Comments on: The strange case of The Pigeon-hole Principle
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle/
Comments on MetaFilter post The strange case of The Pigeon-hole PrincipleSat, 10 Nov 2012 07:10:32 -0800Sat, 10 Nov 2012 07:10:32 -0800en-ushttp://blogs.law.harvard.edu/tech/rss60The strange case of The Pigeon-hole Principle
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle
It's Saturday; why not think about <a href="http://mindyourdecisions.com/blog/2008/11/25/16-fun-applications-of-the-pigeonhole-principle/">the pigeonhole principle</a>? Here are <a href="http://www.cut-the-knot.org/do_you_know/pigeon.shtml">problems</a> and <a href="http://www.artofproblemsolving.com/Wiki/index.php/Pigeonhole_Principle">more problems</a> and what you might call a <a href="http://www.cs.utexas.edu/~EWD/transcriptions/EWD09xx/EWD980.html">problem with the principle itself</a> as it is often stated.post:www.metafilter.com,2012:site.121712Sat, 10 Nov 2012 07:04:29 -0800WolfdogmathpigeonholepigeonholeproblemsproofBy: Wolfdog
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4677170
Among the problem lists, I didn't see one of my very favorite problems, so here: Prove that every positive integer <em>n</em> has a positive multiple which is written with only 1's and 0's.
(For example, for n=3 there is 3*37=111; for n=4 there is 4*25=100; for n=7, there is 7*1443=10101; for n=23, there is 23*4787=110101; and so on.)comment:www.metafilter.com,2012:site.121712-4677170Sat, 10 Nov 2012 07:10:32 -0800WolfdogBy: escabeche
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4677205
Describing the "covering checkerboard with opposite corners removed with dominos" problem as an application of the Pigeonhole is quite a stretch. You might even call it forcing a square pigeon into a round hole.comment:www.metafilter.com,2012:site.121712-4677205Sat, 10 Nov 2012 07:30:49 -0800escabecheBy: Phyllis Harmonic
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4677232
This sounds like it is related to the Pigeon-hole Principle- if someone more skilled with the maths can verify though:
For a given measured quantity, in a sample of five randomly selected members of a population greater than five, there is at least a 93.8% probability that the mean of the entire population will fall between the highest and lowest members of the sample. (From D.Hubbard- <em>How to Measure Anything</em>)comment:www.metafilter.com,2012:site.121712-4677232Sat, 10 Nov 2012 07:57:37 -0800Phyllis HarmonicBy: squorch
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4677244
Oh god, memories of discrete math coming back to haunt me...comment:www.metafilter.com,2012:site.121712-4677244Sat, 10 Nov 2012 08:10:19 -0800squorchBy: wobh
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4677248
If you think this is neat, you might be interested in the "Strong Law of Small Numbers" which is related. I'd provide links but I'm unable to decide which ones.comment:www.metafilter.com,2012:site.121712-4677248Sat, 10 Nov 2012 08:13:35 -0800wobhBy: escabeche
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4677260
<em>For a given measured quantity, in a sample of five randomly selected members of a population greater than five, there is at least a 93.8% probability that the mean of the entire population will fall between the highest and lowest members of the sample.</em>
This had better not follow from the Pigeonhole Principle, because it's false. Suppose there are a million people with a salary of $50K and one person with a salary of $1m. The mean salary is a little over $50K. But if you sample 5 people at random, the probability is very high that all five of them will have salary exactly $50K, so the mean of the population won't be between the highest and lowest members of the sample.
The source you quote is presumably making some assumptions on the distribution of the measured quantity.comment:www.metafilter.com,2012:site.121712-4677260Sat, 10 Nov 2012 08:20:18 -0800escabecheBy: cap11235
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4677263
Phyllis, that is not related. Also, the author is making a big assumption in there: that the population median and mean are the same. If this is true, then the probability of a given datum being on one side or the other of the mean is 0.5. So, the probability of all 5 being on one side is 0.5^5. Since there are two sides, the probability of the mean not being in the interval formed by the samples is 2*0.5^5, so the probability that it is in is 1-2*0.5^2=0.9375.comment:www.metafilter.com,2012:site.121712-4677263Sat, 10 Nov 2012 08:24:19 -0800cap11235By: escabeche
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4677268
But wait, if the median and the mean are the same, isn't the probabilty <em>exactly</em> 15/16? Why would he say "at least"?comment:www.metafilter.com,2012:site.121712-4677268Sat, 10 Nov 2012 08:31:56 -0800escabecheBy: Zalzidrax
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4677276
What I've always wondered is what kind of messed up person stuffs pigeons into holes? Poor birdies.comment:www.metafilter.com,2012:site.121712-4677276Sat, 10 Nov 2012 08:38:13 -0800ZalzidraxBy: dfan
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4677278
Hubbard's actual "Rule of Five" is
<blockquote>There is a 93% chance that the <b>median</b> of a population is between the smallest and largest values in any random sample of five from that population.</blockquote>
Not mean, not "at least 93%". (Actually the real probability is 93.75%, as cap11235 points out.)
