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2 Discrete Applied Matheatics 157 (2009) Relaxation procedures on graphs Elias Wegert a,, Christian Reiher b a Institute of Applied Analysis, TechUniv Freiberg, Gerany b Keble College, University of Oxford, UK, Received 2 February 2007; received in revised for 22 Noveber 2007; accepted 22 Noveber 2007 Available online 4 March 2008 Abstract The procedures studied in this paper originate fro a proble posed at the International Matheatical Olypiad in We present several approaches to the IMO proble and its generalizations. In this context we introduce a signed ean value procedure and study relaxation procedures on graphs. We prove that these processes are always finite, thus confiring a conjecture of Akiyaa, Hosono and Urabe [J. Akiyaa, K. Hosono, M. Urabe, Soe cobinatorial probles. Discrete Matheatics 116 (1993) ]. Moreover, we indicate relations to sorting and to an iterative ethod used in circle packing. c 2008 Elsevier B.V. All rights reserved. Keywords: Graph algoriths; Reflection processes; Sorting; Electrical networks; Discrete haronic functions; Signed ean values; Circle packing; IMO probles 1. The pentagon gae Our starting point is Proble 3 of the International Matheatical Olypiad (IMO) in 1986 which arguably belongs to the hardest challenges this contest has ever seen [7,14,15]. The proble was proposed by the first author of the present paper and originally eerged as a side product of investigations of an old geoetric question concerning partial reflections of non-convex polygons (see [13] pp ). After the copetition it turned out that the proble can be generalized in various directions and has interrelations with several other topics. In this paper we collect soe known facts and present new perspectives of the proble. The pentagon gae: Five integers with positive su are assigned to the vertices of a pentagon. If there is at least one negative nuber, the player ay pick one of the, say y, add it to its two neighbors x and z, and then reverse the sign of y. The gae terinates when all the nubers are nonnegative. Prove that this gae ust always terinate Decreasing quadratic functions A standard approach for proving finiteness of a procedure is to construct a positive integer-valued function which decreases in every step. Corresponding address: Institute of Applied Analysis, TechUniv Freiberg, Prueferstr 9, D Freiberg, Saxony, Gerany. Tel.: ; fax: E-ail addresses: wegert@ath.tu-freiberg.de (E. Wegert), Christian.Reiher@keble.ox.ac.uk (C. Reiher) X/$ - see front atter c 2008 Elsevier B.V. All rights reserved. doi: /j.da

3 2208 E. Wegert, C. Reiher / Discrete Applied Matheatics 157 (2009) First solution. We denote the five nubers in consecutive order by x 1, x 2, x 3, x 4, x 5 and reark that the su s := x 1 +x 2 +x 3 +x 4 +x 5 reains invariant. A siple calculation shows that the function f of x = (x 1, x 2, x 3, x 4, x 5 ), given by f (x) := (x 1 x 3 ) 2 + (x 2 x 4 ) 2 + (x 3 x 5 ) 2 + (x 4 x 1 ) 2 + (x 5 x 2 ) 2, (1) is strictly decreasing in each step of the gae. In fact the value of f changes fro f (x old ) to f (x new ) = f (x old ) + 2ys < f (x old ). Since all values of f are nonnegative integers, the gae ust stop after at ost f (x) 1 steps. This arguent was found by all but one of the eleven students who succeeded in solving the proble during contest, and it coincides with the solution suggested by the proposer, but there are other quadratic functions which work as well. One exaple proposed by Géza Kóz (see [18], p. 321) is the function 2(x 1 x 2 + x 2 x 3 + x 3 x 4 + x 4 x 5 + x 5 x 1 ) + 3(x 1 x 3 + x 2 x 4 + x 3 x 5 + x 4 x 1 + x 5 x 2 ), which is strictly increasing and bounded fro above by s 2. Although the function f provides the siplest solution, it does not iediately generalize to the analogous gae played on an arbitrary polygon. For a square, for instance, the required decreasing quadratic function is not the naively expected expression (x 1 x 3 ) 2 + (x 2 x 4 ) 2, as the counterexaple x old = ( 1, 3, 5, 4) with x new = ( 1, 2, 5, 1) shows. A substitute which works is 3{(x 1 x 3 ) 2 + (x 2 x 4 ) 2 } + {(x 1 x 2 ) 2 + (x 2 x 3 ) 2 + (x 3 x 4 ) 2 + (x 4 x 1 ) 2 }. To handle the hexagon in a siilar way one ay use soething like 6 {(x k x k+3 ) 2 + 2(x k x k+2 ) 2 + 2(x k + x k+1 2x k+3 ) 2 + 6(x k + 2x k+1 3x k+3 ) 2 }. k=1 In the next subsection we describe a construction given by Alon, Krasikov, and Peres [2] which solves the proble for all n Sus of consecutive eleents During the IMO contest all but one participant who solved