The Branhing Fator of Regular Searh Spae Stefan Edelkamp Intitut für Informatik Am Flughafen 17 79110 Freiburg edelkamp@informatik.uni-freiburg.de Rihard E. Korf Computer Siene Department Univerity of California Lo Angele, Ca. 90095 korf@.ula.edu Abtrat Many problem, uh a the liding-tile puzzle, generate earh tree where different node have different number of hildren, in thi ae depending on the poition of the blank. We how how to alulate the aymptoti branhing fator of uh problem, and how to effiiently ompute the exat number of node at a given depth. Thi information i important for determining the omplexity of variou earh algorithm on thee problem. In addition to the liding-tile puzzle, we alo apply our tehnique to Rubik Cube. While our tehnique are fairly traightforward, the literature i full of inorret branhing fator for thee problem, and the error in everal inorret method are fairly ubtle. Introdution Many AI earh algorithm, uh a depth-firt earh (DFS), depth-firt iterative-deepening (DFID), and Iterative-Deepening-A* (IDA*) (Korf 1985) earh a problem-pae tree. While mot problem pae are in fat graph with yle, deteting thee yle in general require toring all generated node in memory, whih i impratial for large problem. Thu, to onerve pae, thee algorithm earh a tree expanion of the graph, rooted at the initial tate. In a tree expanion of a graph, eah ditint path to a given problem tate give rie to a different node in the earh tree. Note that the tree expanion of a graph an be exponentially larger than the underlying graph, and in fat an be infinite even for a finite graph. The time omplexity of earhing a tree depend primarily on the branhing fator b, and theolution depth d. The olution depth i the length of a hortet olution path, and depend on the given problem intane. The branhing fator, however, typially onverge to a ontant value for the entire problem pae. Thu, omputing the branhing fator i an eential tep in Copyright 1998, Amerian Aoiation for Artifiial Intelligene (www.aaai.org). All right reerved. determining the omplexity of a earh algorithm on a given problem, and an be ued for eleting among alternative problem pae for the ame problem. The branhing fator of a node i the number of hildren it ha. In a tree where every node ha the ame branhing fator, thi i alo the branhing fator of the tree. The diffiulty our when different node at the ame level of the tree have different number of hildren. In that ae, we an define the branhing fator at a given depth of the tree a the ratio of the number of node at that depth to the number of node at the next hallower depth. In mot ae, the branhing fator at a given depth onverge to a limit a the depth goe to infinity. Thi i the aymptoti branhing fator, and the bet meaure of the ize of the tree. In the remainder of thi paper, we preent ome imple example of problem pae, inluding liding-tile puzzle and Rubik Cube, and how how to ompute their aymptoti branhing fator, and how not to. We formalize the problem a the olution of a et of imultaneou equation, whih an be quite large in pratie. A an alternative to an analyti olution, we preent an effiient numerial tehnique for determining the exat number of node at a given depth, and for etimating the aymptoti branhing fator to a high degree of preiion. Finally, we preent ome data on the branhing fator of variou liding-tile puzzle. Example Problem Spae The Five Puzzle Our firt example i the Five Puzzle, the 2 3 verion of the well-known liding-tile puzzle (ee Figure 1). There are five numbered quare tile, and one empty poition, alled the blank. Any tile horizontally or vertially adjaent to the blank an be lid into the blank poition. The goal i to rearrange the tile from ome random initial onfiguration into a partiular goal onfiguration, uh a that hown at right in Figure 1.
