Single-Plyer nd Two-Plyer Buttons & Sissors Gmes (Extended Astrt) Kyle Burke 1, Erik D. Demine 2, Hrrison Gregg 3, Roert A. Hern 4, Adm Hestererg 2, Mihel Hoffmnn 5, Hiro Ito 6, Irin Kostitsyn 7, Jody Leonrd 3, Mrten Löffler 8, Aron Sntigo 3, Christine Shmidt 9, Ryuhei Uehr 10, Yushi Uno 11, nd Aron Willims 3 Astrt. We study the omputtionl omplexity of the Buttons & Sissors gme nd otin shrp thresholds with respet to severl prmeters. Speifilly we show tht the gme is NP-omplete for C = 2 olors ut polytime solvle for C = 1. Similrly the gme is NP-omplete if every olor is used y t most F = 4 uttons ut polytime solvle for F 3. We lso onsider restritions on the ord size, ut diretions, nd ut sizes. Finlly, we introdue severl nturl two-plyer versions of the gme nd show tht they re PSPACE-omplete. 1 Introdution Buttons & Sissors is single-plyer puzzle y KyWorks. The gol of eh level is to remove every utton y sequene of horizontl, vertil, nd digonl uts, s illustrted y Fig. 1. It is NP-omplete to deide if given level is solvle [2]. We study severl restrited versions of the gme nd show tht some remin hrd, wheres others n e solved in polynomil time. We lso onsider nturl extensions to two plyer gmes whih turn out to e PSPACE-omplete. Setion 2 egins with preliminries, then we disuss one-plyer puzzles in Setion 3 nd two-plyer gmes in Setion 4. Open prolems pper in Setion 5. Due to spe restritions, some proofs re skethed or omitted. A full version of this rtile n e found on rxiv. 1 Plymouth Stte University, kgurke@plymouth.edu 2 Msshusetts Institute of Tehnology, {edemine,hester}@mit.edu 3 Brd College t Simon s Rok, {hgregg11,jleonrd11,sntigo11,willims}@simons-rok.edu 4 o@hern.to 5 ETH Zürih, hoffmnn@inf.ethz.h 6 The University of Eletro-Communitions, itohiro@ue..jp 7 Tehnishe Universiteit Eindhoven, i.kostitsyn@tue.nl. Supported in prt y NWO projet no. 639.023.208. 8 Universiteit Utreht, m.loffler@uu.nl 9 Linköping University, hristine.shmidt@liu.se. Supported in prt y grnt 2014-03476 from Sweden s innovtion geny VINNOVA. 10 Jpn Advned Institute of Siene nd Tehnology, uehr@jist..jp 11 Osk Prefeture University, uno@mi.s.oskfu-u..jp
2 3 1 4 4 2 1 2 8 6 5 5 8 3 7 6 7 3 8 9 9 9 () () () Fig. 1. () Level 7 in the Buttons & Sissors pp is n m n = 5 5 grid with C = 5 olors, eh used t most F = 7 times; () solution using nine uts with sizes in S = {2,3} nd diretions d = (no vertil ut is used); () gdget used in Theorem 5. 2 Preliminries A Buttons & Sissors ord B is n m n grid, where eh grid position is either empty or oupied y utton with one of C different olors. A ut is given y two distint uttons 1, 2 of the sme olor tht shre either the x-oordinte, the y-oordinte, or re loted on the sme digonl (45 nd 45 ). The size s of ut is the numer of uttons on the line segment 1 2 nd so s 2. A ut is fesile for B if 1 2 only ontins uttons of single olor. When fesile ut is pplied to ord B, the resulting ord B is otined y sustituting the uttons of olor on 1 2 with empty grid entries. A solution to ord B is sequene of ords nd fesile uts B 1,x 1,B 2,x 2,...,B t,x t,b t+1, where B t+1 is empty, nd eh ut x i is fesile for B i nd retes B i+1. Eh instne n e prmeterized s follows (see Fig. 1 for n exmple): 1. The ord size m n. 2. The numer of olors C. 3. The mximum frequeny F of n individul olor. 4. The ut diretions d n e limited to d {,,, }. 5. The ut size set S limits fesile uts to hving size s S. Eh d {,,, } is set of ut diretions (i.e. for horizontl nd vertil). We limit ourselves to these options euse n m n ord n e rotted 90 to n equivlent n m ord, or 45 to n equivlent k k ord for k = m+n 1 with lnk squres. Similrly, we n sher the grid y pdding row i with i 1 lnks on the left nd m i lnks on the right whih onverts d = to d =. We otin the fmily of gmes elow (B&S[n n,,,,{2,3}](b) is the originl): Deision Prolem: B&S[m n,c,f,d,s](b). Input: An m n ord B with uttons of C olors, eh used t most F times. Output: True B is solvle with uts of size s S nd diretions d. Now we provide three oservtions for lter use. First note tht single ut of size s n e omplished y uts of size s 1,s 2,...,s k so long s s = s 1 + s 2 + + s k nd s i 2 for ll i. Seond note tht removing ll uttons of single olor from solvle instne nnot result in n unsolvle instne.
