Using matlab to evolute a simple vector , but not fit the inference

clear
clc
time = 4;
size = 8;
% initialize.
a = ones(time,size);
a(1,6) = 5;
% if a(i)>1,then a(i)=a(i)-1 and a(i-1)=a(i-1)+1.
% Saved in a new row.
for t = 1:time-1
    for i = 2:size
        if a(t,i)>1
            a(t+1,i-1) = a(t,i-1) + 1;
            a(t+1,i) = a(t,i) - 1;
        end
    end
end
ans = a(:,:)

Notice the code above, and I want to achieve :

1 1 1 1 1 5 1 1

1 1 1 1 2 4 1 1

1 1 1 2 2 3 1 1

1 1 2 2 2 2 1 1

but the answer is:

1 1 1 1 1 5 1 1

1 1 1 1 2 4 1 1

1 1 1 2 3 3 1 1

1 1 2 3 4 2 1 1

Where is wrong?

[Jos (10584)]

A tip: do not use names for variables that are also functions, like size ;-)

Answer 1

[Andrei Bobrov]

m = 4;
n = 8;
a = ones(m,n);
a(1,6) = 5;

for ii = 1:m-1
    for jj = 2:n
        if a(ii,jj)>1
            if ii == 1
                I = jj;
                a(ii+1,jj-1) = a(ii,jj-1) + 1;
                a(ii+1,jj) = a(ii,jj) - 1;
                break
            end
            a(ii+1,jj-1:I-1) = a(ii,jj);
            a(ii+1,I) = a(ii,I) - 1;
            break
        end
    end
end

or

m = 4;
n = 8;
a = ones(m,n);
a(1,6) = 5;

[I,J] = find(a>1);
a(:,J - m:J - 1) = a(:,J - m:J - 1) + flip(tril(ones(m),-1),2);
a(:,J) = a(I,J) - (0:m-1)';

[Darcy Chou]

Good job and thanks a lot

Answer 2

[David Goodmanson]

Hi Darcy,

It appears that the rule you want is

a(t+1,i-1) = a(t+1,i-1) + 1

which does give the result you are looking for. For a(t+1,i-1) This allows the lowering effect due to a(t,i-1)>1 to persist until it is raised by by the effect due to a(t,1)>1.

[Darcy Chou]

Thank you . Codes submited by Andrei Bobrov are right.