Archive-name: puzzles/archive/pickover/part1
Last-modified: 17 Aug 1993
Version: 4
See reader questions & answers on this topic! - Help others by sharing your knowledge ==> pickover/pickover.01.p <== Title: Cliff Puzzle 1: Can you beat the numbers game? From: cliff@watson.ibm.com If you respond to this puzzle, if possible please include your name, address, affiliation, e-mail address. If you like, tell me a little bit about yourself. You might also directly mail me a copy of your response in addition to any responding you do in the newsgroup. I will assume it is OK to describe your answer in any article or publication I may write in the future, with attribution to you, unless you state otherwise. Thanks, Cliff Pickover * * * At a recent trip to the Ontario Science Center in Toronto, Canada I came across an interesting puzzle. The center is located minutes from downtown Toronto and it's a vast playground of science with hundreds of exhibits inviting you to touch, try, test, and titillate your curiosity. The puzzle I saw there can be stated as follows. In the 10 boxes below, write a 10-digit number. The digit in the first box indicates the total number of zeros in the entire number. The box marked "1" indicates the total number of 1's in the number. The box marked "2" indicates the total number of 2's in the number, and so on. For example, the "3" in the box labeled "0" would indicate that there must be exactly three 0's in the 10-digit number. ------------------------------- | 0| 1| 2| 3| 4| 5| 6| 7| 8| 9| | 3| | | | | | | | | | ------------------------------- Stop And Think 1. Is there a solution to this problem? Are there many solutions to this problem? 2. A more advanced an interesting problem is to continue to generate a sequence in a recursive fashion such that each row becomes the sequence for the previous. For example, start with the usual 0 through 9 digits in row 1: Row 1: 0 1 2 3 4 5 6 7 8 9 Assume Row 2 is your solution to the puzzle. I've just inserted random digits below so as not to give away the solution: Row 1: 0 1 2 3 4 5 6 7 8 9 S(1) Row 2: 9 3 2 3 3 1 6 7 8 9 S(2) Row 3: S(3) Row 2 is now the starting point, and your next job is to form row 3, row 4, etc. using the same rules. In the previous example, a digit in the first box would indicate how many 9's there are in the next 10-digit number, and so forth. Contest: I am looking for the longest sequence of numbers users can come up with using these rules. Can you find a Row 2 or Row 3? Is it even possible to generate a "row 2" or "row 3"? ==> pickover/pickover.01.s <== 1) 0 1 2 3 4 5 6 7 8 9 2) 6 2 1 0 0 0 1 0 0 0 3) 0 0 0 4 4 4 0 4 4 4 4) 6 6 6 0 0 0 6 0 0 0 5) 0 0 0 4 4 4 0 4 4 4 . . . and so on, repeating rows 3 and 4. I don't know yet whether there are multiple solutions, but I'll keep looking. Mark Hayes Goddard Space Flight Center (GSFC) / Interferometrics, Inc. mwh@gemini.gsfc.nasa.gov GSFC Code 926.9 Greenbelt, MD 20771 ------------------------- In article <1992Sep14.133741.34561@watson.ibm.com>, you write: |> The puzzle I saw there can be stated as follows. In the 10 boxes below, |> write a 10-digit number. The digit in the first box indicates the total |> number of zeros in the entire number. The box marked "1" indicates the |> total number of 1's in the number. The box marked "2" indicates the |> total number of 2's in the number, and so on. For example, the "3" in |> the box labeled "0" would indicate that there must be exactly three 0's |> in the 10-digit number. |> |> ------------------------------- |> | 0| 1| 2| 3| 4| 5| 6| 7| 8| 9| |> | 3| | | | | | | | | | |> ------------------------------- |> |> |> Stop And Think |> |> 1. Is there a solution to this problem? Are there many solutions to this |> problem? This is an old puzzle, but I'll solve it as if it was new because I find your extension below to be interesting. Since all possible digits must be "counted" once, the ten digits must add up to 10. Consider the first digit (= the amount of zeroes used): 9: Impossible, since all the other digits would have to be zero. 8: Also impossible, since we must mark a 1 under the 8, and the other digits then must be zeroes. 7: We must mark a 1 under the 7, and we have one more non-zero digit to assign. We've used a 1, so we must put a non-zero digit under the 1. However, if we put a 1 there, it's wrong because we've used two ones, and if we put a two that's also wrong. So 7 zeroes doesn't work. 6: Begin as before, putting a 1 under the 6. Now we must mark under the 1, but putting a 1 is wrong, so put a 2. Now we have one non-zero digit left to assign, and marking a 1 under the two works. 6210001000 works. 5: Now we take a different approach to analyze this. If there are only five zeroes, then there are five non-zeroes, one of which is the 5 we marked under the zero. Obviously a 1 must be marked under the 5 and zeroes under 6-9, so we have 5----10000, where the dashes contain one zero and three other numbers, which must add up to four (since all ten digits must add up to ten) - so we have two ones and a two. But then the digits we have are described by 5310010000, which is not the set of digits we have, so there is no solution. Similar proofs show that there cannot be 4,3,2, or 1 zero. 0: Impossible, since you would have to use a zero to indicate you didn't have a zero. |> 2. A more advanced an interesting problem is to continue to |> generate a sequence in a recursive fashion such that each row becomes |> the sequence for the previous. For example, start with the usual |> 0 through 9 digits in row 1: |> |> Row 1: 0 1 2 3 4 5 6 7 8 9 |> |> Assume Row 2 is your solution to the puzzle. I've just inserted random |> digits below so as not to give away the solution: |> |> |> Row 1: 0 1 2 3 4 5 6 7 8 9 S(1) |> Row 2: 9 3 2 3 3 1 6 7 8 9 S(2) |> Row 3: S(3) |> |> Row 2 is now the starting point, and your next job is to form row 3, row 4, |> etc. using the same rules. In the previous example, a digit in the |> first box would indicate how many 9's there are in the next 10-digit number, |> and so forth. |> |> Contest: I am looking for the longest sequence of numbers users can come |> up with using these rules. Can you find a Row 2 or Row 3? |> Is it even possible to generate a "row 2" or "row 3"? Well, first off, our handy rule about all the digits adding up to ten no longer applies. Let's see if we can find an answer: Row 1: 0 1 2 3 4 5 6 7 8 9 Row 2: 6 2 1 0 0 0 1 0 0 0 Row 3: ? All the same digits must be placed under all the zeroes in row 2, or some of them would be wrong, and this digit cannot be larger than 4 since six non-zeroes are used under the zeroes in row 2. So, consider the cases: 4: If we put 4's under all the zeroes, we must put zeroes everywhere else. 0004440444 works. 