Channel Assignment Strategies for Cellular Phone Systems

Transcription

1 Channel Assignment Strategies for Cellular Phone Systems Wei Liu Yiping Han Hang Yu Zhejiang University Hangzhou, P. R. China Contat: 000 Mathematial Contest in Modeling (MCM) Meritorious Award Paper

2 Abstrat Nowadays people benefit a lot from mobile ommuniation systems. One of reent major breakthroughs in solving the problem of spetral ongestion is the ellular onept, whih means replaing a large transmitter with many little transmitters, eah providing overage to a small piee of the servie area. Neighboring transmitters are assigned different groups of hannels so that the interferene between transmitters is minimized. Eah ellular transmitter is assigned a radio hannel to be used within a small geographi area alled a ell. By limiting the overage area to within the boundaries of a ell, the same hannel may be used to over different ells that are separated from one another by distanes large enough to keep interferene levels within tolerable limits. The design proess of seleting and alloating hannel groups for all of the ellular transmitters in a system is alled frequeny reuse or frequeny planning. The hexagonal ell shape shown in the piture of the over is oneptual and is a simplisti model of radio overage for eah transmitter, but it has been universally adopted sine the hexagon permits easy and manageable analysis of a ellular system. Our team has made a omprehensive study on a pratial problem of designing mobile ommuniation systems. We know that there are several onstraints on frequeny assignment. First, any two transmitters within a distane four times as muh as the length of the hexagon side annot be assigned the same hannel. Seond, due to spetral spreading, transmitters within distane twie as muh as the length of the hexagon side must be assigned hannels whih differ by at least k. We have developed quite an effiient strategy dependent on k to assign hannels as few as possible to a given servie area aording to these rules. If we put our strategy into pratie under simple situations, we have hannels to be the fewest when k equals, hannels when k equals, and hannels when k equals. When the situation beomes ompliated as k >, we find the smallest number of hannels needed is a linear funtion k +, whih is a pretty simple solution. Thus we have the entire results as follows: k + 5, k =, span = k +, k = k +, k.

3 Patterns of our assignments are as follows: k+ k+ k+5 k+ k+5 k+ (a) (b) () (d) Figure : Channel assignment patterns. (a) k = ; (b) k = ; () k = ; (d) k. We also study several general problems with more onstraints. An algorithm aimed at getting a lower bound is devised through a branh-and-bound strategy. This algorithm is very useful as it suffies to provide a good estimate of the lower bound, whih may be a great help when we apply integer programming to solve this problem. We also developed a method to deal with the situation that the number of the hannels is fixed. A new strategy is devised to redue interferene as greatly as possible. We have first onverted this problem into a nonlinear program. A lot of algorithms are available, suh as SA (Simulated Annealing) and LG (Least Gradient). We hoose GA (Geneti Algorithm) to solve this problem. A simple proedure of the algorithm is given. Introdution The design objetive of early mobile radio systems was to ahieve a large overage area by using a single, high-powered transmitter with an antenna mounted on a tall tower. While this approah ahieved very good overage, it also meant that it was impossible to reuse those same frequenies throughout the system sine any attempts to frequeny reuse would result in interferene. For example, the Bell mobile system in New York City in the 0s ould only support a maximum of simultaneous alls over a thousand square miles []. Faed with the fat that government regulatory agenies ould not make spetrum alloations in proportion to the inreasing demand

4 for mobile servies, it beame imperative to restruture the radiotelephone system to ahieve high apaity with limited radio spetrum, and at the same time over very large areas. The ellular onept has been one of the revolutionary breakthroughs in developments of mobile ommuniation systems. Allowing a group of lower power transmitters (small ells) to replae a single, high power transmitter (large ell), the ellular system may provide good overage to only a small portion of a given area with a portion of the total number of hannels being alloated to eah base station. Here the design proess of seleting and alloating hannel groups for all of the ellular base stations in a system is very ritial, whih is alled frequeny reuse or frequeny planning []. Our task is to model the assignment of hannels to a symmetri network of transmitter loations over a large planar area, whih has been partitioned into hexagons. An interval of the frequeny spetrum is to be allotted for transmitter frequeny. The interval will be divided into regularly spaed hannels, represented by integer,,... Eah transmitter will be assigned one positive integer hannel. The same hannel an be reused at many loations, provided that interferene is avoided. Our goal is to minimize the width of the interval in the frequeny spetrum whih is needed to assign hannels subjet to some onstraints. This is ahieved by the onept of a span. The span is the minimum, over all assignments satisfying the onstraints, of the largest hannel used at any loation. It is not required that every hannel smaller than the span be used in an assignment that attains the span. Assumptions and Definitions Assumptions: The honeyomb-style is hexagon. In Solving Requirement B, C, D, we assume that the grid always extends to the infinity in all diretions. The reason is explained below. The hannels are regularly spaed.

