Two Approaches to Analyzing the Permutations of the 15 Puzzle

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1 Two Approhs to Anlyzin th Prmuttions o th 15 Puzzl Tom How My 2017 Astrt Th prmuttions o th 15 puzzl hv n point o ous sin th 1880 s whn Sm Lloy sin spin-o o th puzzl tht ws impossil to solv. In this ppr, w xplor whih prmuttions o th 15 puzzl r otinl y utilizin proprtis o prmuttions n rsults rom rph thory. W in our invstition y usin rut or to otin lmntry prmuttions o th puzzl ollow y simpl proos to show tht xtly hl o th prmuttions o th 15 puzzl my otin. Whil this pproh is suiint to monstrt th proprtis o th 15 puzzl, mor lnt proo usin proprtis o simpl rphs yils us rsults tht my xtrpolt onto othr similr-styl puzzls tht my rprsnt s simpl rphs. Th portion o this ppr tht ntrs roun rph thory is n xposition o Rihr Wilson s 1973 ppr Grph Puzzls, Homotopy, n th Altrntin Group. 1 Introution This ppr will invstit whih prmuttions o th 15 puzzl r otinl. Th ruls o th puzzl r rltivly simpl: th oriinl squn o th numrs in th puzzl (pitur low) r srml, n th solvr must rstor th oriinl oniurtion y strtilly movin numr squrs into th mpty sp. 1

2 Th 15 puzzl ws invnt y postmstr nm Noys Chpmn in Cnstot, Nw York n ws istriut y th Emossin Compny roun Howvr, th puzzl ros to prominn in 1878 whn prominnt Amrin puzzl nthusist nm Sm Lloy lim tht h h vlop nw puzzl whih ws moii vrsion o th 15 puzzl [5]. Howvr, th xtnt to whih Lloy ontriut to th suss o th puzzl in its rly ys is not rtin. Lloy s puzzl ws titl th puzzl n rustrt mny urious mins who tri to solv it. Th pttion ws intil to th oriinl puzzl xpt tht th numrs 14 n 15 wr swith in orr. As w will xplor in this ppr, rstorin puzzl oniurtion lik this to th oriinl squn s shown in th iur ov is impossil. Lloy ws prpr to or $1, 000 to nyoy who oul solv his triky puzzl. Th hyp roun th 15 puzzl ws so siniint strtin in Jnury 1880 tht th invntor o th puzzl, Noys Chpmn, ppli or ptnt on th puzzl in Mrh Nrly ntury ltr, th invntor o th Ruik s Cu, Erno Ruik, lim to hv n inspir y th suss o Lloy s puzzl to mk nothr snstionl puzzl [5]. Sin Sm Lloy s pitliztion on th populrity o th puzzl, mny othr vritions o th puzzl hv risn. Mny o thm orm n im whn omplt or spll out mss. Extly hl o ll prmuttions o th 15 puzzl my otin. Only thos oniurtions tht rquir n vn numr o trnspositions twn th mpty sp n numr tils my rt. Aron Arhr prov in 1999 with simpl st o loil stps tht only hl o th oniurtions o th puzzl oul vr otin y mrly shulin th tils roun. This short proo monstrt tht Sm Lloy ws lvr opportunist who took ull vnt o th hyp surrounin th puzzl y promisin juiy montry priz to somon or omplishin tsk h knw to impossil. This ppr onsirs two pprohs to invstitin th possil oniurtions o th puzzl: rph thory pproh n n strt lr pproh. Th pproh rom n strt lr prsptiv is spii to th 15 puzzl n involvs th us o prmuttion roups n thorms. Th rph thory pproh will sk th rr to onsir th puzzl s 4 4 rph ontinin 16 vrtis. On vnt to th rph thory pproh is its nrlity rltiv to th strt lr pproh. 2