3 Remrk 1. A ord n e solved with ut sizes S = {2,3,...} if nd only if it n e solved with ut sizes S = {2,3}. Also, {3,4,...} nd {3,4,5} re equivlent. Remrk 2. If ord B is otined from ord B y removing every utton of single olor, then B&S[m n,c,f,d,s](b) = B&S[m n,c,f,d,s](b ). 3 Single-Plyer Puzzle 3.1 Bord Size We solve one row prolems elow, nd give onjeture for two rows in Setion 5. Theorem 1. B&S[1 n,,,,{2,3}](b) is polytime solvle. Proof. Consider the following ontext-free grmmr, S ε SS xsx xsxsx where is n empty squre nd x {1,2,...,C}. By Remrk 1, the solvle 1 n ords re in one-to-one orrespondene with the strings in this lnguge. 3.2 Numer of Colors Hrdness for 2 olors. We egin with strightforwrd redution from 3SAT. The result will e strengthened lter y Theorem 7 using more diffiult proof. Theorem 2. B&S[n n,2,,,{2,3}](b) is NP-omplete. Proof Sketh: A vrile gdget hs its own row with extly three uttons. The middle utton is lone in its olumn, nd must e mthed with t lest one of the other two in the vrile row. If the left utton is not used in this mth, we onsider the vrile set to true. If the right utton is not used, we onsider the vrile set to flse. A utton not used in vrile is n ville output, nd n then serve s n ville input to e used in other gdgets. Every luse gdget hs its own olumn, with extly four uttons. The topmost utton (luse utton) is lone in its row; the others re inputs. If t lest one of these is n ville input, then we n mth the luse utton with ll ville inputs. We onstrut one luse gdget per formul luse, onneting its inputs to the pproprite vrile outputs. Then, we n ler ll the luses just when we hve mde vrile seletions tht stisfy the formul. The vriles re onneted to the luses vi multi-purpose split gdget (Fig. 2()). Unlike the vrile nd the luse, this gdget uses uttons of two olors. The ottom utton is n input; the top two re outputs. If the input utton is ville, we n mth the middle row of the gdget s shown in Fig. 2(), leving the output uttons ville. But if the input is not ville, then the only wy the middle row n e lered is to first ler the red uttons in vertil pirs, s shown in Fig. 2(); then the output uttons re not ville. We provide further desription of the split gdget nd omplete the proof in the full version of this rtile.
