Making Light: All of my opinions about the shooting of Representative Gabrielle Giffords (D-AZ) ::: January 12, 2011, 07:00 PM Win the world's tiniest violin solo, or equivalent no-prize, for the most far-fetched self-pitying historical metaphor. This contest idea comes courtesy of Randal Milholland, who gave us I got pants for Christmas. I guess the store was out of plague-ridden blankets. and I was a dick to my waitress last night and I didn't get a soda refill. Did we learn nothing from the Apartheid? My first hack is We've got to drive to work in the snow? Who do they think they are, Napoleon? but I think the Fluorosphere can do better. --- Making Light: Open thread 152 ::: January 28, 2011, 10:01 AM Sarah S @ 682: Shutting down the Internet by presidential fiat is not possible in the US, yet. But a bill to enable such a shutdown is gaining Joe-mentum. --- Making Light: Open thread 153 ::: February 07, 2011, 01:08 PM pedantic @24 -- Saying that you can "fill a hexagon with 153 evenly spaced points" relies on a somewhat unusual definition of "evenly spaced," as depicted in the Wikipedia article. I would have called the centered hexagonal numbers evenly spaced, but 153 is not one of them. --- Making Light: Open thread 153 ::: February 07, 2011, 04:00 PM B. Durbin @ 38: You can be male (XY) but not female (XX). You can be haploid (X or Y) but not polyploid (XXY etc). --- Explain xkcd Date: February 8th, 2011 From: Dan Hoey Subject: Archimedes sans-serif wrote: The joke is on how the first part of both sentence/proverb resembles common demands made by villains. Yes! And the cute thing is that the translation from the standard "conditional" form to the threatening "imperative" form is the substitution of "and" for "or", which makes this a symbolic logic joke. --- Newsgroups: talk.bizarre From: Dan Hoey Date: Tue, 08 Feb 2011 16:57:48 -0500 Subject: Re: That deodorant commercial nikolai kingsley wrote: >> Try watching the Dove deodorant commercial with > this would entail turning on the television, which i am generally > against, these days. Can your television access http://www.youtube.com/watch?v=04Xy_GEPu3k ? Dan --- Making Light: Open thread 153 ::: February 08, 2011, 04:59 PM Tom Whitmore @ 107 -- I don't think the ill-definedness of "interestingness" is a case of Russell's paradox. The two paradoxes are related, in that they are both based on circular definition (we are considering uninterestingness as one criterion for interestingness). But Russell's paradox is specifically about the property of "non-self-descriptiveness". One way in which the paradoxes differ is that we may assume that all numbers are interesting, and then there is no contradiction. But there is no way to assign a consistent value to non-self- descriptiveness, as applied to itself. --- Making Light: Open thread 153 ::: February 11, 2011, 10:32 AM Bill Stewart @ 184 -- Assuming the axiom of choice, whatever set of numbers you choose can be well-ordered*, so there would still be a least uninteresting number, which would therefore be interesting. The problem is that there is no particular well-ordering we can point out. You always get that kind of problem with the axiom of choice. * For anyone who doesn't bother to follow the link, please be aware that "well-ordered" means "ordered in such a way that every nonempty subset has a least element." People unfamiliar with the term tend to guess it means "assigned an ordering in an unambiguous way." --- Making Light: Today's Joke ::: February 14, 2011, 10:39 AM Bruce Cohen STM @34 -- That would work better with pi or e instead of phi, because phi is algebraic. --- Making Light: Today's Joke ::: February 14, 2011, 10:42 AM So I heard this single entendre, but it wasn't very funny. So I made up another one. Then I put the two together, if you know what I mean. --- Newsgroups: rec.puzzles From: Dan Hoey Date: Tue, 01 Mar 2011 00:09:51 -0500 Subject: Re: Uniqueness of fractional part of square roots. The Qurqirish Dragon wrote: [...] > Intuition tells me "yes" > reasoning: let d_n be defined by d_n := sqrt(n) - sqrt (n-1), for n>1 [...] > iii) for sufficiently large n, d_n is arbitrarily small, so d_n > converges to 0 Since sqrt'(x)=1/sqrt(4x)<1/sqrt(4n-4) for x in [n-1,n], it is clear that d_n <= 1/sqrt(4n-4). It should be easy to prove this without calculus, but I haven't managed it. Dan Hoey --- Making Light: Today's Joke ::: March 04, 2011, 06:04 PM Oh, I almost forgot: Happy imperative day! March forth and celebrate. --- Explain xkcd Date: March 15th, 2011 From: Dan Hoey Subject: Fairy Tales tfogarty wrote: The story is called, The Ant and the Grasshopper, which isn't a fairy tale but one of Aesop's fables. You can learn more about it here: http://en.wikipedia.org/wiki/The_Ant_and_the_Grasshopper So somehow the grasshopper singing/playing/not working is modeled by "contraction to a point on a manifold that is not a 3-sphere. Any idea what the connection is? --- Newsgroups: rec.puzzles From: Dan Hoey Date: Fri, 18 Mar 2011 16:10:10 -0400 Subject: Re: Yet another sequence Mark Brader wrote: > John Jones: >> Spoiler: This is A142958 in http://oeis.org/ >> (I wouldnt have guessed) > And just think, if only it was A14½2958, there'd be a geometrical > progression in the sequence number! > This makes me wonder, in a casual way, if there is any sequence > whose own sequence number in the OEIS contains its leading terms. How about the sequence whose nth term is one plus the nth term of sequence number n? But I doubt any