3: Now we must place one non-zero digit under either the 6 or the 2, since there are two 1's that must stay alike. Putting any non-zero digit under the 6 is wrong since there aren't any sixes, unless you put a 6 under the 6, which is still wrong. Similarly no digit works under the two. 2: Now we must put a non-zero digit under the 2, since we already used 6 of them. We must also have two zeroes, which can only go under the ones. This gives us --02220222. However, we must put a non-zero under the 6, and we can't put a one, since we must have zeroes under the ones. Any number greater than one is wrong, because we don't have that many 6's. 1: OK, we start with ---111-111, and one of the -'s must be a zero. This zero must go under the 2 or the 6, because the ones must be alike (and we've already used some ones). Suppose we put 6's under the ones, and don't use any more ones. Then we need a 2 under the 6, and we need a one under the 2, which breaks what we did before. So, instead put 7's under the ones. Now we must put a 1 and a 0 in the other two spots, but either arrangement is wrong. We can't put a higher number under the ones because there aren't enough spaces left, so there is no solution with 1 zero. 0: Self-contradiction, as in the original problem. So now we have a unique third row. Can we make a fourth? Row 1: 0 1 2 3 4 5 6 7 8 9 Row 2: 6 2 1 0 0 0 1 0 0 0 Row 3: 0 0 0 4 4 4 0 4 4 4 Now there can only be two different digits used in the next number. Consider the possibilities: No zero is used: We need to mark this by putting zeroes under the zeroes Self-contradiction. Some zeroes are used: They can't go under the zeroes, so put zeroes under the fours. Now six zeroes are used, so put 6's under the zeroes. 6660006000 works. The same logic used to find row four shows that row five must be 0004440444 again, and we get into an infinite cycle alternating between these two. -- ----w-w--------------Joseph De Vincentis--jwd2@owlnet.rice.edu---------------- ( ^ ) Disclaimer: My opinions do not represent those of Owlnet. (O O) Owlnet: George R. Brown School of Engineering Educational Network. v-v (Unauthorized use is prohibited.) (Being uwop-ap!sdn is allowed.) Snail mail: Rice U., 6100 S. Main, Houston TX 77005. ------------------------- In rec.puzzles you write: >Title: Cliff Puzzle 1: Can you beat the numbers game? >From: cliff@watson.ibm.com [...] >1. Is there a solution to this problem? Are there many solutions to this >problem? Yes. No. >2. A more advanced an interesting problem is to continue to >generate a sequence in a recursive fashion such that each row becomes >the sequence for the previous. For example, start with the usual >0 through 9 digits in row 1: [...] >Contest: I am looking for the longest sequence of numbers users can come >up with using these rules. Can you find a Row 2 or Row 3? >Is it even possible to generate a "row 2" or "row 3"? My program produces the following output: 0123456789 6210001000 no solutions found So I believe that the result for row 2 is unique and that there is no result for row 3. [ I am including the program at the end of this message just for your interest ] >If you respond to this puzzle, if possible please include your name, >address, affiliation, e-mail address. If you like, tell me a little bit >about yourself. You might also directly mail me a copy of your response >in addition to any responding you do in the newsgroup. I will assume it >is OK to describe your answer in any article or publication I may write >in the future, with attribution to you, unless you state otherwise. >Thanks, Cliff Pickover The name, address etc should appear in my signature. As for myself, I'm a PhD student due to finish much too shortly who likes solving puzzles. Pauli Paul Dale | grue@cs.uq.oz.au Department of Computer Science | +61 7 365 2445 University of Queensland | Australia, 4072 | Did you know that there are 41 two letter | words containing the letter 'a'? The program I used follows: --------------------------------------8<------------------------------ #include <stdio.h> #include <stdlib.h> #define START(in) for(in=0;in<9;in++) { \ if(sum+in > 10) \ break; \ else \ sum = sum+in; \ counts[digits[in]]++; #define STOP(in) counts[digits[in]]--; \ sum -= in; \ } main() { short counts[10]; short i, sum; short i0,i1,i2,i3,i4,i5,i6,i7,i8,i9; static short digits[10] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }; short solns[10][100]; short solcnt=0; printf("0123456789\n"); again: for(i=0;i<10;i++) counts[i]=0; sum = 0; START(i0) START(i1) START(i2) START(i3) START(i4) START(i5) START(i6) START(i7) START(i8) START(i9) if(counts[0]==digits[i0] && counts[1]==digits[i1] && counts[2]==digits[i2] && counts[3]==digits[i3] && counts[4]==digits[i4] && counts[5]==digits[i5] && counts[6]==digits[i6] && counts[7]==digits[i7] && counts[8]==digits[i8] && counts[9]==digits[i9]) { printf("%d%d%d%d%d%d%d%d%d%d\n", i0,i1,i2,i3,i4,i5, i6,i7,i8,i9); for(i=0;i<10;i++) solns[0][solcnt] = i0; solns[1][solcnt] = i1; solns[2][solcnt] = i2; solns[3][solcnt] = i3; solns[4][solcnt] = i4; solns[5][solcnt] = i5; solns[6][solcnt] = i6; solns[7][solcnt] = i7; solns[8][solcnt] = i8; solns[9][solcnt] = i9; solcnt++; } STOP(i9) STOP(i8) STOP(i7) STOP(i6) STOP(i5) STOP(i4) STOP(i3) STOP(i2) STOP(i1) STOP(i0) if(solcnt == 0) { printf("no solutions found\n"); } else if(solcnt == 1) { for(i=0;i<10;i++) digits[i] = solns[i][0]; solcnt = 0; goto again; } else printf("multiple solutions found\n"); } --------------------------------------8<------------------------------ ------------------------- In article <1992Sep14.133741.34561@watson.ibm.com> you write: >Title: Cliff Puzzle 1: Can you beat the numbers game? >From: cliff@watson.ibm.com > >If you respond to this puzzle, if possible please include your name, >address, affiliation, e-mail address. If you like, tell me a little bit >about yourself. You might also directly mail me a copy of your response >in addition to any responding you do in the newsgroup. I will assume it >is OK to