the proble used the function f defined in (1) (see [14], p. 20). The only alternative solution was found by Joseph G. Keane fro the US tea, whose idea had the rare distinction to be honored by a special prize. Instead of a su of squares he considered a function involving absolute values. Second solution. Let the function g be defined by g(x) := 5 ( x j + x j + x j+1 + x j + x j+1 + x j+2 + x j + x j+1 + x j+2 + x j+3 ), where all indices are reduced odulo 5. In each step of the algorith all but one of the suands reain invariant or switch places. Only the ter s y is changed to s + y, where y denotes the negative nuber chosen by the player. Consequently g decreases by the positive integer d := s y s + y. The function g can easily be adapted to the corresponding gae played on polygons with real nubers x 1,..., x n. In order to siplify notations we extend the sequence (x j ) periodically to all integers j and define s i j := x i + x i x i+ j 1, i, j 1. (2) Then the generalized function g is a su of absolute values of the s i j, n 1 g(x) := s i j. i=1 (3)

4 E. Wegert, C. Reiher / Discrete Applied Matheatics 157 (2009) Again g decreases in each step by d := s y s + y, which shows that the algorith stops if the x j are integers. In order to prove that it also terinates if the x j are real nubers, we denote by S the ultiset of all nubers s i j with 1 i n and 1 j n 1. As stated above, if the value of an eleent a in S is changed, then a = s y, so y < 0 is equivalent to a > s. If s < a 2s then a is replaced with the new nuber s + y = 2s a s, which then ust reain constant forever. If a 2s, then s + y = a 2s, i.e. a is reduced by 2s. Since this can happen only a finite nuber of ties, any eleent of S is eventually trapped in the interval [0, s], and then the algorith ust stop. This arguent also shows that the nuber of steps needed to turn every nuber non-negative depends only on the initial configuration and not on the player s choice. Indeed, if we denote by x that integer satisfying x 1 x < x, then any y S ay be reduced y+s 2s any ties in the anner described above. Note that in each step of the algorith exactly one eleent of S is diinished and that there ust still be a negative nuber as long as there reain eleents of S outside the interval [0, s]. Consequently, the forula y + s N = 2s y S (4) gives the total nuber of operations to be perfored. Alon, Krasikov, and Peres [2] derived a siilar forula using the squares s 2 i j instead of the absolute values s i j. In order to show that (4) coincides with their result we denote by T the ultiset of all nubers s i j = x i + x i x i+ j 1, 1 i n, 1 j n 1. Taking into account that x s x aps T bijectively onto itself, we obtain N = s + t + s t = s + t + 2s 2s 2s t T,t>0 t T,t 0 t T,t>0 t T,t s t. 2s As 2s t and t+s 2s give the nuber of even and odd integers in the interval (0, s t ), respectively, both su up to s t. Denoting by x the integer with x x < x + 1, we arrive at the forula given in [2], N = t = s t = t. (5) s s s t T,t>s t T,t<0 t T,t<0 An unbeatably elegant application of sus of consecutive eleents is due to Bernard Chazelle ([5], [7] p. 6). Third solution. Let S be the infinite ultiset of all sus s i j defined in (2) with i, j Z, 1 i n and 1 j. Since the su s = x x n is positive, the nuber of negative eleents in S is finite. In each step of the gae all eleents of S, except one, reain invariant or switch places with others. Only the negative nuber y chosen by the player is changed to y. In order to verify this we arrange the eleents of S in the following table x 1 x 1 + x 2 x 1 + x 2 + x 3... x x n 1 s s + x 1... x 2 x 2 + x 3 x 2 + x 3 + x 4... x x n s s + x 2... x 3 x 3 + x 4 x 3 + x 4 + x 5... x x 1 s s + x x n x n + x 1 x n + x 1 + x 2... x n + + x n 2 s s + x n.... If, without loss of generality, the player chooses the nuber y = x 1 then the eleents in every row, except in the first and the second, are preserved. Apart fro the eleent x 1, the new first row has the sae eleents as the old second row, and the new second row coincides with the old first row without x 1. Hence, in every ove exactly one negative eleent of S is changed to positive. Since the su s is positive, the nuber N of negative eleents in S is finite and the algorith ust terinate after at ost N steps. In fact it cannot stop earlier, since then S would still have negative eleents, which is ipossible if none of the x i is negative. As S ay be constructed as the infinite ultiset of all t + z s, where t runs through T and z through the non-negative integers, we again obtain forula (5).

