June 2004 - mathfun@mailman.xmission.com

[math-fun] trig atrig
by R. William Gosper 23 Jun '04

23 Jun '04

Last week I had the pleasure to meet Albert Rich (of Derive fame), who introduced me to a class of simplifications including sqrt(15) 2 atan(--------) 9 sqrt(15) - sqrt(3) sin(----------------) = ------------------ . 3 8 With Macsyma I then concocted 1/3 9 - 3 4 atan(----------) 1/2 4/3 4/3 1/3 4/3 5 3 (4 + 2 - 7) - 6 4 - 2 + 10 tan(----------------) = ------------------------------------------- 3 5 It isn't even trivial to *verify* this symbolically. Along the way we exposed a bug in my MULTIPLY_CONJUGATES function wherein the imaginary unit, replaced by an explicit sqrt(-1), prematurely reverted to the symbol %I. --rwg

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[math-fun] a property of Bernoulli Numbers (4n+2)
by Simon Plouffe 22 Jun '04

22 Jun '04

hello, There is this puzzling property of the B(4n+2), I consider here only the fractional part, easily computable with the Von-Staudt-Clausen formula. Also, the B(4n+2) are all positives. Property ; the fractional part of the positive Bernoulli Numbers is almost never : 1/6 < {Bernoulli(4*n+2)} < 2/3. ... except for 2 cases (so far), which is {B(2072070)} = .666443506838622... and {B(6216210)} = .6588649656359250... Now, is this known? , my fist thought was that no Bernoulli(4n+2) has a fractional part between 1/6 and 2/3 , so far only 2 cases. I saw no references on this subject. The behaviour of {B(2*n)} when negative is well distributed mod 1 but the positive cases are NOT well distributed at all. Here is a graph of the succ. fractional part of Bernoulli's plotted on a circle, that is x in [-1,1] -> exp(2*Pi*x) a line is drafted between values x(0), x(1), ... http://www.lacim.uqam.ca/~plouffe/SuccFracBernoulli5000points.jpg Does someone has any reference on this? Simon Plouffe

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Re: [math-fun] a property of Bernoulli Numbers (4n+2)
by Simon Plouffe 22 Jun '04

22 Jun '04

Hello, I have found the local minimal point ( I can't explain why yet), Conjecture : there are no fractional part of positive Bernoulli numbers between 1/6 and 0.647247618323605748006387... That number 0.647247618323605748006387 is a local minimum which correspond to numbers of the form 207*10*1001*3^n that is {B(207*10*1001*3^n)} is minimal at 0.647247618323605748006387 and there are apparently no other values between 1/6 and that number. Why is it numbers of that form...I don't know. The big cluster of lines is at 1/6 of course since many B's have a fractional part of 1/6. I did the graph of {B(n)}, n=2,4,6,... to show how the fractional part behave for large n compared to a graph of succ. random lines, it is clearly not random at all. Simon Plouffe

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[math-fun] Seti : not in the top500 but completed exp(50) computations
by Simon Plouffe 21 Jun '04

21 Jun '04

Hello, the top-500 computers in the world list just came out, An interesting article about the top 500 is found here : http://www.newscientist.com/news/news.jsp?id=ns99996045 (partly by Dongarra), he talks about the fact that the Seti project is not in the top500 because of its definition. nevertheless, SETI has reached almost exactly exp(50) elementary computations that is ; 5.176068e+21 at a rate of 24.35 tera-flops (per second) which is not far from the earth similator, 35 tera-flops. http://setiathome.ssl.berkeley.edu/totals.html Simon Plouffe

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Re: [math-fun] SOLUTION to Find solution in Gaussian intgers (SPOILER)
by Daniel Asimov 20 Jun '04

20 Jun '04

My apologies to Michael Reid, who posted in Jan., 2002 an equivalent solution to the Gaussian integer problem I just posted (but his post contained his proof of the result). I didn't see it at the time, and thought I was posting a new problem to Math-Fun. --Dan

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[math-fun] Find solution in Gaussian integers
by Daniel Asimov 18 Jun '04

18 Jun '04

Problem: Find all Gaussian integers u,v,w such that u + v + w = uvw. Dan

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[math-fun] Erdos number for sale -- a postscript
by Richard Schroeppel 16 Jun '04

16 Jun '04

Science News has an article about the Ebay auction of an Erdos number (as an opportunity to collaborate with the offeror on a joint paper). http://sciencenews.org/articles/20040612/bob8.asp The auction was won with a last minute bid of $1031, by a person who wanted to spoil the auction. The offeror is trying to set up more collaborations between amateur mathematicians. Rich rcs(a)cs.arizona.edu

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[math-fun] Re: number of longest common substrings problem
by N. J. A. Sloane 14 Jun '04