3 2 Astrt Alr Approh 2.1 Nin Ky Prmuttions Th 15 puzzl s oniurtions r quivlnt to th st o prmuttions o st o 15 lmnts. Tht is, th st o prmuttions o th 15 puzzl n thouht o s th st S 15. Lt us onsir th puzzl s oniurtion whr th opn sp is in th uppr riht ornr, n th til ll 1 is in th squr to its lt. I numrs tril rom 1 oin rom riht to lt in th irst row, thn lt to riht in th son, thn riht to lt in th thir, n lt to riht in th inl row, thn w hv squn o onsutiv numrs S Fiur 1 low or n illustrtion o this oniurtion Fiur 1: Typil 1-15 rrnmnt o th 15-puzzl. Noti tht oth puzzls tur th sm prmuttion o th itn numr squrs spit th t tht th lnk squr hs irnt lotion in h oniurtion. In this oniurtion o onsutiv numrs, movin th mpty squr horizontlly within row os not hn th prmuttion o th 15 lmnts. Howvr, i th mpty squr is mov twn rows (up n own), thn it n hn th prmuttion o th 15 lmnts. Howvr, th prmuttion nrt y this oprtion is vn. Thr xmpls o ths prmuttions r ivn in Fiurs

4 Fiur 2: Th vn prmuttion (3, 5, 4) tks pl whn th mpty sp swiths pls with th numr 3. 3 ns up in th ol position o 5, 5 tks th pl o 4, n 4 tks th pl o Fiur 3: Th vn prmuttion (2, 6, 5, 4, 3) tks pl whn th mpty sp swiths pls with th numr 2. 2 ns up in th ol position o 6, 6 tks th pl o 5, 5 tks th pl o 4, 4 tks th pl o 3, n 3 tks th pl o Fiur 4: Th vn prmuttion (1, 7, 6, 5, 4, 3, 2) tks pl whn th mpty sp swiths pls with th numr 1. Th squn o th prmuttion is trmin usin th sm ountin mtho rom th prvious two iurs. Sin vry movmnt o th mpty squr ithr prsrvs th urrnt oniurtion or ltrs th oniurtion in wy hrtriz y n o yl, h movmnt o th mpty squr rsults in prmuttion in whih n vn numr o lmnts swp positions. 4

5 2.2 Anlysis o Prmuttions Thr r totl o nin prmuttions tht n our on th oniurtion o th puzzl in Fiur 1. Ths prmuttions r s ollows: p 1 = (3, 5, 4) p 2 = (2, 6, 5, 4, 3) p 3 = (1, 7, 6, 5, 4, 3, 2) p 4 = (7, 9, 8) p 5 = (6, 10, 9, 8, 7) p 6 = (5, 11, 10, 9, 8, 7, 6) p 7 = (11, 13, 12) p 8 = (10, 14, 13, 12, 11) p 9 = (9, 15, 14, 13, 12, 11, 10) Not tht h o ths prmuttions r vn sin thy ontin n o numr o lmnts in thir rsptiv yls. Sin ths r th nin prmuttions tht ris rom movin th mpty sp twn rows o th puzzl, w s tht th roup nrt rom ths nin prmuttions < p 1, p 2, p 3,..., p 8, p 9 > nrt ll prmuttions o th puzzl. Aorin to Josph Gllin s txtook Contmporry Astrt Alr 4th Eition, th prout o ny numr o vn prmuttions is lso vn. Noti tht ths nin prmuttions lso orm th st o onsutiv 3-yls o A 15 : (1, 2, 3) = p 2 3 p 2 1 p 2 3 (2, 3, 4) = p 3 p 2 1p 1 3 (3, 4, 5) = p 2 1 (4, 5, 6) = p 1 2 p2 1p 2 (5, 6, 7) = p 2 6p 2 4p 2 6 (6, 7, 8) = p 6 p 2 1p 1 6 (7, 8, 9) = p 2 4 (8, 9, 10) = p 1 5 p2 4p 5 (9, 10, 11) = p 2 9p 2 7p 2 9 (10, 11, 12) = p 9 p 2 7p 1 9 (11, 12, 13) = p 2 7 (12, 13, 14) = p 1 8 p2 7p 8 (13, 14, 15) = p 1 9 (p 1 8 p2 7p 8 )p 9 So w s tht w hv vry onsutiv 3-yl in our st o prmuttions o th puzzl. W must now prov tht vry 3-yl in A n n otin rom prouts o th onsutiv 3-yls o A n. Tht is, th st o thr-yls (1, 2, 3), (2, 3, 4),...(n 3, n 2, n 1), (n 2, n 1, n) orm ll 3-yls in A n. Aitionlly, w must show tht vry prmuttion o A n n writtn s prout o 3-yls. 5

