The method is: Stationary brick #1 has mass=1. It is hit by sliding brick #2 of mass=100^N, which pushes it to a wall. The total number of bounces off both the wall and off brick #2, experienced by brick #1, is an integer agreeing in base 10 with the first N digits of pi. (Assume Newton laws of motion, rigid balls, no gravity, no friction, instantaneous perfect elastic collisions. He also depends on a conjecture that pi does not ever have a huge number of 9s in a row; if that is untrue then this could occasionally get the last digit off by 1.) Actually, what is quite cool about this is that all the collisions (except perhaps the final one) happen in total time span of order 1 (assuming initial unit speed for brick #2, initial travel distance 1 for brick #1 to wall), no matter how massive brick #2 is. (Hint: think about the final 2 collisions.) Can anybody devise a "physical" way to compute e, depending perhaps on the fact e = lim (1+1/n)^n ? On 6/18/14, math-fun-request@mailman.xmission.com <math-fun-request@mailman.xmission.com> wrote:
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Today's Topics:
1. Rising chain mystery (Dan Asimov) 2. Re: secretary problem (Warren D Smith) 3. Re: secretary problem (Eugene Salamin) 4. Re: secretary problem (Dan Asimov) 5. Re: Rising chain mystery (Whitfield Diffie) 6. Re: keyword frequencies in programming languages (Whitfield Diffie)
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Message: 1 Date: Wed, 18 Jun 2014 16:33:23 -0700 From: Dan Asimov <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Subject: [math-fun] Rising chain mystery Message-ID: <3C7B52FF-81FE-43A0-9F10-C71E0F404D6D@earthlink.net> Content-Type: text/plain; charset=us-ascii
Don't recall seeing this mentioned on math-fun:
In Jan., there was an article in the NY Times about how when a chain stuffed into a container open at the top, and one end is tossed over the side with lots of room to fall (e.g., letting it drop over a balcony), the chain falls into a pattern with the chain initially rising surprisingly high before it falls -- looking like a fountain.
If you don't want to hear the explanation right away so you can think about it, stop this video soon after 1:00. (The explanation begins soon after 1:15.)
http://www.nytimes.com/video/science/100000002747620/fountains-of-chain.html... (1 minute 47 seconds).*
Even with the explanation, I still find this amazing.
--Dan __________________________________________ * Unfortunately, to watch the video you apparently must watch the ad before it, but I didn't find this too painful.
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Message: 2 Date: Wed, 18 Jun 2014 19:44:40 -0400 From: Warren D Smith <warren.wds@gmail.com> To: math-fun@mailman.xmission.com Subject: Re: [math-fun] secretary problem Message-ID: <CAAJP7Y2x1gLG8rdBLQW7s-jwPYuOgoHhNqCG-x3_SuM2NUsapg@mail.gmail.com> Content-Type: text/plain; charset=ISO-8859-1
Brent Meeker: The version of this problem that I've heard is that there are N pieces of paper, each with a positive integer written on its face. They are lying face down on the table. You chose and discard them one at a time until you decide to keep the most recently one chosen. Your objective is to end up with the biggest number possible. Note that the range of numbers is not specified.
WDS: Asimov tried to formulate it where the numbers are chosen from some arbitrary distribution over the reals he called "F" and goal was to maximize expected value. Meeker goal as hereby interpreted by me, is to minimize expected rank (counting from the greatest, with rank=1, down, with rank=2,3,...,N).
The Asimov formulation sucks in the sense that for arbitrary F there may not exist any expectation at all. E.g. Cauchy distribution has no expectation value. So for now let me just consider the rank formulation.
If you've seen n numbers so far, 0<=n<N, the only question facing you is "should I stop now (getting as my reward, the rank of the last number seen, among the full set of N numbers)?" The answer is: Stop if the expected rank of the present number, is less than the expected rank you will get by continuing play.
Let E(m) = expected rank if stop after >=m numbers revealed, using optimal play from m onward.
Let E(m,k) = expected rank if stop after revealing exactly m numbers, and the last one is ranked k among those m, 1<=k<=m.
E(N) = (1+N)/2.
E(m,k) = Sum[ r * Binomial[r-1, k-1] * Binomial[N-r, m-k] / Binomial[N, m-1] , {r, k, N+k-m} ] in the notation of mathics.net . Except mathics.net crashed when I asked it to evaluate this sum. I presume E(m,k) = 1 + N*k/(m-1) at least approximately.
Presumably best play is to stop when m=N (since you have to) or stop if, earlier, E(m,k)<E(m+1).
