Reputation: 460
I have a simple math algo. All it does is it takes an input and finds i,j such that i^2 + j^2 = input with the restriction that j >= i (so that it doesn't print it's counterpart e.g., 2^2 + 3^2 == 3^2 + 2^2 but I only need the latter as j >= i)
For my code, I did the following: I have 2 for loops, first loop for i and second loop for j. Takes both i and j values and test if i^2 + j^2 == input and if j >= i. if yes, print it and update count.
The problem is, with large sums of values, it takes a very long time as it loops twice from 1 to 2000 and then 1 to 2000 again.
def some_mathfn(n):
count = 0
for i in range(1,n+1):
for j in range(1,n+1):
if(i**2 + j**2 == n and j >= i):
g = print(i, '^2 + ', j,'^2')
count += 1
return count
some_mathfn(2001)
Upvotes: 0
Views: 1678
Reputation: 1503839
You've got an O(n2) algorithm for no obvious reason. It's easy to make this O(n1/2)...
n/2
(for variable i
) - because when i
is greater than sqrt(n/2)
then i*i + j*j
will be greater than n
for any j
greater than i
.
n
, because i
j
The last two steps are effectively just checking that the square root of n - i*i
is actually an integer, but in some cases (for very large values of n) finding the nearest integer and then checking the condition could be a more reliable approach, in order to avoid floating point limitations causing issues, where the nearest-representable double to the theoretical result could be an integer, despite that actual result not being an integer. This would only happen for really large values of n, but...
Upvotes: 10