5 2210 E. Wegert, C. Reiher / Discrete Applied Matheatics 157 (2009) Alon, Krasikov, and Peres as well as Chazelle also proved that the final position is uniquely deterined by the initial configuration. A siilar solution given by John M. Capbell ([6], see [12]) uses only one infinite two-sided sequence (v i ) of consecutive sus (here shown for n = 5)..., x 4 x 5, x 5, 0, x 1, x 1 + x 2, x 1 + x 2 + x 3,..., x x 5, 2x 1 + x x 5,..., which is constructed by adding (respectively subtracting) cyclically the nubers x i. Fourth solution Any ove exchanges the eleents v i+nj and v i+1+nj for soe i {1,..., n} with v i > v i+1 and all j Z. The procedure can be perfored as long as there exists i {1,..., n} with v i > v i+1. We denote by N i the (finite) nuber of eleents v j which stand right of v i and are less than v i. It is clear that every ove reduces the su N = N N n by one, and a little thought then shows that the procedure stops after exactly N steps with a strictly increasing sequence (v i ). Capbell s solution reveals that the pentagon gae can be reforulated as a sorting procedure, which clearly explains why the final position is uniquely deterined. We explore this idea further in the next sections Breaking syetry a sorting procedure The next approach is a finite version of Capbell s solution. The first author learned it in 1987 fro Sergej Steinberg [19] during a personal counication in Pushchino. The idea is to represent the nubers x i as differences. Fifth solution. Let y 1 = 0 and define y 2,..., y n by y i = x 1 + x x i 1. Then x i = y i+1 y i for i = 1,..., n 1 and x n = s y n = y 1 y n + s. Rewriting the rules of the gae for the nubers y 1,..., y n we get the following two possible operations. If y i+1 < y i for soe i = 1,..., n 1 it is allowed to interchange y i+1 and y i. If y n > y 1 + s it is allowed to replace the first nuber y 1 with y n s and the last nuber y n with y 1 + s. It is convenient to think of the nubers y i as written on cards which are arranged in a line and are nubered fro left to right. Then the first operation exchanges two neighboring cards, the card carrying the greater nuber oving right and the card with the saller nuber oving left. The second operation is allowed if the difference y n y 1 is greater than s. Here the larger nuber y n is diinished by s, the saller nuber y 1 is increased by s, and the two cards change places. Since the nubers y i are only changed by ultiples of s, all possible values belong to a discrete set R. This set is even finite, since the axial nuber y i never increases and the inial nuber never decreases. Because R is finite, ax y i ust reain constant after a nuber of steps. Let this axiu be. Since no card with y i < can ever reach the value, at least one card ust carry the nuber forever. Now, analogously, each of the reaining n 1 cards can change its value only a finite nuber of ties. Going on by induction, we see that after soe tie the values of all cards reain unchanged. It is now clear that the reaining sorting process ust stop, since it contains no cycle and the nuber of perutations is finite. As an alternative to the above reasoning one ay consider the nonnegative function f (y) = y 2 j. If the second operation is perfored the value of f decreases by d := 2s(y n y 1 ) 2s 2 = 2s(y n y 1 s) > 0. Since y n and y 1 belong to the finite set R, the nuber d is bounded fro below by a positive constant. So the second operation can be applied only a finite nuber of ties. In order to deterine the final configuration x j we reark that the final values y i of y i ust satisfy y 1 y 2 y n y 1 + s. Moreover, the final value on each card differs fro its initial value by a ultiple of s and the sus n i=1 y i and ni=1 y i ust be equal. These observations allow to find y i as follows:

6 E. Wegert, C. Reiher / Discrete Applied Matheatics 157 (2009) Let 0 r j < s be the reainders of y j odulo s. Denote by r j the rearrangeent of the r j such that 0 r1... r n < s. Then y j is given by yi = ri+ j + ks (i = 1,..., n j), y i = ri+ j n + (k + 1)s (i = n j + 1,..., n), where the integers j and k are chosen such that 0 j n 1 and (y i r i ) = (kn + j)s. i=1 For the final values x i of x i we then obtain x 1 = y 2 y 1, x 2 = y 3 y 2,..., x n 1 = y n y n 1, x n = y 1 y n + s Keeping syetry threshold sorting The solution via the above sorting procedure breaks the syetry between the variables. In the sequel we develop a siilar approach keeping syetry. This will lead to a new kind of probles which are treated in ore generality in the next section. Here we start with a siple situation. The threshold sorting procedure. Let d be a positive constant and let y 1,..., y n be a finite sequence of real nubers. If there are u = y i and v = y j with u > v + d then replace u by v + d and v by u d. Repeat this step as long as nubers u and v with u > v + d exist. Deterine whether this procedure always stops. First of all we observe that all nubers y i are changed by ultiples of d, their axiu is not increasing and their iniu is not decreasing. Consequently the set of all possible values is finite. Hence there exists a nuber c such that u v d c > 0 for all u, v to which the operation ight be applied during the whole process. It follows that the nonnegative function f (y) = y 2 i is decreasing in each step of the algorith by u 2 + v 2 (u d) 2 (v + d) 2 = 2d(u v d) 2cd. Therefore the procedure always stops. The function f gives a rather bad estiate of the nuber of steps. A better one can be obtained using the function g given by g(y) := y i y j. In order to prove that g is decreasing, we reark that the new nubers u d and v + d lie in the interval with endpoints u and v. It follows that (u d) (v + d) is less than u v by at least 2 in(c, d), and it is easy to see that for any w the su (u d) w + w (v + d) is less than or equal to u w + w v. The last result can be used to syetrize the fourth solution of the Pentagon gae. Sixth solution. There exist uniquely deterined values y 1 := 0, y 2,..., y n, y n+1 := y 1, such that the x i have the syetric representation x i = y i+1 y i + s, i = 1,..., n. n Reforulating the original algorith for x 1,..., x n in ters of y 1,..., y n we get the following operation: If there are two neighbors u = y i and v = y i+1 such that u > v + s/n then u is replaced with v + s/n and v is replaced with u s/n. Clearly this rule is ore restrictive than that of the Threshold Sorting Procedure, and so the algorith ust stop. It is interesting to see how the function g looks in ters of the x i. In fact it is quite siilar to the one considered earlier in (3), naely g(x) = n 1 i=1 j=i ( j i + 1)s n j x k. k=i

7 2212 E. Wegert, C. Reiher / Discrete Applied Matheatics 157 (2009) The signed ean value procedure It turns out that threshold sorting is just a special case of a ore general algorith which we introduce and investigate in this section. We start with fixing the rules of the gae, which is now played on an arbitrary finite collection of real nubers. The signed ean value procedure. Fix a positive constant d and let y 1,..., y n be real nubers. If there are nubers (signs) η 1, η 2,..., η n { 1, 0, 1} such that s := η 1 y 1 + η 2 y η n y n > d, then set := η1 2 + η η2 n and substitute y j y j 2η j ( s d ), j = 1, 2,..., n. (7) Repeat this as long as nubers η 1, η 2,..., η n { 1, 0, 1} with (6) exist. Note that always satisfies 1 n and can vary fro step to step. The Threshold Sorting Procedure corresponds to the ore restrictive rule where all the η j are zero, except two which are 1 and 1, respectively. Theore 1. The Signed Mean Value Procedure always stops. Proof. In what follows we assue that there is a procedure which does not stop. 1. The function f (y) = y1 2 + y y2 n is strictly decreasing. In fact ( ) f (y old ) f (y new ) = y 2 j s d 2 