14 Jun '04

Hugo asks: > Here you introduce the word "different" for the first time. Are two > substrings "different" if the substrings themselves differ, or is it > sufficient for the letters of the substring to be drawn from different > positions in X or Y? Ie given: > X = "aabaa", Y = "aaabb" > are there 1 or 4 occurrences of the LCS "aaa"? Me: The latter, I think, since the goal of the question is to determine how rapidly these numbers can grow. So in your example, the LCS has length 3, and there seem to be 10 ways to get 3: X = aabaa Y = aaabb --------------------- aab.. aa.b. aab.. aa..b aab.. a.ab. aab.. a.a.b aab.. .aab. aab.. .aa.b aa.a. aaa.. aa..a aaa.. a..aa aaa.. .a.aa aaa.. so a(5) >= 10. NJAS

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[math-fun] Longest common subsequences: summary so far
by N. J. A. Sloane 14 Jun '04

14 Jun '04

Longest common subsequences: summary so far The following is an edited record of the correspondence on this problem so far. NJAS Jun 14 2004 *************************************************************************************** >From njas(a)research.att.com Fri Jun 11 23:40:54 2004 Date: Fri, 11 Jun 2004 23:40:53 -0400 (EDT) From: "N. J. A. Sloane" <njas(a)research.att.com> Message-Id: <200406120340.XAA13275(a)fry.research.att.com> To: math-fun(a)mailman.xmission.com, seqfan(a)ext.jussieu.fr Subject: number of longest common subsequences problem I ran into my old friend Al Aho today for the first time in many years, and he mentioned the following possibly unsolved problem. Given two strings X and Y of length n from some fixed alphabet, let f(X,Y) = number of longest common subsequences of X and Y. What is a(n) = max f(X,Y) over all choices of X and Y? Remarks: S is a subsequence of X if S can be obtained by deleting some of the entries of X (not necessarily adjacent). You can choose any alphabet you like. Here is a definition of "longest common subsequence" from http://www.cs.sunysb.edu/~algorith/files/longest-common-substring.shtml (see also http://www.nist.gov/dads/HTML/longestCommonSubstring.html) Given two strings X and Y, what is the longest sequence S such that the characters of S appear as a scattered subsequence of both X and Y? Then the question is, how many different longest common subsequences can occur? There are two versions of the problem, depending on whether we take into account the location of S inside X and Y (Version 1) or not (Version 2). NJAS *************************************************************************************** >From hv(a)crypt.org Sat Jun 12 10:25:07 2004 Cc: math-fun(a)mailman.xmission.com, seqfan(a)ext.jussieu.fr Subject: Re: number of longest common subsequences problem Here you introduce the word "different" for the first time. Are two subsequences "different" if the subsequences themselves differ, or is it sufficient for the letters of the subsequence to be drawn from different positions in X or Y? Ie given: X = "aabaa", Y = "aaabb" are there 1 or 4 occurrences of the LCS "aaa"? Hugo van der Sanden My reply: There are 2 versions of the problem. In Version 1 the answer would be 4, in Version 2 the answer would be 1. For Version 1: In your example, the LCS has length 3, and there seem to be 10 ways to get 3: X = aabaa Y = aaabb --------------------- aab.. aa.b. aab.. aa..b aab.. a.ab. aab.. a.a.b aab.. .aab. aab.. .aa.b aa.a. aaa.. aa..a aaa.. a..aa aaa.. .a.aa aaa.. so a(5) >= 10. NJAS *************************************************************************************** >From Sterten(a)aol.com Sat Jun 12 11:15:42 2004 Date: Sat, 12 Jun 2004 11:15:06 EDT Subject: re: number of longest common subsequences problem let m be the size of the alphabet. I get : (n=1,2,...) m=2:1,2,2,4,4,6,8,11,15,20, m=3:1,2,3,4,6,10, m=4:1,2,3,4,7, m=5:1,2,3,4,7, I assume that the maximum is usually attained for X=123..m123..m... , except for some small values of n. fixing X this way gives: m=2:1,2,2,3,4,6,7,10,14,20,26,36,51,70,96,141, m=3:1,2,3,4,6,10,14,24,36,58, m=4:1,2,3,4,7,10,19,28,51, m=5:1,2,3,4,6,10,15,28, m=6:1,2,3,4,6,9,14, with hillclimbing my best values were: m=2:1,2,2,4,4,6,8,11,15,20,26,36,50,70,96,141,192, m=3:1,2,3,4,6,10,14,24,36,58,88,145, m=4:1,2,3,4,7,10,19,28,51,78,141,220, m=5:1,2,3,4,7,10,19,28,44,69,112,186, my lower bounds for m=5 are obviously bad, or there could even be a bug in my program. I didn't find the above sequences in the OEIS Guenter Me: I added some of them: %I A094858 %S A094858 1,2,2,4,4,6,8,11,15,20 %N A094858 Maximal number of longest common subsequences between any two binary strings of length n (Version 2). %C A094858 Definitions: S is a subsequence of X if S can be obtained by deleting some (not necessarily adjacent) entries of X. %C A094858 S is a longest common subsequence of X and Y if