Combinatorics and Graph Theory

1911 Submissions

[2] viXra:1911.0203 [pdf] replaced on 2019-11-12 02:43:47

Yes, P = NP=NP-Complete=NP-hard, says the NP-hard Traveling Salesman Problem

Authors: A. A. Frempong
Comments: 38 Pages. Copyright © by A. A. Frempong

The solution of the traveling salesman problem (TSP) in this paper makes this problem no longer an NP-hard problem, but rather, a P problem. The TSP was solved in polynomial time and its solution was also correctly checked in polynomial time. Also solved were an NP-Complete TSP, and six other NP-Complete problems. The TSP solution killed two (three) birds with one stone, because its solution made the NP-hard problems and NP-Complete problems become P problems. The shortest route as well as the longest route for the salesman to visit each of nine cities once and return to the base city was determined. In finding the shortest route, the first step was to arrange the data of the problem in increasing order, since one's interest is in the shortest distances; but in finding the longest route, the first step was to arrange the data of the problem in decreasing order, since one's interest is in the longest distances. For the shortest route, the main principle is that the shortest route is the sum of the shortest distances such that the salesman visits each city once and returns to the starting city; but for the longest route, the main principle is that the longest route is the sum of the longest distances such that the salesman visits each city once and returns to the starting city. Since ten cities are involved, ten distances would be needed for the salesman to visit each of nine cities once and return to the starting city. One started the construction of the shortest route using only the shortest ten distances; and if a needed distance was not among the set of the shortest ten distances, one would consider distances longer than those in the set of the shortest ten distances. For the longest route, the construction began using only the longest ten distances; and if a needed distance was not among the set of the longest ten distances, one would consider distances shorter than those in the set of the longest ten distances. It was found out that even though, the length of the shortest or the longest route is unique, the sequence of the cities involved is not unique. The approach used in this paper can be applied in work-force project management and hiring, as well as in a country's work-force needs and immigration quota determination. Since approaches that solve the TSP and NP-Complete problems can also solve other NP problems, and the TSP and NP-Complete problems have been solved, all NP problems can be solved. If all NP problems can be solved, then all NP problems are P problems, and therefore, P is equal to NP. The CMI Millennium Prize requirements have been satisfied
Category: Combinatorics and Graph Theory

[1] viXra:1911.0056 [pdf] submitted on 2019-11-04 08:50:15

Refutation of the Chromatic Number of Product of Two ℵ1-Chromatic Graphs as Countable

Authors: Colin James III
Comments: 1 Page. © Copyright 2019 by Colin James III All rights reserved. Note that Disqus comments here are not read by the author; reply by email only to: info@cec-services dot com. Include a list publications for veracity. Updated abstract at ersatz-systems.com.

We evaluate the chromatic number of the product of two ℵ1-chromatic graphs as countable. It is not tautologous as are then also the derived consistent and relatively consistent conjectures for k=3. These results form a non tautologous fragment of the universal logic VŁ4.
Category: Combinatorics and Graph Theory