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Hello! I have an interesting problem that seemed simple to me, but I'm unable to solve it on my own.

Suppose I am drawing k numbers out of n numbers labeled from 1 to n. Considering all $\binom{n}{k}$ combinations of numbers drawn, how often does the maximal difference q between two consecutive numbers – but also between zero and the lowest number, or the highest number and n – occur.

I already found an algorithm to compute the sequence, but it's too computationally intense for large n, so I'm looked for an explicit formula. If it's too complicated to find a formula a distribution would also be fine.

I think someone must already have worked on this problem, but I can't find anything.

The resulting series is:
(Read: n,m:: q:number of combinations with q as maximal difference)
2,1:: 1:2
3,1:: 1:1, 2:2
3,2:: 1:3
4,1:: 1:0, 2:2, 3:2
4,2:: 1:3, 2:3
4,3:: 1:4
5,1:: 2:1, 3:2, 4: 2
...
10,1:: 8:2, 9:2, 5:2, 6:2, 7:2
10,2:: 3:3, 4:12, 5:12, 6:9, 7:6, 8:3
10,3:: 2:4, 3:36, 4:40, 5:24, 6:12, 7:4
10,4:: 2:45, 3:90, 4:50, 5:20, 6:5
10,5:: 1:6, 2:120, 3:90, 4:30, 5:6
10,6:: 1:35, 2:126, 3:42, 4:7
10,7:: 1:56, 2:56, 3:8
10,8:: 1:36, 2:9
10,9:: 1:10

Where the problem arose: I'm doing a masters thesis in bioinformatics on a quick clustering algorithm. So I'm looking for shared q-grams (substrings with length q) of pairs of sequences with the length n that differ in at most m sites. I want to find the biggest q possible so that 99% of all sequences of length n with m randomly distributed differences share a substring of length q.

Illustration of the problem: (n=10, m=3; X for mismatch, "." for match)

X....X...X -> 4
..X..X..X. -> 2
.......XXX -> 7

Here's the python script:

import itertools
def f(n,m):
    rdict={}
    for d in itertools.combinations(range(n),m):
        t=[-1]+list(d)+[n]
        m=max([x[0]-x[1] for x in zip(t[1:],t[:-1])])-1
        if not rdict.has_key(m):
            rdict[m]=1
        else:
            rdict[m]+=1
    return rdict
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When you say "or the highest number and $n$," do you mean $n+1$? –  Douglas Zare Jun 19 '12 at 12:25

1 Answer 1

up vote 2 down vote accepted

Choices of $k$ out of $n$ correspond to ordered $k+1$-tuples of nonnegative numbers which add up to $n-k$ by counting the dots between the Xs.

The number of such $k+1$-tuples so that $a$ particular terms are at least $q$, with no restrictions on the others, is $n-aq \choose k$ [edit: when $n-aq \ge 0$, and $0$ otherwise], since subtracting $q$ from each of the terms we know are at least $q$ gives an unrestricted $k+1$-tuple adding to $n-aq$. So, the technique of inclusion-exclusion lets us count $f(n,k,q)$, the number of $k+1$-tuples with no term which is at least $q$:

$$ f(n,k,q) = \sum_{a=0}^{k+1} (-1)^a {k+1\choose a}{n-aq\choose k}.$$

To count sequences where the maximum is exactly $q$, take $f(n,k,q+1)-f(n,k,q)$.


Edit: The above formula is incorrect because it includes the terms where $n-aq$ is negative. For these terms, replace $n-aq \choose k$ with $0$, or change the upper limit of the sum:

$$ f(n,k,q) = \sum_{a=0}^{\lfloor n/q \rfloor} (-1)^a {k+1\choose a}{n-aq\choose k}.$$

share|improve this answer
    
If I implement your f, it always sums up to 0 –  Christoph Jun 21 '12 at 8:39
    
Oops, I should not have included the terms where $n-aq$ is negative. –  Douglas Zare Jun 21 '12 at 9:49

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