Here are solutions to some selected problems for Math 5707 taught by Professor Dennis White.
If 4 divides n-1, divide n-1 vertices into 4 groups as above and draw edges as above. Then add one new vertex v, and draw all vA and vD edges, and no vB or vC edges.
For (c), suppose f is an automorphism and let u and v be two vertices. Let w=f^{-1}(u) and x=f^{-1}(v), so that f(w)=u and f(x)=v. Then wx is an edge iff f(w)f(x)=uv is an edge.
For (a), suppose f and g are two automorphisms and let u and v be two vertices. Then uv is an edge iff f(u)f(v) is an edge iff g(f(u))g(f(v)) is an edge.
If G is connected and has a cut vertex v of degree 1, then the removal of v will create at most 1 component. That is, G-v is still connected, which means v is not a cut vertex. For G not connected, apply this to each connected component of G.
Suppose G has two different bipartitions, (X, Y) and (A, B), with A not equal to either X or Y. Let U = (A intersect X) union (B intersect Y) and V = (A intersect Y) union (B intersect X). Clearly, U and V partition the vertex set of G. Furthermore, since A is neither X nor Y, B cannot be either X or Y, so U and V are both nonempty. Finally, there is no edge between U and V: if there were, then that edge would go between (A intersect X) and (A intersect Y), between (A intersect X) and (B intersect X), between (B intersect Y) and (A intersect Y), or between (B intersect Y) and (B intersect X). These four possibilities violate independence of the sets A, X, Y and B, respectively.
Let a_k=1. Swap a_k with a_{k-1}, then a_{k-1} with a_{k-2}, etc., until the 1 is in the first position. Now repeat for the 2, then the 3, etc.
u=AxByC and
v=AyBxC,
where A, B and C are partial permutations. WLOG, we assume x < y. It is clear any pair (w, z), w, z both not equal to x or y, is an inversion in u if and only if it is an inversion in v. Therefore, we need only consider when one of w or z (say z) is x or y. There are four cases.
If w is in A or C, then (w,x) (resp. (w,y)) will be an inversion in u if and only if it is an inversion in v.
If w is in B and w < x or w> y, then (w,x) (resp. (w,y)) will be an inversion in u if and only if (w,y) (resp. (w,x)) is an inversion in v.
If w is in b and x < w < y, then (w,x) and (w,y) are both inversions in v while neither are inversions in u.
Finally, (x,y) is an inversion in v but not u.
In the first two cases, inversions appear or don't appear in both permutations. In the third case, v gains 2 inversions for each such w. In the last case, v gains one inversion. Therefore, v gains an odd number of inversions, thus changing the parity of the number of inversions. That is, if u were in X, then v must be in Y and vice versa.
Let R be the path of length t from x to y. Except for x and y, R cannot have any vertex in common with either P or Q. For if it did, then there would be a shorter path between a vertex in P and a vertex in Q.
Therefore ARB is a path of length >= m/2 + m/2 + t > m, which contradicts the maximality of m.
To do the construction, construct a graph with m vertices (m odd) with degree sequence (2k+1, 2k+1, ..., 2k+1, 2k). This is clearly possible by Theorem 1.3.31 for m sufficiently large. Duplicate this graph, then draw an edge e between the two vertices of degree 2k. The resulting graph will be 2k+1 regular and e will be a cut-edge.
(n-1-d_{n}, n-1-d_{n-1}, ..., n-1-d_1) = (d_1, d_2, ..., d_{n})
Since n is odd, there is a middle value k in this sequence, so n-1-d_{k}=d_{k}, or d_{k}=(n-1)/2.
There are (n choose 2) choices for non-leaves. Then there are 2 choices for each leaf: either attach it to the smaller non-leaf or to the larger non-leaf. That gives 2^{n-2} possible choices. However, two of these are not allowed: all attached to the smaller non-leaf or all attached to the larger non-leaf. (These cases are not allowed since then one of the non-leaves would become a leaf.) So the total is (n choose 2) (2^{n-2}-2). In the Prufer code, exactly two labels appear. (n choose 2) picks the two labels, then each could appear in each position, giving 2^{n-2}. Again, the case where all the labels are the same is not allowed.