5 All of the transmitter use omni-diretional antennas. All of the ells are idential. The transmitter is loated over a large planar area. Definitions: sp: the value of span. s: the length of a side of one of the hexagons. N: the number of ells in a single tile. tile: pre-assigned network piees, whih are desribed later. f(x): the hannel number assigned to the ell x. Preliminary Analysis 0 s.s 5.s 5 Figure : The first and seond loop. 5

6 . Distane Between Cells First, we aim to alulate the distane of two ertain ells. Look at Fig.. It is easy to work out the distane between ell and its adjaent ells,,, 5,,, whih is s. Furthermore, the distane between ell and ells, 0,,,, is s while the distane between ell and ells,,, 5,, are s. We all ells,,, 5,, surrounding ell the first loop and the ells - surrounding ell the seond loop. We may easily find that only ells in the first loop have a loser distane than s with ell. With the same reason only ells in the seond loop have a loser distane than s with ell. So when we onsider the s(s)-onstraint, we may only onentrate on those ells in the first (seond) loop. This makes our efforts to solve the problem muh easier.. Developments.. Tiling Tehnique It is obvious that the same frequeny annot be used in the adjaent ells beause of the sonstraint. In order to keep the ells assigned with the same frequeny far enough from eah other, several surrounding ells annot be assigned that frequeny too. We all these ells a tile. Only ells in different tiles an be frequeny-reused. Formation of the tile should satisfy the following two onditions:. Tiles and tiles an onnet without gaps between adjaent ells;. The distane between any two ells with the same frequeny should be larger than s. After we have overed the area with tiles, there are only ertain tile sizes and ell layouts whih are possible [], they an only have values whih satisfy equation: N = i + ij + j, where i and j are non-negative integers. All the tiles in our models satisfy the above equation. It is obvious that this tiling tehnique is always effetive no matter whether the region an be overed by a single basi blok or not. We now assume that the grid always extends to the infinity in all diretions.

7 .. sp = When k = If we have k = then we an use integer hannels,,,, 5,, to fill in a infinite grid with the following tile. Refer to Appendix B- for a more detailed piture. 5 Figure : sp = when k =, where N =, i =, j =. It is lear that integers are not enough to fill this tile. If not so, there must be repetitive integers in the tile, whih will surely violate the s-onstraint... sp = When k = It is very easy to prove that we annot satisfy both the onstraints if we try to assign hannels with a span of to the transmitters in a infinite grid. If this is not true, let us suppose that we used a ertain way of assignment with span to fill all the ells. Then we are sure to find an integer in one of the ells whih we alled ell A. If we annot find the integer in any of these ells, simply subtrat from every integer assigned to all the ells. Then we get a better assignment with a new span of that is smaller than the previous one. We know that annot be plaed in any of ell A s adjaent ells (in order to satisfy the s-onstraint). We name the A s adjaent ells B, C, D, E, F, G, shown in Fig.. So,, 5,,, must be used to be filled in these adjaent ells. We may notie that no repetitive integers an be used to fill these ells beause any two of these ells are within the distane of s, so the s-onstraint will not be satisfied if repetitive integers are used. Thus we know that integer must be used in one of the ells adjaent to ell A aompanying integer. But we may still notie that only, 5,,, are available to be filled in the adjaent ells of the ell filled with. Sine we are unable to assign these 5 integers into ells without repetition we learn that the span annot be ahieved. So at least hannels are needed. There are several ways to get the appropriate assignment of

8 5 B E C A G D F Figure : Adjaent ells. these hannels. After a few trials we have suessfully found a right assignment as shown in Fig. 5, a more detailed one also inluded in Appendix B-. Figure 5: sp = when k =, where N =, i = 0, j =... sp = When k = and Lower Bound of Span for k We have got sp = when k = and sp = when k =. Assuming that sp depends on k linearly we may guess that sp = k + 5, But when it applies to the situation k we find that k + 5 is not a tight lower bound for the span and k + is a tighter one. An effetive proof is inluded in Appendix A-. We also have designed a ertain pattern of assignment to show sp = = k + when k = (See Fig., a more detailed one is in Appendix B-). Sine we have got a realisti example we are sure to say sp = k + when k = and sp k + when k.

