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Parity only depends on the number of ones and is therefore a symmetric Boolean function.. The n-variable parity function and its negation are the only Boolean functions for which all disjunctive normal forms have the maximal number of 2 n − 1 monomials of length n and all conjunctive normal forms have the maximal number of 2 n − 1 clauses of length n.
Parity can be generalized to Coxeter groups: one defines a length function ℓ(v), which depends on a choice of generators (for the symmetric group, adjacent transpositions), and then the function v ↦ (−1) ℓ(v) gives a generalized sign map.
For example, −4, 0, and 82 are even numbers, while −3, 5, 7, and 21 are odd numbers. ... The parity function maps a number to the number of 1's in its binary ...
Self-concordant function; Semi-differentiability; Semilinear map; Set function; List of set identities and relations; Shear mapping; Shekel function; Signomial; Similarity invariance; Soboleva modified hyperbolic tangent; Softmax function; Softplus; Splitting lemma (functions) Squeeze theorem; Steiner's calculus problem; Strongly unimodal ...
Define the parity vector function Q acting on as = = (()). The function Q is a 2-adic isometry . [ 27 ] Consequently, every infinite parity sequence occurs for exactly one 2-adic integer, so that almost all trajectories are acyclic in Z 2 {\displaystyle \mathbb {Z} _{2}} .
However, the Levi-Civita symbol is a pseudotensor because under an orthogonal transformation of Jacobian determinant −1, for example, a reflection in an odd number of dimensions, it should acquire a minus sign if it were a tensor. As it does not change at all, the Levi-Civita symbol is, by definition, a pseudotensor.
Parity learning is a problem in machine learning. An algorithm that solves this problem must find a function ƒ, given some samples (x, ƒ(x)) and the assurance that ƒ computes the parity of bits at some fixed locations. The samples are generated using some distribution over the input.
An example of problem in NC 1 is the parity check on a bit string. [6] The problem consists in counting the number of 1s in a string made of 1 and 0. A simple solution consists in summing all the string's bits.