Complete Solutions for a Combinatorial Puzzle in Linear Time and Its Computer Implementation

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1 Appl Math Inf Sci 9, No, (015) 851 Applied Matheatics & Inforation Sciences An International Journal Coplete Solutions for a Cobinatorial Puzzle in Linear Tie and Its Coputer Ipleentation Daxin Zhu 1, Yingjie Wu,, Lei Wang 3 and Xiaodong Wang 1,, 1 Faculty of Matheatics & Coputer Science, Quanzhou Noral University, Quanzhou, China School of Matheatics and Coputer Science, Fuzhou University, Fuzhou, China 3 Microsoft AdCenter, Bellevue, WA 98004, USA Received: 5 Jun 014, Revised: 3 Sep 014, Accepted: 5 Sep 014 Published online: 1 Mar 015 Abstract: In this paper we study a single player gae consisting of n blac checers and white checers, called shifting the checers We have proved that the iniu nuber of steps needed to play the gae for general n and is n+n+ We have also presented an optial algorith to generate an optial ove sequence of the gae consisting of n blac checers and white checers, and finally, we present an explicit solution for the general gae Keywords: single player gae, optial algorith, explicit solutions 1 Introduction Cobinatorial gaes often lead to interesting, clean probles in algoriths and coplexity theory Many classic gaes are nown to be coputationally intractable Solving a puzzle is often a challenge tas lie solving a research proble You ust have a right cleverness to see the proble fro a right angle, and then apply that idea carefully until a solution is found In this paper we study a single player gae called shifting the checers The gae is siilar to the Moving Coins puzzle [,3,7], which is played by re-arranging one configuration of unit diss in the plane into another configuration by a sequence of oves, each repositioning a coin in an epty position that touches at least two other coins In our shifting checers gae, there are n blac checers and white checers put on a table fro left to right in a row The n++1 positions of the row are nubered 1,,n++1 Initially, the n blac checers are put in the position 1,,n, and the white checers are put in the position n+,,n++1 The position n+1 is initially vacant In the final state of the gae, the left ost positions nubered 1,, are occupied by white checers, and the right ost n positions nubered +,,+n+1 are occupied by blac checers, leaving the position + 1 vacant, as shown in Fig 1 Fig 1: Shifting checers There are only two perissible types of oves A ove of the gae consists of sliding one checer into the current vacant position, or juping over the adjacent checer into the current vacant position The goal of the gae is to ae a sall nuber of oves to reach the final state of the gae We are interested in algoriths which, given integers n and, generate the corresponding ove sequences to reach the final state of the gae with the sallest nuber of steps In this paper we present an optial algorith to generate all of the optial ove sequences of the gae consisting of n blac checers and white checers This paper is structured as follows In the following 4 sections we describe the algoriths and our coputing experience with the algoriths for generating optial ove sequence of the general gae Corresponding author e-ail: wangxiaodong@qztceducn; yjwu@fzueducn Natural Sciences Publishing Cor

2 85 D Zhu et al : Coplete Solutions for a Cobinatorial Puzzle in consisting of n blac checers and white checers In Section we describe a new variant tree search based algorith for generating all optial solutions for the shifting checers gaes of the sall size A linear tie recursive construction algorith is proposed in Section 3 Based on the recursive algorith proposed in Section 3, an explicit solution for the optial ove sequence of the general gae is presented in Section 4 we discuss the nuber of optial solutions of the gae in Section 5 Soe concluding rears are in Section 6 A Bactracing Algorith In a row of checers of the gae, if two checers have different colors and the blac checer is on the left of the white checer, then the two checers are called an inversion pair For exaple, in the initial state of the gae consisting of n blac checers and white checers, since all of the n blac checers are on the left of all the white checers, there is total n inversion pairs On the other hand, in the final state of the gae, since all of the n blac checers are on the right of all the white checers, there are no inversion pairs in this case Siilarly, for the vacant position, if a blac checer is on the left of the vacant position, or a white checer is on the right of the vacant position, then the checer and the vacant position are called a vacant inversion pair For exaple, in the initial state of the gae, since all of the n blac checers are on the left of the vacant position, and all of the white checers are on the right of the vacant position, there are total n+ vacant inversion pairs On the other hand, in the final state of the gae, since all of the n blac checers are on the right of the vacant position, and all of the white checers are on the left of the vacant position, there are no vacant inversion pairs in this case Of the two types of checer oves, we can further list 1 different cases of the oves into a table, as shown in Table 1 Sliding a blac checer right into the current vacant position is denoted as slide(b, r) The other three oves slide(b, l), slide(w, r), and slide(w, l) are defined siilarly Juping a blac checer right over the adjacent white checer into the current vacant position is denoted as jup(b, w, r) The other 7 oves