sequences in the OEIS have a 14½2958th term. Dan --- Newsgroups: sci.math From: Dan Hoey Date: Mon, 21 Mar 2011 15:57:32 -0400 Subject: Re: On lists of songs and Russell's paradox Apart from the spurious connection to Russel's paradox, the topic of self-referential songs may be of some interest. However, it is generally the case that only the _lyrics_ of the song contain song references, and a performance of the same music with different lyrics is often spoken of as a performance of the _same_ song. This is the case, for instance, in "Leaves that are Green", which Simon and Garfunkel sing with the lyric I was twenty-one years when I wrote this song. I'm twenty-two now, but I won't be for long. Presumably this lyric was not used when the writer was twenty-one. In song whose lyrics refer to its music, neither the lyrics nor the music is, strictly speaking, self-referential. This prompts me to consider two questions. Are there songs whose lyrics refer explicitly to those same lyrics? Or are there songs in which a self-reference is somehow encoded in the music itself? Dan --- Math Forum From: Dan Hoey Date: Mar 21, 2011 4:24 PM Subject: Re: Proving floor fn equation from Ramanujan On 3/20/11 10:28 PM, Brad Cooper wrote: > Ramanujan derived the formula > [n/3] + [(n+2)/6] + [(n+4)/6] = [n/2] + [(n+3)/6] > where [] is the floor function. > Showing this involves proving (A) > 2n (mod 6) + (n+2) (mod 6) + (n+4) (mod 6) > equals > 3n (mod 6) + (n+3) (mod 6) + 3 (mod 6) [remainder omitted] I don't know where you got (A), but it is not correct. For example, if n=4(mod 6), then 2n (mod 6) + (n+2) (mod 6) + (n+4) (mod 6) = 2+0+2, while 3n (mod 6) + (n+3) (mod 6) + 3 (mod 6) = 4+1+3. The straightforward way of proving this is to take the six possibilities of n (mod 6). n=6k or n=6k+1 => [n/3] + [(n+2)/6] + [(n+4)/6] = 2k + k + k = 3k+k = [n/2] + [(n+3)/6] n=6k+2 => [n/3] + [(n+2)/6] + [(n+4)/6] = 2k + k + k+1 = 3k+1 + k = [n/2] + [(n+3)/6] n=6k+3 => [n/3] + [(n+2)/6] + [(n+4)/6] = 2k+1 + k + k+1 = 3k+1 + k+1 = [n/2] + [(n+3)/6] n=6k+4 or n=6k+5 => [n/3] + [(n+2)/6] + [(n+4)/6] = 2k+1 + k+1 + k+1 = 3k+2 + k+1 = [n/2] + [(n+3)/6] Dan --- Article 1046661 of rec.arts.sf.fandom From: Dan Hoey Newsgroups: rec.arts.sf.fandom Subject: Re: $19 K For Pre-School Tuition? Date: Mon, 21 Mar 2011 16:44:24 -0400 Organization: Naval Research Laboratory, Washington, DC David V. Loewe, Jr wrote: > http://www.washingtonpost.com/nyc-mom-sues-19000-a-year-preschool-says-it-promised-elite-prep-but-is-just-a-playroom/2011/03/14/ABrVhvV_story.html I was impressed by the subtle wordplay: "Some Manhattanites\222 high expectations of preschool are the stuff of urbane legend." Dan --- Math Forum From: Dan Hoey Date: Mar 22, 2011 4:02 PM Subject: Re: 2 questions about 3-SAT problem On 3/22/11 10:41 AM, Joshua Cranmer wrote: > On 03/22/2011 08:57 AM, Peter wrote: >> [...] Or would an algorithm which only decides if a TRUE-yielding >> assignment exists, without finding a particular one, be accepted as >> well? Are we looking for a constructive proof or is an existential >> one sufficient (would both of them imply that P=NP if they ran in >> polynomial time)? > The standard formulation of 3-SAT merely asks whether or not there > exists a satisfying assignment; it does not require that you produce an > assignment if one exists (that is the NP-hard version of the problem). Furthermore, if there is an algorithm of one type that runs in polynomial time, then there is an algorithm of the other type. This is obvious in one direction; in the other direction we go from an existential algorithm to a constructive one as follows: Run the existential algorithm once; if it fails, then there is no satisfying assignment. Otherwise, for each variable v in turn, make two augmented problems, one with an added clause "v or v or v" and the other with an added clause "not v or not v or not v". As long as the existential algorithm is correct, at least one of these must be satisfiable, so use the existential algorithm to find out which, and use that problem for augmenting with the other variables. This entails running the existential algorithm a polynomial number of times on problems polynomially larger than the original problem, so it runs in polynomial time. This sort of approach is can be used in any nondeterministic problem to convert an existential algorithm to a constructive algorithm, with at most polynomial increase in the problem size and number of iterations. Unfortunately, a polynomial increase in the problem size may require more than a polynomial increase in running time unless P=NP. Dan Hoey --- File 770 news of science fiction fandom Ellison Reference in Paul? March 24, 2011 at 12:59 am Dan Hoey says: Phil Nichols (on harlanellison.com) wrote: "A couple of days ago I asked the guy at The Californian about his suggestion that Jeffrey Tambor is playing a thinly disguised Harlan Ellison. In his reply to me, he cites only two things as the basis of his deduction: The hat. The profanity." You may or may not agree that this is specifically about Harlan Ellison. Not having seen PAUL, I have no idea. But always there is controversy. --- Making Light: Open thread 155 ::: March 24, 2011, 11:37 AM @339 Should I worry about spoiling a fifty-year-old movie? I'll take the chance. For me, the most memorable scene is at the end of Butterfield 8 when she's cruising down the road free as a bird and then WHAM the Code jumps out and kills her. You know the Code is going to get her because it's explicitly