S is a subsequence of X, S is a subsequence of Y, and for any T, if T is a subsequence of X and of Y, then |T| <= |S|. Let LCS(X,Y) = length of any longest common subsequence of X and Y. %C A094858 For each common subsequence S of X and Y, there may be several ways to delete entries from X and from Y to get S, but in this version of the problem we do not take this into account (cf. A094837). Let F(X,Y) be the number of different choices for S, without regard to where it appears in X and Y. Sequence gives max F(X,Y) over all choices for binary strings X and Y of length n. %C A094858 For this version of the problem using a larger alphabet helps (cf. A094859, A094863). %C A094858 For an alphabet of size m, the maximum appears to be attained for X=123..m123..m..., except for some small values of n. %Y A094858 Cf. A094859-A094862, A094837, A094824. %O A094858 1,2 %K A094858 nonn,nice,more %A A094858 Guenter Stertenbrink (Sterten(AT)aol.com) Jun 14 2004 %C A094858 Hill-climbing gives the following lower bounds for the next few terms: 26,36,50,70,96,141,192. %I A094859 %S A094859 1,2,3,4,6,10 %N A094858 Maximal number of longest common subsequences between any two ternary strings of length n (Version 2). %C A094858 Same as A094858 but now we can use an alphabet of size 3. %O A094859 1,2 %K A094859 nonn,more %A A094859 Guenter Stertenbrink (Sterten(AT)aol.com) Jun 14 2004 %I A094860 %S A094860 1,2,2,3,4,6,7,10,14,20,26,36,51,70,96,141 %N A094860 Same as A094858, except that we fix X = 121212...12(1). %O A094860 1,2 %K A094860 nonn,more %A A094860 Guenter Stertenbrink (Sterten(AT)aol.com) Jun 14 2004 %I A094861 %S A094861 1,2,3,4,6,10,14,24,36,58 %N A094861 Same as A094858, except that we fix X = 123123123... %O A094861 1,2 %K A094861 nonn,more %A A094861 Guenter Stertenbrink (Sterten(AT)aol.com) Jun 14 2004 %I A094862 %S A094862 1,2,3,4,7,10,19,28,51 %N A094862 Same as A094858, except that we fix X = 123412341234... %O A094862 1,2 %K A094862 nonn,more %A A094862 Guenter Stertenbrink (Sterten(AT)aol.com) Jun 14 2004 %I A094863 %S A094863 1,2,3,4,7 %N A094863 Maximal number of longest common subsequences between any two strings of length n (Version 2). %C A094863 Same as A094858 (which has much more information about the problem), except that we now we allow an arbitrary alphabet. %O A094863 1,2 %K A094863 nonn,more,nice,hard %A A094863 Guenter Stertenbrink (Sterten(AT)aol.com) Jun 14 2004 *************************************************************************************** >From russcox(a)gmail.com Sat Jun 12 12:46:49 2004 Date: Sat, 12 Jun 2004 12:46:17 -0400 From: Russ Cox <russcox(a)gmail.com> To: njas(a)research.att.com Subject: Re: [math-fun] number of longest common subsequences problem This is for Version 1: Is it obvious that the best answer is always going to use a two-character alphabet? It seems to be true, empirically. My search program would be a lot faster if it only checked the two-character alphabet. I think the C program below computes the initial terms of this sequence, for a two-character alphabet or an arbitrary alphabet. For a two-character alphabet: len 1: 1 lcs of length 1 for a a len 2: 2 lcs of length 1 for aa ab len 3: 4 lcs of length 2 for aab abb len 4: 10 lcs of length 2 for abba baab len 5: 24 lcs of length 2 for abbba baaab len 6: 46 lcs of length 3 for aabbba abaaab len 7: 100 lcs of length 4 for aaaaabb aabbbbb len 8: 225 lcs of length 4 for aaaaaabb aabbbbbb len 9: 525 lcs of length 5 for aaaaaaabb aaabbbbbb len 10: 1225 lcs of length 6 for aaaaaaabbb aaabbbbbbb len 11: 3136 lcs of length 6 for aaaaaaaabbb aaabbbbbbbb len 12: 7056 lcs of length 7 for aaaaaaaaabbb aaaabbbbbbbb The search finds the alphabetically earliest pair of strings that exhibit the maximum. It's interesting that len 4, 5, 6 don't follow the same string pattern as all the rest. Over alphabets of arbitrary size, the results match the above through len 6. Beyond that, the search takes longer than I am patient. The sequence 1,2,4,10,24,46 does not appear in the OEIS. I'd be more comfortable submitting it if the terms were corroborated by someone else. Russ #include <stdio.h> #include <stdlib.h> #define MAXN 20 #define debug 1 typedef struct Match Match; struct Match { int len; int count; }; /* * Compute number of lcs for a, b using dynamic programming. * The lcs for the substrings a[0:i] b[0:j] is one of * * if a[i]==b[j], lcs(a[0:i-1], b[0:j-1])+a[i] * lcs(a[0:i-1], b[0:j]) * lcs(a[0:i], b[0:j-1]) * * We can compute the number of lcs by just adding up the * number from each place, except that if we take them from * lcs(a[0:i-1], b[0:j]) and lcs(a[0:i], b[0:j-1]) and those in turn * didn't actually add any new ones, then we'll double-count * the ones from lcs(a[0:i-1], b[0:j-1]), so we need to subtract * those out. */ int numlcs(char *a, char *b, int n, int *plen) { int i, j, d, t, len, count; Match match[MAXN][MAXN]; for(d=0; d<2*n-1; d++){ i = 0; j = d-i; if(j >= n){ j = n-1; i = d-j; } for(; i<n && j>=0; i++, j--){ /* start with nothing */ len = 0; count = 0; /* try adding to a previous match */ if(a[i] == b[j]){ if(i>0 && j>0){ len = match[i-1][j-1].len+1; count = match[i-1][j-1].count; }else{ len = 1; count = 1; } } /* try matches not containing a[i] */ if(i>0){ if((t=match[i-1][j].len) > len){ len = t; count = match[i-1][j].count; }else if(t == len) count += match[i-1][j].count; } /* try matches not containing b[j] */ if(j>0){ if((t=match[i][j-1].len) > len){ len = t; count = match[i][j-1].count; }else if(t == len) count += match[i][j-1].count; } /* we may have double-counted matches not containing either */ if(i>0 && j>0 && match[i-1][j-1].len==len) count -= match[i-1][j-1].count; match[i][j].len = len; match[i][j].count = count; } } if(plen) *plen = match[n-1][n-1].len; return match[n-1][n-1].count; } /* * Enumerate all pairs of strings of length n over an alphabet of size nalpha * and run numlcs on them. buf contains the concatenation of the two strings; * only nbuf of the characters have been filled in so far. The nbuf characters * only use the first nused characters in the alphabet, so we need only * consider the first nused+1 alphabet characters as next possible. * * Accumulate the current max in bestcount, bestlen, besta, bestb. */ int bestcount; int bestlen; char besta[MAXN+1]; char bestb[MAXN+1]; char *alpha = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"; void search(int nalpha, int nused, int n, char *buf, int nbuf) { int count, len; char *p, *ep; if(nbuf == 2*n){ count = numlcs(buf, buf+n, n, &len); if(count > bestcount){ bestcount = count; bestlen = len; memmove(besta, buf, n); memmove(bestb, buf+n, n); } }else{ ep = alpha+nalpha; if(nused < nalpha) ep = alpha+nused+1; for(p=alpha; p<ep; p++){ buf[nbuf] = *p; search(nalpha, (p<alpha+nused ? nused : nused+1), n, buf, nbuf+1); } } } void usage(void) { fprintf(stderr, "usage: lcs count aabaa aaabb\n"); fprintf(stderr, " or lcs search n\n"); fprintf(stderr, " or lcs search2 n\n"); exit(1); } int main(int argc, char **argv) { int i, n, len; char buf[2*MAXN]; if(argc < 2) usage(); if(strcmp(argv[1], "count") == 0){ if(argc != 4) usage(); n = strlen(argv[2]); if(n != strlen(argv[3])){ fprintf(stderr, "string lengths must match\n"); exit(1); } n = numlcs(argv[2], argv[3], n, &len); printf("%d of length %d\n", n, len); }else if(strcmp(argv[1], "search") == 0){ if(argc != 3) usage(); n = atoi(argv[2]); search(2*n, 0, n, buf, 0); printf("len %d: %d lcs of length %d for %s %s (all alphabets)\n", n, bestcount, bestlen,besta, bestb); }else if(strcmp(argv[1], "search2") == 0){ if(argc != 3) usage(); n = atoi(argv[2]); search(2, 0, n, buf, 0); printf("len %d: %d lcs of length %d for %s %s (just alphabet=ab)\n", n, bestcount, bestlen, besta, bestb); }else usage(); return 0; } Me: I added these two sequences: %I A094837 %S A094837 1,2,4,10,24,46,100,225,525,1225,3136,7056 %N A094837 Maximal number of longest common subsequences between any two binary strings of length n (Version 1). %C A094837 Definitions: S is a subsequence of X if S can be obtained by deleting some (not necessarily adjacent) entries of X. %C A094837 S is a longest common subsequence of X and Y if S is a subsequence of X, S is a subsequence of Y, and for any T, if T is a subsequence of X and of Y, then |T| <= |S|. Let LCS(X,Y) = length of any longest common subsequence of X and Y. %C A094837 For each common subsequence S of X and Y, there may be several ways to delete entries from X and from Y to get S: let F(X,Y) be the total number of ways. Sequence gives max F(X,Y) over all choices for binary strings X and Y of length n. %C A094837 It appears that using a larger alphabet does not increase the answers: is this true? %O A094837 1,2 %K A094837 nonn,nice,more %Y A094837 A094838 gives length of S. See A094824 for a related problem involving substrings. %A A094837 Russ Cox (russcox(AT)gmail.com) Jun 13 2004 %e A094837 Example illustrating a(4) = 10: %e A094837 abba baab S %e A094837 ------------ %e A094837 a..a .aa. aa %e A094837 ab.. .a.b ab %e A094837 ab.. ..ab ab %e A094837 a.b. .a.b ab %e A094837 a.b. ..ab ab %e A094837 .bb. b..b bb %e A094837 .b.a ba.. ba %e A094837 .b.a b.a. ba %e A094837 ..ba ba.. ba %e A094837 ..ba b.a. ba %e A094837 Details for lengths 1 through 12 showing lexicographically earliest examples for X and Y: %e A094837 len 1: 1 lcs of length 1 for a a %e A094837 len 2: 2 lcs of length 1 for aa ab %e A094837 len 3: 4 lcs of length 2 for aab abb %e A094837 len 4: 10 lcs of length 2 for abba baab %e A094837 len 5: 24 lcs of length 2 for abbba baaab %e A094837 len 6: 46 lcs of