d(s d) y j 2η j = 4 > 0. (8) Let f k denote the value of f at step k. Since the sequence ( f k ) is onotone and bounded it converges to a liit f. 2. Let s k,, η j,k and y j,k denote the values of s,, η j and y j in the k-th step, respectively. Since 1 n it follows fro (8) that 0 < s k d n 4d ( f k f k+1 ). We write f 1 f as a (absolutely) convergent telescopic series, f 1 f = ( f k f k+1 ). k=1 Together with (9) this shows the (absolute) convergence of k=1 (s k d), and in particular we have s k d. Further, by (7), y j,k+1 y j,k 2(s k d), j = 1,..., n. Consequently, the convergent series y j,1 + 2 (s k d) k=1 serves as a ajorant for the representation y j,k+1 = y j,1 + k y j,i+1 y j,i. i=1 This iplies that all sequences (y j,k ) converge to certain liits Y j as k tends to infinity. 3. We now consider the difference (η 1,k y 1,k + + η n,k y n,k ) (η 1,k Y η n,k Y n ), (6) (9)

8 E. Wegert, C. Reiher / Discrete Applied Matheatics 157 (2009) which converges to zero, since y j,k Y j as k. The first ter in parentheses is s k, which has been shown to converge to d. Consequently also η 1,k Y 1 + η 2,k Y η n,k Y n d. However, the set {η 1 Y 1 + η 2 Y η n Y n : η j { 1, 0, +1}} contains only a finite nuber of eleents, which together with (10) then iplies that η 1,k Y 1 + η 2,k Y η n,k Y n = d for all sufficiently large k, say for all k K. 4. Finally, we observe that the values I k := (y j,k Y j ) 2 all converge to zero, since y j,k Y j as k. In fact all I k are equal for k K, naely, ( ) s k d 2 ( ) 2 I k+1 I k = y j,k 2η j,k Y j y j,k Y j = 4 η 2 j,k = 4 s k d = 4 s k d ( ) sk d 2 4 [ [ η 2 j,k s k d s k d η j,k s k d (y j,k Y j ) η j,k y j,k + η j,k y j,k + ] η j,k Y j ] η j,k Y j = 0, where we used (6) and (11) in the last step. Since I k 0 it follows that I k = 0 for all k K, which iplies y j,k = Y j. But then all variables y j would be constant after the K -th step, a contradiction. 3. Relaxation procedures In this section we return to the IMO Pentagon Gae and forulate a natural generalization to graphs. A relaxation procedure on graphs: Let G be a connected graph with at least two vertices. To each vertex v j of G a real nuber x j, called a label, is assigned. Assue that s := x j > 0. If the label x associated with a vertex v is negative then it is allowed to add 2x/ to each of the labels at the vertices adjacent to v, and then to replace x by x. This step is perfored repeatedly as long as negative labels exist. In their paper [1], Akiyaa, Hosono and Urabe asked if this procedure necessarily terinates for regular graphs. We shall prove that this indeed always happens for arbitrary graphs. Note that connectedness can be oitted if we assue that s > 0 holds in every coponent of the graph. The nae Relaxation Procedure is otivated by the following interpretation. If we consider the x j as charges sitting at the vertices v j, their distribution induces a tension of the graph G. We do not describe precisely what this eans but vaguely speaking, the ore an edge contributes to the tension, the greater the differences of charges at its incident vertices is. So tension is a easure for non-unifority of a charge distribution. The rules of a relaxation procedure are such that the charges are allowed to be shifted along the edges so that the tension is reduced. Quite recently we learned that procedures of this kind are basic for iterative ethods in circle packing, as described in Kenneth Stephenson s beautiful and inspiring book [20]. Here the vertices of the graph correspond to the circles involved in the packing and the edges are defined by the prescribed tangency structure of the packing. Each circle (vertex) carries two labels, one is the radius, the other one is the angle su (which easures the local curvature ). The angle su at a circle C is expressed by the radii of C and all circles adjacent to C. (10) (11)