1 2 3 4 5 Figure 1: Side and orner tate in the Five Puzzle. The branhing fator of a node in thi pae depend on the poition of the blank. There are two different type of loation in thi puzzle, ide or poition, and orner or poition (ee Figure 1). Similarly, we refer to a node or tate where the blank i in an or poition a an or node or tate. For impliity at firt, we aume that the parent of a node i alo generated a one of it hildren, ine all operator are invertible. Thu, the branhing fator of an node i three, and the branhing fator of a node i two. Clearly, the aymptoti branhing fator will be between two and three. The exat value of the branhing fator will depend on f, the fration of total node at a given level of the tree that are node, with f 1 f being the fration of node. For a given level, f will depend on whether the initial tate i an node or a node, but we are intereted in the limiting value of thi ratio a the depth goe to infinity, whih i independent of the initial tate. Equal Likelihood The implet hypothei i that f i equal to 2/6 or1/3, ine there are two different poition for the blank, out of a total of ix poible poition. Thi give an aymptoti branhing fator of 3 1/3 + 2 2/3 2.333. Unfortunately, thi aume that all poible poition of the blank are equally likely, whih i inorret. Intuitively, the poition are more entrally loated in the puzzle, and hene overrepreented in the earh tree. Random-Walk Model A better hypothei i the following. Conider the ix-node graph at left in Figure 1. In a long random walk of the blank over thi graph, the fration of time that the blank pend in any partiular node will eventually onverge to an equilibrium value, ubjet to a minor tehniality. If we divide the ix poition into two et, oniting of 1, 3, and 5 vere 2, 4, and the blank in Figure 1, every move take the blank from a poition in one et to a poition in the other et. Thu, at even depth the blank will be in one et, and at odd depth in the other. Sine the two et are ompletely ymmetri, however, we an ignore thi iue in thi ae. The equilibrium fration from the random walk i eay to ompute. Sine node have degree three, and node have degree two, the equilibrium fration of time pent in an individual tate veru an individual tate mut be in the ratio of three to two(motwani & Raghavan 1995). Sine there are twie a many tate a tate, tate are oupied 4/7 of the time and tate are oupied 3/7 of the time. Thi give a branhing fator of 3 3/7+2 4/7 2.42857, whih differ from the value of 2.3333 obtained above. Unfortunately, thi alulation i inorret a well. While the random-walk model aurately predit the probability of being in a partiular tate given a long enough random walk down the earh tree, if the tree ha non-uniform branhing fator, thi i not the ame a the relative frequenie of the different tate at that level of the tree. For example, onider the imple tree fragment in Figure 2. If we randomly ample the three leaf node at the bottom, eah i equally likely to appear. However, in a random walk down thi tree, the leftmot leaf node will be reahed with probability 1/2, and the remaining two node with probability 1/4 eah..5 1.5.25.5.5.5.25 Figure 2: Tree with Nonuniform Branhing Fator. The Corret Anwer The orret way to ompute the equilibrium fration f i a follow. A node at one level generate an node and another node at the next level. Similarly, an node at one level generate another node and two node at the next level. Thu, the number of node at a given level i two time the number of node plu the number of node at the previou level, and the number of node i the number of node plu the number of node at the previou level. Thu, if there are nf node and nf node at one level, then at the next level we will have 2nf + nf node and nf + nf node at the next level. Next we aume that the fration f onverge to an equilibrium value, and hene mut be the ame at the next level, or f nf + nf nf + nf +2nf + nf f +1 f f +1 f +2f +1 f 1 f +2
