I am trying to solve the following problem and got stuck if you can give some guidance

Problem 713. Find the maximum number of decimal places in a set of numbers

Find the maximum number of decimal places in a set of numbers after decimal point that are significant

function y = find_max_sigdec(x)

y = max(max(length(rem(x,1))));

end

[Dyuman Joshi]

Note - this question is about MATLAB Cody problem no 713 - https://in.mathworks.com/matlabcentral/cody/problems/713-find-the-maximum-number-of-decimal-places-in-a-set-of-numbers

Firstly, the author of that question does not seem to know/understand how double precision numbers are stored. (It is a not a good question imo.) The values stored is not the same as value displayed.

Nobody has pointed out the problem with question yet (I am still waiting for someone to point it out, after I figured it - which, took quite awhile; even after submitting a solution that works for the given test suite, thinking I "solved" the problem).

I'll give you a hint for now (going AFK) - See what the output of rem(x,1) is and how would length() interpret it.

Also, see - https://in.mathworks.com/matlabcentral/answers/57444-why-is-0-3-0-2-0-1-not-equal-to-zero. This will give you a better understanding/idea of double precision numbers.

[dareen]

thank u much appreciated

[dareen]

what a save i dont understand how i missed this up so bad with length i figured something is wrong but kept changing the elements of a matrix of the same size and some how got really fitting answers 4 and the max was always 4 digits ,so when i used length it gave me the length of the matrix not the elements ,and the elements are not strings they are numebrs whatever i substitue has a length of 1 which is somehow confusing in other languages the length of the number was the num of digits it has so it seems like i have to change the input to a string to be able to deal with this the link you added really made the double precision topic more understandable thanks , i will try to learn more about strigns so that i can reach a solution of my own and just to check

if i want to fix the code is there a simpler approach that you would reccomend

thanks again @Dyuman Joshi

Answer 1

[Hassaan]

Looks like a MATLAB Cody Problem. One of the many approaches that may exist:

x = [1.000 1.04 0.22 10.1;
  2.05 2.33 4.1 1000.31;
  5.00010 6.429 7.492 8.0];

y = find_max_sigdec(x)

y = 4

%y = find_max_sigdec(3.123456789)

disp(y)

     4

function y = find_max_sigdec(x)
    maxDecimals = 0; % Initialize the maximum number of decimal places
    x = x(:); % Linearize the matrix to iterate over all elements
    for i = 1:numel(x)
        % Extract the decimal part by subtracting the integer part
        decimalPart = abs(x(i) - floor(x(i)));
        if decimalPart > 0
            % Convert the decimal part to a string
            str = num2str(decimalPart);
            % Find the index of the decimal point
            idx = strfind(str, '.');
            % Calculate the number of significant decimal places
            numDecimals = length(str) - idx;
            % Update maxDecimals if this number is larger
            maxDecimals = max(maxDecimals, numDecimals);
        end
    end
    y = maxDecimals; % Return the maximum number of decimal places found
end

-----------------------------------------------------------------------------------------------------------------------------------------------------

If you find the solution helpful and it resolves your issue, it would be greatly appreciated if you could accept the answer. Also, leaving an upvote and a comment are also wonderful ways to provide feedback.

It's important to note that the advice and code are based on limited information and meant for educational purposes. Users should verify and adapt the code to their specific needs, ensuring compatibility and adherence to ethical standards.

Professional Interests

• Technical Services and Consulting
• Embedded Systems | Firmware Developement | Simulations
• Electrical and Electronics Engineering

Feel free to contact me.

[Dyuman Joshi]

Does not work for many cases -

y = find_max_sigdec(3.123456789123456789123456789)

y = 16

y = find_max_sigdec(3.010203040506070809)

y = 16

function y = find_max_sigdec(x)
    maxDecimals = 0; % Initialize the maximum number of decimal places
    for i = 1:length(x)
        decimalPart = abs(x(i) - floor(x(i))); % Get the decimal part
        if decimalPart ~= 0 % If there is a decimal part
            decimalStr = num2str(decimalPart, '%.16f'); % Convert to string with sufficient precision
            decimalStr = regexprep(decimalStr, '0+$', ''); % Remove trailing zeros
            dotIndex = find(decimalStr == '.', 1); % Find the position of the decimal point
            numDecimals = length(decimalStr) - dotIndex; % Count the number of significant decimal places
            maxDecimals = max(maxDecimals, numDecimals); % Update the maximum if needed
        end
    end
    y = maxDecimals; % Return the maximum number of decimal places found
end

[dareen]

really helpful did not know most of the functions here so learnt a lot , yet it still fails at passing the cases

[Hassaan]

@dareen Check my updated answer. It passes the test cases though it still faces issues on the points highligted by @Dyuman Joshi a good catch.

-----------------------------------------------------------------------------------------------------------------------------------------------------

If you find the solution helpful and it resolves your issue, it would be greatly appreciated if you could accept the answer. Also, leaving an upvote and a comment are also wonderful ways to provide feedback.

It's important to note that the advice and code are based on limited information and meant for educational purposes. Users should verify and adapt the code to their specific needs, ensuring compatibility and adherence to ethical standards.

Professional Interests

• Technical Services and Consulting
• Embedded Systems | Firmware Developement | Simulations
• Electrical and Electronics Engineering

Feel free to contact me.

[dareen]

thank u this was helpful ,and pardon me i will just wait for a while before accepting an answer so that there is a possibility to recive new comments or answers

[Hassaan]

Sure. You are welcome.