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Re: Re: random matrix from row and column sums
*To*: mathgroup at smc.vnet.net
*Subject*: [mg53609] Re: [mg53598] Re: random matrix from row and column sums
*From*: DrBob <drbob at bigfoot.com>
*Date*: Thu, 20 Jan 2005 03:47:50 -0500 (EST)
*References*: <csinpt$njk$1@smc.vnet.net> <200501190700.CAA06903@smc.vnet.net>
*Reply-to*: drbob at bigfoot.com
*Sender*: owner-wri-mathgroup at wolfram.com
Random procedures that don't first identify ALL solutions can certainly be faster.
This works well, for instance:
Clear@randomFill
randomFill[rowSums:{__Integer?Positive},
colSums:{__Integer?Positive}]:=Module[{r=Length@rowSums,c=
Length@colSums,m,gaps,p,i,j},
m=Table[0,{r},{c}];zeroes=0m;
While[(gaps=Table[Min[rowSums[[i]]-Tr@m[[i]],
colSums[[j]]-Tr@m[[All,j]]],{i,r},{j,c}])!=zeroes,
p=Position[gaps,_?Positive];{i,j}=p[[Random[Integer,{1,Length@p}]]];
m[[i,j]]+=Random[Integer,{1,gaps[[i,j]]}]
];m
]
randomFill[{7,3,2,1},{2,9,2}]
{{0,6,1},{2,1,0},{0,1,1},{0,1,0}}
However, I don't think solutions are chosen with equal frequency, under any such scheme.
For instance:
Timing[f=Frequencies[Array[randomFill[Range@4,Range@4]&,{20000}]][[All,1]]]
Through[{Length,Min,Max}@f]
{36.734 Second,{103,204,45,72,
155,102,105,105,77,69,157,55,57,45,18,58,67,51,74,37,73,36,28,
71,143,96,60,55,91,110,97,100,49,73,127,40,26,82,74,18,41,67,31,44,56,43,
82,28,41,43,24,55,52,23,70,35,39,184,73,78,196,147,79,105,85,73,158,88,
30,50,144,77,113,53,94,62,36,67,85,82,116,28,43,115,93,34,55,93,98,94,
35,96,103,103,103,76,101,60,51,107,94,44,30,32,65,71,58,58,33,42,71,146,
62,125,120,83,171,58,30,53,49,25,54,57,23,60,44,39,67,36,65,
43,35,41,49,96,171,91,107,115,98,178,132,54,105,44,28,55,94,76,102,70,30,
48,80,155,80,39,58,47,34,111,64,72,139,117,57,76,130,75,91,110,70,101,
35,51,51,31,59,66,30,60,42,43,145,95,80,90,83,81,92,125,120,47,50,112,80,
41,59,64,39,67,67,43,91,127,42,48,121,87,43,65,91,47,67,67,54,107,116,
56,56,151,108,96,169,113,46,65,175,111,107,45,77,34,22,47,73,75,145,
119,75,58,112,158,115,33,30,107,73,36,42,73,32,54,58,35,90,190,88,73,182}}
{261,18,204}
#.# &[f - Mean@f]/Mean@f
ChiSquarePValue[%,261]
50351983/10000
OneSidedPValue -> 0.
Bobby
On Wed, 19 Jan 2005 02:00:27 -0500 (EST), Paul Abbott <paul at physics.uwa.edu.au> wrote:
> In article <csinpt$njk$1 at smc.vnet.net>, adiggle at agric.wa.gov.au wrote:
>
>> Is there an efficient method in Mathematica to generate random tables
>> of nonnegative integers from a list of row sums and a list of column
>> sums?
>>
>> For example for row sums of {7,3,2} and column sums of {2,9,1} the
>> following two tables satisfy the constraints:
>>
>> {{2, 5, 0}, {0, 2, 1}, {0, 2, 0}}
>> and
>>
>> {{1, 6, 0}, {1, 2, 0}, {0, 1, 1}}
>
> Here is one way to do this. The following module
>
> [1] constructs an arbitrary matrix of dimension implied by the row and
> column sums (need not be square);
>
> [2] uses Reduce to apply the conditions that the entries are nonnegative
> integers to determine all possible solutions;
>
> [3] turns this output into a list of replacement rules; and
>
> [4] returns a list of _all_ matrices satisfying the constraints.
>
> NonnegativeIntegerMatrices[rows_, cols_] := Module[
> {mat, r, c, a, vars, cond, red, m = Length[rows], n = Length[cols]},
> mat = Table[a[i, j], {i, m}, {j, n}];
> r = Thread[Plus @@ Transpose[mat] == rows];
> c = Thread[Plus @@ mat == cols];
> vars = Flatten[mat];
> cond = Thread[Flatten[mat] >= 0];
> red = {ToRules[Reduce[Join[r, c, cond], vars, Integers]]};
> If[red != {}, mat /. red, {}]
> ]
>
> Once you have the complete list of solutions, say
>
> sols = NonnegativeIntegerMatrices[{7,3,2}, {2,9,1}]
>
> it is easy to select one at random
>
> sols[[ Random[Integer, {1, Length[sols]}] ]]
>
> However, this approach is not optimal if you have to work with large
> numbers of different row and column sums. Also, computing all solutions
> can be slow even for rather small matrices, e.g., try
>
> sols = NonnegativeIntegerMatrices[Range[4], Range[4]]
>
> In such cases a random search method will be more efficient -- and I'm
> confident that other posters will help you here.
>
> Cheers,
> Paul
>
--
DrBob at bigfoot.com
www.eclecticdreams.net
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