[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"doc-detail-85387-en":3,"doc-seo-85387-105":30,"detail-sidebar-cat-0-en-105":91},{"code":4,"msg":5,"data":6},0,"success",{"doc_id":7,"user_id":8,"nickname":9,"user_avatar":10,"doc_module":4,"category_id":11,"category_name":12,"doc_title":13,"doc_description":14,"doc_content":15,"file_id":16,"file_url":17,"file_type":18,"file_size":19,"view_count":20,"is_deleted":4,"is_public":21,"is_downloadable":21,"audit_status":21,"page_count":22,"language":23,"language_code":24,"site_id":25,"html_lang":24,"table_of_contents":26,"faqs":27,"seo_title":13,"seo_description":14,"update_tm":28,"read_time":29},85387,1099513958607,"Jiven","https://ap-avatar.wpscdn.com/avatar/100002390cf8733938c?x-image-process=image/resize,m_fixed,w_180,h_180&k=1778829742770036399",8,"Research & Report","Maker-Breaker Solved in Polynomial Time on Hypergraphs of Rank 3","Maker-Breaker positional games are studied on hypergraphs of rank 3. Maker and Breaker alternately choose vertices, and Maker wins exactly when she claims all vertices of some edge. While deciding the winner is PSPACE-complete even for 4-uniform hypergraphs, earlier work resolved the linear subcase for rank 3. This paper confirms a conjecture for general rank-3 hypergraphs by giving a structural characterization of outcomes and both players’ optimal strategies, leading to a polynomial-time algorithm and logarithmic-round bounds for Maker wins.","Maker-Breaker is solved in polynomial time on hypergraphs of rank 3  \nFlorian Galliot 1 , Sylvain Gravier 2 , and Isabelle Sivignon3  \n1 Aix-Marseille Université, CNRS, I2M, UMR 7373, 13453 Marseille, France  \n2 Univ. Grenoble Alpes, CNRS, Institut Fourier, 38000 Grenoble, France  \n3 Univ. Grenoble Alpes, CNRS, Grenoble INP, GIPSA-lab, 38000 Grenoble, France  \narXiv :2209 . 12819v4 [ cs .DM] 11 Jul 2026  \nAbstract  \nIn the Maker-Breaker positional game, Maker and Breaker take turns picking vertices of ahypergraph H, and Maker wins if and only if she possesses all the vertices of some edge of  \nH. Deciding the outcome (i.e., which player has a winning strategy) is a PSPACE-complete problem even when restricted to 4-uniform hypergraphs [Gal25] . As for hypergraphs of rank 3, Kutz [Kut05] has solved the linear subcase (i.e., any two distinct edges intersect on at most one vertex), by obtaining a structural characterization of the outcome and a polynomial-time algorithm to decide it. A conjecture by Rahman and Watson [RW20] implies that the same results can be obtained for general hypergraphs of rank 3, which we confirm in this paper. We provide a structural characterization of the outcome anda description of both players’ optimal strategies, all based on intersections of some key subhypergraph collections. From this, we derive a polynomial-time algorithm, thus closing the complexity gap for Maker-Breaker games relative to the size of the edges. Another corollary of our structural result is that, if Maker has a winning strategy on a hypergraph of rank 3, then she can ensure to win the game within a number of rounds that is logarithmic in the number of vertices.  \n1 Introduction  \nMaker-Breaker games. A positional game is played on a hypergraph H, with two players who take turns picking vertices of H. There are several possible conventions of play, which determine the winner of the game. Let us mention two major ones:  \n• In the Maker-Maker convention, the winner is the player who first possesses all vertices of some edge of H. If no player achieves this, then the game ends in a draw. The first general formulation of this convention goes back to Hales and Jewett [HJ63], while the first general results are due to Erdős and Selfridge [ES73] . The game of tic-tac-toe and its generalizations [HJ63 , Bec08] are the most famous examples of Maker-Maker positional games.  \n• In the Maker-Breaker convention, one player (\"Maker\") wins if she possesses all vertices of some edge of H, while the other (\"Breaker\") wins if he can prevent this from happening. No draw is possible here. The first general formulation of this convention is due to Chvátaland Erdős [CE78] . The board game Hex [Hei42 , Nas52] and the Shannon switching game [Gar61, Leh64 , CE78] are the most famous examples of Maker-Breaker positional games.  \nThis paper deals with the Maker-Breaker convention, which is the most studied in the literature as it possesses some nice properties (e.g., monotonicity properties) . We always assume that Maker plays first: if Breaker plays first, then we can consider all possibilities of his first pick to reduce to the case where Maker plays first. Given a hypergraph H, there are two possibilities for the outcome of the game: either Maker or Breaker has a winning strategy on H , and we say H is a Maker win or a Breaker win accordingly. From an algorithmic point of view, the natural decision problem MakerBreaker takes an input hypergraph H and returns \"yes\"if H is a Maker win or \"no\" if H is a Breaker win.  \nThe Maker-Breaker game can be interpreted as a propositional logic problem. Indeed, it is directly linked with the Quantified Boolean Formula problem (QBF), also known as Quantified SAT (QSAT), which differs from SAT in that each quantifier may be existential or universal. These types can be assumed to alternate, so that QBF comes down to a game played on a formula in conjunctive normal form (CNF) where two players take turns choosing truth values for the","cbCairNRS2qWmM7n","https://ap.wps.com/l/cbCairNRS2qWmM7n","pdf",1084974,4,1,66,"English","en",105,"# Introduction\n## Maker-Breaker games and conventions\n## Logic and QBF connections\n## Problem focus and structural approach","[{\"question\":\"What determines the winner in the Maker-Breaker hypergraph game?\",\"answer\":\"Players alternately pick vertices of a hypergraph H. Maker wins if and only if she owns all vertices of some hyperedge of H; Breaker aims to prevent this, and no draw is possible.\"},{\"question\":\"What is known about the computational complexity of deciding the winner?\",\"answer\":\"Deciding the outcome of Maker-Breaker is PSPACE-complete even when restricted to 4-uniform hypergraphs.\"},{\"question\":\"What does the paper contribute for hypergraphs of rank 3?\",\"answer\":\"It proves a conjecture for general rank-3 hypergraphs by providing a structural characterization of the outcome and descriptions of both players’ optimal strategies. Using this, it derives a polynomial-time decision algorithm and shows Maker can win within logarithmically many rounds when she has a winning strategy.\"}]",1784203071,166,{"code":4,"msg":31,"data":32},"ok",{"site_id":25,"language":24,"slug":33,"title":13,"keywords":34,"description":14,"schema_data":35,"social_meta":86,"head_meta":88,"extra_data":90,"updated_unix":28},"maker-breaker-solved-in-polynomial-time-on-hypergraphs-of-rank-3","",{"@graph":36,"@context":85},[37,53,68],{"@type":38,"itemListElement":39},"BreadcrumbList",[40,44,48,51],{"item":41,"name":42,"@type":43,"position":21},"https://docshare.wps.com","Home","ListItem",{"item":45,"name":46,"@type":43,"position":47},"https://docshare.wps.com/document/","Document",2,{"item":49,"name":12,"@type":43,"position":50},"https://docshare.wps.com/document/research-report/",3,{"item":52,"name":13,"@type":43,"position":20},"https://docshare.wps.com/document/maker-breaker-solved-in-polynomial-time-on-hypergraphs-of-rank-3/85387/",{"url":52,"name":13,"@type":54,"author":55,"headline":13,"publisher":57,"fileFormat":60,"inLanguage":24,"description":14,"dateModified":61,"datePublished":62,"encodingFormat":60,"isAccessibleForFree":63,"interactionStatistic":64},"DigitalDocument",{"name":9,"@type":56},"Person",{"url":41,"name":58,"@type":59},"DocShare","Organization","application/pdf","2026-07-25","2026-07-16",true,{"@type":65,"interactionType":66,"userInteractionCount":20},"InteractionCounter",{"@type":67},"ViewAction",{"@type":69,"mainEntity":70},"FAQPage",[71,77,81],{"name":72,"@type":73,"acceptedAnswer":74},"What determines the winner in the Maker-Breaker hypergraph game?","Question",{"text":75,"@type":76},"Players alternately pick vertices of a hypergraph H. Maker wins if and only if she owns all vertices of some hyperedge of H; Breaker aims to prevent this, and no draw is possible.","Answer",{"name":78,"@type":73,"acceptedAnswer":79},"What is known about the computational complexity of deciding the winner?",{"text":80,"@type":76},"Deciding the outcome of Maker-Breaker is PSPACE-complete even when restricted to 4-uniform hypergraphs.",{"name":82,"@type":73,"acceptedAnswer":83},"What does the paper contribute for hypergraphs of rank 3?",{"text":84,"@type":76},"It proves a conjecture for general rank-3 hypergraphs by providing a structural characterization of the outcome and descriptions of both players’ optimal strategies. 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