The mathematics of Spinpossible

THE MA THEMA TICS OF SPINPOSSIBLE ALEX SUTHERLAND AND ANDREW SUTHERLA N D 1. Introduction Spinpo ssible TM is play ed on 3 × 3 b oard of tiles num ber ed from 1 to 9, each of which m y be r ight-side-up or up-s ide-down. One poss ible starting p osition is the bo ard: 5 4 9 2 1 6 7 8 3 The ob jective o f the ga me is to return the b oa rd to the standard configuration: 1 2 3 4 5 6 7 8 9 This is accomplished by a seq ue nc e o f spins , each of which ro ta tes a rectangular region of the b o ard by 180 ◦ . The goa l is to minimize the num b er of spins used. The starting b oar d ab ov e ma y b e so lved using t wo s pins : 5 4 9 2 1 6 7 8 3 = ⇒ 1 2 9 4 5 6 7 8 3 = ⇒ 1 2 3 4 5 6 7 8 9 In this example, the first s pin rotates the the 2 × 2 rectangle in the top left, and the second spin rotates the 3 × 1 rectangle alo ng the rig h t edg e. Y ou ca n play the game online at ht tp://s pinpo ssible.com In these notes w e give a mathematical description of this game, and some of its generaliza tions, and co nsider v arious questions that natura lly arise. Perhaps the most obvious is this: is it a lwa ys p ossible to return a g iven boa rd to the standard configuratio n with a sequence o f spins? W e shall se e sho rtly that the a ns wer is yes. A more difficult question is the following: wha t is the maximum n um b er of spins required to s olve an y b oar d? An ex haustive search has found that 9 spins ar e alwa ys sufficient (and sometimes necessar y ), but no short pro of of this fact is known. 2. A ma thema tical description of th e game W e b egin by defining the group Spin m × n , for a fixed pair of p os itive int eger s m and n with pro duct N = mn . Let S N denote the symmetric gr oup on N letters with the action on the r ight (so the p er mu tation αβ applies α and then β ), and let V N = ( Z / 2 Z ) N denote the a dditive group of N -bit vectors. F o r an y vector v = ( v 1 , . . . , v N ) ∈ V N and p ermutation α ∈ S N , we use v α = ( v α − 1 (1) , . . . , v α − 1 ( N ) ) to denote the v ector obtained by applying α to v . 1 2 ALEX SUTHERLAND AND ANDREW SUTHERLAND Definition 1 . The gr oup Spin m × n is the set { ( α, u ) : α ∈ S mn , u ∈ V mn } under t he op er at ion ( α, u )( β , v ) = ( αβ , u β + v ) . Equivalently, Spin m × n is the wr e ath pr o duct Z / 2 Z ≀ S N . Readers familiar with Coxeter g roups will recog niz e Spin m × n as the hypero c- tahedral g roup of degree N (the symmetry gro up of bo th the N -cub e and the N -o ctahedron), equiv alen tly , the W eyl gr oup of t yp e B N (and C N ). This group can also b e represented using signed p ermutation matrices, but the representation as a wrea th pro duct is better suited to our purp oses here. The definition of the group Spin m × n depe nds only on N = mn , ho wev er the integers m and n determine the set of generator s we will be defining shortly , and these play a key role in the game (playing Spinpossible on a 1 × 9 b oa rd w ould be muc h less interesting!). A b o ar d is an m × n ar ray of uniquely identified elemen ts called t iles , which we num ber from 1 to mn . Eac h tile may b e or iented positively (r ight-side-up), or negatively (upside-down). The p o sitions of the b oard are fixed loca tions, whic h for conv enience we regard as unit squares in the plane, also n umbered from 1 to mn , starting at the top left a nd pro ceeding left to right, to p to b ottom. The standar d b o a r d has tile i in p osition i , with po sitive orien tation. There is a 1- to-1 corr esp ondence b etw een m × n b oar ds and elements o f Spin m × n , but we will g e nerally think of elemen ts of Spin m × n as acting on the set of all m × n bo ards as follows: the element ( α, v ) first p er mut es the tiles b y mo ving the tile in po sition i to p osition α ( i ), and then re verses the orientiation of the tile in the i th po sition if and only if v i = 1. Of course this is just the actio n of Spin m × n on itself. The pro jection map π : Spin m × n → S N that sends ( α, u ) to α is a g roup homo- morphism, and we ha ve the short exact seque nce 1 − → V N − → Spin m × n − → S N − → 1 . It is worth empha sizing that the pr o jection from Spin m × n to V N is not a group homomorphism (for N > 1). W e no w distinguish the elements of Spin m × n that corresp ond to spins , the moves per mitted in the game. A r e c tangle R sp ecifies a rectangular subset of the p os itions on an m × n b oar d, and ha s dimensions i × j , with 1 ≤ i ≤ m and 1 ≤ j ≤ n . A spin rotates some rectangle b y 180 ◦ . It is r easona bly clear what this means, but to make it mo re precise we define a notion of distance that will b