Any spanning tree of G_n must use two of the three edges (e_n, f_n, g_n) or use all three of them. Those spanning trees which use two of them are in 1-1 correspondence to spanning trees of G_{n-1}, giving 3A_{n-1}. Those that use all three are in 1-1 corrspondence to spanning trees of G_{n-1} which use e_{n-1}: simply swap the edge e_{n-1} with the three edges (e_n, f_n, g_n). The spanning trees of G_{n-1} which do not use e_{n-1} must use f_{n-1} and g_{n-1}, and are therefore counted by A_{n-2}.
The sets {e_1,...,e_{k-1}} and {e_1',...,e_{k-1}'} must be different since otherwise Kruskal would always choose e_k' at the k-th step. Let e_i be the first edge in T not in T'. Then T'+e_i has a cycle and all the edge costs on that cycle are less than or equal to the cost of e_i, since otherwise a costlier edge could be removed from the cycle, giving a cheaper spanning tree. Since T and T' agree in all edges cheaper than e_i, there must be one edge on that cycle not in T which has cost equal to e_i. Therefore, it has cost c_i. We may assume it is e_i'. Then T'+e_i - e_i' is a minimal spanning tree whose edges agree with T one step further. Repeating this, eventually the edges of T and T' agree up to e_{k-1} and e_{k-1}'.
If there is no perfect matching, let M be a maximum matching. Player 1 picks an unmatched vertex v0. Every vertex adjacent to v0 is M-saturated. Otherwise, if w is not M-saturated, then v0-w could be added to M to form a larger matching. Therefore, no matter what vertex Player 2 selects, it is M-saturated and so is paired with another M-saturated vertex.
For the general step, suppose Player 2 has selected an M-saturated vertex v, and vw is in M, with w not picked. Then Player 1 picks w. We show all other vertices adjacent to w are M-saturated. Suppose not, suppose u adjacent to w and u not M-saturated. Then the path from u back to v0, determined by the players' choices up to this point, is augmenting, contradicting maximality of M. Thus, Player 2 is forced to select an M-saturated vertex u which is matched to an unpicked vertex.
Now suppose G is bipartite (X, Y) and the condition holds. In particular, the condition holds for S subset of X, and so there is an X-saturating matching in G, so that |X|<=|Y|. On the other hand, it holds for S subset of Y, so that there is also a Y-saturating matching in G, and |Y|<=|X|. Therefore, |X|=|Y|, and so an X-saturating match is also Y-saturating.
Odd cycles are an obvious infinite class of counterexamples. They satisfy the condition, but there is no perfect match.
Any matching in this matrix corresponds to a permutation of {1,2,...,n}. We want to show that the identity permutation minimizes the matching. Any permutation can be transformed into the identity permutation by a sequence of adjacent swaps (cf. Ex 1.2.17). Therefore, we need only show that if A>=B are two morning durations and C<=D are two afternoon durations, then the pairings (A, C) and (B, D) are better than the pairings (A, D), (B, C). That is, any matching which does not correspond to the identity permutation can be converted to the identity by a series of swaps, each of which improves the matching.
We consider four cases. Suppose A+C, B+D both <= t. Then the cost of ((A, C), (B, D)) is 0, which is clearly at least as good as the cost of ((A, D), (B, C)).
Suppose the cost of A+C, B+D both >= t. Then the cost of ((A, C), (B, D)) is A+B+C+D-2t. The cost of ((A, D), (B, C)) is at least A+D-t, since A+D>=A+C. If B+C>=t, then the cost of ((A, D), (B, C)) is also A+B+C+D-2t. If B+C<t, then we have A+B+C+D-2t=(A+D-t)+(B+C-t)< A+D-t.
Suppose A+C< t < B+D. Then the cost of ((A, C), (B, D)) is B+D-t. However, the cost of ((A, D), (B, C)) is at least A+D-t, which is greater.
Suppose B+D< t < A+C. Then the cost of ((A, C), (B, D)) is A+C-t. However, the cost of ((A, D), (B, C)) is again at least A+D-t, which is greater.
