# how to count the total number of recursion ?

조회 수: 47(최근 30일)
Kratos 2015년 3월 27일
댓글: Andrew Newell 2015년 3월 27일
Hello
I have a function
function [out num] = collatz(val)
num = 0;
if val == 1
out = 1;
elseif mod(val, 2) == 0
out = collatz(val/2);
else
out= collatz(3*val+1);
end
k = num + 1;
num = [num k]
end
For the number of counts, num I keep getting num = [0 1] How do I count the total number of recursion that occurs? I heard that we can make a helper function but I have no idea how to make one. Thank you.

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### 채택된 답변

Andrew Newell 2015년 3월 27일
편집: Andrew Newell 2015년 3월 27일
The problem is that you're not passing num to collatz, so how can it know how many times it has been called? Here is a version that makes num an optional argument, so you don't need to provide it in your call:
function [out, num] = collatz(val,varargin)
if nargin >1
num = varargin{1};
else
num = 0;
end
if val == 1
out = 1;
elseif mod(val, 2) == 0
[out,num] = collatz(val/2,num);
else
[out,num] = collatz(3*val+1,num);
end
num = num+1;
Now if you try
[out,num] = collatz(4)
you get
out =
1
num =
3
##### 댓글 수: 2표시숨기기 이전 댓글 수: 1
Andrew Newell 2015년 3월 27일
They must be defining it differently than I am. Just replace
num = 0;
by
num = -1;

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