written into the Code, but the sudden artificiality of the ending is just the naked Code at its most brutal and unreasoning. I'm more than half persuaded it was a piece of satire ahead of its time. --- Newsgroups: sci.math, rec.puzzles From: Dan Hoey Date: Thu, 24 Mar 2011 12:40:58 -0400 Subject: Re: One Through Nine, With Eight Missing, And Two Ones Ilan Mayer wrote: > Leroy Quet wrote: >> Leroy Quet wrote: >>> (The number that is the sum in any particular 3-square/cell row, >>> column, or main diagonal may be in the square/cell on either end or >>> may be the central square/cell.) >> This is confusing. I am talking about, in any particular row, column, >> or main diagonal, the largest integer (the sum) may be any one of the >> 3 integers of that particular row, column, or main diagonal. >> Thanks, >> Leroy Quet > SPOILER > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > 312 > 167 > 459 I got the same answer, but with some reasoning. I mostly rely on the fact that there are only three triples containing 9: {2,7,9}, {3,6,9}, and {4,5,9}. There are four lines through the center, so the 9 can't be in the center. If the 9 is on a side, the other two numbers on that side form one of the three triples, then there are four possibilities for the center. In those twelve cases, fill in lines with the sum or difference to reach a quick contradiction. If the 9 is on the corner, the diagonal not containing that corner must have one each of {2,7}, {3,6}, and {4,5}, so must be either {2,3,5}, {2,4,6}, or {3,4,7}, then in each case we have three possibilities for the center. The three triples then determine all the numbers in a line with 9, and the other two numbers must be 1. The above solution is the only one that works. Is there a solution method that involves fewer than these 21 cases? Dan --- Making Light: Open thread 155 ::: March 24, 2011, 04:32 PM abi @ 404 -- You used a Phrase of Power The coming r*** w**? I wouldn't have thought such a phrase would be common enough to target, but if the polluters are using it, I'm glad they're filtered out. --- Newsgroups: sci.math, rec.puzzles From: Dan Hoey Date: Thu, 24 Mar 2011 21:12:21 -0400 Subject: Re: One Through Nine, With Eight Missing, And Two Ones Leroy Quet wrote: > The solution -- except reflections and rotations -- can be seen to be > unique by first noting that, due to the evenness and oddness of the > numbers, the three even numbers must be in some order along a main > diagonal. Neat. So combining the two ideas (even diagonal and 9 triples) there are only nine cases, all but one of which yield quick contradictions. 2 9 7 2 9 7 4 9 5 4 9 5 6 9 3 6 9 3 9 3 6 9 3 6 9 5 4 . 4 . . 6 . . 2 . . 6 . . 2 . . 4 . 5 2 . 7 4 . 7 6 . . 5 6 . 3 4 . 7 6 . 3 2 . 7 4 . 5 2 4 . 7 2 . 5 2 . 3 Dan --- Article 1046959 of rec.arts.sf.fandom From: Dan Hoey Newsgroups: rec.arts.sf.fandom Subject: Re: The news about Luna. Date: Mon, 28 Mar 2011 21:40:51 -0400 Keith F. Lynch wrote: > Ben Yalow wrote: >> "David V. Loewe, Jr" writes: >>> "Keith F. Lynch" wrote: >>>> Everyone calls it the 14th Street Bridge, but its official name is >>>> the Arland D. Williams Jr. Memorial Bridge. >>> According to Wikipedia, that is a complex of five different bridges >>> http://en.wikipedia.org/wiki/14th_Street_Bridge_%28Potomac_River%29. > The locals call just one of them the 14th Street bridge.[...] That's just wrong. Almost all of us locals refer to the three roadway spans as the 14th Street Bridge. The two rail spans are not usually included, though. Almost all discussion of the 14th Street Bridge is in the context of automotive travel. When a distinction needs to be made, we talk of the northbound lanes, the southbound lanes, or the express lanes. Dan --- Making Light: Open thread 155 ::: March 29, 2011, 04:22 PM According to a WSJ blog article, thh fhthrh hf Hnglhsh hs dhshmvhwhllhd. --- Making Light: Open thread 155 ::: March 29, 2011, 04:55 PM Terry @ 729 -- Yes, it's the latest viral career trainwreck, the tastiest Schadenfreude in days. I thought the best part was that the unwise author's first few salvos consisted of glowing reviews from amazon. Apparently she meant them as refutations of the lukewarm review. Reality-challenged, a little? The glowing reviews, though, are fairly amazing. Excerpts: I paricularly enjoyed being made a part of the story (or the feeling that I was) and also how discriptively it is written so that you can see the places and feel like you have visited them, which I find very important in a book. The story is about a seaman with his younger wife that use to be a ballet dancer. The descriptions of each new daily adventure she encountered seemed to effortlessly find themselves on each page for the reader to enjoy. Of course, it may be only that readers with a tenuous grasp of English are best equipped to appreciate haphazard writing. But I can't help suspecting a bit of pseudonymous self-reviewing. The most mystifying bits of the author's rant come when she is declaring victory over her detractors: My first book is great! and I intend to promote now without your ball. and Just look at your ball all of you....look at you. Can anyone shed light on this turn of phrase? --- Making Light: Open thread 155 ::: March 29, 2011, 08:41 PM Xopher @747 -- I don't think the word 'as' is missing. The way it looks to me, she took the grammatically correct "D&K watched Gino place more coffees...." and shoved 'hypnotically' in. She's trying to modify 'watched', but doesn't know that English doesn't