describe your answer in any article or publication I may write >in the future, with attribution to you, unless you state otherwise. >Thanks, Cliff Pickover > >* * * >At a recent trip to the Ontario Science Center in Toronto, Canada I came >across an interesting puzzle. The center is located minutes from >downtown Toronto and it's a vast playground of science with hundreds of >exhibits inviting you to touch, try, test, and titillate your curiosity. >The puzzle I saw there can be stated as follows. In the 10 boxes below, >write a 10-digit number. The digit in the first box indicates the total >number of zeros in the entire number. The box marked "1" indicates the >total number of 1's in the number. The box marked "2" indicates the >total number of 2's in the number, and so on. For example, the "3" in >the box labeled "0" would indicate that there must be exactly three 0's >in the 10-digit number. > >------------------------------- >| 0| 1| 2| 3| 4| 5| 6| 7| 8| 9| >| 3| | | | | | | | | | >------------------------------- > > >Stop And Think > >1. Is there a solution to this problem? Are there many solutions to this >problem? A. Since there are ten digits in the number, the sum of the digits in the bottom row must be 10. B. If x appears under y there must be x appearences of y, hence x*y<10 So, the MAXIMUM that can appear under each number is: --------------------- |0|1|2|3|4|5|6|7|8|9| |9|9|4|3|2|1|1|1|1|1| max --------------------- C. In fact, under the numbers 5..9 there can be AT MOST one non-zero (1) answer since otherwise two numbers of the 5..9 veriaty would appear and violate rule A. D. So there must be at least 4 zeros. If there were exactly 4 zeros, then the numbers 1..4 will all have under them non-zeros (as the zeros are used up for the 5..9 group). There is also at least one number that is 5 or greater. Well, there is a 5 (or more), a 4 (under zero), a 1 (under the 5..9 category) and something above zero under the other 1..4 digits for a total above 10. This violates rule A. E. So there must be at least 5 zeros. So a (exactly one) number that is at least 5 has a 1 under it. (since under zero would appear a >=5 number). F. Under 1 there must be at least 1 since the solution has at least one 1 (the one under a 5..9 number). However it could not be exactly 1 as then there would be 2 (or more) 1's in the solution. G. If there were 3 or more ones, then they must be under 2..9 . But then there would be a 5 (or more) under zero + a 3 (or more) under one + a 1 under three (or more) other places for a total above 10. H. So there must be at exactly 2 ones in the solution. And hence, at least 1 under two. We can summerize: --------------------- |0|1|2|3|4|5|6|7|8|9| |5|2|1|0|0|----1----| min |6|2|2|1|1|----1----| max --------------------- where the maximum under each digit is 10 - SUM(minimum of all others) I. Since no 3 or 4 is now possible, those two numbers must have a zero under them. J. So there are six zeros. Hence: --------------------- |0|1|2|3|4|5|6|7|8|9| |6|2|1|0|0|0|1|0|0|0| min |6|2|2|0|0|0|1|0|0|0| max --------------------- K. Notice that "min" is a solution, while "max" is not. Hence, "min is the *ONLY* solution! My name is Dan Shoham. This is the only fact about me I care to make public. You are free to attribute it, but provide me a note when you do so. shoham@ll.mit.edu ------------------------- >From clong@romulus.rutgers.edu (Chris Long) Tue Sep 15 06:08:45 1992 Path: igor.rutgers.edu!romulus.rutgers.edu!clong From: clong@romulus.rutgers.edu (Chris Long) Newsgroups: rec.puzzles Subject: Re: Puzzle 1 (SPOILER) Message-ID: <Sep.15.06.08.45.1992.9569@romulus.rutgers.edu> Date: 15 Sep 92 10:08:45 GMT References: <1992Sep14.133741.34561@watson.ibm.com> <1992Sep15.052438.12478@questrel.com> Organization: Rutgers Univ., New Brunswick, N.J. Lines: 62 In article <1992Sep15.052438.12478@questrel.com>, Chris Cole writes: Chris, don't forget to include my name on my solutions in the FAQ, please. My old article should be replaced with the following in the FAQ, anyway: --Cut here-- Solution prepared by Chris Long. Unfortunately, this isn't completely new, since I believe a similar puzzle I posted and answered are in the FAQ. However, it *is* different enough to be interesting. In article <1992Mar3.164702.428@hls.com>, ravi@hls.com writes: > Here's a small number puzzle : > Generate numbers such that the each digit in the number specifies > the number of the occurences of the position of the digit ( postions starting > with 0 from the left ). Example > The number 1210 ... My guess is only: 1210 21200 3211000 42101000 521001000 6210001000 No 1, 2, or 3 digit numbers are possible. Letting x_i be the ith digit, starting with 0, we see that (1) x_0 + ... + x_n = n+1 and (2) 0*x_0 + ... + n*x_n = n+1, where n+1 is the number of digits. I'll first prove that x_0 > n-3 if n>4. Assume not, then this implies that at least four of the x_i with i>0 are non-zero. But then we would have \sum_i i*x_i >= 10 by (2), impossible unless n=9, but it isn't possible in this case (51111100000 isn't valid). Now I'll prove that x_0 < n-1. x_0 clearly can't equal n; assume x_0 = n-1 ==> x_{n-1} = 1 by (2) if n>3. Now only one of the remaining x_i may be non-zero, and we must have that x_0 + ... + x_n = n+1, but since x_0 + x_{n-1} = n ==> the remaining x_i = 1 ==> by (2) that x_2 = 1. But this can't be, since x_{n-1} = 1 ==> x_1>0. Now assuming x_0 = n-2 we conclude that x_{n-2} = 1 by (2) if n>5 ==> x_1 + ... + x_{n-3} + x_{n-1} + x_n = 2 and 1*x_1 + ... + (n-3)*x_{n-3} + (n-1)*x_{n-1} + n*x_n = 3 ==> x_1=1 and x_2=1, contradiction. Case n>5: We have that x_0 = n-3 and if n>=7 ==> x_{n-3}=1 ==> x_1=2 and x_2=1 by (1) and (2). For the case n=6 we see that x_{n-3}=2 leads to an easy contradiction, and we get the same result. The cases n=4,5 are easy enough to handle, and lead to the two solutions above. -- Chris Long, 265 Old York Rd., Bridgewater, NJ 08807-2618 -- Chris Long, 265 Old York Rd., Bridgewater, NJ 08807-2618 ------------------------- The number "2020" was left off my list by mistake ... sorry. -Chris ------------------------- > * * * > At a recent trip to the Ontario Science Center in Toronto, Canada I came > across an interesting puzzle. The center is located minutes from > downtown Toronto and it's a vast playground of science with hundreds of > exhibits inviting you to touch, try, test, and