So escabeche and cap11235's issues with the misstated rule are correct but don't apply to the actual one.
Now let's get back to the pigeonhole principle!comment:www.metafilter.com,2012:site.121712-4677278Sat, 10 Nov 2012 08:39:44 -0800dfanBy: philipy
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4677281
Yes, why does no-one ever think of the poor pigeons?
Still at least we've established that 5 is not a great sample size.comment:www.metafilter.com,2012:site.121712-4677281Sat, 10 Nov 2012 08:42:26 -0800philipyBy: Wolfdog
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4677285
<i>What I've always wondered is what kind of messed up person stuffs pigeons into holes?</i>
In Hungarian, it's the <abbr title="Skatulyaelv">matchbox principle</abbr>, if that makes you feel any better about it. Had the nomenclature been decided in the internet age, I'm certain it would be universally known as the catbox principle.comment:www.metafilter.com,2012:site.121712-4677285Sat, 10 Nov 2012 08:48:18 -0800WolfdogBy: charlie don't surf
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4677288
<a href="http://en.wikipedia.org/wiki/Hilbert's_paradox_of_the_Grand_Hotel">Hilbert</a> totally demolished this concept.comment:www.metafilter.com,2012:site.121712-4677288Sat, 10 Nov 2012 08:50:32 -0800charlie don't surfBy: Wolfdog
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4677290
Not really! I have uncountably many guests who want to visit Hilbert's hotel. (IF HE CAN HAVE AN INFINITE HOTEL, I CAN HAVE UNCOUNTABLY MANY GUESTS.) There <i>will</i> be multiple occupancies.comment:www.metafilter.com,2012:site.121712-4677290Sat, 10 Nov 2012 08:52:07 -0800WolfdogBy: philipy
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4677309
I demand uncountably many Schrodinger's Catboxes.comment:www.metafilter.com,2012:site.121712-4677309Sat, 10 Nov 2012 09:10:22 -0800philipyBy: charlie don't surf
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4677318
<em>I have uncountably many guests who want to visit Hilbert's hotel.</em>
Yeah, <a href="http://en.wikipedia.org/wiki/Cantor's_diagonal_argument">Cantor</a> took care of that one.comment:www.metafilter.com,2012:site.121712-4677318Sat, 10 Nov 2012 09:22:12 -0800charlie don't surfBy: Phyllis Harmonic
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4677356
Thanks for clarifying my earlier post, guys. I confess to relying on recall- which is getting more risky all the time, dammit.comment:www.metafilter.com,2012:site.121712-4677356Sat, 10 Nov 2012 09:59:06 -0800Phyllis HarmonicBy: cschneid
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4677360
My favorite variation: <a href="http://en.wikipedia.org/wiki/The_Library_of_Babel">http://en.wikipedia.org/wiki/The_Library_of_Babel</a>comment:www.metafilter.com,2012:site.121712-4677360Sat, 10 Nov 2012 10:03:00 -0800cschneidBy: irrelephant
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4677386
Further links on that page led me to this <a href="http://en.wikipedia.org/wiki/Hairy_ball_theorem">wonderfully named theorem</a>. I propose that it be renamed the <a href="http://www.youtube.com/watch?feature=player_detailpage&v=Ah6WEo2y7do#t=621s">Gervais Theorem</a>.comment:www.metafilter.com,2012:site.121712-4677386Sat, 10 Nov 2012 10:24:49 -0800irrelephantBy: iotic
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4677470
There's also this <a href="http://en.m.wikipedia.org/wiki/Cox–Zucker_machine">handy algorithm</a>. Might come in useful when you're combing hairy balls.comment:www.metafilter.com,2012:site.121712-4677470Sat, 10 Nov 2012 11:34:19 -0800ioticBy: hattifattener
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4677506
<i>What I've always wondered is what kind of messed up person stuffs pigeons into holes?</i>
Originally referred to the nesting boxes in dovecotes— pigeons actually like to have a nook to nest in. Desk cubbyholes reminded people of dovecots (or possibly reminded people of ossuaries, which reminded people of dovecotes) and the mathematical idea reminded people of desk cubbyholes.comment:www.metafilter.com,2012:site.121712-4677506Sat, 10 Nov 2012 12:00:36 -0800hattifattenerBy: j.edwards
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4677650
Years ago, in Siena, I saw <a href="http://www.flickr.com/photos/jedwards/2956683460/">a vivid demonstration of the Principle</a>.comment:www.metafilter.com,2012:site.121712-4677650Sat, 10 Nov 2012 13:44:04 -0800j.edwardsBy: Brian B.