e useful later. The distanc e ρ ( p 1 , p 2 ) betw een po sitions p 1 and p 2 is mea sured by applying the ℓ 1 -norm to the centers of the corresp o nding unit sq ua res. Tw o p ositions a re adjac ent when they hav e a s ingle edge in co mmo n, equiv a lently , when the distance betw een them is 1. F or a p os ition p and a rectangle R , we use ρ ( p, R ) to denote the distance from the cen ter of p to the center of R (again using the ℓ 1 -norm). Definition 2 . The spin ab out R is t he element of Spin m × n that tra nsp oses the tiles in p ositions p 1 , p 2 ∈ R if and only if ρ ( p 1 , p 2 ) = 2 ρ ( p 1 , R ) = 2 ρ ( p 2 , R ) , and then re verses the orientation of e a ch tile in R . W e say that an element o f Spin m × n is a spin if it is a spin abo ut some r ectan- gle R . The pro po sition b elow records some useful facts ab out spins. The proo fs are straight-forward, but for the sake o f completenes s w e fill in the details. They can (and probably sho uld) b e skipped on a first rea ding. Prop ositi o n 1. L et s 1 and s 2 b e spins ab out r e ctangles R 1 and R 2 r esp e ctively. (1) s 1 is its own inverse (as is s 2 ). THE MA THEMA TICS OF SPINPOS SIBLE 3 (2) s 1 s 2 is not a spin. (3) s 1 s 2 = s 2 s 1 if and only if R 1 and R 2 ar e disjoi nt or have a c ommon c enter. (4) s 1 s 2 s 1 is a spin s 3 if and only if either s 1 and s 2 c ommut e or R 1 c on- tains R 2 . The r e ctangle of s 3 has the same shap e as R 2 . Pr o of. (1) is clear. F or (2), suppose s 3 = s 1 s 2 is a spin about so me rectang le R 3 . Then s 1 6 = s 2 , and therefo re R 1 6 = R 2 , since the iden tity elemen t is no t a spin. Now suppo se there exist positions p 1 ∈ R 1 − R 2 and p 2 ∈ R 2 − R 1 . Then s 3 mov es the tile in p ositio n p i to the s ame lo cation that s i do es, which implies that R i and R 3 hav e a common center, for i = 1 , 2 . But then there is a p osition in R 3 containing this co mmon center (either in its cen ter or along an edge) and s 3 = s 1 s 2 do es not change the orientation of the tile in this p osition, which is a contradiction. Now assume witho ut loss of gener ality that R 1 prop erly con tains R 2 . Let p 1 and p 2 be corners of R 1 not contained in R 2 (let p 1 = p 2 if R 1 has width or heig ht 1, and p 1 6 = p 2 otherwise). Then s 3 acts on the tiles in positions p 1 and p 2 the sa me wa y that s 1 do es, and this implies that R 3 and R 1 hav e a co mmon center and that R 1 ⊂ R 3 . But R 3 m ust lie in the union of R 1 and R 2 , so R 3 = R 1 and s 3 = s 1 , but then s 2 is the identit y , which is aga in a con tradictio n. So s 1 s 2 is not a spin. W e now a ddr ess (3). F or any p os ition p not in the int eres ection of R 1 and R 2 , bo th s 1 s 2 and s 2 s 1 hav e the same effect on the tile t in p osition p . Now supp ose p is in the intersection o f R 1 and R 2 . Then the orientation o f t is preser ved by bo th s 1 s 2 and s 2 s 1 , so w e need only consider the position to whic h t is mov e d. The pro duct of tw o rotations b y π is a translatio n (poss ibly trivial). Reversing the or der of the ro tations yields the inv erse translation, th us t is moved to the sa me positio n if and only if the translation is trivial, whic h o ccurs prec isely when R 1 and R 2 hav e a common center. This proves (3). F or (4), it is clear that if s 1 and s 2 commute then s 3 = s 2 is a spin. Now suppose R 1 contains R 2 . Let R 3 be the in verse image of R 2 under the per mu tation π ( s 1 ), and let s 3 = s 1 s 2 s 1 . F or tiles in R 3 , the action of s 3 is the pro duct of three ro tations by π , which is again a rotatio n by π , and the center of this rotatio n is the cen ter of R 3 . Thus s 3 is a spin ab o ut R 3 , which has the same shap e as R 2 . T o prov e the other direction o f (4), suppose for the sake of contradiction that s 3 is a spin ab out so me recta ng le R 3 , that R 1 and R 2 are not disjoin t, do not have a common center, a nd that R 1 do es not con tain R 2 . These assumptions guarantee the existence of a position p ∈ R 2 − R 1 whose image under π ( s 2 ) is in R 1 . The ac tio n of s 3 on the tile t in position p is the same as s 2 s 1 , whic h is t wo ro tations by π , hence a tra nslation. But R 1 and R 2 do not hav e a co mmon c e nter, so this tra nslation is non-trivial and s 3 mov es tile t without changing its orientation, co nt ra dicting our assumption that s 3 is a spin.  