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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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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 comentario
Roger Stafford
el 22 de En. de 2014
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".)
Respuestas (1)
AJ von Alt
el 21 de En. de 2014
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 comentarios
Walter Roberson
el 21 de En. de 2014
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
el 22 de En. de 2014
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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