4 () () () Fig. 2. Split gdget () nd the two possile wys to ler it () nd (). = e v w v w u 1 u 2 = u 3 u 4 u 5 u 6 e u 1 u 2 v w v w p 1 p 2 p 3 p 1 p 2 p 3 d = d d = d Fig. 3. Top-left: splitting 3-yles when there re no djent tringles to edge e; top-right: splitting 3-yles when e hs djent tringles (shded). Bottom-left: onstruting G from four uts loking eh other in yle; ottom-right: onstruting G from the sme uts fter ressigning the loking uttons Polynomil-Time Algorithm for 1-olor nd ny ut diretions. Given n instne B with C = 1 olor nd ut diretions d {,,, }, we onstrut hypergrph G tht hs one node per utton in B. A set of nodes is onneted with hyperedge if the orresponding uttons lie on the sme horizontl, vertil, or digonl line whose diretion is in d, i.e., they n potentilly e removed y the sme ut. By Remrk 1 it is suffiient to onsider hypergrph with only 2- nd 3-edges. A solution to B orresponds to perfet mthing in G. For lrity, we shll ll 3-edge in G tringle, nd 2-edge simply n edge. Cornuéjols et l. [1] showed how to ompute perfet K 2 nd K 3 mthing in grph in polynomil time. However, their result is not diretly pplile to our grph G yet, s we need to find mthing tht onsists only of edges nd proper tringles, nd voids K 3 s formed y yles of three edges. To pply [1] we onstrut grph G y dding verties to eliminte ll yles of three edges s follows (see top of Fig. 3). Strt with G = G. Consider n e = (v,w) G in 3-yle ( yle of three edges). There re two ses: e is not djent to ny tringle in G, or e is djent to some tringles in G. In the first se we dd verties u 1 nd u 2 tht split e into three edges (v,u 1 ), (u 1,u 2 ), nd (u 2,w). In the seond se, when e is djent to k tringles, we dd 2k verties u 1,u 2...,u 2k long e, nd reple every p i vw with p i vu 2i 1.
5 Lemm 1. There exists perfet edge- nd tringle-mthing in G iff there exists perfet edge- nd tringle-mthing in G. Proof. Given perfet mthing M in G, we onstrut perfet mthing M in G. Consider e = (v,w) in G. If e is not djent to ny tringles in G, then if e M then dd edges (v,u 1 ) nd (u 2,w) of G to M (oth v nd w re overed y e, nd ll v, w, u 1, nd u 2 re overed y M ); if e M then dd edge (u 1,u 2 ) of G to M (v nd w re not overed y e, nd u 1 nd u 2 re overed y M ). In oth ses ove the extr nodes in G re overed y edges in M, nd if v nd w in G re overed y e in M then v nd w re overed y (v,u 1 ) nd (u 2,w) in G. If e is djent to some tringles in G, if e M then in G dd edges (v,u 1 ), (u 2k,w), nd (u 2j,u 2j+1 ) to M, for 1 j < k; if p i vw M for some i then dd p i vu 2i 1, edges (u 2j 1,u 2j ) for 1 j < i, (u 2j,u 2j+1 ) for i j < k, nd (u 2k,w) of G to M ; if neither e nor ny tringle djent to e is in M then dd edges (u 2j 1,u 2j ) of G to M, for 1 j k. In ll the ove ses the extr nodes in G re overed y edges in M, nd if v nd w in G re overed y e or tringle in M then v nd w re lso overed y (v,u 1 ) nd (u 2,w) or y orresponding tringle in G. Refer to the full version of this rtile for the detils on how to rete perfet mthing in G from one in G. Thus, perfet edge- nd tringle-mthing in G tht does not use 3-yle (if it exists) n e found y first onverting G to G nd pplying the result in [1] to G. A solution of B onsisting of 2- nd 3-uts n e redued to perfet edge- nd tringle-mthing in G; however, the opposite is not trivil tsk. A perfet mthing in G n orrespond to set of uts C M in B tht re loking eh other (see ottom of Fig. 3). To extrt proper order of the uts we uild nother grph G tht hs node per ut in C M nd direted edge etween two nodes if the ut orresponding to the seond node is loking the ut orresponding to the first node. If G does not hve yles, then there is prtil order on the uts. The uts tht orrespond to the nodes with no outgoing edges n e pplied first, nd the orresponding nodes n e removed from G. However, if G ontins yles, there is no order in whih the uts n e pplied to ler up ord B. In this se we will need to modify some of the uts in order to remove yles from G. We provide the detils in the full version of this rtile. By Lemm 1 nd y the onstrution ove we otin the following theorem. Theorem 3. B&S[n n,1,,d,{2,3}](b) is polytime solvle for ll d {,,, }. 3.3 Frequeny of Colors Theorem 4. B&S[n n,,3,,{2,3}](b) is polytime solvle.