titillate your curiosity. > The puzzle I saw there can be stated as follows. In the 10 boxes below, > write a 10-digit number. The digit in the first box indicates the total > number of zeros in the entire number. The box marked "1" indicates the > total number of 1's in the number. The box marked "2" indicates the > total number of 2's in the number, and so on. For example, the "3" in > the box labeled "0" would indicate that there must be exactly three 0's > in the 10-digit number. > > ------------------------------- > | 0| 1| 2| 3| 4| 5| 6| 7| 8| 9| > | 3| | | | | | | | | | > ------------------------------- > > > Stop And Think > > 1. Is there a solution to this problem? Are there many solutions to this > problem? > [Second question and contest problem omitted] Good puzzle! I am wondering though whether the second question (which I have not tried to solve yet) is moe amenable to computer search. It seems to me that there should not be so many cases to consider, so that even exhaustive search should work. So, here is my ten minutes work on the first question. I think there is a unique solution which is: 6210001000. Here is the reasoning. Let the number be (in Tex notation) d_0 d_1 d_2 d_3 d_4 d_5 d_6 d_7 d_8 d_9. By definition d_0 + d_1 + d_2 + d_3 + d_4 + d_5 + d_6 + d_7 + d_8 + d_9 = 10. (1) Moreover, d_0 > 0, since d_0 = 0 contradicts itself. Let d_0 = c for some integer 9 >= c >= 1. If c = 9, then d_9 = 1, contradiction since d_1 should both be 0 and 1 then. If 9 > c >= 1, we rewrite (1) removing all d_i s that are zeros c + d_(i_1) + d_(i_2) + ... + d_(i_(9-c)) = 10 <=> d_(i_1) + d_(i_2) + ... + d_(i_(9-c)) = 10 -c (2) where all the d_(i_j) >= 1, j=1,...,9-c (3) (2) & (3) imply that the d_(i_j)s are 8-c 1s and one 2. Since there exists ONE 2, then there exists at least one 1. So the only digits in the number are 0, 1, 2, and c (if different than 1 and 2). If c is either 1 or 2, we have 3 different digits in the number, which implies d_1 <= 3, impossible since d_1 = 8 - c >= 6. If c> 2, we have four different digits in the number, and in fact d_0 = c, d_1 = 8-c, d_2 = 1, d_c = 1, which leaves us with 6 0s. QED I hope I did not miss any other cases. Do you plan to post answers or comments later? Leonidas -------------------------------------------------------------------------------- Leonidas Palios The Geometry Center 1300 South Second Str palios@geom.umn.edu Minneapolis, Minnesota 55454 ------------------------- ------------------------------- | 0| 1| 2| 3| 4| 5| 6| 7| 8| 9| ------------------------------- | 6| 2| 1| 0| 0| 0| 1| 0| 0| 0| | 0| 0| 0| 4| 4| 4| 0| 4| 4| 4| <- | 6| 6| 6| 0| 0| 0| 6| 0| 0| 0| | | 0| 0| 0| 4| 4| 4| 0| 4| 4| 4| <- . . . I must be missing something in my understanding of your rules. I found the second row by imagining that I'd need lots of zeros and putting nine in the 0 column, then skipping back and forth adjusting things. I had to put a tic in the 9 column, then I had to put one in the 1 column, then I realized that had to change that to a two since now there were two ones, and at the same time another required tic in the 2 column balanced the change of one to two in the 1 column, and then of course there weren't nine zeros anymore, but there were still six and so by changing the nine in the 1 column to a six, the one in the 9 column sould just migrate down to the 6 column. But it almost seems like cheating to use fours in the second row when there were none in the second row to necessitate this kind of adjusting. *shrug* If this is right, the series is infinite, obviously. Please let me know if I'm interpreting something wrong. Thanks, and nice puzzle. :) Grant Culbertson grant@minos.nmt.edu dgray@sirius.nmt.edu ==> pickover/pickover.02.p <== Title: Cliff Puzzle 2: Grid of the Gods From: cliff@watson.ibm.com If you respond to this puzzle, if possible please include your name, address, affiliation, e-mail address. If you like, tell me a little bit about yourself. You might also directly mail me a copy of your response in addition to any responding you do in the newsgroup. I will assume it is OK to describe your answer in any article or publication I may write in the future, with attribution to you, unless you state otherwise. Thanks, Cliff Pickover * * * Consider a grid of infinitesimal dots spaced 1 inch apart in a cube with an edge equal in length to the diameter of the sun (4.5x10**9 feet). For conceptual purposes, you can think of the dots as having unit spacing, being precisely placed at 1.00000...., 2.00000...., 3.00000...., etc. Next choose one of the dots and draw a line through it which extends from that dot to the edge of the huge cube in both directions. Stop And Think 1. What is the probability that your line will intersect another dot in the fine grid of dots within the cube the size of the sun? Would your answer be different if the cube were the size of the solar system? 2. Could a computer program be written to simulate this process? 3. Answer the two questions above, but this time assume the line to have some finite thickness, T. ==> pickover/pickover.02.s <== ------------------------- In article <1992Sep14.141551.42075@watson.ibm.com> you write: >Title: Cliff Puzzle 2: Grid of the Gods >From: cliff@watson.ibm.com > >If you respond to this puzzle, if possible please include your name, >address, affiliation, e-mail address. If you like, tell me a little bit >about yourself. You might also directly mail me a copy of your response >in addition to any responding you do in the newsgroup. I will assume it >is OK to describe your answer in any article or publication I may write >in the future, with attribution to you, unless you state otherwise. >Thanks, Cliff Pickover > >* * * > >Consider a grid of infinitesimal dots spaced 1 inch apart in a cube with >an edge equal in length to the diameter of the sun (4.5x10**9 feet). >For conceptual purposes, you can think of the dots as having unit >spacing, being precisely placed at 1.00000...., 2.00000...., >3.00000...., etc. Next choose one of the dots and draw a line through it >which extends from that dot to the edge of the huge cube in both >directions. > >Stop And Think > >1. What is the probability that your line will intersect another dot >in the fine grid of dots within the cube the size of the sun? >Would your answer be different if the cube were the size of the >solar system? That depends on