Each spin is uniquely determined by its rectangle R , th us we may sp ecify a s pin in the for m [ p 1 , p 2 ], where p 1 and p 2 ident ify the p ositions o f the upper left and low e r right corne r s of R (resp ectively). F or e x ample, o n a 3 × 3 b oar d the spin ab out the 2 × 2 r ectangle in the upper r ight c orner is [2 , 6] =  (2 6)(3 5) , 0 1101 1 000  . The moves p er mitted in a game of Spinpo ssible on an m × n boa rd ar e precise ly the set S = S ( m, n ) of all spins [ p 1 , p 2 ], where 1 ≤ p 1 ≤ p 2 ≤ mn (we consider v aria tions of the game that pla ce restrictions o n S in § 3). 4 ALEX SUTHERLAND AND ANDREW SUTHERLAND In mathematical terms, the ga me works as follows: given an element b ∈ Spin m × n (the starting b oa r d), write b − 1 as a pr o duct s 1 s 2 s 3 · · · , s k of elemen ts in S , with k as sma ll as p ossible (a solution ). Applying the spins s 1 , s 2 , . . . , s k to b then yields the identit y (the standar d boa rd). In ge neral there will be man y solutions to b , but some b oards hav e a unique solution; this topic is discussed further in § 4. Let R i × j denote the s ubset of Spin m × n that are s pins ab out an i × j rectangle. The set R i × j is necessarily empty if i > m or j > n , and we may hav e R i × j = ∅ even when R j × i 6 = ∅ (although this can o ccur only when m 6 = n ). In these notes we shall alwa ys consider the sets R i × j and R j × i together, thus we de fine S i × j = R i × j ∪R j × i . The set of spins S in Spin m × n is then the union of the sets S i × j , each of which w e refer to as a spin typ e . Prop ositi o n 2. Assume 1 ≤ i ≤ m , 1 ≤ j ≤ n , and m ≤ n . The fol lowing hold: (1) |R i × j | = ( m + 1 − i )( n + 1 − j ) . (2) |S i × j | = ( |R i × j | + |R j × i | for i 6 = j, |R i × j | for i = j. (3) |S | =  m +1 2  n +1 2  . (4) Ther e ar e 1 2 m (2 n − m + 1) distinct spin typ es S i × j in Spin m × n . Pr o of. F or (1), w e note that there are ( m + 1 − i )( n + 1 − j ) p oss ible lo cations for the upp er left cor ner of a n i × j rectangle on an m × n b oar d. The for mula in (2) is immediate. F or (3) we hav e m X i =1 n X j =1 ( m + 1 − i )( n + 1 − j ) = m X i =1 n X j =1 ij =  m + 1 2  n + 1 2  , and for (4) we hav e m X i =1 n X j = i 1 = m X i =1 ( n + 1 − i ) = m ( n + 1 ) −  m + 1 2  = m (2 n − m + 1) 2 .  It is well k nown that the symmetric group S N is g enerated b y the set of all transp ositions (per m utations that swap tw o elements and leav e the r est fixed). Slightly less well known is the fact that S N is generated b y an y set of transpositio ns that form a co nnected graph, as des c rib ed in the following lemma. Lemma 1. L et E ⊆ S N b e a set of tr ansp ositions ( v i , v j ) acting on a set of vertic es V = { v 1 , . . . , v N } . L et G b e the undir e cte d gr aph on V wi th e dge set E . Then E gener ates S N if and only if G is c onne cte d. Pr o of. It suffices to show that E gener ates every tr ansp osition in S N . If the se- quence of edge s ( e 1 , . . . , e k ) is a path from v i to v j in G , then the per m utation e 1 e 2 . . . e k − 2 e k − 1 e k e k − 1 e k − 2 . . . e 2 e 1 is the tra nsp osition ( v i , v j ). Let H b e the subgro up of S N generated b y E . The H -o rbits of V corresp o nd to connected comp onents of G , and H can achiev e any per mutation of the vertices in a g iven comp onent, since it can transp ose any pair of vertices connected b y a path. Thus H = S N if and only if G is connected.  Corollary 1. S 1 × 1 ∪ S 1 × 2 gener ates Spin m × n . THE MA THEMA TICS OF SPINPOS SIBLE 5 Pr o of. The se t S 1 × 2 consists of transp ositions that form a connected g r aph whose vertices a re the positions o n an m × n b oar d with edges betw e en adjacent p os itions. If follows from Lemma 1 that, given any ele ment ( α, u ) in Spin m × n , there is a vector v ∈ B mn for whic h we can construct ( α, v ) as a product of elemen ts in S 1 × 2 . By applying appro priate elemen ts of S 1 × 1 to ( α, v ) w e ca n o btain ( α, u ).  The corollar y implies that every star ting boar d in the Spinp ossible ga me has a solution. W e now g ive a n upper b o und on the le ng th of any solution. Theorem 1. Every element of Spin m × n c an b e expr esse d as a pr o duct of at most 3 mn − ( m + n ) spins. Pr o of. Let ( α, u ) b e an element of Spin m × n . F or any v ∈ V N we ma y wr ite ( α, u ) as ( α, v )( ι, u + v ), where ι denotes the tr ivial p ermutation. It is clear that we can express ( ι, u + v ) as the product of at most mn e le men ts in S 1 × 1 . Thus it suffices to show that we can c o nstruct an element of the form ( α, v ), for some v ∈ V N , a s a pro duct of at most 2 mn − ( m + n ) spins. Since we may use an y v we like, we now ignore the orientation of tiles and fo cus on the p ermutation α . Ra ther than constructing α , w e shall constr uc t α − 1 (whic h is equiv alent, since α is a rbitrary). W e now pro ceed b y induction on N to show that we