Meeting times of Taboo random walks on bipartite graphs

Transcription

1 Meeting times of Taboo random walks on bipartite graphs By Xianwu Zhang MS candidate: Applied and Computational Mathematics Advisor: Barry James Co-advisor: Kang Ling James Department of Mathematics and Statistics University of Minnesota Duluth 1

2 ACKNOWLEDGEMENTS First I want to acknowledge all of the help from my advisor, Professor Barry James and my co-advisor Professor Kang James. Without their help I wouldn t be where I am today. I also would like to give my thanks to Professor Qi, who is my committee member for his help in writing this paper and in course study. I also give my special thanks to Yun Peng for his help on the programming. Finally I give my thanks to the statistics seminar class which helps me to prepare the talk better and better.

3 DEDICATION This dissertation is dedicated to my parents. For your everlasting love 3

4 Abstract In this paper we tried to determine the distribution of the first time T that two indepent taboo random walks meet on a bipartite graph. Firstly, we did simulations for complete and incomplete bipartite graphs to support our conjecture that T should have a geometric distribution deping on the even or odd steps. Secondly, for complete graphs, we figured out a formula with proof for theoretical conditional probabilities which are consistent with the simulation results; then we gave a pdf fort. Then, for one example from incomplete graphs, from valid paths we calculated the theoretical probabilities which are very consistent with frequencies from the simulation results fort. Last, we tried a new definition of meeting to avoid the even or odd steps analysis and obtained some conjectures. 4

5 Contents Meeting times of Taboo random walks on bipartite graphs Introduction Review of some related literature and some preliminaries Related literature Some terminologies and properties Quick review of Markov chains A Markov chain associated with our taboo random walk Results for a complete bipartite graph Simulation results Formula for complete graphs Conjecture for incomplete graphs Simulation based on Bernoulli's law of large numbers Theoretical conditional probabilities An idea for using Markov chain to get E(T) Further research References... 4 Appices... 6 Program # Program #... 6 Program # Program # Program # Program # Program # Program # Program # Program # Program #

6 1. Introduction In graph theory, a bipartite graph is a graph whose vertices are in two disjoint groups and each vertex in one group is connected only to some other vertices in the other group. The degree of one vertex is the number of its neighbors in the other group. A complete graph means that each left vertex is connected to each right vertex like fig.1 and an incomplete graph means that each left vertex is connected to some of right vertices but not to all right vertices like fig.. Our graph in this paper goes like this: Let G denote a bipartite graph with left vertex set of size L and right vertex set of size R. In general both L and R could be very large. Assume that all the left vertices have the same degree k, all the right vertices have the same degree q and the transitions from any vertex to its neighbors are equiprobable. A taboo random walk in this paper means that you can t go back to the vertex where you came from. Now consider the following problem: we select two left vertices uniformly at random and start two indepent taboo random walks from each vertex. Our question is how to determine the distribution of the first time T when the walks meet. The following two graphs are examples of a complete graph and an incomplete graph, with k R, q L 3in the first graph and k, q4 in the second graph. Fig.1 Fig. 6

7 D.J. Aldous and others have done the research on the meeting time of Markov chains and have also given upper bounds for the expected meeting time. Their results inspire us to look at the distribution of meeting times of random walks and Markov chains. There is still future work to do, but we do get some interesting results in this paper. In section we review some related papers and explain some preliminaries. In section 3 we get some results for complete bipartite graphs, and in section 4 we get some conjectures for incomplete graphs. In section 5 we will try an idea to use Markov chain to get E (T). In section 6 we mention future research topics.. Review of some related literature and some preliminaries.1 Related literature The literature on this problem is sparse. But we do have some papers which focus on the related topic: an upper bound for the expected meeting time. Those papers are inspiring and beneficial for potential research. In his 1990 paper Meeting times for indepent Markov chains, D.J. Aldous researched the expected time two indepent continuous-time reversible Markov chains first meet. Suppose X, Y are two chains in question, then t t T min t : X Y. M t t T M is the first meeting time for X t and Y t. The worst-case expected meeting time is max E(T X i, Y j). M i, j M 0 0 D.J. Aldous got an upper bound for M : τ i M K i τ1v EπHi 1 π, 7

