Discrete Mathematics, Spring 2004 Homework 8 Sample Solutions

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1 Discrete Mathematics, Spring 4 Homework 8 Sample Solutions 6.4 #. Find the length of a shortest path and a shortest path between the vertices h and d in the following graph: b c d a f 4 6 e g h i j z Solution. The shortest path is h, f, c, d, with total length. To see this, we use Dijkstra s algorithm, modified as shown in Example in the text. Let V (j) denote the vertex chosen in the beginning of the jth iteration of the while loop, let L(j) denote the weight of the least weight path from h to V (j), and let P(j) denote the vertex from which V (j) was labeled. The table below indicates the order of selection; the last column of the table indicates the vertices which have not yet been made permanent at the end of the jth iteration (but have weight < ). j V (j) L(j) P(j) Temporary (V, L, P) h (a,4,h), (e,7,h), (f,,h), (i,,h) i h (a,4,h), (e,7,h), (f,,h), (j,8,i) a 4 h (e,7,h), (f,,h), (j,8,i), (b,7,a) 4 f h (e,7,h), (j,8,i), (b,7,a), (c,7,f), (g,9,f) b 7 a (e,7,h), (j,8,i), (c,7,f), (g,9,f) 6 c 7 f (e,7,h), (j,8,i), (g,9,f), (d,,c) 7 e 7 h (j,8,i), (g,9,f), (d,,c) 8 j 8 i (g,9,f), (d,,c), (z,,j) 9 g 9 f (d,,c), (z,,j) d c done

2 6.4 #8. Write an algorithm that finds the lengths of the shortest paths between all vertex pairs in a simple, connected, weighted graph having n vertices in time O(n ). Solution. Probably the easiest approach is to take Dijkstra s algorithm as shown in the text, and then make a few minor modifications. The input to our algorithm will now simply be the n n matrix w of edge weights of a simple connected graph G whose vertices are labeled to n. (For convenience, we assume that w(i,j) = if i and j are not adjacent.) The output will be the n n matrix L, where L(i,j) is the length of the shortest path from i to j. A possible solution is given below (I omit the begin and end statements to save space).. procedure dijkstra (w, n). for i = to n do. L(i,i) := 4. for all j i do. L(i,j) := 6. T := {,...,n} 7. while T is nonempty do 8. choose y T with minimum L(i, y) 9. T := T {y}. for j = to n do. L(i,j) := min {L(i,j), L(i,y) + w(y,j)}. end dijkstra Note that since the original Dijkstra s algorithm is O(n ), this algorithm is evidently O(n ). 6.4 #. True or false? When a connected, weighted graph and vertices a and z are input to the following algorithm, it returns the length of a shortest path from a to z. If the algorithm is correct, prove it; otherwise, give an example of a connected, weighted graph and vertices

3 a and z for which it fails.. procedure algor (w, a, z). length :=. v := a 4. T := set of all vertices. while (v z) do 6. begin 7. T := T {v} 8. choose x T with minimum w(v, x) 9. length := length + w(v,x). v := x. end. return(length). end algor Solution. Here is a sample graph for which the above algorithm fails to produce the length of the shortest path from a to z (it will traverse the path a,b,z instead): a b c z 6. #. Write the incidence matrix of the complete bipartite graph K,. Solution. Below is a possible matrix (where we assume that V = {v,v 4 } and V = {v,v,v }

4 are the disjoint sets of vertices in the natural partition): v v v v 4 v e e e e 4 e e 6 6. #7. Draw the graph represented by the 7 7 adjacency matrix whose ijth entry is if i + divides j + or j + divides i +, i j, whose ijth entry is if i = j, and whose ijth entry is otherwise. Solution. Label the vertices from to 7. From the conditions given above we conclude that there is a loop at each vertex, and there is a single edge incident on each of the following pairs of vertices: {,}, {,}, {,7}, {,}, {,7}. Therefore the graph below has the given adjacency matrix:

5 6. #. Suppose that a graph has an adjacency matrix of the form ( A ) A = where all entries of the submatrices A and A are zero. What must the graph look like? A Solution. First observe that if A has k rows and l columns, then A has l rows and k columns (since the matrix A is symmetric). But then we must have k + l = n since the number of rows in A and A clearly sum to the number of rows in A. Now let (v,...,v n ) be the ordering of vertices in the graph corresponding to the numbering of the rows and columns of A; then it is clear that for any pair of vertices (v i,v j ) with i k and k + j n there is no edge joining them. But then the graph must be disconnected, for if not we could find a path (w,e,w,...,e m,w m ) from v to v n, for example, and at least one edge on this path would have to be incident on some vertices v i and v j with i k and k + j n, which gives us a contradiction. 6. #9. Let A be the adjacency matrix of the complete graph K. Let d n be the common value of the diagonal elements of A n and let a n be the common value of the off-diagonal elements of A n. Show that d n+ = 4a n, a n+ = d n + a n, a n+ = a n + 4a n. Solution. The third equation follows immediately by substituting the first equation d n = 4a n in the second, so we merely need to prove the first two formulas. There are two ways in which one can approach this problem. From a graph theory perspective, we observe that d n represents the number of paths of length n from any vertex v K to itself, while a n represents the number of paths of length n from any vertex v to any vertex w v. So let v,...,v be the vertices of K. The number of paths of length n + from v to itself is precisely the number of paths of length n from v to any of v,v,v 4,v. Now there are a n paths of length n from v to v, and similarly a n paths of length n from v to v, etc. Hence d n+ = 4a n. Any path of length n + from v to v, on the other hand, is either a path of length n from v to itself followed by the edge (v,v ), or is a path of length n from v to v j, (j =, 4, or ), followed by the edge (v j,v ). Hence a n+ = d n + a n, by the Addition Principle. We can view this problem as simply one of matrix multiplication as well. Since d m and a m represent the diagonal and off-diagonal elements of the matrix A m, from the matrix

6 formula A n+ = A n A we have d n+ a n+ a n+ a n+ a n+ d n a n a n a n a n a n+ d n+ a n+ a n+ a n+ a n+ a n+ d n+ a n+ a n+ a n+ a n+ a n+ d n+ a n+ = a n d n a n a n a n a n a n d n a n a n a n a n a n d n a n. a n+ a n+ a n+ a n+ d n+ a n a n a n a n d n Using the definition of matrix multiplication on the right hand side, we easily obtain the desired two formulas (it suffices to find the (,)-th and (,)-th entries of the matrix product). 7. #. Construct an optimal Huffman code for the set of letters in the table. Letter Frequency I 7. U. B. S 7. C. H. M. P. Solution. We proceed by combining the smallest two frequencies until only two remain, and then building the tree by expanding the frequencies backwards:.,.,., 7.,.,.,., 7..,., 7.,.,.,., 7. 7.,.,.,.,., 7.., 7.,.,., 7..,., 7., 7. 7., 7., 4 4, We illustrate the tree being constructed step-by-step on the following page. The Huffman code associated to the final tree structure is given by Note: this code is certainly not unique. I, U, B, S, C, H, M, P. 6

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