6 Proof. A single ut in ny solution removes olor. By Remrk 2, these uts do not mke solvle ord unsolvle. Thus, greedy lgorithm suffies. t u Hrdness ws estlished for mximum frequeny F = 7 in [2]. We strengthen this to F = 4 vi the modified luse gdget in Fig. 1 (). In this gdget the leftmost irulr utton n e removed if nd only if t lest one of the three non-irulr uttons is removed y vertil ut. Thus, it n reple the luse gdget in Setion 4.1 of [2]. Theorem 5 is proven in the full version of this rtile. Theorem 5. B&S[n n,, 4,, {2, 3}](B) is NP-omplete. 3.4 Cut Sizes Setion 3.2 provided polytime lgorithm for 1-olor. However, if we redue the ut size set from {2, 3, 4} to {3, 4} then it is NP-omplete. We lso strengthen Theorem 2 y showing tht 2-olor puzzles re hrd with ut size set {2}. Hrdness for Cut Sizes {3, 4} nd 1-Color Theorem 6. B&S[n n, 1,,, {3, 4}](B) is NP-omplete. Proof. We show B&S[n n,1,,,{3,4}](b) to e NP-hrd y redution from PLANAR 3-SAT, whih ws shown to e NP-omplete y Lihtenstein [3]. An instne F of the PLANAR 3-SAT prolem is Boolen formul in 3CNF onsisting of set C of m luses over n vriles V. The vrile-luse inidene grph G = (C V, E) is plnr, nd ll vriles re onneted in () () () Fig. 4. () The only two ut possiilities in the vrile gdget (shown in lk nd gry), orresponding to truth ssignments of true nd flse, respetively. () The end gdget for the 1-olor se. () The luse gdget for the 1-olor se.
7 yle. The PLANAR 3-SAT prolem is to deide whether there exists truth ssignment to the vriles suh tht t lest one literl per luse is true. We turn the plnr emedding of G into Buttons & Sissors ord, i.e., we present vriles, luses nd edges y single-olor uttons tht need to e ut. We provide detiled desriptions of eh gdget in the full version of this rtile. The vrile gdget, shown in Fig. 4(), enles us to ssoite horizontl nd digonl ut ptterns with true nd flse vlues, respetively. The end gdget, shown in Fig. 4(), enles us to end wire to mth the ends in G s emedding while enforing tht the sme vlues re propgted through the ent wire. () () Fig. 5. () The not gdget, negting the input truth ssignment, for the 1-olor se. () The split gdget for the 1-olor se. The split gdget, shown in Fig. 5(), enles us to inrese the numer of wires leving vrile nd propgting its truth ssignment. The not gdget, shown in Fig. 5(), enles us to reverse the truth ssignment in vrile wire. The luse gdget is shown in Fig. 4(). This gdget simultes onjuntion of literls. Thus, the resulting Buttons & Sissors ord hs solution if nd only if t lest one of the literls per luse is set to true, tht is, if nd only if the originl PLANAR 3-SAT formul F is stisfile. It is esy to see tht this redution is possile in polynomil time. In ddition, given Buttons & Sissors ord nd sequene of uts, it is esy to hek whether those onstitute solution, i.e., whether ll uts re fesile nd result in ord with only empty grid entries. Hene, B&S[n n, 1,,, {3, 4}](B) is in the lss NP. Consequently, B&S[n n, 1,,, {3, 4}](B) is NP-omplete. t u Hrdness for Cut Size {2} nd 2-Colors An intermedite prolem is elow. Deision Prolem: Grph Deyling on (G, S). Input: Direted grph G = (V,E) nd set of disjoint pirs of verties S V V. Output: True, if we n mke G yli y removing either s or s0 from G for every pir (s, s0 ) S. Otherwise, Flse.