8 where a v b means the maximum number of a and b; EH means the expected time that the chain goes to state k from state i ; k i EH πeh the weighted expected hitting time from state i to the other states and is π i k k i k stationary distribution. In 008, Boaz Nadler concluded a sharper bound for the meeting time. The meeting time of two random walks is O (L). The bigo notation contains a multiplicative constant that deps on the parameters k and q, and the exact degrees of the left and right vertices. Intuitively he thought the larger k, the faster the random walk can spread to far away vertices. He also did simulation to support the conclusion. His graph is very similar to my project. As predicted theoretically, the meeting time is linear in L and it decreases with k. In 01, Roberto Imbuzeiro Oliveira proved that the expected value of T is at most a constant multiple of the largest hitting time of an element in the state space, which is a sharper bound compared with Aldous s results.. Some terminologies and properties..1 Quick review of Markov chains If we want to model this random walk using Markov chains, we have to make sure that the chain has the Markov property. The Markov property is that the future of the process deps only on present states and not on the past. Suppose X 0, X1, X, is a Markov chain, then for all n 0,1,, and states i, j, the Markov property is P n1 n n1 n n1 n (X j X i) P(X j X i,x i,,x i,x i ). Let n p ij denote the probability that the chain goes from state i to state j in n steps. If the number of states in state space S is N, P 1 [p ij ] NN is the one-step transition matrix and 8

9 P [p ] P [] ij NN is the two-step transition matrix. P is stochastic because # of states pij 1. j1 And P is doubly stochastic if we have # of states pij 1. i1 n State i is accessible from state j if p 0 for some n 0 and we say state i leads to state j and i If i j. ij jand j i, then i communicates with j and i j. If i j for all states, then the chain is irreducible. Let fi denote the probability of a return to i starting from i. For an irreducible chain, if fi 1, the chain is recurrent and if fi 1, then the chain is transient... A Markov chain associated with our taboo random walk The taboo random walk on a bipartite graph is not a Markov chain because of the taboo condition. The taboo condition means that the future step of the taboo random walk deps on the previous one step not the current step. We define a state by a pair of vertices of the taboo random walk in two consecutive steps. The Markov property is guaranteed because of the taboo condition and the future state now only deps on the current state, which is the Markov property. Fact 1: the chains for complete graphs and for the incomplete example in this paper are irreducible and have a finite number of states. Suppose i and j are any two states in the state space S, obviously i jand j i,which means i communicates with j for any state i and j. So the whole chain is irreducible. Then the chain has one communicating class and has a finite number of states; the number of states is Rq or Lk. 9

10 Fact : the chains above are positive recurrent. The chain in question has a finite number of states, which means there is at least one recurrent state in the state space. In addition, the chain has only one communicating class. So the chain is positive recurrent. Lemma 1: the transition matrix for the chain of complete graph is doubly stochastic. Proof: j th i th p ij Theith row is stochastic because # of states j1 p ij 1. Suppose state j is mn or mn If j is mn, its direction is from L to R and then the previous step should be from R to L. And because of the taboo condition the possible previous states are 1 m, m, ( n 1) m,( n 1) m, Rm. Then the corresponding entries in the jth column are nonzero but the rest entries are zeros. 1 pij *(R 1) 1. R 1 # of states i1 If j is mn, its direction is from R to L and then the previous step should be from L to R. And because of the taboo condition the possible previous states are 1 m, m, ( n 1) m,( n 1) m, Lm. Then the corresponding entries in the jth column are nonzero but the rest entries are zeros. Therefore, 10

11 1 pij *( L 1) 1. L 1 # of states i1 Lemma: the above chains have a stationary distribution. Proof: the chain is irreducible, the chain has a finite number of states and the chain is doubly stochastic. Then the chain has a stationary distribution, which is in this case a discrete uniform distribution. Since Rq denotes the number of the states, (,,, ) Rq Rq Rq. Lemma3: the above chains are periodic and the period is or 4 deping on the graphs. Proof: the chain has only one communicating class and we only have to find the period of one state because period is a class property. Because of the bipartite graph structure, the chain will come back only in an even step. According to the definition of period So n is an even number and d i is or 4. i ii n d gcd n1; P 0, Case 1: for a complete graph, the period is as long as k, q The following example is the simplest complete bipartite graph We focus on state n 4 11

