Robin Hankin wrote:
> Hello Amin
>
> The partitions library does this.
>
> If N=4 and k=3:
>
> > library(partitions)
> > blockparts(rep(4,3),4)
>
> [1,] 4 3 2 1 0 3 2 1 0 2 1 0 1 0 0
> [2,] 0 1 2 3 4 0 1 2 3 0 1 2 0 1 0
> [3,] 0 0 0 0 0 1 1 1 1 2 2 2 3 3 4
> >
>
> The solutions are enumerated in the columns.
>
> HTH
>
>
> rksh
>
>
Nice. You can also do it with a straightforward piece of recursive
programming:
splitsum <- function(N,k)
if (k==1) N else
do.call(rbind, lapply(0:N, function(i) cbind(i,splitsum(N-i, k-1))))
> splitsum(4,3)
i i
[1,] 0 0 4
[2,] 0 1 3
[3,] 0 2 2
[4,] 0 3 1
[5,] 0 4 0
[6,] 1 0 3
[7,] 1 1 2
[8,] 1 2 1
[9,] 1 3 0
[10,] 2 0 2
[11,] 2 1 1
[12,] 2 2 0
[13,] 3 0 1
[14,] 3 1 0
[15,] 4 0 0
Or (I think there was an old Stats 1 combinatorics exercise to a similar
effect)
> splitsum2 <- function(N,k)
t(apply(rbind(0,combn(N+k-1,k-1),N+k),2,diff)-1)
> splitsum2(4,3)
[,1] [,2] [,3]
[1,] 0 0 4
[2,] 0 1 3
[3,] 0 2 2
[4,] 0 3 1
[5,] 0 4 0
[6,] 1 0 3
[7,] 1 1 2
[8,] 1 2 1
[9,] 1 3 0
[10,] 2 0 2
[11,] 2 1 1
[12,] 2 2 0
[13,] 3 0 1
[14,] 3 1 0
[15,] 4 0 0
Exercise for the reader: Explain!
>
>
>
> On 19 Nov 2007, at 09:57, [EMAIL PROTECTED] wrote:
>
>
>> Dear all,
>> Is there any method in R to find all possible nonnegative integer
>> solutions to the linear equation with unit coefficients as follow:
>> X1+X2+...+Xk=N
>> Thank you,
>> Amin Zollanvari
>>
>>
--
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