9 5 0 Figure : sp = when k =, where N =, i =, j =...5 Preliminary Estimate of Upper Bound of The Span In order to avoid the situation that integers assigned to adjaent ells whih may violate the sonstraint we generate an integer sequene, k +, k +, k +, k +, 5k +, k +, in whih any two integers satisfy the s-onstraint beause the minimal differene of them is k. Assign these integers into the basi tile shown in Fig. (More detailed one is in Appendix B-), we may notie that the s-onstraint is satisfied. Sine we also adopt the tiling tehnique, we may be quite sure that this assignment satisfies the s-onstraint too. Thus we get a preliminary estimate of the upper bound of the span (sp k + ). +k +k +k +5k +k +k Figure : Upper bound of the span... sp = k + When k We an learly see that there are a great few of hannels wasted when we apply the tile in Fig.. So we managed to lower it to some extent. Considering the geometry atahresti of hexagon and

10 sp = k + when k =, we find that oloring algorithm in the graph theory an be used to prove that k + is the span when k. Before we proved it, we first use this algorithm to find the strategy that span = k + when k. (See Fig., a more detailed one is in Appendix B-5.) k+ k+ k+5 k+ k+5 k+ Figure : sp = k + when k. This speial ase tell us that k + is the upper bound of the span when k. So we only need to prove that k + is also the lower bound of span. The proof is inluded in Appendix A-. Now we know that k + is not only the upper bound of span but also the lower bound. Hene, sp = k + when k... Solution of The Span and Assignment Strategies Up to now we have ome up to the following onlusions:. sp = = k + 5 when k = (in setion..).. sp = = k + 5 when k = (in setion..).. sp k + when k, sp = when k = (in setion..).. sp = k + when k (in setion..). With these onlusions, we derive the span as a funtion of k as follows: k + 5, k =, span = k +, k = k +, k. () 0

11 Using tiling tehnique, we develop the strategies, whih are: When k =, we use the tile in Fig. (also inluded in Appendix B-). When k =, we use the tile in Fig. 5 (also inluded in Appendix B-). When k =, we use the tile in Fig. (also inluded in Appendix B-). When k, we use the tile in Fig. (also inluded in Appendix B-5). The above are the solutions for Requirement C, while Requirement A and B have been solved in the previous disussion. The span of Requirement A is, and the strategy is shown in Fig.. Sine we have adopted the tiling tehnique, The result when k = an also be applied to Requirement B assuming that the grid extends arbitrarily far in all diretions. The span of Requirement B is also Figure : Channel assignment strategy for requirement A.

12 Generalized Model Analysis. Generalize The Two-Level-Interferene Model As a most diret extension of the former hannel assignment problem, we have a more generalized s-onstraint that hannels for transmitters within distane s must differ by at least given integer e. So we have: s-onstraint: if d(u, v) s, f(u) f(v) e s-onstraint: if d(u, v) s, f(u) f(v) k k, e are integers. If d(u, v) s is satisfied then d(u, v) s will be surely satisfied so k e. Making a few adjustments to the former model we have the following results as the upper bound for the this problem: When k > e, sp = k + e +. When e < k e, sp = k + 5e +. When e < k e, sp = k + e +. The orresponding tiles are shown in Fig. 0. Figure 0: Two level interferene model. (a) e < k e; (b) k > e; () e < k e.