Case 1: M = {(a,b), (c,d)} (b,c) is unstable.
Case 2: M = {(a,c), (b,d)} (a,b) is unstable.
Case 3: M = {(a,d), (b,c)} (a,c) is unstable.
Now suppose r divides k. Induction on k/r. Clearly true if r=k. From Corollary 3.1.13, G has a perfect match. Let M_1 be this match. Remove M_1 from G to form a (k-1)-regular bipartite graph. Now repeat until r perfect matches, M_1, M_2, ..., M_r, have been removed. The graph H that remains is (k-r)-regular and r divides k-r. By induction, H can be decomposed into r-factors. Furthermore, the union of M_1, M_2, ..., M_r is itself an r-factor M. Then M with the decomposition of H is a decomposition of G into r-factors.
For k = 2m + 1, first form H by removing a perfect match from K_{2m+2}, then form H' by adding a new vertex v which is attached to all but one of the vertices in H. Then H' is simple, has 2m+3 vertices, and all degrees are 2m+1 except one which is 2m. Now form G by attaching each of 2m+1 copies of H' to a new vertex v at the vertex of degree 2m. G is then (2m+1)-regular. Furthermore, removing v creates 2m+1 components, each with 2m+3 vertices, i.e., 2m+1 odd components. Therefore, by Tutte's theorem, no 1-factor can exist.
For the second part, take two copies of the graph H' from 3.3.7 above (when m=1) and draw an edge between the two vertices of degree 2. This is a 3-regular graph with a cut-vertex and a 1-factor.
On the other hand, sum the vertex degrees in an odd component H. This sum will be kn(H)-t, where t is the number of edges from H to S. This number must be even and n(H) is odd. Therefore, either k is even and t is even or k is odd and t is odd. Since the removal of k-2 or fewer edges will not disconnect, we know that t>k-2. Thus, in either case, t>=k. Summing over the odd components gives m>=ko(G-S). Therefore there is a 1-factor by Tutte's theorem. Note that the non-empty condition on S is needed to apply this connectivity argument.
The case where S is empty is handled by the fact that G has even order.
Suppose H has a 1-factor. Then the vertices in X are saturated by the corresponding perfect match M. The edges in M saturating the vertices in X thus form an X-saturating match in G.
(b) Since the subgraph induced by Y* is a clique, odd components can arise from the removal of T in only two ways. First, if all the neighbors of a vertex x in X are in T, then x becomes isolated. Let B be the set of these isolated vertices. Second, the remaining component, whose size is n(H)-|T|-|B|, could be odd. Therefore, since |A| <= |N(A)| for every A, we have, in particular, |B|<=|N(B)|. But N(B) must be a subset of T, so |B|<=|T|. We have o(H-T)=|B| or |B|+1. In the former case, we are done. In the latter, n(H)-|T|-|B| is odd and n(H) is even. Therefore |T| and |B| have opposite parity and so |B|<|T|.
(c) All that is required is to show the "hard" part of Hall's theorem. Suppose the matching condition is satisfied. Then by (b) H satisfies Tutte's condition, so by Tutte's Theorem, there is a 1-factor in H. By (a) there is an X-saturating matching in G.
Suppose G is k-connected. Let B be a separating set of H. But V(K_r) must be a subset of B, since otherwise H-B would still be connected by using paths through K_r. Therefore B=(V(K_r) union A) and A must be a separating set of G. Then |A| >=k, so |B| >= k+r, making H (k+r)-connected.
Suppose d = deg_H(s) > r. Then the subgraph of G induced by s and one neighbor in each odd component to which it is adjacent is K_{1, d}, which is not allowed. Therefore, deg_H(s) <= r for every s in S.
On the other hand, suppose C is an odd component in Y. Then deg_H(C) >= r, since the vertices in S to which C is adjacent is a separating set in G, and G is r-connected.
Now let m be the number of edges in H. By summing over vertices in S, we get m <= r|S|. By summing over components in Y, we get m >= r|Y|. Therefore, |Y| <= |S|, which is Tutte's condition. The case when S is empty follows from the fact that G has even degree.