length 3 for aabbba abaaab %e A094837 len 7: 100 lcs of length 4 for aaaaabb aabbbbb %e A094837 len 8: 225 lcs of length 4 for aaaaaabb aabbbbbb %e A094837 len 9: 525 lcs of length 5 for aaaaaaabb aaabbbbbb %e A094837 len 10: 1225 lcs of length 6 for aaaaaaabbb aaabbbbbbb %e A094837 len 11: 3136 lcs of length 6 for aaaaaaaabbb aaabbbbbbbb %e A094837 len 12: 7056 lcs of length 7 for aaaaaaaaabbb aaaabbbbbbbb %I A094838 %S A094838 1,1,2,2,2,3,4,4,5,6,6,7 %N A094838 Length of longest common subsequence corresponding to answers in A094837. %O A094838 1,3 %K A094838 nonn,more %A A094838 Russ Cox (russcox(AT)gmail.com) Jun 13 2004 *************************************************************************************** Date: Sat, 12 Jun 2004 13:00:19 -0400 To: math-fun <math-fun(a)mailman.xmission.com> From: "Allan C. Wechsler" <acw(a)alum.mit.edu> Subject: [math-fun] Most repeated longest subsequence I'm having a little bit of trouble understanding the problem. It's a matter of how the quantifiers are nested, and how the counting is to be done. I would greatly appreciate it if Neil would go back to Al Aho and get a really clear description of the problem. My difficulties begin with the definition of f(X,Y). It is not clear that we are permitted to count multiple occurrences of the same subsequence; it seems equally probable to me that Aho wants to count only subsequences with distinct content. I have other similar hairsplitting problems, but let's start with that one. And ... remember the Al Aho. --ACW Me: I hope the edited messages above have now answered these questions. *************************************************************************************** From: "Thane Plambeck" <thane(a)best.com> Subject: Re: [math-fun] subsequence, substrings, np completeness 1) Just a nit, but usually a string obtained by possibly non adjacent deletions is called a "subsequence," not a "substring." Me: Yes, sorry, I have tried to correct that in this edited file. 2) I'll contribute also the following NP-complete problem reference from Garey and Johnson's book (pg 228 in my copy): [SR10] Longest Common Subsequence Instance: Finite alpha \Sigma, finite set R of strings from \Sigma^{*} and a positive integer K. Question: Is there a string w with |w| \geq K such that w is a subsequence of each x \in R? Reference: Transformation from vertex cover [maier, 1978]. Remains NP complete even if |\Sigma| = 2. Solvable in polynomial time for any fixed K or for fixed |R| (by dynamic programming). The analogous longest common substring problem is triviallly solvable in polynomial time Thane Plambeck 650 321 4884 office 650 323 4928 fax http://www.plambeck.org *************************************************************************************** Date: Sat, 12 Jun 2004 16:28:37 -0400 (EDT) From: Edwin Clark <eclark(a)math.usf.edu> Subject: Re: number of longest common subsequences problem There seem to be many different ways to interpret this problem. In particular internet seaches indicate there is a difference between common substrings and common subsequences. From Neil's description it seems he is referring to what some call common "subsequences". Maple, for example, in its StringTools package has the two procedures LongestCommonSubString(s1,s2) and LongestCommonSubSequence(s1,s2) where s1 and s2 are strings. These have the following descriptions (which seem to agree with other definitions found by googling around.) -------------------------------------------------------------------- A substring of a string S is a contiguous sequence of the characters appearing in S. The empty string is a substring of every string. A subsequence of a string S is a sequence of characters from S, which may not be contiguous in S. Every substring of S is a subsequence of S. For example, "bc" is a substring of "abc", and "ac" is a subsequence of "abc" which is not a substring. The LongestCommonSubString(s1, s2) command returns from its input strings, s1 and s2, a common substring of maximum length. Many common substrings of maximum length may exist. Which among the candidates is returned depends upon the suffix structure of the pair of strings, but is deterministic. The LongestCommonSubSequence(s1, s2) command is similar, but searches for subsequences of the pair of input strings rather than substrings. -------------------------------------------------------------------- Once we know what a longest common subsequence or substring is, there is the problem of how to count them: with multiplicities or not? Anyhow, I did one of the simplest interpretations, presumably NOT one that Aho intended, nevertheless perhaps new and of interest: I'm counting (contiguous) substrings and not counting multiplicity: What I did was to find the set CS(s1,s2) of common substrings (in the above sense) for binary strings s1 and s2 of length n and count the number A(s1,s2) of strings in CS(s1,s2) that are the longest. Then a(n) is the max of A(s1,s2) over all such strings. If I haven't made an error this gives the sequence a(n), n = 1,...,11: 1, 2, 2, 2, 3, 3, 4, 6, 6, 7, 7 For example the fact that a(7) = 4 is given by the two strings s = 0001011,t = 0011010 for which the set of maximum length common substrings is {011,001,101,010} and the fact that no other pair of strings of length 7 has more than 4 common maximum length substrings. Apparently this sequence is not is the OEIS. But it seems simple enough that maybe someone can find a formula for it. And maybe someone using something faster than Maple can compute more terms even if no one can figure out a formula. --Edwin Here is the new sequence: %I A094824 %S A094824 1,2,2,2,3,3,4,6,6,7,7,8 %N A094824 Maximum number of longest common substrings of two binary sequences of length n. %C A094824 A substring of a string is a subsequence of contiguous symbols in the string. For example, 00 is a substring of 001 but not of 010. For this sequence we do not count the multiplicity of occurrence of common substrings. %e A094824 a(7) = 4 since the two strings 0001011 and 0011010 have as maximum length common substrings the 4 strings 011,001,101,010, and computer search shows that no other pair of strings of length 7 has more than 4 common maximum length substrings. %Y A094824 See A094837 for a related sequence. %K A094824 hard,more,nonn,nice,new %O A094824 1,2 %A A094824 W. Edwin Clark (eclark(AT)math.usf.edu) Jun 12 2004 %Y A094837 A094838 gives length of S. See A094824 for a related problem involving substrings. %Y A094858 Cf. A094859-A094862, A094837, A094824. *************************************************************************************** >From njas(a)research.att.com Sat Jun 12 17:17:45 2004 To: seqfan(a)ext.jussieu.fr, math-fun(a)mailman.xmission.com Subject: Re: longest common subsequence - clarification of problem Al Aho tells me that the problem he is interested in is Version 2: find the maximum number of longest common subsequences of two sequences X and Y of length n, without regard to where the subsequence occurs in X and Y. So for X = aabaa, Y = aaabb, the length of the lcs is 3, and there are two subsequences of length 3 that are possible, aaa and aab. So here F(X,Y) = 2. Some definitions: S is a subsequence of X if S can be obtained by deleting some of the entries of X (not necessarily adjacent). S is a longest common subsequence of X and Y if S is a subsequence of X, S is a subsequence of Y, and for any T, if T is a subsequence of X and of Y, then |T| <= |S|. (Thanks to Michael Kleber for a correction here) Given sequences X and Y of length n, let LCS(X,Y) be the length of the longest common subsequence. Let F(X,Y) = no of sequences S such that S is a longest common subsequence of X and Y. So S has length LCS(X,Y). Given n, what is the maximal value of F(X,Y) over all choices of X and Y over any alphabet? That's Version 2. The version of the problem that i stated earlier, Version 1, where you also take into account the location of S inside X and Y, is also interesting, it is just not Al's original question. NJAS *************************************************************************************** >From wouter.meeussen(a)pandora.be Sat Jun 12 18:25:25 2004 Subject: Re: [math-fun] Re: longest common subsequence - clarification of problem Date: Sun, 13 Jun 2004 00:25:01 +0200 my earlier program stopped too soon, better: LongestCommonSubstring[keep : {{__} ..}] := Module[{done = False, uit = {}, it = keep, alfa = {1}, z}, While[Intersection @@ it =!= {} && ! done, z = 1; While[ ! (done = ( {0} === (alfa ))) && (alfa = Intersection @@ Take[it, All, {1, z}]) === {}, z++]; If[done, Break[]];(* Print[{z, alfa[[1]]}] *); AppendTo[uit, temp = Flatten[ Thread[w[Take[it, All, {1, z}], alfa[[1]], 1, 1]] /. w -> Position]]; it = Thread[rop[it, temp]] /. rop -> Drop ; it = PadRight[#, Max[Length /@ keep]] & /@ it;(* Print[it] *)]; Drop[Rest@FoldList[Plus, 0, uit], -1]] other lines as before: keep = Table[Random[Integer, {1, 4}], {8}, {36}] soln = LongestCommonSubstring[keep] MapAt[w, keep, Flatten[Transpose[{Range[Length[#]], #}] & /@ soln, 1]] /. w[i_] :> (j = (i - 1)/4.; StyleForm[i, FontColor -> Hue[j, 1, 1], FontWeight -> "Bold"] ) Length @ soln First[keep][[First /@ soln]] W. *************************************************************************************** >From rschroe(a)sandia.gov Sun Jun 13 22:01:17 2004 From: "Schroeppel, Richard" <rschroe(a)sandia.gov> To: "'Russ Cox '" <russcox(a)gmail.com>, "'njas(a)research.att.com '" <njas(a)research.att.com> Cc: "Schroeppel, Richard" <rschroe(a)sandia.gov>, "'math-fun '" <math-fun(a)mailman.xmission.com>, "'seqfan(a)ext.jussieu.fr '" <seqfan(a)ext.jussieu.fr> Subject: RE: [math-fun] number of longest common substrings problem Curious that so many of Russ's counts are squares: 1,-,4,-,-,-,100,225,-,1225,3136,7056 Rich *************************************************************************************** >From hv(a)crypt.org Mon Jun 14 07:46:09 2004 Cc: seqfan(a)ext.jussieu.fr, math-fun(a)mailman.xmission.com Subject: Re: number of longest common substrings problem "N. J. A. Sloane" <njas(a)research.att.com> wrote: :Hugo asks: : :> Here you introduce the word "different" for the first time. Are two :> substrings "different" if the substrings themselves differ, or is it :> sufficient for the letters of the substring to be drawn from different :> positions in X or Y? Ie given: :> X = "aabaa", Y = "aaabb" :> are there 1 or 4 occurrences of the LCS "aaa"? : :Me: The latter, I think, since the goal of the question is to :determine how rapidly these numbers can grow. In that case, here are some initial thoughts: - a(n) has a lower bound of A001405 (C(n, floor(n/2))), since we can always construct an LH string of all 'a's, and a RH string of floor(n/2) a's followed by ceil(n/2) b's. - a couple of hand-crafted examples of a(n) exceeding that lower bound: (aab, abb) => a(3) >= 4 > C(3,1) (abbbc, aabcc) => a(5) >= 12 > C(5,2) (aaabb, abbbb) => a(5) >= 18 (abbbbc, aabccc) => a(6) >= 24 > C(6,3) (abbbbc, aabbcc) => a(6) >= 24 (aaaaab, aabbbb) => a(6) >= 40 I think this serves to demonstrate the style of string pairs that will maximise the count; I'm not yet sure whether it is ever beneficial to have non-monotonic strings such as "abababab"; if not, this essentially boils down to finding <x> and <y>, two ordered partitions of n such that prod(C(max(x_i,y_i), min(x_i,y_i))) is maximised; that product would then be a(n). If I've calculated correctly, that gives a sequence starting: 1 2 4 9 18 40 100 225 525 ... .. but I don't see that currently in EIS: have I miscalculated? Those figures can be generated by the following strings: n a(n)? strings 1 1 a a 2 2 aa ab 3 4 aab abb 4 9 aaab abbb 5 18 aaaab aabbb 6 40 aaaaab aabbbb 7 100 aaaaabb aabbbbb 8 225 aaaaaabb aabbbbbb 9 525 aaaaaaabb aaabbbbbb By eye, it appears that the maximum product is always generated by two-element partitions, in which case it further simplifies to selecting 0 <= x <= y <= n such that C(y, x) * C(n-x, n-y) is maximised, which lets me quickly extend the above sequence: 1 2 4 9 18 40 100 225 525 1225 3136 7056 17640 44100 108900 261360 637065 1656369 4008004 10020010 25050025 64128064 155739584 393853824 1012766976 2538950544 6347376360 15868440900 41408180100 102252852900 261312846300 667799496100 1709566710016 4273916775040 10851741811625 28214528710225 71170904601225 181162302621300 461140406672400 This sequence exhibits some fascinating patterns: if we look at the factorisation of these as a vector of powers of p_n, we get this: 1: (0) 2: 1 (1) 3: 2 (2) 4: 0 2 (2) 5: 1 2 (3) 6: 3 0 1 (4) # ie f(6) = 40 = 2^3 * 3^0 * 5^1 (3 + 0 + 1 = 4) 7: 2 0 2 (4) 8: 0 2 2 (4) 9: 0 1 2 1 (4) 10: 0 0 2 2 (4) 11: 6 0 0 2 (8) 12: 4 2 0 2 (8) 13: 3 2 1 2 (8) 14: 2 2 2 2 (8) 15: 2 2 2 0 2 (8) 16: 4 3 1 0 2 (10) 17: 0 4 1 0 2 1 (8) 18: 0 4 0 0 2 2 (8) 19: 2 0 0 2 2 2 (8) 20: 1 0 1 2 2 2 (8) 21: 0 0 2 2 2 2 (8) 22: 6 0 0 2 2 2 (12) 23: 6 0 0 1 2 2 1 (12) 24: 7 2 0 1 0 2 2 (14) 25: 8 4 0 0 0 2 2 (16) 26: 4 2 0 0 0 2 2 2 (12) 27: 3 2 1 0 0 2 2 2 (12) 28: 2 2 2 0 0 2 2 2 (12) 29: 2 4 2 2 0 0 2 2 (14) 30: 2 4 2 0 2 0 2 2 (14) 31: 2 2 2 0 2 0 2 2 1 (13) 32: 2 0 2 0 2 0 2 2 2 (12) 33: 8 0 0 0 2 0 2 2 2 (16) 34: 7 0 1 0 2 0 2 2 2 (16) 35: 0 0 3 0 2 1 2 2 2 (12) 36: 0 0 2 0 2 2 2 2 2 (12) 37: 0 6 2 0 2 2 0 2 2 (16) 38: 2 6 2 1 1 2 0 2 2 (18) 39: 4 6 2 2 0 2 0 2 2 (20) .. and this could generate further interesting new sequences (eg f(n) is odd; f(n) is not square (do the first differences of this form a cyclce of period 7?); f(n) has an odd number of prime factors). In case others find it useful, I've put up a perl program to calculate the LCS count for two strings at <http://www.crypt.org/hv/puzzle/lcsc>. (It requires the additional perl module Algorithm::Loops, found at <http://search.cpan.org/~tyemq/Algorithm-Loops>.) This program is fast for strings up to a certain complexity, but beyond that consumes large amounts of memory as the number of different subsequences undergoes a combinatorial explosion. Hugo ME: I DID NOT ADD THE ABOVE SEQUENCE YET, because I don't understand why it differs from Russ Cox's sequence A094837 (see earlier in this message), which has a(4) = 10 not 9. Is Hugo's sequence wrong, or is it the answer to a different question? NJAS *************************************************************************************** (end of message)

