Integer partitioning problem

Integer division, refers to a positive integer n is written in the following form:

N=m1+m2+...+mi; (where Mi is a positive integer and 1 <= mi <= n), then {m1,m2,..., mi} is a division of N.

If the maximum value of {m1,m2,..., mi} is not greater than m, that is, Max (m1,m2,..., mi) <=m, it is said to belong to a M division of N. Here we remember that the number of M division of N is f (n,m);

For example But when n=4, he had 5 divisions,

4=4,

4=3+1,

4=2+1+1,4=2+2,

4=1+1+1+1.

The problem is to find out the number of all partitions of n, i.e. f (n, N). Here we consider the method of F (n,m);

1. Recursive method:

Depending on the relationship between N and M, consider the following scenarios:

(1) When n=1, no matter what the value of M (m>0), there is only one division of {1};

(2) When m=1, no matter what the value of N, there is only one division of n 1,{1,1,1,..., 1};

(3) When n=m, according to whether the division contains N, can be divided into two situations:

(a) The case where n is included in the division, and only one is {n};

(b) Where the division does not contain n, then the largest number in the division must also be smaller than n, i.e. all (n-1) division of N (F (n,n-1)).

So f (n,n) =1 + f (n,n-1);

(4) When n<m, because there is no possibility of negative numbers in the division, so it is equivalent to F (n,n);

(5) But when n>m, it can be divided into two cases according to whether the maximum value M is included in the division:

(a) the case where M is included in the division, that is {m, {x1,x2,... XI}, where {x1,x2,... XI} and is n-m, so this case

is f (n-m,m)

(b) The division does not contain the case of M, then all the values in the division are smaller than m, i.e. N (m-1), the number is F (n,m-1);

So f (n, m) = f (n-m, M) +f (n,m-1);

Sum up:

F (n, m) = 1; (N=1, M=1)

F (n,m) = f (n, N); (n<m)

1+ f (n, m-1); (n=m)

F (n-m,m) +f (n,m-1); (n>m>1)

Java implementations:

1  Public classDivideinteger {2     3      Public Static intFintNintm) {4         if(n<1| | M&LT;1)return0;5         if(n==1| | M==1)return1;6         if(n<m)returnf (n,n);7         if(n==m)return1+f (n,n-1);8         returnF (n-m,m) +f (n,m-1);9     }Ten      One      Public Static voidMain (string[] args) { ASystem.out.println (f (6,6)); -     } -      the}

Integer partitioning problem