8 w u t u v s t u v w t u v s v t x () () () (d) Fig. 6. Three types of nodes: () in-degree 1 (tu) nd out-degree 1 (uv); () in-degree 2 (su nd tu) nd out-degree 1 (uv); () in-degree 1 (tu) nd out-degree 2 (uv nd uw). In (d) the nodes u nd v re linked in S nd we n hoose to remove u or v. Lemm 2. Grph Deyling redues to Buttons & Sissors with 2 olors. Proof. Consider n instne (G, S) to grph deyling. First, we oserve tht we n ssume tht every vertex in G hs degree 2 or 3, nd more speifilly, in-degree 1 or 2, nd out-degree 1 or 2. Indeed, we n sfely remove ny verties with in- or out-degree 0 without hnging the outome of the prolem. Also, we n reple node with out-degree k y inry tree of nodes with out-degree 2. The sme pplies to nodes with in-degree k. Furthermore, we n ssume tht every vertex tht ppers in S hs degree 2. Indeed, we n reple ny degree 3 vertex y two verties of degree 2 nd 3, nd use the degree 2 vertex in S without hnging the outome. Similrly, we n ssume tht no two verties of degree 3 re djent. Finlly, we n ssume tht G is iprtite, nd furthermore, tht ll verties tht our in S re in the sme hlf of V, sine we n reple ny edge y pth of two edges. Now, we disuss how to model suh grph in Buttons & Sissors instne. Eh node will orrespond to pir of uttons, either red or green pir ording to iprtition of V. These pirs of uttons will e mpped to lotions in the plne on ommon (horizontl for red, vertil for green) line, nd suh tht ny two uttons of the sme olor tht re not pir re not on ommon (horizontl, vertil, or digonl) line (unless otherwise speified). If two nodes of opposite olors u nd v re onneted y n edge in G, we sy tht u loks v. In this se, one of the uttons of u will e on the sme line s the uttons of v, nd more speifilly, it will e etween the two uttons of v. Tht is, v n only e ut if u is ut first. Buttons of opposite olors tht re not onneted y n edge will not e on ny ommon lines either. As disussed ove, we n ssume we hve only three possile types of nodes. Fig. 6() illustrtes the simplest se, of node u with one inoming edge tu nd one outgoing edge uv. Clerly, t loks u nd u loks v. To model node with in-degree 2, we need to put two uttons of different sme-olored nodes on the sme line (see Fig. 6()). As long s the other endpoints of these two edges re not on ommon line this is no prolem: we never wnt to rete ut tht removes one utton of s nd one of t, sine tht would rete n unsolvle
9 instne. Finlly, to model node with out-degree 2, we simply ple vertil edge on oth ends of u (see Fig. 6()). Note tht is it importnt here tht we do not onnet two nodes with out-degree 2 to the sme two nodes with in-degree 2, sine then we would hve oth pirs of endpoints on ommon line; however, we ssumed tht nodes of degree 3 re never djent so this does not our. It remins to rete mehnism to remove verties from G s ditted y S. This is illustrted in Fig. 6(d) with detils in the full version of this rtile. A proof of Lemm 3 is in the full version of this rtile. Lemms 2 nd 3 give Theorem 7. Lemm 3. SAT redues to Grph Deyling. Theorem 7. B&S[n n, 2,,,{2}](B) is NP-omplete. 4 Two-Plyer Gmes We onsider three two-plyer Buttons & Sissors vrints. First we onsider olor restrited gmes where () eh plyer n only ut speifi olors, nd () plyers re not restrited to speifi olors. For () plyer lue my only ut Blue uttons, while the red plyer my only ut Red uttons. For () we distinguish y winning riterion: for (Imprtil) the lst plyer who mkes fesile ut wins; for (Soring) plyers keep trk of the totl numer of uttons they ve ut. When no uts n e mde, the plyer with the most uttons ut wins. In the following setions, we show tht ll vrints re PSPACE-omplete. 