did not adopt the Gregorian calendar until 1752. My figures, said to reflect England and colonies, indicate that the Christmases of 1593 through 1992 fell on 58 Sundays, 56 Mondays, 60 Tuesdays, 58 Wednesdays, 57 Thursdays, 57 Fridays, and 54 Saturdays. But knowing the days of Christmas does not help us count all the thirteenths during a time of calendar reform. In 1593 through 1992 I calculate there were 685 Sunday the thirteenths, 684 Monday the thirteenths, 685 Tuesday the thirteenths, 688 Wednesday the thirteenths, 684 Thursday the thirteenths, 691 Friday the thirteenths, and 682 Saturday the thirteenths, for a total of 4799 thirteenths in all. The reason there are only 4799 thirteenths in those 400 years is that September 13, 1752 did not occur. We see that Friday was the most common thirteenth in the last 400 years, but to answer the question question as posed, it would make sense to answer it for the Gregorian calendar, since that will continue for quite a few centuries to come (and no one has any good plans for its successor.) In any four hundred consecutive years of the Gregorian calendar there will be 687 Sunday the thirteenths, 685 Monday the thirteenths, 685 Tuesday the thirteenths, 687 Wednesday the thirteenths, 684 Thursday the thirteenths, 688 Friday the thirteenths, and 684 Saturday the thirteenths. So Friday will continue to be the most popular thirteenth, but not to such an extreme extent as it was recently. To summarize these results, if the date is the thirteenth, there will be a 14 1/4% chance of it being Thursday, and likewise for Saturday, a 14 13/48% chance of it being Monday, and likewise for Tuesday, and a 14 5/16% chance of it being Wednesday, and likewise for Sunday. But the chance of it being Friday is 14 1/3%, an entire twelfth of a percent more likely than the least probable day. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: alt.folklore.urban From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Tue, 23 Mar 1993 02:13:25 GMT Subject: Re: Costly Cookie Recipe As long as alt.best.of.internet is out there to reprint the best postings from newsgroups such as AFU, it behooves us to include selections from ABOI when they are relevant. And so: } From: la...@labrador.fishkill.ibm.com (Larry Pieniazek) } Date: Fri, 19 Mar 93 21:42:54 GMT } Reply-To: l...@vnet.ibm.com } Organization: Ya right! } Keywords: urban legend } Subject: Re: Costly Cookie Recipe } } This NM cookie recipe is an urban legend. How do I know??? } } Ann Landers told me (and x million other readers) that she gets lots of } letters about this, usually starting out "a friend of a friend said" etc... } } And, she claims the cookies aren't even all that good. I can't say, never } tried them myself. } } If Ann Landers said so, that's good enough for me! Yes folks, it's a new AFU slogan-of-the-week. ObUL: Ann Landers is paid by the Secret Masters of Urban Legends to regularly issue denials of Urban Legends, thus increasing their spread. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: news.admin.policy, alt.folklore.computers Followup-To: alt.folklore.computers From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Fri, 2 Apr 1993 16:06:03 GMT Subject: Re: ARMM: ARMM: >>>>Ad Infinitum ebra...@jarthur.claremont.edu (Eli Brandt) writes: > You know, that should have been > alt.fan.dick-depew.armm.armm.armm Given the ramifications of the recent escapade, alt.fan.dick-depew.armm:armm:armm: would be more appropriate. If you don't know why, look for alt.fan.enya.puke.puke.pukeSender: in the alt-config-guide FAQ. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles.crosswords From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Tue, 4 May 1993 23:34:42 GMT Subject: R.I.P. Will Weng I heard on the radio this morning that Will Weng passed away this week. He will be remembered fondly by a generation of New York Times crosswords solvers. He was in his eighties. Knowing that his puzzles were a source of amusement to hospital patients, he never defined CANCER as a disease in his puzzles. According to NPR, he always clued it as a constellation or sign of the Zodiac. Unfortunately, he was unable to evade the disease in his own life. Weng, a long-time cigar smoker, died of throat cancer. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: comp.programming From: hoey@zogwarg.tec.army.mil (Dan Hoey) Date: 17 May 93 19:32:42 GMT Subject: Re: area and centroid of polygon gol...@u.washington.edu (Lloyd P. Goldwasser) writes: >There is a handy formula for calculating the area of a polygon, >given the coordinates of its n vertices: > 1/2 * abs(X1*Y2 - X2*Y1 + X2*Y3 - X3*Y2 + ... + XnY1 - X1*Yn). >I'm wondering whether there is a similarly straightforward formula >for calculating the _centroid_ of the polygon from the same coordinates. >The obvious thing to try, the above formula divided by the four times >the area, and with all the Xi squared, misses somewhat: it gives a >number that changes properly as the polygon is translated, but does >not give the correct answer in the first place. dod...@convex.COM (Dave Dodson) writes: >Break the polygon into triangles.... Unless you are including triangles of both positive and negative area, I think triangulating the polygon is the sort of thing he's trying to avoid. I would tend to break the polygon into trapezoids (x[i],0) (x[i],y[i]) (x[i+1]),y[i+1]) (x[i+1],0). Each has (signed) area (x[i]+x[i+1])(y[i+1]-y[i])/2 and centroid ((x[i]+x[i+1])/4, (y[i+1]+y[i])/2). I think the normalized, weighted sum of the centroids should do it. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Wed, 19 May 1993 17:02:06 GMT Subject: Re: Stamp Folding legen...@ens.fr (Stephane Legendre) writes: > How many ways of folding a strip of n stamps ?... and flajo...@nuits.inria.fr (Philippe Flajolet) remarks: > This is a rather famous unsolved problem in combinatorial > enumerations. I would say that it is more than one unsolved problem, depending on how one defines it. For instance, the position _____1_________________ / _____6_______________ / (_______5___________ \ ( _________4_________) \ \ (___________3_______ ) \_______________2_____) / _________________7_____/ cannot be achieved simply by folding the strip of stamps into smaller and smaller strips. At some point, one of the folded strips must be partially unfolded. I'm not sure which of the definitions is more common in the literature. The situation becomes much harder in two dimensions, when there are folded positions in which a sheet could exist (without stretching or self-intersection), but which might be difficult or impossible to create starting with a flat sheet. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.physics From: hoey@zogwarg.tec.army.mil (Dan Hoey) Date: 2 Jun 93 18:42:15 GMT Subject: Re: Mirrors and their reflections cs047...@brando.uwasa.fi (Jari Ylinen) writes: > A simple experiment: > What we need is a sunny day and an ordinary mirror. Put the mirror > so that the reflected sunbeam hits the wall and look at the result. > It is a circle what you see on the wall.... > 1. Why the shape of the reflection will become circle ? zo...@daedalus.stanford.edu (Craig "Powderkeg" DeForest) answers: >Because the shape of the sun is a circle. If you do the experiment during, >say, a partial solar eclipse, the reflection will become crescent-shaped. The setup is a sort of pin-hole camera: The mirror is the pin-hole/lens and the wall is the retina/film. So if you get the mirror small enough, and the wall far enough away and dark enough, you should be able to see sunspots. Also, note that the circularity is only approached as the scale of the image increases. If you have a mirror two inches longer than it is wide, the image will be two inches longer than it is wide, no matter how far away you go (until the aperture effects start making it impossible to measure). Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: news.groups From: hoey@zogwarg.tec.army.mil (Dan Hoey) Date: 9 Jun 93 19:06:34 GMT Subject: Re: RFD: rec.food.spicy khammond@eng.auburn.edu writes: >I feel there should definitely be a separate newsgroup for us who >dearly love spicy foods. Some people on the net have complained that >there are only a few "spicy food" topics come up on rec.food.cooking. >I scan the articles on rec.food.cooking looking for anything >pertaining to "hot" and "spicy", and will usually find a few good >ones. I have never posted any articles to this newsgroup because I >did not think many people would be interested in very, very hot an >spicy foods that I love. That is ridiculous. Other people post about spicy food on the existing group, and you find it interesting. If there aren't enough there to discuss spicy food with, creation of rec.food.spicy won't make them more numerous. The rationale for this group looks slimmer with every message. You should start using r.f.c. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.nude From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Fri, 11 Jun 1993 16:45:59 GMT Subject: Nude News from DC City Paper The Washington, DC City Paper has printed news of rec.nude interest for the last two weeks. The May 28 City Desk has: We Love a Man Out of Uniform There are conventional ways to mark Memorial Day: parades, wreath-laying, speeches. And then there's the ``Armed Forces Recognition Weekend'' sponsored by Pine Tree Associates Family Nudist Park in Crownsville, Md., this Saturday through Monday. ``We'd like to honor our men and women in uniform by getting them out of uniform,'' says club spokesman William Pacer. Families with a member possessing a valid military ID will be admitted to the park without charge (and without uniform, of course). Pacer points to Operation Desert Storm, where troops fought to defend American freedoms, one of which, he says, is ``the freedom to be nude when appropriate--or to disagree with us.'' And in the June 4 edition: Grin and Bare It Virginia may be for lovers[*], but nudists aren't having an easy time of it. For the last few years, the commonwealth has had just two campgrounds to serve its half-dozen clubs affiliated with the Eastern Sunbathing Association. One of those facilities is in Zuni, not far from Richmond off Interstate 95, and the owner reports that it's a welcomed and integral part of the community. (For example, White Tail Park participates in the state's adopt-a-highway program.) But the Avalon Conservation Club, in Madison, was recently shut down by order of the Madison County Circuit Court. It seems that the county's planning board issued more than 900 building permits over an eight-year period without the proper authority. When the state ordered that all the permits be reissued, only one was denied--White Tail's. The park's owners brought their battle to court, where the case has bounced around on appeal. White Tail stayed open during that time, but now the circuit court has ordered that visitors be clothed. Arne Eriksen, executive director of the American Sunbathing Association, says one county supervisor was irked about White Tail because the property has no fences and is on a hill where it can be spied on. He also says that in his eight years with the association, this is only the second time a nudist park has been ordered closed. I don't know whether they have someone on the staff investigating these things, or whether the are getting faxed from some nude information organization. There's no nude news this week, but I can hope for the future. [* For the sake of distant readers, the Virginia tourist motto is ``Virginia is for lovers''.] Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.games.programmer, rec.games.design From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Fri, 11 Jun 1993 20:40:28 GMT Subject: Re: Name generators cah...@nyx.cs.du.edu (Chris Hall) writes: >Suppose the text contained the sequences BIG and BIT, once each, and no >other sequences beginning with BI. Also, suppose that they occur very ^^^^^^^^^ ^^^^ >close in the text, relative to the total size, and in that order (i.e. >with BIG first). Then, for the majority of starting positions, BIG will >be found first. Only for the few positions in between the two will BIT be >selected. However, the probabilities are clearly the same. Actually, the example requires that no other sequence has BI anywhere in it, not just at the beginning, but the idea is right. s...@aero.org (Scott Turner) writes: > What you really want to do is to jump randomly to one of the starting > letters in the text. (That is, in Chris's example, jump to one of the > Bs with equal probability.) Check that, and if the second letter > doesn't match, jump ahead to the next B. And so on. > It's a little harder to program than the first algorithm, but I think > it can still be done using very little extra storage... I don't see how, but there is an easy way to make sure you keep the original probability distribution: Just jump into the text at random until you land at the digraph you're looking for. This requires only about as many probes as the initial scanning algorithm scans over characters. But if you've got a large text, the nonsequentiality can make you thrash pretty badly. But I think you would get better names by dividing the text into alternating strings of consonants and strings of vowels. With each string you keep the number of occurrences at the beginning of a word, the number at the end of a word, and the total number. Then you pick the first string weighted by the frequency at the beginning of a word, and following strings (alternating types) weighted by the total frequency, and use the (number at end)/(total number) as the probability to terminate at each string. In Websters 2nd International Dictionary (235,000 words), there are only about 4100 different strings of consonants and 200 different strings of vowels (20,000 total characters), which makes for reasonable storage requirements. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Wed, 23 Jun 1993 14:10:07 GMT Subject: Re: problem in number theory t...@informatik.uni-kiel.dbp.de (Thomas Wilke) asks > Question A: Are there relatively prime positive integers M1...Mr such that > 1/M1 + 1/M2 + ... + 1/Mr = 1 ? If by ``relatively prime'' you mean gcd(M1,...,Mr)=1, then yes. 1/3 + 1/4 + 1/5 + 1/9 + 1/22 + 1/25 + 1/99 + 1/100 = 1 1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/21 + 1/22 + 1/45 + 1/154 = 1 See The American Mathematical Monthly, vol 98, #2, Feb 91, pg 168 in the article "Coprime Sequences Summing to Unity". [ bb...@dseg.ti.com (Brian Borowicz) writes: > Followup-To: problem in number theory Don't use the ``Followup-To:'' field for subject headings. It's supposed to be the name of the newsgroup for further discussion. > Answer: No. If the numbers are relatively prime, the lcd must be > M1*M2*...*Mr. If he meant pairwise relatively prime, he should have said so. ] > Question B) > If the answer to A) is 'no', do there exist positive integers > M1, ..., Mr (r>1) and N1, ..., Ns (s>1) such that > - the sequences of the Mi and Ni are distinct modulo rearrangement, > - the gcd of all M1, ..., Mr is 1, > - the gcd of all N1, ..., Ns is 1, > - Mi does not divide Mj (i and j distinct), > - Ni does not divide Nj (i and j distinct), > - the equation > 1 1 1 1 > ---- + ... + ---- = ---- + ... + ---- > M1 Mr N1 Ns > holds? Yes, vacuously. But if we leave out the false antecedent, note that the answers to question A do not satisfy the fourth and fifth conditions of question B. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.arts.comics.strips Followup-To: alt.conspiracy From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Fri, 25 Jun 1993 16:07:56 GMT Subject: Unawarded prize [ was Calvin and Hobbes...] v122r...@ubvms.cc.buffalo.edu (Jason E Yungbluth) writes: > ... Are there any other college cartoonists out there who > participated in the Scripps/Howards Foundation College Cartoon > Contest (I'm paraphrasing the name here)? ... > Well, after the ceremony date came and went with no reply, I > assumed that I hadn't won. Turns out that NO ONE did, as they deemed > none of the submissions worthy. They did not even bother to send > their response in the SASE that I included per instructions.... I'm sure the contest was just a scam to get people to send them valuable stamps. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.physics From: hoey@zogwarg.tec.army.mil (Dan Hoey) Date: 25 Jun 93 19:41:08 GMT Subject: Re: Windsreen of car as lens and...@rentec.com (Andrew Mullhaupt) writes: >But the lens effect of the windshield is miniscule, either way; and >much more important is the effect of convection, where the windshield >tends to prevent convection currents in a highly car-dependent way. Certainly the lens idea is bogus. But sunlight alone, even insulated from convection, seems unlikely to start a fire. I wonder if the light could have been reflectively focused by a glass or chrome surface. Or maybe there was something with a lower flash point around, like matches. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: alt.folklore.urban From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Fri, 25 Jun 1993 20:57:23 GMT Subject: Twin Tumor Rumor [ Re: MacDo Paper ... ] writes: > So, how did the doctors know that the thing in the guy's head was > his twin? The way I heard the story, the tumor had the same birthmark. Or maybe it had a similar tumor inside its head. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: alt.folklore.urban From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Fri, 25 Jun 1993 21:17:14 GMT Subject: Re: Semen Curry d...@cs.tcd.ie (David Roe) writes: > What I always find strange about this story is the "getting sick" > part. Why? Is semen somehow likely to make you sick, more so, than > say, large amounts of chicken soup.... Semen is known to be a leading cause of morning sickness. But rarely when administered orally. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: comp.arch From: hoey@zogwarg.tec.army.mil (Dan Hoey) Date: 25 Jun 93 21:44:39 GMT Subject: Re: why different endians (was Re: --- Endians) >d...@silverton.berkeley.edu (D. J. Bernstein) writes: >| Is there any _disadvantage_? No. Very little code depends on byte order. >| So the chip designers just take whichever order is convenient. Well, actually you can use big-endian machine instructions for sorting strings faster. This is a relic of our habit of sorting strings using big-endian lexicographic order instead of little-endian. Little-endian can be just a little easier for arithmetic, because the mapping N->2^N is more useful than N->2^(Wordsize-N). But that advantage is pretty far down in the noise. This is all pretty minimal, since converting one way or the other is quick. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: comp.arch From: hoey@zogwarg.tec.army.mil (Dan Hoey) Date: 30 Jun 93 16:05:42 GMT Subject: Re: why different endians (was Re: --- Endians) hoey@zogwarg.tec.army.mil (I) wrote >Well, actually you can use big-endian machine instructions for sorting >strings faster. This is a relic of our habit of sorting strings using >big-endian lexicographic order instead of little-endian. ed...@chinet.chinet.com (Edward Lee) writes: >I think Mr. Hoey meant to say that big-endian strings or integers can >be sorted faster given a specific implementation of a radix sort.... Actually, I only meant that multiple characters of a string can be compared using the instructions for comparing integers. This is only relevant to comparison-based sorting. hru...@pop.stat.purdue.edu (Herman Rubin) writes: > But no matter how you look at it, sorting numbers must be big-endian. Only if the numbers are represented in a big-endian way, such as a text string. Integers can be compared on a little-endian machine using little-endian multiple-word compares. > Compare instructions on the little-endian machines are still big-endian. No, big endian means ``most significant first.'' Compare instructions always compare ``most significant most significant.'' Certainly you have to let the comparison of more significant parts override the results of comparing less significant parts, but this doesn't imply any temporal ordering or sequential organization. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.games.abstract From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Fri, 2 Jul 1993 15:24:52 GMT Subject: Re: Hex [ repost ] [ This is a copy of an article I sent last night, which seems to have been dropped. If you read the other one, don't bother with this one, it's the same. ] torb...