12 n n 1 Any return time is a multiple of 4 since 4 and 5 alternate on the right-hand side because there are just two right vertices. So the period for 14 is 4 and the period of the chain is 4. We focus on n n n 8 So the period for 15 is and the period of the chain is. As long as k, q, we can find a path in 6 steps like the above example, which means the period is. Case : the period for our incomplete graph example is We focus on 17 : 1

13 P n n 8 So the period for 17 is and the period of the chain is. Fact 3: if a random variable T ~ GEO( p) and P(T n 1 T n) p P(T n T n 1) p 1 P(T k) (1 p) k p, then PDF for GEO(0.) T fig.3 PDF of GEO(0.) 3. Results for a complete bipartite graph Complete graph means each left vertex is connected to each right vertex. For these connected cases, we have got formulas for the distribution of T when the two walks meet on even steps or odd steps. We have a conjecture that T has a geometric distribution given that T is even or odd. We did simulation for two examples first and found support for our conjecture and then we derived the formula. 3.1 Simulation results 13

14 Let E denote an event and N ( ) n E denote the number of times the event E happens in n trials. According to Bernoulli s law of large numbers, N ( ) n E n p n where p is the probability that event E happens in one trial and the convergence is convergence in probability. Our simulation has two steps. If we get almost identical conditional probabilities from the simulation, it means that our conjecture at least gets some positive support from the simulation. Step1: I did 90,000,000 indepent trials by using MATLAB. In each trial two indepent taboo random walks start from the left side uniformly at random. Then I calculated the proportion of trials for the different steps where the two walks first meet. Step: I used the frequencies or proportions to estimate the real probabilities and used those estimates to calculate the conditional probabilities P(T n 1 T n) and P(T n T n 1) n 0,1,,3 For the first example of a complete graph ( L3, R ), we calculated the conditional probabilities (condp) using the frequencies. T condp T condp T condp T condp Table.1 Conditional probabilities for odd steps T condp

15 T condp T condp T condp Table. Conditional probabilities for even steps For the second example of a complete graph ( L4, R 3 ) T condp T condp T condp Table.3 Conditional probabilities for odd steps T condp T condp T condp Table.4 Conditional probabilities for even steps The simulation results are positively supportive for the geometric distribution conjecture because of almost constant conditional probabilities. 3. Formula for complete graphs Lemma4: for the general complete graph with L and R vertices, the conditional probabilities are R PT n 1 T n p R1 1 L L1, PT n T>n 1 p, n 1,,3 15

16 Proof: Lk Rq, in this case k R, q L. The two walks have to meet the taboo condition. We do not have to pay attention to the previous steps except the previous one step to determine the next step. We prove the following formula and the other one has the similar logic. 1 1 L P(T n T n 1) ( L )* * L1 L1 (L1) If T n, the two walks meet on the left-hand side. Because of the taboo condition, they can meet on L left vertices. Each walk finished the n 1th step with a probability of 1 L 1, and the two walks are indepent. In the same way, we can get 1 1 R P(T n 1 T n) (R )* * R 1 R 1 ( R 1) According to the formula and simulation results, we can get the comparison table and the simulation results are consistent with the formula results. Graph 50,000,000 trials P(T=n+1 T>n) P(T=n T>n-1) fig.a Formula results Simulation results fig.b Formula results Simulation results Table.5 the comparison between simulation and theoretical results Theorem: for the general complete graph with L left vertices and R right vertices, the pdf for T is ( ) ( 1)* ( 1) (1 ) n n P T n P T n T n P T n p p (1 p ) C, 1 P T n P T n T n P T n p p p C, 1 ( 1) ( 1 )* ( ) 1(1 1) n n (1 ) 16

