How Complex is to Solve a Hard Problem with Accepting Splicing Systems

Victor Mitrana, Andrei Păun, Mihaela Păun

2019

Abstract

We define a variant of accepting splicing system that can be used as a problem solver. A condition for halting the computation on a given input as well as a condition for making a decision as soon as the computation has stopped is considered. An algorithm based on this accepting splicing system that solves a well-known NP-complete problem, namely the 3-colorability problem is presented. We discuss an efficient solution in terms of running time and additional resources (axioms, supplementary symbols, number of splicing rules. More precisely, for a given graph with n vertices and m edges, our solution runs in O(nm) time, and needs O(mn2) other resources. Two variants of this algorithm of a reduced time complexity at an exponential increase of the other resources are finally discussed.

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Paper Citation


in Harvard Style

Mitrana V., Păun A. and Păun M. (2019). How Complex is to Solve a Hard Problem with Accepting Splicing Systems.In Proceedings of the 4th International Conference on Complexity, Future Information Systems and Risk - Volume 1: COMPLEXIS, ISBN 978-989-758-366-7, pages 27-35. DOI: 10.5220/0007715900270035


in Bibtex Style

@conference{complexis19,
author={Victor Mitrana and Andrei Păun and Mihaela Păun},
title={How Complex is to Solve a Hard Problem with Accepting Splicing Systems},
booktitle={Proceedings of the 4th International Conference on Complexity, Future Information Systems and Risk - Volume 1: COMPLEXIS,},
year={2019},
pages={27-35},
publisher={SciTePress},
organization={INSTICC},
doi={10.5220/0007715900270035},
isbn={978-989-758-366-7},
}


in EndNote Style

TY - CONF

JO - Proceedings of the 4th International Conference on Complexity, Future Information Systems and Risk - Volume 1: COMPLEXIS,
TI - How Complex is to Solve a Hard Problem with Accepting Splicing Systems
SN - 978-989-758-366-7
AU - Mitrana V.
AU - Păun A.
AU - Păun M.
PY - 2019
SP - 27
EP - 35
DO - 10.5220/0007715900270035