allow an adverb there. And of course, she's got hypnotic watching backwards; if a cobra watches you hypnotically you're the one getting charmed. Sandy B. @748 -- Thanks. I had thought of that, but I can't see her telling her opponents to take their ball home. After considering it, I'm beginning to think it's a mistake for 'bull' as in 'logical error' or 'prattle'. Perhaps there's an ESL issue; I think 'ball' and 'bull' are more similar in some other languages. --- Making Light: Open thread 155 ::: March 29, 2011, 08:53 PM Lee @751: Yes, "used to could" is definitely a Southernism; I think it's a contraction from "it used to be that you could" Almost. I'm pretty sure that "used to could" is its own construction, based on the false analogy I go to the store. I could go to the store. I used to go to the store. I used to could go to the store.* "It used to be that you could" is a construction to express the idea in a grammatical way, but I don't believe it precedes "used to could". *incorrect. --- Making Light: Open thread 155 ::: March 29, 2011, 09:48 PM Xopher @747: The opportunities for truly vicious satire... Check out the Amazon reviews. It is slow and awkward at the beginning but the tempo quickly builds. Gur Terrx Frnzna then explodes in white hot excitement. I was so happy at the conclusion that I fell into a contented sleep. and When I first approached Gur Terrx Frnzna I'll admit I was a bit apprehensive, but soon I was lapping it up! --- Newsgroups: sci.math, rec.puzzles From: Dan Hoey Date: Tue, 29 Mar 2011 22:07:27 -0400 Subject: Re: One Through Nine, With Eight Missing, And Two Ones James Van Buskirk wrote: > Given the O.P. I was hoping for something like the sum of the > reciprocals of all the positive integers that don't have the digit > '8'. I like this more than Leroy Quet's puzzle. Do you have any ideas about how to calculate this number? --- Making Light: Open thread 155 ::: March 30, 2011, 09:12 AM TexAnne @777: Please do not say that the way I've spoken all my life is "incorrect." Sorry, I meant what linguists mean when they say it's "not standard English." "Incorrect" is just the word I use when I talk the way I've spoken all my life. I don't intend any offense. --- Making Light: Open thread 155 ::: March 30, 2011, 09:47 AM Elliott Mason @779 -- Yep, respectfully, that's me. Contemptuous and all. I'm not a tin god, but I'm on the side of the tin angels. --- Making Light: Open thread 155 ::: March 30, 2011, 12:16 PM David Harmon @ 782: a cultural dominance that doesn't really exist Damn straight. Nowadays you can talk like an upland-American and ain't nobody can say you no like they used to could. --- Making Light: Open thread 155 ::: March 30, 2011, 04:47 PM I'm sorry that Texanne takes my comments as a personal insult. If any other Southerners take them as an insult, I regret that, too. I've said I didn't intend to insult anyone, and I mean it. My tongue- in-cheek comments were meant to rebut Dave Harmon's assertion that cultural dominance by standard English does not exist. I am quite sorry that in my efforts to be humorous about it, I referred to a popular stereotype of a disadvantaged group. --- Newsgroups: sci.math, rec.puzzles From: Dan Hoey Date: Wed, 30 Mar 2011 16:48:41 -0400 Subject: Re: One Through Nine, With Eight Missing, And Two Ones James Van Buskirk wrote: > "Dan Hoey" wrote: >> James Van Buskirk wrote: >>> Given the O.P. I was hoping for something like the sum of the >>> reciprocals of all the positive integers that don't have the digit >>> '8'. >> I like this more than Leroy Quet's puzzle. Do you have any ideas >> about how to calculate this number? > I calculated up to double or quadruple precision once. The version > with 9 as the missing digit was from the Olympiad problem book and > you can start with the proof that the series converges and apply > a little more effort and get to something you can evaluate. No one > has ever given an independent calculation that I can check against, > however. According to http://oeis.org/A082837, the sum of reciprocals of positive integers that don't have the digit 8 is 22.726365402679370602833644156742557889210702616360219843536376162. For those without the digit 9, http://oeis.org/A082838 has 22.920676619264150348163657094375931914944762436998481568541998356. Robert Baillie has a paper at http://arxiv.org/abs/0806.4410v2 that covers these and related topics and includes Mathematica code. Dan --- Newsgroups: sci.math, rec.puzzles From: Dan Hoey Date: Thu, 31 Mar 2011 17:11:43 -0400 Subject: Re: One Through Nine, With Eight Missing, And Two Ones James Van Buskirk wrote: > "Dan Hoey" wrote >> According to http://oeis.org/A082837, the sum of reciprocals of >> positive integers that don't have the digit 8 is >> 22.726365402679370602833644156742557889210702616360219843536376162. >> For those without the digit 9, http://oeis.org/A082838 has >> 22.920676619264150348163657094375931914944762436998481568541998356. >> Robert Baillie has a paper at http://arxiv.org/abs/0806.4410v2 that >> covers these and related topics and includes Mathematica code. > Jeez, I did these about a quarter centruy ago. Maybe I should have > published. You might want to look at that paper. It covers a lot of related topics. Also, there wasn't an arxiv 25 years ago. Dan --- Explain xkcd Date: April 25th, 2011 From: Dan Hoey Subject: Etymology Dena wrote: Star Wars Wiki explains that Han Solo wasn't using the parsec as time unit. Obviously the Kessel run is a smuggling route from point A