13 . Irregular Loations and Several Levels of Interferene.. One Level of Interferene and Irregular Loated Transmitters We use graph theory to model this hannel assignment problem with irregular loations of the transmitters. Let G(V ) be a graph with a vertex set V of points in the plane denoting the atual transmitters and with distint verties u and v adjaent whenever there may be interferene if the same hannel is assigned to u and v. The most simplified and basi version of the hannel assignment problem involves oloring suh proximity graphs. We need to assign radio hannels (olors) to transmitters (points in V ) to avoid interferene... Clique Configuration Method We use graph theory to model a ertain type of the hannel assignment problems with irregular loations of the transmitters. Given a set V of points in the plane and given d > 0, we use a funtion f to represent the assignment of hannels f: V {,,,, t}. If d(u, v) < d, we have f(u) f(v) k and sp = min f (t). Let G(V, d) be a graph with vertex set V of points in the plane denoting the atual transmitters and with distint verties u and v adjaent if the Eulidean distane d(u, v) between them is less than d. The weight of the edge (u, v) is the least allowed frequeny differene of u and v. We develop an algorithm based on the CLIQUE question as follows: I. Find the largest lique of the given graph. II. Work out the length of the shortest Hamiltonian path in this lique. Assume the number of the verties of the largest lique found is α. Find one of its Hamilton Path. Suppose that the optimal strategy has been found and the vertex sequene A α on this Hamilton Path are assigned with a weight f(a i )( i α). Sort f(a i ) in an asending way to get a new vertex sequene A i,, A i,,, A i,α. Sine the lique is a omplete sub graph, A i,j and A i,j+ must be adjaent. So sp f(a j,α ) f(a j,α )+k f(a j,α )+k f(a j, )+(α ) k +(α ) k. () Thus we get a lower bound. This result has beome known as the Traveling Salesman Bound. This is often tight for ellular problems, but may need improvement for other types of problem.

14 It is important that this bound is applied to a suitable lique or near lique and not to the whole graph, or the bound may be very weak. One the lique or near lique that give the best bound has been identified, a frequeny assignment algorithm may be more effetive if it is applied to this sub graph first. The assignment of the sub graph an then be fixed and the assignment an be extended to the full graph. If there are h levels of interferene, we have a vetor d = (d,, d h ) T and a vetor k = (k,, k h ) T. Apply the above algorithm to every (d i, k i ) to get a distane graph G i (V, d i ), then we have sp + (α i ) k i. Therefore we get: sp max i h { + (α i ) k i }. () A 5 A A 5 =5 A A Figure : An example of a lique (α = 5). The number adjaent to the vertex is f(a i ). This figure shows the Hamilton Path after sorting the verties by f(a i )... Lower Bounding with Integer Programming Lower bounding tehniques an be used both to give a measure of the quality of any given solution and to give a meaningful omparison of the quality of different algorithms. The hannel assignment problem an be formulated as an integer program and analyzed using polyhedral methods to obtain lower bounds. Lower bounds are important for the evaluation of heuristi methods, to identify bottleneks, and to provide uts for linear relaxations of integer programs.

15 We now establish a model for hannel assignment problems using integer programming. There exists a matrix of separation onstraint C = ( i,j ) n n (n is the number of transmitters). We still use a funtion f to represent the assignment of hannels f : V {0,, t} f(v i ) f(v j ) i,j. Obviously C is a symmetri matrix and all the elements on the diagonal are zeros. Suppose x i = f(v i )( i n), we formulate the following integer program Min z = t + Subjet to x i x j i,j, i < j n x i t, i n () x i, t 0, i n x i, t are integers There are n(n )/ separation onstraints x i x j i,j in this integer programming. Eah x i x j i,j an be replaed by two inequalities: x i x j i,j and x j x i i,j. Considering the binary tree in Fig., we notie that eah leaf is orresponded with one lower bound for the problem, and eah level of nodes in the tree are assoiated with a onstraint respetively. To every node of the tree, we use the above two inequalities to ut the urrent available-domain into two available-domains for its two son nodes respetively. We apply linear searh algorithm to eah node of the tree to get a non-integer lower bound respetively. This tree is generated by the sequene shown in the matrix below. 0, n,,, n, 0,, n, n,, 0 0 n. n n, n, n, n n, n 0 n, n, n, n n, n n, n 0 We searh the entire tree adopting the Branh-and-Bound strategy to find the largest lower bound Φ of the entire solution spae, whih is non-integer. At last we get an integral lower bound Ψ = Φ for the problem. 5

