Algorithmic Discrete Mathematics 4. Exercise Sheet

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1 Algorithmic Dicrete Mathematic. Exercie Sheet Department of Mathematic SS 0 PD Dr. Ulf Lorenz 0. and. May 0 Dipl.-Math. David Meffert Verion of May, 0 Groupwork Exercie G (Shortet path I) (a) Calculate the hortet path from to all other vertice by uing the Dijtra algorithm. Determine the hortet path tree. (b) I the hortet path tree unique? (c) Now change the weight of the edge (, ) to. Show that the Dijktra algorithm doe not work in thi cae. (a) The olution i given by the followig picture. The number next to vertice are the ditance to the tarting vertex and croing out a number mean that there ha been an update. The number in quare indicate the final ditance (hortet ditance). The hortet path tree i indicated by the thick line. The order of viiting the vertice i,,,,,, or,,,,,. 0 7 (b) No the hortet path tree i not unique. By ubtituting the edge (, ) with the edge (, ) one get another poible hortet path tree. (c) The Dijktra algorithm will give the ame olution a in part (a) although the path (,,,, ) (dit ) ha a horter ditance than the path (,, ) (dit ). Thi i what happen: After viiting the vertice,, the algorithm will look for the hortet path to a not yet viited vertex. Thi will be vertex with ditance. After viiting vertex the algorithm will not make an update on vertex becaue it wa already viited and for thi reaon not find the hortet path (ditance ). 7

2 Exercie G (Shortet path II) Show that the problem finding a hortet odd cycle in a imple digraph G = (V, E) with nonnegative weight c e on the edge can be olved by uing a hortet path algorithm. We look at the graph G = (V, E) with weight c e for e E and contruct another copy of it vertex et to get the graph G = (V, E ). The et V i definde by V := V {0} V {}. We connect two vertice (v, σ ) and (v, σ ) by a directed edge iff (σ = σ ) and (v, v ) E. Thi mean v and v are in different copie of V and they are adjacent in G. For the weight of thee edge we chooe the original weight of the correponding edge in E. The problem to find a hortet odd cycle in the orginal graph G can now be olved by looking at hortet path from (v, 0) to (v, ) in the graph G for every vertex v V. All thee path in G yield odd cycle in the original graph G by contruction of G. Thi hold becaue we have an odd number of changing between the two component V {0} and V {}. Hence by taking the hortet path we found we get the hortet odd cycle. Exercie G (Croing the river) A man ha to tranport a wolf, a goat and a cabbage to the other ide of a river. He ha one boat to do thi but it i o mall that he can only take one of the three thing with him each time. I it poible to bring all three thing to the other ide of the river afely? Notice that the wolf and the goat or goat and the cabbage mut never be on the ame ide of the river without urveillance of the man. At leat the wolf i no vegeterian and doe not like to eat cabbage. We contruct a graph for thi problem where every legal tate i repreented by a vertex in the graph and look for a hortet path in that graph. So let S = {M, W, G, C} where M, W, G, C repreent the man, the wolf, the goat and the cabbage. A tate for thi problem i a pair (X, Y ) where {X, Y } i a partition of S. The element in X are till on the tarting ide of the river and the element in Y are already on the other ide. A tate i a legal tate if W, G X, Y M X, Y and G, C X, Y M X, Y. Thi repreent that you may not leave alone the cabbage and the wolf or the goat and the cabbage without urveillance. Now contruct a graph G = (V, E) which ha a vertex for every legal tate of thi problem. We have the directed edge (v, v ) E if we can get from the tate v = (X, Y ) to the tate v = (X, Y ) by jut one boat trip acro the river. All edge get the weight c e =. Now the problem can be olved by calculating the hortet path from the tate (vertex) = (S, ) to the tate (vertex) t = (, S). We get 7 a the olution for the hortet path. Homework Exercie H0 (Algorithm) (0 point) We are looking for an algorithm which get an undirected, connected graph G given a an adjacency lit and determine whether the graph i bipartite or not in runtime O( V + E ). (a) Contruct uch an algorithm. (b) Prove it correctne and analye it runtime. (a) We modify the depth firt earch algorithm uch that it aign every vertex to one of the et V or V. The tarting vertex will be aigned to the et V. One of the neighbour, which i not yet viited, will be aigned to the et V and o on. If the algorithm get to a neighbour of the actual vertex which ha already been viited it check whether the actual vertex and the neighbour are in the ame et. If thi i the cae it return fale otherwie nothing happen. The algorithm keep on doing thi till all vertice are viited once.

