Homework 13 Solutions. X(n) = 3X(n 1) + 5X(n 2) : n 2 X(0) = 1 X(1) = 2. Solution: First we find the characteristic equation

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1 Homework 13 Solutions PROBLEM ONE 1 Solve the recurrence relation X(n) = 3X(n 1) + 5X(n ) : n X(0) = 1 X(1) = Solution: First we find the characteristic equation which has roots r = 3r + 5 r 3r 5 = 0, r 1 = 3 + 9, r = 3 9 For convenience of notation, let us write these roots as r 1 and r until the end of the problem Since r 1 r, the general solution is X(n) = c 1 r n 1 + c r n Using the initial conditions, we get the system of equations Multiply the first equation by r 1 to get and add both equations to get 1 = X(0) = c 1 r c r 0 = X(1) = c 1 r c r 1 1 = c 1 + c = c 1 r 1 + c r (1) r 1 = r 1 c 1 r 1 c = c 1 r 1 + c r r 1 = c r c r 1 = c ( ) c = r 1 Using the first equation from (1), we get 1

2 Thus the specific solution is c 1 = 1 c = r 1 = r X(n) = r r n 1 + r 1 r n Now we plug back in r 1 and r Notice that = 9 Also, r = 1 9, r 1 = 1 9 Therefore, the specific solution is X(n) = ( Solve the recurrence relation 3 + ) n 9 1 ( 9 3 ) n 9 9 X(n) = 4X(n 1) 4X(n ) : n X(0) = 0 X(1) = 1 Solution: The characteristic equation is r = 4r 4 r + 4r + 4 = 0 (r + ) = 0 We have a double root, r = ; therefore, the general solution is X(n) = c 1 r n + c nr n = c 1 ( ) n + c n( ) n To solve for the constants, we use the initial conditions: 0 = X(0) = c 1 ( ) 0 + c (0)( ) 0 1 = X(1) = c 1 ( ) 1 + c (1)( ) 1 0 = c 1 1 = c 1 c ()

3 Plugging the first equation into the second, we get Therefore, the specific solution is X(n) = n( )n c = 1/ = ( )n( )n 1 = n( ) n 1 3 Let N(n) be the number of length n strings each of whose digits are either 1,, 3, or 4, and which do not contain two adjacent 4s (a) Find N(1) and N() Solution: The total number of length 1 strings (with no restriction) is 4 None of these strings have adjacent 4s (they only have one character), so N(1) = 4 The number of length strings (with no restriction) is 16 Only one of these has adjacent 4s: the string 44 Therefore, N() = 15 (b) Find a recurrence relation that N(n) satisfies for n Solution: We will count the number of length n strings which have no adjacent 4s Consider the first character in the string it is either a 4 or it is not If it is a 4, then the next character cannot be a 4 Therefore, the second character is either a 1,, or 3 In each one of these cases, the remaining n characters (from the third onward) can be anything as long as there are no adjacent 4s This means that in each case (1,, or 3 for the second character, but 4 for the first character), there are N(n ) possibilities This gives a total of 3N(n ) possibilities if the first character is a 4 If the first character is not a 4, it is either a 1,, or 3 In each case, the remaining n 1 characters can be anything, so long as there are no adjacent 4s This means that in each case (1,, or 3 for the first character), there are N(n 1) possibilities This gives a total of 3N(n 1) possibilities if the first character is not a 4 Putting together the above answers, we get a recurrence (with initial conditions) of N(n) = 3N(n 1) + 3N(n ) n 3 N(1) = 4 N() = 15 3

4 (c) Solve the recurrence to find N(n) as a function of n Solution: Again, we find the characteristic equation: which has roots r = 3r + 3 r 3r 3 = 0, r 1 = 3 + 1, r = 3 1 Since r 1 r, the general solution is of the form N(n) = c 1 r n 1 + c r n We use the initial conditions to find the constants; however, the initial conditions make the problem very complicated It turns out that things are easier if we use initial conditions for N(0) and N(1) instead of for N(1) and N() Notice that if we use the initial condition N(0) = 1, then the sequence that we get (1, 4, 15, ) is the same as our previous sequence, except that we start with the zeroth term instead of the first term Hence, we will use 1 = N(0) = c 1 r c r 0 4 = N(1) = c 1 r c r 1 1 = c 1 + c 4 = c 1 r 1 + c r Multiply the first equation by r 1 to get and add them to get r 1 = c 1 r 1 c r 1 4 = c 1 r 1 + c r (3) 4 r 1 = c ( ) c = 4 r 1 Plugging this into the first equation of (3), we get c 1 = 1 c = 4 r 1 4

5 This gives a specific solution of Using the fact that the specific solution becomes N(n) = = r 4 N(n) = r 4 r n r 1 r n ( 4 Solve the recurrence relation X(n) = X(0) = X(1) = 3 = 1, 3 + ) n 1 5 ( 1 3 ) n 1 1 ( X(n 1) (X(n )) ) 8 : n Solution: First use the natural logarithm: for n 0, let The recurrence becomes Y (n) = ln X(n) Y (n) = 8Y (n 1) 16Y (n ) : n Y (0) = ln Y (1) = ln 3 The characteristic equation is which has a double root at Therefore, the general solution is r 8r + 16 = 0, r = 4 Y (n) = c 1 4 n + c n4 n To solve for the constants, use the initial conditions: ln = Y (0) = c c (0)4 0 ln 3 = Y (1) = c c (1)4 1 5

