Challenges
Datasets
Workspaces
Papers
Discussions
Leaderboard
Log inSign up
Challenges
Datasets
Workspaces
Papers
Discussions
Leaderboard
Blog
Job Board
Q3AS

© 2026 Aqora Quantum S.A.S.

TermsPrivacyLegal Notice

arXiv:1008.5121 · Physics Letters A

Parrondoʼs game using a discrete-time quantum walk

C. M. Chandrashekar, Subhashish Banerjee

Published Mar 5, 2011• 40 citations
arXivHTMLPDF
DOI
10.1016/j.physleta.2011.02.071

Related papers

Native quantum games from interacting discrete-time quantum walks

Rashid Ahmad

Quantum Information Processing•May 18, 2026•0 citations

Parrondo’s game of quantum search based on quantum walk

Taisuke Hosaka, Norio Konno

arXiv:2303.06579•Quantum Information Processing•Jun 24, 2024•3 citations

A Quantum Analogue of Parrondo's Game

Clement Boateng Ampadu

We consider the discrete-time quantum walk in the plane, and present a quantum implementation of Parrondo's game for four players. Physical significance of the game strategies are also discussed.

arXiv:1104.5275•Apr 28, 2011•1 citation

Quantum models of Parrondo's games

Adrian P. Flitney, Derek Abbott

Physica A Statistical Mechanics and its Applications•May 1, 2003•57 citations

Quantum Parrondo’s Games Under Decoherence

Salman Khan, M. Ramzan, M. K. Khan

arXiv:0909.2897•International Journal of Theoretical Physics•Nov 3, 2009•22 citations

Qunatum Parrondo's games constructed by quantum random walk

Min Li, Yongsheng Zhang, Guang‐Can Guo

We construct a Parrondo's game using discrete time quantum walks. Two lossing games are represented by two different coin operators. By mixing the two coin operators $U_{A}(α_{A},β_{A},γ_{A})$ and $U_{B}(α_{B},β_{B},γ_{B})$, we may win the game. Here we mix the two games in position instead of time. With a number of selections of the parameters, we can win the game with sequences ABB, ABBB, \emph{et al}. If we set $β_{A}=45^{\circ},γ_{A}=0,α_{B}=0,β_{B}=88^{\circ}$, we find the game 1\emph{}with {\normalsize $U_{A}^{S}=U^{S}(-51^{\circ},45^{\circ},0)$, $U_{B}^{S}=U^{S}(0,88^{\circ},-16^{\circ})$ will win and get the most profit.}If we set $α_{A}=0,β_{A}=45^{\circ},α_{B}=0,β_{B}=88^{\circ}$ and{\normalsize{} the game 2 with $U_{A}^{S}=U^{S}(0,45^{\circ},-51^{\circ})$, $U_{B}^{S}=U^{S}(0,88^{\circ},-67^{\circ})$, will win most. And}game 1\emph{}{\normalsize is equivalent to the}game\emph{}2\emph{}with the changes of sequences and steps. But at a large enough steps, the game will loss at last.

arXiv:1303.6831•Mar 27, 2013•0 citations
We construct a Parrondo's game using discrete time quantum walks. Two lossing games are represented by two different coin operators. By mixing the two coin operators UA(αA,βA,γA)U_{A}(α_{A},β_{A},γ_{A})UA​(αA​,βA​,γA​) and UB(αB,βB,γB)U_{B}(α_{B},β_{B},γ_{B})UB​(αB​,βB​,γB​), we may win the game. Here we mix the two games in position instead of time. With a number of selections of the parameters, we can win the game with sequences ABB, ABBB, \emph{et al}. If we set βA=45∘,γA=0,αB=0,βB=88∘β_{A}=45^{\circ},γ_{A}=0,α_{B}=0,β_{B}=88^{\circ}βA​=45∘,γA​=0,αB​=0,βB​=88∘, we find the game 1\emph{}with {\normalsize UAS=US(−51∘,45∘,0)U_{A}^{S}=U^{S}(-51^{\circ},45^{\circ},0)UAS​=US(−51∘,45∘,0), UBS=US(0,88∘,−16∘)U_{B}^{S}=U^{S}(0,88^{\circ},-16^{\circ})UBS​=US(0,88∘,−16∘) will win and get the most profit.}If we set αA=0,βA=45∘,αB=0,βB=88∘α_{A}=0,β_{A}=45^{\circ},α_{B}=0,β_{B}=88^{\circ}αA​=0,βA​=45∘,αB​=0,βB​=88∘ and{\normalsize{} the game 2 with UAS=US(0,45∘,−51∘)U_{A}^{S}=U^{S}(0,45^{\circ},-51^{\circ})UAS​=US(0,45∘,−51∘), UBS=US(0,88∘,−67∘)U_{B}^{S}=U^{S}(0,88^{\circ},-67^{\circ})UBS​=US(0,88∘,−67∘), will win most. And}game 1\emph{}{\normalsize is equivalent to the}game\emph{}2\emph{}with the changes of sequences and steps. But at a large enough steps, the game will loss at last.