@diku.dk (Torben AEgidius Mogensen) writes: >>>...with the modification that the board is of size Nx(N-1) and the >>>second player has to connect the shortest distance, the game is a >>>sure win for player two. >>>The strategy works as follows: divide the board into two equal-sized >>>parts along the short diagonal. Since the board is a non-rhombic parallelogram, this terminology is a little vague. The diagonals of the parallelogram touch only two hex centers each, their endpoints at opposite corners of the parallelogram. I suspect you are talking about lines that cut the board in half parallel to the sides. The line parallel to the odd side touches no hex centers, the line parallel to the even side touches a row of hex centers. >>>Every time player one places a stone in one half of the board, >>>player two places a stone in the other half at the position 180 >>>degrees opposite (imagine rotating one board half on top of the >>>other). Thus ``at the centrally symmetric point.'' It doesn't matter how you cut the board in half, this is about reflection through the *center* of the parallelogram. But db...@csugrad.cs.vt.edu (David Bush) showed the strategy doesn't work. torb...@diku.dk (Torben AEgidius Mogensen) responds: >My mistake. My recollection was a bit vague: you shouldn't *rotate* >the other half of the board, you should *mirror* it across the >dividing line. Well, the non-rhombic parallelogram isn't mirror-symmetric. If we allow a sort of distorted mirror-symmetry by pretending it's a rectangle, we can reflect through a line parallel to a pair of sides. If this is what you mean, then there is an obvious problem if the shorter side is even, because then the shorter line of symmetry goes through hex centers, and what is the response to a move on this line? Perhaps mirror-symmetrically through the center of that line? It doesn't matter, as we see below. Well, even then it doesn't work, whether the short side is even or odd. Look at these two partial games, in which each move by X has been answered by O (distorted) mirror-symmetrically through the marked line. \ \ - - - - - - - - - - - - - - O O X X - - - - - - - - - X X O O - - O O - X X - - - - - - - - X X - O O - - - - - - - - - - - - - - \ - - - - - - - \ In either case it is now X to move and win even if O abandons this strategy, even if X passes (and this is a game without zugzwang). Both games generalize to larger boards. There may be a winning strategy for the second player on the Nx(N-1) board, but I would be quite surprised if it's a simple one. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Article 5740 of dc.general: Newsgroups: dc.general,alt.english From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Subject: Re: Math in Wash Post Sender: usenet@ra.nrl.navy.mil Organization: Naval Research Laboratory, Washington, DC Date: Fri, 2 Jul 1993 22:57:02 GMT mheney@access.digex.net (Michael K. Heney) writes: > Yes, that's very nice, but WHERE was it, and in WHICH issue? All the Washington Post had on June 24 was an AP wireservice story on page A15. cradle@wam.umd.edu (Dave Eisner) writes: > The story was on the front page of the NYT on June 24. I remember > seeing it through the window of a vending machine. I didn't want to > pay 75 cents for the Times, so I bought the Post, instead. > Unfortunately, the Post didn't have the story. They may have > eventually reported it, but the Times got the "Scoop." The Post had it on June 24, see above. Perhaps you didn't get the final edition. The Times story was much better, anyway. And the Times's biographical story on Wiles on June 29 was nice, too. Digressing toward alt.english, I noticed the Times mentioned "Wiles'" proof on page 1 and "Wiles's" proof on page D22. I wonder which is the Times's style. If they followed the AP style, I guess it could possibly be that the front-page editor thought "Wiles" had two syllables. But note that Strunk and White would have you put "'s" after every singular--even "Charles's"--except for "the names of certain ancients". (I think the exception is a craven cavein because they didn't want to take on the Bible's use of "Jesus'".) And there are plenty of people who will tell you never to put "'s" after a proper name ending in "s", but I haven't seen that in a style book. I've even seen "Max'" and "Jeez'", but I think they are simply the usual work of illiterates. (And no, I don't want to hear your or your English teacher's vote unless it's a different criterion, so for Jeez's sake forbear, like Willy the Shakes might have said.) Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Article 5749 of dc.general: Newsgroups: dc.general,alt.usage.english From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Subject: Possessives (was Re: Math in Wash Post) Followup-To: alt.usage.english Sender: usenet@ra.nrl.navy.mil Organization: Naval Research Laboratory, Washington, DC Date: Sat, 3 Jul 1993 18:43:29 GMT Yesterday, about a New York Times article on the Fermat's Last Theorem proof, I wrote: > Digressing toward alt.english, Which should have been alt.usage.english, of course > I noticed the Times mentioned "Wiles'" proof on page 1 and "Wiles's" > proof on page D22. I wonder which is the Times's style. If they > followed the AP style, I guess it could possibly be that the > front-page editor thought "Wiles" had two syllables. But note that > Strunk and White would have you put "'s" after every singular--even > "Charles's"--except for "the names of certain ancients". (I think > the exception is a craven cavein because they didn't want to take on > the Bible's use of "Jesus'".) And there are plenty of people who > will tell you never to put "'s" after a proper name ending in "s", > but I haven't seen that in a style book.... Looking for my S&W I ran across Fowler, whom I should have consulted in the first place: The former prescribe, the latter explains: It was formerly customary, when a word ended in "-s", to write its possessive with an apostrophe but no additional "s", e.g. Mars' hill, Venus' Bath, Achilles' thews. In verse and in poetic or reverential contexts, this custom is retained.... But elsewhere we now usually add the "s" and the syllable--always when the word is monosyllabic, and preferably when it is longer, Charles's Wain, St. James's Street, Jones's children, the Rev. Septimus's surplice, Pythagoras's doctrines. (So it's clear that the front page editor was simply being properly worshipful.) But Fowler still doesn't justify the syllabic criterion. > (And no, I don't want to hear your or your English teacher's vote > unless it's a different criterion, so for Jeez's sake forbear, like > Willy the Shakes might have said.) And Fowler has another section on constructs involving "sake", which follow their own, stranger rules. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.games.abstract From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Tue, 6 Jul 1993 08:50:36 GMT Subject: Re: Hex torb...@diku.dk (Torben AEgidius Mogensen) writes of Hex >>>...with the modification that the board is of size Nx(N-1) and the >>>second player has to connect the shortest distance, the game is a >>>sure win for player two. >>>The strategy works as follows: divide the board into two >>>equal-sized parts along the short diagonal. Every time player one >>>places a stone in one half of the board, player two places a stone >>>in the other half [at a point mirrored across the dividing line]. Last week I took issue with his formulation, ending: > There may be a winning strategy for the second player on the Nx(N-1) > board, but I would be quite surprised if it's a simple one. At that point a...@gen.cam.ac.uk (Aubrey de Grey) was kind enough to surprise me with the news that I had misinterpreted Torben's remarks (which are approximately correct, though somewhat vague) and provided a proof which I present in somewhat modified form below. Torben's idea is to divide the board in half along (almost, but not quite) the short diagonal of the board, and match up the halves as (slightly offset) mirror images. For example, the 7x6 board is matched up as in the left diagram below: / a b c d e f/ g h i j k F a b c d e f l m n o K E g h i j k p q r O J D l m n o s t R N I C p q r u T Q M H B s t /U S P L G A u / To prove that this is a second player win, let us identify the two halves of the board and let the first player play solitaire on the diagram at right. He puts down a red marker whenever he would play on the northwest (lower case) half of the original board, and a blue marker whenever he would play on the southeast (upper case) half. The second player's strategy ensures he cannot put both colors down on the same point. Now to see if the first player has won, we connect any two adjacent markers of the same color, and to model connections across the mirror, we connect pairs along the southeast (utrokf) edge whenever a red marker has a blue neighbor to its northeast. (No connection is made when the blue neightbor is to the southwest, because for example positions "o" and "R" are not adjacent in the original diagram, while positions "r" and "O" are adjacent.) The object is for the first player to make a path from a red marker on the north (abcdef) edge to a blue marker on the southwest (aglpsu) edge. But the only place that the path can change color is on the SE edge, and a red path from from the N edge to the SE edge blocks the blue NE neighbor from continuing on to the SW edge. So the first player cannot win, and the impossibility of draws implies that the second player wins. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Date: Tue, 06 Jul 93 17:28:55 -0400 (EDT) From: Dan Hoey To: Cube Lovers Subject: Rubik's Shrimp I hear on Usenet that there's a show on BBC1 called Wildlife 100, where people saw a Mantis Shrimp playing with a Rubik's Cube. Reports are inconclusive as to whether it was able to actually turn faces, or whether it just waved it around, or even just took it apart. Now if they could get it to turn faces, presumably they could film it and play it back in reverse.... Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Wed, 7 Jul 1993 16:30:32 GMT Subject: Re: Floor Tiling aa...@eecs.nwu.edu (Aaron Feigelson) writes: > Of the twenty four distinct (up to symmetry) heptadiamonds ^^^^^^^^^^^^^ heptiamonds > (shapes composed of seven equilateral triangles with adjacent > edges) EXACTLY ONE cannot be used as the prototile of a monohedral > planar tiling. Well, it's been nearly two months since this problem was last posted in rec.puzzles, so I guess it's time again. Of course, I turned immediately to George E. Martin's Polyominoes_ _A_Guide_to_Puzzles_and_Problems_in_Tiling. He tells us it's the V heptiamond, but unfortunately his figure 10.8 was printed upside- down, so it's not immediate what he means. But it's pretty easy. Twenty-one of the heptiamonds satisfy Conway's criterion, and two form easy satisfying constellations, so the odd man out is *---* *---* \ \ / / * * * \ / *---* which looks like it should be called the V heptiamond, anyway. I wonder how many other of them have names? It's easy to show the V doesn't tile: The concavity must be filled, and there's only one way up to symmetry (A) then the right concavity must be filled, in the only way doesn't fail immediately (B). Then the top concavity must be filled in one of two ways, each of which has an unfillable concavity (C1,C2). (A) (B) *---*---* *---*---* \ \ / / *---*---* *---* *---* * * *---* \ \ / \ (C1) / / \ \ (C2) *---* * * * *---*---* *---* *---*---* *---* / / / \ \ \ \ / \ \ \ / \ *---* *---* *---* *---* * * * *---* * * * \ \ / / / / / \ \ / / / \ \ * * * * *---* *---* * *---* *---* \ / \ / \ / *---* * * Aaron's use of `heptadiamonds' brings up some other questions. Obviously, the heptadiamond should refer to a special kind of 14-iamond, one that can have its triangles paired into seven diamonds. Since all the diamonds, tetriamonds, hexiamonds, and octiamonds tile the plane, the first n-diamond candidate for non-tiling is the pentadiamond. Can anyone find a non-tiling pentadiamond? The two with holes doesn't count! Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Thu, 8 Jul 1993 21:00:46 GMT Subject: Tiling: The Conway Criterion Yesterday I mentioned that all but three of the heptiamonds satisfy the Conway criterion, but I failed to mention what the criterion is. Since the criterion is a must-know for anyone who wants to (monohedrally, periodically) tile the plane, I'll describe it for you. A piece satisfies the Conway criterion if you can divide its perimeter into stretches 0,1,2,3,4,5 reading clockwise, such that stretches 1,2,4,5 each have "S"-symmetry, while 0 and 3 are translated copies of each other (with the piece on the side of 0 opposite to the side it is of 3, so the stretches fit together). Conway suggests marking stretches 0,3 with heavy lines, and separating the 1,2 and 4,5 stretches with big dots. I've done that for the following 30-omino, with numbers and arrows added to emphasize the stretches. >>>>>>>>>>> *--+--+--+--+ # | 1 0 *==* *==* +--+--+ # # # | * *==*==* +--+--+--+--@ <<<<<< # | >>>>> * +--+--+ 2 | | +--+ * <<<<<<<<<<<< | # +--+ *==* *==* 5 | # # # 3 +--+ * *==*==* | # >>>>>>>>>>>>>> @--* <<<<<<<<<<<<<<<<<< 4 The piece needn't be an n-omino, or even a polygon--the stretches may be arbitrary curves as long as they satisfy the symmetry requirements, and any of the stretches may be as small as a point. A piece satisfying the Conway criterion tiles the plane by putting copies of it together with stretches 0-3, 1-1, 2-2, 4-4, and 5-5 of adjacent copies touching. Note that the Conway criterion is a sufficient condition to prove that a piece tiles the plane but not a necessary one. Some pieces fail the criterion and still tile the plane. The criterion is useful because such counterexamples are relatively scarce, at least among the small -ominoes and -iamonds. If a piece fails, you can often put a few copies of it together to form a constellation that satisfies the criterion. Otherwise you can attempt to prove that the piece can or can't tile by some other means. If you want to practice--or just an Obpuzzle--here is the set of twenty-four heptiamonds from George E. Martin's book. Find the three that *don't* satisfy Conway's criterion. Then with each of two others, find the smallest monohedral constellation that *does* satisfy the criterion. *---*---* / \ *---* * * / \ / \ / \ *---*---*---* * * *---*---* / / \ / \ \ * *---*---* *---* *---*---*---* / / / / \ \ / \ * *---* *---* * *---* * * \ / \ / / \ \ / \ / * * * *---* *---* * * * \ \ \ / \ \ / \ \ / * *---* *---*---* * * * / / / / / \ \ \ * * *---* *---*---*---* * * / / \ \ \ / \ \ *---* * *---* * *---*---* *---* \ \ / \ / / \ \ / / * *---*---* *---*---* * * * \ \ \ / \ / *---*---*---*---* * *---*---* \ / \ / * *---* * \ \ / *---*---* Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: soc.veterans, soc.motss, soc.bi, talk.bizarre Followup-To: soc.veterans From: hoey@zogwarg.tec.army.mil (Dan Hoey) Date: 15 Jul 93 23:54:53 GMT Subject: Re: CLINTON TO MILITARY "TAKE GAYS AND LIKE IT!" rpwh...@cs.nps.navy.mil writes: > ...[excerpted without permission from The Washington Herald]... jeddi@next16pg2.wam.umd.edu, allegedly posting from UMd College Park, writes: >The Washington Herald?? man!!.. this paper is not in Washington DC is it?? >Couldn't you quote a more respectable paper?? Must be some kind of tourist. Real Washingtoonians know where the Herald is. Dan --- Newsgroups: comp.programming From: hoey@zogwarg.tec.army.mil (Dan Hoey) Date: 16 Jul 93 20:42:09 GMT Subject: Re: Measuring Time for First Output d...@ccr-p.ida.org (David desJardins) writes: >If there is a problem with quicksort in an application, it is more >likely to be due to some other problem, like its variable running time >and poor worst-case behavior. m...@hubcap.clemson.edu (M. J. Saltzman) writes: >Defeating poor worst-case running time isn't that hard. You can use >a median-of-three rule to select the partition element, which makes >the worst case extremely unlikely. This seems to miss the point completely. You can't defeat worst-case behavior by making it less likely. Changing the likelihood affects average-case behavior, and perhaps reduces the variance, and may affect other statistical measures. But if you want good worst-case behavior, you need to minimize the maximum running time. That's the definition of worst-case behavior. Fortunately, this is possible for quicksort. There are known linear- time median-finding algorithms that can be used to partition the set exactly in half. I think you can save some time if you use a algorithm that guarantees finding an r-tile to partition on--i.e., an element with at least rn elements on either side. As long as the r-tile algorithm is O(n) time, you are guaranteed O(n log n) worst case time. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles, rec.games.abstract Followup-To: rec.puzzles From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Mon, 26 Jul 1993 22:34:49 GMT Subject: Matching digits (was Re: A puzzle) [ Note that I am crossposting this to rec.games.abstract. While this game does not fall under the letter of r.g.a's charter (being non- deterministic) it certainly is an abstract game subject to theoretical analysis. Please followup to rec.puzzles, though. ] A little over three weeks ago, r...@att.com (Ramakrishna Kandarpa) posed a problem which I would rephrase: Two players, the chooser and the matcher, each pick a sequence of six decimal digits. If the sequences share a pair of adjacent digits, the matcher wins; if not, the chooser wins. Find optimal strategies and the probability of a match. st...@hal.nta.no (Stein Kulseth) noticed that the problem statement is ambiguous, in that it is not stated whether an adjacent pair matches its reversal, so there are two different games to analyze. Stein correctly notes that if we count reversed matches, the chooser's only pure undominated strategies are to pick any pair as his only pair, by playing either XYXYXY for a heterogeneous pair or XXXXXX for a homogeneous pair. To prevent the matcher from taking advantage of regularities in his play, the chooser should pick his pair uniformly from the 55 possible pairs. The matcher's pure undominated strategies are to pick 5 different pairs. As long as any of the 55 possible pairs is equally likely to be used, he guarantees a 1/11 probability of winning. This analysis generalizes to sequences of any length of digits in any radix. If a heterogeneous pair does not match its reversal, the problem is much harder. Stein's very good analysis of the problem was marred by two errors: a trivial mistake that caused him to get the wrong answer, and a deep mistake that had no effect on the outcome! The trivial mistake was counting 90 pairs of digits (10 homogeneous pairs and 80 heterogeneous pairs) rather than 100 pairs of digits (10 homogeneous, 90 heterogeneous). The deep mistake was to assume independence of events that were correlated, in order to prove an inequality that turns out to be satisfied anyway. It is clear at first that the chooser's only undominated strategies are to pick either six different digits, or a cycle of one to five digits repeated as many times as necessary to fill out the sequence. For if the chooser's sequence contains any digit twice, d[i] d[i+1] ... d[k] d[i] the entire sequence may consist of d[i]...d[k] without any increased exposure to matches. Let us first analyze the game if the chooser's strategy is a combination of the strategies C1=XXXXXX and C2=XYXYXY. The matcher should then avoid any sequence in which a heterogeneous pair and a its reverse both appear. It should be clear that this game is effectively identical to the reverse-matching variant, since the chooser includes the reverse of each pair. The chooser should follow strategy C1 2/11 of the time and strategy C2 9/11 of the time, so that each homogeneous pairs is chosen with the same frequency as each unordered heterogeneous pair. The matcher follows the same frequency, by averaging 10/11 of a heterogeneous pair and 45/11 homogeneous pairs in each sequence. The probability of a match is 1/11. Now suppose the chooser were to use one of strategies C3=XYZXYZ, C4=XYZWXY, C5=XYZVWX, and C6=XYZUVW some fraction of the time. Stein's important insight is to show that these increase the probability of matching heterogeneous pairs above the probability of C2 matching, no matter how many heterogeneous pairs the matcher picks. Suppose the matcher picks N heterogeneous pairs. Stein reasoned that since the probability of one of the chooser's pairs fails to match one of the matcher's is (1-N/90), the probability of three failing to match is (1-N/90)^3, leading to a match probability of 1-(1-N/90)^3. But the chooser's three pairs are not independent, so (1-N/90)^3 is not the correct joint probability of failure to match. I have determined the correct probabilities by computer search, and they are given in the following table. N Stein XYZX XYZW XYX 2 5941/91125~.0652 1/16 =.0625 23/360 ~.0639 2/45~.0444 3 2611/27000~.0967 11/120~.0917 95/1008~.0942 1/15~.0667 4 631/3375 ~.1275 29/240~.1208 313/2520~.1242 4/45~.0889 5 919/5832 ~.1576 3/20 =.15 155/1008~.1538 1/9 ~.1111 The table assumes that the chooser's sequence does not repeat any digit except in homogeneous pairs; this can be done without impairing the chooser's strategy against C1 and C2. The "Stein" column has Stein's incorrect lower bound on the match probability. The "XYZX" column has the correct match probability with strategy C3. The "XYZW" column has the correct match probability with three of the pairs in strategies C4-C6. The "XYX" column has the correct match probability with strategy C2. Note that although Stein's lower bound is not in fact correct, the match probabilities for C3-C6 are still greater than the match probability for C2. Thus a competent chooser will never use C3-C6, and the analysis with the chooser restricted to C1 and C2 applies to the unrestricted game as well. I have done some preliminary investigation of the game with different length of sequence and digits of different radix. For radix 2 and 3 I calculated complete tables of optimal strategies for chooser and matcher in the following table (though I have not thoroughly checked my case analysis). For the chooser's strategy I only list the first cycle. The table ends when the matcher is able to achieve an unavoidable match. Radix 2 Length Chooser Matcher Prob 1 X any 0 2 (1/2)XX+(1/2)XY (1/2)XX+(1/2)XY 1/4 3 XX XXY 1/2 4 any XXYY 1 Radix 3 Len Chooser Matcher Prob 1 X any 0 2 (1/3)XX+(2/3)XYX (1/3)XX+(2/3)XY 1/9 3 (1/2)XX+(1/2)XYX XXY 1/3 4 (1/2)XX+(1/2)XYX (1/2)XXYY+(1/2)XXYZ 1/2 5 (3/8)XX+(3/8)XYX+(1/4)XYZX (3/4)XXYYZ+(1/8)XXYYX+(1/8)XXYZY 5/8 6 (3/8)XX+(3/8)XYX+(1/4)XYZX (1/4)XXYYZZ+(1/4)XXYYZX+(1/2)XXYYZY 3/4 7 (3/8)XX+(3/8)XYX+(1/4)XYZX (1/4)XXYYZZX+(3/8)XXYYZZY+(3/8)XXYYZXZ 7/8 8 any XXYYZZX 1 The regularities in the radix 3 table and the piecewise linear probability are most intriguing, and I wish I had time to construct the radix 4 table. I know only the following portion of that table: Radix 4 Len Chooser Matcher Prob 1 X any 0 2 (1/4)XX+(3/4)XYX (1/4)XX+(3/4)XY 1/16 3 (2/5)XX+(3/5)XYX (4/5)XXY+(1/5)XYZ 1/5 4 (2/5)XX+(3/5)XYX (1/5)XXYY+(4/5)XXYZ 3/10 .. 