6 Plugging the first equation into the second gives ln = c 1 ln 3 = 4c 1 + 4c (4) 4c = ln 3 4 ln = ln 3 ln 16 = ln( 3 16 ) c = 1 4 ln( 3 16 ) Therefore, the specific solution for Y (n) is Y (n) = ln (4) n ln( 3 16 )n(4)n To find the equation for X(n), we take X(n) = e Y (n) = e ln (4)n + 1 ln( )n(4) n = ( e ln ) ( 4 n e ln( 16) ) 3 n4 n 4 ( ) n(4 n 1 = (4n ) 3 ) 16 PROBLEM TWO 1 For which values of m and n does the complete bipartite graph K m,n have an Euler cycle? A Hamiltonian cycle? Solution: We saw in class that a graph has an Euler cycle if and only if each vertex has even degree Split the vertices of K m,n into two groups: put the vertices from the group of m vertices on the left and put the vertices from the group of n vertices on the right Each vertex on the left is connected to each vertex on the right; therefore, the degree of each vertex on the left is n Similarly, the degree of each vertex on the right is m Using the fact from class, this means that K m,n has an Euler cycle if and only if m and n are both even To answer the next question, we keep the same setup, with m vertices on the left and n vertices on the right If m = n, then we can explicitly construct a Hamiltonian cycle (try this for m = n = 4) If m n then K m,n cannot have a Hamiltonian cycle To show this, we may assume that m < n (otherwise we can just flip the graph) Recall that all edges connect vertices on the left only with vertices on the right If there were a Hamiltonian cycle, traveling on it would force us to alternate sides But 6

7 this means that we would use up all vertices on the left before we touch all the vertices on the right (since m < n) Therefore we would have to visit at least one vertex on the left more than once This is a contradiction and consequently there cannot be a Hamiltonian cycle To sum up, the graph K m,n has a Hamiltonian cycle if and only if m = n Determine whether or not each of the following graphs has a Hamiltonian cycle (a) Solution: I will refer to each edge by its endpoints, for instance (1,), or (7,10) Suppose that the graph has a Hamiltonian cycle Every vertex must be visited exactly once, and therefore each vertex must have degree exactly in the cycle This means, in particular, that the edges (9,10) and (7,10) must be in the cycle But this creates a sub-cycle (7,9,10,7) and we saw in class that this is not allowed Therefore, the graph cannot have a Hamiltonian cycle (b) Solution: We will reason in a way similar to that which we just did Suppose that the graph has a Hamiltonian cycle There are two edges from to 4 If both of them were in the cycle, then they would create a sub-cycle (,4,) Therefore, only one can be in the Hamiltonian cycle This means that (1,) and exactly one of the edges from to 4 must be in the cycle Next consider the vertex 7 If both edges 7

8 (1,7) and (7,4) were in the Hamiltonian cycle, we would have a subcycle (1,,4,7,1) Therefore at most one of these edges may be in the Hamiltonian cycle This means that the edge (7,6) must be in the cycle At the vertex 5, we may use a similar argument: if both edges (1,5) and (5,4) were in the Hamiltonian cycle, then we would have a sub-cycle (1,,4,5,1) Therefore, the edge (6,5) must be in the cycle Last, consider the vertex 3 If both edges (1,3) and (3,4) were in the Hamiltonian cycle, then we would have a sub-cycle (1,,3,4,1) Therefore the edge (3,6) must be in the cycle To recap, the above reasoning gives that if there is a Hamiltonian cycle, then the edges (6,7), (3,6), and (5,6) must be in it But this would give 3 edges coming out of 6, a contradiction Therefore, the graph does not have a Hamiltonian cycle (c) Solution: This graph has a Hamiltonian cycle: (1,, 3, 4, 10, 5, 6, 9, 8, 7, 1) 3 For each of (a), (b), (c), draw a graph with the properties listed, if one such exists (a) The graph is simple and has six vertices which have degrees 1,,3,4,5,5 Solution: Impossible Each vertex which has degree 5 must be connected to all other vertices (recall that since the graph is simple, no vertex may be connected to itself) But this means that all vertices must be connected to at least others This is impossible if one of the vertices is to have degree 1 (b) The graph has four edges and four vertices which have degrees 1,,3,4 Solution: Impossible Let D be the sum of the degrees of all vertices in the graph We know that D = = 10 8

9 On the other hand, notice that any edge in the graph necessarily contributes exactly to the sum of all degrees in the graph This means D = E, where E is the total number of edges in the graph This is impossible, since we are to have E = 4 and so D = 10 8 = E (c) The graph has 5 vertices, is bipartite, has no Euler cycle, but has a Hamiltonian cycle Solution: Impossible We saw earlier that the graph K m,n will have a Hamiltonian cycle if and only if m = n This graph is not necessarily K m,n for some values of m and n, but the same reasoning applies here Since there is no way to split 5 vertices into two equal groups, there cannot be a Hamiltonian cycle 9

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