4.1 Cut-By-Color Gmes In this setion the first plyer n only ut lue uttons, the seond plyer n only ut red uttons, nd the lst plyer to mke ut wins. Theorem 8. The prtisn LAST two-plyer Buttons & Sissors gme, where one plyer uts lue uttons, the other red uttons, is PSPACE-omplete. Proof. The proof is y redution from G %free (CNF) [5]: given oolen formul Φ(x 1,...,x n ) in CNF nd prtition of the vriles into two disjoint susets of equl size V nd V r, two plyers tke turns in setting vlues of the vriles, the first (Blue) plyer sets the vlues of vriles in V, nd the seond (Red) plyer sets the vlues of vriles in V r. Blue wins if, fter ll vriles hve een ssigned some vlues, formul Φ is stisfied, nd loses otherwise. For given instne of formul Φ we onstrut Buttons & Sissors ord B, suh tht Blue n win the gme on B if nd only if he n stisfy formul Φ. We will prove this sttement in different formultion: Red wins the gme on B if nd only if formul Φ nnot e stisfied. For omplete exmple see the full version of this rtile. The red vrile gdget is shown in Fig. 7(). Red sets the vlue of the orresponding vrile y hoosing the first ut to e (flse) or (true), nd
10 () () () (d) (e) Fig. 7. () The red (dshed) vrile gdget, () the lue (solid) vrile gdget, () the split gdget, (d) the OR gdget, nd (e) the AND gdget. Lines (or rs used for lrity) indite whih uttons re ligned. thus unloking one of the two uts, or d, respetively, for Blue to follow up (nd to propgte the vlue of the vrile). The lue vrile gdget is shown in Fig. 7(). Blue sets the vlue of the orresponding vrile y hoosing the first ut to e (flse) or (true), nd thus unloking one of the two uts, d or e, respetively, for the red plyer to follow up. Blue hs one extr ut tht is used to pss the turn to Red. Alterntively, Blue n hoose to strt with the 3-utton ut nd disllow Red from mking ny uts in the gdget. In tht se the orresponding vrile nnot e used to stisfy Φ. Fig. 7(d) depits the OR gdget: if Blue uts or (or oth), Red n leve the gdget with ut h. Cuts nd unlok uts nd d, respetively, whih in turn unlok e nd f, respetively. Fig. 7(e) depits the AND gdget for two inputs. The proper wy of pssing the gdget: Blue mkes oth uts nd, nd Red mkes uts nd d when they get unloked, thus enling Blue to mke ut g nd exit the gdget. However, Red ould lso tke n illegl ut x, thus, unloking two extr uts, e nd f, for the lue plyer, nd, hene, putting Red t disdvntge. Thus, if t ny point in the gme Red hooses (or is fored to) mke ut x in ny of the AND gdgets, the gme result is predetermined, nd Red nnot win on B. Fig. 7() shows the split gdget; it enles us to inrese the numer of uts leving vrile nd propgting its truth ssignment. Blue s ut unloks Red s ut, whih unloks oth nd d. If Blue uts nd d this enles Red to ut e nd f, respetively. The gdget lso exists with Blue nd Red reversed. A vrint of the split gdget evlutes the formul Φ: uts e nd f re deleted. If the vrile vlues re propgted to this gdget nd Red is fored to mke the ut, Blue then gets extr uts whih Red will not e le to follow up. The gme progresses s follows: Blue selets n ssignment to lue vrile. This unloks pth of red-lue uts tht goes through some AND nd OR gdgets nd leds to the finl gdget. As the order of the uts in suh pth is
