how to check for a perfect square in a Matrix

i have a problem in which i have to return a value of true if any element of the matrix is a pefect square

for example

A = [ 2,4,7 ; 8,7,3 ]

b is true

otherwise

b is false

Answer 1

[KSSV]

A = [ 2,4,7 ; 8,7,3 ]; 
b = (mod(sqrt(A), 1) == 0) 
A(b)

[Pallavi Chaudhary]

This will not work in this case

a = [20:30];

[Pallavi Chaudhary]

%%you can use this function
function b = isItSquared(a)

   b = (mod(sqrt(a), 1) == 0)
   cc = a(b);
   
   if (~isempty(cc))
        b = any(a==sqrt(cc));
            if (b ==1)
                b =1;
            end
   else
       b=0;
   end
end

[Dawud Adam Rabiu]

Can you explain the function please

[Image Analyst]

@Dawud Adam Rabiu if the number is a perfect square then the square root of it will be an integer, so that's what the mod checks for. If it's not a perfect square, there will be a fractional part so the mode of the number with 1 will be that fractional part, not 0. Only numbers with mod(sqrt(n))==0 will be perfect squares.

@Pallavi Chaudhary it seems to work fine if you use capital A:

A = [20:30];
b = (mod(sqrt(A), 1) == 0) 

b = 1×11 logical array
   0   0   0   0   0   1   0   0   0   0   0

A(b)

ans = 25

It correctly identifies 25 as the only perfect square in that vector (it's 5 squared). I'm not sure why you said it didn't work, but perhaps you used lower case "a" like you did. MATLAB is case sensitive so a is not the same as A.

[Nick Austinos]

The idea was to find if a matrix cotains two or more numbers for which num_x=(num_y)^2. or vice versa this code (mod(sqrt(A), 1) == 0) works perfectly for A = [ 2,4,7 ; 8,7,3 ] but it doesnt work for A=20:30; For that case the suggested solution below can be used, though it uses more memory.

for ii=1:length(a)
    x(ii)=nnz(a==a(ii)^2);
    if nnz(x)>0
        b=true;
    else
        b=false;
    end
end

Answer 2

[Ameer Hamza]

Use sqrt() and floor() functions to see which elements are perfect squares. Then use any(__, 'all') function to check if any element is a perfect square.

Answer 3

[Ahmed Thawhid Sabit]

function b = isItSquared(a)

if a(1)^2 == a(end)

b = true;

else

b=false;

end

end

this solution has some drawnbacks... i couldnt find anysolution

[John D'Errico]

Note this code is not in fact a solution. Not even close. It compares only the square of the first element, compared then for some reason to the final element in the matrix.

Had you wanteed to use a loop, that would be possible, but this attempt is not close to that.

[Ahmed Thawhid Sabit]

yes iu got the solution.. i just didn't understand the preoblem well

Answer 4

[aimane nedjima]

function b = isItSquared(a)
M = size(a,1);
N = size(a,2);
b = false ;
for p=1:M
 for i=1:N
  % to compare one element with other in the same colomn
     for j=1:N
     if (a(p,i))^2 == a(p,j)
      b  = true ;
      break % if we get one we stop
     end
   end
end

end