6 Thorm 2.1. Evry prmuttion in A n is prout o 3-yls Proo. W will irst show tht vry prmuttion o A n n writtn s prout o 3-yls. First, suppos tht α A n. W not tht w n writ vry prmuttion s prout o 2-yls (Gllin 103). Lt α writtn s prout o 2-yls (x 1, y 1 )(x 2, y 2 ),...(x k, y k ). k must n vn numr sin it is xprssin n vn prmuttion α s prout o o yls. Thror, it is possil to roup th trnspositions into pirs n onsir ny potntil pirs o 2-yls. Thr r thr possil ss tht ris in h pir. Cs 1: Th two 2-yls in th pir shr th sm lmnts- I this is th s, thn th pir o 2-yls my rmov rom th prout sin () () =. Cs 2: Th two 2-yls in th pir shr on lmnt, ut h 2-yl hs istint lmnt- In this instn, th pir o 2-yl n writtn s 3-yl lik so: () () = (). Cs 3: Th two 2-yls in th pir o not shr ommon lmnt- I th two 2-yls r o th orm () () = ()(). Thus, vry prmuttion α A n n writtn s prout o 3-yls. Now, lt us turn our ttntion to provin tht th st o onsutiv 3-yls o A n n nrt ll othrs. Thorm 2.2. Th st o onsutiv 3-yls o A n n orm ll othr 3-yls o A n Proo. W will prov tht th st o onsutiv 3-yls o A n nrts ll othr 3-yls o A n y inution. Lt us onsir th s s A 3. W in with A 3 us this is th smllst roup o vn prmuttions with 3-yl. Lookin t th 3-yl (1 2 3), w osrv tht w my otin th othr 3-yl in A 3 : (1 2 3) 2 =(1 3 2). W hv otin ll 3-yls in A 3. Lt us prov nothr s s, A 4, or oo msur. Th onsutiv 3-yls in A 4 r (1 2 3) n (2 3 4). W my otin ll othr 3-yls in A 4 vi th ollowin oprtions: (1, 2, 3) = (1, 2, 3) (1, 3, 2) = (1, 2, 3) 2 (1, 2, 4) = (1, 2, 3)(2, 3, 4) 2 (1, 4, 2) = (2, 3, 4)(1, 2, 3) 2 (1, 4, 3) = (2, 3, 4) 2 (1, 2, 3) (1, 3, 4) = (1, 2, 3) 2 (2, 3, 4) (2, 3, 4) = (2, 3, 4) (2, 4, 3) = (2, 3, 4) 2 6

7 Lt us now ssum tht th thorm hols tru or som positiv intr k whr k 3. Lt us thn onsir th k + 1 st s: W hv th st o onsutiv 3-yls (1, 2, 3), (2, 3, 4),..., (k 3, k 2, k 1), (k 2, k 1, k), (k 1, k, k + 1) Our mission is to show tht w my rt ny 3-yl with k + 1 in it rom this list. Not tht our inutiv hypothsis llows us to ssum tht th 3-yls, xluin (k 2, k 1, k) nrt ll 3-yls in A k. Lt us onsir thr ss to prov tht ny 3-yl ontinin k + 1 is otinl. Cs 1: Gnrt th 3-yl with th orm (,k+1,k): (, k, k 1) (k 1, k, k + 1) = (, k, k + 1) Whr < k 1. Not tht our otin 3-yl my squr to iv us (, k + 1, k). Cs 2: Gnrt th 3-yl with th orm (, k + 1, k 1): (, k 1, k) (k, k 1, k + 1) = (, k 1, k + 1) Whr < k 1. not tht our otin 3-yl my squr to iv us (,k+1,k-1). Cs 3: Gnrt th 3-yl with th orm (,,k+1): (,, k) (, k + 1, k) = (, k + 1, ) Whr, < k 1. W orrow our 3-yl (, k + 1, k) rom th rsult o s 1. Also osrv tht our otin 3-yl my squr to iv us (,, k +1). W hv now prov tht or ny A n whr n 3, th st o onsutiv 3-yls nrts vry 3-yl in Th rsults ov init tht th st o prmuttions otin y movin th lnk sp twn rows in th puzzl yil ll possil prmuttions o A 15. Thus, ll vn prmuttions o th 15 puzzl r possil to otin. Howvr, non o th o prmuttions my otin. 3 Grph Thory Anothr wy o thinkin out th 15 puzzl is throuh rph thory. Th puzzl is ssntilly 4 4 ri with 16 vrtis. Th ppr Grph Puzzls, Homotopy, n th Altrntin Group, uthor y Rihr M. Wilson provis sris o proos to thorms whih lv into th ntur o th 15 puzzl. Th ppr ouss on mthos o llin rph with 15 lmnts n lnk vrtx. Sliin lmnts into th lnk vrtx yils lnk vrtx in th ormrly oupi vrtx. 7