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4677741
I use it to explain why polygamy is a terrible idea for most men.comment:www.metafilter.com,2012:site.121712-4677741Sat, 10 Nov 2012 15:03:46 -0800Brian B.By: valrus
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4677837
OK, I've been working on <a href="http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4677170">Wolfdog's problem</a> off and on for the better part of the day, and I think I've got it. If someone could check my proof that would be cool. I haven't done a rigorous proof in a while.
<strong>POSSIBLE SPOILERS IF I'M RIGHT</strong> ——————————————————
Ok, so we're given <i>n</i>. Any other number m, when divided by <i>n</i>, must have a remainder that is between 0 and <i>n</i> - 1 inclusive. (In more concise but also more jargon-y terms, consider <i>m</i> mod <i>n</i>.) These remainders will be our pigeonholes.
For the pigeons, consider the first <i>n</i> terms of the sequence 1, 11, 111, 1111, ...
Because there are <i>n</i> of them and only <i>n</i> - 1 possible remainders, the pigeonhole principle says two of them must have the same remainder when divided by <i>n</i>.
If you take these two terms and subtract the smaller from the larger, you have a number whose digits are some number of 1s followed by some number of 0s, and it is divisible by <i>n</i>. QED.
That last assertion—that the difference of two numbers with the same remainder mod <i>n</i> is divisible by <i>n</i>—is a pretty basic fact of number theory but here is a proof if you need it: let the two numbers be <i>a</i> and <i>b</i> and assume <i>a</i> > <i>b</i>. Then—this is basically the definition of a remainder—for some integers <i>q<sub>1</sub></i> > <i>q<sub>2</sub></i>,
<i>a</i> = <i>q<sub>1</sub>n + r</i> and <i>b</i> = <i>q<sub>2</sub>n + r</i>.
So <i>a - b</i> = <i>q<sub>1</sub>n + r</i> - (<i>q<sub>2</sub>n + r</i>) = (<i>q<sub>1</sub> - q<sub>2</sub></i>)<i>n</i>
which proves it's divisible by <i>n</i>.
<strong>END POSSIBLE SPOILERS</strong> ———————————————————————
If I'm right in thinking that proof works, I can see why you like that problem so much, <b>Wolfdog</b>. I have to say I was extraordinarily pleased with myself when I solved it, and it's a nice, elegant little solution and use of the pigeonhole principle.comment:www.metafilter.com,2012:site.121712-4677837Sat, 10 Nov 2012 16:55:41 -0800valrusBy: dfan
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4677845
There are <em>n</em> possible remainders (you said 0 to <em>n</em>-1 inclusive), but I guess all you have to do is take the first <em>n</em>+1 terms instead. The rest seems reasonable to me at first glance.