9 2214 E. Wegert, C. Reiher / Discrete Applied Matheatics 157 (2009) The goal is to find appropriate radii which flatten the packing, which happens if all interior angle sus are 2π (or a ultiple thereof for branched packings). This is achieved by an iterative procedure, where in each step one circle is chosen and its radius is adjusted such that the local curvature becoes zero (or sall ). This changes the curvature labels of C and of the adjacent circles. In effect the curvature overhead of C is distributed aong its neighbors according to a rule which resebles the setting of the above relaxation procedure. The iteration is stopped if the absolute values of all local curvatures are less then a positive threshold. For details we refer to [20], especially pages Does a relaxation procedure necessarily terinate? If yes, how any steps can be (or ust be) perfored and what are the possible final configurations? For the procedure defined above the following theore gives an affirative answer to the first question. Theore 2. For each graph the above relaxation procedure stops. Proof. 1. We reduce the proble to a special case of the Signed Mean Value Procedure. In order to do so, G is first converted to a digraph by choosing arbitrary directions of its edges. Further, we double the nuber n of vertices of G by associating with any vertex v j a new vertex v j which is adjacent (exactly) to v j by an edge e j directed fro v j to v j. The resulting digraph is denoted by G. To each vertex v j of G which does not belong to G we assign the label d, where d := s/n and n denotes the nuber of vertices of G. Then the total su of all vertex labels of G is zero. 2. With any directed edge e i of G we associate a label ( the conductance ) y i, such that the vertex labels x j are equal to the su of the edge labels at the incident incoing edges inus the su of the edge labels at the incident outgoing edges. The existence of such labels follows fro Kirchhoff s law, using the fact that the su of all vertex labels is zero. More directly, to find appropriate edge labels one can select a spanning tree T of G and choose arbitrary labels at those edges of G which do not belong to T. The reaining labels at the edges of T are then uniquely deterined and can be found by successively deleting onovalent vertices of T together with the corresponding edges. Note that all edges fro v j to v j belong to any spanning tree and carry the label d. 3. We investigate the edge labels during a step of the relaxation procedure. Each vertex label x has the representation x = d η i y i where y i are the labels of the incident edges belonging to G, with η i = 1 for incoing, and η i = +1 for outgoing edges. Let x be the (negative) label of a vertex v with valency selected in a step of the procedure. Then x < 0 is equivalent to s := η i y i > d. If the labels y i of the incident edges belonging to G are replaced by y i 2η i (s d)/ and the label d of the edge between v and v reains unchanged, these new values are copatible with the new vertex labels. In fact, ( ) x new = d η i y i 2 s d = d s + 2s 2d = x old, i=1 and changing the labels y i of all edges incident with v according to the rule s d y i y i 2η i y x i + 2η i has the sae effect like adding 2x/ to the labels at all vertices adjacent to v. So the relaxation procedure for the vertex labels induces a special Signed Mean Value Procedure (with preselected η i ) for the edge labels. By Theore 1 the latter ust terinate. In contrast to the proble for polygons neither the final configuration nor the nuber of steps is uniquely deterined. For instance, if the labels 1, 2, 3, 4 are attached to the vertices of a coplete graph of order four we get the following results (scaled by a coon factor of 27), depending on whether one starts with 1 or 2 in the initial step: ( 27, 54, 81, 108) (27, 72, 63, 90) ( 21, 72, 15, 42) (21, 58, 1, 28) ( 27, 54, 81, 108) ( 63, 54, 45, 72) (63, 12, 3, 30).