Presumably when 0<=m<N E(m) obeys the recurrence
E(m) = (1/m) * Sum[ min( E(m+1), E(m,k) ), {k, 1, m} ]
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Message: 3 Date: Wed, 18 Jun 2014 17:20:25 -0700 From: Eugene Salamin <gene_salamin@yahoo.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] secretary problem Message-ID: <1403137225.51249.YahooMailNeo@web162102.mail.bf1.yahoo.com> Content-Type: text/plain; charset=iso-8859-1
A proper formulation of this problem must include the cost of picking up another piece of paper.? Otherwise, since there are only finitely many pieces of paper, if there's no cost, just pick up all of them.
? --? Gene
________________________________ From: Warren D Smith <warren.wds@gmail.com> To: math-fun@mailman.xmission.com Sent: Wednesday, June 18, 2014 4:44 PM Subject: Re: [math-fun] secretary problem
Brent Meeker: The version of this problem that I've heard is that there are N pieces of paper, each with a positive integer written on its face.? They are lying face down on the table.? You chose and discard them one at a time until you decide to keep the most recently one chosen. Your objective is to end up with the biggest number possible.? Note that the range of numbers is not specified.
WDS: Asimov tried to formulate it where the numbers are chosen from some arbitrary distribution over the reals he called "F" and goal was to maximize expected value. Meeker goal as hereby interpreted by me, is to minimize expected rank (counting from the greatest, with rank=1, down, with rank=2,3,...,N).
The Asimov formulation sucks in the sense that for arbitrary F there may not exist any expectation at all.? E.g. Cauchy distribution has no expectation value. So for now let me just consider the rank formulation.
If you've seen n numbers so far, 0<=n<N, the only question facing you is "should I stop now (getting as my reward, the rank of the last number seen, among the full set of N numbers)?" The answer is: ? Stop if the expected rank of the present number, is less than the expected rank you will get by continuing play.
Let E(m) = expected rank if stop after >=m numbers revealed, using optimal play from m onward.
Let E(m,k) = expected rank if stop after revealing exactly m numbers, and the last one is ranked k among those m, 1<=k<=m.
E(N) = (1+N)/2.
E(m,k) = Sum[ r * Binomial[r-1, k-1] *? Binomial[N-r, m-k] / Binomial[N, m-1] ,? {r, k, N+k-m} ] in the notation of? mathics.net . Except mathics.net crashed when I asked it to evaluate this sum. I presume? E(m,k) = 1 + N*k/(m-1)? at least approximately.
Presumably best play is to stop when m=N (since you have to) or stop if, earlier, E(m,k)<E(m+1).
Presumably when 0<=m<N E(m) obeys the recurrence
? ? E(m) = (1/m) *? Sum[ min( E(m+1), E(m,k) ),? ? {k, 1, m} ]
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Message: 4 Date: Wed, 18 Jun 2014 18:21:50 -0700 From: Dan Asimov <dasimov@earthlink.net> To: Eugene Salamin <gene_salamin@yahoo.com>, math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] secretary problem Message-ID: <EC5930C6-337C-4710-A6F7-0894645F88F9@earthlink.net> Content-Type: text/plain; charset=iso-8859-1
Just to be clear, I didn't formulate any problem, just stated some facts about distributions.
--Dan
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Message: 5 Date: Wed, 18 Jun 2014 18:24:54 -0700 From: Whitfield Diffie <whitfield.diffie@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Rising chain mystery Message-ID: <CAF+O-CXn6Lm2ijQ8jzcMOWwGPEV+JcTBe3cKP9_+Tqbj8AhTjQ@mail.gmail.com> Content-Type: text/plain; charset=ISO-8859-1
Even with the explanation, I still find this amazing.
This is indeed wonderful. I wonder if it accounts for cases of chains hopping off of pulleys. It seems that if the explanation is correct, the phenomenon shold depend on the compressibility of the chain; suppose it had long thing links? Or rubber links?
Whit
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Message: 6 Date: Wed, 18 Jun 2014 18:44:15 -0700 From: Whitfield Diffie <whitfield.diffie@gmail.com> To: Warren D Smith <warren.wds@gmail.com> Cc: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] keyword frequencies in programming languages Message-ID: <CAF+O-CUPRcy0wERBi-jM6arfRXvHiSEUWiswOV8MHf0JssA+4A@mail.gmail.com> Content-Type: text/plain; charset=ISO-8859-1
... all the counts by me and by Diffie are somewhat flawed due to our laziness about not really counting things "right".
It wasn't laziness on my part (although I have plenty); I didn't know what you wanted. It wouldn't be difficult to produce counts of two sorts of primitives. I could do this study on my code, counting only the lisp commands I did not write myself or I could do it on the system code and count only the ones defined in C. That is easy to get from the documentation system and I believe I can send a list without much trouble.
I believe that Allan Wechsler is right in thinking that lots of the Elisp primitives are about the data structures. Weeding those out would take more judgement. Presumably plus and times are about the fact that a language does arithmetic. As I said earlier, I think a count of words in Common Lisp would be rather different.
Whit
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