I guess that proves an even stronger result, in that the multiple of <em>n</em> is some number of 1s followed by some number of 0s, not just some combination of 1s and 0s.comment:www.metafilter.com,2012:site.121712-4677845Sat, 10 Nov 2012 17:08:31 -0800dfanBy: valrus
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4677848
Oh, you're right. Oops. I think I've been prone to off-by-one errors all my life. :/comment:www.metafilter.com,2012:site.121712-4677848Sat, 10 Nov 2012 17:13:26 -0800valrusBy: King Bee
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4677905
Impressive, valrus. I really like your proof. I had been thinking of it off and on for most of the day also!comment:www.metafilter.com,2012:site.121712-4677905Sat, 10 Nov 2012 17:44:20 -0800King BeeBy: bongo_x
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4677983
I'm not much of a mather, but none of the examples or statements seemed surprising or unintuitive. Were they supposed to be, or am I missing the point?comment:www.metafilter.com,2012:site.121712-4677983Sat, 10 Nov 2012 19:26:59 -0800bongo_xBy: valrus
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4678084
dfan: it occurs to me that the result could be made even stronger: there are an infinite number of such multiples. I guess I don't know if there's a formulation of the pigeonhole principle that applies if you have infinite pigeons, but it seems obvious that if there are finitely many pigeonholes, one's got to have infinitely many pigeons in it.comment:www.metafilter.com,2012:site.121712-4678084Sat, 10 Nov 2012 22:57:56 -0800valrusBy: iotic
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4678122
If you have infinite pigeons there's going to be a lot of poop. I hope you haven't parked your car anywhere nearby.comment:www.metafilter.com,2012:site.121712-4678122Sun, 11 Nov 2012 00:27:35 -0800ioticBy: dfan
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4678223
<blockquote>dfan: it occurs to me that the result could be made even stronger: there are an infinite number of such multiples.</blockquote>
True, but that was clear already; once you get your first one, you can add as many zeroes as you want on the end and it'll still be a multiple of the original number.comment:www.metafilter.com,2012:site.121712-4678223Sun, 11 Nov 2012 05:58:29 -0800dfanBy: valrus
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4678298
derpcomment:www.metafilter.com,2012:site.121712-4678298Sun, 11 Nov 2012 07:26:17 -0800valrusBy: the quidnunc kid
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4678373
Nice work, valrus!comment:www.metafilter.com,2012:site.121712-4678373Sun, 11 Nov 2012 09:03:29 -0800the quidnunc kidBy: Wolfdog
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4678441
Nice work indeed! Also, you're right that you only need to look at the first <i>n</i> terms of your sequence. If any of those is zero (mod <em>n</em>), then it's your solution. Otherwise, there's a pair with the same nonzero remainder mod <em>n</em>. So you weren't off by one after all! And if you wanted to strengthen the result just a little bit, you could add "Furthermore, if n is relatively prime to 10, then <em>n</em> has a multiple with only 1's."
Tricky as the problem is, it is much harder when it doesn't come in a thread about the pigeonhole principle.comment:www.metafilter.com,2012:site.121712-4678441Sun, 11 Nov 2012 10:34:30 -0800WolfdogBy: Obscure Reference
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4678507
I solved it when I had to solve a different problem: that any integer prime to 10 has a multiple, all of whose digits are 7. And, yes, I did it that same way.comment:www.metafilter.com,2012:site.121712-4678507Sun, 11 Nov 2012 11:55:42 -0800Obscure ReferenceBy: iotic
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4678520
Now, can we find a method for finding the lowest such solution? I.e. the lowest multiple of any n that consists of only zeros and ones.comment:www.metafilter.com,2012:site.121712-4678520Sun, 11 Nov 2012 12:08:28 -0800ioticBy: the quidnunc kid
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4678654
I was really hoping someone would say: "write n in binary, then QED".comment:www.metafilter.com,2012:site.121712-4678654Sun, 11 Nov 2012 13:53:19 -0800the quidnunc kidBy: Nioate
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4678989
Brute force can find the lowest, now that we know that algorithm will eventually terminate. The question is whether we can find a method with worst-case run time better than O(n).comment:www.metafilter.com,2012:site.121712-4678989Sun, 11 Nov 2012 18:20:28 -0800NioateBy: Elementary Penguin
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4682620
Now that I have my undergrad set theory book in front of me, I can confidently state the following infinite version of the pigeonhole principle:
Let κ ≤ λ and let μ be the largest number such that λ > κ × μ. Then if A is a set of size λ and P is a partition of A into κ subsets, at least one subset in the partition has size greater than μ.
So, for example, if you divide 8 things up into three piles, at least one of the piles has at least 3 things in it. But what's great is that this works even when both κ and λ are infinite cardinals, as well. And since ℵ<sub>α</sub>×ℵ<sub>β</sub> = ℵ<sub>β</sub> whenever α ≤ β, this means that if you divide an uncountable set into countably many piles, at least one of the piles has uncountably many things in it.comment:www.metafilter.com,2012:site.121712-4682620Tue, 13 Nov 2012 14:45:56 -0800Elementary PenguinBy: Elementary Penguin
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4682846
Argh. Let μ be the smallest number such that λ ≤ κ × μ ... has size greater than or equal to μ.comment:www.metafilter.com,2012:site.121712-4682846Tue, 13 Nov 2012 16:19:09 -0800Elementary PenguinBy: philipy
http://www.metafilter.com/121712/The-strange-case-of-The-Pigeonhole-Principle#4684126
A fun generalization of valrus' proof would seem to be that you can take any n,m whatever, and there is some multiple of m which consists of repetitions of the digit string of n followed by a bunch of zeroes.
For example you could take your phone number, and for any n, there is some multiple of n that is your phone number repeated umpteen times and then lots of zeroes.comment:www.metafilter.com,2012:site.121712-4684126Wed, 14 Nov 2012 09:36:28 -0800philipy