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An algorithm is said to be exponential time, if T(n) is upper bounded by 2 poly(n), where poly(n) is some polynomial in n. More formally, an algorithm is exponential time if T(n) is bounded by O(2 n k) for some constant k. Problems which admit exponential time algorithms on a deterministic Turing machine form the complexity class known as EXP.
In computational complexity theory, the complexity class 2-EXPTIME (sometimes called 2-EXP) is the set of all decision problems solvable by a deterministic Turing machine in O(2 2 p(n)) time, where p(n) is a polynomial function of n.
In computational complexity theory, the complexity class EXPTIME (sometimes called EXP or DEXPTIME) is the set of all decision problems that are solvable by a deterministic Turing machine in exponential time, i.e., in O(2 p(n)) time, where p(n) is a polynomial function of n.
For example, the amount of time it takes to solve problems in the complexity class P grows at a polynomial rate as the input size increases, which is comparatively slow compared to problems in the exponential complexity class EXPTIME (or more accurately, for problems in EXPTIME that are outside of P, since ).
The empirical average-case complexity (time vs. problem size) of such algorithms can be surprisingly low. An example is the simplex algorithm in linear programming, which works surprisingly well in practice; despite having exponential worst-case time complexity, it runs on par with the best known polynomial-time algorithms. [27]
For every n, there exist n-state NFAs such that every subset of states is reachable from the initial subset, so that the converted DFA has exactly 2 n states, giving Θ(2 n) worst-case time complexity. [8] [9] A simple example requiring nearly this many states is the language of strings over the alphabet {0,1} in which there are at least n ...
Also, when implemented with the "shortest first" policy, the worst-case space complexity is instead bounded by O(log(n)). Heapsort has O(n) time when all elements are the same. Heapify takes O(n) time and then removing elements from the heap is O(1) time for each of the n elements. The run time grows to O(nlog(n)) if all elements must be distinct.
The sort has a known time complexity of O(n 2), and after the subroutine runs the algorithm must take an additional 55n 3 + 2n + 10 steps before it terminates. Thus the overall time complexity of the algorithm can be expressed as T(n) = 55n 3 + O(n 2). Here the terms 2n + 10 are subsumed within the faster-growing O(n 2). Again, this usage ...