Cro multiplying reult in the quadrati equation f 2 +2f 1 0, whih ha poitive root 2 1.4142. Thi give an aymptoti branhing fator of 3f + 2(1 f )3( 2 1) + 2(2 2) 2+1 2.4142. The aumption we made here i that the parent of a node i generated a one of it hildren. In pratie, we wouldn t generate the parent a one of the hildren, reduing the branhing fator by approximately one. It i important to note that the redution i not exatly one, ine pruning the tree in thi way hange the equilibrium fration of and tate. In fat, the branhing fator of the five puzzle without generating the parent a a hild i 1.3532, a we will ee below. Rubik Cube A another example, onider Rubik Cube, hown in Figure 3. In thi problem, we define any 90, 180, or 270 degree twit of a fae a a ingle move. Sine there are ix different fae, thi ugget a branhing fator of 18. However, it i immediately obviou that we houldn t twit the ame fae twie in a row, ine the ame reult an be obtained with a ingle twit. Thi redue the branhing fator to 5 3 15 after the firt move. The next thing to notie i that twit of oppoite fae are independent of one another, and hene ommutative. Thu, if two oppoite fae are twited in equene, we retrit them to be twited in one partiular order, to eliminate the idential tate reulting from twiting them in the oppoite order. For eah pair of oppoite fae, we label one a firt fae, and the other a eond fae, depending on an arbitrary order. Thu, Left, Up and Front might be the firt fae, in whih ae Right, Down, and Bak would be the eond fae. After a firt fae i twited, there are three poible twit of eah of the remaining five fae, for a branhing fator of 15. After a eond fae i twited, however, there are three poible twit of only four remaining fae, leaving out the fae jut twited and it orreponding firt fae, for a branhing fator of 12. Thu, the aympoti branhing fator i between 12 and 15. To ompute it exatly, we need to determine the equilibrium frequenie of firt (f) and eond () node, where an f node i one where the lat move made wa a twit of a firt fae. Eah f node generate ix f node and nine node a hildren, the differene being that you an t twit the ame fae again. Eah node generate ix f node and ix node, ine you an t twit the ame fae or the orreponding firt fae immediately thereafter. Let f f be the equilibrium fration of f node at a given level, and f 1 f f the equilibrium fration of node. Sine we aume that thi equilibrium fration eventually onverge to a ontant, the fration of f node at equilibrium mut be 6f f +6f f f 6f f +6f +9f f +6f 6f f + 6(1 f f ) 15f f + 12(1 f f ) 6 3f f +12 2 f f +4 Cro multiplying give u the quadrati equation ff 2+4f f 2 0, whih ha a poitive root at f f 6 2.44949. Thi give u an aymptoti branhing fator of 15f f + 12(1 f f ) 13.34847. Figure 3: Rubik Cube. The Sytem of Equation The above example required only the olution of a ingle quadrati equation. In general, a ytem of imultaneou equation i generated. A a more repreentative example, we ue the Five Puzzle with predeeor elimination, meaning that the parent of a node i not generated a one of it hildren. To eliminate the invere of the lat operator applied, we have to keep trak of the lat two poition of the blank. Let denote a tate or node where the urrent poition of the blank i on the ide, and the immediately previou poition of the blank wa in an adjaent orner. Define, and node analogouly. Figure 4 how thee different type of tate, and the arrow indiate the hildren they generate in the earh tree. For example, the double arrow from to indiate that eah node in the earh tree generate two node. Figure 4: The graph of the Five Puzzle with predeeor elimination.
Let n(t, d) be the number of node of type t at depth d in the earh tree. Then, we an write the following reurrene relation diretly from the graph in figure 4. For example, the lat equation ome from the fat that there are two arrow from to, and one arrow from to. n(, d +1) n(, d) n(, d +1) n(, d) n(, d +1) n(, d) n(, d +1) 2n(, d)+n(, d) Note that we have left out the initial ondition. The firt move will either generate an node and two node, or a node and a node, depending on whether the blank tart on the ide or in a orner, repetively. The next quetion i how to olve thee reurrene. Numerial Solution The implet way i to iteratively ompute the value of ueive term, until the relative frequenie of the different type of tate onverge. At a given earh depth, let f, f,f and f be the number of node of the given type divided by the total number of node at that level. Then we ompute the ratio between the total node at two ueive level to get the branhing fator. After about a hundred iteration of the equation above we get the equilibrium fration f.274854, f.203113, f.150097, and f.371936. Sine the branhing fator of and tate i two, and the branhing fator of the other i one, thi give u the aymptoti branhing fator f +2f +2f +1f.274854 +.406226 +.300194 +.371936 1.35321. If q i the number of different type of tate, four in thi ae, and d i the depth to whih we iterate, the running time of thi algorithm i O(dq). Analytial Solution To olve for the branhing fator analytially, we aume that the fration onverge to a et of equilibrium fration that remain the ame from one level to the next. Thi fixed point aumption give rie to a et