14 any WWXXWYYZZYXZWZ 1 I would be interested if anyone knows of or would like to investigate further results on this game. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Date: Thu, 29 Jul 93 08:36:14 -0400 (EDT) From: Dan Hoey To: Cube Lovers Subject: Re: Hint wanted for 4x4x4 ronnie@cisco.com (Ronnie B. Kon) asks for hints for exchanging a pair of edges: I tend to solve using commutators, but I don't see a way here.... The key is that commutators are always odd permutations. So do the move that is an odd permutation of the edges, then use commutators. Dik.Winter@cwi.nl (dik t. winter) shows a neat way of moving most of the cubies affected by the odd permutation into the top slice, where they can be cycled using Ronnie's commutator, which cycles the TB(R), TR(F), and TF(L) cubies: . . 0 . . . . . . . . 0 . 0 . . (I'm naming them by their edge and their near side.) I suspect Ronnie is using something like (F Ti F') T (F Ti' F) T (F Ti F') T^2 (F Ti' F). (For this I'm using "i" to mark inside slabs). But you can cycle the FL(T), FR(T), RB(T) cubies directly, using a different commutator. With more effort, there is a commutator that doesn't mess up face centers. We are getting to the part where it's hard to distinguish between the hintable and the obvious, but if people send me email about not being able to figure out what commutators I'm talking about I'll answer, and post them if such nobility is common. My solution for the 4x4x4 always was: first corners, next edges and finally centers. Because there are many identical pieces for the centers those are reasonably simple. It would be much more difficult if each center had its own place. As I mentioned years ago, I've made places for mine by cutting corners of to clusters of face centers and their neighboring edges on each face. +----+----+----+----+ | | | | | | | | | | +----+---( )---+----+ | | | | | | | | | | +---( )---+----+----+ | | | | | | | | | | +----+----+----+----+ | | | | | | | | | | +----+----+----+----+ It's not that hard to fix the face centers, just time-consuming. It's a good thing we do the edges first, though, or it would be hard to figure where the cuts go. Dan Hoey Hoey@AIC.NRL.Navy.Mil ( So much discussion on this quiescent list will probably flush out someone who wants to unsubscribe. Remember to send your note to cube-lovers-request@ai.ai.mit.edu to avoid annoyance.) --- Date: Thu, 29 Jul 93 10:11:13 -0400 (EDT) From: Dan Hoey To: Cube Lovers Subject: Oops... Re: Hint wanted for 4x4x4 I wrote: ? The key is that commutators are always odd permutations. So do the ? move that is an odd permutation of the edges, then use commutators. But I *Meant*: ! The key observation is that commutators are always even ! permutations. So you to perform an odd permutation on edge, you ! should do the move that is an odd permutation of the edges, then ! use commutators. Dan --- Newsgroups: alt.humor.best-of-usenet From: Dan Hoey Date: Sat, 7 Aug 1993 06:01:29 GMT Subject: [talk.bizarre] Corny but amusing From: cur...@snake.CS.Berkeley.EDU (Curtis Yarvin) Newsgroups: talk.bizarre,sci.med.dentistry,misc.kids Subject: Re: Short Shameful Cornfession bout...@isis.cshl.org (Justice of the Peach) writes: >After I finished gnawing, I discovered I had a little bit o' cornstuff >wedged between my teeth. >I spent several minutes determinedly fingering away in an attempt >to remove it. >I finally got it out. Lucky bugger. As anyone who's ever made grits or dropped a corncob in the Mississippi knows, CORN EXPANDS IN WATER. Yes, boys and girls, this includes saliva. Corn is a fine American food but you must treat it with the respect it deserves. ALWAYS SWALLOW YOUR CORN! Do NOT let it loiter about in your mouth as you suck out the last tender wisps of flavor; do NOT get it stuck between your teeth. If you do, remove it with any available tools or contact a certified dental practitioner AT ONCE. Even the TINIEST SHRED of corn, it stays wet long enough, will expand to a size greater than your entire HEAD. By this time you will almost certainly be dead. If you have an especially quick tongue, you may be able to flick it out of your mouth right after it rips your teeth out, but do NOT count on this possibility. IMMEDIATE AMPUTATION is the ONLY SAFE COURSE! So be careful out there, boys and girls. Food is NOT a game. -- -- alt.humor.best-of-usenet -- -- Funniest postings from USENET, altnet, and the worlds beyond -- -- Moderator's address: b...@polaris.async.vt.edu -- --- Date: Fri, 13 Aug 93 18:26:09 -0400 (EDT) From: Dan Hoey To: Cube Lovers Subject: Re: Squares group Mark Longridge has some interesting things to say about the antipodes of the group generated by half-turns: If we define "symmetry level" as the number of distinct patterns generated by rotating the cube through it's 24 different orientations in space then most known antipodes are symmetry level 6. Thus the lower the number the higher the level of symmetry. The least symmetric positions have level 24, and this is very common. This approach is somewhat unfortunate in two ways. First, it would be better to use the full 48-element symmetry group M of the cube, because some patterns are not recognized as transformed images of each other if you only use the 24-element group C of rotations. For instance, the positions reached by processes F2R2T2 and F2T2R2 cannot be related with C, so you would see four classes of positions at distance three rather than three. But the antipodes you give are all mirror-symmetric, so there is no new coalescence there. Relating processes that are conjugates by a reflection is usually somewhat tricky, since the moves of the process must be changed in direction (replacing clockwise by counterclockwise) but in the squares group this is a nonproblem. The second deficiency of your approach is that you lose information by specifying only the index of the symmetry subgroup (the ``number of distinct patterns generated ...''). It makes sense to find out exactly which subgroup of M is the symmetry group of your positions. I've done that, below. Each of these symmetry groups comes in three conjugates, so I've transformed some of the processes (marked x) so they all use the same particular symmetry group(s). The group elements are given as cycles of the cube faces, so (TD)(FRBL) means to reflect T<->D and rotate F->R->B->L->F. Cases with symmetry level 6: These are cases where the symmetry group has order 8. p66x Double 4 corner sw L2 D2 R2 T2 L2 T2 F2 R2 (F2 B2 T2 F2) T2 L2 B2 p80 2 DOT, Invert T's R2 B2 D2 R2 B2 L2 B2 L2 (T2 D2 F2 T2) F2 L2 T2 p99 2 DOT, 4 ARM R2 B2 D2 L2 B2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2 p100 2 Cross, 4 ARCH 1 R2 B2 T2 R2 F2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2 p66x, p80, p99, and p100 have symmetry group P=<(TD)(FRBL),(FB),(LR)>. p67x Antipode 2 R2 D2 B2 T2 B2 T2 F2 L2 (F2 B2 T2 F2) L2 F2 D2 p130 2 Cross, 4 ARCH 2 L2 B2 D2 B2 L2 D2 F2 L2 (T2 D2 F2 L2) F2 L2 T2 p67x and p130 have symmetry group Q=<(TD),(FRBL)>. p135x 2 X, 4 T D2 B2 L2 F2 R2 F2 R2 D2 (R2 L2 F2 R2) D2 L2 F2 p137 2 X, 4 ARM L2 F2 T2 B2 T2 F2 T2 L2 (T2 D2 F2 T2) L2 D2 F2 p135x and p137 have symmetry group S=<(TD),(FB)(LR),(FR)(BL)>. Cases with symmetry level 12: These have 4-element symmetry groups. p108 2 DOT, 2 T, 2 ARM L2 F2 T2 R2 B2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2 p128x 2 H, 2 T, 2 CRN L2 D2 R2 T2 L2 T2 F2 R2 (F2 B2 T2 F2) T2 L2 F2 p129x 2 H, 2 T, 2 ARCH R2 T2 L2 T2 L2 T2 F2 R2 (F2 B2 T2 F2) T2 L2 F2 p131x 2 H, 2 ARM, 2 ARCH L2 T2 R2 B2 D2 L2 B2 L2 (F2 B2 T2 F2) T2 L2 T2 p132 2 Cross,2 ARCH,2CRN L2 F2 D2 R2 F2 L2 B2 L2 (T2 D2 F2 T2) F2 L2 D2 p136x 2 H, 2 ARM, 2 CRN R2 T2 L2 F2 D2 L2 F2 L2 (F2 D2 T2 F2) T2 L2 D2 p108, p128x, p129x, p131x, p132, and p136x have symmetry group HP=<(FB),(LR)>. p133x 2 Cross, 2 T, 2 ARM L2 T2 B2 T2 B2 T2 F2 L2 (F2 B2 T2 F2) L2 F2 D2 p134x 2 CRN, 2 X, 2 ARCH L2 T2 B2 D2 F2 T2 F2 L2 (F2 B2 T2 F2) L2 F2 D2 p133x and p134x have symmetry group HS=<(TD),(FB)(LR)>. In case you have trouble forming the closure of these groups: P = {I, (FB)(LR), (TD)(FRBL), (TD)(FLBR), (FB), (LR), (TD)(FR)(BL), (TD)(FL)(BR)} Q = {I, (FB)(LR), (TD), (TD)(FB)(LR), (TD)(FRBL), (TD)(FLBR), (FLBR), (FRBL)} S = {I, (FB)(LR), (TD), (TD)(FB)(LR), (TD)(FR)(BL), (TD)(FL)(BR), (FR)(BL), (FL)(BR)} HP = {I, (FB)(LR), (FB), (LR)} HS = {I, (FB)(LR), (TD), (TD)(FB)(LR)}. I should note that the subgroup names M, C, P, Q, S, HP, and HS are part of a general classification of subgroups of M that I worked out some time ago. I have a chart of them I can send; just ask by email. A few observations... - It is not possible to swap just 1 pair of edges and corners Certainly, all the generators are even permutations on the edges and on the corners. - It is only possible to have 4, 6 or 8 corners out of place That is a nice, concise way of putting it. To elaborate, if you permute one of the corner orbits in a 3-cycle, the other will also be permuted in a 3-cycle; otherwise, any pair of cycle structures of the same parity is possible. - In reaching an antipode one may start with any of the 6 turns (since antipodes are global maxima, any turn will get you one move closer) Careful! This also relies on the fact you call a conjecture, below. Otherwise you could have two neighboring global maxima, and their inverses would be antipodes that do not have this property. For instance, consider the corner group as generated by the 24 pairs of neighboring squares (F2R2, etc). This is a 48-element group with diameter 2, trivial enough to be analyzed by hand. Antipodes (L2B2)(D2R2) and (D2R2)(T2F2) are neighbors, because (L2B2)(D2R2)(F2T2)=(D2R2)(T2F2). So there is no length-2 process equivalent to (F2T2)(R2D2) that starts with T2F2. - If the corners are fixed, the position is NOT an antipode - All known (probably all!) antipodes have symmetry level 6 or 12 I presume these comments are left over from before you found them all. - Longest order appears to be 12 Appears? The orbits are all of size 4 (two orbits of corners, three orbits of edges), so 12=LCM(2,3,4) is an easy upper bound. Finding one is easy given the processes Singmaster lists. - Although only conjectural, it is now believed that one turn of a face MUST lead to a new state which is either 1 move closer or 1 move farther from START Conjectural? It's immediate from the fact that each generator is an odd permutation of the corner orbit {FTR,FDL,BTL,BDR}. Question: Are there any irreducible square's group sequences that are longer then 10 moves? Are these truly irreducible or only irreducible under Dik Winter's Kociemba inspired program? Well, that could be searched for; a matter of checking 600K positions for each of the 15K or so pattern representatives. I hope I can find the time to hack it up. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Date: Mon, 16 Aug 93 18:05:47 -0400 (EDT) From: Dan Hoey To: Cube Lovers Subject: Squares group, correction I should proofread these things better. I got the processes for p130, p135x, p137, and p136x wrong in my last message. Here is the corrected list of squares-group antipodes and their symmetry groups. SG Pos Name Process P p66x Double 4 corner sw L2 D2 R2 T2 L2 T2 F2 R2 (F2 B2 T2 F2) T2 L2 B2 P p80 2 DOT, Invert T's R2 B2 D2 R2 B2 L2 B2 L2 (T2 D2 F2 T2) F2 L2 T2 P p99 2 DOT, 4 ARM R2 B2 D2 L2 B2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2 P p100 2 Cross, 4 ARCH 1 R2 B2 T2 R2 F2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2 Q p67x Antipode 2 R2 D2 B2 T2 B2 T2 F2 L2 (F2 B2 T2 F2) L2 F2 D2 Q p130x 2 Cross, 4 ARCH 2 T2 B2 R2 B2 T2 R2 F2 T2 (L2 R2 F2 T2) F2 T2 L2 S p135x 2 X, 4 T L2 D2 B2 T2 F2 T2 F2 L2 (F2 B2 T2 F2) L2 B2 T2 S p137x 2 X, 4 ARM L2 T2 F2 D2 F2 T2 F2 L2 (F2 B2 T2 F2) L2 B2 T2 HP p108 2 DOT, 2 T, 2 ARM L2 F2 T2 R2 B2 L2 F2 L2 (T2 D2 F2 T2) F2 L2 T2 HP p128x 2 H, 2 T, 2 CRN L2 D2 R2 T2 L2 T2 F2 R2 (F2 B2 T2 F2) T2 L2 F2 HP p129x 2 H, 2 T, 2 ARCH R2 T2 L2 T2 L2 T2 F2 R2 (F2 B2 T2 F2) T2 L2 F2 HP p131x 2 H, 2 ARM, 2 ARCH L2 T2 R2 B2 D2 L2 B2 L2 (F2 B2 T2 F2) T2 L2 T2 HP p132 2 Cross,2 ARCH,2CRN L2 F2 D2 R2 F2 L2 B2 L2 (T2 D2 F2 T2) F2 L2 D2 HP p136x 2 H, 2 ARM, 2 CRN R2 T2 L2 F2 D2 L2 F2 L2 (F2 B2 T2 F2) T2 L2 D2 HS p133x 2 Cross, 2 T, 2 ARM L2 T2 B2 T2 B2 T2 F2 L2 (F2 B2 T2 F2) L2 F2 D2 HS p134x 2 CRN, 2 X, 2 ARCH L2 T2 B2 D2 F2 T2 F2 L2 (F2 B2 T2 F2) L2 F2 D2 Sorry if anyone was led astray. Dan --- Newsgroups: news.admin.policy From: hoey@zogwarg.tec.army.mil (Dan Hoey) Date: 20 Aug 93 19:20:29 GMT Subject: Re: WARREN THE MASS POSTER b...@cyberspace.com (Brian Cartmell) writes: > Brian Cartmell (b...@cyberspace.com) wrote: >: Ok, I have received about 1200+ messages on this user and expect about >: another 1000+. I'm still in the process of checking all the facts out >Its me again, well now I have received over 3000 messages from some really So why haven't you issued cancels for the original postings? People won't complain (or have a reason to complain) about messages they don't see. I know cancels have been disabled in a lot of places, but there are a lot of places where they are still useful. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.arts.comics.misc From: hoey@zogwarg.tec.army.mil (Dan Hoey) Date: 20 Aug 93 23:15:58 GMT Subject: Re: PEOPLE: Alison Bechdel (was Re: Searching for Wimmin's Comix) j...@lynx.frame.com (James Drew) writes: >Alison Bechdel is some kind of lesbian comix goddess. Since 1985, she >has produced the strip "Dykes To Watch Out For," now syndicated in >more than 40 gay/lesbian, feminist, and progressive newspapers around >the world. For a brief time, she also contributed a monthly strip to >The Advocate, entitled "Servants to the Cause.".... I'll second the lesbian comix goddess. But be sure to mention the latest ish of Gay Comics, which is an all Bechdel book. ``Featuring absolutely no Dykes to Watch out for!'' It's got a selection of ``Servants'' and a long autobiographical piece. Not to be missed. Dan --- Newsgroups: rec.arts.comics.misc From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Tue, 24 Aug 1993 21:42:55 GMT Subject: Re: PEOPLE: Alison Bechdel Jym Dyer writes > ... It's a pity that the all-Bechdel issue is just going to > reprint a (different) strip.... Perhaps it would have been a pity. But as I said in the message you quoted another line from, it doesn't just repost that strip, it also contains a long autobiographical piece. Or more precisely, as James Drew, j...@lynx.frame.com mentioned: > ...autobiographical strips entitled "Coming Out Story," The Power of > Prayer," and "True Confession".... And while I'd always like to see more new stuff, I don't feel the need for her to break out of any ruts. The plot synopsis you propose doesn't fit the strips I remember. Mileage varies, and all that: I expect you remember a few that annoyed you, while I remember a few that amused me. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: misc.entrepreneurs, misc.forsale, misc.misc, misc.wanted, alt.folklore.urban, alt.folklore.science, sci.chem Followup-To: alt.folklore.science From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Wed, 25 Aug 1993 22:33:48 GMT Subject: Red mercury is back! (Was: Help information needed) fche...@ucsb.edu (Francis Cheung) twitters: > I would like to know if anyone out there knows of companies, > corporations, research institutions, etc. which uses, buys, > or sells metals such as Osmium, Scandium, Red Mercury, > Cobalt, Titanium, and Tungsten. > I have an overseas source for these metals and would like > to know where I should start to contact firms that would use > these metals. The metals from my source are legitamate and > legal with all records and certificates of origin. I was told > that the metals that I listed above come in either powder or > ingot. .. > fche...@cs.ucsb.edu > or > sophi...@cs.ucsb.edu > voice: 805 968 4354 Much of this sounds normal, if unusual. But ``Red Mercury'' is some mythical ingredient of nuclear weapons that makes them beelions of times more powerful. Either that or it's a code word for plutonium, depends on which supermarket tabloid or spy thriller you read. Personally, I much prefer Red Mars. Followups to alt.folklore.science, I suppose. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Date: Thu, 26 Aug 93 10:30:57 -0400 (EDT) From: Dan Hoey To: Cube Lovers Subject: Tartan reborn (Re: Tools lost in the mists of time...) Alan Bawden mentioned the joy of rediscovering his lost cube-solving techniques. This happened to me about three years ago for an unusual reason. I've become active in science fiction fandom, and fans determine where the World Science Fiction Convention (Worldcon) is held each year by running miniature political campaigns. A friend of mine was bidding for Glasgow, and she asked if I had any `plaid things'. I told her I had a plaid Rubik's cube, and a political strategy was born. The plaid cube is of course the Tartan, which Jim Saxe and I discovered and described in this group on 16 February 1981 (see archives). I blanked some old cubes, and figured out how to use spray paint to efficiently create Tartan cubes. I produced a half dozen or so, and they make good conversation pieces at conventions. Unfortunately, I seem to be the only convention-going science fiction fan who can *solve* a Tartan (with the possible exception of Phil Servita who as I recall figured out an effective method but wearied in its execution). So I would see a scrambled Tartan at a convention party, and fix it, and put it down, and five minutes later it would be scrambled again. I quickly found out how rusty I was, and through the enforced practice I've gotten about as good as I was a decade ago. But some of the Glasgow promoters took Tartan cubes over to the UK, and those cubes just never get solved. I sent them instructions for solving it, but I don't know if any of them have figured out the instructions. Well, eventually they told me they really wanted something mere mortals could deal with, and I painted some pieces of wood plaid that they could use for doorstops. I was surprised, though, to find that to make a plaid pattern going around a corner, if you only have four colors of paint, it seems the *only* thing you can do is use a coloring locally identical to the Tartan. As for the cubes in the UK, I expect to get there in 1995. For it seems the clever ploy worked, and the fans voted to have the 1995 Worldcon in Glasgow. I'm sure they owe it all to the Tartan. Sure. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Article 3176 of rec.arts.sf.fandom: Newsgroups: rec.arts.sf.written,rec.arts.sf.fandom From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Subject: Re: 1993 Hugo Winners Followup-To: rec.arts.sf.fandom Sender: usenet@ra.nrl.navy.mil Organization: Naval Research Laboratory, Washington, DC Date: Thu, 16 Sep 1993 21:16:19 GMT In reply to > The rumor that I heard is that Charlie Brown forgot to vote.... and other rumors, sgoldberg@davidsys.com (Seth Goldberg) writes: > ...I can not swear my memory is 100% accurate, but I do have a good > memory of ballots from Charles N Brown and Andrew Porter. How and > what they voted for is not part of the public record. And is the fact that they voted at all considered part of the public record? Not that I can imagine them complaining in this case, but was thought given to setting a precedent? Given the concern over Brad Templeton's second vote, I can imagine the result if you reported, for instance, that J. Random Nominee's infant offspring submitted a Hugo ballot. Incidentally, I still prefer the version that says Charlie abstained because of a premonition he'd have to share the stage with Andy. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Article 3180 of rec.arts.sf.fandom: Newsgroups: alt.folklore.computers From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Thu, 16 Sep 1993 21:19:37 GMT Subject: Re: C macros (was Incompetent ...) pdsm...@bbn.com (Peter D. Smith) writes: > #define SEVEN 5+2 > main () > { > printf ("8 * 7 is %d\n", 8 * SEVEN); > } Right answer, wrong question. To match the Hitchhiker's Guide, this should be #define SIX 1+5 #define NINE 8+1 Then the question to Life, the Universe, and Everything is ``What is SIX * NINE?'' This makes about as much sense than the base-13 explanation. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.arts.sf.fandom From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Subject: The Bermuda Triangle Upset theory (Re: RISC Worldcon) Sender: usenet@ra.nrl.navy.mil Organization: Naval Research Laboratory, Washington, DC Date: Fri, 17 Sep 1993 02:16:56 GMT With regard to the original subject, I think these people were actually talking about a RASC (reduced _at_traction set con), but.... On the subject of the Bermuda Triangle Upset theory, dlow@svale.hp.com (Danny Low) writes: > According to someone who analyzed the voting for the Bermuda Triangle > bid, it came very close to winning. Basically it was everyone's SECOND > choice. If the first round of votes had not come up with a majority, > the Bermuda Triangle would have won on the second round. However > there was a majority on the first round but it was only by 2 votes. Yes, the first ballot was indeed only a two vote majority. But transferrable ballot mechanics make the BTU theory implausible. I've heard the rumor repeatedly--I even considered it myself when the vote totals came out--but it just doesn't hold water. If you get close to a majority on the first ballot, you almost always get enough transferred votes in the runoffs to go over the top. The later-ballot upsets happen when neither of the leaders gets terribly close to a majority, or when both the leaders are very close to the majority and there are only a handful of transfer votes to distribute. In this case there were several hundred votes for St. Louis and Columbinatti, and even a few Hold-over-funds and random writeins. If we suppose a dozen of the Nawlins voters had forgotten to vote, you would still need all but ten of the transferred votes to prefer Bermuda to Nawlins for the upset to occur. There may have been a lot of second-choice boat people, but even second-choice sentiment wasn't running anything like the 95% you'd need on this topic. There were a lot of people opposed to the NO2 bid (say, for being a little too heavy on the booze-swilling bon temps rouler) who were even more opposed to the excessively elitist Bermudans. I'm pretty sure I was one of them (while eschewing no swilling nor elitism on my own behalf, of course). This another of those probability-zero rumors that spreads for the what-iffiness of it, but just doesn't suspend past the disbelief. I recognize the genre well--I made one up earlier this evening. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Article 3184 of rec.arts.sf.fandom: Newsgroups: rec.arts.sf.fandom From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Subject: The Bermuda Triangle Upset theory, v2 (Re: RISC Worldcon) Sender: usenet@ra.nrl.navy.mil Supersedes: <9309162256.hoey@aic.nrl.navy.mil> Organization: Naval Research Laboratory, Washington, DC Date: Fri, 17 Sep 1993 11:19:23 GMT [ My apologies to those of you who read last night's flawed version of this message. ] With regard to the original subject, I think these people were actually talking about a RASC (reduced _attraction_ set con), but I'm really here to discuss the Bermuda Triangle Upset theory, of which dlow@svale.hp.com (Danny Low) writes: > According to someone who analyzed the voting for the Bermuda Triangle > bid, it came very close to winning. Basically it was everyone's SECOND > choice. If the first round of votes had not come up with a majority, > the Bermuda Triangle would have won on the second round. However > there was a majority on the first round but it was only by 2 votes. Yes, the first ballot was indeed only a two vote majority. But transferrable ballot mechanics make the BTU theory implausible. I've heard the rumor repeatedly--I even considered it myself when the vote totals came out--but it just doesn't hold water. If you get close to a majority on the first ballot, you almost always get enough transferred votes in the runoffs to go over the top. The later-ballot upsets happen when neither of the leaders gets terribly close to a majority, or when both the leaders are very close to the majority and there are only a handful of minority votes to transfer. In this case there were several hundred votes for St. Louis and Columbinatti[*], and even a few Hold-over-funds and random writeins. Let's imagine that a dozen of the Nawlins voters were lost at sea. Each of them reduces the the NO2 total by one vote, but also reduces the majority level by a half vote. So the margin only shifts by six, to Nawlins four votes away from a majority. For the upset to occur, all but four of the votes transferred from the minorities would have to have preferred the Triangle to the Delta. Also, there are always people who don't vote a second preference, and they count as a half vote against the BTU. You can't make this plausible without actually counting up all the minority preferences, something that was not done. There may have been a lot of second-choice boat people, but even second-choice sentiment wasn't running anything like the 99% you'd need on this topic. There were a lot of people opposed to the New Orleans low-data bon-temps-rouler bid style who were even more opposed to the excessively elitist Bermudans. I'm pretty sure I was one of them (while eschewing neither bon temps nor elitism on my own behalf, of course). This another of those probability-zero rumors that spreads for the what-iffiness of it, but just doesn't suspend past the disbelief. I recognize the genre well--I made one up yesterday. Dan Hoey Hoey@AIC.NRL.Navy.Mil [*] A respondent to last night's version notes that this should be `Columbinnati'. Seriously, though, I mean no slur on the bidders here, this is just a bidnomen I use for whatever humorous effect it may still have. --- Newsgroups: rec.arts.comics.misc Followup-To: alt.followup.to.alt.followup.to From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Fri, 17 Sep 1993 14:28:45 GMT Subject: Re: New Weekly Releases: September 13, 1993 d...@po.CWRU.Edu (Damon B. Crumpler) writes!! > me, too!! > great book that doesn't get enough press on the net!! Talk about not getting enough press on the net!! You're lucky you can get the book at all!! I go into the local comix shop and ask for it all the time!! They look at me like they don't know what I'm talking about!! They just don't get it!!!! Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.physics, sci.misc, alt.folklore.urban Followup-To: alt.folklore.urban From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Fri, 17 Sep 1993 21:32:06 GMT Subject: Re: Will foil in hubcaps foil police radar? las...@watsun.cc.columbia.edu (Charles Lasner) wrote: > Then explain the empirical evidence about trucks going 68 MPH, not > less or more. Moreover, my vehicle is rated at 23.5 MPG by EPA > standards which I understand are run with a speed of 44 MPH, yet I > routinely get much more than that at higher speeds, and... the car > is very quiet at 80 MPH, and it doesn't really feel like some cars > do when traveling almost that fast, etc. Yes, `etc' about sums it up. Maybe it's time to change the subject heading on this thread. It's been three weeks since Ed Zotti asked about foil in hubcaps, and the `Straight Dope' column he was researching has been published, at least in DC. ObUL: `Chaff' is a term that was invented in World War II when the British electronics engineers noticed how loquacious the Yanks got when you started them out on some innocent topic like radar jamming. It's a contraction of `chat forever'. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: alt.folklore.computers From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Fri, 17 Sep 1993 22:04:24 GMT Subject: Re: Why americans read # as pound ? Chris Marriott wrote: >We have the pound sign on shift-3, the hash sign where a US-keyboard has >"\", and the "\" and an extra key between the left shift and the "z" key. and glr...@afterlife.ncsc.mil (Gerald Reno) asked: >Really? You mean there's a STANDARD place for the backslash? Sure, doesn't everyone have a backslash on Shift-L? Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: alt.usage.english From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Fri, 17 Sep 1993 23:54:55 GMT Subject: Chadlessness (was Re: Foobar) David Casseres goes: > Aha! Thank you, I think you are the first to post a *first-person* > account of the word chadless in actual usage. This tells me that > such tales as the acronym "Card Hole Aggregate Debris" are > after-the-fact inventions, and chad is a back-formation from > chadless. and anno4...@zrz.tu-berlin.de (Anno Siegel) chimes in: : Yes, I liked that one too. The fact that Donald apparently wasn't aware : that those devices were named for their manufacturer Chadless makes it : even more convincing. and dra...@netcom.com (Donald Ravey) explains: > ... I'm sure I paid no attention to who manufactured the equipment. > My experience was in 1952-53, and at that time I recall that we used > both "chad" and "chadless", but I have no idea how those terms > originated. That was just what we called it/them. Well, the acronym explanation was pretty clearly a crock all along. But I still think the Chadless->chad back-formation theory is pretty far-fetched, too. It's a nice story, but I heard it from someone who is good at making up nice stories. THND3.00 says: Historical note: One correspondent believes `chad' (sense 2) derives from the Chadless keypunch (named for its inventor).... But a lot of people believe a lot of things. So has anyone got a citation to an advertisement or patent or any historical record asserting that there ever existed a punch manufacturer or inventor named Chadless? That would be a strong bit of circumstantial evidence. The fact that people called chad-free punches `chadless' is consistent with any etymology of `chad'. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math Followup-To: alt.fan.magister.ludi From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Tue, 21 Sep 1993 23:49:24 GMT Subject: Re: FAQ ON LP Benjamin.J.Ti...@dartmouth.edu (Benjamin J. Tilly) says the Luudster > ... is not a mathematician, he has no mathematical training and he > is presently a dishwasher at a local inn. And of course this leads Ludi-Poo to threaten > IF YOU HAD SAID THAT TO ME IN PRIVATE THAT IS FINE, BUT WHEN YOU > BLAIR IT OUT TO THE WORLD ON SUCH A NEWS NETWORK THAT IS A DIFFERENT > MATTER, BECAUSE THIS STATEMENT IS A LIE. > DO YOU UNDERSTAND WHAT SERIOUS TROUBLE YOU ARE IN! Yes, I FEAR that Ben may be subjected to a CLASS-ACTION LAWSUIT from THOSE who have been SLANDERED by his vicious ATTACK. To suggest that this pathetic crank is a DISHWASHER defames everyone who has ever GOTTEN their HANDS wet. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: alt.usage.english From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Wed, 22 Sep 1993 15:53:54 GMT Subject: Re: Odd sins c...@merit.edu (Chris Weider) writes: >Everyone knows what sodomy is, fl...@dgp.toronto.edu (Alan J Rosenthal) writes: >Not so. Some people say it's anal sex (most common), some say it's >bestiality (e.g. Woody Allen's Everything You Wanted to Know About >Sex), and I've also heard it used to refer to manual-genital sex. klu...@grissom.larc.nasa.gov (Scott Dorsey) mentions: >The legal definition in Virginia applies to all of the above. In addition, >there are several positions of conventional intercourse which the law >defines as sodomy. Oral sex also is defined as sodomy in all cases. The Washington, DC sodomy law included at least anal-genital and oral-genital, but omitted any restriction on manual-anal or oral-anal acts. That's moot now, though, because the occupational government has permitted us to repeal the law entirely. Thank you, Massa Congress. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Fri, 24 Sep 1993 21:05:39 GMT Subject: Re: Q: Row,col of element in upper T matrix? j...@cac.psu.edu (JAMES J VINCENT) asks: > Can anyone suggest a straightforward way to determine the row and > column number of a given element of a matrix while only counting > elements of the upper triangle, including the diagonal? For instance > in a 6x6 matrix if I want the 8th element it would be row 2, column 3 > by counting across rows, and only counting the upper triangle members. In an NxN matrix, if the I'th element is Row R, Column C, then I = (R-1)(2N-R+2)/2 + (C-R+1) The inverse is given as R = N - floor((sqrt((2N+1)^2 - 8I + 4) - 1)/2) C = I - (2N-R)(R-1)/2 Be sure to credit your sources. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.arts.comics.misc, alt.comics.superman From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Fri, 24 Sep 1993 21:16:15 GMT Subject: Re: Why is DC Ruining Clark Kent bu...@westford.ccur.com (Bill 'gdfbm' Burks) blithers: > What's next for Clark Kent an earring? Then a dress? Well, it would probably have to be a clip-on earring, because how are you going to pierce his super-earlobe? Or has that changed in the less super era? If SM (love the name) can be pierced, I don't see why you'd go for anything as boring as an earring, anyway. Now, a guiche would make sense. That way if he goes for the dress, he'll have a place to hang his super-tights. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.arts.comics.misc From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Mon, 27 Sep 1993 23:08:18 GMT Subject: Going to San Francisco! tor...@edb.tih.no (Tore Busch) write: > Yes! We are going to take a trip over to San Francisco, AND we are > wondering if there are any GREAT Comic stores there. The Comic Relief store is outstanding. It's at 2138 University Avenue, about two blocks from the Berkeley BART station. They have a wall of bookshelves full of graphic novels and cartoon books, and a huge selection of undergrounds. They had a bunch of other stuff too, but I didn't get to see it--I ran out of time and money after three hours and over three hundred dollars. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: alt.folklore.urban From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Thu, 30 Sep 1993 17:03:17 GMT Subject: Folklore radio, again The public radio show Fresh Air is going to talk about African- American legends tonight. They mentioned the one about the CIA developing AIDS to kill blacks, and the one about the Klan owning Reebok, but who knows where they could lead. I'm not sure if they are planning on being credulous or not, I've heard some pretty flaky things on Fresh Air. If you want to catch it in the Washington, DC area, it will be on at 10:00 PM on WAMU-FM, 88.5 MHz. Dan "Your cognomen here" Hoey Hoey@AIC.NRL.Navy.Mil ObUL: Hormel is covertly owned by Nike. The secret agenda is that eating Spam will make people Bush on their shoes, increasing sales. --- Newsgroups: rec.puzzles From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Fri, 1 Oct 1993 15:44:54 GMT Subject: Re: Folding a strip of stamps a...@pact.srf.ac.uk (Andy Pepperdine) notes a discrepancy in reported numbers of ways of folding a strip of stamps: > stamps Me Wells Sloane > 1 1 1 . > 2 1 1 1 > 3 2 2 2 > 4 5 5 5 > 5 14 14 14 > 6 38 38 39 !! > 7 120 120 120 > 8 353 353 358 !! > 9 1148 1148 1176 !! > 10 3527 3527 The problem with the stamp folding problem is that there is more than one problem. I'll call them the sticky-stamp folding problem and the dry-stamp folding problem. In the sticky-stamp folding problem, everytime you make a fold, you have to flatten it completely, and the stamps on the two sides of the fold stick together and can't be unfolded later. In the dry-stamp folding problem you can partially unfold a previous fold to make a new fold. The fundamental dry-stamp fold that you can't make with sticky stamps is the extended S: ________________ ________) (________________ because whichever fold you make first, the second is prohibited. The smallest completely folded strip you can't form with sticky stamps is: _____1_________________ / _____6_______________ / (_______5___________ \ ( _________4_________) \ \ (___________3_______ ) \_______________2_____) / _________________7_____/ If you unfold it carefully, using only the inverses of sticky-stamp folds, you'll find yourself with some variation of the extended S. This is why you found the discrepancy with a strip of seven stamps. The situation becomes much more complicated in two dimensions, when there are folded positions in which a sheet could exist (without stretching or self-intersection), but which might be difficult or impossible to create starting with a flat sheet and manipulating it in three diminesions. Consider a flat tube that swallows its tail ouroborously, for instance. I have never been able to put together even a 3x4 example of this, though I imagine it's possible with the sufficiently smooth paper. With a 3x5 sheet, you can have interlocking cuffs that may be impossible to build without some sort of smart paper. legen...@ens.fr (Stephane Legendre) sent sci.math an article about the dry-stamp problem in July. He lists the following values: n | 2 3 4 5 6 7 8 9 10 -----+----------------------------------------------- b(n) | 1 2 5 14 39 120 358 1176 3527 -----+----------------------------------------------- n | 11 12 13 14 15 16 -----+------------------------------------------------ b(n) | 11622 36627 121622 389560 1301140 4215748 -----+------------------------------------------------ citing J. E. Koehler 1968, "Folding a strip of stamps", Journal of Combinatorial Theory 5 (1968) 135-152 I hadn't heard of published material describing the sticky-stamp problem or the difference between the two models. I am surprised to hear that _Martin_Gardner's_Sixth_Book_of_Mathematical_Recreations_ uses the sticky numbers, because his book _Wheels,_Life_and_Other_ _Mathematical_Amusements_ uses W. F. Lunnon's results on folding dry two-dimensional maps. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math, sci.physics, sci.bio, sci.chem, sci.geo.geology, sci.astro, news.misc Followup-To: news.software.readers From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Mon, 4 Oct 1993 21:20:28 GMT Subject: Careful kill files (was Re: PHYSICS AND MATH EXPLANATION ...) cyeom...