can construct α − 1 using at most 2 mn − ( m + n ) spins. F or N = 1 we necessa rily hav e α − 1 = ι , which is the pro duct of 0 = 2 m n − ( m + n ) spins. F or N > 1, a ssume without lo ss of generality that m ≤ n (in terchange the role of rows and colunmns in wha t follo ws if not). W e first use the spin ab out the r ectangle [1 , α (1)] to restor e tile 1 to its correct p osition in the upp er left corner . Now let i and j b e the vertical and ho rizontal distances, resp ectively , betw een p ositions n + 1 and α ( n + 1), so that ρ ( n + 1 , α ( n + 1 )) = i + j . T o mov e tile n + 1 to p osition n + 1 (the sec o nd row of the leftmost column) we first apply an elemen t of S ( i +1) × 1 to mov e tile n + 1 to the correct row, and then apply an e le men t of S 1 × ( j +1) to move tile n + 1 to the corr ect co lumn (w e ca n omit spins in S 1 × 1 , which a rise when i o r j is zero ). Neither of these spins affects p osition 1. In a similar fashion, we can successively move each tile k n + 1 for 2 ≤ k < m from po sition α ( k n + 1) to pos ition k n + 1 using a t mo st tw o s pins per tile, without disturbing any of the tiles in p os itions j n + 1 for j < k . The total num b er of spins used to correctly positio n all the tiles in the leftmost column is 2 m − 1. By the inductive hypothesis, we can correctly p ositio n the remaining tiles in the ( m − 1) × n bo ard obtained b y ignoring the leftmost column using at most 2 ( m − 1) n − ( m − 1 + n ) spins. The total num b er of spins used is 2 m − 1 + 2( m − 1 ) n − ( m − 1 + n ) = 2 m n + m − 3 n − 1 , and since m ≤ n this is less than 2 mn − ( m + n ).  The upper b ound in Theorem 1 can b e improved for mn > 1. A more detailed analysis of the c a se Spin 3 × 3 shows that one can mov e every tile to its co rrect po sition using at most 9 spins 1 , and then or ient ev ery tile correctly using at most 7 spins, yielding a n upp er b o und o f 16, versus the b ound of 21 given b y The o rem 1. W e als o note that the leading constant 3 is not the b est p ossible: for m, n ≥ 3 the techn ique used to orient tiles in the 3 × 3 case can b e generalized to achiev e 25 / 9 . Let k ( m, n ) denote the maximum length of a solution to a boa rd in Spin m × n . Theorem 1 gives an upper bo und on k ( m, n ). W e now prov e a low er b ound. 1 It is known tha t 7 spins alw ays suffice, and are sometimes nece ssary . 6 ALEX SUTHERLAND AND ANDREW SUTHERLAND Theorem 2. A ssume N = mn > 1 . Then k ( m, n ) ≥ ln(2 N N !) ln  m +1 2  n +1 2  + 1  . This implies the b ound k ( m, n ) ≥ 1 2 mn − (1 − ln 2) 2 · mn ln mn + 1 4 . Pr o of. Let c = |S | + 1. The n umber of distinct expressions of the fo rm s 1 · · · s j with s 1 , . . . , s j ∈ S and j ≤ k is at most c k . Not all of these expressio ns yield dis tinct elements of Spin m × n , but in any case it is clear that they corresp ond to at most c k distinct elements of Spin m × n . The car dinality of Spin m × n is 2 N N !, thu s in order to express ev ery ele ment of Spin m × n as a product of at most k spins w e m ust hav e (1) c k ≥ 2 N N ! F rom Prop ositio n 2 we hav e c =  m +1 2  n +1 2  + 1. T ak ing logar ithms in (1) a nd dividing by ln c yields the first b o und for k ( m, n ). F or the se c ond bo und, w e no te that N 2 > c for all N = mn > 1 , th us we can replace the LHS of (1) by N 2 k . By b ounding the erro r term in Stirling’s approximation o ne can show that ln N ! ≥ N ln N − N + 1 2 ln N , for all N ≥ 1. Applying N 2 k > c k and taking loga rithms in (1) yields 2 k ln N ≥ N ln N − (1 − ln 2) N + 1 2 ln N . Dividing by 2 ln N gives k ≥ 1 2 N − 1 − ln 2 2 · N ln N + 1 4 , which pr oves the second bo und.  F or m = n = 3, Theorem 2 give the low er bound k (3 , 3) ≥ 6, which is not far below the kno wn v alue k (3 , 3) = 9. Asymptotically , we ha ve the following c o rollar y . Corollary 2. The asymptotic gr owth of k ( m, n ) is line ar in N = mn . Mor e pr e- cisely, for every ǫ > 0 ther e is an N 0 such that  1 2 + ǫ  N < k ( m, n ) ≤ 3 N for al l N > N 0 . Recall that for a g roup G generated by a s e t S , the Cayley graph Cay( G, S ) is the g raph with vertex set G and e dg e ( g , h ) lab elled by s whenever sg = h , w her e s ∈ S and g , h ∈ G . A solution to a bo ard b ∈ Spin m × n corres p o nds to a shor test path fro m b to the identit y in the graph Cay(Spin m × n , S ). The quantit y k ( m, n ) is the diameter of this graph. THE MA THEMA TICS OF SPINPOS SIBLE 7 3. Restricted Spin Sets Spinpo ssible includes v ar iations of the standard game that place restr ictions on the t yp es of spins that are allow ed. F or exa mple, the “no singles/doubles ” puzzle levels prohibit the use o f