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

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

Answer 1

[Dyuman Joshi]

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

[Sergio Rojas Blanco]

You are a genius! You have a new friend for ever :D

Thanks a lot.

[Dyuman Joshi]

Hahaha, I'm not a genius, but yeah happy to have a new friend :)

Glad to have helped!

[Stephen23]

Note that this function will result in an infinite recursion for some inputs, e.g.:

A = {1,2};
a = 2;
b = 1;
nested(A,a,b)

[Sergio Rojas Blanco]

You are right. I have found this situation.

[Dyuman Joshi]

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]

It works! Thanks to @Stephen23 and my friend :D