@ms.uky.edu (Charles Yeomans) asked how to put authors in a KILL file, and b...@guitar.ucr.edu (john baez) replies: > This is probably newsreader-dependent, but for rn I use, e.g., /plutonium/hj > This kills articles that he posts, but not that merely contain his name. Unfortunately, that also kills articles that mention plutonium (as the crank, the element, or even the hostname) in any header. I don't know of any hosts named plutonium, but given that several recent "sci.math" articles mention "magnesium.club.cc.cmu.edu" in the "Path:" header, the chance of deleting some innocuous or even valuable posting is very high with your method. Even narrowing to the "From:" header is dangerous--I don't know if anyone posts from a host named plutonium, but there's no particular reason to suppose it's trash for that reason. I advise using /^from:.*plutonium@/hj instead. This will only delete postings by a user whose name ends in the string "plutonium". I believe this pretty much narrows it down to Ludi, though anyone else who wants to crawl under his rock with him may do so. Anyone else who needs help writing kill files should take it to news.software.readers instead of a half-dozen `science' groups. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Tue, 5 Oct 1993 16:46:15 GMT Subject: Correction Re: Folding a strip of stamps a...@pact.srf.ac.uk (Andy Pepperdine) notes a discrepancy in reported numbers of ways of folding an unmarked strip of stamps, for 6, 8, and 9 stamps. I wrote: > The problem with the stamp folding problem is that there is more than > one problem. I'll call them the sticky-stamp folding problem and the > dry-stamp folding problem. which is true as far as it goes, but it is not the source of the discrepancy. For instance, it does not make a difference until the number of stamps reaches 7, as careful readers of my message may have noticed. It seems everyone here has indeed been dealing with the dry-stamp problem. According to Stephane Legendre's survey of the (dry) stamp-folding literature, posted to sci.math three months ago, the following sequences appear in Sloane's handbook of integer sequences: n | 2 3 4 5 6 7 8 9 10 -----+------------------------------------------------ b(n) | 1 2 5 14 39? 120 358? 1176? 3527 -----+------------------------------------------------ k(n) | 0 2 4 6 8 18 20 56 48 -----+------------------------------------------------ r(n) | 2 2 4 10 24 66 174 504 1406 n | 11 12 13 14 15 16 -----+------------------------------------------------ b(n) | 11622 36627 121622 389560 1301140 4215748 -----+------------------------------------------------ k(n) | 178 132 574 348 1870 1008 -----+------------------------------------------------ r(n) | 4210 12198 37378 111278 346846 1053874 Where b(n) (Sloane's #576) is the number of ways of folding a strip of n blank stamps, k(n) (#120) is the number of top-bottom symmetric ways of folding n marked stamps, and r(n) (#464) is the number of ways of folding n marked stamps with the first stamp at the top. He remarks that b(n) = (n r(n) + k(n))/4, which is not satisfied by the columns in which question marks appear. In fact, the numbers for b(n) seem to be in error, as Andy Pepperdine noticed. Andy's corrected numbers are confirmed by David Gibson, and reconfirmed by a program I wrote independently. In fact, they are `confirmed' by the other sequences in Sloane, assuming they are all listed that far. So the following seems to be the corrected table. n | 2 3 4 5 6 7 8 9 10 -----+------------------------------------------------ b(n) | 1 2 5 14 38 120 353 1148 3527 -----+------------------------------------------------ k(n) | 0 2 4 6 8 18 20 56 48 -----+------------------------------------------------ r(n) | 2 2 4 10 24 66 174 504 1406 n | 11 12 13 14 15 16 -----+------------------------------------------------ b(n) | 11622 36627 121622 389560 1301140 4215748 -----+------------------------------------------------ k(n) | 178 132 574 348 1870 1008 -----+------------------------------------------------ r(n) | 4210 12198 37378 111278 346846 1053874 Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: alt.folklore.urban From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Wed, 6 Oct 1993 15:54:05 GMT Subject: Coca-cola and Guinness and Fizz hatu...@netcom.com (DaveHatunen) writes: > ...and, for cans at least, lower carbonation due to problems with the > tensile strength of the cans. It does seem to me that Coke from a glass > bottle (can you find them anymore?) tastes better than from cans or > plastic. and PA...@kcgl1.eng.ohio-state.edu (Jonathan Papai) claims: > The aluminium (sic) cans also impart a taste into the soda. Ask folks > who've had chemo-therapy if Coke from cans is drinkable. Something > about the treatment makes for enhanced sensitivity. But is it the aliumiuninnium (ick) or the carbonation? Coke in cans is carbonated with phosphoric acid, rather than carbonic acid, and it definitely has a different taste. You can even taste the difference in the evolved gas if you capture it in a jar or a stomach or just sip it off the top of a newly-opened container. I guess a good way to check is to compare with Coke from plastic bottles, where they also use phosphoric acid. I guess I don't believe there's enough Al in Coke to taste. de...@nezsdc.icl.co.nz (Derek Tearne) writes: > All guiness brewed outside the UK is manufactured by taking a local > beer (any beer or lager will do) and adding 'Guinness extract'. > Guinness extract is basically concentrated Guinness, it is supposed > to look just like vegemite. (And taste just like chicken?) Somehow I don't find this is as credible as the Coke extract story, but then I never thought they brewed it outside the UK at all. Back to the fizz issue, david.ett...@geologi.uio.no (David Ettner) writes of the > tapping device found only in Ireland. The Guinness tap there has 2 > speeds, a fast speed for the initial fill and a slow speed for final > filling, creating the fameous head. And in the US there is a canned product called ``Guinness Draft''. This has a plastic device on the inside at the bottom, containing pressurized CO2. When you pop the can, a spray of fine bubbles is released forming a creamy head. In order to keep the bubble spray from initiating the ``fire extinguisher in a can'' effect they have taken two steps: they carbonate the stout less, and they warn you to chill it before opening. Maybe real Guinness lovers like it, but I end up with a flatter, colder drink than I wanted. I guess someone's mileage must vary, though, or they wouldn't bother. Dan Hoey Hoey@AIC.NRL.Navy.Mil ObUL: Nicotine is added to Marmite and exported to third-world countries to open new markets for the US tobacco industry. Skol! OOBUL: I dreamt I found a wax tadpole in my ancestor's Maidenform Bra, but she made me give it back. --- Newsgroups: misc.misc From: hoey@zogwarg.tec.army.mil (Dan Hoey) Date: 6 Oct 93 23:15:00 GMT Subject: Re: Fermat a...@stretch.polymer.uakron.edu (The Fish) writes: {...@saul.cis.upenn.edu (Mark-Jason Dominus) writes: {|> The statement of Fermat's Last Theorem is that when N is bigger {|> than 2 then for all A,B,C each bigger than 1 there is no solution in {|> integers to the equation {|> {|> A B C {|>>>>> N + N = N {i thought fermat's last theorem was that there are no real solutions ^^^^ read as 'integer' {to the equation: { { n n n { a + b = c { {when n > 2 and when a =/ b =/ c =/ 1 { ^^ { read as 'not equal to' ^^^^^^^^ read as 'less than' { {the case when n=2 is the pythagores theorem for a right angle triangle. ^^^^^^^^^^ might as well read as 'Pythagoras' while you're at it. --- Newsgroups: alt.folklore.urban From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Thu, 7 Oct 1993 15:56:02 GMT Subject: Re: Bungee Jumping UL C527...@mizzou1.missouri.edu (Greg Foster) writes: > When the first person jumped off, the company realized > there was no such thing as the 13th floor. . . .. There was a cartoon a few months back with Bill Clinton jumping off a bridge, thinking ``Length of bungee cord, 60 feet; amount of stretch, 742%; height of bridge, 449 feet, no wait, make that 439 feet....'' The _really_ funny thing is that the two guesses _both_ end up way underground, by 56.2 and 66.2 feet, respectively. So it's the cartoonist who can't do percentages, not the president who can't get the numbers to add up. (The actual numbers have been changed to protect my plausible deniability of having the cartoon with me, but the point's the same.) Dan ``will proof editorial cartoons for scientific acucracy'' Heoy Hoey@AIC.NRL.Navy.Mil (but will carefully debate the hoox) ObUL: Nina Hartburg became a porno star because she was brainwashed by subliminal advertising hidden in the Gerber's Baby Food baby's hair. [I disclaim! On the off chance there's someone out there actually named Nina Hartburg, it really _is_ a coincidence, explanation on request, get away from me with that that tarhook, Missa Hartley. Remember, there's a Dan Hoey Drive in Dexter, Michigan, so coincidences really do concatenate. ] --- Newsgroups: alt.folklore.urban From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Thu, 7 Oct 1993 16:09:46 GMT Subject: Re: Cars and their Evil Twins gmoul...@epas.utoronto.ca (Glenn Moulaison) writes: >The new Chevy Nova, by the way, is a Corolla. What does that mean >in Spanish? f...@netcom.com (Keith Fichtemaier) writes: > 'Corolla' is spanish for secretary. But Corolla is English for ``Car tips over going around curve, killing six''. Dan ``details at 11'' Hoey Hoey@AIC.NRL.Navy.Mil ObUL: In the late 40's there was a Japanese make of car called the ``Nogovelisolli''. It was never imported because of complaints that the bloken Engrish joke was determined to be lacist and tlite. --- Newsgroups: rec.puzzles From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Thu, 7 Oct 1993 21:53:21 GMT Subject: Rare Vintage Whine [ Was Re: squares puzzle (no computer) ] lu...@fy.chalmers.se (Dror Lubin) writes: > fe...@Dartmouth.EDU (Ted Schuerzinger) writes: > > > > [endlessly requoting the same old problem > [ whinge, whinge, whinge, can we have more non computer puzzles. ] [whinge, whinge, whinge, can we have fewer dumb computer word puzzles.] > How many of the words which you "grep" out of /usr/dict/web2 would > you use in an "on the board" game of scrabble? % sort -m OSPD/* web2|uniq -d|wc -l 47224 > Computers are a fact. The real challenge, I believe, is to invent > puzzles which are impossible, or at least not trivial, to solve on a > computer. The first real challenge is to pose a puzzle that doesn't lead to five people posting the same eighty-digit answers without showing something of substance, like how they found the answers, or that they are the correct and only answers, or at least something they couldn't have greeped off each other's postings. The second real challenge is to let people know about an old chestnut that yet another popular press book has gotten wrong WITHOUT leading fifty-eight people to argue over whether it is really right [it isn't], or feeling the need to show why it's really wrong [it's in the FAQ, better], or how you get the FAQ [read news.newusers.questions, isn't it about time? Yes, I'm talking to you.], or asking how you figure out probabilities [read a book] or how to learn to read [Furr. Fu.]. The third real challenge is to is to take a puzzle you don't like and instead of reposting it and driveling about why it's the perfect example of a puzzle you don't like, how about _creating_ a puzzle that isn't in every puzzle book from Sam Loyd to Winning Ways, and has a sensible answer. Then posting that puzzle, and maybe you won't get someone to go on about how your whine is the perfect example of the kind of whine that really should be sent as email to /dev/null/part/of/the/problem. The fourth real challenge is to find all the rectangular prisms of integer side whose surface (in square units) is equal to its volume (in cubic units). With references to the literature. Spell check it, this time, do you think I have time to proofread every message that comes to this bulletin board? The fifth real challenge is to figure out just which real challenges are real challenges worth solving and which are really exercises in banality. Thank you for sharing, Dan Hoey@AIC.NRL.Navy.Mil ObFAQt: Back in the Old Days, before Usenet became a stinking sewer of BART D00DY, before every kid and his president had a Usenet account, you know, back then. When people posted an article that wasn't a puzzle, they used to feel this sort of like an OBLIGATION to do something to justify the putrid existence of their purulent post on a puzzles newsgroup when it didn't have a PUZZLE in it. And they added a thing with a name made up of the first syllable of OBLIGATORY and the last two syllables of PUZZLE, that expressed their fervent need to actually provide some EXCUSE for the aforedisparaginglymentioned p.e.o.t.p.p.o.a.p.n.w.i.d.h.a.p.i.i., which might go something like ObPuzzle: There are three doors: behind one is a goat, behind the second is two goats, and behind the third is three goats. You are going to pick a door, after which the host is going to pick a door, and you are going to divide the number of goats behind the your door by the number beind the host's door, and that is how many dollars you get to pay per goat for all the goats no one picked. What do you expect it will cost? Show your work. Case analyses are dull, figure out a jazzy shortcut. How much for N doors? Do I have to ask you all the questions? Were you raised in a barn? Shut the door quietly on your way out. That's better. --- Newsgroups: alt.folklore.urban From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Thu, 7 Oct 1993 22:23:39 GMT Subject: Re: Bungee Jumping UL I wrote: :There was a cartoon a few months back with Bill Clinton jumping off a :bridge, thinking ``Length of bungee cord, 60 feet; amount of stretch, :742%; height of bridge, 449 feet, no wait, make that 439 feet....'' :The _really_ funny thing is that the two guesses _both_ end up :way underground, by 56.2 and 66.2 feet, respectively. So it's the :cartoonist who can't do percentages, not the president who can't get :the numbers to add up. t...@kimbark.uchicago.edu (Ted Frank) answers: : The cartoon was by Tom Toles in the New Republic. He had the numbers : right when I did the calculations. Thanks for the name, but the numbers were indeed wrong. The point is that when a cord stretches by 742%, you have to multiply the length by 8.42, not 7.42. Toles multiplied by 7.42, and so did you, so of course you got the same wrong answer. Problem: A worker uses a Barrel-O'-Bricks^TM to stretch a fifty-foot iron chain by 0.002%. Does the chain get shorter or longer? Answer: Read the letter to his boss requesting sick leave. Problem: John Thomas stretches his four-inch elastic by one hundred per cent. How long is it now? Answer: That's not funny. j...@netcom.com (John Switzer) cavils: > Perhaps that was the point of the cartoon? Or do you think only you > have a calculator handy? I didn't have a calculator, but the numbers were easy enough to multiply in my head. Tom Toles may have had a calculator, but it couldn't help him any more than it could help Ted Frank. If you don't know how to translate words into button pushes, your calculator is a dumb Nintendo. Dan "Knows What Button to Push" Hoey Hoey@AIC.NRL.Navy.Mil ObUL: The Bureau of Preconceived Notation is a little-known part of the Department of Commerce whose function it is to determine such things as that Noon is 12PM, that 2000 is in the twenty- first century, and that MIM does not mean what MCMXCIX means. Instead of issuing regulations (which the populace might take it upon themselves to flout) they communicate in 200-minute spot ads that replace carefully chosen words spoken by people on Tee Vee. The next big standards war concerns whether they will start reprocessing the video, too, or will define lip-reading to be use of a Regulated Wartime Munitions Device requiring form DD-5083-PNTT-HTD (Promise not to tell and Hope to die). --- Article 2461 of alt.fandom.cons: Newsgroups: rec.arts.sf.fandom,alt.fandom.cons From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Subject: Counterfeit vs countercounterfeit vs .... Sender: usenet@ra.nrl.navy.mil Organization: Naval Research Laboratory, Washington, DC Date: Fri, 22 Oct 1993 00:57:33 GMT matt@access.digex.net (Matt Lawrence) writes: > If ConFrancisco had used "photosensitive badges", I would have created a > stencil saying (probably) "ConFiasco" and borrowed a photoflash to > permenantly add it to my badge. and hazelton@mizar.usc.edu (John W. Hazelton) adds: > An amusingly creative response, if indeed a photoflash would blacken > the badge. Point taken, they did promise a photoflash wouldn't mess it up, didn't they? How about just using a photocopier? If indeed a photocopier would blacken this mythical plastic (and for the sake of word-mincers, until someone comes up with the repeatedly requested and pointedly unmentioned manufacturer's name, this stuff is a myth). dante@shakala.com (Charlie Prael) mentions: > Nevermind if we thought things through. Nevermind if we had > contingency planning in place. Nevermind if we never got a sample to > test. It *might* have happened -- and that's what we (ConFrancisco) > have been getting dumped on about. But I thought it was you who said you didn't have a sample, that was your excuse for not knowing the specifications, or whether it would really turn or not turn black. And the examples I've seen of you thinking things through sound awfully naive. You couldn't handle the crowds you actually had at the door, what were the contingency plans for thousand replacement badges? And the solutions proposed to the black badge problem are far too member-unfriendly. "Explain how it happened" my foot, I was photocopying it to send in a letter home, why not? Well, maybe there were plans to make it unavoidably known that you shouldn't try to photocopy the badge, but it's still an imposition not to be able to do so. I would also find it quite annoying were the badge to turn black over the years, as I cherish it as a souvenir of the con (and I never say ``ConFiasco'' save with the fondest of doting memories. It was surely the most wonderful ConFiasco I've ever seen.) How were you supposed to know the badge would last after the con? This is not nevermind-if-you-had-thought-things-through, this is a question: What *did* you think through? (As far as many other proposed technological solutions go, I would also be really annoyed if my badge set off alarms at bookstores or libraries or nuclear reactors. I just feel that, unlike my employee ID, the con badge is my property, and should be built to handle the ordinary conditions of life.) ConFrancisco managed to trip, stumble, and fall upon the best solution mentioned yet: the diffraction film on the badge was good enough to stop everyone who was not willing and able to circumvent anything you could reasonably produce (including photo badges as a triviality.) Before you go into your ``we're losing billions if not trillions of dollars'' estimates yet again, try estimating the number of counterfeit badges actually used at the con in question. Is there any evidence that even a single counterfeit badge was ever used by a non-member of the con? (As an official ConFrancisco badge counterfeiter myself, I of course was a member. And I only counterfeited Kevin Standlee's badge and my own.) Speaking of which: Does anyone from the concom know how many people actually ponied up the $145 to replace a lost badge, and how many were returned for the refund less $35 processing fee? I'm just wondering how much money they got to make up for the ill will generated. No, I'm not interested in yet another round of everyone's estimate of the hypothetical monies not lost by having a less restrictive policy, I'd just like to know how much CF really made on the deal. (And I guess I'd better find a parenthetical remark to end this paragraph with, too). Fisoo, Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Article 144 of alt.humor.best-of-usenet: Newsgroups: alt.humor.best-of-usenet From: hoey@AIC.NRL.Navy.Mil Subject: [comp.admin.policy] PC vs Macintosh Sender: thespian@netcom.com (Stephanie M. Clarkson-Aines) Organization: NETCOM On-line Communication Services (408 241-9760 guest) Date: Tue, 9 Nov 1993 14:14:20 GMT Approved: thespian@netcom.com From: kherron@ms.uky.edu (Kenneth Herron) Newsgroups: comp.admin.policy Subject: Re: A question Date: 22 Oct 93 20:39:30 GMT Lin Y. C. wrote: | Which of pc and Mac is best choice ? | I do not know really..... It really depends on what you want to do with it. PCs with their metal construction are better at holding up cars so you can work under them. Macintoshes, on the other hand, won't tire your arm as much when tenderizing meat or mashing potatoes. PCs will hold more charcoal and give you a bigger grilling surface, but some of the high-end Macs are properly shaped for this as well. And for a doorstop, you'd either want a Timex-Sinclair 1000 or and Amiga 500. -- Kenneth Herron kherron@ms.uky.edu University of Kentucky +1 606 257 1429 Dept. of Mathematics "If you ever drop your keys in a river of molten lava, let 'em go because, man, they're gone." -- -- alt.humor.best-of-usenet -- -- Funniest postings from USENET, altnet, and the worlds beyond -- -- Moderator's address: best@polaris.async.vt.edu -- Article 4449 of rec.arts.sf.fandom: Newsgroups: rec.arts.sf.fandom From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Subject: Re: Cons Cashing Checks Sender: usenet@ra.nrl.navy.mil Organization: Naval Research Laboratory, Washington, DC Date: Wed, 17 Nov 1993 22:04:00 GMT ddb@tdkt.kksys.com (David Dyer-bennet) wrote: DDB> The best scheme I've seen has the envelopes delivered directly DDB> to registration, who then do data entry, stamp the checks, make DDB> a deposit, and turn over a report to treasury. This scheme has DDB> minimal handling and minimal delay to deposit. Treasury people DDB> often seem nervous about it, but since they're willing to let DDB> registration handle the actual checks, I don't see how any risk DDB> is being added. Beth.Friedman@p5.f341.n282.z1.tdkt.kksys.com (Beth Friedman) replies: BF> The only problem I have with this scheme is one that is inherent BF> in the software, rather than the scheme per se. Occasionally I BF> have to put an amount in the computer that makes the batch BF> balance, rather than the real number. This results in an BF> incorrect treasury transmittal report. For normal conventions, BF> this only happens when someone sends in a check for the wrong BF> amount, but at WFC, it occurred when people combined banquet and BF> registration on the same check. OTOH, if the database had BF> banquet tickets included as an item, that wouldn't be problem. Yech. I had this problem entering Disclave registration for three years, and I figured out what to do about it. The real problem is that your database needs an overpayment (OP) field (use negative values for underpayment). If you don't do banquet tickets in your database, record it as an overpayment. When you transmit the list of people who paid for banquet tickets, you subtract out their overpayments. At some point, you have to do something with each OP (not just zero it, you have to collect or refund money, or at least record that you're accepting a gift or forgiving an underpayment). There is still a problem with this method, when you have (e.g.) someone who underpaid and then sent an increment check. So you also need a balance brought forward (BBF) field. At the beginning of each batch of checks, you have a program copy the current registration amount plus OP of each record into the BBF. For the treasury report, you add Registration amount plus OP minus BBF, and that should be what the check was for. If it isn't, you've got data entry problems. You need to fix these problems before you start a new batch of checks (because then you lose the BBF). What I really wanted was an account record for each person, so I could have a record with dates for each transaction, like on a hotel bill. That allows you to figure out the really weird errors. But OP and BBF are enough to deal with treasury reports. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math Followup-To: poster From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Wed, 24 Nov 1993 22:26:55 GMT Subject: Re: MvS erred on Monty Hall. was: dumping on Marilyn vos Savant I should note that this problem appears in sci.math and rec.puzzles with unpleasant frequency--I know I've been watching it bob up every few months for the past five years--so I have strong reservations about adding more discussion on the topic. This time, though, my concern about the accuracy of the discussion has overcome my usual willingness to let the problem die of its own accord. The sci.math readership should not, however expect me to inflict\\\\\\\favor you with such clarification often; in particular, followup posts are not invited and will generally not elicit my response. I am considerably more forthcoming in email. Just so we know what we're discussing, the problem as printed in Marilyn vos Savant's column goes: ] "Suppose you're on a game show, and you're given a choice of three ] doors. Behind one door is a car; behind the others, goats. You pick ] a door--say, No.1--and the host, who knows what's behind the doors, ] opens another door--say, No.3--which has a goat. He then says to you, ] 'Do you want to pick door No.2?' Is it to your advantage to switch ] your choice?" tycc...@kronecker.mit.edu (Timothy Y. Chow) writes: > I have yet to meet someone who unwittingly chooses unusual tacit > assumptions, correctly reasons from those tacit assumptions to a > probability of 1/2, and is unaware that the tacit assumptions most > people make lead to a probability of 2/3.... Marilyn vos Savant > certainly did not provide enough context from the letters she quoted > to enable readers to tell which way those Ph.D.'s were thinking, and > extrapolating from my own experience I strongly suspect that the > Ph.D.'s were way off base. Well, I am someone who wittingly refuses to ascribe tacit assumptions to the problem, whether usual or unusual. I have had to conclude that the problem as stated does not determine a probability of winning by switching, because it does not specify whether the game show host always offers the contestant a chance to switch or not, and if not under what circumstances the offer is made. I will say that my understanding of the problem statement is that there is indeed one door with a valuable prize and two without, and that the host will award whatever is behind the door the contestant ends up picking. Perhaps that is tacit, but it is the way I understand the everyday English of the problem. I deny that my understanding of this is comparable to the assumption that the host always offers a chance to switch. In August, ruln...@nimbus.seas.ucla.edu (John M. Rulnick) posted nine letter-excerpts from Marilyn vos Savant's column taking issue with her analysis. Even if we reject the one that suggested a probability textbook and the one that blamed her mistake on her gender, there are still seven for which the excerpt printed is consistent with an objection on these grounds. > In short, the problem was indeed ambiguously worded.... Once again, I deny that the problem was ambiguously worded. It is an unambiguous presentation of a problem that does not provide enough information to determine the kind of answer you and Marilyn are looking for. Reread the problem statement: there is no support for the assumption that the host's offer of a second choice would have been made had the contestant chosen a different door. Hundreds of people posting arguments and programs with this tacit assumption do not provide such support. f...@cc.gatech.edu (Gary Peterson) goes on > In addition, some people assume the host does not want to give out > the big prize (which is the opposite of how game shows really work) > and therefore suspect the offer to switch is made more frequently > when your initial choice contains the big prize. First, it has been repeatedly noted that the problem is inconsistent with the way the ``Let's Make A Deal'' game show worked. On that game show the contestant was never offered a second chance to choose a previously rejected prize (and if you are talking about some other game show, you had best say which one.) So arguments about ``how game shows really work'' are not relevant to the problem. Second, you have misstated the position of everyone I know who mentions the possibility of the host ``not wanting to give out the big prize''. We are not making the assumption that the host has that strategy. We are simply trying to make it clear to you that the assumption you make has changed the problem. There are other possible strategies of the host, including the strategy of always offering a choice as you have assumed. There are many other strategies, as well. It is just not specified which possible strategy occurs in the case of the problem as stated. I will note, however, that the contestant can guarantee a 1/3 probability of winning the prize by choosing a door at random and refusing all offers to switch. Any contestant strategy that includes accepting an offer to switch will have a lower payoff under some host strategies. Thus, refusing to switch maximizes the minimum payoff. There is no way to maximize the average payoff without information about the probability distribution of host strategies. > The discussion of odds after this point becomes absurd. Well, I discussed probability in the previous paragraph, and I don't happen to think it is absurd. > Hence: Always shows goat and always makes offer are critical > for many people's understanding of the problem. (Although they may > choose to ignore this data.) There is no logical support for your conclusion. Just because you can't compute probability without changing the problem doesn't mean the problem really means what you are changing it into. The problem as stated simply does not determine the probability of winning by switching. > That it still comes up, and still people both state it and solve > it wrong despite even having the FAQ now, is terribly depressing. It is also depressing having a FAQ that has the answer wrong. Note that at least the rec.puzzles archive article (to which the sci.math FAQ refers) mentions that the problem is underspecified, and notes the assumptions used to derive the probabilistic solution. > (But the most depressing aspect are the TOTAL BOZOS who post > programs to do simulations to "prove" their answer....) I have to agree. It convinces them of an incorrect answer, based on an unfaithful model of the problem. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Fri, 26 Nov 1993 17:05:17 GMT Subject: Re: Picking a tiling of the M*N rectangle with dominos t...@mantis.co.uk (Tony Lezard) wrote: >Perhaps if you have a quick determinant finder (which can easily be >modified to find permanents, as far as I can tell) and followed up: > On second thoughts, perhaps it isn't so easy.... Indeed, Valiant showed that finding the permanent is "#P-complete" [Garey and Johnson, problem AN13], which implies it's NP-hard. So it's very, very unlikely to be that easy. > It doesn't matter though. Set elements in the adjacency matrix that > correspond to _vertical_ adjacency to i instead of 1 and the > determinant of this matrix is the permanent of the original one, times > some power of i (which depends on the dimensions of the grid and the > way you number the squares of the grid). I don't think this approach works, but I can't be sure. Have you, for instance, verified this for fairly large adjacency matrices? Have you a proof? I would you would need to relate permutation parity to the subset of vertical dominoes, and I doubt there is a relation there. I do, however, have an approach that will greatly speed up the exhaustive-search calculation of the number of tilings of large regions: 1. Cut the region to be tiled approximately in half with a straight cut along the edges of squares. (I advise you to cut the short way if there is a difference). The two parts, while not necessarily equal, will be called the `halves' below. 2. In every domino tiling, each edge of this straight cut will either bisect a domino, or will lie along the edge of the domino. Choose each subset of the edges that will bisect dominoes (you can just about cut the work in half by choosing only the subsets of the correct cardinality modulo 2.) 3. For each subset, separate the two halves of the region minus those dominoes chosen in step 2. We are going to compute the number of tilings of each half. 4. If either half has forced dominoes (squares with only one neighbor left) take them out at this time. If isolated squares show up in either half, that half will be untileable, and the number of tilings of the opposite half can be ignored. 5. Count the number of tilings of each half (using this divide-and- conquer technique recursively if the halves are large enough, or brute force if not. Exactly where the cutoff is depends on your program.) The number of tilings of the entire region is the sum of the products of the number of half-tilings over all the subsets of step 2. Adapting the algorithm to finding the N'th tiling (using an ordering appropriate to the algorithm) is straightforward. I should warn James A. Cherry (and anyone else who has a hack attack over this one) that I believe you will have to use gmp to make effective use of this technique. It gives you the power to count tilings in the range where longs (and probably double floats) are just not big enough. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: alt.folklore.urban From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Mon, 29 Nov 1993 17:48:05 GMT Subject: Re: Bungee Jumping cartoon, researched Back in October, when we were discussing legends of bungee jumpers who miscalculate and go splat, I mentioned a cartoon on the topic. There was some disagreement about what, exactly, the cartoon said, so I've gone to the trouble of looking it up. I found the cartoon in COMIC RELIEF #54 (Late July, 1993) where it was reprinted from the NEW REPUBLIC. It's a Toles cartoon entitled ``Bungee Presidency'', and shows Clinton bungeeing down the valley of the shadow of death while thinking: Distance to rocks: 3214 feet, length of cord: 500 feet, percent of cord expansion: 642% ... or was that 644% .... As I pointed out (explicitly using different figures, since I didn't have the cartoon with me then), a 500-foot cord that undergoes an expansion of 642%-or-was-that-644% will end up being 3710-or-was- that-3720 feet long--both well into the splat range. Toles presumably just multiplied by 6.42-owt-6.44 and thought the cord length would be either 3210-owt-3220 feet, making the humorous point that a 2% error in Clinton's figures might make a big difference in his plans' outcome. Instead, it's a not so much a case of a president who can't keep his numbers straight as a cartoonist who can't do percentages. The cutest part of this is the marginal note, ``Better stick to jogging''. Should that be ``jotting''? For providing motivation to research this question, I would like to thank t...@kimbark.uchicago.edu (Ted Frank) who contested my analysis: > Except he didn't say "cord stretches *by* 742%." He said "cord > stretches *to* 742% of original length", or words to that effect. demonstrating that even a noted AFU legal information provider will occasionally talk though his hat when speaking _ex_camaro_, > Only one with a wrong answer here is you, Bubba. and that that calling people names doesn't improve his ferocity. Dan ``or was that _a_capella_'' Hoey Hoey@AIC.NRL.Navy.Mil ObUL: Lorena Bobbitt's lawyer is going to try to convince the jury that her clever cleaver work expanded John's Thomas by 10%. --- Newsgroups: alt.usage.english, alt.folklore.urban Followup-To: alt.usage.english From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Mon, 29 Nov 1993 23:52:42 GMT Subject: Re: Bungee Jumping cartoon, researched I wrote about the gaffe in the Toles cartoon that shows Clinton bungeeing down the valley of the shadow of death while thinking: > Distance to rocks: 3214 feet, length of cord: 500 feet, percent of > cord expansion: 642% ... or was that 644% .... Amazingly, t...@kimbark.uchicago.edu (Ted Frank) still argues that this statement might mean the cord would stretch to 3210 feet or 3220 feet, depending on the percentage: > I stand by my statement. It's ambiguous and doesn't say the cord > stretches *by* anything. There is no ambiguity in the caption. You merely have to consider what a percent of expansion is. The word ``expansion'' expresses an increase. An ``expansion of 0%'' or a ``0% expansion'' or (to quote faithfully) a ``percentage of expansion: 0%'' means no expansion at all, leaving the initial amount unchanged. Similarly, a 100% expansion is a doubling of the initial amount, a 200% expansion is a tripling, and so forth. It is easy to see that a 642% expansion is a 7.42-fold increase in size. I can't believe you can otherwise competently use the English language and still not understand the meanings of these words. If you were not so inexplicably wedded to your (and Toles's) error, you would never argue with this. > While less clear than one would want writing an engineering manual, > to say the least, my remembrance of my interpretation is at least as > legitimate as yours.... I disagree. The problem is not that the writing is unclear, it is just that you have not read it with care. [It is true that if I read this in an engineering manual, I would look for corroboration that the writer of the manual did not make the same error you and Toles made. But that is because there's a lot of sloppy arithmetic and writing in engineering manuals.] It is telling that you argue on the strength of your remembrance of your interpretation. Apparently, your initial interpretation was based on a careless reading of the text, so you didn't notice Toles's blunder. When I saw it, I noticed the error, and so I re-examined it carefully to make sure that the error was one of his misdenotation rather than my misinterpretation. I still have not heard anything other than your preposterous claims of ambiguity to argue against my conclusion. > ... especially given the fact that all I was disputing was your > unreserved claim that the preposition "by" was present. I never claimed that the preposition ``by'' appeared in the caption. What I do claim is that a ``percentage of expansion'' of 642% happens to mean the same as an ``642% expansion or an ``expansion by 642%''. The meaning you argued for, of ``expansion to 642% of the original size'' means something different, in the same way ``expanding two meters'' means something different from ``expanding to two meters''. Dan Hoey Hoey@AIC.NRL.Navy.Mil GratNonUL: A budding young law student's old hat expands 10 percent. He files suit based on the possibility that a 10 percent expansion may mean that it became one tenth of its former size, crushing his skull. Regrettably, we must dismiss the suit because nothing of the sort has occurred. The hat has merely enlarged to contain a modest increment in the fatuity of his arguments. --- Newsgroups: alt.folklore.urban Followup-To: poster From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Tue, 30 Nov 1993 02:19:07 GMT Subject: Smileys seaw...@vm2.cis.pitt.edu writes: > Yes, it says in the FAQ that smileys are frowned on... > Well, I've seen more than one person posting with smileys. > YOU think they're weaselly; *I* think they're courteous. Well, you should stick to telling people how to express themselves in alt.invertebrates.noxious, where you perhaps know what you're talking about. Here in alt.folklore.urban we have a different tradition, and it is mentioned in the FAQ for people who can't figure it out on their own. But for your information, smileys are subtitles for dolts. They are for the sole benefit of people who can't tell the difference between literal statements and satire. Those of us who write for non-dolts don't use them. And it's rude of you to presume that we are too stupid to avoid taking you seriously without littering your message with simulated line noise. Dan Hoey Hoey@AIC.NRL.Navy.Mil ObUL: Salvador Dali included in each of his paintings a small round design symbolizing a jellyfish, the legendary Portuguese Lord-O'-War. --- Newsgroups: sci.electronics, talk.bizarre, rec.autos, alt.conspiracy Followup-To: alt.conspiracy From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Tue, 30 Nov 1993 23:10:23 GMT Subject: Re: Efficient Electric Car m...