to point, where the easier route is much longer, and the shorter is more dangerous. Thus he was referring to the maneuverability of the ship, rather than his speed. According to https://secure.wikimedia.org/wikipedia/en/wiki/Millennium_Falcon#Depiction, George Lucas has advanced this explanation in the A New Hope DVD audio commentary. According to me, this explanation is a factitious rationalization used to paper over a glaring error in the movie. --- Explain xkcd Date: April 25th, 2011 From: Dan Hoey Subject: Etymology StarFreak1138 wrote: There's a line in the Star Wars Novelization (ghost written by Alan Dean Foster) where Obi-Wan comments, "... even a duck has to be taught to swim." Luke queries, "What's a duck?" Which in turn is an homage to "why a duck" from the Marx brothers. --- Explain xkcd Date: May 17th, 2011 From: Dan Hoey Subject: Number Line Chrisl wrote: Nope, its easily possible to create a number smaller than that with single-precision floats. MIN_FLOAT is 1.4E-45, and the difference (1.0f - 0.9999999f) (the longest 0.99... float for which this difference is bigger than 0.0f) evaluates to 1.1920929E-7. Is there a float between 1.0f and 0.9999999f? What's 1.0f-x, where x is the largest single-precision float strictly less than 1.0f? --- Explain xkcd Date: May 24th, 2011 From: Dan Hoey Subject: Darmok and Jalad That doesn't look too promising. How about Belindi Kalenda from the Star Wars universe? Playing with Star Trek/Star Wars mashups is a frequent source of humor, though I don't recall whether I've seen it on xkcd. Just to admit I'm not the world's biggest fan/nerd, I got the Belindi Kalenda ID from the xkcd forum. --- Explain xkcd Date: May 25th, 2011 From: Dan Hoey Subject: Extended Mind Shishir wrote: You hit an infinite loop from http://en.wikipedia.org/wiki/Mathematics. And yes I know, I have not hit anything in italics or parenthesis. It takes just 3 clicks to get back to Mathematics :) Wrong. See above: mathematics -> quantity -> property -> modern philosophy -> philosophy . --- Explain xkcd Date: May 25th, 2011 From: Dan Hoey Subject: Extended Mind AlanB wrote: AlanB wrote: AlanB wrote: Okay, who's gone and changed the Wiki page for "Quantity"?? "property" is not a link but "magnitude" is, which gives a loop. (Quantity->Magnitude->Ordering->Mathematics->Quantity). I know it worked *before* I started reading these comments (around 12:30pm Eastern) Okay. It's back now. Loop gone. (And I did try refresh a number of times.) Aha! It all makes a lot more sense if you look at the revision history for the Quantity page. This is the way the Wikipedia ends. Not with determined vandals, but with bored hackers. --- Explain xkcd Date: May 25th, 2011 From: Dan Hoey Subject: Extended Mind Seegtease wrote: If you go from the Panpsychism article, you will get rickrolled. That one lasted 8 minutes. --- From haoyuep@aol.com Thu Jun 9 20:04:55 2011 Date: Thu, 09 Jun 2011 20:04:31 -0400 From: Dan Hoey To: [redacted] CC: [redacted] Subject: Re: The fall picnic Speaking of tomorrow's meeting, could someone me if it's at my place? I know I signed up for a meeting this summer, but I'm not sure which month. Thanks, Dan --- Explain xkcd Date: July 25th, 2011 From: Dan Hoey Subject: Speculation Harm wrote: In this respect, can the ball be metaphorical in the sense of "passing the ball" in a conversation? Or possibly "catching" as in "sexually submissive" is metaphorical for "didactically submissive." Black Hat writes his own rules: he pitches. --- Making Light: Open Thread 161 ::: July 26, 2011, 04:08 PM Jacque @342: The two terms are independent: I don't believe either derives from the other. Air quotes are gestures made with the fingers to signify quotes, like air guitar. Scare quotes are quotes used to discredit (or at least withhold support from) the utterance, as opposed to attributing the utterance to a third person. They are used to warn that the utterance is misleading. Certainly written quotes may be used either to attribute or discredit a phrase. On the other hand, while air quotes are usually used as scare quotes, they may denote attribution. For instance, the statements He asked, "Who is innocent?" during wartime. and He asked, "Who is innocent during wartime?" may be hard to distinguish when spoken. I have seen people raise their fingers in air quotes to indicate the quoted phrase when saying such a sentence. Spoken attributive quotation is sometimes indicated by using the words quote and unquote to surround the quotation. These words may also be used as oral scare quotes, but they are usually spoken together, preceding the scare-quoted statement. See, for example, explanations by The Free Dictionary or John Lawler. Hope you appreciate this "educational" interlude. --- Newsgroups: sci.math From: Dan Hoey Date: Fri, 29 Jul 2011 15:10:02 -0400 Subject: Re: An cyclic inequality quasi wrote: > Presumably, the intended problem was as follows: > Prove that if x,y,z are positive real numbers such that > x + y + z = 3 > then > (x+1)/(y^2+1) + (y+1)/(z^2+1) + (z+1)/(x^2+1)>= 3 It still fails when x=y=z=1. Dan --- Newsgroups: sci.math From: Dan Hoey Date: Fri, 29 Jul 2011 17:01:52 -0400 Subject: Re: An cyclic inequality quasi wrote: > Dan Hoey wrote: >> >On 7/28/11 5:42 PM, quasi wrote: >>> >> Presumably, the intended problem was as follows: >>> >> Prove that if x,y,z are positive real numbers such that >>> >> x + y + z = 3 >>> >> then >>> >> (x+1)/(y^2+1) + (y+1)/(z^2+1) + (z+1)/(x^2+1)>= 3 >> >It still fails when x=y=z=1. > The symbol>= means "greater than or equal to". > Thus the claimed inequality holds for x = y = z = 1. > quasi Oops, my arithmetic took a hiccup. Of course you're right. Dan --- Newsgroups: rec.puzzles, sci.math, rec.games.abstract From: Dan Hoey Date: Wed, 03 Aug 2011 13:37:47 -0400 Subject: Re: Sum Of Products, Number Of Numbers Game Leroy Quet wrote: > Player 1 gets the sum of the scores for each row. The score for a row > is (the number of 1's in the row) * (the number of 2's in the row) * > (the number of 3's in the row) *...