Using matlab to evolute a simple vector , but not fit the inference
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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?
1 Comment
Jos (10584)
on 28 Feb 2019
A tip: do not use names for variables that are also functions, like size ;-)
Accepted Answer
Andrei Bobrov
on 28 Feb 2019
Edited: Andrei Bobrov
on 28 Feb 2019
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)';
More Answers (1)
David Goodmanson
on 28 Feb 2019
Edited: David Goodmanson
on 28 Feb 2019
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.
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