11 deterministi, nd does not ffet the hoie of vlues of other vriles, w.l.o.g., we ssume tht Red nd Blue mke ll the uts in this pth (until it gets stuk ) efore setting the next vrile. The pth gets stuk when it rehes some AND gdget for whih the other input hs not een lered. The lst ut in suh pth ws mde y Red, thus fterwrds it will e Blue s turn, nd he my hoose to mke the leftover ut from the vrile gdget to pss the turn to Red. If the finl gdget is not unloked yet, Red lwys hs ut to mke fter Blue mkes move, s there is the sme numer of lue nd red vriles. However, if Blue n fore Red to mke moves until the finl gdget is rehed, then Blue gets extr uts; thus, Red will run out of moves nd lose the gme. Otherwise, if Blue nnot fulfill some AND or OR gdgets, the Red plyer will mke the lst move nd win. Therefore, if Φ nnot e stisfied, Red wins. 4.2 Any Color Gmes Theorem 9. Imprtil two-plyer Buttons & Sissors is PSPACE-omplete. Theorem 10. Soring two-plyer Buttons & Sissors is PSPACE-omplete. We show tht Imprtil is PSPACE-omplete, then use one more gdget to show Soring is PSPACE-omplete. We redue from Geogrphy 1, (PSPACEomplete [4]). We use Lemm 4 to strt with low-degree Geogrphy instnes. Lemm 4. Geogrphy is PSPACE-omplete even when verties hve mx degree 3 nd the mx in-degree nd out-degree of eh vertex is 2. The full version of this rtile proves Lemm 4 nd Theorem 9 with these gdgets: In-degree 1, out-degree 1: The gdget for this is pir of uttons suh tht removing the first pir frees up the seond, s in Fig. 8(). In-degree 1, out-degree 2: See Fig. 8() nd the full version of this rtile. In-degree 2, out-degree 1: See Fig. 8() nd the full version of this rtile. In-degree 0: The gdgets for this look just like the gdgets for the nlgous in-degree 1 gdgets, ut without the utton pir for the inoming edge. Out-degree 0: Eh edge is utton pir tht won t free up other uttons. To show Soring is hrd, we rete redution where fter eh turn, tht plyer will hve ut the most uttons; the lst plyer to move wins. This lternting-dvntge sitution is used y n initil gdget. The optiml ply sequene egins y utting two uttons, then three, then three, then three finl time. After these four moves, the first plyer will hve five points nd the seond plyer six. Eh susequent ut removes two uttons so eh turn ends with the urrent plyer hed. Fig. 8(d) shows the strting gdget tht sets up this initil k-nd-forth. The olor-f uttons will e the lst two ut; the right-hnd f utton must e loking the next gdget. Lemm 5 postultes tht f will e lst. 1 Speifilly, Direted Vertex Geogrphy, usully lled Geogrphy.
12 z y x x z () y z () () y x x y x d e f Fig. 8. Redution gdgets for vertex with () one inoming r nd one outgoing r, () one inoming r nd two outgoing rs, nd () two inoming rs nd one outgoing rs. (d) The strting gdget for Soring. (d) Lemm 5. If plyer hs winning strtegy, then prt of tht winning strtegy inludes utting ll possile uttons of olors,,, d, nd e efore utting f. The full version of this rtile proves Lemm 5, nd lso shows how these lemms provide Theorem 10. 5 Open Prolems Interesting prolems for ords with onstnt numer of rows re still open. A onjeture for m = 2 rows ppers elow. Conjeture 1. There is polynomil time lgorithm tht removes ll ut s uttons from ny full 2 n ord with C = 2 olors for some onstnt s. Referenes 1. G. Cornuéjols, D. Hrtvigsen, nd W. Pulleylnk. Pking sugrphs in grph. Opertions Reserh Letters, 1(4):139 143, 1982. 2. H. Gregg, J. Leonrd, A. Sntigo, nd A. Willims. Buttons & Sissors is NPomplete. In Pro. 27th Cnd. Conf. Comput. Geom., 2015. 3. D. Lihtenstein. Plnr formule nd their uses. SIAM J. Comput., 11(2):329 343, 1982. 4. D. Lihtenstein nd M. Sipser. Go is polynomil-spe hrd. J. ACM, 27(2):393 401, 1980. 5. T. J. Shefer. On the omplexity of some two-person perfet-informtion gmes. Journl of Computer nd System Sienes, 16(2):185 225, 1978.