MATLAB Answers

Problem with factorial recursive function

116 views (last 30 days)
Stu145
Stu145 on 28 Jan 2015
Edited: Walter Roberson on 26 Dec 2017
Whenever I run the code for a matrix of n values, like n=1:10, only the last 2 factorials are displayed while the rest are 0's. Would anyone be able to tell me where I have gone wrong?
Code:
function x = fact(n)
if n<=1
x = 1;
else
x = n .* fact(n-1) ;
end
end

  0 Comments

Sign in to comment.

Answers (4)

Titus Edelhofer
Titus Edelhofer on 28 Jan 2015
Hi Arun,
your code does not work for arrays because of the "if n<=1". You would need to do something like
idx = (n>1);
x(~idx) = 1;
x(idx) = n(idx).* ...
Titus

  3 Comments

Stu145
Stu145 on 28 Jan 2015
Hi Tidus,
I am now getting a 'Maximum recursion limit error....' for even single values now when using the following code. I have tried to implement what you said, but it's the final equation that is now causing me an issue.
function x = fact(n)
idx = (n>1);
x(~idx)=1;
x(idx) = n(idx).*fact(n(idx)-1);
end
John D'Errico
John D'Errico on 28 Jan 2015
For how large of a value are you trying to compute the factorial? Be careful, as this could be important, since that too would blow up the recursion limits.
Regardless, as you have written it, the code will still run forever. What happens when all values of n are <= 1?
The function as you wrote it will still call fact, but with an empty array for n. so you will have fact calling fact, each time with an empty argument n.
I would add a test to stop that from happening.
Shubham Tyagi
Shubham Tyagi on 15 Aug 2016
I'm compiling this code with a 2*4 matrix but (dimensions doesn't match in val(i)=r(i).*fact(r(i)-1);) type error is coming.
function val=fact(r)
i=(r>1); val(~i)=1; if r==1 return end val(i)=r(i).*fact(r(i)-1); end

Sign in to comment.


Image Analyst
Image Analyst on 31 Jan 2015
Not sure what you did, but it worked fine for me. I put all this (both functions) into test.m
function test
clc;
for k = 1 : 10
kFactorial = fact(k);
fprintf('%d! = %d\n', k, kFactorial);
end
function x = fact(n)
if n<=1
x = 1;
else
x = n .* fact(n-1) ;
end
In the command window you'll see:
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
7! = 5040
8! = 40320
9! = 362880
10! = 3628800

  4 Comments

Show 1 older comment
Shubham Tyagi
Shubham Tyagi on 15 Aug 2016
what about m*n matrix rather than a vector??
Image Analyst
Image Analyst on 15 Aug 2016
I don't know what matrix you're talking about. Please give the code you are going to run to call your function, and show what output you'd expect to see.
If you want fact n to print out an array from the first integer to the last integer, the fact() function will need to be modified since now it's just expecting a single number, not a vector or matrix.
Torsten
Torsten on 25 Aug 2016
Reshape everything to vectors and reshape back to matrices, if necessary.
Best wishes
Torsten.

Sign in to comment.


Titus Edelhofer
Titus Edelhofer on 25 Aug 2016
Hi,
two minor changes are needed:
function x = fact(n)
x = ones(size(n));
idx = (n>1);
if any(idx)
x(idx) = n(idx).*fact(n(idx)-1);
end
First, you need to initialize x as an array of the same size as n (because for fact(n-1) otherwise the array could be smaller), giving the error you observed.
Second, protect the recursive call to fact(n-1) by doing this only if there is anything to do...
Titus

  1 Comment

Titus Edelhofer
Titus Edelhofer on 26 Aug 2016
Calling for a matrix works fine, btw.:
fact([2 3 4; 5 6 7])
ans =
2 6 24
120 720 5040

Sign in to comment.


Anil Sarode
Anil Sarode on 26 Dec 2017
Edited: Walter Roberson on 26 Dec 2017
sorry i copied your code :
this is the one way to solve your problem :
function file: from where you make function call each time using for loop :
n=1:10;
y=[];
for i=1:length(n)
y(i)=fact(i) ;
end
function: which do desire task
function [x] = fact(n)
if n<=1
x = 1;
else
x = n .* fact(n-1) ;
end
end

  0 Comments

Sign in to comment.

Sign in to answer this question.