spins in S 1 × 1 and S 1 × 2 . This raises the question of whether it is still possible to solve every boar d under s uch a restriction. Mo re generally , we may ask : which subsets of the full set of spins S = S ( m, n ) gener ate Spin m × n ? W e b egin by defining three subsets of S that cannot generate Spin m × n when mn > 1, using three different notions of parity . (1) The even ar e a spins S a are the spins whose re ctangles hav e ev en a r ea. S a is the unio n of the S i × j for which ij ≡ 0 mo d 2. (2) The even p ermut ation spins S p are the spins that contain an even n umber of tr a nsp ositions. S p is the union of the S i × j for which ij ≡ 0 or 1 mo d 4. (3) The even distanc e spins S d are the spins that transp ose p ositions at ev en distances. S d is the union o f the S i × j for which i + j ≡ 0 mod 2. W e now consider the co rresp onding subg roups of Spin m × n . In these definitions α is a p er mut ation in S N , v is a vector in V N , and wt( v ) deno tes the Hamming weigh t of v (the n umber of 1s it con tains). The g roup A N is the alternating group in S N , and w e define the p er mu tation g r oup D N ∼ = S ⌈ N/ 2 ⌉ × S ⌊ N/ 2 ⌋ as follows: D N = { α : ρ ( i, α ( i )) ≡ 0 mod 2 for 1 ≤ i ≤ N } . Here i and α ( i ) identify p ositions on an m × n board and ρ ( i, α ( i )) is the ℓ 1 -distance. The subgr o ups Spin ∗ m × n , where ∗ is a , p , or d , a re defined as follows: (1) Spin a m × n = { ( α, v ) : wt( v ) ≡ 0 mo d 2 } (index 2). (2) Spin p m × n = { ( α, v ) : α ∈ A N } (index 2). (3) Spin d m × n = { ( α, v ) : α ∈ D N } (index  N ⌊ N/ 2 ⌋  ). It is not necess a rily the ca se that S ∗ generates Spin ∗ m × n , but we alwa ys have S ∗ = S ∩ Spin ∗ m × n . In pa rticular, it is clea r that hS ∗ i ⊆ Spin ∗ m × n . The following prop ositions give some conditions under which equality holds. Prop ositi o n 3. Ass u me m, n ≥ 2 and mn > 4 . Then S 1 × 2 ∪ S 2 × 2 (and ther e- for e S a ) gener ates Spin a m × n . Pr o of. Let G = hS 1 × 2 ∪ S 2 × 2 i , and let ˆ π denote the restrictio n of π to G . The fact that G contains S 1 × 2 implies tha t π ( G ) = S N , by Lemma 1. It thus suffices to show that the kernel of ˆ π has index 2 in ker π ∼ = V N . The following product of spins in G transpo ses the tiles in positio ns 1 and 2: (2) [2 , 3][1 , n + 1][1 , n + 2][2 , 3][1 , n + 2][1 , n + 1][2 , 3] =  (1 2) , 0  W e can transfor m the iden tity ab ov e by applying any square-pres erving isometry of Z 2 (the group gener ated by unit translations and reflections abo ut the lines y = 0 and y = x ). Suc h a transfor mation may change the lo cation and/or orientation of the rec ta ngles identif ying the s pins that appea r in the pro duct, but it do es not change their spin type (the set S i × j to which they b elong). This allows us to transp ose any pair of adjace n t tiles o n the m × n b oar d using a pro duct of spins in G . It follows fro m L e mma 1 tha t G contains the subg roup H = { ( α, 0 ) : α ∈ S N } . F or each even int eger w from 0 to N , we can construct so me g w = ( β , v ) with wt( v ) = w , as a pro duct of elements in S 1 × 2 . The coset g w H ⊂ G then contains elements of the for m ( β , v ) for every vector v with wt( v ) = w . Multiplying each 8 ALEX SUTHERLAND AND ANDREW SUTHERLAND ( β , v ) on the left b y ( β − 1 , 0 ), we see that G contains elements ( ι, v ) for every even weigh t v ector v . Therefore ker ˆ π ha s index 2 in ker π .  W e note that S a do es not generate Spin a m × n when m = n = 2, nor when exactly one of m or n is 1. Prop ositi o n 4. Assume mn 6 = 4 . Then S 1 × 1 ∪ S 2 × 2 ∪ S 1 × 3 (and ther efor e S d ) gener ates Spin d m × n . Pr o of. Let G = hS 1 × 1 ∪ S 2 × 2 ∪ S 1 × 3 i . Then G con tains ker π = hS 1 × 1 i ∼ = V N . It remains to show that π ( G ) = D N . Assume for the moment that m ≥ 2 and n ≥ 3. The following pro duct of spins in G transp ose the tiles in p ositions 1 and n+2: (3) [2 , n + 3][1 , 3 ][2 , n + 3][ n + 3 , n + 3] =  (1 n + 2) , 0  As in the pro o f of Pr op osition 3, we may transform this identit y by applying any square-pr eserving iso metry of Z 2 . Thu s we can transp ose any pair of tiles that share exactly one common vertex (i.e., that are “dia g onally adjacent”), and w e can also handle the case m ≥ 3 and n ≥ 2. It then follows fro m Lemma 1 that these transp ositions gener a te D N . W e no w c o nsider the case where m or n is 1 . If mn ≤ 2 the prop osition clearly holds (we o nly need spins in S 1 × 1 ), so assume without lo ss of generality that m = 1 and n ≥ 3. W e no w repla c e (3) with [1 , 1][1 , 3][2 , 2][1 , 1] =  (1 3) , 0  , and apply the sa me argument.  