I'm trying to create a code that determines whether a word (or phrase) is a palindrome. The code only needs to look at the actual letters within the input string, but I'm not sure how to do that.

The jist of the problem is to create a function that determines whether a string input is a palindrome by returning the logical true or false. The function needs to disregard any non-letter (i.e. punctuation, spaces, capitalization, etc.), but I'm not sure how to do this.

Here is my code:

function palindrome = hw4_problem2(v)
palindrome = true;
l = lower(v);
n = length(v);
if mod(n, 2) == 0
    for ii = 1:n
        for jj = n:-1:1
            if l(ii) ~= l(jj)
                palindrome = false;
            else
                palindrome = true;
            end
        end
    end
end
end

I was able to use "lower" to get everything into lower letter form, but how do I disregard nonletters? Thanks.

Answer 1

[John D'Errico]

Well, you did show some effort, actually a credible one at that. :) You already know how to use tools like lower, so try a few others.

str = 'The quick Brown fox jumped over the lazy dog.';

str = lower(str);

str = str(ismember(str,'abcdefghijklmnopqrstuvwxyz'))
str =
    'thequickbrownfoxjumpedoverthelazydog'    

Now how can you tell if it is a palindrome? Have you considered flip?

isequal(str,flip(str))
ans =
  logical
   0

[James Metz]

I have tried using flip, but I thought it was essentially the same as what I already had in my code with the

for ii 1:n 
    for jj = n:-1:1

But I'll try using flip and see what happens.

As of now, my function is stuck on 'Madam, I am Adam'

I think it's probably due to the coding above.

[James Metz]

It worked with flip! Thanks so much.

[John D'Errico]

You wrote a double loop, which is not the same. You might think of that double loop as essentially comparing all elements of the string with all other elements. When you have nested for loops like that, they are not running concurrently. One runs inside the other, so you have n^2 comparisons, even though one of them is running backwards.

[Guillaume]

I would strongly recommend using isstrprop as I answered instead of ismember(str,'abcdefghijklmnopqrstuvwxyz'), at least if you don't want to offend the majority of languages (pretty much anything but english) that have more letters than just a:z.

>> s = char(224:246)
s =
    'àáâãäåæçèéêëìíîïðñòóôõö'
>> isstrprop(s, 'alpha')  %Indeed they're all letters
ans =
  1×23 logical array
   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1

Answer 2

[Guillaume]

See isstrprop.

[James Metz]

This operator converts the string to a vector of logical values. Then I have to somehow convert it back to the original values for my function to work at all, but thanks for the recommendation.

[Guillaume]

"This operator converts the string to a vector of logical values"

Yes, and you can use the logical vector as an index into your char vector (not string) to discard the non-letters:

>> str = 'abc12cba34';
>> str(isstrprop(str, 'alpha'))
ans =
    'abccba'

As demonstrated in the very first example of the doc I linked to.

[James Metz]

gotcha! that makes sense. Sorry I'm new to MATLAB so most of this stuff just goes right over my head.

Ok so this is what I have so far, but its still providing a false answer when I input... 'was it a car or a cat I saw?'

function palindrome = hw4_problem2(v)
palindrome = true;
l = lower(v);
m = l(isstrprop(l, 'alpha'));
n = length(m);
    for ii = 1:n
        for jj = n:-1:1
            if m(ii) ~= m(jj)
                palindrome = false;
            else
                palindrome = true;
            end
        end
    end
end

What am I missing?

[Steven Lord]

That checks each element of m against each element of m, but then throws away each result in turn as it checks the next pair of elements. This is only really checking the last element (ii equal to n) and the first element (jj equal to 1.)

You don't want that; you only want to check each element against the corresponding element "at the other end" of m. You only need one for loop, but you do need to stop the loop or treat the value of palindrome as somewhat "sticky" (once you fail one check and know that the string is not a palindrome, no future check can undo that failure and make it a palindrome again.)

The Tips on the documentation page for the for keyword may be of use to you, as might the functions any, all, and, and/or or. Of if the assignment doesn't require you to use a loop, see John D'Errico's answer.

Answer 3

[James Metz]

For anyone looking for the right code for this question I have attached the code that worked:

function palindrome = hw4_problem2(v)
l = lower(v);
m = l(ismember(l, 'abcdefghijklmnopqrstuvwxyz'));
    if isequal(m, flip(m)) == 0
       palindrome = false;
    else
       palindrome = true;
    end
end

Thanks to everyone who helped me get there!

Best.