of equation, eah being derived from the orreponding reurrene. Let b be the aymptoti branhing fator. If we view, for example, f a the normalized number of node at depth d, then the number of node at depth d + 1 will be bf. Thi allow u to diretly rewrite the reurrene above a the following et of equation. The lat one expree the fat that all the normalized fration mut um to one. bf f bf f bf f bf 2f + f 1 f + f + f + f We have five equation in five unknown. A we try to olve thee equation by repeated ubtitution to eliminate variable, we get larger power of b. Eventually we an redue thi ytem to the ingle quarti equation, b 4 b 2 0. It i eay to hek that b 1.35321 i a olution to thi equation. While quarti equation an be olved in general, thi i not true of higher degree polynomial. In general, the degree of the polynomial will be the number of different type of tate. The Fifteen Puzzle, for example, ha ix different type of tate. General Formulation In thi etion we abtrat from the above example to exhibit the general truture of the equation and their fixed point. We begin with an adjaeny matrix repreentation P of the underlying graph G (V,E). For Figure 4, the row P j of P, with j {,,, }, are P (0,1,0,0), P (0,0,1,1), P (0,0,0,2) and P (1,0,0,0). Without lo of generality, we label the vertie V by the firt V integer, tarting from zero. We repreent the fration of eah type of tate a a ditribution vetor F. In our example, F (f,f,f,f ). We aume that thi vetor onverge in the limit of large depth, reulting in the equation bf FP, where b i the aymptoti branhing fator. In addition, we have the equation i V f i 1, ine the fration um to one. Thu, we have a et of V + 1 equation in V + 1 unknown variable. The underlying mathematial iue i an eigenvalue problem. Tranforming bf FP lead to 0 F (P bi) for the identy matrix I. The olution for b are the root of the harateriti equation det(p bi) 0 where det i the determinant of the matrix. In the ae of the Five Puzzle we have to alulate b 1 0 0 det 0 b 1 1 0 0 b 2 0 1 0 0 b whih implifie to b 4 b 20. Note that the aumption of onvergene of the fration vetor and the aymptoti branhing fator i not true in general, ine for example the aymptoti branhing fator in the Eight Puzzle of Figure
5 alternate between two value, a we will ee below. Thu, here we examine the truture of the reurrene in detail. Let n d i bethenumberofnode of type i at depth d in the tree, and n d be the total number of node at depth d. Let N d be the ount vetor (n d 0,n d 1,..., n d V 1 ). Similarly, let f i d be the fration of node of type i out of the total node at depth d in the tree, and let F d be the ditribution vetor (f0 d,f1, d..., f V d 1 ) at level d in the tree. In other word, fi d n d i /nd, for all i V. We arbitrarily et the initial ount and ditribution vetor, F 0 and N 0 to one for i equal to zero, and to zero otherwie. Let the node branhing fator b k bethenumberofhildrenofanode of type k, and let B be the vetor of node branhing fator, (b 0,b 1,..., b V 1 ). In term of P the value b k equal j V p k,j, with the matrix element p k,j in row k and olumn j denoting the number of edge going from tate k to tate j. We will derive the iteration formula F d F d 1 P/F d 1 B to determine the ditribution F d given F d 1. For all i V we have f d i n d i /n d N d 1 (P T ) i j V N d 1 (P T ) j j V nd 1 j p j,i j V k V nd 1 k p k,j j V fd 1 j k V n d 1 p j,i j V f d 1 k n d 1 p k,j F d 1 (P T ) i k V fd 1 k j V p k,j F d 1 (P T ) i /F d 1 B. It i not diffiult to prove that the branhing fator of depth d + 1 equal F d B. Therefore, if the iteration formula reahe equilibrium F the branhing fator b reahe equilibrium a well. In thi ae b equal F B and we get bak to the formula bf PF a ited above. Even though we have etablihed a neat reurrene formula, up to now we have not found a full anwer to the onvergene of the imulation proe to determine the aymptoti branhing fator. A olution to thi problem might be found in onnetion to homogenou Markov proee (Norri 1997), where we have a imilar iteration formula F d QF d 1, for a well defined tohati tranition matrix Q. Experiment Here we apply our tehnique to derive the branhing fator for quare liding-tile puzzle up to 10 10. Table 1 plot the odd and even aymptoti branhing fator in the (n 2 1)-puzzle with predeeor elimination. A n goe to infinity, the value in both olumn will onverge to three, the aymptoti branhing fator of an infinitely large liding-tile puzzle, with predeeor elimination. n n 2 1 even depth odd depth 3 8 1.5 2 4 15 2.1304 2.1304 5 24 2.30278 2.43426 6 35 2.51964 2.51964 7 48 2.59927 2.64649 8 63 2.6959 