It is easy to check that when mn = 4 the set S d do es not genera te Spin d m × n . W e leav e ope n the question of when S p generates Spin p m × n , but for Spin 3 × 3 we note that S p ⊂ S a (see b e low), th us S p do es not ge nerate Spin m × n in this ca se. F or refer e nce, we list the s pin types S i × j contained in S a , S p , and S d for all i, j ≤ 3 ≤ m, n : (1) S 1 × 2 , S 2 × 2 , S 2 × 3 ⊂ S a . (2) S 2 × 2 , S 3 × 3 ⊂ S p . (3) S 1 × 1 , S 1 × 3 , S 2 × 2 , S 3 × 3 ⊂ S d . T o simplify our analysis of the subsets of S that g enerate Spin m × n , we introduce an equiv alence relation on spin types. Definition 3 . Two s pin t yp es S i × j and S i ′ × j ′ ar e e quivalent, denote d S i × j ∼ S i ′ × j ′ , whenever hS i × j i = hS i ′ × j ′ i . Prop ositi o n 5. L et m, n ≥ 3 . F or 1 ≤ i , i ′ , j, j ′ ≤ 3 ther e is exactly one n on-trivial e quivalenc e of spin typ es S i × j ∼ S i ′ × j ′ , namely, S 1 × 2 ∼ S 2 × 3 Pr o of. If a particular spin type is contained in S ∗ (where ∗ is a , p , or d ), then so is every equiv alent spin type. Examining the list of s pin t yp es for S ∗ , we ca n use this cr iterion to r ule out all but t wo p oss ible equiv a lences a mo ng the 6 spin t yp es S i × j with 1 ≤ i , j ≤ 3. The first is the pair S 1 × 1 and S 3 × 3 , but these cannot b e equiv alent bec ause hS 1 × 1 i lies in ker π 1 ∼ = V N but S 3 × 3 do es not. The second is the pair S 1 × 2 and S 2 × 3 , which w e now sho w are equiv a lent . F or simplicity we sha ll wr ite spins in ter ms of r ectangles with c o ordinates on a 3 × 3 boa rd, but these can b e genera lized to an m × n boar d by repla cing p ositions THE MA THEMA TICS OF SPINPOS SIBLE 9 4, 5, 6, 7 , 8 a nd 9 with po s itions n + 1, n + 2 , n + 3, 2 n + 1, 2 n + 2 , and 2 n + 3, resp ectively . W e c an write the spin [1 , 6] as a pro duct of spins in S 1 × 2 as follows: [1 , 6] = [2 , 5][2 , 3][4 , 5 ][5 , 6][1 , 2 ][4 , 5][2 , 3][3 , 6][1 , 4] . As in the pro ofs of Prop ositions 3 a nd 4, we can transfo rm this identit y via a square-pr eserving iso metry of Z 2 to express any s pin in S 2 × 3 as a pro duct of spins in S 1 × 2 . Thu s hS 2 × 3 i ⊂ hS 1 × 2 i . F or the o ther inclusion, we may write the spins [1 , 2] and [4 , 5] as pro ducts of spins in S 2 × 3 as follows: [1 , 2] = [1 , 6][4 , 9][1 , 8 ][4 , 9][1 , 8 ][1 , 6][1 , 8][4 , 9][1 , 8][1 , 6][1 , 8][4 , 9][1 , 8][1 , 6][1 , 8] , [4 , 5] = [1 , 6][2 , 9][1 , 6 ][2 , 9][4 , 9 ][2 , 9][1 , 6][2 , 9][4 , 9][2 , 9][1 , 6][2 , 9][4 , 9][2 , 9][4 , 9] By transforming one of these t wo identit ies with a suitable isometry we ca n express any spin in S 1 × 2 as a pro duct of spins in S 2 × 3 . Thus hS 1 × 2 i ⊂ hS 2 × 3 i .  W e are now r e ady to pro ve our main theorem, which completely deter mines the combinations o f spin types tha t genera te Spin 3 × 3 . Theorem 3. Assume t hat m, n ≥ 3 . L et T b e a union of spin typ es S i × j , wher e 1 ≤ i, j ≤ 3 . F or T t o gener ate Spin m × n , it is su fficient for T to c ontain one of the fol lowing six sets: S 1 × 2 ∪ S 1 × 1 , S 1 × 2 ∪ S 1 × 3 S 1 × 2 ∪ S 2 × 2 ∪ S 3 × 3 , S 2 × 3 ∪ S 1 × 1 , S 2 × 3 ∪ S 1 × 3 S 2 × 3 ∪ S 2 × 2 ∪ S 3 × 3 . When m = n = 3 , this c ondition is also ne c essary. Pr o of. W e fir st prove sufficiency . By P r op osition 5, S 1 × 2 ∼ S 2 × 3 , so it is enough to prov e that eac h of the first three sets listed in the theorem g e ne r ates Spin m × n . As ab ov e, we spe c ify spins using co ordinates on a 3 × 3 bo ard, but these can co ordinates can b e trans ferred to an m × n boa rd as noted in the pro of of Prop o s ition 5. By Corolla ry 1 , the set S 1 × 1 ∪ S 1 × 2 generates Spin m × n . F o r S 1 × 2 ∪ S 1 × 3 , it is enough to sho w that S 1 × 1 ⊂ hS 1 × 2 ∪ S 1 × 3 i . W e note that eac h element of S 1 × 1 has the for m s i = ( ι , e i ), wher e e i is the weigh t 1 v ector in V N with the i th bit set. If ( α, u ) is a ny element o f Spin m × n with α (1) = i , then w e hav e ( α, u ) − 1 ( ι, e 1 )( α, u ) = ( α − 1 , u α − 1 )( ι, e 1 )( α, u ) = ( α − 1 , u α − 1 + e 1 )( α, u ) = ( ι, e α 1 ) = ( ι, e i ) . Since π ( hS 1 × 2 i ) = S N , b y Lemma 1, w e can generate a suitable ( α, u ) for each i from 1 to N . Thus it is enough to s how ho w to express the spin [1 , 1] as a pro duct of spins in S 1 × 2 ∪ S 1 × 3 : [1 , 1] = [1 , 2][1 , 4][1 , 3 ][1 , 4][1 , 2 ][4 , 6][3 , 6][4 , 6] . The s ame ar guments apply to the third set S 1 × 2 ∪ S 2 × 2 ∪ S 3 × 3 , thus it suffices to note that: [1 , 1] = [2 , 3][1 , 5][1 , 2 ][3 , 6][4 , 7 ][1 , 5][2 , 3][5 , 8][1 , 9][5 , 9][1 , 5][5 , 8][8 , 9] . W e now prov e the necessit y of the c ondition in the theorem, under the assump- tion m = n = 3. The se t T is the union of so me subset of the six spin types U =  S 1 × 1 , S 1 × 2 , S 1 × 3 , S 2 × 2 , S 2 × 3 , S 3 × 3  . 