jup(b, w, l), jup(w, b, r), jup(w, b, l), jup(b, b, r), jup(b, b, l), jup(w, w, r), and jup(w, w, l) are defined siilarly These 1 cases of oves are nubered fro 1 to 1 The colun Inversions of Table 1 denote the inversion increent of the checer row when the corresponding case of oves applied Siilarly, the colun V-Inversions of Table 1 denotes the vacant inversion increent of the checer row when the corresponding case of oves applied It is not difficult to verify the following facts on the optial solutions to play the gae Lea 1Any optial solution for playing the gae of shifting the checers with iniu nuber of oves consists of only the classes of oves nubered fro 1 to 4 in Table 1 Proof We first notice that in an optial solution, when juping, a checer ay only jup a single checer of the opposite color If a checers jup over the adjacent checer of the sae color into the vacant position, then we will arrive an unfavorable status For exaple, if a step of jup(b, b, r) is applied, then the white checers located in the right of the vacant position will be stuc unless a step of slide(b, l) is applied iediately But these two steps can be substituted by only one step slide(b, r) The other cases can be analyzed siilarly Therefore, we now that the classes of oves nubered fro 9 to 1 in Table 1 will not appear in an optial solution with the iniu nuber of oves We have nown that fro the initial state of the gae to the final state of the gae, there are total of n inversions and n + vacant inversions to be reduced Fro Table 1 we see that for each step of oves nubered fro 1 to 8 in, at ost 1 inversion or vacant inversion can be reduced Therefore it requires at least n+n+ steps to play the gae consisting of n blac checers and white checers In other words, n+n+ is a lower bound for solving the gae In the next section, we will present an optial solution for the gae in exactly n+n+ steps If a solution for the gae contains any steps of oves nubered fro 5 to 8 in Table 1, then these steps will increase the inversions or the vacant inversions of the checerboard, and thus the nuber of steps to play the gae ust be no less than n + n + + Therefor, a solution for the gae containing any steps of oves nubered fro 5 to 8 in Table 1 cannot be an optial solution of the gae Suing up, an optial solution for playing the gae of shifting the checers with iniu nuber of oves consists of only the steps of oves nubered fro 1 to 4 in Table 1 Fro Lea 1, we can conclude that the following theore holds Theore 1For the general gae of shifting the checers consisting of n blac checers and white checers, it needs at least n+n+ steps to reach the final state of the gae fro its initial state According to Theore 1, if we can find a ove sequence to reach the final state of the gae with n+n+ steps, then the sequence will be an optial ove sequence, since no ove sequence can reach the final state of the gae in less than n+n+ steps In order to study the structures of the optial solutions for the general gae of shifting the checers, we first present a bactracing algorith [1, 5, 6] to generate all optial solutions of the gaes with sall size In the algorith described above, the paraeter i is the current nuber of steps and the paraeter e is the currently vacant position The current solution is stored in array x For i=1,,,n+n+, the ove of step i is Natural Sciences Publishing Cor

3 Appl Math Inf Sci 9, No, (015) / wwwnaturalspublishingco/journalsasp 853 Table 1: All cases of checer oves No Move Change Inversions V-Inversions 1 slide(b,r) ¼ ¼ 0-1 slide(w,l) ¼ ¼ jup(b,w,r) ¼ ¼ jup(b,w,l) ¼ ¼ jup(w,b,r) ¼ ¼ jup(w,b,l) ¼ ¼ slide(b,l) ¼ ¼ slide(w,r) ¼ ¼ jup(w,w,r) ¼ ¼ 0 10 jup(b,b,l) ¼ ¼ 0 11 jup(w,w,l) ¼ ¼ 0-1 jup(b,b,r) ¼ ¼ 0 - stored in x[i 1] This eans that we ove the checer located at positions x[i 1] to the current vacant position and leaving the positions x[i 1] the new vacant position A recursive function call BACKTRACK(1, n + 1) will generate all optial solutions which ove checers fro initial state to a final state in n+n+ steps It is not difficult to generate all optial solutions of the gae with sall size by the bactracing algorith described above Algorith 1: BACKTRACK(i, e) coent: Generate all optial solutions if i>n+n+ { if e=+1 and final state reached then then output current solution if e> and jup(b,w,r) feasible x[i 1] e ove checer at position e to vacant then BACKTRACK(i+1,e ) ove checer at position e to vacant if e<n+ and jup(b,w,l) feasible x[i 1] e+ ove checer at position e + to vacant then BACKTRACK(i+1,e+) ove checer at position e to vacant else if e > 1 and slide(b, r) feasible x[i 1] e 1 ove checer at position e 1 to vacant then BACKTRACK(i+1,e 1) ove checer at position e to vacant if e<n++1 and slide(w,l) feasible x[i 1] e+1 ove checer at position e + 1 to vacant then BACKTRACK(i+1,e+1) ove checer at position e to vacant 3 A Linear Tie Construction Algorith The bactracing algorith described in the previous section can produce all optial solutions for the gae with fixed size It generally wors only for sall size In this section, we will present a linear tie construction algorith which can produce all optial solutions in linear tie for very large size n + The Decrease-and-Conquer strategy [4] for algorith design is exploited to design our new algorith Without loss of generality, we assue n in the following discussion Since there are only 4 possible oves slide(w, l), slide(b, r), jup(b, w, l), and jup(b,w,r), we can siplify our notation for these 4 oves to slide(l), slide(r), jup(l), and jup(r) in the following discussion 31 A special case of the proble We first focus on the special case of n= If we denote a blac checer by b, a white checer by w, and the vacant position by O, then any status of the checer board can be specified by a sequence consisting of characters b, w and O The special case of our proble is then equivalent to transforing the initial sequence b bow w to the sequence w wob b in the iniu nuber of steps We have noticed that a ey status of the checer board can be reached fro the initial status with iniu nuber of steps Lea The initial status of the checer board b bow w can be transfored to one of the status of the checer board Obw bw or bw bwo in (+1) steps Natural Sciences Publishing Cor

