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The problem for graphs is NP-complete if the edge lengths are assumed integers. The problem for points on the plane is NP-complete with the discretized Euclidean metric and rectilinear metric. The problem is known to be NP-hard with the (non-discretized) Euclidean metric. [3]: ND22, ND23
Hermite's problem is an open problem in mathematics posed by Charles Hermite in 1848. He asked for a way of expressing real numbers as sequences of natural numbers , such that the sequence is eventually periodic precisely when the original number is a cubic irrational .
In operations research, Johnson's rule is a method of scheduling jobs in two work centers. Its primary objective is to find an optimal sequence of jobs to reduce makespan (the total amount of time it takes to complete all jobs).
The key to solving a problem recursively is to recognize that it can be broken down into a collection of smaller sub-problems, to each of which that same general solving procedure that we are seeking applies [citation needed], and the total solution is then found in some simple way from those sub-problems' solutions. Each of these created sub ...
It concerns sequences of integers in which each term is obtained from the previous term as follows: if a term is even, the next term is one half of it. If a term is odd, the next term is 3 times the previous term plus 1. The conjecture is that these sequences always reach 1, no matter which positive integer is chosen to start the sequence.
In mathematics, the Skolem problem is the problem of determining whether the values of a constant-recursive sequence include the number zero. The problem can be formulated for recurrences over different types of numbers, including integers, rational numbers, and algebraic numbers. It is not known whether there exists an algorithm that can solve ...
A topological problem with a fresh twist, and eight other new recreational puzzles 1972 May: Challenging chess tasks for puzzle buffs and answers to the recreational problems 1972 Jun: A miscellany of transcendental problems: simple to state but not at all easy to solve 1972 Jul: Amazing mathematical card tricks that do not require prestidigitation
Separate the counting sequences according to the first vote. Any sequence that begins with a vote for B must reach a tie at some point, because A eventually wins. For any sequence that begins with A and reaches a tie, reflect the votes up to the point of the first tie (so any A becomes a B, and vice versa) to obtain a sequence that begins with B.
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