write a function called palindrome that takes one input argument a char vector and recursively determine whether that argument is a palindrome you are not allowed to use loops not built in function like srtcmp etc. the function returns true or false

function ok=palindrome(txt)

iflength(txt)<=1

ok=true;

else

ok=(txt(1)==txt(end)&&palindrome(txt(2:end-1)));

end

end

hi,

It shows error in else statment in the above program, kindly give me a solution to solve the above problem.

Answer 1

[Ameer Hamza]

There should be a space between if keyword and the condition

if length(txt)<=1
 %^ insert a space here

Apart from that, the logic is correct, but you just have a misplaced bracket. Following is correct

ok=(txt(1)==txt(end))&&palindrome(txt(2:end-1));

[JAYANTHI SANKARALINGAM]

Thank u for the reply

Answer 2

[Sandeep Kumar Patel]

If we apply the edits recommended by @Ameer Hamza, we get this:

function ok=palindrome(txt)
    if length(txt)<=1 % added space
        ok = true;
    else
        ok = (txt(1)==txt(end)) && palindrome(txt(2:end-1));
        % close parentheses --^
    end
end

[DGM]

Editor's note: I added the comments here so that readers know the two lone characters that were actually changed.

Answer 3

[Black Woods]

function ans=palindrome(v)
if v(1)==v(end)
    ans=true;
    if length(v)==1 || length(v)==2
        return
    else
        palindrome(v(2:length(v)-1));
    end
else
    ans=false;
end
end
    

Answer 4

[Tarun Sangwan]

function p = palindrome(n)
if length(n)/2<1
    p = 2
else
    p = n(1) == n(end)
    p = [p   palindrome(n(2:end-1))]
end
p = ~ismember(0,p)