16 x-x, x-x, x -x, n n x-x, x -x, x-x, x-x, x-x, x -x, x-x, x-x, x -x, x-x, xn--xn n-,n Figure : Binary searh tree... Geneti Algorithm: What If Channels Are Fixed In some situations there may not be enough hannels to be assigned to the mobile phone network. A bandwidth of a ertain size is alloated for the mobile phone use. What s more, a real hannel will take up some ertain spae in the spetrum so the maximal number of the hannels available must be a fixed one. For instane if a total bandwidth of MHZ is alloated for mobile phone usage and a single hannel oupies 5KHZ, then we may work out that at most 00 hannels are available. But applying our model we may get a span of 0, so under suh irumstanes, the above models seem a little weak. Considering this, we must sarifie the quality of the system bearing a little interferene to fulfill the fixed-hannels onstraint. Under suh irumstanes, we want to find an assignment with whih the interferene an be redued as greatly as possible. Suppose there are N transmitters dotted throughout a distrit. We have only M( N) frequeny hannels available to utilize and we need to assign eah transmitter with one hannel. Our requirement is that their mutual interferene be minimized. The further away we reuse a ertain frequeny hannel the better, but this is not always the ase. The harateristi of the whole transmitter network an be represented by a Mutual Interferene Matrix A = (a i,j )(i, j N) where a i,j is the mutual interferene between transmitters i and j if they are assigned with the same hannel. Usually this number is getting larger as the transmitters are plaed loser. We an use some figures of merit to evaluate a ertain

17 frequeny assignment sheme (for instane, sum up the interferene values in the matrix for all o-frequeny pairs), and try to minimize that figure by hanging the frequeny assignments. An iterative algorithm is needed to give us the best hannel assignment (global minimum of that figure of merit). Let x i,j equals if transmitter i is assigned to frequeny f and if not x i,j equals 0. The objetivity funtion F is defined below: i,j=n,f=m F = min a i,j x i,f x j,f (5) i,j=,f= Subjet to: for i =,, N, for f =,, M, Mf= x i,f = Ni= x i,f K f for i =,, N, f =,, M, x i,j {0, }. K f is the maximum number of transmitters you an assign to frequeny f, it may be vauous. In our ase this onstraint may not work. Having set up the above model, we now adopt the Geneti Algorithm taking the following steps: Step. Coding: We use a binary number to denote a ertain way of assignment. Call it an individual. Step. Initiate a group ontaining n individuals. Create every individual D i randomly. Notie that D i is a matrix. Step. Calulate F (D i ) for every i. Let f old be the smallest value in this group. Step. Generate new individuals by Reprodution and Mutation to get n offsprings. (Reprodution and Mutation are ertain binary operations suh as Or, And, ExlusiveOr, et.) Step 5. Selet the best n of the n offsprings whih have the lowest values of F. Let f new be the smallest value in this new group. Step. If f new f old > ε (whih is a pre-set little non-negative number ), then goto Step. Step. End.

18 5 Strengths and Weaknesses Our model works exellently under the onditions of this problem. The strategies of our basi model depend heavily on the speial geometri harater of hexagons and the s, s onstraints. Our basi model gives an aurate desription of the relationship between k and the span as a funtion of k, and it also gives us a very simple and effetive strategy to assign hannels, whih is learly demonstrated in Appendix B. Furthermore we have also onsidered some more general forms of the hannel assignment problem. As most of them are NP-Complete, we developed a way to find a lower bound, whih uses linear programming instead of the traditional way of integer programming to find the lower bound fast, and these lower bounds are ritial to the branh-and-ut algorithm. When the situations hange, suh as that the interferene level exeeds two or that the transmitters are plaed irregularly, it is lear that the omplexities of these problems are not polynomial, beause they are NP-Complete. So all of our algorithms dealing with these problems will only give sub-optimal answers in most ases. But we an use these algorithms to find lower bounds for ertain ases. With the limitation of time we annot arry on a deeper study on this problem. For instane Simulated Annealing and Heuristi Searh are also good methods to solve the fixed hannel problem, but it is a pity that we don t have enough time to implement them. Summary We develop a model for solving the hannel assignment problem. With a tiling tehnique a simple yet effiient strategy is provided to assign hannels to the given regions. Our model is based on number theory and oloring theory. We try to onstrain the span in an interval as small as possible, by inreasing the lower bound and dereasing the upper bound for the span. We find that the upper bound and the lower bound meet at some ertain value whih is a funtion of k. Our model an be suessfully applied to the situation in whih there are only two levels of interferene and the transmitters are loated regularly. We also examine some general situations, in whih the hannel assignment problem beomes a NP-Complete problem. Be aware of this, we design