the manner the dot and the direction of the line were choosen. If both process used uniform (or any other continous) distribution, then - of course - the probability would be zero. If, for instance, the direction of the line is always choosen to be parallel to one of the cube's edges, then the probability is one. > >2. Could a computer program be written to simulate this process? Not a meaningfull question. Simple minded programs could never simulate infinitesimal points, but well thought out algorithm could express anything that can be shown analytically. > >3. Answer the two questions above, but this time assume the line >to have some finite thickness, T. > This is equivelent to making each dot of diameter T, and keeping the line thin. For T> (1 inch / 4.5*10^9 ft) inches, the probability -> 1. A simple minded computer program could simulate this. Dan Shoham shoham@ll.mit.edu ------------------------- In article <1992Sep14.141551.42075@watson.ibm.com> you write: >1. What is the probability that your line will intersect another dot >in the fine grid of dots within the cube the size of the sun? About 50%, because I usually draw horizontal lines. I.e., YOU DIDN'T GIVE THE DISTRIBUTION OF "lines". cf the puzzle of "what is the probability that a randomly selected chord of a circle is longer than the radius of that circle." The answer depends on how you "randomly select." _________________________________________________________ Matt Crawford crawdad@fncent.fnal.gov Fermilab ==> pickover/pickover.03.p <== Title: Cliff Puzzle 3: Too many 3's From: cliff@watson.ibm.com If you respond to this puzzle, if possible please include your name, address, affiliation, e-mail address. If you like, tell me a little bit about yourself. You might also directly mail me a copy of your response in addition to any responding you do in the newsgroup. I will assume it is OK to describe your answer in any article or publication I may write in the future, with attribution to you, unless you state otherwise. Thanks, Cliff Pickover * * * How many numbers have at least one digit -- a three? In the first 10 numbers, 1,2,3,4,5,6,7,8,9,10 there is only one number which contain the digit 3. This means that 1/10 or 10% of the numbers have the number 1 in the first 10 numbers. In the first 100 numbers the occurrence of numbers with at least one three seems to be growing. In fact there are 19 numbers: 3,13,23,33,43,53,63,73,83,93, 30,31,32,34,35,36,37,38,39. This means that about 19% of the digits contain the number 3 in the first 100 numbers. We can make a table showing the percentage of numbers with at least one 3-digit for the first N numbers. N % 10 1 100 19 1000 27 10000 34 The percentages rapidly increase to 100% indicating that almost all of the numbers have a 3 in them! In fact, a formula describing the proportion of 3's can be written: 1-(9/10)**N. The proportion gets very close to 1 as N increases. Stop And Think 1. How can it be that almost all of the numbers have a 3 in them? ==> pickover/pickover.03.s <== ------------------------- You wrote (in article <1992Sep14.141704.26532@watson.ibm.com>): >Title: Cliff Puzzle 3: Too many 3's >1. How can it be that almost all of the numbers have a 3 in them? Because as the numbers get larger, they contain more digits, increasing the probability that one of the digits in them might be a 3. In fact, the probability that a 3 will _not_ appear in a very long number is very low. I like this puzzle. Simple, but it made me think for a moment. A three in every number? Preposterous! ;) As for the other information you requested from responders: You have my name and email address now, I don't give out my home address unless it's necessary, and what sort of 'affiliation' are you seeking -- religious, business, or what? << Brian >> -- _/_/_/ Brian Kendig Macintosh Jedi Live never to be ashamed _/_/ Starfleet Captain Oracle Employee if anything you do or say _/ Intrepid Adventurer Saturn SL2 Owner is published around the world bskendig@netcom.com Wizard of Frobozz -- even if what is published Princeton '92! BSE/CS Writer/Actor/Singer is not true. ------------------------- In article <1992Sep14.141704.26532@watson.ibm.com> you write: >Title: Cliff Puzzle 3: Too many 3's >From: cliff@watson.ibm.com > >If you respond to this puzzle, if possible please include your name, >address, affiliation, e-mail address. If you like, tell me a little bit >about yourself. You might also directly mail me a copy of your response >in addition to any responding you do in the newsgroup. I will assume it >is OK to describe your answer in any article or publication I may write >in the future, with attribution to you, unless you state otherwise. >Thanks, Cliff Pickover > >* * * > >How many numbers have at least one digit -- a three? > >In the first 10 numbers, 1,2,3,4,5,6,7,8,9,10 there is only one number >which contains the digit 3. This means that 1/10 or 10% of the numbers >have the number 1 in the first 10 numbers. In the first 100 numbers the >occurrence of numbers with at least one three seems to be growing. In >fact there are 19 numbers: 3,13,23,33,43,53,63,73,83,93, >30,31,32,34,35,36,37,38,39. This means that about 19% of the digits >contain the number 3 in the first 100 numbers. > >We can make a table showing the percentage of numbers with >at least one 3-digit for the first N numbers. >N % >10 1 >100 19 >1000 27 >10000 34 > >The percentages rapidly increase to 100% indicating that almost all of >the numbers have a 3 in them! In fact, a formula describing the >proportion of 3's can be written: 1-(9/10)**N. The proportion gets >very close to 1 as N increases. > >Stop And Think > >1. How can it be that almost all of the numbers have a 3 in them? > Thaddeus Crews, 509 Windsor Green Blvd., Goodlettsville, TN, 37072 Graduate Student (Ph.D.) @ Vanderbilt University, Computer Science thaddeus@vuse.vanderbilt.edu This problem seems a little bit simple to me, but I was never that great at math problems so I am not betting the farm on this answer. The percentages you show for # of the first N numbers with at least one 3-digit is also true (about) for the # of the first N numbers with at least one 4-digit, at least one 5-digit, etc... Basically, as N increases, so does the number of digits in N, and therefore so does the number of chances for the digit 3 to appear (as well as all other digits). Given a number