O(10^N): trying to break a password by testing every possible combination (assuming numerical password of length N) p.s. why is your last example is of complexity O(infinity) ? it’s linear search O(N) .. there are less than 7 billion people in the world.
It is certainly not possible with a plain singly-linked list. Sketch proof: to examine the last node of a singly-linked list, we must perform n-1 operations of following a “next” pointer [proof by induction on the fact that there is only one reference to the k+1th node, and it is in the kth node, and … Read more
Since the big-O notation is not about comparing absolute complexity, but only relative one, O(n+n) is in fact the same as O(n). Each time you double x, your code will take twice as long as it did before and that means O(n). Whether your code runs through 2, 4 or 20 loops doesn’t matter, since … Read more
It sounds like what you are asking for is an extention of the Halting Problem. I do not believe that such a thing is possible, even in theory. Just answering the question “Will this line of code ever run?” would be very difficult if not impossible to do in the general case. Edited to add: … Read more
Make the graph bigger and you’ll see that O(n logn) isn’t quite a straight line. But yes, it is pretty near to linear behaviour. To see why, just take the logarithm of a few very large numbers. For example (base 10): log(1000000) = 6 log(1000000000) = 9 … So, to sort 1,000,000 numbers, an O(n … Read more
there is a very detailed table on python wiki which answers your question. However, in your particular example you should use enumerate to get an index of an iterable within a loop. like so: for i, item in enumerate(some_seq): bar(item, i)
In short: Hamiltonian cycle : O(N!) in the worst case WordBreak and StringSegment : O(2^N) NQueens : O(N!) Note: For WordBreak there is an O(N^2) dynamic programming solution. More details: In Hamiltonian cycle, in each recursive call one of the remaining vertices is selected in the worst case. In each recursive call the branch factor … Read more
Decision Tree implementations differ primarily along these axes: the splitting criterion (i.e., how “variance” is calculated) whether it builds models for regression (continuous variables, e.g., a score) as well as classification (discrete variables, e.g., a class label) technique to eliminate/reduce over-fitting whether it can handle incomplete data The major Decision Tree implementations are: ID3, or … Read more
The confusion is actually caused by your “remove” function. In java, there are two remove functions. remove() -> This is to remove the head/root, it takes O(logN) time. remove(Object o) -> This is to remove an arbitrary object. Finding this object takes O(N) time, and removing it takes O(logN) time.
Short answer: No. Long answer: Games actually aren’t all that complicated. It depends on what you’re talking about when you say “games” but the two contenders for most complex games would be 3D games and online games (particularly massively online games). The complication in 3D games comes from taking a model of a world and … Read more