(the number of m's in the row), > where m is the largest integer such that all integers 1 through m > occur in that row. > Player 2 gets the sum of the scores for each column. The score for a > column is (the number of 1's in the column) * (the number of 2's in > the column) * (the number of 3's in the column) *...(the number of m's > in the column), where m is the largest integer such that all integers > 1 through m occur in that column. > A player gets no points for a row/column if there are no 1's in that > row/column. The last stipulation seems gratuitously irregular. It would be more consistent for rows without any 1s to score as the empty product, 1. Dan --- Math Forum From: Dan Hoey Date: Aug 3, 2011 3:16 PM Subject: Re: [Graph algorithm] Does this problem have a name ? On 8/3/11 1:30 PM, candide wrote: > Consider the following question relative to graph theory : > Let G a bipartite graph. To make the problem more concrete suppose > G is the disjoint union of two sets, say I and S. Suppose > *) I represents Individuals with name 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 > *) S represents Skills with name a,b, c, d, e, f, g, h. > So, each individual has _some_ skills, for instance, > -- individual 1 has skills b, d, g, and h, > -- individual 2 has skills a, f, and h, > -- etc. > [in the example, datas are randomly given]. > We aim to build a team composed of the _minimum_ number of > individuals from I in such a way that _every_ skill in S will be > represented in the team, that is for each skill s in S, there exists > a member of the team having the skill s. > Does this problem have a name ? Does an efficient algorithm for > solving it is known ? You can transform this from a graph problem to a set problem by providing as input the list of all the skill sets. The set problem is called MINIMUM COVER in Garey and Johnson. It is NP-complete even if each individual has exactly three skills, and even if we only want to know whether there is an exact cover (a cover in which no skill is duplicated). Dan --- Making Light: Crooked Timbre ::: August 04, 2011, 01:50 PM The spam I accused vanished while I was posting. Just so you know I'm not making any judgment on wyzeone1. --- Making Light: Open thread 162 ::: August 08, 2011, 03:27 PM Tim Walters @ 245: Just for the heck of it, I looked at their terms of service and poked around their FAQ, and I couldn't find anything about one-time events. --- Making Light: Open thread 162 ::: August 08, 2011, 04:18 PM On the Lemurian coins, I found a lot of interesting comments, but not the two that leapt out at me. First, the stipulation that unidentified coins are "no more likely to be silver than gold" suggests to me that we should consider the possibility that the unidentified coins may be less likely to be silver than gold. That leads to solutions parameterized by the probability of the unidentified coins being silver, which makes the restriction moot. The second consideration is one I've discussed before, perhaps even in the dim past of Making Light if such a problem has come up here. Suppose in an alternate universe Mr. Jones comes to the meeting with two silver coins. I propose that it's most likely he'd say, I have two Lemurian coins in my hand. At least one is silver. given what we know of the sort of utterances he is prone to. So what happens if he finds that he has one silver coin and one gold? I don't see any reason why he should prefer proclaiming his possession of a gold coin to one of silver, or vice versa. So my interpretation of the problem is that half of the time he has two matching coins, determining which statement he makes, and the other half of the time he has one of each, and chooses equiprobably between the two statements. Given that in the given instance he makes a statement about having a gold coin, he is equally likely to have two gold coins as he is to have one of each metal. --- Newsgroups: rec.puzzles From: Dan Hoey Date: Tue, 09 Aug 2011 10:09:18 -0400 Subject: Re: cotpi 16 - Chessboard of averages Mark Brader wrote: >> Each square of a chessboard contains a number that is equal to the >> average of the numbers in the squares adjacent to it. What is the >> maximum possible difference between the maximum number and the minimum >> number in the chessboard? > Usually the response to this question would be "define 'adjacent'", > but here it's not necessary, if we are allowed to assume that a > solution exists. If it does, then the answer is 0, i.e. all 64 > numbers are equal. > This is obvious because in a chessboard layout as described, if all > the numbers are multiplied by N then it is still a valid layout, but > the difference between the maximum and minimum numbers has also been > multiplied by N. So either there is no maximum possible difference > or it's 0. More directly, note that the average of a collection of numbers must be either equal to every number in the collection or strictly less then the greatest number in the