Let x and y be two vertices in K-A. Choose z to be a vertex in (H intersect H') which is not in A. Such a z exists because A has size k-1 and (H intersect H') has size k. WLOG, assume x is in H. Since z is also in H, there is an (x, z)-path in H-A (since H is k-connected), and therefore in K-A. Similarly, there is a (y,z)-path in K-A. Therefore, there is an (x,y)-path in K-A, so A cannot separate K.
A->C: Suppose e=uv and f=wx are two edges. Menger's Theorem implies that there are two edge-disjoint uw paths. Together these paths form a closed trail T. Menger's Theorem also implies that there are two edge-disjoint vx paths, which, together, form a closed trail T'.
If e and f are both on T or T', we are done. Otherwise, one of the paths making up T, say P, contains neither e nor f, and also one of the paths making up T', say P', contains neither e nor f. Then PeP'f is a closed trail.
C->D: Let u and v be two distinct vertices. Pick an edge e incident upon u and an edge f incident upon v. Note that e and f can be chosen differently (otherwise, u and v have degree 1, and G is K_2). Now the trail containing e and f contains u and v.
D->A: Pick any edge e. Then e is on a closed trail, which means it is on a cycle, a contradiction.
Suppose G is 2-edge connected. Then kappa'(u,v)>=2 for every pair of vertices u and v. By Menger's Theorem, there will be at least two pairwise edge-disjoint u,v-paths for every u and v. Pick u and v and let P and Q be two edge-disjoint u,v-paths. Suppose P and Q intersect at vertices u=v_0,v_1,...,v_k, v_{k+1}=v and suppose v_1, ..., v_k appear in that order along P. If v_1,...,v_k also appear in that order along Q, then let P_i be the portion of P between v_i and v_{i+1} and Q_i the portion of P between v_i and v_{i+1}. Then P_iQ_i forms a cycle and these cycles intersect at v_1,...,v_k, thus forming a necklace.
We now show that if the vertices do not appear in order along Q, we can find two other edge-disjoint u,v-paths in which the common vertices do appear in the same order along the paths.
Suppose V={v_1, ...,v_k} do not appear in that order along Q, and P and Q have been chosen so that k is minimal with respect to this property. WLOG we assume v_1 is out of order along Q, for, otherwise, we could replace u with v_1. Let v_j be the first vertex in V along Q. Since v_1 does not appear first along Q, j>1. Also, let v_r be the first vertex in V along P after v_j on P which is also after v_1 along Q. Such an r exists since v appears after v_1 on both P and Q. Let A be the portion of P between u and v_1, B the portion of Q between v_1 and v_r, C the portion of P between v_j and v_r, and D the portion of Q between u and v_j. Let E be the portion of P from v_r to v and let F be the portion of Q from v_r to v. Then AB and DC are two paths from u to v_r which use a subset of the edges of P and Q. Thus ABE and DCF are two edge-disjoint u,v-paths. ABE and DCF intersect in a subset of V-{v_1,v_j}. By the minimality of P and Q, these intersections must appear in the same order along both paths.
(b) Let e=uv, f=wx, g=yz. Suppose e and f are on C1 and f and g are on C2. Let P1 be the portion from u to w on C1. WLOG assume e and f not on P1. Let P2 be the portion from v to x on C1. Then e and f will not be on P2 either. Similarly, define Q1 as the portion of C2 from w to y and Q2 as the portion of C2 from x to z, again with neither f nor g on Q1 or Q2. Then the cycle e-P1-Q1-g-Q2-P2 contains e and g.
To show the inequality is sharp, just join u and v by k vertex disjoint paths of length d. The number of vertices in such a graph is clearly (d-1)k+2.
For the second inequality, pick u and v such that d(u,v)=d and let P be a u,v-path of length d. Let S be every other vertex along P, so |S| = ceiling of (d+1)/2. S must be independent, since if two vertices in S were adjacent, there would be a shorter u,v-path. Therefore, alpha(G)>=|S|.