10 E. Wegert, C. Reiher / Discrete Applied Matheatics 157 (2009) Open proble: Find a characterization of all graphs where the final configuration and/or the nuber of steps are unique. There are any possibilities to change the rules of the gae. One option is to adit weighted shifts of the charges, which leads to relaxation procedures on weighted digraphs. To define these procedures we assue that every edge e i j of a digraph G with n vertices v i carries a non-negative label c i j, its edge conductance, such that for all i = 1,..., n c i := c i j > 0. j i (12) To siplify notations we assue that G is copletely bi-oriented, which can be achieved by adding virtual edges with conductance zero, and define the weights w i j for i, j = 1,..., n by w i j := { 1 if i = j c i j /c i if i j. (13) A relaxation procedure on weighted digraphs. Let G be a digraph endowed with edge conductances c i j, let the weights w i j be defined by (12) and (13), and fix a relaxation paraeter λ R +. Assue further that any vertex v i of G carries a label x i, its charge, so that the total charge x x n is positive. If there is at least one negative charge, say x i, it is allowed to replace all charges according to the rule x j x j + 2λw i j x i, j = 1,..., n. (14) Repeat this step as long as there are negative charges. Reark. Condition (12) guarantees that any negative charge has the option to be distributed aong their neighboring vertices. If one adits that c i vanishes for soe indices i, it is natural to set all corresponding weights w i j, including w ii, to zero and to odify the procedure by not allowing the substitution (14) for those values of i. Open proble: Assue that the edge conductances of G are given. Describe the set of all relaxation paraeters λ such that, for any initial distribution of the charges, the relaxation procedure terinates. The rules of the relaxation procedure are designed so that the total charge reains invariant, which was otivated by our physical interpretation. One should not think that this condition is natural to get a reasonable procedure on the contrary. In 1987 Shahar Mozes invented what is now called Mozes Nubers Gae (see [16]), which corresponds to the substitution rule { x x j j if i = j (15) x j + w i j x i if i j with integer weights w i j. Using Weyl groups and Kac-Moody algebras, Mozes gave an algebraic characterization of the initial positions giving rise to finite gaes and proved that for those the nuber of steps and the finial configuration do not depend on the oves of the player. For w i j {0, 1} (which includes the original Pentagon proble) Anders Björner [3] (see also [4], Section 4.3, and [8]) gave an eleentary proof (without characterizing the initial configurations for which the gae terinates). Kio Eriksson [9] showed that (for a subclass of probles) one can decide which positions are reachable fro a given initial configuration, and his subsequent investigations [10, 11] revealed deep connections to Coxeter groups and greedoids. Proctor [17] discovered that Mozes Nubers Gae is related to Bruhat lattices. Further odifications of the proble arise if one adits siultaneous substitutions of (soe or all) negative labels. Another direction is to allow substitutions without sign restrictions, and to ask for the set of all reachable configurations. Does this set contain inial configurations? Of, course one can also replace the real nubers by other (partially ordered) algebraic structures... Isn t it fascinating to find so uch atheatics hidden in a siple gae?

11 2216 E. Wegert, C. Reiher / Discrete Applied Matheatics 157 (2009) References [1] J. Akiyaa, K. Hosono, M. Urabe, Soe cobinatorial probles, Discrete Matheatics 116 (1993) [2] N. Alon, I. Krasikov, Y. Peres, Reflection sequences, Aer. Math. Monthly 96 (1989) [3] A. Björner, On a cobinatorial gae of S. Mozes, Preprint KTH Stockhol, [4] A. Björner, F. Brenti, Cobinatorics of Coxeter Groups, Springer, [5] B. Chazelle, personal counication, [6] J.M. Capbell, personal counication, [7] A. Engel, Proble-solving strategies, in: Proble Books in Matheatics, Springer, New York, [8] K. Eriksson, Convergence of Mozes s gae of nubers, Linear Algebra and its Applications 166 (1992) [9] K. Eriksson, Reachability is decidable in the nubers gae, Theoretical Coputer Science 131 (2) (1994) [10] K. Eriksson, The nubers gae and coxeter groups, Discrete Matheatics 139 (1 3) (1995) [11] K. Eriksson, Strong convergence and a gae of nubers, Eur. J. Cobin. 17 (4) (1996) [12] G. Burosch, H.-D. Gronau, W. Moldenhauer, IMO-Übungsaufgaben (Geran) 23. Universität Rostock, [13] N.D. Kazarinoff, Analytic Inequalities, Holt, Rinehard and Winston, New York, 1961, (quoted after 2nd edition, Dover Publications, 2003). [14] M.E. Kucza, International Matheatical Olypiads , The Matheatical Association of Aerica, [15] Matheatics Magazine 59 (1986) [16] S. Mozes, Reflection processes on graphs and Weyl groups, J. Cobin. Theory, Ser. A 53 (1) (1990) [17] R.A. Proctor, Minuscule eleents of Weyl groups, the nubers gae, and d-coplete Posets, Journal of Algebra 213 (1) (1999) [18] I. Reian, International Matheatical Olypiad , Wibledon Publ. Cop., [19] S. Steinberg, personal counication, [20] K. Stephenson, Introduction to Circle Packing, Cabridge University Press, 2005.

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