collection. Thus every square with the maximum value must be equal to all of its neighbors. By induction, every connected component containing the maximum value must be constant. Since the neighborhood graph of the chessboard is connected, every square must be equal. Dan --- Article 1053696 of rec.arts.sf.fandom From: Dan Hoey Newsgroups: rec.arts.sf.fandom Subject: Re: Remember doing fanzines with mimeo? Date: Tue, 09 Aug 2011 11:39:11 -0400 Organization: Naval Research Laboratory, Washington, DC Michael Stemper wrote: > prd@pauldormer.cix.co.uk (Paul Dormer) writes: >> mkeesan@post.harvard.edu (Morris Keesan) wrote: >>> Dorothy J Heydt wrote: >>>>> But that's okay, because those old typewriters didn't have a period >>>>> (UK: "full stop") key. We had to use the decimal point instead. >>>> Decimal points aren't period either. >>> No, but they look alike. >> Do people still write decimal points above the line? I know I do. > Am I to infer from your question that some people put decimal points > (or decimal commas in Europe) *below* the line? The line for the > bottoms of letters (excluding their descenders)? I've never seen > or heard of that before. No, the distinction is between decimal points at the character baseline and decimal points above the baseline. The latter is somewhat old-fashioned and usually British, in my experience. Dan --- Making Light: Open thread 162 ::: August 09, 2011, 04:28 PM Abi @ 384: Who has standing and ability to lecture the looters? Does vox pop count? I expect we'll hear interviews from people whose home or workplace is burned or looted and whose communities become wastelands from the lack of commercial services and economic opportunity. I'm not saying they were living in a capitalist paradise to begin with, but pillage doesn't improve the situation. I live in Washington, DC, and there are still areas depressed from the riots of 1968. A lot of neighborhoods have recovered greatly in the last couple of decades, but we are still paying the price. I'm pretty torn by the whole demonstration/mass action dynamic. I really want to see justice done in the case of police, business, and political misconduct, and I'm unsure how that can be accomplished without taking it to the streets. On the other hand, there is a lot of room for riots to make things worse than they were before. --- Article 1054132 of rec.arts.sf.fandom From: Dan Hoey Newsgroups: rec.arts.sf.fandom Subject: Re: Remember doing fanzines with mimeo? Date: Mon, 15 Aug 2011 16:46:56 -0400 Organization: Naval Research Laboratory, Washington, DC Daniel R. Reitman wrote: > "Keith F. Lynch" wrote: >> Daniel R. Reitman wrote: >>> Jette Goldie wrote: >>>> I worked in the bank when we went decimal (well, very shortly after >>>> that) and we were advised that the amount in words should be in >>>> words and that the words trump figures if the two amounts differ. >>>> maybe that's Scots law? >>> Definitely the American rule. >> Then what's the point in writing numbers, if they're always ignored no >> matter what? >> I thought the whole point of writing down the amount in two different >> ways was to catch any typos (writeos?), ambiguities, or other >> mistakes. But for that to work, a check where the two amounts >> differ should be rejected. > Let's look at this practically. > You are a payee. You receive a check by mail which has a one-cent > discrepancy between text and figures. If your bank is required to > reject it, do you really want to have to go chase down the issuer? Within the last few months, I deposited a check that had a significant discrepancy between the text and the figures. The (US) bank said their policy is to honor the check for the lesser amount. My hazy recollection is that they said this was a relatively new policy, that some years ago they used to bounce such a check. Dan --- Article 1359485 of sci.math From: Dan Hoey Newsgroups: sci.math Subject: Re: Equivalent Spaces Date: Mon, 15 Aug 2011 22:58:00 -0400 Organization: Aioe.org NNTP Server William Elliot wrote: > Stephen J. Herschkorn wrote: >> Stephen J. Herschkorn wrote: >>> William Elliot wrote: >>>> Let p,d be two topologically equivalent metrics for S. >>>> If (S,d) is complete metric space, is (S,p) complete? >>> Consider the metric p(x,y) = |1/x - 1/y| on the interval (0, 1). >> Oops. Not quite right. Call that metric d, and put it on the interval >> (0, 1]. > It seems S = ((0,1),d) is homemorphic to (1,oo) which isn't complete, > by the isometry x -> 1/x. But then S wouldn't be complete. Actually, S and (1,oo) are isomorphic (as metric spaces) under the reciprocal map. So of course neither is complete. Consider instead T=((0,1],d) which is homeomorphic to (0,1]. (0,1] fails to be complete, because sequences {s_n} approaching zero do not converge; these are the only counterexamples to completeness. But T is complete because those sequences {s_n} are not Cauchy in metric d. Alternatively, T is isomorphic to [1,oo), which is complete; the sequences {1/s_n} approach oo, showing that they aren't Cauchy. Dan --- Article 278670 of rec.puzzles From: Dan Hoey Newsgroups: rec.puzzles Subject: Re: cotpi 17 - Pawns Date: Tue, 16 Aug 2011 23:21:51 -0400 Organization: Aioe.org NNTP Server cotpi wrote: > Jeffrey Turner wrote: >> cotpi wrote: >>> There are 64 pawns placed on a chessboard. Each square contains a pawn. >>> Two players are on the same side of the chessboard. Each player takes >>> turns and removes a pawn along with all pawns above it and to its right. >>> The player who