17 where ( R1)( L1) C 1 P(0) P(1) RL Proof: Even steps P(T 0) L* * L L L P( T ) P( T T 1)* P( T 1) p *(1 P(0) P(1)) p C P( T 4) P( T 4 T 3)* P( T 3) p *(1 P(0) P(1) P() P(3)) p (1 p )(1 p ) C 1 P( T 6) P( T 6 T 5)* P( T 5) p(1 p1) (1 p ) C p *(1 P(0) P(1) P(5)) P( T 8) P( T 8 T 7)* P( T 7) p *(1 P(0) P(1) P(7)) P( T n ) p(1 p1) (1 p ) C : : P( T n T n 1)* P( T n 1) (1 ) n n p p (1 p ) C Odd steps L( L 1) L 1 P( T 1) R* ** * * * L L R R R* L P( T 3) P( T 3 T )* P( T ) p *(1 P(0) P(1) P()) 1 p (1 p ) C 1 1 P( T 5) P( T 5 T 4)* P( T 4) p1(1 p1)(1 p ) C p *(1 P(0) P(1) P(4)) P( T 7) P( T 7 T 6)* P( T 6) p *(1 P(0) P(1) P(6)) 1 3 p1(1 p1) (1 p ) C P( T 9) P( T 9 T 8)* P( T 8) p *(1 P(0) P(1) P(8)) 1 P( T n 1) 3 4 p1(1 p1) (1 p ) C : : P( T n 1 T n)* P( T n) p p p C 1 1(1 1) n n (1 ) 4. Conjecture for incomplete graphs 17

18 frequencies for T We did simulation for the incomplete graph example. In this example L 6, R 3, k, q 4. The construction of graph is the following: 4.1 Simulation based on Bernoulli's law of large numbers The conditional probabilities are based on the simulation results. From those conditional probabilities, we can conjecture that T has a geometric distribution given that T is even or odd histogram for frequencies T(1-80) fig.4 histogram of trials 18

19 T condp T condp Table.6 Conditional probabilities for odd steps T condp T condp Table.7 Conditional probabilities for even steps We got almost constant estimates for conditional probabilities for odd and even steps from the simulations. Because of large number of trials, I believe that these estimates are very close to the real conditional probabilities. 4. Theoretical conditional probabilities Can we get the real probabilities for T and then the real conditional probabilities? To confirm our conjecture and to answer the last question, we also tried another way to give some support. The idea here is to calculate the theoretical probabilities for different meeting times. Step1: find all the valid meeting pairs that the two indepent random walks can meet on different steps. Step : add all the probabilities of each meeting pair to get the theoretical probability for 19

20 T 1,,3, and calculate the conditional probabilities A valid meeting pair satisfies two properties. The pairs meet the taboo conditions; the two walks meet for the very first time at the very of each pair. The calculation goes like this. We use one meeting pair for T to explain the calculation This is a meeting pair. One walk can go 7 1 and the other walk can go The probability for one walk is and the probability for one meeting pair is because they are indepent. Each pair has the same probability So the probability for T is The following table is part of calculations T Number of meeting pairs Theoretical probabilities Simulation results Table.8 part of calculations for comparison The following table is the comparison between the simulation results and probabilities from the meeting pairs. 0

21 T Frequency by simulation # of valid meeting pairs Probabilities from pairs T Frequency by simulation # of valid meeting pairs Probabilities from pairs Table.9 comparison results between theoretical results and simulation results We can tell that the simulation results are consistent with the theoretical probabilities calculated from the valid pairs. Even though we don t have a specific formula for the conditional probabilities, we still can conjecture that T given on T is even or odd has a geometric distribution. And we can calculate the pdf by using the valid pairs. 5. An idea for using Markov chain to get E(T) A new definition of meeting for two Markov chains can avoid the analysis for even and odd steps. At first, we define a state as a pair of vertices like before. Then we only consider the Markov chain in every two steps. The new definition for meeting is they have same first vertex or same second vertex in the same step. Let T ' denote the first time the two random walks meet. T ' 0,1, and in the diagram above ' T 1,which corresponds to T or T 3. We then have 1

22 Proportion * T ' T * T' 1 *E( T ') E( T) *E( T ') 1 We did the simulation for the following example estimates for conditional probabilities for different steps from the simulation p= T Fig.5 conditional probabilities for new idea in trials

23 ' T condp T condp T condp T condp T condp Table.10 Conditional probabilities for new idea in trials From the conditional probabilities based on the simulation results we can say T has at least approximately a geometric distribution. From the simulation for the new idea, ET '.739, and from the previous simulation in terms of random walk, ET , so it is true that * E T ' E T * E T ' 1. From the simulation results based on the new idea, it is quite possible that geometric distribution. If T ' ~ GEO(p), then T ' has a * T ' T * T ' 1 i1 P T ' i (1 p) * p, P{T' i} PT i or i 1 i1 P{T i} P{T i1} (1 p) * p (a) Also, from the previous conjecture that T has a geometric distribution when restricted to even or odd steps, 3

