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Start your free trialJames Estrada
Full Stack JavaScript Techdegree Student 25,866 PointsPolynomial Runtime (O(n^k)) vs Exponential Runtime(O(k^n))
How do you choose n and k when determining if an algorithm's complexity is of O(n^k) or O(k^n)? In the padlock example, when the size is 2, there're 10^2 possible combinations. Isn't this running in quadratic time?
2 Answers
Alexander Davison
65,469 PointsNo is the short answer.
10^2 seems "quadratic-y", but notice that the varying variable is not the base (10), but is the exponent!
n^2 (a quadratic) looks like this for the varying n
s:
1^2 2^2 3^2 4^2 5^2 ...
But the exponential 10^n look like:
10^1 10^2 10^3 10^4 10^5 ...
Notice how the exponent, not the base, is varying.
Exponentials explode off even more than quadratics/cubics, so generally it is very bad to have an exponential algorithm.
G H Mahimaanvita
16,234 PointsHello James Estrada, I hope this helps:
There are 2 boxes to be filled using 10 numbers(0-9).
number of combinations =(number of ways in which the box1 can be filled)x(number of ways in which box 2 can be filled)...(number of ways in which the box n can be filled)
either box can be filled by 0 or 1 or 2 ....or 9 (10 ways).
therefore, total number of combinations is (10) X (10) or 10^2.
when number of boxes increase to 3, '2' changes to '3' but the range of numbers will still be (0-9). So, '10' does not change. (the answer would then be 10 x 10 x 10 = 10^3)
So, k=10(the number of items in range ) and n=2(number of boxes)
James Estrada
Full Stack JavaScript Techdegree Student 25,866 PointsJames Estrada
Full Stack JavaScript Techdegree Student 25,866 PointsThank you for clarifying how an exponential algorithm can grow more than a polynomial one, but I still have the question: How do you know if n or k should be the varying variable? How do you choose which is which?
Alexander Davison
65,469 PointsAlexander Davison
65,469 PointsUsually
n
is the varying variable, andk
refers to some constant