4 854 D Zhu et al : Coplete Solutions for a Cobinatorial Puzzle in Proof We can design a recursive algorith to solve this proble as follows Algorith 31: MOVE1(t) if t > 1 then MOVE1(t 1) coent: t 1 jups for i 1 to t 1 coent: 1 slide SLIDE(dir) coent: change oving direction CHANGE(dir) In the algorith described above, the paraeter t is the recursion depth, or the nuber of blac checers to be treated The variable dir is used to deterine the current oving direction Its value l indicates the checer should be oved left, otherwise the checer should be oved right The initial value of dir can be set to l or r, which will lead to different oving sequences The current direction dir can be changed in O(1) tie by the algorith CHANGE(dir) as follows Algorith 3: CHANGE(dir) if dir=r then dir l else dir r Based on the algorith above, the Lea can be proved by induction The oving steps generated by the algorith MOVE1() for the first two easy cases of = 1 and = are shown in Table and Table 3 The nuber of steps for these two cases are 1 and 3 respectively The lea is correct for the base cases Assue that, the Lea is true for < t For the case of = t, the algorith MOVE1(t 1) is applied first and the status of the checerboard is transfored to t t bobw bww or bbw bwow depending on the initial value of dir Then, t 1 jups followed by 1 slide of the algorith MOVE1() will transfor the status of the t t checerboard to bw bwo or Obw bw The algorith MOVE1(t 1) needs (t 1)t/ steps by the induction hypothesis, so the nuber of steps used by the algorith MOVE1(t) is (t 1)t/+ t 1+1=(t 1)t/+ t = t(t+ 1)/ The proof is copleted The ey status of the checerboard bw bwo or Obw bw can be transfored to another ey status of the checerboard Owb wb or wb wb O readily by jups Any of these two status of the checerboard can then be transfored to the final status w wob b This proble is exactly the inverse proble of Lea Lea 3The ey status of the checerboard Owb wb or wb wb O can be transfored to the final status w wob b in (+1) steps Proof We can design a recursive algorith to solve this proble, which is exactly a reversed algorith of the algorith MOVE1() Algorith 33: MOVE4(t) coent: change oving direction CHANGE(dir) coent: 1 slide SLIDE(dir) coent: t 1 jups for i 1to t 1 coent: recursive call if t > 1 then MOVE4(t 1) In the algorith described above, the paraeter t is the recursion depth, or the nuber of blac checers to be treated The variable dir is used to deterine the current oving direction Its initial value is retained fro previous coputation Based on the algorith above, the Lea can be proved by induction The first two easy cases of = 1 and = are siilar to the cases of Table and Table 3 The nuber of steps for these two cases are 1 and 3 respectively The lea is correct for the base cases Assue that, the Lea is true for < t In the case of = t, the algorith MOVE4(t) ipleent 1 slide and t 1 jups first The ey status of the checerboard Owb wb or wb wbo can then be transfored to the status of the checerboard wwb wb Ob or wo wb wb b respectively Then a recursive call Natural Sciences Publishing Cor

5 Appl Math Inf Sci 9, No, (015) / wwwnaturalspublishingco/journalsasp 855 Table : Move1 for the easy case of =1 Direction Step Move Status dir=1 dir= 1 0 ¼ 1 slide(l) ¼ 0 ¼ 1 slide(r) ¼ Table 3: Move1 for the easy case of = Direction Step Move Status dir=1 dir= 1 0 ¼ 1 slide(l) ¼ jup(r) ¼ 3 slide(r) ¼ 0 ¼ 1 slide(r) ¼ jup(l) ¼ 3 slide(l) ¼ MOVE4(t 1) is applied to transfor the checerboard to the final status w wob b The algorith MOVE4(t 1) needs(t 1)t/ steps by the induction hypothesis, so the nuber of steps used by the algorith MOVE4(t) is 1+ t 1+(t 1)t/=t+(t 1)t/=t(t+ 1)/ The proof is copleted The 3 stages of the algoriths can now be cobined into a new algorith to solve our proble for the special case of n= as follows Algorith 34: MOVE(, d) coent: initial oving direction dir d coent: first stage MOVE1() coent: jups for i 1 to coent: last stage MOVE4() The algorith requires (+1)/++(+1)/= + steps It has been nown that + is a lower bound to solve the gae consisting of blac checers and white checers by Theore 1 Therefore, our algorith is optial to solve the gae for the special case of n = Fro this point, we can also clai that the algoriths MOVE1() and MOVE4() presented in the proofs of Lea and Lea 3 are also optial Otherwise, there ust be an algorith to solve the proble in less than + steps and this is ipossible 3 The algorith for the general case of the proble We have discussed the special case of n = In this subsection, we will discuss the general cases n>of the proble In these general cases, n > 0 We can first use the algorith MOVE1() to transfor n the checerboard to the status b bobw bw or n b bbw bwo in (+1) steps Then jups are applied to transfor the checerboard to the status n n b bowb wb or b bwb wbo At this point, we have to try to ove the leftost n blac checers to the rightost n positions It is not difficult to do this by a siple algorith siilar to the algorith MOVE1() Lea 4The ey status of the checerboard n n b bowb wb or b bwb wbo can be transfored n n to the status wb wbob b or Owb wbb b in (n )(+1) steps Proof We can design a recursive algorith to solve this proble as follows Algorith 35: MOVE3(t) coent: 1 slide to right SLIDE(r) coent: change juping direction CHANGE(dir) coent: jups for i 1to coent: recursive call if t > 1 then MOVE3(t 1) In the algorith described above, the paraeter t is the recursion depth, or the nuber of blac checers to be oved to the right ost positions The variable dir is used to deterine the current juping direction Its initial value is retained fro previous coputation Natural Sciences Publishing Cor