24 P{T i1} c P{T i} (b) We actually can get the P{T i} and P{T i 1} by solving (a) and (b). Even if T has not a geometric distribution, we can t get P{T i} and P{T i 1} but we still can get a good approximation for ET. This should be a very useful approximation when L and R are large. Because the larger the L and R, the easier the Markov chain can spread away. It means that T ' and T will be larger and the larger T means *E( T ') or *E( T ') 1will be a better approximation for ET ( ) in terms of relative error. In all three reference papers they obtained a upper bound for ET ( ). 6. Further research Actually this new idea could give us a very good approximation for ET ( ). One concern about this: In Aldous s paper Meeting times for indepent Markov chains, continuous time reversible Markov chains are studied and the upper bound is for this kind of chain. For the complete graph, the two step transition matrix is symmetric and therefore two step Markov chain is time reversible. Perhaps an analogous result can be obtained for incomplete graph, and Aldous s result could be compared with our approximation. One more topic for future research is to develop a program to calculate the probabilities as long as you have some information like L, R, k, q. 7. References D. J. Aldous, Meeting times for indepent Markov chains, Stochastic Processes and their Applications 38 (1991) Boaz Nadler, Taboo Random Walk on an Expander Graph, Sept Roberto Imbuzeiro Oliveira, On the coalescence time of reversible random walks, Trans. Amer. Math. Soc. 364(01),

25 Sheldon M. Ross, Introduction to Probability Models 10 th edition, Academic Press (Elsevier), Amsterdam, 010 5

26 Appices Program #1 %%==simulation for the second example===== clear clc counter=zeros(1,100000); records=[]; for n=1: a1=rand; b1=rand; a=translate(a1);b=translate(b1); if a==b counter(1)=counter(1)+1; if a~=b recorda=[];recordb=[]; T=1; flaga=1;flagb=1; recorda=[recorda a];recordb=[recordb b]; while a~=b if a<=6 && b<=6 y1=rand;[a,c]=translate1a(a,y1,flaga,recorda);recorda=[recorda a];flaga=c; y=rand;[b,c]=translate1b(b,y,flagb,recordb);recordb=[recordb b];flagb=c;t=t+1; else y3=rand;a=translatea(a,y3,recorda);recorda=[recorda a]; y4=rand;b=translateb(b,y4,recordb);recordb=[recordb b];t=t+1; counter(t)=counter(t)+1; counter(1:6)/ Program # function y=translate(x) if x<=1/6 y=1; elseif x<=/6 y=; elseif x<=3/6 y=3; elseif x<=4/6 y=4; elseif x<=5/6 y=5; elseif x<=1 6

27 y=6; Program #3 function [w,f]=translate1a(x,y,flaga,z) if flaga==1 if y<=1/ && x==1 w=7; elseif y>=1/ && x==1 w=8; if y<=1/ && x== w=7; elseif y>=1/ && x== w=9; if y<=1/ && x==3 w=8; elseif y>=1/ && x==3 w=9; if y<=1/ && x==4 w=7; elseif y>=1/ && x==4 w=9; if y<=1/ && x==5 w=8; elseif y>=1/ && x==5 w=9; if y<=1/ && x==6 w=7; elseif y>=1/ && x==6 w=8; if flaga== 7

28 if x==1 && z(-1)==7 w=8; elseif x==1 && z(-1)==8 w=7; if x== && z(-1)==9 w=7; elseif x== && z(-1)==7 w=9; if x==3 && z(-1)==9 w=8; elseif x==3 && z(-1)==8 w=9; if x==4 && z(-1)==9 w=7; elseif x==4 && z(-1)==7 w=9; if x==5 && z(-1)==9 w=8; elseif x==5 && z(-1)==8 w=9; if x==6 && z(-1)==8 w=7; elseif x==6 && z(-1)==7 w=8; if flaga==1 f=; if flaga== f=; Program #4 function w=translatea(x,y,z) if (x==7) && (z(-1)==1) 8