8 Dinition 3.1 (Llin). A llin on G is ijtiv mppin : V (G) {1, 2,..., n, }. Th vrtx x with (x) = is si to unoupi in. Tht is, x is th mpty vrtx. [4] Dinition 3.2 (Ajnt Llin). Llins, on G r jnt i n only i n otin rom y sliin ll lon n o G onto th unoupi vrtx. This rltion o jny o llins is symmtri n irrlxiv n ins simpl rph not puz(g). Th vrtis o puz(g) r llins o G, n two vrtis o puz(g) r onnt y n i n only i thy r jnt. Eh omponnt o puz(g) is ompos o llins tht my otin rom sris o slis whih swp th mpty vrtx with n jnt vrtx s ll. W now in oupl itionl trms or proin with th proo. Dinition 3.3 (Biprtit Grph). A rph in whih th vrtis my prtition into two prts not A n B suh tht no two vrtis within A (B) r jnt. Dinition 3.4 (Non-Sprl Grph). Any rph tht my not isonnt y rmovin on vrtx. Thorm 3.1. Lt G init simpl non-sprl rph othr thn polyon or tht rph. Thn puz(g) is onnt unlss G is iprtit, in whih s puz(g) hs xtly two omponnts. In this lttr s, llins, on G hvin unoupi vrtis t vn (or rsptivly, o) istn in G r in th sm omponnt o puz(g) i n only i 1 is n vn (or rsptivly, o) prmuttion o V (G). [4] Th proo o this thorm will support our prvious inin tht xtly hl o th prmuttions o th 15 puzzl my otin strtin rom ny ivn rrnmnt. W irst outlin ky initions n put orwr two propositions. Lt us in pth p in rph G s squn p = (x 0, x 1, x 2,..., x n ) o vrtis o G whr x i 1 n x i r jnt in G, i = 1, 2..., n. W ll p pth rom th initil vrtx, x 0, to th inl vrtx x n. A pth p is si to simpl whn x 0, x 1, x 2,...x n r istint with th possil xption tht x 0 = x n in whih s p is simpl los pth. Also in ρ = (x n,..., x 2, x 1, x 0 ) to th rvrs o p. For h p = (x 0, x 1, x 2,..., x n ) in G, lt us in prmuttion σ p s th prout o 2-yls shown low: σ p = (x n 1, x n ),..., (x 2, x 3, )(x 1, x 2 ), (x 0, x 1 ) 8