6 856 D Zhu et al : Coplete Solutions for a Cobinatorial Puzzle in Based on the algorith above, the Lea can be proved by induction on n When n =1, we have to ove the leftost blac checer to rightost We first ae a slide right, then jups followed as described by the algorith MOVE3() The status of the checerboard will be changed to wb wbob or Owb wbb It costs +1 steps The lea is correct for the base case of n = 1 Assue that, the Lea is true for n <t For the case of n = t, the algorith MOVE3(t) ipleent 1 slide and jups first The ey status of the t t checerboard b bowb wb or b bwb wbo can then be transfored to the status of the checerboard t 1 t 1 b bwb wbob or b bowb wbb respectively Then a recursive call MOVE3(t 1) is applied to t transfor the checerboard to the status wb wbob b t or Owb wbb b The algorith MOVE3(t 1) needs(t 1)(+1) steps by the induction hypothesis, so the nuber of steps used by the algorith MOVE3(t) is +1+(t 1)(+1)=t(+1) The proof is copleted The 4 stages of the algoriths can now be cobined into a new algorith to solve our proble for the general cases of n as follows Algorith 36: MOVE(n,, d) coent: initial oving direction dir d coent: first stage MOVE1() coent: second stage for i 1 to coent: third stage if n > 0 then MOVE3(n ) coent: last stage MOVE4() algorith MOVE3() is also optial Otherwise, there ust be an algorith to solve the proble in less than n+n+ steps and this is ipossible Theore The algorith MOVE(n,,d) requires n+n+ steps to solve the general oving checers gae consisting of n blac checers and white checers, and the algorith is optial 33 Reove recursions The algoriths MOVE1(), MOVE3() and MOVE4() are all recursive algoriths The recursions of these algoriths can be easily reoved by only one for loop The equivalent iterative algorith for solving the general oving checers gae consisting of n blac checers and white checers can be described as follows Algorith 37: ITERATIVE MOVE(n,, d) coent: initial oving direction dir d coent: stage 1 for i 1to for j 1 to i 1 do SLIDE(dir) CHANGE(dir) coent: stage for i 1to coent: stage 3 for i 1to n SLIDE(r) CHANGE(dir) do for j 1 to coent: stage 4 for i downto 1 CHANGE(dir) SLIDE(dir) do for j 1 to i 1 By Lea, Lea 3 and Lea 4 we now that the algorith requires (+1)/++(n )(+1)+ (+1)/=n+n+ steps It has been nown fro Theore 1 that n+n+ is a lower bound to solve the gae consisting of n blac checers and white checers Therefore, our algorith is optial to solve the gae for the general cases of n We can also clai that the 4 The Explicit Solutions to the Proble The optial solution found by the algorith MOVE() or ITERATIVE MOVE() can be presented by a vector x For i = 1,,,n+n+, the step i of the optial ove sequence is given by x i This eans that the checer located at position x i will be oved in step i to the current Natural Sciences Publishing Cor

7 Appl Math Inf Sci 9, No, (015) / wwwnaturalspublishingco/journalsasp 857 vacant positions and leaving the positions x i the new vacant positions This can also be viewed that x is a function of i, which is called a ove function In this section we will discuss the explicit expression of function x If we denote x 0 = n+1 and d i = x i 1 x i,1 i n+n+ (1) then the vector d will be a ove direction function of the corresponding ove sequence A related function t can then be defined as t i = i j=1 d j,1 i n+n+ Since t i = i j=1 we have d j = i j=1 (x j 1 x j )=x 0 x i = n+1 x i x i = n+1 t i,1 i n+n+ () Therefore, our tas is equivalent to copute the function t efficiently In this section, the functions x and t will be divided into three parts The first part is corresponding to the first two stages of the algorith ITERATIVE MOVE() presented in the last section The second part is corresponding to the stage 3 of the algorith ITERATIVE MOVE() and the third part is corresponding to the stage 4 41 The first part of the solution Siilar to the initial value of dir which can be set to l or r, the first ove direction d 1 can be set to 1 or -1 If we set d 1 = 1, then fro the algorith ITERATIVE MOVE() presented in the last section, the ove direction sequence for the stage 1 and ust be 1,, 1,,,1,,( 1) 1, ( 1),,( 1) This ove direction sequence can be divided into sections as ,, 1,,,, ( 1) 1,( 1),,( 1) The section j consists of 1 slide and j jups and thus has a size of j+ 1 The total length of the sequence is therefore s 1 = j=1 ( j+ 1) = (+3)/ Our tas is now to find t i = i j=1 d j quicly for each 1 i s 1 If we denote the j+ 1 eleents of the section j as a t j,1 t j + 1, and the su of section j as a j = j+1 t=1 a t j, j = 1,,, then it is not difficult to see that for each j = 1,,, { ( 1) a t j = j 1 t = 1 ( 1) j (3) t > 1 and for 1 j+ 1, t=1 a t j =( 1) j ( 3) (4) Therefore, a j =( 1) j ( j 1), j = 1,, These sus for an alternating sequence 1,3, 5,,( 1) ( 1) For each 1, we have, j=1 a j = j=1 The steps in each section ust be ( 1) j ( j 1)=( 1) (5) 1 1,, 3,4,5,, ( 