29 if y<=1/3 w=; elseif y<=/3 w=4; elseif y<=1 w=6; if (x==7) && (z(-1)==) if y<=1/3 w=1; elseif y<=/3 w=4; elseif y<=1 w=6; if (x==7) && (z(-1)==4) if y<=1/3 w=1; elseif y<=/3 w=; elseif y<=1 w=6; if (x==7) && (z(-1)==6) if y<=1/3 w=1; elseif y<=/3 w=; elseif y<=1 w=4; if (x==8) && (z(-1)==1) if y<=1/3 w=3; elseif y<=/3 w=5; elseif y<=1 w=6; if (x==8) && (z(-1)==3) if y<=1/3 w=1; elseif y<=/3 w=5; elseif y<=1 w=6; if (x==8) && (z(-1)==5) 9

30 if y<=1/3 w=1; elseif y<=/3 w=3; elseif y<=1 w=6; if (x==8) && (z(-1)==6) if y<=1/3 w=1; elseif y<=/3 w=3; elseif y<=1 w=5; if (x==9)&& (z(-1)==) if y<=1/3 w=3; elseif y<=/3 w=4; elseif y<=1 w=5; if (x==9) && (z(-1)==3) if y<=1/3 w=; elseif y<=/3 w=4; elseif y<=1 w=5; if (x==9) && (z(-1)==4) if y<=1/3 w=; elseif y<=/3 w=3; elseif y<=1 w=5; if (x==9) && (z(-1)==5) if y<=1/3 w=; elseif y<=/3 w=3; elseif y<=1 w=4; 30

31 Program #5 %%%---- use recursive algorithm to get all the possible qualified routes in %%%----- pair for A and B-- clear clc global M M=[]; T=input('PLEASE INPUT THE STEPS YOU WANT:'); n=t; if T%==0 A=creatfromlastspot(1);L=size(A); for i=1:l() b=[1;1];a=[a(:,i) b]; while length(a)<t+1 a=creatfrom(a,t); B=creatfromlastspot();L=size(B); for i=1:l() b=[;];a=[b(:,i) b]; while length(a)<t+1 a=creatfrom(a,t); C=creatfromlastspot(3);L=size(C); for i=1:l() b=[3;3];a=[c(:,i) b]; while length(a)<t+1 a=creatfrom(a,t); D=creatfromlastspot(4);L=size(D); for i=1:l() b=[4;4];a=[d(:,i) b]; while length(a)<t+1 a=creatfrom(a,t); E=creatfromlastspot(5);L=size(E); for i=1:l() b=[5;5];a=[e(:,i) b]; while length(a)<t+1 a=creatfrom(a,t); F=creatfromlastspot(6);L=size(F); for i=1:l() b=[6;6];a=[f(:,i) b]; while length(a)<t+1 a=creatfrom(a,t); 31

32 if T%==1 A=creatfromlastspot(7); for i=1:length(a) b=[7;7];a=[a(:,i) b]; while length(a)<t+1 a=creatfrom(a,t); if T==1 % disp(a); M=[M;a]; B=creatfromlastspot(8); for i=1:length(b) b=[8;8];a=[b(:,i) b]; while length(a)<t+1 a=creatfrom(a,t); if T==1 % disp(a); M=[M;a]; C=creatfromlastspot(9); for i=1:length(c) b=[9;9];a=[c(:,i) b]; while length(a)<t+1 a=creatfrom(a,t); if T==1 % disp(a); M=[M;a]; disp(' all qualified routes in pair for pig A and pig B are') M; disp('the # of the all qualified routes in pair for pig A and pig B is') s=size(m);s(1)/ Program #6 %%--to test if there is any invalid route in all the possible routes-- A=M; s=size(a);m=s(1)-1;n=s()-1;flag=0; for i=1:m for j=1:n if A(i,j)==A(i,j+1) if mod(i,)==1 disp(i) else disp(i-1) 3

33 disp('th is an invalid route') flag=flag+1; for i=1::m for j=1:n if A(i,j)==A(i+1,j) disp(i) disp('th is an invalid route') flag=flag+1; if flag==0 disp('awesome!!! all routes are valid routes') Program #7 function routes=creatfrom(a,t) global T T=t; global M c=a(:,1:);b=zeros(,1);a=a; % T=t; if c(1,1)<=6 if c(1,1)==1 if c(1,)==7 b(1,1)=8; if c(1,)==8 b(1,1)=7; if c(1,1)== if c(1,)==7 b(1,1)=9; if c(1,)==9 b(1,1)=7; if c(1,1)==3 if c(1,)==8 b(1,1)=9; if c(1,)==9 b(1,1)=8; if c(1,1)==4 if c(1,)==7 b(1,1)=9; if c(1,)==9 b(1,1)=7; if c(1,1)==5 if c(1,)==8 b(1,1)=9; if c(1,)==9 b(1,1)=8; if c(1,1)==6 33