3 Algorithm DFS(G) for each vertex u V do do color[u] white G-bipartit true for each vertex u V do if color[u]= white then BIP-et[u] V if DFS-Viit(u)=fale then G-bipartit fale return G-bipartit Algorithm DFS-Viit(u) color[u] GRAY for each vertex v Adj[u] do if color[v] = white then if BIP-et[u] = V then BIP-et[v]=V ele BIP-et[v]=V DFS-Viit(v) ele if BIP-et[u] = BIP-et[v] then return fale return true (b) Now be prove the correctne of the algorithm and claim that G i bipartite iff the algorithm return true. firt cae: G i bipartite. We have to how that the algorithm return true. The firt vertex will be aigned to V. Every neighbour of that vertex will be aigned to V (ee line - in DFS-Viit). So two adjacent vertice will never be aigned to ame et. In the end the algorithm will correctly return true, ince G i bipartite and aigning all neighbour to other et a the vertex itelf i neceary for getting a bipartition. econd cae: G i not bipartite. We have to how that the algorithm doe not return true. Since G i not bipartite we know that it contain a cycle C of odd length. Let ay C conit of n + vertice with n being even. The algorithm will aign n vertice correctly by alway aigning the adjacent vertex in the cycle to the oppoite et. When it viit the n + vertex in the cycle it will realize that one adjacent vertex wa already viited and wa aigned to the ame et a itelf. Therefore it will return fale a deired. The runtime of the algorithm i the ame a for the DFS algorithm becaue we only made modification adding contant running time in each relevant part of the algorithm (for each part). So the runtime i O( V + E ).

4 Exercie H (Graph) Given the graph G in Figure. (0 point) Figure : a graph (a) I the graph G eulerian? Jutify your anwer. (b) Doe the graph G contain a Hamiltonian cycle? Thi a cycle that contain each vertex exactly once. Jutify your anwer. (c) Determine the adjacency matrix and adjacency lit of the graph G. (a) The graph i not eulerian ince it contain vertice not having an even degree. (b) The graph contain no Hamiltonian cycle ince the vertex number ha to be viiited at leat two time to viit the vertex. (c) The adjacency matrix look a follow: A = For the adjacency lit we have the following information in the given order. Number of vertice Number of edge For each vertex the degree and the incident edge So the lit look a follow: 0 ; ;,,, ;,, ;,7, ;,9 ;,,9 ;,0 ;,0

5 ;, ;7, Exercie H (Shortet path) (0 point) (a) Let G = (V, E) be a directed graph. i. Aume G ha no negative cycle and ( = i 0, i,..., i k = t) i a hortet imple path from to t. Then every ubpath from to i j with j {0,..., k} i a hortet imple path from to i j. ii. Show that thi tatement i wrong for general graph which may contain negative cycle. (b) Given a directed graph G = (V, E) we want to calculate the hortet path from the tart to the detination. The problem i that tart conit of n vertice and the detination conit of m vertice. So we are looking for the hortet path poibly tarting at any i with i {,..., n} and poibly going to any detination t j with j {,..., m}. How can thi be done efficiently? Jutify your anwer. Hint: (a) i. Aume there i a horter imple path P from to i j. By connecting the path P and (i j, i j+,..., t) we get a path from to t which i horter than the original hortet imple path from to t. Thi path doe not need to be imple. By eliminating all cycle in thi path, which make it ever horter becaue we aumed no negative cycle, we get a horter imple path from to t than the original one. Thi i a contradiction ii. Here i an example how the tatement i not true for negative cycle. The hortet imple path from to t i (, v, t) with ditance. Taking the ubpath to (, v ) with ditance we find a horter imple path from to v namely (, t, v ) with ditance. t v (b) We modify the given graph G = (V, E) to a graph G = (V, E ). The vertex et i definded by V := V {, t} and the edge et i defined by E := E {(, i ) : i n} {(t j, t) : j m}. All the weight for the additional edge are et to zero. Notice that any weight would do the job. It i jut important that they all get the ame weight. Now we determine the hortet path from to t in the graph G which we will be (, i,..., t j, t) for ome i {,..., n} and j {,..., m}. The ubpath ( i,..., t j ) i the deired hortet path we wanted to find. If there would be a horter path than thi one we would have found it by uing the hortet path algorithm on the graph G, ince the edge (, i ) and (t j, t) will not add any weight to the path (or add the ame amount for every path from to t if we have choen nonzero weight).