1)(+)/+1,,(+3)/ If we denote the j+ 1 steps of the section j as b t j,1 t j+1, and the boundary of section j as b j = b ( j+1) j, j= 1,,, then it is not difficult to see that for each j = 1,,, { b j = j( j+ 3)/ 1 j b t j = b j 1 + t 1 t j+ 1,1 j For any integer 1 i b, the corresponding integer such that the integer i falls into the section can be coputed by a function f 1 (x) as follows Lea 5Let f 1 (x) = 8x+1 1 For any integer 1 i b, it ust be a step nuber in the section = f 1 (i) Proof It can be seen readily that function f 1 (x) is a strictly increasing function on(0,+ ) For each section,1, its first step nuber is b 1 + 1=( 1)(+)/+1 and it satisfies f 1 ( ( 1)(+) + 1)= (6) 4( 1)(+)+9 1 (+1) = 1 = Therefore, for each integer i in the section, we have, f 1 (i)<+1 This eans f 1 (i) = The proof is copleted Fro Lea 5 and forula (4) and (5), we can now copute t i = i j=1 d j,1 i s 1 as follows Let α = f 1 (i), then, t i = i j=1 d j = α 1 j=1 a j + i j=b α 1 +1 d j = ( 1) α 1 (α 1)+( 1) α ((i b α 1 ) 3) Natural Sciences Publishing Cor

8 858 D Zhu et al : Coplete Solutions for a Cobinatorial Puzzle in It follows that for each 1 i s 1, t i =( 1) α (i α(α+ )) (7) where, α = 8i+1 1 It follows fro forula () that for each 1 i s 1, x i = n+1 ( 1) α (i α(α+ )) (8) If we set d 1 = 1, a siilar result will be obtained In this case, we have, x i = n+1+( 1) α (i α(α+ )) (9) Cobine these two cases, we conclude that, x i = n+1 d 1 ( 1) α (i α(α+ )) (10) 4 The second part of the solution If we set d 1 = 1, then according to the algorith ITERATIVE MOVE() presented in the last section, the ove direction sequence for the stage 3 ust be in the for of (11) This ove direction sequence can be divided naturally into n sections The section j consists of 1 slide and jups and thus has a size of +1 The total length of the sequence is therefore s = (n )(+1) Our tas for this part is now to find t i = i j=1 d j quicly for each s i s 1 + s If we denote the +1 eleents of the section j as a t j,1 t + 1, and the su of section j as a j = t=1 +1 a t j, j = 1,,n, then it is not difficult to see that for each j = 1,,n, { 1 t = 1 a t j = ( 1) + j (1) t > 1 and for 1 +1, t=1 a t j = 1+( 1) + j ( ) (13) Therefore, a j = 1 + ( 1) + j, j = 1,,n These n sus for an alternating sequence j=1 1+( 1) +1,1+( 1) +,,1+( 1) n For each 1, we have, a j = + j=1 ( 1) + j = +( 1) + ( 1) (14) If we set j=i s 1, then the steps in each section ust be in the for of (15) If we denote the +1 steps of the section j as b t j,1 t +1, and the boundary of section j as b j = b (+1) j, j = 1,,n, then it is not difficult to see that for each j = 1,,n, { b j = j(+1) 1 j n b t j = b j 1 + t 1 t +1,1 j n (16) For any integer 1 j b, the corresponding integer such that the integer j falls into the section can be coputed by a function f (x) as follows Lea 6Let f (x)= x+ +1 For any integer 1 j b, it ust be a step nuber in the section = f ( j) Proof It can be seen readily that function f (x) is a strictly increasing function on(0,+ ) For each section,1 n, its first step nuber is b 1 +1=( 1)(+1)+1 and it satisfies f (( 1)(+1)+1)= ( 1)(+1) = (+1) +1 = Therefore, for each integer j in the section, we have, f ( j)<+1 This eans f ( j) = The proof is copleted Fro Lea 5 and forula (13) and (14), we can now copute t i = i j=1 d j,s i s 1 + s as follows Let r= i s 1, β = f (r), and p=r (β 1)(+1) then, t i = i j=1 d j = t s1 + β 1 j=1 a j+ p j=1 a jβ Therefore t i t s1 = β 1+( 1) +β 1 ( 1) + 1+( 1) +β (p ) = β +( 1) +β (p ) (( 1) +β +( 1) +β ) = β +( 1) +β (p (1+( 1) β )) It follows that for each s i s 1 + s, t i = t s1 + β+( 1) +β (p (1+( 1) β )) (17) where, β = i s 1+ +1, and p=i s 1 (β 1)(+1) It follows fro forula () that for each s i s 1 + s, x i = n+1 t s1 β ( 1) +β (p (1+( 1) β )) (18) If we set d 1 = 1, a siilar result will be obtained In this case, we have, x i = n+1 t s1 β+( 1) +β (p (1+( 1) β )) (19) Cobine these two cases, we conclude that, x i = x s1 β d 1 ( 1) +β (p (1+( 1) β )) (0) Natural Sciences Publishing Cor

9 Appl Math Inf Sci 9, No, (015) / wwwnaturalspublishingco/journalsasp 859 1, ( 1) +1,,( 1) +1,1, ( 1) +,,( 1) +,,1, ( 1) n,,( 1) n (11) ,,+1, +,,+,, (n 1)(+1)+1,,(n )(+1) (15) 43 The third part of the solution According to the algorith ITERATIVE MOVE() presented in the last section, if d 1 = 1, then the ove direction sequence for the stage 4 ust be 1 ( 1) n+1 (1,,,, 1,,,,,( 1) 1 ) This ove direction sequence can be divided naturally into sections The section j consists of 1 slide and j jups and thus has a size of j+1 The total length of the sequence is therefore s 3 = (+1)/ Our tas for this part is now to find t i = i j=1 d j quicly for each s 1 + s + 1 i s 1 + s + s 3 = n+n+ If we denote the j+1 eleents of the section j as a t j,1 t j+1, and the su of section j as a j = j+1 t=1 a t j, j= 1,,, then it is not difficulty to see that for each j = 1,,, { ( 1) n+ j t = 1 a t j = ( 1) n+ j (1) t > 1 and for 1 j+ 1, t=1 a t j =( 1) n+ j ( 1) () Therefore, a j = ( 1) n+ j (( j)+1), j = 1,, These sus for an alternating sequence ( 1) n+1 (( 1), ( 3),,( 1) 1 ) For each 1, we have, j=1 a j = j=1 ( 1)n+ j ( ( j 1)) =( 1) n ((( 1) 1)/) ( 1) ) Therefore, j=1 a j =( 1) n (( 1) ( ) ) (3) If we set j = i s 1 s, then the step nubers in each sections ust be 1 1,,, +1,, 1,, (+1)/ If we denote the j+ 1 steps of the section j as b t j,1 t j+ 1, and the boundary of section j as 1 b j = b ( j+1) j, j = 1,,, then it is not difficulty to see that for each j = 1,,, { b j = j(+1) j( j+ 1)/ 1 j b t j = b j 1 + t 1 t j+ 1,1 j (4) For any integer 1 j b, the corresponding integer such that the integer j falls into the