BIBLIOS

  Sistema de Gestão de Referências Bibliográficas de Ciências

Modo Visitante (Login)
Need help?


Voltar

Detalhes Referência

Tipo
Artigos em Revista

Tipo de Documento
Artigo Completo

Título
Combinatorics of Jenga

Participantes na publicação
Alda Carvalho (Author)
João Pedro Neto (Author)
Dep. Informática
BioISI
Carlos Pereira dos Santos (Author)

Resumo
Jenga, a very popular game of physical skill, when played by perfect players, can be seen as a pure combinatorial ruleset. Taking that into account, it is possible to play with more than one tower; a move is made by choosing one of the towers, removing a block from there, that is, a disjunctive sum. jenga is an impartial combinatorial ruleset, i.e., Left options and Right options are the same for any position and all its followers. In this paper, we illustrate how to determine the Grundy value of a Jenga tower by showing that it may be seen as a bidimensional vector addition game. Also, we propose a class of impartial rulesets, the clock nim games, jenga being an example of that class.

Data de Publicação
2019-12-10

Instituição
FACULDADE DE CIÊNCIAS DA UNIVERSIDADE DE LISBOA

Suporte
The Australasian Journal of Combinatorics

Identificadores da Publicação
ISSN - 2202-3518

Volume
75
Fascículo
1

Número de Páginas
18
Página Inicial
87
Página Final
104

Identificadores de Qualidade
SCIMAGO Q2 (2019) - 0.55 - Discrete Mathematics and Combinatorics

Keywords
Combinatorial Game Theory


Exportar referência

APA
Alda Carvalho, João Pedro Neto, Carlos Pereira dos Santos, (2019). Combinatorics of Jenga. The Australasian Journal of Combinatorics, 75, 87-104. ISSN 2202-3518. eISSN .

IEEE
Alda Carvalho, João Pedro Neto, Carlos Pereira dos Santos, "Combinatorics of Jenga" in The Australasian Journal of Combinatorics, vol. 75, pp. 87-104, 2019.

BIBTEX
@article{42134, author = {Alda Carvalho and João Pedro Neto and Carlos Pereira dos Santos}, title = {Combinatorics of Jenga}, journal = {The Australasian Journal of Combinatorics}, year = 2019, pages = {87-104}, volume = 75 }