2.6959 9 80 2.73922 2.76008 10 99 2.79026 2.79026 Table 1: The aymptoti branhing fator for the (n 2 1)-Puzzle. To undertand the even-odd effet, onider the Eight Puzzle, hown in Figure 5. At every other level of the earh tree, all tate will be tate, and all thee tate will have branhing fator two, one the parent of the tate ha been eliminated. The remaining level of the tree will onit of a mixture of tate and m tate, whih have branhing fator of one and three, repetively. We leave the analyti determination of the branhing fator at thee level a an exerie for the reader. m 1 2 3 Figure 5: Side and Corner and Middle State in the Eight Puzzle. In general, if we olor the quare of a liding-tile puzzle in a hekerboard pattern, the blank alway move from a quare of one olor to one of the other olor. For example, in the Eight Puzzle, the tate will all be one olor, and the ret will be the other olor. If the two different et of olored quare are entirely equivalent to eah other, a in the five and fifteen puzzle, there will be a ingle branhing fator at all level. If the different olored et of quare are different however, a in the Eight Puzzle, there will be different odd and even branhing fator. In general, a retangular liding-tile puzzle will have a ingle branhing fator if 8 7 6 4 5
at leat one of it dimenion i even, and alternating branhing fator if both dimenion are odd. Appliation to FSM Pruning So far, we have pruned dupliate node in the lidingtile puzzle earh tree by eliminating the invere of the lat operator applied. Thi pruning proe an be repreented and implemented by the finite tate mahine (fm) hown in Figure 6. A node repreent the lat operator applied, and the ar inlude all legal operator, exept for the invere of the lat operator applied. Thu, the FSM give the legal move in the earh pae, and an be ued to prune the earh. However, even more dupliate an be eliminated by the ue of a more omplex FSM. For example, there i yle of twelve move in the liding-tile puzzle that ome from rotating the ame three tile in a two by two quare pattern. Taylor and Korf (Taylor & Korf 1993) how how to automatially learn uh dupliate pattern and expre them in an FSM for pruning the earh pae. For example, they generate an FSM with 55,441 tate for pruning dupliate node in the Fifteen Puzzle. An inremental learning trategy for FSM pruning i addreed by Edelkamp (Edelkamp 1997). The tehnique deribed here an be readily applied to determine the aymptoti branhing fator of thee pruned pae. Sine the number of different type of node i o large, only the numerial imulation method i pratial for olving the reulting ytem of equation. For example, we omputed a branhing fator of 1.98 for the above mentioned FSM, after about 50 iteration of the reurrene relation. Thi ompare with a branhing fator of 2.13 for the Fifteen Puzzle with jut invere operator eliminated. Left Up Start Right Conluion We howed how to ompute the aymptoti branhing fator of earh tree where different type of node have different number of hildren. We begin by writing a et of reurrene relation for the generation of the different node type. Thee reurrene relation an then be ued to determine the exat number of node at a given depth of the earh tree, in time linear in the depth. They an alo be ued to etimate the aymptoti branhing fator very aurately. Alternatively, we an rewrite the et of reurrene relation a a et of imultaneou equation involving the relative frequenie of the different type of node. The number of equation i one greater than the number of different node type. For relatively mall number of node type, we an olve thee equation analytially, by finding the root of the harateriti equation of a matrix, to derive the exat aymptoti branhing fator. We give aymptoti branhing fator for Rubik Cube, the Five Puzzle, and the firt ten quare lidingtile puzzle. Aknowledgment S. Edelkamp i upported by DFG within graduate program on human and mahine intelligene. R. Korf i upported by NSF grant IRI- 9619447. Thank to Eli Gafni and Elia Koutoupia for helpful diuion onerning thi reearh. Referene Edelkamp, S. 1997. Suffix tree automata in tate pae earh. In KI 97, 381 385. Korf, R. E. 1985. Depth-firt iterative-deepening: An optimal admiible tree earh. Artifiial Intelligene 27:97 109. Motwani, R., and Raghavan, P. 1995. Randomized Algorithm. Cambridge Univerity Pre, Cambridge, UK. Norri, J. R. 1997. Markov Chain. Cambridge Univerity Pre, Cambridge, UK. Taylor, L. A., and Korf, R. E. 1993. Pruning dupliate node in depth-firt earh. In AAAI 93, 756 761. Down Figure 6: An automaton for predeeor elimination in the liding tile puzzle.