How can I avoid infinite while loop?
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N = 1; while (N<11) n = ((N*2)+(N+1))/N; N = n end
I wish to assign the value of 'n' to 'N' and compare if it's less than '11' and again enter the loop until 'N' is equal to 11. Please help.
采纳的回答
dpb
2014-7-11
Well, since
((N*2)+(N+1))/N = (2*N+N+1)/N = (3*N+1)/N
n=3+1/N
and the expression approaches 3+ as a limit, never approaching anything near 11, the answer to the question posed is "you can't".
Need a different problem statement; perhaps you're looking for the point at which the change is less than some epsilon or something of that nature?
I'll wait for clarification of the actual problem nature before guessing further...
1 个评论
Shashanka
2014-7-11
Thank you for the response! The example was a simplified version of the code I'm trying to run, which has decimal inputs. I have attached an image of the actual code below for your reference. The 'AxIFN' is to be updated with the latest value of 'AxIFN1' which is calculated with each iteration. Finally giving a value of 'AxIFN' which is less than '0.4'.
Thank you!
更多回答（2 个）
Daniel
2014-7-11
Set a number equal to the number of times you want to loop (this case I chose 11), subtract each time you're in the loop, and add an and condition
LOOP_LIMIT = 11;
N = 1;
while (N < 11 && LOOP_LIMIT > 0)
n = ((N*2)+(N+1))/N;
N = n;
LOOP_LIMIT = LOOP_LIMIT-1;
end
1 个评论
Shashanka
2014-7-11
Thank you for the response! The example was a simplified version of the code I'm trying to run, which has decimal inputs. I have attached an image of the actual code below for your reference. The 'AxIFN' is to be updated with the latest value of 'AxIFN1' which is calculated with each iteration. Finally giving a value of 'AxIFN' which is less than '0.4'.
Thank you!
C.J. Harris
2014-7-11
编辑：C.J. Harris
2014-7-11
0 个投票
You have an infinity loop because your seed (N) is starting at one. Note that your equation ((N*2)+(N+1))/N is in fact equal to 3+(1/N), thereby meaning you'll only get 11 (or greater) for values less than 0.125.
Not sure why you even need a loop, can't you just rearrange, therefore getting n = 1/(11-3) = 0.125.
1 个评论
Shashanka
2014-7-11
Thank you for the response! The example was a simplified version of the code I'm trying to run, which has decimal inputs. I have attached an image of the actual code below for your reference. The 'AxIFN' is to be updated with the latest value of 'AxIFN1' which is calculated with each iteration. Finally giving a value of 'AxIFN' which is less than '0.4'.
Thank you!
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