10 ALEX SUTHERLAND AND ANDREW SUTHERLAND Of the 64 subsets of U , one finds that 22 o f them hav e unions that are contained in S a or S d , thus T cannot b e the union of a ny of thes e 22 subsets. Conversely , one finds that 3 9 of the r emaining 4 2 subsets of U hav e unions containing one o f the 6 sets listed in the prop osition. The 3 r emaining subsets of U all hav e unions contained in S 1 × 2 ∪ S 2 × 3 ∪ S 3 × 3 , which w e now argue does no t generate Spin m × n . Since S 1 × 2 ∼ S 2 × 3 , it is eno ug h to show that S 1 × 2 ∪ S 3 × 3 do es not generate Spin m × n . By Lemma 6 b elow, a ny pro duct of elements in S 1 × 2 ∪ S 3 × 3 is equiv a lent to a pro duct in which the unique element of S 3 × 3 app ears only o nce, in the rig ht most po sition. It follows that the cardinality of hS 1 × 2 ∪ S 3 × 3 i is at most (in fact, exactly) t wice that of hS 1 × 2 i . But b y Lemma 2 b elow, the subgroup hS 1 × 2 i has trivial int eres e ction with ker π and th us ha s index 2 9 = 512 in Spin 3 × 3 . So hS 1 × 2 ∪ S 3 × 3 i is a prop er subg roup of Spin 3 × 3 .  Lemma 2. The r estriction of the pr oje ction map π : Spin m × n → S N to the gr oup G = hS 1 × 2 i is an isomorphism fr om G to S N . Pr o of. Let ˆ π : G → S N be the restrictio n o f π to G . The fact that ˆ π is surjective follows from Lemma 1, so we only need to show that ˆ π is injective. Let h b e any element of the kernel of ˆ π . T hen h = s 1 · · · s k is a pro duct of spins in S 1 × 2 , and h fixes the po sition of every tile on the m × n boar d. W e will show that h also fixes the orientation of ev ery tile, and therefore h is the iden tity . Consider tile t in p osition t on the standa rd bo ard b . If we apply h to b , ea ch spin s i po tent ially moves the tile t , but if it does , it alwa ys moves t to a n adjacent po sition on the boa rd, since s i ∈ S 1 × 2 . Thus t is mov ed a lo ng some path o n the m × n boa rd (p ossibly trivial) that must ev entually return t to its original p os itio n. The length o f this path is necessarily a n even in teger, therefore t is a lso returned to its or ig inal orientation.  Let T b e a subset of the spins in S . Generalizing o ur definition o f k ( m, n ), we define k ( m, n, T ) a s the dia meter of the Cayley graph Cay(Spin m × n , T ), and consider upper and low er bo unds for k ( m, n, T ). T o do so, we in tro duce a notion of weight for a spin, defined the total distance c overd by all the tiles it mov es. Definition 4. The weight of a r e ctangle R is wt( R ) = 2 P p ∈ R ρ ( p, R ) , and the weight of a spin s ab out R is wt( s ) = wt( R ) . W e ma y de no te the weigh t of a n i × j rectangle R by w ( i, j ), since it dep ends only on the dimensio ns of R , no t its lo c a tion. Lemma 3. L et ε : Z → { 0 , 1 } b e the p arity map. Then w ( m, n ) = 1 2  mn 2 + nm 2 − ε ( m ) n − ε ( n ) m  . Pr o of. When m and n are b oth even w e hav e w ( m, n ) = 4 · 2   m / 2 X i =1 n / 2 X j =1 ( i + j − 1)   = 1 2  mn 2 + nm 2  . THE MA THEMA TICS OF SPINPOS SIBLE 11 When m and n a re b oth o dd w e hav e w ( m, n ) = 4 · 2   m − 1 2 X i =1 n − 1 2 X j =1 ( i + j )   + 2 · 2   m − 1 2 X i =1 i + n − 1 2 X j =1 j   = 1 2  mn 2 + nm 2 − m − n )  . The ca ses wher e m and n hav e o ppo site parit y ar e similar and left to the rea de r .  Lemma 4. L et T b e any set of spins in Spin m × n . Then k ( m, n, T ) ≥ w ( m, n ) max { wt( s ) : s ∈ T } for al l m, n ≥ 1 . Pr o of. Let b ∈ S m × n , with wt( b ) = w ( m, n ). If s 1 · · · s k is a pro duct of spins in T equiv alent to b , then w ( m, n ) ≤ P wt( s i ) ≤ k w max . The lemma follows.  Lemma 5. Every element of Spin m × n c an b e expr esse d as the pr o duct of at most w ( m, n ) + mn spins in S 1 × 1 ∪ S 1 × 2 . Pr o of. Let b ∈ Spin m × n . W e will construct b − 1 = ( α, u ) by constr ucting an ele ment ( α, v ) as a pro duct of at mos t w ( m, n ) s pins in S 1 × 2 , to which we may then apply at most mn spins in S 1 × 1 to obtain ( α, v ). Let d = m + n − 2. Then d is the maximum ( ℓ 1 ) distance betw een any po sition and the center of the m × n r e ctangle R containing all the p ositions o n the boa rd. F or each position i at distance d fro m the cen ter (the 4 corners when mn > 1), we can mov e tile i to p o sition i using at mos t 2 d s pins in S 1 × 2 . Next we place the correct tiles in po sitions at distance d − 1 fro m the cent er, and ea ch o f these tiles can currently lie at most d − 1 p ositions awa y from the center (since the distance d p ositions ar e already filled with the correct tiles), thus we use at most 2( d − 1 ) spins in S 1 × 2 to place the correct tiles in the p os itions at distance d − 1 from the center. Note that we can do this by mo ving each tile along a path tha t do e s not disturb an y tiles that ha ve alrea dy b een pla ced. Contin uing in this fas hion, we use at most 2 ρ ( p, R ) spins to place the corre c t tile in positio n p , and the to ta l n umber of spins is at most wt( R ) = w ( m, n ).  