How to add a cumulative constant to all values on each row of a matrix, starting from the bottom row and ending at the top row?

I have a matrix that is very large. A small example is:

1 2 3
1 2 3
1 2 3       

I need to add a constant to each row, starting at the bottom row. The additon should be cumulative. For example, if the constant is 0.5, the matrix should become:

2.5 3.5 4.5
2.0 3.0 4.0
1.5 2.5 3.5

I know to use "cumsum" for cumulative addition in Matlab but the indexing required in this problem exceeds my skills. I would be vey grateful for any help.

Answer 1

[the cyclist]

There are many ways to do this. Here is one way:

% Inputs
in = [1 2 3;
      1 2 3;
      1 2 3];
constant = 0.5;

% Algorithm
[m,n] = size(in);
out = in + kron(constant*(m:-1:1)',ones(1,n));

Answer 2

[Voss]

c = 0.5;
A = [1,2,3;1,2,3;1,2,3]

A = 3x3

1 2 3 1 2 3 1 2 3

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One way:

N = size(A,1);
result = c*(N:-1:1).'+A

result = 3x3

2.5000 3.5000 4.5000 2.0000 3.0000 4.0000 1.5000 2.5000 3.5000

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Another way:

N = size(A,1);
result = cumsum(c*ones(N,1),'reverse')+A

result = 3x3

2.5000 3.5000 4.5000 2.0000 3.0000 4.0000 1.5000 2.5000 3.5000

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[Srh Fwl]

Thank you very much indeed, Voss. I wish I could accept two answers. the cyclists worked better because my matrices are so large.

[Voss]

You're welcome!

[Voss]

I wonder in what sense using kron is better than the approaches I showed, for large matrices.

I guess you don't use the term "better" to mean "faster", because kron seems to be the slowest.

c = 0.5;
N = 4;
T = array2table(zeros(N,4),'VariableNames',{'Size','cyclist','Voss1','Voss2'});
for n = 1:N
    A = rand(10^n);
    tc = timeit(@()cyclist(A,c));
    tv1 = timeit(@()Voss1(A,c));
    tv2 = timeit(@()Voss2(A,c));
    T{n,:} = [10^n,tc,tv1,tv2];
end
T

T = 4x4 table
    Size      cyclist        Voss1         Voss2   
    _____    __________    __________    __________

       10    3.5819e-05    4.8595e-06    5.0667e-06
      100    2.0085e-05    1.2872e-05    7.8766e-06
     1000    0.00028163    0.00019339    0.00011734
    10000       0.12138        0.0996      0.096075

function B = cyclist(A,c)
[m,n] = size(A);
B = A + kron(c*(m:-1:1)',ones(1,n));
end

function B = Voss1(A,c)
B = c*(size(A,1):-1:1).'+A;
end

function B = Voss2(A,c)
B = cumsum(c*ones(size(A,1),1),'reverse')+A;
end

[Srh Fwl]

Thank you so much, Voss. I am currently on the road, but look forward to checking this out in more detail later. I did mean "faster" so will think about why that is an issue for me but not you. Thank you very much for thinking about this!

[Voss]

No problem. I was mostly just curious what you meant by "better". Thanks for clarifying you meant faster. Anyway, the time difference is not much.