34 if c(1,)==7 b(1,1)=8; if c(1,)==8 b(1,1)=7; if c(,1)==1 if c(,)==7 b(,1)=8; if c(,)==8 b(,1)=7; if c(,1)== if c(,)==7 b(,1)=9; if c(,)==9 b(,1)=7; if c(,1)==3 if c(,)==8 b(,1)=9; if c(,)==9 b(,1)=8; if c(,1)==4 if c(,)==7 b(,1)=9; if c(,)==9 b(,1)=7; if c(,1)==5 if c(,)==8 b(,1)=9; if c(,)==9 b(,1)=8; if c(,1)==6 if c(,)==7 b(,1)=8; if c(,)==8 b(,1)=7; if c(1,1)>6 [B,f]=creatfromrightmiddlespot(c);m=length(B); if f==1 for i=1:m b=b(:,i);routes=[b A]; while length(routes)<t+1 routes=creatfrom(routes,t); if length(routes)==t+1 && length([b A])==T+1 % disp(routes); M=[M;routes]; routes=[b A]; 34

35 Program #8 function initialroutes=creatfromlastspot(n) n=n; if n==7 initialroutes=combntns([1 4 6],)'; if n==8 initialroutes=combntns([ ],)'; if n==9 initialroutes=combntns([ 3 4 5],)'; if n==1 initialroutes=combntns([7 8],)'; if n== initialroutes=combntns([7 9],)'; if n==3 initialroutes=combntns([8 9],)'; if n==4 initialroutes=combntns([7 9],)'; if n==5 initialroutes=combntns([8 9],)'; if n==6 initialroutes=combntns([7 8],)'; Program #9 function [routes,flag]=creatfromrightmiddlespot(cc) c=cc;flag=1; if c(1,1)==c(,1) flag=0; if c(1,1)==7 if c(1,)==1 B=[ 4 6]; if c(1,)== B=[1 4 6]; if c(1,)==4 B=[1 6]; if c(1,)==6 35

36 B=[1 4]; if c(1,1)==8 if c(1,)==1 B=[3 5 6]; if c(1,)==3 B=[1 5 6]; if c(1,)==5 B=[1 3 6]; if c(1,)==6 B=[1 3 5]; if c(1,1)==9 if c(1,)== B=[3 4 5]; if c(1,)==3 B=[ 4 5]; if c(1,)==4 B=[ 3 5]; if c(1,)==5 B=[ 3 4]; if c(,1)==7 if c(,)==1 C=[ 4 6]; if c(,)== C=[1 4 6]; if c(,)==4 C=[1 6]; if c(,)==6 C=[1 4]; if c(,1)==8 if c(,)==1 C=[3 5 6]; if c(,)==3 C=[1 5 6]; if c(,)==5 C=[1 3 6]; if c(,)==6 C=[1 3 5]; if c(,1)==9 if c(,)== C=[3 4 5]; if c(,)==3 C=[ 4 5]; if c(,)==4 36

37 C=[ 3 5]; if c(,)==5 C=[ 3 4]; D=[]; for i=1:3 for j=1:3 if C(j)~=B(i) d=[b(i);c(j)];d=[d d]; routes=d; Program #10 %%--- to eliminate the replication in the routes--- L=length(M); for i=1::l for j=i+::l- if j+<=l if M(j,:)==M(i,:) if M(j+1,:)==M(i+1,:) M=[M(1:j-1,:);M(j+:,:)];L=length(M); Program #11 %%--to test if there is any invalid route in all the possible routes-- A=M; s=size(a);m=s(1)-1;n=s()-1;flag=0; for i=1:m for j=1:n if A(i,j)==A(i,j+1) if mod(i,)==1 disp(i) else disp(i-1) disp('th is an invalid route') 37

38 flag=flag+1; for i=1::m for j=1:n if A(i,j)==A(i+1,j) disp(i) disp('th is an invalid route') flag=flag+1; if flag==0 disp('awesome!!! all routes are valid routes') 38