This inequality is sharp for any d and any k. As above, join u and v by k vertex disjoint paths of length d. But form a clique with all the vertices at a fixed distance away from u, and join all vertices distance j away with all vertices distance j+1 away from u.
Suppose vu is a directed edge and u is labeled (*, epsilon) and v is not labeled. Suppose f(vu)>l(vu). Then v can be labeled (u-, min{epsilon,f(vu)-l(vu)}).
The minimum cut in D is clearly a set of s,t-separating edges. Therefore, by the max-flow, min-cut theorem, the minimum number of s,t-separating edges is the maximum number of edge disjoint s,t-paths.
(b) Let X = {x_1,...,x_p} and Y={y_1,...,y_q}. Let (U,V) be any cut. Let k=|U intersect X| and l=|U intersect Y|. Then cap(U,V) will have three contributions. First is the sum of the m's corresponding to vertices in (V intersect X). Second is the sum of the n's corresponding to vertices in (U intersect Y). Third is from the edges between (U intersect X) and (V intersect Y). The first sum has has p-k m's in it and the second has l n's in it. So the first sum (A) is >= last p-k m's (A'), since these are the (p-k) smallest m's. The second sum (B) is >= first l n's (B'), since these are the l smallest n's. The third part (C) is k(q-l).
Now suppose the inequality in the exercise holds. Then by adding the last (p-k) m's to both sides, we get that A' + B' + C >= sum of all the m's, which we will call D. From above, cap(U, V) = A + B + C >= A' + B' + C, so cap(U, V) >= D. Since D is the capacity of one cut, it must be the minimum cut, so D is also the maximum flow.
On the other hand, if the maximum flow is D, then the minimum cut is D. A' + B' + C is the capacity of a special cut in which U includes the vertices with the largest m's and the vertices with the smallest n's. Thus, A' + B' + C >= D, from which the inequality in the exercise follows.
Now suppose alpha(H)>=n(G). Since each vertex in H projects onto a vertex in G, and all the vertices in H that project onto the same vertex in G form a clique, it follows that alpha(H)=n(G). For if the inequality were strict, then two vertices in an independent set must project to the same vertex in G.
Let S be the largest independent set in H, and let S_i be the portion of S in the i-th copy of G that makes up H. Each S_i must be independent in its respective copy of G. Now project each S_i onto G. These S_i then form m independent sets in G. Furthermore, from the previous paragraph, none of these independent sets can overlap. Since |S|=n(G), these independent sets must partition all the vertices of G. This gives a proper m-coloring of G.
Suppose the statement is true for all values less than k and k > 1. Suppose G is the union of k bipartite graphs, G_1, ..., G_k. Let (X,Y) be the bipartition of the vertices of G_k. Let G(X) and G(Y) be the subgraphs of G induced by X and Y. Then G_1,...,G_{k-1} must cover both G(X) and G(Y), so G(X) and G(Y) are both unions of k-1 bipartite graphs. By induction, then, both G(X) and G(Y) are 2^{k-1}-colorable. Then G is 2^k-colorable by using one set of 2^{k-1} colors for X and another set for Y.
Now suppose G is 2^k colorable. Color G using 2^k colors, and partition the vertices of G into X and Y, where X is the set of vertices colored with the first 2^{k-1} colors and Y is the set colored with the last 2^{k-1} colors. Let G(X) and G(Y) be the subgraphs of G induced by X and Y. By induction, G(X) and G(Y) each are unions of k-1 bipartite graphs, say G_1,...,G_{k-1} and H_1,...,H_{k-1}. Let F_i=(G_i union H_i). The F_i are also bipartite. Thus (G(X) union G(Y)) is the union of F_1,...,F_{k-1}. Now let F_k be the bipartite graph whose vertex partition is (X,Y) and whose edges are the edges between X and Y in G. Then G is the union of F_1, ..., F_k.