Corollary 3. F or al l m, n ≥ 1 let T = T ( m, n ) b e a set of spins with weight b ounde d by some c onstant W . Than as N = mn → ∞ we have the asymptotic b ound k ( m, n, T ) = Θ( mn 2 + nm 2 ) . Mor e pr e cisely, for every ǫ > 0 ther e is an N 0 such t hat  1 2 W + ǫ  ( mn 2 + nm 2 ) < k ( m, n, T ) < (1 + ǫ )( mn 2 + nm 2 ) , for al l N > N 0 . F or m = n this giv es a Θ( N 1 . 5 ) bound, whic h ma y be contrasted with the Θ( N ) bo und of Cor ollary 2, wher e the weigh t of the spins was unrestricted. W e note that in the c ase o f Spin 3 × 3 and T = S 1 × 1 ∪ S 1 × 2 , Lemmas 4 and 5 give the b ounds 12 < k (3 , 3 , T ) < 33, compar ed to the actual v alue k (3 , 3 , T ) = 25. 12 ALEX SUTHERLAND AND ANDREW SUTHERLAND 4. Unique So lutions Certain elements of Spin m × n are disting uished by the fa ct that they ha ve a unique solution (a unique shortest expres sion as a pr o duct of spins). This is clearly the ca se, for example, when b ∈ S . There a re many le s s trivia l ex amples, some 2,203,4 01 of them in Spin 3 × 3 . Thes e include wha t appea r to b e the most difficult puzzles in the game, so me of whic h are featured in separ ate puzzle levels designated as “unique s ”. While these can b e quite challenging, knowing that the solution is unique can b e an aid to solving such a puzzle. W e b egin with a lemma used in the pro o f of Theo r em 3 , which also allows us to rule out many possible candidates for a unique solution. Lemma 6. L et b = s 1 · · · s i · · · s k b e a pr o duct of spins in Spin m × n , with i < k and s i ∈ S 1 × 1 ∪ S m × n . If s i ∈ S 1 × 1 , then b c an b e written as b = s 1 · · · s i − 1 s i +1 · · · s k t i with t i ∈ S 1 × 1 . If s i ∈ S m × n , then b c an b e written as b = s i · · · s i − 1 t i +1 · · · t k s i , with e ach t j a spin of t he same typ e as s j , for i ≤ j ≤ k . Pr o of. W e first supp ose that s i ∈ S 1 × 1 . Then the recta ngle R i of s i contains just a single p osition. Let R i +1 be the recta ngle of s i +1 . If R i is contained in R i +1 , then b y Pro p o sition 1 , we hav e s i +1 s i s i +1 = t with t ∈ S 1 × 1 . Multiplying on the left b y s i +1 , we have s i s i +1 = s i +1 t i , allowing us to “ shift” the spin s i to the right, potentially c hanging the loc a tion of its rectangle but not its t yp e. If R i is not contained in R i +1 then R i and R i +1 are disjoin t and we simply let t = s i , since then t and s i +1 commute. Applying the same pro cedure to s i +2 , . . . , s k , w e even tually obtain a pro duct b = s 1 · · · s i − 1 s i +1 · · · s k t i of the desired form (using a p otentially different t at each s tep). W e now suppo se that s i ∈ S m × n . Then the rectangle R i of s i cov ers the entire m × n b oard. Let R i +1 be the rectangle of s i +1 , whic h is necess arily c ontained in R i . W e then have s i s i +1 s i = t i +1 , where t i +1 is a spin of the s ame t yp e as s i +1 , and therefor e s i s i +1 = t i +1 s i . W e may pro ce ed in the same fashion to compute t i +2 , . . . , t k , even tually obtaining the desired pro duct b = s i · · · s i − 1 t i +1 · · · t k s i .  Prop ositi o n 6. Supp ose that s 1 · · · s k is the unique solut ion to a b o ar d b in Spin m × n . Then the fol lowing hold: (1) None of the s i ar e c ont aine d in S 1 × 1 or S m × n . (2) Conse cutive p airs s i and s i +1 have r e ctangles R i and R i +1 that overlap and do not shar e a c ommon c enter, with n either c ontaine d in the other. Pr o of. (1) follows from Lemma 6 and its pro of: if s i were an element of S 1 × 1 or S m × n we could o btain a different expression for b a s a pro duct of spins of the same length by “ s hifting” s i either to the left or rig ht. (2) follows from parts (3) and (4 ) of Prop os itio n 1.  W e co nc lude with a list o f so me op en pro ble ms : 1. Give a short pro of that k (3 , 3) = 9. 2. Determine k (4 , 4). 3. Determine whether lim n →∞ k ( n, n ) /n 2 exists, and if so, its v alue. 4. Analyze the distribution of solution leng ths in Spin m × n . 5. Determine which spin types S i × j are equiv alent. 6. Determine which combinations of spin types genera te Spin 4 × 4 . 7. Give bounds o n the num b er of b o a rds with unique so lutions in Spin m × n .