19 several models to find a good lower bound. Integer programming and lique onfiguration are proposed. A situation is also onerned that there might not be enough hannels to be assigned to satisfy all the separation onstraints in real pratie. Under suh irumstanes, we want to find an assignment with whih the interferene an be redued as greatly as possible. We think that the Geneti Algorithm is an effetive alternative, whih is inluded in our models. Finally, strengths and weaknesses are disussed. Referenes [] A. Gibbons, Algorithmi Graph Theory, Cambridge University Press, 5. [] G. Calhoun, Digital Cellular Radio, Arteh House,. [] C. Berge, Graphs, North-Holland, 5. [] H. J. Walther, Ten Appliations of Graph Theory, Reidel,. [5] J. D. Parsons, Propagation and Interferene in Cellular Radio Systems, In IEE Conferene Publiations, -5, Sep.. [] V. H. MaDonald, The Cellular Conepts, The Bell System Tehnial Journal, 5():5-, Jan.. [] T. Rappaport, Wireless Communiations: Priniple and Pratie, st edition, IEEE Press,.

20 Appendix A: Rationales and Proofs A-. k + Is A Lower Bound for The Span When k Proof. Definition of symbol: Ω(i): Define the available set of integer i as a set that holds all the available integers that an be assigned adjaent to the ell of i. In other words any n is in the available set of i if it satisfies the onstraint n i k. Denote this set as Ω(i). Obviously, if integer i is used, Ω(i) must have elements. d min (i) = min f(p j ) f(p k ) where p j, p k are two adjaent ells around ell i. If integer i is used, there are different numbers adjaent to ell i. The s-onstraint requires d min (i) k. First let us suppose that integer k + is used in this assignment. We remove any integer n in the sequene,,, k + 5 if it satisfies n (k + ) < k, leaving those integers as the available set of k +. Ω(k +) = {,,,, 5, }. The s-onstraint requires that the integers used to be plaed exatly around k + to be different from eah other. Obviously integer must be plaed adjaent to k +. But only an be plaed adjaent to sine,,, 5 all have too lose a distane with. Sine two different integers are needed to fill the two ells adjaent to integer, a single integer is not enough. So we may assert that integer k + is not used in this assignment. Seond we suppose that integer is used in this assignment. We also remove any integer n in the sequene,,, k + 5 if it satisfies n < k, leaving those integers as the available set of. If k then Ω() = {k +, k +, k +,, k + 5}(k + is not used as we have just proved in the above paragraph). This means d min () (k + 5) (k + ) = k, it ontradits with d min (i) k. So is not used. If k then Ω() = {,, k, k +, k +,, k + 5}. Notie that there are atually 5 elements in Ω() whih is not enough to be plaed around integer onsidering the s-onstraint. So we may assert that is not used in this assignment. Third we suppose that integer k + is used in this assignment. Ω(k + ) = {,,,, 5, } ( is not used as we have just proved). Still the s-onstraint requires that the integers whih are used to be plaed exatly around k + to be different from eah other. Obviously is used. But only an be plaed adjaent to sine,,, 5 all have too lose a distane with. There are two adjaent 0

