AMS 550.472/672: Graph Theory Homework Problems - Week X
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AMS 550.472/672: Graph Theory Homework Problems - Week X Problems to be handed in on Wednesday, April 5: 3,6,11. You may use results from class or previous HWs without proof. 1. Show that the block-cutvertex graph of any graph is a forest. 2. Suppose G be a graph with no isolated vertices. Show that if G has no even cycles, then every block of G is an edge or an odd cycle. Solution: Since G has no isolated vertices, every block B is either an edge, or is 2-connected. In the latter case, let P0 [ P1 [ ::: [ Pk be an ear decomposition of B. Since G has no even cycles, P0 must be odd. Now, if the ear P1 is not empty, then we have a path between two vertices in P0 that shares only the endpoints, say x; y, with P0. Consider the two paths in P0 between x and y. Since P0 has odd length, one of these paths must have odd length and one of them must have even length. Depending on the parity of the length of the path P1, this would create an even cycle in B. This is a contradiction. Thus, B cannot have any ears - it only consists of the odd cycle P0. 3. Let k ≥ 2. Show in a k-connected graph any k vertices lie on a common cycle. [Hint: Induction] Solution: By induction on k. If k = 2, then the result follow from the characterization of 2- connected graphs. For the induction step, consider any k vertices x1; : : : ; xk. By the induction hypothesis, since G is also k − 1-connected, there is a cycle C that contains x1; : : : ; xk−1. If these are the only vertices in C, i.e., C has length k − 1, then by the Fan Lemma (Theorem 4.2.23 in the textbook), we have k − 1 vertex disjoint paths from xk to C and appending the paths from xk to x1 and x2, we get a cycle containing all x1; : : : ; xk. Otherwise, there are at least k vertices in C, and thus again by the Fan Lemma, one can construct k vertex disjoint paths to C. By the pigeonhole principle, two of these paths have endpoints in C that are contained on the path from xi and xi+1 (inclusive) for some i = 1; : : : ; k − 2, or are contained on the path from xk−1 to x1. In either case, we can traverse the cycle C and make a detour to xk through these two vertices. 4. Recall that a graph is said to be even if every vertex has even degree. Show that a graph is even if and only if each block is even. Solution: (() If every block is even, then since the degree of any vertex is the sum of its degrees in each block (which is counted as 0 if it does not belong to a block), every vertex in the graph has even degree. Thus, the graph is even. ()) We prove this by induction on the number of blocks. Since the block-cutvertex graph of any graph is a forest, these exists a block B which shares at most one vertex with any other block (this would correspond to a degree 1 vertex in the block-cutvertex graph). Remove all the vertices in this block that is not shared with any other block and call this graph G0. We now have two cases: Case 1: There is no vertex in B that is shared by any other block. This means B is a connected component of the graph. Since the graph is even, this component must be even and hence B is even. By the induction hypothesis applied to G0, the remaining blocks also are even. Case 2: There is a vertex v that is shared by B and at least one other block. Since the graph is even, all vertices in B except v have even degree, since their degree in B is the same as their degree in the graph. Therefore, since any graph has an even number of odd degree vertices, 1 v must have even degree in B (otherwise B would have only 1 odd degree vertex). By the induction hypothesis applied to G0, all other blocks are also even. Thus, we are done. 5. Show that a graph is bipartite if and only if each block is bipartite. Solution: ()) If the graph is bipartite, then the same bipartition restricted to the blocks show that the blocks are bipartite. (( We show that there are no odd cycles. Consider any cycle C in the graph. Since C is two-connected, it must be contained in a block. Since this block is bipartite by hypothesis, the cycles must be even. Therefore, the graph has no odd cycles and is therefore bipartite. 