section can be coputed by a function f 3 (x) as follows Lea 7Let f 3 (x) = (+1) x+9/4+ 3/ For any integer 1 j b, it ust be a step nuber in the section = f 3 ( j) Proof It can be seen readily that function f (x) is a strictly increasing function on (0, ( + 1)/] For each section,1, its first step nuber is b 1 +1=( 1)(+ 1) ( 1)/+1 and it satisfies f 3 (( 1)(+1) ( 1)/+1)= +3/ (+1) ( 1)(+1)+ ( 1) +9/4 = +3/ ( 3/) = Therefore, for each integer j in the section, we have, f 3 ( j)<+1 This eans f 3 ( j) = The proof is copleted Fro Lea 7 and forula (1) and (), we can now copute t i = i j=1 d j,s 1 + s + 1 i n+n+ as follows Let r=i s 1 s, γ = f 3 (r), and q=r (γ 1)(+ 1)+γ(γ 1)/ then, t i = i j=1 d j = t s + γ 1 j=1 a j+ q j=1 a jγ Therefore t i t s =( 1) n (( 1) γ 1 ( γ+ 1) )+( 1) n+γ (q 1) =( 1) n+γ (γ+ q ) ( 1) n It follows that for each s 1 + s + 1 i n+n+, t i = t s +( 1) n+γ (γ+ q ) ( 1) n (5) where, γ = (+1) (i s 1 s )+9/4+ 3/, and q=i s 1 s (γ 1)(+1)+γ(γ 1)/ Natural Sciences Publishing Cor

10 860 D Zhu et al : Coplete Solutions for a Cobinatorial Puzzle in It follows fro forula () that for each s 1 + s + 1 i n+n+, x i = n+1 t s ( 1) n+γ (γ+q )+( 1) n (6) If we set d 1 = 1, a siilar result will be obtained In this case, we have, x i = n+1 t s +( 1) n+γ (γ+q ) ( 1) n (7) Cobine these two cases, we conclude that, x i = x s d 1 (( 1) n+γ (γ+ q ) ( 1) n ) (8) Suing up, the explicit optial solutions for solving the general gae of shifting the checers consisting of n blac checers and white checers can be given in three parts as shown in the following Theore Theore 3In the general gae of shifting the checers consisting of n blac checers and white checers, its optial ove steps x i,1 i n + n +, can be expressed explicitly in forulas (9) and (30) where, d is the first ove direction It requires O(1) tie to copute( 1) for any positive integer, since { ( 1) 1 if odd = 1 if even Therefore, for each 1 i n+n+, x i can be coputed in O(1) tie by using the forula (7), and then the optial ove sequence of the general gae consisting of n blac checers and white checers can be easily coputed in optial O(n + n + ) tie 5 The Nuber of Optial Solutions In this section we will use the state space graph of a gae as a tool to discuss the nuber of optial solutions of our proble A state refers to the status of a gae at a given oent In our proble it ust be the positions of the checers on the checerboard In solving a proble one starts fro soe initial state and tries to reach a goal state by passing through a series of interediate states In gae playing, each ove on the gae board is a transition fro one state to another If we thin of each state being connected to those states which can follow fro it, we have a graph Such a collection of interconnected states is called a state space graph For exaple, the initial state and the goal state in our proble n n are b bow w and w wob b A state space graph of the easy case of n==1 is shown in Fig In state based search, a coputer progra ay start fro an initial state, then loo at one of its successor or children states and so on until it reaches a goal state It Fig : A state space graph of the easy case of n==1 ay reach a dead end state fro where it cannot proceed further In such a situation the progra ay bactrac, ie undo its last ove and try an alternative successor to its previous state A path fro the initial state to the goal state constitutes a solution An optial solution of the proble corresponds to a shortest path fro the initial state to the goal state in the state space graph of the proble Our tas in this section is to count the nuber of different optial solutions of the proble, which is equivalent to count the nuber of different shortest paths fro the initial state to the goal state in the state space graph of the proble For exaple, in the easy case of n = = 1, we have two different optial solutions of the proble, as shown in Fig In any optial solutions, the following 4 special states are especially iportant: n ξ 0 = b bow w ξ g = w wob b n ξ 1 = b bwow w n 1 n 1 ξ = b bobw w The state ξ 0 is the initial state, and ξ g is the goal state of the gae Fro Lea 1 we now that in any optial ove sequence, only the classes of oves nubered fro 1 to 4 in Table 1 are possible With this restriction, our first ove fro the initial state ust be a slide in one direction If the first ove is slide(l), then the initial state ξ 0 will be changed to ξ 1 Otherwise, the first ove ust be slide(r), and the initial state ξ 0 will be changed to ξ In other words, the shortest paths fro the initial state ξ 0 to the goal state ξ g ust be in the fors ξ 0,ξ 1,P 1,ξ g or ξ 0,ξ,P,ξ g, where ξ 1,P 1,ξ g is a shortest path fro ξ 1 to the goal state ξ g and ξ,p,ξ g is a shortest path fro ξ to the goal state ξ g If we have ade a first ove, the Natural Sciences Publishing Cor