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RE: [math-fun] number of longest common substrings problem
by Schroeppel, Richard 13 Jun '04

13 Jun '04

Curious that so many of Russ's counts are squares: 1,-,4,-,-,-,100,225,-,1225,3136,7056 Rich -----Original Message----- From: Russ Cox To: njas(a)research.att.com Cc: math-fun; seqfan(a)ext.jussieu.fr Sent: 6/12/2004 10:46 AM Subject: Re: [math-fun] number of longest common substrings problem I think the C program below computes the initial terms of this sequence, for a two-character alphabet or an arbitrary alphabet. For a two-character alphabet: len 1: 1 lcs of length 1 for a a len 2: 2 lcs of length 1 for aa ab len 3: 4 lcs of length 2 for aab abb len 4: 10 lcs of length 2 for abba baab len 5: 24 lcs of length 2 for abbba baaab len 6: 46 lcs of length 3 for aabbba abaaab len 7: 100 lcs of length 4 for aaaaabb aabbbbb len 8: 225 lcs of length 4 for aaaaaabb aabbbbbb len 9: 525 lcs of length 5 for aaaaaaabb aaabbbbbb len 10: 1225 lcs of length 6 for aaaaaaabbb aaabbbbbbb len 11: 3136 lcs of length 6 for aaaaaaaabbb aaabbbbbbbb len 12: 7056 lcs of length 7 for aaaaaaaaabbb aaaabbbbbbbb <snip>

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