I am having an array with positive integers, and i want to know all the possibility that the sum of elements of the array is close to or equal to a number say (N))

Question

Ashish Verma 2023-1-5

1
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1888832-i-am-having-an-array-with-positive-integers-and-i-want-to-know-all-the-possibility-that-the-sum-of

评论： Matt J 2023-1-7

I am having an array with positive integers, and i want to know all the possibility that the sum of elements of the array is close to or equal to a number say (N)) and no index (element in the array) will be repeated

A = [ 0 7 30 16 23 11 19 15 28 8 8 7 14 6 19 11 12 26 17 6 15 5 10 ]

N = 40 (or close to 40 like 36,37,38,39)

the possible sum of the numbers are like

8+6+26 = 40

23+15 = 38

16+17+6=39

15+5+19=39

and so on

4 个评论
显示 2更早的评论隐藏 2更早的评论

Matt J 2023-1-5

编辑：Matt J 2023-1-5

Why is 23+15 = 38 an acceptable selection when it is still possible to add further A(i) to the sequence without exceeding N=40? In particular: 23+15+0 = 38

Torsten 2023-1-5

23+15+0 = 39

?

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Matt J 2023-1-5

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1888832-i-am-having-an-array-with-positive-integers-and-i-want-to-know-all-the-possibility-that-the-sum-of#answer_1142352

编辑：Matt J 2023-1-6

在 MATLAB Online 中打开

N=40;
A = [1 7 13 20 38 39 40];
A=unique(A);
n=find(cumsum(A)<=N,1,'last');
result=cell(n,1);
for i=1:n
    tmp=subCollection(N,A,i);
    tmp(:,end+1:n)=nan;
    result{i}=tmp;
end
result=cell2mat(result) %final result
result = 7×3
    40   NaN   NaN
     1    38   NaN
     1    39   NaN
     1     7    13
     1     7    20
     1    13    20
     7    13    20
function T=subCollection(N,A,m)
  
  J=nchoosek(1:numel(A),m);
  I=repmat( (1:height(J))'  ,1,m);
  %S=sparse(I,J,1);
  S=accumarray([I(:),J(:)],1);
  delta=N-S*A(:);
  C=(1./(1-S));
  mincomp=min( C.*A(:).' ,[],2);
  idx=delta<mincomp & delta>=0;
  T=(1./S).*A(:).';
  T=sort(T(idx,:),2);
  T(isinf(T))=nan;
  T=T(:,1:m);
end

2 个评论
显示无隐藏无

Ashish Verma 2023-1-7

According to me results will be some matrices, where the sum will be close to 40 and the matrix contains distinct elements or no index of array will be repeated (because of duplicacy). For eg, possibility of 3 elements sum will be

[8+6+26 = 40

23+15 = 38

16+17+6=39

15+5+19=39

...]