How to compute sum of binomial more efficiently?

问题:

I must calculate an equation as follows:

where k1,k2 are given. I am using MATLAB to compute P. I think I have a correct implementation for the above equation. However, my implementation is so slow. I think the issue is from binomial coefficient. From the equation, could I have an efficient way to speed up the time? Thank all.

For k1=150; k2=150; D=200;, it takes 11.6 seconds

function main
warning ('off');
  function test_binom()
      k1=150; k2=150; D=200; P=0;
      for i=0:D-1
          for j=0:i
              if (i-j>k2||j>k1) 
                  continue;
              end
              P=P+nchoosek(k1,j)*nchoosek(k2,i-j)/nchoosek((k1+k2),i);          
          end 
      end
  end
f = @()test_binom(); 
timeit(f)
end

Update: For measure time, I found that nchoosek is the reason for large computational time. Hence, I rewrite the function as follows

function re=choose(n, k)
    if (k == 0)
        re=1;
    else
        re=(n * choose(n - 1, k - 1)) / k;
    end
end

Now, the computational time is reduced as 0.25 second. Is has any better way?

回答1:

You can vectorize all the process and it will make it super-fast without any need for mex.

First the nchoosek function:

function C = nCk(n,k)
% use smaller k if available
k(k>n/2) = n-k(k>n/2);
k = k(:);
kmat = ones(numel(k),1)*(1:max(n-k));
kmat = kmat.*bsxfun(@le,kmat,(n-k));
pw = bsxfun(@power,kmat,-1./(n-k));
pw(kmat==0) = 1;
kleft = ones(numel(k),1)*(min(k):n);
kleft = kleft.*bsxfun(@gt,kleft,k);
t = bsxfun(@times,kleft,prod(pw,2));
t (kleft==0) = 1;
C = prod(t,2);
end

Then beta and P computation:

function P = binomial_coefficient(k1,k2,D)
warning ('off','MATLAB:nchoosek:LargeCoefficient');
i_ind = nonzeros(triu(ones(D,1)*(1:D)))-1;
j_ind = nonzeros(tril(ones(D,1)*(1:D+1)).')-1;
valid = ~(i_ind-j_ind>=k2 | j_ind>=k1);
i_ind = i_ind(valid);
j_ind = j_ind(valid);
beta = @(ii,jj) nCk(k1,jj).*nCk(k2,ii-jj)./nCk((k1+k2),ii);
b = beta(i_ind,j_ind);
P = sum(b(:));
end

and execution time drops from 10.674 to 0.49696 seconds.

EDIT:

Taking the idea of @rahnema1, I managed to make this even faster, using a table for all unique nCk computations, so none of them will be done more than once. Using the same nCk function from above, this is how the new binomial_coefficient function will look:

function P = binomial_coefficient(k1,k2,D)
warning ('off','MATLAB:nchoosek:LargeCoefficient');
i_ind = nonzeros(triu(ones(D,1)*(1:D)))-1;
j_ind = nonzeros(tril(ones(D,1)*(1:D+1)).')-1;
valid = ~(i_ind-j_ind>=k2 | j_ind>=k1);
i_ind = i_ind(valid);
j_ind = j_ind(valid);
ni = numel(i_ind);
all_n = repelem([k1; k2; k1+k2],ni); % all n's to be used in thier order
all_k = [j_ind; i_ind-j_ind; i_ind]; % all k's to be used in thier order
% get all unique sets of 'n' and 'k':
sets_tbl = unique([all_n all_k],'rows');
uq_n = unique(sets_tbl(:,1));
nCk_tbl = zeros([max(all_n) max(all_k)+1]);
% compute all the needed values of nCk:
for s = 1:numel(uq_n)
    curret_n = uq_n(s);
    curret_k = sets_tbl(sets_tbl(:,1)==curret_n,2);
    nCk_tbl(curret_n,curret_k+1) = nCk(curret_n,curret_k).';
end
beta = @(ii,jj) nCk_tbl(k1,jj+1).*nCk_tbl(k2,ii-jj+1)./nCk_tbl((k1+k2),ii+1);
b = beta(i_ind,j_ind);
P = sum(b(:));
end

and now, when it takes only 0.01212 second to run, it's not just super-fast code, it's a flying-talking-super-fast code!

回答2:

You can save results of nchoosek to a table to prevent repeated evaluation of the function, also an implementation of binomial coefficients provided:

%binomial coefficients
function nk=nchoosek2(n, k)
    if n-k > k
        nk = prod((k+1:n) .* prod((1:n-k).^ (-1/(n-k))));
    else
        nk = prod((n-k+1:n) .* prod((1:k).^ (-1/k)) ) ;
    end
end
%function to store and retrieve results of nchoosek to/from a table
function ret = choose (n,k, D, K1, K2)
    persistent binTable = zeros(max([D+1,K1+K2+1]) , D+1);
    if binTable(n+1,k+1) == 0
        binTable(n+1,k+1) = nchoosek2(n,k);
    end
    ret = binTable(n+1,k+1);
end

function P = tst()
    P=0;k1=150; k2=150; D=200; P=0;
    choose(1,0,D,k1,k2);
    for i = 0:D-1
        for j = j=max(i - k2 , 0):min (i,k1-1)
            P=P+choose(k1,j)*choose(k2,i-j)/choose((k1+k2),i);
        end
    end
end

Your code with nchoosek2 compared with this: online demo

回答3:

rahnema1's answer has a very good approach: create a table of values that you generate once and access later (as well as some other clever optimizations).

One thing I would change is the way the binomial coefficients are calculated. If you look at calculating the factorials for nchoosek(n, k) and nchoosek(n, k+1), you're recalculating n! both times, and for k+1, you're recalculating k! and multiplying it by k+1. (Similarly for (n-k)!.)

Rather than throw away the computations each time, we can iteratively compute nchoosek(n, k+1) based on the value of nchoosek(n, k).

function L=combList(n, maxk)
% Create a vector of length maxk containing
%   [nchoosek(n, 1), nchoosek(n, 2), ..., nchoosek(n, maxk)]
% Note: nchoosek(n, 0) == nchoosek(n, n) == 1

assert(maxk<=n, 'maxk must be less than or equal to n');

L = zeros(1,maxk);
L(1) = n;                    % nchoosek(n, 1) == n
for k = 2:maxk
   L(k) = L(k-1)*(n-k+1)/k;
end

In your program, you would just create 3 lists for k1, k2, and k1+k2 with the appropriate limits, and then index into those lists to generate the sums.