# Check if a value exists in a cell array starting from a value whose index is the desired value

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Sergio Rojas Blanco on 14 Aug 2023
Let's suppose:
A={5, 15, 0, [11, 7], 9, 0, [1, 20, 21], 18, [22,15], 0, 16, [13, 14], 17, 4, 6 };
a=6; b=1;
Now, I would like to know if I can get to b from a using the values as an index. In this case:
Value Index
6 -> 15
15 -> [2, 9]
for
2 -> Nan
9 -> 5
and
5 -> 1
Thanks

Dyuman Joshi on 14 Aug 2023
I have unsuppressed the output of variable temp to show the path that is followed (as you have demonstrated above) -
A={5, 15, 0, [11, 7], 9, 0, [1, 20, 21], 18, [22,15], 0, 16, [13, 14], 17, 4, 6 };
a=6; b=1;
B = [A{:}];
%maximum value present in A
m = max(B);
z1 = nested(A,6,1)
temp = 15
temp = 1×2
2 9
temp = 1×0 empty double row vector
temp = 5
temp = 1
z1 = 1
%Any value greater than the maximum value in the cell array will not be present
%So for any value of b, the output will be zero
z2 = nested(A,m+randi(100),randi(numel(A)))
z2 = 0
z3 = nested(A,9,4)
temp = 5
temp = 1
temp = 7
temp = 4
z3 = 1
%3rd value corresponds to zero, so the search is terminated for any value in A
z4 = nested(A,3,B(randi(numel(B))))
temp = 1×0 empty double row vector
z4 = 0
function z = nested(A,a,b);
%Returns 0 if not related, 1 if related
%Initialize z as 0
z=0;
nA=numel(A);
%Values corresponding to indices of elements in A
idx=1:numel(A);
%Check if b is present in a or not
if ismember(b,a)
z=1;
return;
%Return 1 if present, else continue checking
else
for k=1:numel(a)
%Check only if the values is within the range of number of elements of B
if a(k)>0 && a(k)<=nA
%Finding the index of the value
y=cellfun(@(x) ismember(a(k),x), A);
%Corresponding values
temp=idx(y)
z = nested(A,temp,b);
else
%If the value is not in the range, skip to the next iteration
continue;
end
end
end
end
Dyuman Joshi on 14 Aug 2023
Edited: Dyuman Joshi on 14 Aug 2023
Thanks for pointing out the flaw in my code, @Stephen23.
I have edited it, this should work -
A = {1, 2};
a = 2;
b = 1;
z = nested(A,a,b)
z = 0
function z = nested(A,a,b);
%Returns 0 if not related, 1 if related
%Initialize z as 0
z=0;
nA=numel(A);
%Values corresponding to indices of elements in A
idx=1:numel(A);
%Check if b is present in a or not
if ismember(b,a)
z=1;
return;
%Return 1 if present, else continue checking
else
for k=1:numel(a)
%Check only if the values is within the range of number of elements of B
if a(k)>0 && a(k)<=nA
%Finding the index of the value
y=cellfun(@(x) ismember(a(k),x), A);
%Corresponding values
temp=idx(y);
%If temp is equal to a, return 0, avoiding infinite recurison
if isequal(temp,a)
return;
end
z = nested(A,temp,b);
else
%If the value is not in the range, skip to the next iteration
continue;
end
end
end
end
Sergio Rojas Blanco on 14 Aug 2023
It works! Thanks to @Stephen23 and my friend :D

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