9 Th ollowin two propositions r vint rom ths initions. Proposition 1: Llins, on rph G r in th sm omponnt o puz(g) i = σ p or som pth p in G rom 1 ( ) to 1 ( ). Lt us in Γ(x, y) = Γ G (x, y) to th st o ll prmuttions σ p o V (G) whr p is pth rom x to y in G. W will rvit Γ(x, y) = Γ G (x, y) with th xprssion Γ(x) = Γ G (x) i x = y. Not tht pths n writtn trnsitivly. For instn, i p is pth rom x to y n q pth rom y to z, thn on o th pths rom x to z is th prout qp. In othr wors, σ p σ q = σ qp. W r now ry to onsir th son proposition. Proposition 2: For h vrtx x o G, Γ(x) is roup o prmuttions o V (G), h ixin x. I p is pth rom x to y in G, thn Γ(x, y) = σ p Γ(x) = Γ(y)σ ρ n Γ(y) = σ p Γ(x)σp 1. Ths propositions, onjuntion with Thorm 3.2 (low) n n ssortmnt o lmms prov Thorm 3.1. Thorm 3.2. Lt G init simpl rph othr thn polyon or th rph θ 0. Thn, or ny vrtx x o G, unlss G is iprtit, in whih s Γ(x) =sym(v (G) {x}), Γ(x) =lt(v (G) {x}). Not tht sym(x) nots th symmtri roup o llins on th st X whil lt(x) nots th ltrntin roup o llins on th st X whr X is th st o vrtis o G. Bus o our propositions, it will sui to show tht lt(v (G) {x}) Γ(x). Lt us now prov thorm whih will hlp us prov Thorm 3.2 y inution. Dinition 3.5 (Ar). An r is init tr with xtly two monovlnt vrtis nmly, its ns. Thorm 3.3 (Th Hnl Thorm). Lt G non-sprl, simpl rph with t lst thr vrtis n suppos tht K is non-sprl propr surph o G with non-mpty st. Thn w n writ G = H A, whr H is non-sprl surph o G ontinin K, A n r-surph o G, n H A onsists only o th ns o A. 9

10 Proo. This proo will rquir th onsirtion o two ss: Cs 1: V (H) = V (G) In this instn, w noti tht th surph H ontins vry vrtx tht G ontins. Sin H is propr surph o G n ontins ll vrtis o G, it my only omit on o G t most. I H omitt mor thn on o G, thn w woul l to in lrr surph o G n rn th nw surph H. Cs 2: V (H) < V (G) In this instn, w noti tht th surph H os not ontin vry vrtx tht G ontins. Suppos tht vrtx v is in V (G) ut not V (H). Lt u vrtx o H. Sin G is 2-onnt, thr is yl C ontinin v n u. Followin this yl rom v to u, Lt w th irst vrtx in H. Continuin on th yl rom u to v, lt x th lst vrtx in H. I x w, lt A th pth (x, v 1, v 2,..., v k, v = v k+1, v k+2,..., v m, w), tht is, th portion o th yl twn x n w ontinin no vrtis o H xpt x n w. Sin H tothr with A is 2-onnt, it is G, s sir [2]. I x = w thn x = w = u. Lt y vrtx o H othr thn u. Sin G is 2-onnt, thr is pth P rom v to y tht os not inlu u. Lt v j th lst vrtx on P tht is in {v 1,..., v,..., v m }; without loss o nrlity, suppos j k + 1. Lt z th irst vrtx on P tr v j tht is in H. Thn lt A th pth (u, v 1,..., v = v k+1,..., v j,..., z), whr rom v j to z w ollow pth P. Now H A is 2-onnt surph o G, ut it is not G, s it os not ontin th {u, v m }, ontritin th mximlity o H. Thus x w [2]. Th Hnl Thorm provis th sis o inution tht w prorm on simpl, non-sprl rphs. W mk th s tht ny non-sprl rph is ithr iprtit or not iprtit. Thus, vry simpl, non-sprl rph G is suh tht ll llins my otin rom n initil llin or only hl my otin. Th only two xptions to this r th polyon rphs n th rph θ 0 s shown low. Howvr, th purpos o th ollowin stion is to monstrt tht whn pullin hnls o o non-sprl, simpl rphs, w n lwys rmov hnl suh tht θ 0 is not otin. Bor w lunh into this inution, lt us in msur o omplxity or rphs. Dinition 3.6 (Btti numr). Th Btti numr is msur o th omplxity o rphs. A ivn rph G hs Btti numr not β n is in s ollows: β(g) = E(G) V (G)