21 ells of integer whih means that one available integer is not enough. So we may assert that integer k + is not used in this assignment. Fourth we suppose that the integer is used in this assignment. If k then Ω() = {k +, k +,, k + 5}(k + is not used as we have just proved before). This means d min () (k + 5) (k + ) = k, it ontradits with d min (i) k. So is not used. If k then Ω() = {,, k, k +,, k + 5}. Notie that there are atually 5 elements in Ω() whih is not enough to be plaed around integer onsidering the s-onstraint. So we may assert that is not used in this assignment. Using the same tehnique we may assert that none of k+,, k+,, k+0, 0,, k+5, k+5 exist in the assignment, so only integers,,, or 5 an be used in this assignment whih is, in fat, impossible. The only reason for this is that our presumption is wrong. We annot fill the grid with only,, k +, k + 5. In other words at least k + integers are needed to fill the grid. So at least a span of k + is neessary when k. A-. k + Is A Lower Bound for The Span When k Proof. Consider an arbitrary ell a and another ells a, a,, a nearby on an infinite grid, as shown in piture A-. a a a a a a a a5 a Piture A- We onstrut a adjaent graph G with these ells. V (G) = {a, a,, a }. If a i and a j are adjaent, there is an edge in G between the vertex a i and a j. G is shown in piture A-. Aording to Coloring Algorithm we need three olors to olor the graph G. Here we olor a, a, a with olor A, olor a, a 5, a with olor B and olor a, a, a with olor C. Cells of the

22 C a a B a A A a C a C a a B B a5 A Piture A- a same olor annot be assigned with repetitive hannels beause of the s-onstraint. We an use onseutive integers to assign the ells of the same olor in order not to waste any integer hannel among the ells of the same olor. There are three olors and eah olor is assigned to three ells, so at least hannels are needed to fill these ells. Obviously, a is adjaent to a,, a, so we know that a has a different olor from all its adjaent ells. Sine a is an arbitrary ell on the infinite grid, every ell on the infinite grid should have the same harater with a aording to the symmetry. We an say that a,, a, also have this harater. That means every ell has a different olor from its adjaent ells. If two ells a i, a j are painted different olors then it is true f(a j ) f(a i ) k. Divide a,, a aording to their olors: set A = {a, a, a 5 }, set B = {a, a, a }, and set C = {a, a, a }. We suppose: f(a ) < f(a ) < f(a 5 ), f(a ) < f(a ) < f(a ), f(a ) < f(a ) < f(a ), and f(a ) < f(a ) < f(a ). There exist three situations: (i) If f(a ) < f(a ) < f(a ), then f(a ) < f(a ) < f(a ) < f(a ) < f(a ) < f(a 5 ).

23 Otherwise we may get an inequality: f(a ) < f(a ) < f(a ) < f(a i ), i = or. Insert a into this inequality and we now have a new one in an asending way. Sine any two onseutive integer hannels belong to two different olor sets, so sp f(a ) + k + k + k + k > + k + = k +, k >. It is known that sp k +, it leads to a ontradition. Insert f(a ) to (i), there are 5 plaes for f(a ), suh as f(a ) < f(a ) < f(a ) < f(a ) < f(a ) < f(a 5 ) < f(a ). Thus, sp f(a ) k + f(a 5 ) k + + f(a ) + f(a ) + k > + k + f(a ) + k + + f(a ) + k k + + = k + + k > k + +, k >. The result is idential if we assert a into the other 5 plaes. Therefore the assumption a < a < a leads to sp > k +. It ontradits with sp k +, so the assumption is atually wrong. (ii) If a < a < a 5, the same proof as above an be given again to show that the assumption is wrong. (iii) a 5 < a. This inequality holds forever. The proof as above shows f(a ) < f(a ). Then we have f(a ) < f(a ) < f(a 5 ) < f(a ) < f(a ) < f(a ) < f(a ) < f(a ) < f(a ), sp f(a ) f(a ) + f(a ) + k + f(a ) + k + f(a 5 ) + k + f(a ) + k + k +. So sp = k +. A feasible assignment strategy is: f(a ) =, f(a ) =, f(a 5 ) =, f(a ) = k +, f(a ) = k +, f(a ) = k + 5, f(a ) = k + 5, f(a ) = k +, f(a ) = k +.

24 Appendix B: Strategies N= i= j= k= 5 Appendix B- N= i=0 j= k= Appendix B-

25 N= i= j= k= Appendix B k +k +k +k +k +k +5k +k +k +k +5k +k +k +k +k +k +k +k +5k +k +k +5k +k +k Appendix B- N= i= j= +k +k +k +5k +k +k 5

26 k+ k+ k+ k+5 k+ k+ k+ k+5 k+ k+ k+5 k+ k+ k+5 k+ k+5 k+ k+ k+5 k+ k+5 k+ k+5 k+ Appendix B- 5 N= i= j=0 k> k+ k+ k+5 k+ k+5 k+