Filling an array with recursive function

2019 1 7
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更新时间：2019 1 9
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Dear MATLAB Users,
I am trying to fill in the empty array using a recursive formula. The formula is shown as follows:
(This expression is the base case for resursive function)
For all
In this formula, I already know the B array which is .
So far, I had many attempts but non of them worked. My function is shown below. When I run it, I get the error message "Assignment has more non-singleton rhs dimensions than non-singleton subscripts". Could you please help me to solve this problem?
function I_matrix = fill_I_recursively(B_matrix, u_ind, v_ind, w_ind)
M = size(B_matrix,1);
if u_ind == 1 && v_ind == 1 && w_ind == 1
I_matrix(u_ind,v_ind,w_ind) = 0;
else
if w_ind > 1
I_matrix(u_ind,v_ind,w_ind) = fill_I_recursively(B_matrix, u_ind, v_ind, w_ind-1) + B_matrix(u_ind,v_ind,w_ind-1);
elseif v_ind > 1
I_matrix(u_ind,v_ind,w_ind) = fill_I_recursively(B_matrix, u_ind, v_ind-1, M-1) + B_matrix(u_ind,v_ind-1,M-1);
else
I_matrix(u_ind,v_ind,w_ind) = fill_I_recursively(B_matrix, u_ind-1, M-1, M-1) + B_matrix(u_ind-1,M-1,M-1);
end
end
I am looking forward to hearing from you

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Luna
Luna 2019-1-7
Could you please share the below inputs while you are calling the function?
fill_I_recursively(B_matrix, u_ind, v_ind, w_ind)
mustafa ozendi
mustafa ozendi 2019-1-7
B_matrix is the array computed before and I know all elements of it. Dimensions of B_matirx is MxMxM which is in my case 3X3X3. The other inputs u_ind, v_ind, w_ind indices. I call this function in my program as shown below:
I_matrix = NaN(max_boxes,max_boxes,max_boxes);
%%fill the I matrix using recursive function
for u_ind = 1:max_boxes
for v_ind = 1:max_boxes
for w_ind = 1:max_boxes
I_matrix(u_ind,v_ind,w_ind) = fill_I_recursively(B_matrix, u_ind, v_ind, w_ind);
end
end
end
I initialize the I_matrix which is to be filled (max_boxes =3). B_matrix is loaded in my workspace which is 3x3x3.

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Stephen23
Stephen23 2019-1-7
编辑：Stephen23 2019-1-7

1 个投票

I don't see why you need to use a recursive function:
M = 3;
B = randi(9,M,M,M);
I = zeros(M,M,M);
for ku = 1:M
for kv = 1:M
for kw = 1:M
if kw>1
I(ku,kv,kw) = I(ku,kv,kw-1) + B(ku,kv,kw-1);
elseif kv>1
I(ku,kv,kw) = I(ku,kv-1,M-1) + B(ku,kv-1,M-1);
elseif kw>1
I(ku,kv,kw) = I(ku-1,M-1,M-1) + B(ku-1,M-1,M-1);
end
end
end
end
Note that the formula you gave will give different results depending on the order that you fill the dimensions in.

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mustafa ozendi
mustafa ozendi 2019-1-7
Dear Stephan
As far as I know the more for loops in a matlab program means the more execution time. That is the reason why I am triyng to implement it as a recursive function. Moreover, when I was working on the problem using a sheet of paper and pen I realized that this problem can only be solved with a recursive fucntion but I will definitely try your solution.
Thank you for helps
Stephen23
Stephen23 2019-1-7
"As far as I know the more for loops in a matlab program means the more execution time"
Not really. This might have been true fifteen years ago, but MATLAB from fifteen years ago is definitely not the same as MATLAB today. Loops are very effiicient and there is absolutely no reason to avoid using them. Just make sure that you preallocate as required!
"That is the reason why I am triyng to implement it as a recursive function"
Recursive functions have a lot more overhead than a few loops, so your approach is likely to be much slower. Test it yourself to find out!
"I realized that this problem can only be solved with a recursive fucntion but I will definitely try your solution"
I am confused: if it can only be solved using a recursive function, why bother trying my loops?

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