11 I w hnl (n r) to polyon, w otin roup o rphs ll th θ rphs whih r non-sprl rphs with two vrtis o r thr whih hv xtly thr pths onntin thm. θ-rphs lso hv Btti numr o 2. Th most simpl o ths rphs, w will ll it U, is pitur low. Evry θ-rph is suivision o U. In othr wors, vry θ-rph my otin y iviin th thr rs o U into multipl s y insrtin vrtis on h r. W will irst not tht Wilson provs tht Thorm 3.2 is tru or θ-rphs. Thus, Thorm 3.2 is tru or rphs with Btti numr o 2. W now suppos tht Thorm 3.2 is tru or som rph G with β(g) 3. Lt us writ G s H A whr H is non-sprl surph o G with β(h) = β(g) 1 n A is n r with only its n vrtis in H. Blow, w monstrt tht w my hoos H to rph othr thn θ 0 (pitur low). Not tht w r only t risk o pikin θ 0 to H i G hs Btti numr 3 n is o th orm o on o iht rphs. Four o ths rphs pitur low. W show urthr low tht w my voi pikin θ 0 to H y strtilly hoosin whih r w sint to A. 11

12 Furthrmor, h o ths rphs n hv on r rmov to yil θ rph tht is not θ 0. W will rss h o th our rphs on s t tim. Cs 1: Th rsultin rph is θ rph tht is not θ 0. This rph is sint s H whil th r onntin vrtis n is sint s th r A. Cs 2: Th rsultin rph is θ rph tht is not θ 0. This rph is sint s H whil th r onntin vrtis n is sint s th r A. 12

13 Cs 3: Th rsultin rph is θ rph tht is not θ 0. This rph is sint s H whil th r onntin vrtis n is sint s th r A. Cs 4: Th rsultin rph is θ rph tht is not θ 0. This rph is sint s H whil th r onntin vrtis n is sint s th r A. Th rminin our ss r similr to th our shown, n th pross o rmovin n r to yil θ-rph tht is not θ 0 is lso similr to th prosss shown ov. Now tht w hv stlish tht H n not θ 0, w my pro with inution ssumin tht Thorm 2 hols or H. With Thorm 2 stlish or th rph H, w n writ lt(v (G) {x}) Γ H (x) or h x V (H). Thr is rsult in rph thory nloous to our rlir proo tht th roup o thr yls o st o prmuttions orm th whol roup A n. Wilson ivs short proo o this in his ppr. W now orrow on mor thorm rom Wilson. Thorm 3.4 (Thorm 4). Lt Γ trnsitiv prmuttion roup on X n suppos tht Γ ontins 3-yl. I Γ is primitiv (in prtiulr, i Γ is oul trnsitiv), thn lt(x) Γ. Sin Γ G (x) ontins Γ H (x) whih ontins 3-yls, w n invok th rsult o Thorm 4 n mrly show tht Γ G (x) is ouly trnsitiv on V (G) {x}. Wilson monstrts tht this t is tru, n thus w hv provn Thorm 2 or ny rph G or whih β(g) 2 xluin θ 0. Not tht rphs with 13

14 Btti numr o 1 r polyon rphs whih r xlu in th hypothss or Thorms 3.1 n 3.2. Thorm 3.1 ollows losly rom Thorm 3.2. Bus th 15-puzzl my rprsnt s olltion o 16 vrtis rrn in 4 4 ri whih is iprtit rph, only hl o its vrtx llins my otin ivn strtin oniurtion. This urthrs our inins rom th strt lr stion o this ppr. Bus rsolvin th puzzl woul rquir sinl trnsposition (whih mns tht this woul n o prmuttion o th tils), th oriinl oniurtion o th 15 puzzl my not otin rom th oriinl oniurtion o th puzzl. Rrns [1] Josph A. Gllin. Contmporry Astrt Alr 7th.. Cn Lrnin, Blmont, Cliorni, Print. [2] Dvi Guihr. Communition. Wll, Wll, WA. [3] Dvi Guihr. An Introution to Comintoris n Grph Thory Crtiv Commons, Sn Frniso, Cliorni, W. [4] Rihr M. Wilson. Grph Puzzls, Homotopy, n th Altrntin Groups. Journl o Comintoril Thory 16, 86-96, W. [5] Th history o th 15 puzzl. W. [6] Arhr, A. F. A Morn Trtmnt o th 15 Puzzl. Th Amrin Mthmtil Monthly 106, , W. 14

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