811312A Data Structures and Algorithms, , Exercise 6, solution

Transcription

1 811312A Data Structures and Algorithms, , Exercise 6, solution The topics of this exercise are breadth-first and depth-first search in graphs. Cormen, Ch. 22 Task 6.1 Find using breadth-first search, the shortest path through the following maze. One enters the maze through the upper left corner and exits through the lower right corner. Construct a graph (node in each conjunction, denoted by letters) and run the algorithm in detail. a b e c h i d k j f g Solution. We get a graph, whose adjacency list representation is a b h b a c c b d e d c e c f f e g g f h a i i h j j i k k j ow execute breadth-first search starting from the vertex a. Below is shown, how the arrays color, p and d are handled. In the array color, G means gray and B black. color G d 0 p IL

2 The neighbors of a are b and h which are colored gray, the arrays p and d are updated, and the vertices are inserted into queue. Thereafter a is colored black. Queue: b,h. a b c d e f g H i j k color B G G d p IL a a ow take b from the queue and consider its neighbors a and c. The vertex a is white no more, hence it requires no operations. Queue: h,c color B B G G d p IL a b a Vertex h is taken from queue and its neighbors a and i are dealt with. Queue: c,i. color B B G B G d p IL a b a h Vertex c is taken from queue and its neighbors b,d and e are dealt with. Queue: i,d,e. color B B B G G B G d p IL a b c c a h Vertex i is taken from queue and its neighbors h and j are dealt with. Queue: d,e,j. color B B B G G B B G d p IL a b c c a h i Vertex d is taken from queue and its neighbor c (not white) is dealt with. Then vertex e is taken from queue and its neighbors c and f are dealt with. Queue: j,f. color B B B B B G B B G d p IL a b c c e a h i

3 Vertex j is taken from queue and its neighbors i and k are dealt with. Queue: f,k. I j k color B B B B B G B B B G d p IL a b c c e a h i j Vertex f is taken from queue and its neighbors e and g are dealt with. Queue: k,g. color B B B B B B G B B B G d p IL a b c c e f a h i j Finally the vertices k and g are taken from queue and colored black. The algorithm terminates as the queue gets empty. color B B B B B B B B B B B d p IL a b c c e f a h i j ow we can read the path from a to g from the predecessor table p. Starting from g we get the path in reverse order: g,p[g]=f,p[f]=e,p[e]=c,p[c]=b,p[b]=a. Hence the actual path is a,b,c,e,f,g. Task 6.2 As we recall, the algorithm for depth-first search is as follows: DFS(G) 1.for each u in V 2. color[u] = WHITE 3. p[u] = IL 4. time = 0 DFS_VISIT(u) 1. color[u] = GRAY 2. time = time+1 3. d[u] = time 4. for each v in Adj[u] 5.for each u in V 5. if color[v]==white 6. if color[u]==white 6. p[v] = u 7. DFS_VISIT(u) 7. DFS_VISIT(v) 8.return 8. color[u] = BLACK 9. time = time+1 10.f[u] = time 11.return

4 Show how depth-first search works on the following graph. a b c d e f h g Show the discovery and finishing times for each vertex, and show the classification of each edge (as tree edge, forward edge, back edge or cross edge Check also, whether one can detect a cycle in the graph during the execution of the algorithm. Solution. The adjacency representation of G is a d e h b c g c g d f e g f h g h d Start the depth-first search from the vertex a, whose neighbor d is considered first. The stack is d, a. The tables contain color G G p IL a d 1 2 f

5 Then move to d s neighbor f, and its neighbor h. The stack is now h,f,d,a and the tables contain color G G G G p IL a d f d f The vertex h has no more white neighbors; instead we notice that it has a grey neighbor d, and we detect a cycle. The vertex h is finished and we return to vertex f that has no white neighbors. Return to d, whose neighbors have been detected already. ow the stack only contains a: color G B B B p IL a d f d f We are now handling a, who still has a white neighbor e, that we take into consideration. Stack is a, e. color G B G B B p IL a a d f d f Vertex e has a white neighbor g that is pushed into the stack: color G B G B G B p IL a a d e f d f The vertices g, e, and a do not have any white neighbors at this point. Hence the stack becomes empty and the tables are updated color B B B B B B p IL a a d e f d f

6 ext we move to white vertex b, whose only white neighbor is c. These are handled and the algorithm terminates: color B B B B B B B B p IL IL b a a d e f d f The depth-first forest consists of the trees a b d e c f g h ow we can classify the edges of the graph. The edges (a,d),(d,f),(f,h),(a,e),(e,g) and (b,c) are tree edges, because new vertices are found inspecting these edges. They are also shown in the trees above. The edge (a,h) is a forward edge, because a is a predecessor of h and the edge is not a tree edge. The edge (h,d) is a back edge, because h is a successor of d. The edges (b,g) and (c,g) are cross edges because they do not belong to tree edges, forward edges or back edges.

7 Task 6.3 Implement (in C or Python) a modified breadth-first search algorithm such that it detects whether the graph is bipartite or not. An undirected graph is called bipartite if its nodes can be colored red and blue in such a way that the node in one end of each edge is red and the other is blue. Algorithm should also do the coloring if it is possible. Apply your program to the following graphs Solution. When we apply breadth-first search on the graph, starting from vertex s, it can be colored red. ow the neighbors of s, which the algorithm finds first, must be colored blue. The neighbors of these vertices, which have not yet been detected, have distance 2 from s and must again be colored red. Continuing like this, we color vertices with even distance from s red, and the vertices with odd distance from s blue. aturally, it might happen that two vertices with even distance from s are neighbors, or that two vertices with odd distance are neighbors. In these cases, coloring is not possible. Let us modify the algorithm of breadth-first search such that, when a vertex u is taken from the queue, we check if some of its neighbors are already colored with the same color that we assign to u. If this happens, coloring is not possible; otherwise we color the neighbors. Add table redblue, containing the colors of vertices (RED/BLUE). Moreover, every vertex must be handled in order to check all the components of the graph.

8 Following is the modified algorithm: Input: Graph G=(V,E). Use adjacency list representation. Output: Otherwise like breadth-first search, except that all vertices are considered. The table redblue contains coloring if it is possible. Returns TRUE if coloring possible, FALSE otherwise. BIPARTITE(G) 1. for each u in V 2. color[u] = WHITE 3. d[u] = IF 4. p[u] = IL 5. redblue[u]=x // Undefined color 6. bipart = TRUE 7. for each s in V 8. if color[s]==white 9. d[s] = color[s] = GRAY 11. redblue[s] = RED 12. QUEUE Q = EMPTY // Empty queue 13. EQUEUE(Q,s) 14. while Q!= EMPTY() // Repeat until queue empty 15. u = DEQUEUE(Q) 16. for each v in Adj[u] 17. if color[v] == WHITE 18. color[v] = GRAY 19. d[v] = d[u] if d[v] mod 2 == reblue[v] = RED 22. else 23. reblue[v] = BLUE 24. p[v] = u 25. EQUEUE(Q,v) 26. else 27. if redblue[v] == redblue[u] 28. bipart = FALSE 29. color[u] = BLACK 30. return bipart Compared to breadth-first search, only one table and assignments to it is added. However, algorithm checks every vertex. Hence, the complexity class is here ( M ). The solution is implemented in the link below. The function that performs the coloring is bicoloring. In the C program, the table redblue that will contain the coloring is given as a parameter to the function, which will return 0 if the coloring is not possible. In the Python implementation, the table is returned from the function and the return value is one, if the coloring cannot be done. When applied to the example graphs, one notices that the first graph can be colored and the second cannot.