11 Appl Math Inf Sci 9, No, (015) / wwwnaturalspublishingco/journalsasp 861 n+1 ( 1) α d(i α(α+ )) 1 i s 1 x i = x s1 β ( 1) +β d(p (1+( 1) β )) s i s 1 + s x s d(( 1) n+γ (γ+ q ) ( 1) n ) s 1 + s + 1 i n+n+ s 1 = (+3)/ s =(n )(+1) α = 8i+1 1 β = i s γ = (+1) (i s 1 s )+9/4+ 3/ p=i s 1 (β 1)(+1) q=i s 1 s (γ 1)(+1)+γ(γ 1)/ (9) (30) following paths P 1 or P can be deterined by the oving rules of Lea 1 There are two cases to be distinguished 51 The general case of n >1 Without loss of generality, let the first ove be slide(l), and the initial state ξ 0 be changed to ξ 1 We now consider the shortest path P 1 We have noticed in the proof of Lea 1 that in any optial ove sequence no two or ore pieces of the sae color can coe together, unless the two ends of the sequence Fro this point of view, we can prove the following facts by induction Lea 8The following special states λ 1,λ,,λ ust be in the path P 1, and for all i such that 1 i 1, the shortest paths between the states λ i and λ i+1 are unique n 1 1 λ 1 = b bowbw w n λ = b bwbwbow w n t t t b bwb wbow w t even λ t = n t t t b bowb wb w w t odd n b bwb wbo even λ = n b bowb wb odd Proof It is clear that the only way to ove optially fro the state ξ 1 is a ove of jup(r), which changes the state ξ 1 to λ 1 Siilarly, fro the state λ 1, the unique choice aong the 4 possible oves is a ove of slide(r), otherwise two pieces of the sae color will coe together, which is forbidden Following this ove, two jup(l) are forced for the sae reason At this point, the state λ is reached, and the shortest path fro λ 1 to λ is unique Suppose the clai is true for i < t In the case of i=t <, we have to ove fro the state λ t to λ t+1 If t n t t t is even, then λ t = b bwb wb Ow w It is clear that fro the state λ t, the oves jup(l) and jup(r) will ae two pieces of the sae color coe together If a slide(r) is applied first, then no jups are possible in the following oves The gae will reach a dead end to this case Therefore, the only choice for the first ove is slide(l) Following this ove, t + 1 jup(r) are forced for the sae reason, and then we reached the state λ t+1 The shortest path fro λ t to λ t+1 is clearly unique The clai is therefore true by induction If t is odd, the proof is siilar The proof is copleted Lea 9The following special states µ 1, µ,, µ n ust be in the path P 1, and for all i such that 1 i n 1, the shortest paths between the states µ i and µ i+1 are unique n 1 b bowb wbb even µ 1 = n 1 b bwb wb Ob odd n b bwb wbobb even µ = n b bowb wbbb odd n t t b bwb wbob b + t even µ t = n t t b bowb wbb b + t odd Natural Sciences Publishing Cor

12 86 D Zhu et al : Coplete Solutions for a Cobinatorial Puzzle in n wb wbob b n even µ n = n Owb wbb b n odd Proof n If is even, then λ = b bwb wbo It is clear that fro the state λ, only two oves jup(r) and slide(r) are possible If a jup(r) is applied first, then no slides are possible in the following oves The gae will reach a dead end to this case Therefore, the only choice for the first ove is slide(r) Following this ove, jup(r) are forced for the sae reason, and then we reached the state µ 1 The shortest path fro λ to µ 1 is clearly unique In the case of odd, the analysis is siilar Suppose the clai is true for i<t For the case of i= t < n, we have to ove fro the state µ t to µ t+1 n t If + t is odd, then µ t = b bowb wbb b It is clear that fro the state λ t, the oves jup(l) will go bac and thus not optial If a jup(r) is applied first, then the gae will reach a dead end n t t Ob bwb wb b b for this case If a slide(l) is applied first, then the next ove ust be a jup(r), n t 1 which will lead to the state b b Owbbwb wb b b In this state, the two blac pieces coe together, and they are not at the right end since > 1 This is ipossible Therefore, the only choice for the first ove is slide(r) Following this ove, jup(l) are forced for the sae reason, and then we reached the state µ t+1 The shortest path fro µ t to µ t+1 is clearly unique The clai is therefore true by induction n t If +t is even, then µ t = b bwb wbob b It is clear that fro the state λ t, the oves jup(l) and jup(r) will go bac and thus not optial If a slide(l) is applied first, then the gae will reach a dead end n t t b bwb wbb bo for this case Therefore, the only choice for the first ove is slide(r) Following this ove, jup(l) are forced for the sae reason, and then we reached the state µ t+1 The shortest path fro µ t to µ t+1 is clearly unique The clai is therefore true by induction The proof is copleted Lea 10The following special states ν 1,ν,,ν ust be in the path P 1, and for all i such that 1 i 1, the shortest paths between the states ν i and ν i+1 are unique ( 1) n +1 wowb wb b b n even ν 1 = ( 1) n +1 wwb wbo b b n odd t t t ( ) n + wwwb wbo b b n even ν = ( ) n + wwowb wb b b n odd t ( t) n +t w wwb wbo b b n+ t even ν t = t ( t) n +t w wowb wb b b n+ t odd ν = w wob b Proof The clai of this lea is syetric to Lea 8, and therefore the proof is also syetric Cobining Lea 8,9 and 10, we conclude that in the shortest path ξ 0,ξ 1,P 1,ξ g fro the initial state ξ 0 to the goal state ξ g, the shortest path P 1 ust be unique The analysis for the case of first ove slide(r) is siilar, and we can conclude also that in the shortest path ξ 0,ξ,P,ξ g fro the initial state ξ 0 to the goal state ξ g, the shortest path P ust be unique Finally we conclude that in the general case of n > 1, there are only two different shortest paths fro the initial state ξ 0 to the goal state ξ g Therefore, in this case, the nuber of optial solutions of the gae is The two different optial solutions of the gae can be coputed by the forula (9) of Theore 3 in linear tie 5 The special case of n =1 In this special case, Lea 8 and 10 are also true Therefore, the shortest paths fro the initial state ξ 0 to the state λ 1, and the shortest path fro the state µ n 1 to the state ν 1 = ξ g are still unique The special states of Lea 9 becoe coplicated in the case of =1, since the shortest paths between any two consecutive states of these special states are no longer unique In this special case, we will expand the special states µ 1, µ,, µ n of Lea 9 further to µ i j,0 i n 1,1 j as follows n 1 µ 01 = b bwbo n 1 µ 0 = b bowb n µ 11 = b bwbob n µ 1 = b bowbb n Natural Sciences Publishing Cor

