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plotting the number of actions for ex i=i+1 ( action), pgd=n/i ( action), pgd=n(action)

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ime sem
ime sem 2013-11-9
关闭： MATLAB Answer Bot 2021-8-20
function pgd = entier(n)
i = 2;
while i≤n/2 & mod(n,i)~-0
i = i+1;
end
if mod(n,i)==0
pgd = n/i;
else
pgd = n;
end
end
  1 个评论
Roger Stafford
Roger Stafford 2014-1-22
In this example you are seeking the smallest divisor of n which is greater than one. However, it isn't necessary to go all the way to n/2. You can stop at sqrt(n) because if there isn't a divisor by that point, you won't find any others less than n. (Of course none of this has anything to do with counting "actions".)

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回答（1 个）

AJ von Alt
AJ von Alt 2014-1-21
You can create a variable to count the number of actions taken and increment it every time an action is take. Have your function return that variable and store it for later plotting.
function [pgd , nActions] = entier(n)
nActions = 0;
i = 2;
while i<=n/2 && mod(n,i)~=0
i = i+1;
nActions = nActions + 1;
end
if mod(n,i)==0
pgd = n/i;
nActions = nActions + 1;
else
pgd = n;
nActions = nActions + 1;
end
end
  2 个评论
Walter Roberson
Walter Roberson 2014-1-21
The line
while i<=n/2 && mod(n,i)~=0
requires at least two actions, one for the division and one for the mod() calculation. I would also suggest that if assignment is to be treated as an action, that comparison would have to be an action as well, remembering the && comparison should also be an action. Then one needs to take into account that && "short circuits" and so the mod() and comparison to 0 would not require actions if i<=n/2 is false.
AJ von Alt
AJ von Alt 2014-1-22
Good catch! If the user wants to count the modulo and logical operations in addition to the ones listed in the title, some rework would be in order. Ime used & instead of && in his original implementation, so if he sticks with elementwise AND rather than my short circuit AND, he at least won't have to worry about short circuiting.

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