6. Let G1;:::;Gk be the blocks of a simple graph G. Show that χ(G) = maxi=1;:::;k χ(Gi). Solution: By induction on the number k of blocks. The base case with k = 1 is trivial. For the induction step, consider the block-cutpoint graph of G. Since this is a tree, there is a leaf which corresponds to a block; without loss of generality let this be block Gk. Color the 0 graph G formed by G1 [ ::: [ Gk−1 using maxi=1;:::;k−1 χ(Gi) (by the induction hypothesis), 0 and color Gk using χ(Gk) colors. Then, if the colors of the common vertex of G and Gk in the two colorings are different, then just switch two of the colors in the coloring in Gk so that the common vertex gets the same color as in G0. Then we are done. 7. Suppose that G has no even cycles. Show that χ(G) ≤ 3. [Hint: Use Problem 2] Solution: By Problem 2 above, each block is an edge or an odd cycle. Thus, the chromatic number of each block is at most 3. We then have the result, from Problem 6. 8. Suppose every edge in a graph G appears in at most one cycle. Show that χ(G) ≤ 3. Solution: Each block is either an edge or a cycle; otherwise, if there is block which contains a cycle C and an edge e not on this cycle, we can take any edge f from the cycle and by the characterization of 2-connected graphs obtain a cycle C0 through e and f. C0 has to be different from C because e 62 C. Thus, f is on two different cycles C and C0 contradicting the hypothesis. The result then follows from Problem 6, because the chromatic number of any cycle is at most 3. 9. Construct a graph G that is not a complete graph, nor an odd cycle, but has a vertex ordering according to which the greedy coloring algorithm uses ∆(G) + 1 colors. [Note that Brook's theorem tells us that χ(G) ≤ ∆(G).] Solution: Consider the path on 4 vertices: x1; x2; x3; x4. Greedy coloring with the vertex ordering x1; x4; x2; x3 uses 3 colors and the maximum degree is 2. 10. For each k ≥ 2, construct a tree Tk with maximum degree k such that Tk has a vertex ordering with respect to which the greedy coloring gives k + 1 colors. Solution: We give a recursive construction. For k = 2, use the path from Problem 5 above along with its vertex ordering. For k ≥ 3, create Tk by taking k copies of Tk−1. For each copy, there is a vertex ordering which uses k colors. Let vi be the vertex in copy i which receives color i. Add a new vertex v and connect it to each vi. The vertex ordering for Tk is obtained by concatenating the vertex orderings for the k copies of Tk−1 and then v as the final vertex. This forces greedy to use color k + 1 on vertex v. 11. For every n ≥ 2, find a bipartite graph with 2n vertices such that there exists a vertex ordering relative to which the greedy coloring algorithm uses n colors rather than 2 colors. Solution: The bipartite graph be with V1 = fv1; v2; : : : ; vng and V2 = fu1; u2; : : : ; ung. vi is connected to uj if and only if i 6= j. Then the ordering v1; u1; v2; u2; : : : ; vi; ui; : : : ; vn; un uses n colors. 2 12. Show that every graph G has a vertex ordering according to which the greedy coloring algo- rithm uses χ(G) colors. Solution: Consider any coloring with χ(G) colors. Order the vertices, so that all vertices of color 1 appear first, followed by vertices of color 2, then color 3 and so forth. The greedy coloring then produces an optimal coloring. 13. A graph with χ(G) = k is called k-chromatic. A k-chromatic graph is called critically k- chromatic if for every vertex v, G n fvg is k − 1-colorable. (i) Show that every k-chromatic graph has an induced subgraph that is critically k-chromatic. Solution: Iteratively remove vertices that doesn't decrease the chromatic number. (ii) Show that every critically k-chromatic graph has minimum degree at least k−1. Solution: Suppose to the contrary that there is a vertex v with deg(v) ≤ k −2. Since G is critically k-chromatic, then G n fvg is k − 1 colorable. But since v has at most k − 2 neighbors, we can color it with one of the remaining colors, and get a k − 1 coloring of G which contradicts the fact that G is k-chromatic.