picks the pawn at the bottom-left corner loses. Who can >>> definitely win the game? >> If the lower left corner is (0,0), and numbers go up to the right and >> up, then if the first player picks the pawn at (1,1) they will win. > If the first player picks the pawn at (1, 1) and all pawns at (x, y) > such that x > 1 and y > 1, what forces the second player to pick the > pawn at (0, 0)? I think Jeffrey Turner is assuming this is the game of Chomp, where a move at (1,1) removes all pawns at (x,y) where x >= 1 and y >= 1. In that case the only pawns left are at (x,0) and (0,y), and the first player wins by mirroring the second player's strategy. > The second player is free to choose a pawn at (7, 7) and the game > would continue for a while. Can you show that the second player > would be forced to pick the pawn at (0, 0) in the end? No, the first player already removed the pawn at (7, 7), even in the situation you described, since 7 > 1. I think that in the game you describe, where taking the pawn at (x0, y0) also removes pawns at (x, y) where x>x0 and y>y0, the first player can continue by responding to the second player's move (xn, yn) with a move at (yn, xn). If that works, the first player still wins. Dan --- Article 278714 of rec.puzzles From: Dan Hoey Newsgroups: rec.puzzles Subject: Re: Enigma 1654 - Tricky cards Date: Fri, 19 Aug 2011 13:14:33 -0400 Organization: Naval Research Laboratory, Washington, DC Chappy wrote: > Enigma 1654 - Tricky cards > New Scientist magazine, 09 July 2011. > By Bob walker. > Shown below are two of the cards Joe asked > Penny to make for him. Joe specified four > colours (A, B, C and D) and Penny could > choose any one of them for any of the four > areas. > |\ A /| |\ A /| > | \ / | | \ / | > |A X A| |D X B| > | / \ | | / \ | > |/ A \| |/ C \| > +---------+ +---------+ > Penny had only made a few cards when she > turned one round and noticed that it was the > same as the one she had just made. She had > assumed that all the cards must be different, > so she set about calculating how many > different cards made up a full set. > How many cards make up the set? > Ciao, > Chappy. > PS: Here's a link to the actual picture in case you > find my ASCII representation ambiguous. > http://www.newscientist.com/data/images/archive/2820/28201501.jpg We've had several correct answers, so let's try a slightly harder variant. Suppose the cards are made of colored glass, so that they can be turned over, changing to their mirror image. How many distinct cards are there in this case? Dan --- Article 278717 of rec.puzzles From: Dan Hoey Newsgroups: rec.puzzles Subject: Re: Enigma 1654 - Tricky cards Date: Fri, 19 Aug 2011 16:14:42 -0400 Organization: Naval Research Laboratory, Washington, DC Mark Brader wrote: > Dan Hoey: >> We've had several correct answers, so let's try a slightly harder >> variant. > I don't think it's harder, just different. >> Suppose the cards are made of colored glass, so that they >> can be turned over, changing to their mirror image. How many >> distinct cards are there in this case? > The only cards where a reflection produces a variant that cannot also > be produced by a rotation are those of types AABC and ABCD. So we lose > half of each of those types, or 12 + 3, and the new total is 55. Mark has the right answer. I calculated it using the Lemma that is not Burnside's. Of the 8 symmetry operations on a square, The identity leaves 256 cards fixed, the 180-degree rotation leaves 16 cards fixed, each of the two 90-degree rotations leaves 4 cards fixed, each of the two orthogonal reflections leaves 16 cards fixed, and each of the two diagonal reflections leaves 64 cards fixed. (256+16+4+4+16+16+64+64)/8 = 440/8 = 55. The original problem is solved with (256+16+4+4)/4 = 280/4 = 70. Dan --- Article 278731 of rec.puzzles From: Dan Hoey Newsgroups: rec.puzzles Subject: Re: cotpi 17 - Product Date: Mon, 22 Aug 2011 16:36:29 -0400 Organization: Naval Research Laboratory, Washington, DC cotpi wrote: > There is a list of 7 unique positive integers less than 10. One can > not deduce whether their sum is a multiple of 3 from their product. > What is their product? The word "unique" is makes no sense here. I assume you mean "distinct." It is easier to solve this problem by considering the two positive integers from {1,2,3,...9} that are not among the seven given. The product of the two numbers is in one-to-one correspondence with the product of the seven given, and the sum of the two numbers mod 3 is the negative of the sum of the sum of the seven given. Therefore, a necessary and sufficient condition is one cannot deduce whether the sum of the two is a multiple of three from their product. If the product were 1 mod 3, then the two numbers would either both be 1 mod 3 or both be 2 mod 3, so one could deduce that the sum is not a multiple of 3. If the product were 2 mod 3, then the two numbers would be 1 mod 3 and two mod 3, so one could deduce that the sum is a multiple of 3. Therefore, the product is divisible by 3. If the product were not divisible by 9, then one of the numbers would be a multiple of 3 and the other a non-multiple of 3, so one could deduce that the sum is not a multiple of 3. Therefore the product is divisible by 9, and we can't tell from their product whether the two numbers are {x,9y} or {3x,3y}. In order that the numbers be less than 10 and distinct, x=2 and y=1. Therefore, the seven given integers are either {1,3,4,5,6,7,8} or {1,2,4,5,7,8,9} and their product is 20160. Dan ---