13 Appl Math Inf Sci 9, No, (015) / wwwnaturalspublishingco/journalsasp 863 n t 1 t µ t1 = b bwbob b n t 1 t µ t = b bowbb b n 1 µ (n 1)1 = wbob b n 1 µ (n 1) = Owbb b The clai of Lea 9 will odified to the following Lea 11 Lea 11The special states µ i j,1 i n 1,1 j ust be in the path P 1 or P For each i such that 0 i n, there is only one shortest path fro the state µ i1 to the state µ (i+1) ; there are two shortest paths fro the state µ i, one to the state µ (i+1)1, and the other to the state µ (i+1) Proof It is clear that µ 01 = ξ 1 and µ 0 = ξ Two oves slide(r) and jup(r) change the state µ 01 to µ 1 Two oves slide(r) and jup(l) change the state µ 0 to µ 11, and another two oves slide(l) and jup(r) change the state µ 0 to µ 1 For the general case of 0 i n 1, two oves slide(r) and jup(r) change the state µ i1 to µ (i+1) ; two oves slide(r) and jup(l) change the state µ i to µ (i+1)1, and another two oves slide(l) and jup(r) change the state µ i to µ (i+1) Notice that the length of the shortest paths fro µ 01 = ξ 1 and µ 0 = ξ to µ (n 1)1 and µ (n 1) is (n 1) by Lea 4, we conclude that The special states µ i j,1 i n 1,1 j ust be in the path P 1 or P The proof is copleted Denote the nuber of different shortest paths fro the state µ i j to the states µ (n 1)1 or µ (n 1) as ρ(i, j), then fro Lea 11 we have, ρ(i,1)=ρ(i+1,) ρ(i,)=ρ(i+1,1)+ρ(i+1,) (31) ρ(n,1)= 1 ρ(n,)= The solution of this recurrence is { ρ(i,1)=fn i (3) ρ(i,)=f n i+1 where F n is the nth Fibonacci nuber The sequence F n of Fibonacci nubers is defined by the recurrence relation F n = F n 1 + F n, with seed values F 0 = 0,F 1 = 1 The Fibonacci (( nubers have a closed-for solution F n = 1 ) 1+ 5 n ( ) n ) Therefore, in this case, the nuber of different shortest paths fro the initial state ξ 0 to the goal state ξ g is ρ(0,1)+ρ(0,)= F n + F n+1 = F n+ Suing up, the nuber of optial solutions for solving the general gae of shifting the checers consisting of n blac checers and white checers can be given in the following Theore Theore 4For the general gae of shifting the checers consisting of n blac checers and white checers, let ϕ(n, ) be the nuber of optial solutions for solving the gae, then ϕ(n,) can be expressed explicitly as follows ( ( ) n+ ( ) ) n+ 1 5 ϕ(n,)= 5 n =1 n >1 (33) 6 Concluding Rears We have studied the general shifting the checers gae consisting of n blac checers and white checers It has been proved in the section that the iniu nuber of steps needed to play the gae for general n and is n+n+ All of the optial solutions for the oving checers gae of sall size can be found by a bactracing algorith presented in section In the section 3, a linear tie recursive construction algorith which can produce an optial solution in linear tie for very large size n and is presented The tie cost of the new algorith is O(n) and O(n+) space is used In Section 4, an extreely siple explicit solution for the optial oving sequences of the general gae is given The forula gives for each individual step i, its optial ove in O(1) tie Finally, in Section 5 we give the coplete optial solutions for the gae in general cases Another siilar gae is to reverse the n checers nubered 1,,n by two perissible types of oves slide and jup It is not clear whether our ethods presented in this paper can be applied to this gae We will investigate the proble further Acnowledgent This wor was supported by the Natural Science Foundation of Fujian under Grant No013J0147, Fujian Provincial Key Laboratory of Data-Intensive Coputing and Fujian University Laboratory of Intelligent Coputing and Inforation Processing References [1] R Bird, Pearls of Functional Algorith Design, 58-74, Cabridge University Press, 010 Natural Sciences Publishing Cor

14 864 D Zhu et al : Coplete Solutions for a Cobinatorial Puzzle in [] Eri D Deaine, Playing gaes with algoriths, Algorithic cobinatorial gae theory Proceedings of the 6th Syposiu on Matheatical Foundations in Coputer Science, LNCS 136, 18-3, 001 [3] Eri D Deaine and Martin L Deaine, Puzzles, Art, and Magic with Algoriths, Theory of Coputing Systes, 39, , 006 [4] A Levitin and M Levitin, Algorithic Puzzles, 3-31, Oxford University Press, New Yor, 011 [5] J Kleinberg, E Tardos Algorith Design, 3-38, Addison Wesley, 005 [6] DL Kreher and D Stinson, Cobinatorial Algoriths: Generation, Enueration and Search, , CRC Press, 1998 [7] John S Gray, The shuttle puzzle ł A lesson in proble solving, Journal of Coputing in Higher Education, 10, 56-70, 1998 [8] Hazewinel, Michiel, ed (001), Fibonacci nubers, Encyclopedia of Matheatics, Springer, Daxin Zhu received his MSc degree in Coputer Science fro Huaqiao University of China in 003 He is now an associate professor in Quanzhou Noral University of China His current research interests include design and analysis of algoriths, networ architecture and data intensive coputing Lei Wang PhD in Coputer Science fro Georgia Institute of Technology 011 Applied researcher at Microsoft Has experience in coputer science with ephasis in algorith design The areas of interest are approxiation and randoized algoriths,echanis design, aret equilibriu coputation Xiaodong Wang is currently a professor in Coputer Science Departent of Quanzhou Noral University and Fuzhou University,China Has experience in coputer science and applied atheatics The areas of interest are design and analysis of algoriths,exponential-tie algoriths for NP-hard probles,strategy gae prograing Yingjie Wu received the PhD degree in coputer science fro Southeast University of China in 01 He is currently an associate professor in the Departent of Coputer Science at Fuzhou University, China His research interests include data privacy and data ining Natural Sciences Publishing Cor