@rottweiler.ece.nd.edu (Melvin M. Gladstone) writes: > Recent study and investigation have led me to a new idea for an > efficient electric car.... > My solution is to connect the [deleted] to the [deleted].... > Eventually, the efficiency will be such that only a small battery, > e.g. a C-cell, will be required.... Dear Mr. Gladstone: Perhaps we should remind you of your message of Sun, 21 Nov 1993 00:09:57 GMT, in which you wrote: > I will restrict my posting to non-big-brother related discoveries. Perhaps it did not occur to you that the inefficiency of electric cars is a vital economic asset to our nation's petroleum-based industries. Security considerations prevent us from detailing exactly how your theories might be considered prejudicial to these industries. But if you don't think petroleum is a vital part of the complex popularly known as ``big brother'', it is time for you to start thinking about it. Think very hard. Note that even if you heed this warning, it is impossible to rule out the possibility that we will have to kill you. Dan Hoey :: What are two things that the government wants Hoey@AIC.NRL.Navy.Mil :: information on? Smoking, drugs, and sex, that's :: what. How can such information be gathered? --- Newsgroups: news.groups Followup-To: alt.philosophy.objectivism From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Tue, 30 Nov 1993 23:12:02 GMT Subject: Re: comp.dcom.telecom.tech ... More from Mr. Townson John_David_Galt@cup.portal.com writes: > I just got more flamage from one of the pro-c.d.t.t folks, who says > that there is a second vote going on. I don't see it here, but > please enter my NO vote. Who is this guy? Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: alt.fan.john-palmer From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Wed, 1 Dec 1993 01:00:36 GMT Subject: How many unwanted JPs? (was Re: How many unwanted JP newsgroups?) t...@cbs.ksu.ksu.edu (Tim Ramsey) writes: > ...When your neighbors, your bankers, your customers are reading > alt.fan.john-palmer? That doesn't seem to be a problem in Palmer's case--all his neighbors, bankers, customers, employers, lawyers, parole officers, nurses are aliases for the same wascally wabbit. If he weren't getting a newsfeed from uunet, he would be in a different world entirely. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Date: Mon, 06 Dec 93 10:19:00 -0500 (EST) From: Dan Hoey To: Cube Lovers Subject: Re: God's Algorithm, 24-fold Symmetry, Edges of 3x3x3 These results look very interesting, though I haven't had time to examine them closely, nor even (in a few cases) quite understand them. I especially like to see the categorization by symmetry class. I was somewhat startled to see the unique antipode of the 3x3x3 edges in the quarter-turn metric. Do you know what pattern that is? Dan --- Date: Tue, 07 Dec 93 20:13:08 -0500 (EST) From: Dan Hoey To: Cube Lovers Subject: Re: Unique antipode of edges only "Jerry Bryan" writes: I spoke too quickly when I said the antipodal was simply Start with the edges flipped. I stared at it, flipped the edges in my mind, and it "looked" solved, so I assumed it was Start. It's interesting to note that this is All-Edges-Flipped composed with a mirror reflection of Start. Begging the question: *which* mirror reflection? The answer is, it doesn't matter: since these are the edges of a cube without centers, all reflections are the same position. As long as we get to choose which reflection, the canonical one would be the central reflection. When composed with All-Edges- Flipped, it makes the following antipode. (I think using BFTDLR notation instead of 123456 makes these diagrams a lot easier to read). + T + + F + T T R L + T + + B + + L + + F + + R + + D + + D + + D + L L F F R R => F B R L B F + L + + F + + R + + T + + T + + T + + D + + B + D D R L + D + + F + + B + + T + B B R L + B + + D + I am not yet for sure what they look like, but of the other two states with order-24 equivalence classes, one is at level 9 and the other is at level 12. Since the only one at an even level is at level 12, I am assuming that will be the one which is Start with the edges all flipped. The one at level 9 will probably be the mirror image of Start. If an order-24 equivalence class means what I think it does, I'm pretty sure those two states have to be Mirror-Start and All-Edges- Flipped, being the only sufficiently symmetric positions. But as to their depth, the parity argument (which Chris Worrell also cited) is not valid here. Remember that the cube has no face centers, so the position is not changed by rotating the assemblage of edges in space (i.e., with respect to the absent face centers). But a quarter-turn of the cube in space is an odd permutation of the edges. So permuta- tion parity is not an intrinsic property of edge positions. We can show that there is no sort of parity here by explicitly constructing an odd cycle. Just use a process that would permute the edges of a cube with faces as (FR,FT,FL,FD)(BR,BT,BL,BD)(RT,TL,LD,DR). This has to be an odd process, but it's an identity on the faceless cube. My (very cheap) guess for where we will find the other two M-symmetric positions is opposite to Jerry Bryan's. On a cube with faces, the central reflection of the edges with respect to the faces is Pons Asinorum, which has the easy 12-qt tight lower bound we've seen before (or if not, you can of course get it from me with email). I'm guessing that this bound happens to be tight on the cube without faces, as well. But I have no proof of this guess, and I'm very grateful we won't have to settle for guesses for very long. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math, news.software.readers Followup-To: news.software.readers From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Thu, 9 Dec 1993 23:01:50 GMT Subject: Careful kill files, again (Re: Fermat status) br...@ccr-p.ida.org (Bradley Brock) writes: > I didn't see this because I have > /plutonium/h:j > in my KILL file. You might want to try it. As I mentioned back in October, that is a Bad Idea. It will kill articles that mention plutonium in innocuous places like the Path header, so you are likely to kill some innocuous or even valuable article. If you want to avoid Ludi Toons, /^from:.*plutonium@/h:j is more precise. If your news reader is a version of xrn that calls this a "malformed kill file entry", leaving off the "h:" should work. Anyone else who needs further help writing kill files should take it to news.software.readers. Dan Hoey Hoey@AIC.NRL.Navy.Mil ObMath: Up to symmetry, there are ten ways of cutting a regular dodecahedron into two simply-connected six-pentagon pieces. One of them is not like the others! --- Date: Mon, 13 Dec 93 22:31:31 -0500 (EST) From: Dan Hoey To: Cube Lovers Subject: Symmetry I suppose it's time for a few observations on symmetry. After all, tomorrow is the thirteenth anniversary of "Symmetry and Local Maxima." As Jerry Bryan notes, we can perform the "R" turn by rotating the cube to put the R face in front, performing "F", and undoing the rotation. But we can also perform "R'" by reflecting the cube in a left-to-right mirror, performing "L", and undoing the reflection. Thus conjugation can be extended to use the 48-element group of rotations and reflections, which we call M. In the absence of face centers, there is another kind of reduction that takes account of the 24 possible positions of the resulting collection of edges in space. So two positions X and Y are considered equivalent if X = m' Y m c where m is a rotation or reflection in M, and c is a rotation. My understanding of Jerry Bryan's method is that he combines "m c" into a single rotation or reflection, and factors out the reflection on both sides. It seems to me that what he calls a a "color rotation" is premultiplication, while a "cube rotation" is postmultiplication. [I am somewhat uncertain of this, because it doesn't explain how there can be a 1252-element symmetry group when face centers are present, so perhaps I'm missing something crucial.] But I think we are at least conceptually better off dealing with M when dealing with conjugation, because it takes account of the essential similarity between clockwise and anticlockwise turns. Alan Bawden mentioned back in 1980 that certain positions with sufficient symmetry were local maxima (in terms of distance from start), on the grounds that any clockwise or anticlockwise turn gives us essentially the same position. Jim Saxe and I formalized the notion in a paper entitled "Symmetry and Local Maxima" that we posted on 14 December 1980. [You can find it in /pub/cube-lovers/cube-mail-1.Z on FTP.AI.MIT.Edu]. We had some hope that some of these local maxima might turn out to be global maxima. My hopes for that have been somewhat low in recent years. That is perhaps my best excuse for not noticing immediately that the single global maximum for the edge group turns out to be one of these symmetric local maxima. In fact, all four of the positions with 24-element equivalence classes appear in the list of M-symmetric positions. The paper on Symmetry and Local Maxima also catalogues the positions that have 48-element equivalence classes and 72-element equivalence classes. The The former are the H-symmetric positions, "Six-H" and "Six-H with all edges flipped". The latter are the twelve T-symmetric positions. For T-symmetry, the set of flipped edges may be any of {none, girdle-edges, off-girdle-edges, or all}; the set of edges exchanged with their antipodes may be any of the four as well. But if we choose "none" or "all" for all both choices we get one of the four M-symmetric positions with 24-element equivalence classes, so only twelve of the sixteen possibilities have 72-element equivalence classes. With regard to the edge cube, I should mention that no one has mentioned a 9 QT process for the all-flip nor a 15 QT process for the pons-asinorum-all-flip. Of course, the latter would be somewhat more interesting, being the longest optimal sequence. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Tue, 14 Dec 1993 18:54:19 GMT Subject: Concise proofs (was Re: Elegance in Mathematics) all...@ugcs.caltech.edu (Allen Knutson) writes: > In a talk I went to a few years ago by L\'aszlo Lov\'asz, he > mentioned the following breed of problem. Given a bunch of knights > you wish to sit at a round table, and a list of who's willing to sit > next to whom, does there exist a seating arrangement? And isr...@math.ubc.ca (Robert Israel) correctly notes that this is the Hamiltonian circuit problem in fancy dress. > Obviously any problem with a positive answer has a short proof: > exhibit the arrangement. But it had been proven (I probably > shouldn't post on this, never knowing attributions! Oh well) that > most problems of this sort with negative answers only had proofs > exponentially long in the specification (i.e. as bad as checking all > seatings and pointing out why they don't work). Robert Israel notes that this cannot have been proven without disproving P=NP, which has not been done. But it is more precise to say that the property of having a concise proof is equivalent to membership in NP, so the existence of a concise proof in the negative case is equivalent to NP=co-NP, which or might not be true even if P can be shown unequal to NP. > Is there any hope that there exist a finite set of axioms that could > shorten all of the nonexistence proofs for this sort of problem? > Or are such things only able to kill off some vanishingly small > fraction of them? Perhaps I do not understand Robert Israel's comments on this question. He seems to argue that if P=NP there would always be short proofs, which I suppose is a ``hope'' for such nonexistence proofs. But we should realize how faint that hope is: It cannot be true unless NP=co-NP. For suppose we have an axiom system that admits concise proofs of a problem in co-NP. These proofs can be found using nondeterministic search just as easily as ever. So if the problem has concise proofs, then the nondeterministic algorithm that finds the proofs is an NP recognizer. If proofs in the conjectured axiom system are consistent with the brute-force proofs in the axiom system we know, then the NP recognizer recognizes the co-NP language. Thus any consistent way of forcing concise proofs on a co-NP-complete language will prove NP=co-NP. None of this, however, addresses the question of whether ``most problems of this sort'' have or could have concise proofs. I don't know any probabilistic results in this area. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Thu, 16 Dec 1993 20:23:18 GMT Subject: Re: Wanted: Rubik's Revenge nich...@bbn.com (Nichael Cramer) writes: > ... For "even-order" cubes (like the 4X) the basic mechanism is > fundamentally different in that there's no corresponding central > stem. The cubies just have to hold each other together, almost by > mutual attraction, as it were. I'm afraid this description has been simplified beyond recognizability. In the 4^3's I've seen, the face centers have a protrusion in the shape of a quarter-mushroom, and quadruples of them fit into a mushroom-shaped socket in the central sphere. Pairs of them are allowed to ride in tracks (of a half-mushroom cross section) from socket to socket around the sphere. Flanges on the face centers hold the edges in, and the edges have flanges to hold the corners, much as in a 3^3. > As a consequence, the 4X's that I've had fall apart didn't do so > because a piece broke, rather they just fell apart because they were > held together too loosely and, once a piece had popped-out, the > whole thing came apart. It sounds like your cubies were a little small, or the central sphere too large. In the 4^3's I've seen, though, the mechanism is quite reliable, and the pieces have even less of a tendency to pop out than in the 3^3. However, the quarter-mushroom stem of the face centers is a piece of plastic with about a 2mm square cross section, and it can break with vigorous twisting. I've broken one, and I've read on the Cube-Lovers mailing list of a couple of others breaking. > Ob puzzle: As a practical matter, the only other case that needs to > be worried about is the 6X, because you can't make a 7X (8X?) or > higher order cube. Why not? I thought this piece of misinformation had been laid to rest a long time ago. It's true that if you cut the 7^3 with cubies of equal size, then turn the top slab 45 degrees, that the corner cubie won't touch the slab beneath it. So what? Several people have suggested ``smart cubes'' with little microprocessors that decide when a cubie should hold onto its neighbors, or virtual reality solutions, but they are really far too complicated. All you really need is a linkage that can hold a cubie when it's adjacent to only two others. Curved, dovetailed tracks would be sufficient. There is an aesthetic concern that the ends of the tracks would have to go through the surface of the cube, but I think they can be colored black and will look like the space between the cubies, just a little wider. But you can build a higher-order cube without changing the technology at all, if you want. Just make the outer slabs thick enough that the corners stay in contact with the inner linkage. Some of the ``cubies'' will be rectangular, and of various sizes, but that doesn't change the puzzle a bit. +------+-+-+-+-+-+------+ | | | | | | | | | | | | | | | | | | | | | | | | +------+-+-+-+-+-+------+ +------+-+-+-+-+-+------+ +------+-+-+-+-+-+------+ +------+-+-+-+-+-+------+ +------+-+-+-+-+-+------+ +------+-+-+-+-+-+------+ | | | | | | | | | | | | | | | | | | | | | | | | +------+-+-+-+-+-+------+ I think having 7x7x7 cubes would be a good thing, for a reason Chris Worrell noted here a couple of years ago: there is a sense in which the 7^3 is the largest interesting cube. People who didn't see it then might be interested in figuring it out for themselves as an ObPuzzle. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Date: Wed, 22 Dec 93 13:58:42 -0500 (EST) From: Dan Hoey To: Cube Lovers Subject: The 4^3 and 3^4 Rubik puzzles [ Cube-Lovers, There has recently been a discussion on Usenet group rec.puzzles about some cube topics. There were a few pieces of new information, such as that you can now get Ishi's 5^3 cubes in a lot of places (I got mine in Learningsmith's) for about $35. Here's a message I sent that's relevant to some Cube-Lovers topics. By the way, I'm still working through Jerry Bryan's articles on his brute-force program and his approach to symmetry. I hope to get a reply out soon. ] eric@gsb002.cs.ualberta.ca (Holleman Eric) wrote: By the way, I found the Revenge somewhat easier than the Cube, and I don't think that it was because of my familiarity with the earlier puzzle. x87bennett@gw.wmich.edu (Joe) wrote: From my experience, if you can solve a Rubik's Revenge, you can solve the Cube very easily. Once you get each of the middle 2 cubes on each edge to match, and all 4 center cubes on each face to match, it works exactly like a Rubik's cube. and alan@saturn.cs.swin.oz.au (Alan Christiansen) wrote: I have both. I solved both. The 4x4x4 is a superset of the 3x3x3. ie by fixing all the face centres and then pairing all edges you are left with a 3x3x3 cube, except that when you have solved this 3x3x3 there may be a single pair of edges flipped. This is impossible on a real 3x3x3. Fixing this requires a middle layer to be rotated 1/4 revolution and then all the bits put back. I cant see how it can be [any] easier than a 3x3x3. I, too, found the 3^3 easier than the 4^3. But I can imagine ways in which a solver could find the 4^3 easier. Let us first consider a 4^3 with the faces fixed, the edges together, and the correct simulated edge flip parity. I would solve this as if it were a 3^3, and a lot of people do. But another solver might find it easier to take advantage of the extra moves that are not possible on a 3^3. To take a concrete example, it could be that the solver has a hard time with flipping edges by pairs, as is needed to solve the 3^3. On the 4^3 you can flip one edge at a time. So the solver would find the 4^3 position easier than the corresponding position on a 3^3. If the solver finds this so much easier that it overcomes the difficulty of putting the faces and edges together--or in fact puts the faces and edges together in the course of solving the corners and the edge positions--then the 4^3 could be easier. It depends on the solution procedure. alan@saturn.cs.swin.oz.au (Alan Christiansen) continues: ANyway the real reason I am writing this is that I have written a cube simulator. It can simulate 3x3x3 4x4x4 5x5x5 .... cubes. I am working on 4x4x4x4 cube simulation. This is interesting, as there is more than one way to model the four-dimensional cube problem. Consider the 3^4 cube. It has eight hyper-faces, each in the shape of a cube. One model of this puzzle is that you could turn any face of any hyper-face as if it were a face of a 3^3 Rubik's cube. In a second model, you cannot move part of a hyper-face, but can turn each hyper-face as if it were a solid cube in space. A third model allows either kind of move. These models are different from each other. The second model permits the face centers of the hyper-faces to move around, whereas in the first model only edges and corners move. In the first model, odd permutations of corners are possible, which is not true in the second model. Of course, the third model is the closure of the first two. According to Hofstatder's column reprinted in _Metamagical_Themas_, there is an unpublished 1982 manuscript by H J Kamack and T R Keane entitled ``The Rubik Tesseract''. They calculated the size of the group of the 3^4 puzzle, but I don't know which model was used. Alan Christiansen indicates he has gone directly to the 4^4 puzzle. I don't know which model he plans, or if the models become more similar with the extra possibilities inherent in the larger cube. I don't even know whether he plans to figure out how big the groups are or whether they are identical. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Wed, 22 Dec 1993 16:57:39 GMT Subject: The 4^3 and 3^4 Rubik puzzles (was Re: Wanted: Rubik's Revenge) e...@gsb002.cs.ualberta.ca (Holleman Eric) wrote: > By the way, I found the Revenge somewhat easier than the Cube, and I > don't think that it was because of my familiarity with the earlier > puzzle. x87benn...@gw.wmich.edu (Joe) wrote: > From my experience, if you can solve a Rubik's Revenge, you can > solve the Cube very easily. Once you get each of the middle 2 cubes > on each edge to match, and all 4 center cubes on each face to match, > it works exactly like a Rubik's cube. and a...