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The shuffle product was introduced by Eilenberg & Mac Lane (1953). The name "shuffle product" refers to the fact that the product can be thought of as a sum over all ways of riffle shuffling two words together: this is the riffle shuffle permutation. The product is commutative and associative. [2]
The Fisher–Yates shuffle is named after Ronald Fisher and Frank Yates, who first described it. It is also known as the Knuth shuffle after Donald Knuth. [2] A variant of the Fisher–Yates shuffle, known as Sattolo's algorithm, may be used to generate random cyclic permutations of length n instead of random permutations.
A riffle shuffle permutation of a sequence of elements is obtained by partitioning the elements into two contiguous subsequences, and then arbitrarily interleaving the two subsequences. For instance, this describes many common ways of shuffling a deck of playing cards, by cutting the deck into two piles of cards that are then riffled together.
In the mathematics of permutations and the study of shuffling playing cards, a riffle shuffle permutation is one of the permutations of a set of items that can be obtained by a single riffle shuffle, in which a sorted deck of cards is cut into two packets and then the two packets are interleaved (e.g. by moving cards one at a time from the bottom of one or the other of the packets to the top ...
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Cards lifted after a riffle shuffle, forming what is called a bridge which puts the cards back into place After a riffle shuffle, the cards cascade. A common shuffling technique is called the riffle, or dovetail shuffle or leafing the cards, in which half of the deck is held in each hand with the thumbs inward, then cards are released by the thumbs so that they fall to the table interleaved.
A simple algorithm to generate a permutation of n items uniformly at random without retries, known as the Fisher–Yates shuffle, is to start with any permutation (for example, the identity permutation), and then go through the positions 0 through n − 2 (we use a convention where the first element has index 0, and the last element has index n − 1), and for each position i swap the element ...
The formula is valid for all index values, and for any n (when n = 0 or n = 1, this is the empty product). However, computing the formula above naively has a time complexity of O(n 2), whereas the sign can be computed from the parity of the permutation from its disjoint cycles in only O(n log(n)) cost.
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