Let v be a vertex of maximum degree m > 2sqrt(n). N(v) must be an independent set (triangle-free). Remove v and N(v) from G to form H. H is triangle-free with n - m - 1 vertices. By induction, H is colorable using 2sqrt(n-m-1) colors. But since m > 2sqrt(n), n - m - 1 < n - 2sqrt(n) - 1 = (sqrt(n) - 1)^2 - 2. So H can be colored using 2sqrt((sqrt(n)-1)^2-2) colors. Therefore it can be colored using 2sqrt((sqrt(n)-1)^2)=2sqrt(n)-2 colors. Since v is not adjacent to any vertex in H, v can be colored any of the colors used in H, and the vertices in N(v) can be colored something different, giving a coloring of G with 2sqrt(n)-1 colors. Thus, G can be colored using 2sqrt(n) colors.
(b) Note misprint (F should be H). The number of proper colorings of H with k colors is k times the number of proper colorings of G with (k-1) colors: first color the new vertex using one of k colors. That color is now forbidden for all the vertices of G, so color G in the remaining (k-1) colors. Thus, for the wheel W_n, chi(W_n;k) = k (k-2)^n + (-1)^n k (k-2).
When k=2, strict inequality fails for all connected bipartite graphs, since there are 2 ways to properly color every such graph using 2 colors.
By induction on n. When n=k, then G must be a k-clique, and p(G)=1=k^0.
Suppose n>k. Pick any vertex v. Let H = G - v. If ch(H)=k, then v can be placed in any of the parts of any partition of V(H) into k parts. p(H) <=k^{n-1-k}, so p(G)<=kp(H)<=k^{n-k} as required, where the first inequality is an equality only when v is isolated.
Now suppose chi(H)=k-1. Then there is some partition of G into k independent sets where v is in a part by itself. For we could simply color H with k-1 colors, and use color k for v. Furthermore, for this partition, v is adjacent to vertices in each of the k-1 parts of V(H). For otherwise, v could be colored the color not represented among its neighbors, giving a (k-1)-coloring of G. Let X be this set of k-1 vertices among the neighbors of G.
Now pick an arbitrary partition of V(G) into k independent sets, and suppose the part containing v is R+v and has size 1+r. Clearly, chi(G-R-v)=k-1. Also, R contains no elements of X (since vertices in X are adjacent to v). And by induction, a(G-R-v)<=(k-1)^{n-r-k}.
Now choose R arbitrarily among all subsets of G-X-v. The number of ways to do this is (n-k) choose r. Summing over r gives an upper bound for a(G) of sum_{r=0}^{n-k} ((n-k) choose r) (k-1)^{n-r-k} = k^{n-k} by the binomial theorem. For equality to hold, N(v)=X and G-X-v must be n-k isolated vertices for each choice of v.
For equality, the graph described clearly has chi = k. Any partition into independent sets must have the k vertices of the K_k in separate blocks. Each of the isolated vertices can be placed into one of these blocks arbitrarily, giving k^{n-k} choices. This is the only such graph from the remarks about equality in the previous paragraphs.
(b) Suppose a face of G* contains two vertices of G. Since there are no edges separating the two vertices in G*, there is no path between the two vertices in G, so G is not connected.
(c) If G is connected, then there is exactly one face in G* for each vertex in G. Since G* is also connected, there is exactly one face in G for each vertex in G*. If two faces are adjacent in G, the corresponding vertices are adjacent in G*, and then the corresponding faces in G**. Thus, G=G**. On the other hand, if G=G**, then G is connected (since any graph * is connected).
We have n(T*) = n(G*) = e(G) - n(G) + 2, since n(G*) is the number of faces of G. Also, e(T*) = e(G) - e(T) = e(G) - (n(G) - 1) (since T is a tree). Thus, e(T*) = n(T*) - 1. Therefore, T* is a tree.
(b) Euler implies A + B + C + n = e + 2, and handshake implies e = 3n/2. Therefore n/4 + n/6 + n/8 + n = 3n/2 + 2, so n = 48, which means A=12, B=8 and C=6. (This graph is actually polyhedral, and can be realized as the semiregular truncated cuboctahedron.)
(b) See 6.1.30 above.
(b) Now suppose G is not planar. By Kuratowski, G has a Kuratowski subgraph, say H. By deleting the edges of G not in H, the resulting graph is H. But then the corresponding K_{3,3} or K_5 can be obtained from H by contracting along the subdivided edges.
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