@saturn.cs.swin.oz.au (Alan Christiansen) wrote: > I have both. I solved both. The 4x4x4 is a superset of the 3x3x3. > ie by fixing all the face centres and then pairing all edges you are > left with a 3x3x3 cube, except that when you have solved this 3x3x3 > there may be a single pair of edges flipped. This is impossible > on a real 3x3x3. Fixing this requires a middle layer to be rotated > 1/4 revolution and then all the bits put back. > I cant see how it can be [any] easier than a 3x3x3. I, too, found the 3^3 easier than the 4^3. But I can imagine ways in which a solver could find the 4^3 easier. Let us first consider a 4^3 with the faces fixed, the edges together, and the correct simulated edge flip parity. I would solve this as if it were a 3^3, and a lot of people do. But another solver might find it easier to take advantage of the extra moves that are not possible on a 3^3. To take a concrete example, it could be that the solver has a hard time with flipping edges by pairs, as is needed to solve the 3^3. On the 4^3 you can flip one edge at a time. So the solver would find the 4^3 position easier than the corresponding position on a 3^3. If the solver finds this so much easier that it overcomes the difficulty of putting the faces and edges together--or in fact puts the faces and edges together in the course of solving the corners and the edge positions--then the 4^3 could be easier. It depends on the solution procedure. a...@saturn.cs.swin.oz.au (Alan Christiansen) continues: > ANyway the real reason I am writing this is that I have written > a cube simulator. > It can simulate 3x3x3 4x4x4 5x5x5 .... cubes. > I am working on 4x4x4x4 cube simulation. This is interesting, as there is more than one way to model the four-dimensional cube problem. Consider the 3^4 cube. It has eight hyper-faces, each in the shape of a cube. One model of this puzzle is that you could turn any face of any hyper-face as if it were a 3^3 Rubik's cube. In a second model, you cannot move part of a hyper-face, but can turn each hyper-face as if it were a solid cube in space. A third model allows either kind of move. These models are different from each other. The second model permits the face centers of the hyper-faces to move around, whereas in the first model only edges and corners move. In the first model, odd permutations of corners are possible, which is not true in the second model. Of course, the third model is the closure of the first two. According to Hofstatder's column reprinted in _Metamagical_Themas_, there is an unpublished 1982 manuscript by H J Kamack and T R Keane entitled ``The Rubik Tesseract''. They calculated the size of the group of the 3^4 puzzle, but I don't know which model was used. Alan Christiansen indicates he has gone directly to the 4^4 puzzle. I don't know which model he plans, or if the models become more similar with the extra possibilities inherent in the larger cube. I don't even know whether he plans to figure out how big the groups are or whether they are identical. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math From: hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: Mon, 27 Dec 1993 17:23:57 GMT Subject: Re: unequal halves Two weeks ago k...@spdcc.com (Karl Heuer) asked: > Can the vertices of the n-cube (n >= 1) be partitioned into two > equal-sized subsets which are not congruent to each other? Not for n=1,2,3, but yes for n>=4. For n=4, labelling the vertices by binary coordinates: 0000--0010--0011 1000--1010--1110 | | | | | 0100--0110--0111 1101--1001--1011 | | | | 1100 1111 0101--0001 The adjacency graphs are not even isomorphic. Also, the former contains a square of diagonal 2 (0000,0011,1111,1100) while the rectangle of diagonal 2 in the latter is 1 by sqrt(3) (0001,0101,1110,1010). I (not extremely carefully) enumerated the ways of partitioning the vertices of the 4-cube into connected 8-point subsets, and counted that four of the twenty-four partitions were into non-congruent parts. I suspect that the partitions into congruent parts become asymptotically scarce, but I haven't figured out how to prove it. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Date: Mon, 27 Dec 93 17:52:16 -0500 (EST) From: Dan Hoey To: Cube Lovers Subject: Group theory basics (Re: Symmetry) Jerry Bryan asked a bunch of questions a couple of weeks ago, and I'll try to get to them all. The first bunch has to do with some fairly basic stuff, that I thought had been pretty well understood since the beginning of the mailing list, but maybe we need a refresher, or an explicit statement. In the message of Tue, 14 Dec 1993 20:50:51 EST, Jerry describes his representation of cube positions and transformations. In my computer model, the corner facelets are simply numbered from 1 to 24, and any configuration of the corners is an order-24 row vector. The rotation and reflection operators are also order-24 row vectors, again with each cell simply containing a number from 1 to 24. That is the most usual way of doing it, but it's important to specify what you represent by those vectors. When I do it, I number the corner facelet locations from 1 to 24, and these locations retain their numbers through manipulations of the cube. I use a vector A to specify a position in which the facelet whose home location is i has been moved to location A(i), for each i. I use a vector P to specify the transformation that moves the facelet in location i to location A(i), for each i. I'll assume you're doing the same, though you could, for instance, be representing the inverse of the operators, or the locations from which the facelets originate. Note that a position is represented by the same vector that represents the transformation that takes SOLVED to that position. Well, if P is a rotation operator, you could perform a rotation two ways. I guess one is pre-multiplication and one is post-multiplication. 1) For i = 1 to 24 B(i) = A(P(i)) I would write this as B = P A, and say that A is premultiplied by P, or equivalently that P is postmultiplied by A. In a general group, we could have B = P A where the multiplication is not considered to be the composition of permutations. But it turns out we can restrict our attention to permutation groups without loss of generality. For instance, when we are dealing with the supergroup, we can consider the orientation of a face center to be a permutation of the corners of the face center. 2) For i = 1 to 24 B(i) = P(A(i)) Here B = A P, A is postmultiplied by P, and P is premultiplied by A. (Note that the operator or position name appears in the reverse order from the prefix format. Algebraists sometimes avoid this by writing (i)B = ((i)A)P. I kid you not.) (As an aside, this illustrates the question I raised in my previous post about "which is the operator and which is the thing being operated on?" Is P operating on A, or is A operating on P?) Well, the answer is ``both''. I agree it's easy to get confused, which is why proofs are a good idea. Finally, if Q is a reflection (actually, if Q1 is the identity and Q2 is the reflection), then we have For j = 1 to 24 for k = 1 to 24 for m = 1 to 2 for i = 1 to 24 Bj,k,m(i) = Qm(Pj(A(Qm(Pk(i))))) I believe this loop calculates Dan Hoey's M. On the the theory that proofs are a good idea, let's see what this loop calculates. I'm going to put brackets around the subscripts. Then I'll substitute "R" for "Q", because I use Q for the set of quarter-turns of faces. Furthermore, I'll use "C" instead of "P", because the P[j] are just the elements of C, the group of cube rotations. So you are computing B[j,k,m] = C[k] R[m] A C[j] R[m] (1) for j in {1,...,24}, k in {1,...,24}, and m in {1,2}. Now every member of M (the group of cube rotations and reflections) has a unique representation as M[n] = C[k] R[m]. Let us define Cind() and Rind() as the functions for which M[n]=C[Cind(M[n])] R[Rind(M[n])]. So we can write (1) as B[j,k,m] = M[n] A M'[n] (M[n] C[j] R[Rind(M[n])]) Note that (M[n] C[j] R[Rind(M(n))] must be an element of C. So B is a set of elements of the form M[n] A M'[n] C[o]. To see that we have all such elements, first observe that (M[n]' C[o] R[Rind(M[n])]') is an element of C, say C[j]. So equation (1) includes: C[Cind(M[n])] R[Rind(M[n])] A C[j] R[Rind(M[n])] = M[n] A (M[n]' C[o] R[Rind(M[n])]') R[Rind(M[n])] = M[n] A M'[n] C[o]. Thus the set of all B[j,k,m] is the set of all M[n] A M'[n] C[o]. Or in English, that's the set of all M-conjugates of A, operated on by all whole-cube rotations. In my data base, I store the minimum of Bj,k,m over j = 1 to 24, k = 1 to 24, and m = 1 to 2. I tend to call the minimum of Bj,k,m a canonical form. I am not sure if that is the best terminology. The minimal element is not any simpler than any other. It is just that I need a function to choose an element from a set, and picking the minimal element seems very natural. Any other element would do as well, provided I could always be sure of picking the same element. It's pretty common terminology. You might be slightly better off calling it a ``representative element,'' as that connotes that the element is ordinary except in that it represents the equivalence class (like representatives in the U.S. Congress). Also, my criterion for equivalence is slightly different (but isomorphic, I think) than the one described by Dan Hoey. Suppose A and B are two cubes. Rather than mapping A to B or B to A in M, I map both A and B to their respective canonical forms. A and B are equivalent if their respective canonical forms are equal. This is straightforward once we show that M-conjugacy is an equivalence relation, and B[j,k,m] is an equivalence class. If A ~ Representative[A] = Representative[B] ~ B, then by transitivity A ~ B. Conversely, if A ~ B, then Class[A] = Class[B], and therefore Representative[A] = Representative[B]. This shows that the criteria are equivalent. Now, as to the centers. I still sometimes have a certain doubt about the centers. They are fixed, so how can you reduce the problem (i.e., increase the size of the equivalence classes) by both rotating the cube and rotating the colors (by both pre- and post-multiplication)? What you have done is to increase the size of the whole cube problem by a factor of 24, by dealing with all rotations of the cube, and the equivalence classes expand by the same factor, from 48 to 1152. This has allowed you to calculate something like M-conjugacy classes for cube problems that lack face centers. But the size of the equivalence classes doesn't shrink the problem for cubes that have face centers. You could have just calculated M-conjugates and got the same answer. I am not sure if this answers Dan's question about my model with centers added. It's clear now. I hadn't realized you were rotating the cube in space when the face centers were present. I expected that to be a wasted effort. But I am impressed by the way it allows you to shrink the database by storing positions together that differ only by whole-cube moves of the face centers. I think it should be possible to shrink the database without the effort, though. In your message of Thu, 16 Dec 1993 15:36:58 EST, on the ``Duality of Operators and Operatees'': I have mentioned several times my discomfort about "an operator" as opposed to "the thing being operated on" when it comes to groups. I am never quite sure just which of the two it is that people are talking about, even (or especially) when I am listening to myself talk. It is hard to keep it straight. Sometimes we all get it wrong. The best way to avoid errors, as far as possible, is to avoid such language and talk about group multiplication. But then we have to explain what is going on with the cube, so we get caught into talking about operators again. It's a discomfort that must be endured. The code to translate between the ASCII string X and the EBCDIC string Y is something like for i = 1 to n Y(i) = T(X(i)) where T is the translate table. Yes, or Y = X T as above. I am going to continue reading, but perhaps I could pose a question to Dan Hoey anyway: is reversing the role of X and T in the TRANSLATE function above essentially the same thing as switching between pre-multiplication and post-multiplication? Yes. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Date: Tue, 28 Dec 93 14:16:49 -0500 (EST) From: Dan Hoey To: Cube Lovers Subject: Re: Some Additional Distances in the Edge Group In his message of Fri, 17 Dec 1993 00:54:00 EST, Jerry Bryan makes some observations on the distances between the following positions in the edge group: I = Solved, P = Pons Asinorum (or Mirror), E = All edges flipped, and PE = P E = Pons Asinorum with all edges flipped. [I _will_ continue to use permutation multiplication as we have done so in this group since its inception. I realize that this agrees with some textbooks and is backwards from others, but it would be far more confusing to write these functionally all the time.] Jerry's brute-force search has shown that d(I,PE)=15, and he notes that conjugation by E shows us that d(P,E)=15 as well. He concludes: I have the sensation in describing this that the Edge group is square, with Start and Mirror-Image-of-Edges-Flipped 180 degrees apart, and Mirror-Image-of-Start and Edges-Flipped at the other two corners of the square. Well, it's not quite a square, since d(I,P)=12 and d(I,E)=9, according to Jerry's message of Wed, 8 Dec 1993 10:02:15 EST. Conjugation will similarly show that d(E,PE)=12 and d(P,PE)=9. So we are dealing with a rectangle. The sides of the rectangle are 9 and 12, and the diagonal is 15: a most fortuitous set of numbers, in that we can actually embed such a rectangle in the Euclidean plane! We can map the positions of the edge group to 4-tuples of distances. For any position X, let f(X)=(d(I,X), d(E,X), d(P,X), d(PE,X)). If f(X)=(a,b,c,d), then conjugation shows us that f(X E)=(b,a,d,c), f(X P)=(c,d,a,b), and F(X PE)=(d,c,b,a). So the set of quadruples has the symmetries of the rectangle. We know f(I)=(0,9,12,15). What is more, the earlier results on symmetry show us that I is at a local maximum distance from E, P, and PE. So, letting I1 be the unique (up to M-conjugacy) position adjacent to I, we have F(I1)=(1,8,11,14). (This destroys Euclidean embeddability.) An analogous result holds for the unique neighbor of each corner of the rectangle. We also have Jerry's results of Wed, 8 Dec 1993 22:41:28 EST and 23:16:50 EST that H (the 6-H pattern) and HE=H E are at distances 8 and 13 from start, respectively. Since H is an M-conjugate of P H, this gives us f(H)=(8,13,8,13). [Note: there are two distinct M-conjugates of H, call them H and Hbar. This distinction is important when we compose permutations: H H = I, but H Hbar = P. So we have to be careful when conflating M-conjugates.] We can by symmetry find f(H1)=(7,12,7,12) for H's unique neighbor H1. What quadruples are possible? If f(X)=(a,b,c,d), and X is not one of the eight corners and neighbors, we have max(2,9-b,12-c,15-d) <= a <= min(14,9+b,12+c) with constraints on b, c, and d from symmetry. A quick hack tells me there are 7836 such quadruples. I wonder how many of them are realized? If it's fairly few, I would like to see a diagram of quadruples, with lines between those quadruples that represent adjacent positions (adjacent quadruples differ by at most one in each coordinate). Maybe with the number of positions for each quadruple, too. I have an idea that such a diagram might tell us something about the problem, or at least look pretty. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Date: Tue, 28 Dec 93 18:40:52 -0500 (EST) From: Dan Hoey To: Cube Lovers Subject: Re: Cube Rotations mark.longridge@canrem.com (Mark Longridge) writes: Perhaps my description of the rotations was unclear... Yes. ...Perhaps it is better to use the form old FACE A -> new FACE A old FACE B -> new FACE B Where the faces A & B are adjacent. That will serve to uniquely identify a rotation, but it's somewhat verbose. Worse, it does not suffice to uniquely identify a symmetry from the group of rotations and reflections, M. I find it's far more informative to identify a rotation or reflection as a permutation of the faces, in cycle format. There are only ten kinds: Even rotations: I=Identity (1), (FRT)(BLD)=120-degree rotation (8), (FB)(RL)=180-degree orthogonal rotation (3). Odd rotations: (FRBL)=90-degree rotation (6), (FB)(TR)(DL)=180-degree diagonal rotation (6). Even reflections: (FR)(BL)=diagonal reflection (6), (FRBL)(TD)=90-degree glide reflection (6), Odd reflections: (FB)=orthogonal reflection (3), (FRTBLD)=60-degree glide reflection (8), (FB)(RL)(TD)=central reflection (1). In case it isn't clear, the cycle notation for (e.g.) a 120-degree rotation (FTL)(BDR) means that the F, T, L, B, D, and R faces move to the T, L, F, D, R, and B, locations, respectively. The only thing I'm afraid of with this notation is that someone will think I'm describing a magic-cube process rather than a whole-cube move. So when you say Top->Down, Front->Left, I would say (TD)(FL)(BR) for the 180-degree diagonal rotation, to distinguish it from (TD)(FLBR) the 90-degree glide reflection. ....wait a second, I don't think faces A & B have to be adjacent for the rotation to be unambiguous. Any 2 faces should do! No, you're back to your original bogosity. Knowing the destinations of two opposite faces doesn't give you any more information than knowing the destination of one (unless you go breaking the axles). Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Date: Wed, 29 Dec 93 17:43:28 -0500 (EST) From: Dan Hoey To: Cube Lovers Subject: Correction Re: Some Additional Distances in the Edge Group A couple of days ago, I said that proofs are a good idea. I'll say it again today with a redder face. Yesterday I discussed the edge group positions I = Solved, P = Pons Asinorum (or Mirror), E = All edges flipped, and PE = P E = Pons Asinorum with all edges flipped and the function from the edge group to 4-tuples of distances f(X)=(d(I,X), d(E,X), d(P,X), d(PE,X)). I wrote: ?? If f(X)=(a,b,c,d), then conjugation shows us that ?? ?? f(X E)=(b,a,d,c), f(X P)=(c,d,a,b), and F(X PE)=(d,c,b,a). ?? ?? So the set of quadruples has the symmetries of the rectangle. ?? The first sentence is incorrect, though the argument as a whole is reparable. First, I'll do what I should have done yesterday, and define the distance function d(X,Y). We want the minimum length process Z such that X Z = Y. But premultiplying both sides by X', we have Z = X' Y. So I define d(X,Y)=Length(X' Y). From the properties of the length function (Length(I)=0, Length(X)=Length(X'), and Length(X Y)<=Length(X) + Length(Y)) we can conclude that d(X,Y) is a metric. Suppose f(X)=(a,b,c,d). I claim f(E X)=(b,a,d,c), f(P X)=(c,d,a,b), and F(PE X)=(d,c,b,a). Proof: To show f(E X)=(b,a,d,c), first observe that I=I', E=E', and P E = E P. d(I,E X) = Length(I' E X) = Length(E' X) = d(E,X), so d(E,E X) = d(I, E E X) = d(I,X); d(P,E X) = Length(P' E X) = Length((PE)' X) = d(PE,X) so d(PE,E X)=d(P,E E X)=d(P,X). To show that f(P X)=(c,d,a,b), exchange P and E in the above argument. To show that f(PE X)=(d,c,b,a), use both occurrences of the argument. QED. So the idea of yesterday's message is correct, but I had X E, X P, and X PE instead of E X, P X, and PE X, respectively. I would show you a counterexample to yesterday's formulation, but it turns out there is none. I claim that f(X,E)=f(E,X), f(X,P)=f(P,X), and f(X,PE)=f(PE,X). Proof: Recall that E commutes with every element of the Rubik cube group, so f(X E)=f(E X). It turns out that ``up to M-conjugacy'', P commutes with every element of the edge group as well. For P performs a mirror-reflection of the edges, and so can be regarded as an element of M acting on the edge group. So P' X P = Xbar is an M-conjugate of X, and X P = P Xbar. Since Length(X) agrees on M-conjugates, so does d(X,Y), and so f(X), so f(X P)=f(P Xbar) = f(P X). Finally, f(X PE) = f(X P E) = f(E X P) = f(P E X) = f(PE X). QED. So it turns out it that the statement about f was true. But I am no less embarrassed for asserting it, for I had no reason to think it would be true. It's only rescued by the surprising commutativity of the Pons Asinorum. Finally, I would like to note something that I nearly included in yesterday's message, but yanked when I decided it was false: f(X')=f(X). Now I'll prove it: Proof: For W among {I,E,P,PE}, we have X W = W Xbar, for Xbar an M-conjugate of X. So d(X,W)=Length(X'W)=Length(W'Xbar')=Length(W'X')=d(W,X'). QED. Dan Hoey Hoey@AIC.NRL.Navy.Mil ---