N with enough (?) digits, there is a 100% chance of all digits 0-9 appearing in that number (of course, 1.0E10000000000) does not have a 3 in it, but if you take the next 1.0E10000000000 numbers the percent that has a 3 will be (I suspect) 100%. My proof is clearly weak, but the claim is this: as N increases, the number of digits in N also increases. As N approaches infinity, the number of digits in N approaches infinity (at a slower rate, however). As the number of digits approaches infinity, the likelyhood of any specific digit appearing at least once approaches 100%. I think the real question (to be answered by someone with a better math training) would be "At what number N does the statistical likelyhood become 100% of at least one 3-digit appearing in the first N numbers." Hope this helps.... -- -- Thad Crews (email thaddeus@vuse.vanderbilt.edu) -------------------------------------------------------------------------- "Some people have a way with words, and some people, ... oh ... *not* have a way, I suppose..." -- Steve Martin ------------------------- Heh. As the numbers get larger, they have more digits. Assuming a random occu various digits in the larger numbers (not unreasonable when n-> infinity) the pr number NOT having a 3 is very low. -john 'I know it's not a proof...' karakash- ------------------------- >Title: Cliff Puzzle 3: Too many 3's Seth Breidbart PO Box 5157 New York, NY 10185 Morgan Stanley & Co. >1. How can it be that almost all of the numbers have a 3 in them? The probability that a random sequence of n digits does not contain a 3 is .9^n; as n->infinity, this probability -> 0. Since almost all numbers have a lot of digits (there are only a finite number of integers with <n digits, and infinitely many with >n), the limiting probability is 0. ------------------------- In article <1992Sep14.141704.26532@watson.ibm.com> you write: >Title: Cliff Puzzle 3: Too many 3's >From: cliff@watson.ibm.com > >If you respond to this puzzle, if possible please include your name, >address, affiliation, e-mail address. If you like, tell me a little bit >about yourself. You might also directly mail me a copy of your response >in addition to any responding you do in the newsgroup. I will assume it >is OK to describe your answer in any article or publication I may write >in the future, with attribution to you, unless you state otherwise. >Thanks, Cliff Pickover > >* * * > >How many numbers have at least one digit -- a three? > >In the first 10 numbers, 1,2,3,4,5,6,7,8,9,10 there is only one number >which contains the digit 3. This means that 1/10 or 10% of the numbers >have the number 1 in the first 10 numbers. In the first 100 numbers the >occurrence of numbers with at least one three seems to be growing. In >fact there are 19 numbers: 3,13,23,33,43,53,63,73,83,93, >30,31,32,34,35,36,37,38,39. This means that about 19% of the digits >contain the number 3 in the first 100 numbers. > >We can make a table showing the percentage of numbers with >at least one 3-digit for the first N numbers. >N % >10 1 >100 19 >1000 27 >10000 34 > >The percentages rapidly increase to 100% indicating that almost all of >the numbers have a 3 in them! In fact, a formula describing the >proportion of 3's can be written: 1-(9/10)**N. The proportion gets >very close to 1 as N increases. > >Stop And Think > >1. How can it be that almost all of the numbers have a 3 in them? > No problem. In fact almost all very large numbers have all digits in them. It is rather hard for a number with zillions of digits to avoid "3"s (or any other digit). It fact, the sequences "15", "172", and "666" (and any other finite sequence) are also contained (in order) within almost all numbers. Dan Shoham shoham@ll.mit.edu ------------------------- Before I forget: Chris Long, 265 Old York Rd., Bridgewater, NJ 08807-2618 clong@remus.rutgers.edu -- Chris Long, 265 Old York Rd., Bridgewater, NJ 08807-2618 ------------------------- >Title: Cliff Puzzle 3: Too many 3's >From: cliff@watson.ibm.com >If you respond to this puzzle, if possible please include your name, >address, affiliation, e-mail address. If you like, tell me a little bit >about yourself. You might also directly mail me a copy of your response >in addition to any responding you do in the newsgroup. I will assume it >is OK to describe your answer in any article or publication I may write >in the future, with attribution to you, unless you state otherwise. >Thanks, Cliff Pickover >* * * >How many numbers have at least one digit -- a three? >In the first 10 numbers, 1,2,3,4,5,6,7,8,9,10 there is only one number >which contains the digit 3. This means that 1/10 or 10% of the numbers >have the number 1 in the first 10 numbers. In the first 100 numbers the >occurrence of numbers with at least one three seems to be growing. In >fact there are 19 numbers: 3,13,23,33,43,53,63,73,83,93, >30,31,32,34,35,36,37,38,39. This means that about 19% of the digits >contain the number 3 in the first 100 numbers. > >We can make a table showing the percentage of numbers with >at least one 3-digit for the first N numbers. >N % >10 1 >100 19 >1000 27 >10000 34 >The percentages rapidly increase to 100% indicating that almost all of >the numbers have a 3 in them! In fact, a formula describing the >proportion of 3's can be written: 1-(9/10)**N. The proportion gets >very close to 1 as N increases. >Stop And Think >1. How can it be that almost all of the numbers have a 3 in them? I'm not sure this is the answer you are looking for, but: 9 = 9 9*9 = 81 9*9*9 = 729 9*9*9*9 = 6561 etc. The probability of having 3 as the digit in a one-digit number is 1/10. " of not having 3 " is 9/10. In a two-digit number, the prob. of NOT having 3 as the first digit or the second digit, ie. not having 3 in the two-digit number, is simply the product of (NOT having 3 in first digit) times (NOT having 3 in second): (9/10)*(9/10) = 81/100 = 0.81 For a three-digit number: (9/10)*(9/10)*(9/10) = 729/1000 = 0.729 For an n-digit number: (9/10)**n = probability. We can see that as "n" becomes larger and larger, the probability of NOT having a three at all in the number becomes smaller and smaller. Indeed, as "n" approaches infinity, this probability approaches zero. In other words, it is very rare for a large number NOT to have 3 as one of its digits. In fact, it is very rare for a large number NOT to have any of the ten