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If the array contains all non-negative numbers, then the problem is trivial; a maximum subarray is the entire array. If the array contains all non-positive numbers, then a solution is any subarray of size 1 containing the maximal value of the array (or the empty subarray, if it is permitted).
The loop at the center of the function only works for palindromes where the length is an odd number. The function works for even-length palindromes by modifying the input string. The character '|' is inserted between every character in the inputs string, and at both ends. So the input "book" becomes "|b|o|o|k|". The even-length palindrome "oo ...
Induction: If the claim is true for arrays of length l ≥ 1, then we show that the claim is true for arrays of length l +1 (together with the base case this proves that the claim is true for arrays of all lengths). Since the claim depends on whether l is odd or even, we prove each case separately.
The problem is NP-hard even when all input integers are positive (and the target-sum T is a part of the input). This can be proved by a direct reduction from 3SAT. [2] It can also be proved by reduction from 3-dimensional matching (3DM): [3] We are given an instance of 3DM, where the vertex sets are W, X, Y.
For shared items: define a 2-dimensional array such that (,) = iff there exists a solution giving a total weight of w i to agent i. It is possible to enumerate all possible utility profiles in time O ( n ⋅ c 2 ) {\displaystyle O(n\cdot c^{2})} where n is the number of items and c is the maximum size of an item.
is how one would use Fortran to create arrays from the even and odd entries of an array. Another common use of vectorized indices is a filtering operation. Consider a clipping operation of a sine wave where amplitudes larger than 0.5 are to be set to 0.5. Using S-Lang, this can be done by
The length of is more than the length of but it is possible that not all elements in this array are used by the algorithm (in fact, if the longest increasing sequence has length then only [], …, [] are used by the
The number of perfect matchings of the complete graph K n (with n even) is given by the double factorial (n – 1)!!. [12] The crossing numbers up to K 27 are known, with K 28 requiring either 7233 or 7234 crossings. Further values are collected by the Rectilinear Crossing Number project. [13] Rectilinear Crossing numbers for K n are