possible integers represented at least once. [Aside, N % 10 1 = (10 - 9)/1 times 100 100 19 = (100 - 81)/100 times 100 1000 27 = (1000 - 729)/1000 times 100 10000 34 = (10000 - 6561)/10000 times 100 etc. ] Kumar kumar@ug.cs.dal.ca ps: I'll leave it as a small exercise to tie up the loose ends. ==> pickover/pickover.04.p <== Title: Cliff Puzzle 4: Time in a Bottle From: cliff@watson.ibm.com If you respond to this puzzle, if possible please include your name, address, affiliation, e-mail address. If you like, tell me a little bit about yourself. PLEASE ALSO directly mail me a copy of your response in addition to any responding you do in the newsgroup. I will assume it is OK to describe your answer in any article or publication I may write in the future, with attribution to you, unless you state otherwise. Thanks, Cliff Pickover * * * Consider a chain of bottles (B) each connected to one another by a thin tube. A marble is placed in bottle 1. Each tube contains a one-way valve so marbles can only go from left to right in the tubes which are symbolized with "-" marks: 1 2 3 4 B - B - B - B - The tubes are thin so it takes 1 hour of constant random shaking to get the marble from B1 to B2. Likewise for each bottle. I have not fully described the bottle collection. Each bottle has a backward 1-way tube to bottle 1. I've tried to diagram these with "*" symbols. Each time the marble enters bottle B(N) it has a 50% probability of going back to bottle 1 via these tubes. ****<******** * * ***<***** * * * * * * * * * 1 2 3 4 B - B - B - B - Stop And Think 1. In how many hours will you expect to get the marble out of bottle 10 after placing the marble in bottle 1? 2. Is there a general formula for the amount of time required to get the ball out of bottle N into bottle N+1 given a probability P of backwards motion (given as 50% in this problem)? 3. In how many hours will you expect to get the marble out of bottle 10 after placing the marble in bottle 1 given two backward tubes for each bottle instead of one backward tube? ==> pickover/pickover.04.s <== ------------------------- Subject: Re: Cliff Puzzle 4 (SPOILER) Newsgroups: rec.puzzles References: <1992Sep15.205532.4172@watson.ibm.com> In article <1992Sep15.205532.4172@watson.ibm.com>, Cliff writes: > 1. In how many hours will you expect to get the marble out of bottle 10 > after placing the marble in bottle 1? The expected amount of time to go from state n-1 to n (state 11 is an absorbing state) is E(n-1,n) = 1 + E(1,n)/2 for 1<n<11; also E(n-1,n+1) = E(n-1,n) + E(n,n+1) for 1<n<11. If n=2 then E(1,2) = 1 + E(1,2)/2 ==> E(1,2) = 2 (it should be clear that no E is infinite for this problem). E(2,3) = 1 + E(1,3)/2 = 1 + E(1,2)/2 + E(2,3)/2 ==> E(2,3)/2 = 2 ==> E(1,3) = 6. I claim that in general E(1,n) = 2^n - 2 and E(n-1,n) = 2^(n-1). Assume true for n, then E(n,n+1) = 1 + E(1,n+1)/2 = 1 + E(1,n)/2 + E(n,n+1)/2 ==> E(n,n+1)/2 = 1 + E(1,n)/2 ==> E(n,n+1) = 2 + E(1,n) ==> E(n,n+1) = 2 + 2^n - 2 = 2^n. Furthermore E(1,n+1) = E(1,n) + E(n,n+1) = 2^n - 2 + 2^n = 2^(n+1) - 2. Therefore by induction the result is established. Now E(1,11) = E(1,10) + 1 (ball can't go back to bottle 1 after leaving bottle 10) = 2^10 - 1. > 2. Is there a general formula for the amount of time > required to get the ball out of bottle N into bottle N+1 given > a probability P of backwards motion (given as 50% in this problem)? I'd have to check my work, but I get E(n,n+1) = q^n, where q = 1/(1-p). > 3. In how many hours will you expect to get the marble out of bottle 10 > after placing the marble in bottle 1 given two backward tubes for each > bottle instead of one backward tube? I get E(1,n) = (q^n - q)/(q-1), so E(1,11) = E(1,10) + 1 = (3^10 - 3)/2 + 1. ------------------------- In regards to your fourth problem, the following comments (marked with a ">") should be added. I thought the answer was quite surprising! --- The expected amount of time to go from state n-1 to n (state 11 is an absorbing state) is E(n-1,n) = 1 + E(1,n)/2 for 1<n<11 > since we expect it'll take an hour for the ball to leave bottle n-1, > and it then has a probability of 1/2 of returning to the first bottle; also E(n-1,n+1) = E(n-1,n) + E(n,n+1) for 1<n<11 > since the only way of getting to state n+1 from n-1 is to first > go from state n-1 to n, and then from n to n+1; the total expected > time is the sum of the two. ==> pickover/pickover.05.p <== Title: Cliff Puzzle 5: Mystery Sequence From: cliff@watson.ibm.com If you respond to this puzzle, if possible please send me your name, address, affiliation, e-mail address. If you like, tell me a little bit about yourself so I can cite you appropriately if you provide unique information. PLEASE ALSO directly mail me a copy of your response in addition to any responding you do in the newsgroup. I will assume it is OK to describe your answer in any article or publication I may write in the future, with attribution to you, unless you state otherwise. Thanks, Cliff Pickover * * * What is the next term in the Mystery Sequence: 22.45906, 17600.22, 0.34714E+12, ==> pickover/pickover.05.s <== ------------------------- Some serious roundoff error going on here, but... The sequence 22.45906, 17600.22, 0.34714E+22 is clearly meant to be: Pi^e, (Pi^e)^Pi, ((Pi^e)^Pi)^e, so the next term should be (((Pi^e)^pi)^e)^pi = 1.80169E36. Actually, it looks like you were using "pi" = 3.14159 and "e" = 2.71828, possibl with other intermediate rounding off, so you may have been looking for something more like 1.8011E36. James jlayland@grissom.jpl.nasa.gov ------------------------- In article <+M_UUYZ8!@linac.fnal.gov> you write: >cliff@watson.ibm.com (cliff) writes: >>What is the next term in the Mystery Sequence: >> >>22.45906, 17600.22, 0.34714E+12, > >I disagree about the last couple of significant digits, but otherwise >the series is pi^e, (pi^e)^pi, ((pi^e)^pi)^e, ... and the next term >is about 1.8017E+36. >_________________________________________________________ >Matt Crawford crawdad@fncent.fnal.gov Fermilab > > Background: I recognized the approximate value of the first term from figuring out (during high school, about 20 years ago) the old question of which is larger, e^pi or pi^e. After that it was a mater of taking ratios of logs of the terms. _________________________________________________________ Matt Crawford crawdad@fncent.fnal.gov Fermilab BS 1978 Applied Math & Physics; PhD 1985 Physics ------------------------- Before I go barking up a wrong tree, I notice that e Pi = 22.45916 >22.45906, 17600.22, 0.34714E+12, which seems suspiciously close to your first constant. Which should I assume? "Typo. It should read 22.45916", "Coincidence.", or "No Comment -- no clues." ??? /Alan Paeth ------------------------- In article <1992Sep17.132745.21035@watson.ibm.com> you write: >What is the next term in the Mystery Sequence: >22.45906, 17600.22, 0.34714E+12, As a one-time math major, I thought I recognized that telltale 22.45906 ... The sequence continues with 1.8016851E+36 Steve -- -- monson@diablo.amd.com -- (512) 462-4013 __ | signature designed by | One thing about kneading that pizza dough by (_ | (and ripped off from) | hand -- it sure gets your fingernails clean! __)teve | Stephen Wayne Miller | Pizzaria Friend of Danny ==> pickover/pickover.06.p <== Title: Cliff Puzzle 6: Star Chambers From: cliff@watson.ibm.com If you respond to this puzzle, if possible please send me your name, address, affiliation, e-mail address. If you like, tell me a little bit about yourself so I can cite you appropriately if you provide unique information. PLEASE ALSO directly mail me a copy of your response in addition to any responding you do in the newsgroup. I will assume it is OK to describe your answer in any article or publication I may write in the future, with attribution to you, unless you state otherwise. Thanks, Cliff Pickover * * * As many of you probably know, 5-sided stars produced by drawing a continuous line with your pencil can nest inside each other. (One star can sit inside the pentagon produced by the larger star. Each of the 5 points of the small star coincide with the 5 points of the internal pentagon of the large star.) Start with a five sided star formed with 5 line segments, each 1 inch long. Continually nest stars so that the assembly of stars gets bigger and bigger. Questions: 1. How many nestings N are required to make star N have an edge-length equal to the diameter of the sun (4.5E9 feet)? 2. How many nestings N are required to make the cumulative length of lines of all the nested stars equal to the diameter of the sun? ==> pickover/pickover.06.s <== ------------------------- Cliff Pickover, So here I am, waiting to see if one of my long Grobner basis calculations is going to finish before the machine goes down. This is a good time to read news, and I came across this trivial problem in rec.games.puzzles. I'm not sure why I'm responding, perhaps the hour, or perhaps curiousity to see what will come of this, but I could have done this the day in high school when I learned how to compute cos(pi/5). The ratio between side lengths of successive pentagrams is r = (3+sqrt(5))/2 = 1 + golden ratio = 2.618... . The smallest N for which r^N > 5.48e10 (slightly more accurate value for sun's diameter in inches) is 26, with r^26 = 7.37e10. The smallest N for which 5[r^(N+1)-1]/(r-1) > 5.48e10 is 24, with 5(r^25 - 1)/(r-1) = 8.70e10. This seems too trivial to post, but do with this response as you like. Bob Holt ------------------------- I just started reading 'rec.puzzles', so have just seen this one and the one before (#5)... and to be honest I'm not sure why you put this one out, since it is pretty straightforward. >Start with a five sided star formed with 5 line segments, each 1 inch >long. Continually nest stars so that the assembly of stars gets bigger >and bigger. The analytical (and general) answer to this problem comes from the basic relationship of a "chord" of a regular pentagon, which is defined as follows: _=*=_ _=/ / \=_ _=/ | \=_ _=/ | \=_ * / * | | <-- "chord" | \ | / | / | \ | / | / | *-------------* compared to the length of one of the sides is the golden ratio, i.e. _ 1 + \/5 --------- . 2 It can then be derived that the length of the "chord" (i.e. segment length) of the next bigger Star compared to the length of the "chord" of its incribed Star is the square of the golden ratio, or the golden ratio plus one, same thing. >Questions: >1. How many nestings N are required to make star N >have an edge-length equal to the diameter of the sun (4.5E9 feet)? Back-of-envelope calculations as follows: 4.5E9 * 12 = total of 5.22E10 inches. ratio of Star sizes approx. 2.618. The best answer is 27 nested Stars, although it produces a Star with a "chord" length of 7.366E10 inches, i.e. a bit bigger. The first, and smallest Star, is assumed to be the one with "chord" length of 1 inch. >2. How many nestings N are required to make the cumulative length >of lines of all the nested stars equal to the diameter of the sun? This is just the sum of a geometric sequence with the ratio being the golden ratio squared (or the golden ratio plus one). _ / 1 + \/5 \ 2 So, S = 1 inch, and S = S | --------- | 0 n n-1 \ 2 / The sum is just the standard geometric summation, which I can't remember offhand, but the contributing terms in the sum (other than the last term), are less than one 1.6th of the total (by conincidence the inverse of the golden ratio ;-). This means that the 25th Star (term) is the determining factor, and in this case is the answer with a total length of 8.694E10 inches amoung all of them, and 5.373E10 inches for just the sum of the segments of the 25th Star (again, counting the first one as side length of 1 inch, or sum of 5 inches). Well, that's it, I guess. Sorry to be so exhaustive, but I like to use analytical methods to be sure I have the right answer. My .signature explains most of what you need to know. What I mean by "Honorary Grad Student" is that I have been taking Grad math classes since my sophomore year, and for all intensive purposes might as well be one. My Snail-mail address is 1521 S.W. 66th Ave., Portland, OR 97225. As to info about myself... I love learning about things, and mathematics and consequently computers tend to be a great focus. Anyway, if you have any more puzzles, pass them along... I am still pondering on that sequence (puzzle #5) that you posted. Thanks for your time. Erich -- "I haven't lost my mind; I know exactly where it is." / -- Erich Stefan Boleyn -- \ --=> *Mad Genius wanna-be* <=-- { Honorary Grad. Student (Math) } Internet E-mail: <erich@gemini.mth.pdx.edu> \ Portland State University / WARNING: INTERESTED AND EXCITABLE ## User Contributions: |

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