COMPSCI 575/MATH 513: Combinatorics and Graph Theory

Solutions to Practice Final Exam, Fall 2016

David Mix Barrington

20 December 2016

Directions:

• Answer the problems on the exam pages.
• There are six problems (most with multiple parts) for 125 total points. Estimated scale is A = 110, C = 70.
• If you need extra space use the back of a page.
• No books, notes, calculators, or collaboration.
• Numerical answers may be given as expressions involving arithmetic operations (including exponentiation and factorial) or the functions P(n, k) and C(n, k). In general a summation is a complete answer to a counting problem, but a recurrence or a generating function is not.

  Q1: 20 points
  Q2: 20 points
  Q3: 20 points
  Q4: 20 points
  Q5: 30 points
  Q6: 15 points
Total: 125 points
  

Question text is in black, solutions in blue.

• Question 1 (20): Identify each of these statements as true or false. No explanation is needed or wanted, and there is no partial credit or penalty for guessing (2 points each).
  • (a) Every graph that has a Hamilton circuit also has a Hamilton path that is not a circuit.

    TRUE. Just delete any one edge from the circuit and you have a path that visits each vertex exactly once.

  • (b) Every graph that has an Euler circuit also has an Euler path that is not a circuit.

    FALSE. The former is possible if and only the graph is connected and has no vertices of odd degree, and the latter is possible if and only if it is connected and has exactly two such vertices.

  • (c) Every connected planar graph with 10 vertices has at most 24 edges.

    TRUE. We derived the formula E ≤ 3V - 6 from Euler's Formula for planar graphs and the fact that each face has at least three edges.

  • (d) The chromatic polynomial of the graph H of Question 5 is r⁴ - 3r³.

    FALSE. It is r(r-1)³. The given polynomial might be obtained by an incomplete (and thus invalid) I/E argument. You can take the r⁴ total colorings, and subtract the r³ that violate each edge condition, but you then have to correct for multiply counting colorings that violate more than one edge condition.

  • (e) If every vertex in a tree has odd degree, the number of edges in the tree may be either odd or even.

    FALSE. There must be an even number of vertices of odd degree in any graph, since the sum of the degrees is twice the number of edges. So the number of edges, which is one less than the number of vertices, must be odd.

  • (f) There are exactly ten ways to distribute eleven identical treats to three identical dogs, assuming that each dog gets at least one treat.

    TRUE. Such a partition is given by a non-increasing sequence of three positive integers adding to eleven. We have 9-1-1, 8-2-1, 7-3-1, 7-2-2, 6-4-1, 6-3-2, 5-5-1, 5-4-2, 5-3-3, and 4-4-3. This is an instance of the partition problem, which does not have a simple solution in terms of binomial coefficients, so we use an exhaustive search.

  • (g) There are exactly 44 permutations of a five-element set that fix none of the elements, that is, 44 derangements of the set.

    TRUE. There are 24 elements of S₅ that are 5-cycles, and 20 that are the products of a disjoint 2-cycle and 3-cycle. This is also the derangement number D(5), which is given by 5! times (1 - 1 + 1/2! - 1/3! + 1/4! - 1/5!) which is 44.

  • (h) For any non-negative integer n, the sum for i from 0 to n of C(n, i)(-1)ⁱ is equal to 0.

    FALSE. The is the binomial expansion for (1-1)ⁿ, which is 0 for all positive integers n, but not for n = 0. I made this mistake myself in setting the second midterm.

  • (i) There is a game with a finite number of positions whose value, in Conway's terminology, is the number 1/3.

    FALSE. All such games whose values are numbers are dyadic rationals, with denominator a power of two. We saw a Hackenbush game in lecture with value 1/3, but it had infinitely many positions.

  • (j) Every position in the game of Domineering has a value that is a number, in Conway's terminology.

    FALSE. The L triomino has a value equal to that of a one-pile Nim game with one stone, since the first player's only move is to create a zero game. This value, called * or *1 by Conway, is a "nimber" rather than a number. It violates the condition for a number because it has a left option, 0, that is not less than one of its right options, 0.

• Question 2 (20): Recall that an (a, b) cut of a graph G is a partition of the vertices with vertex a in one set and vertex b in the other. In this problem you may quote standard results about network flow without proof.
  • (a, 10) Prove that for any positive integer k, there are k edge-disjoint paths from a to b if and only if every a-b cut contains at least k edges crossing the cut (with one endpoint in each set).

    Clearly if the k edge-disjoint paths exist, there cannot be a cut-set with fewer than k edges, since deleting fewer than k edges will leave at least one of the paths intact.

    For the other direction, Consider the network flow problem where every edge has capacity one in either direction. We know that the optimal flow is equal to the size of the minimum cut, and that the augmenting path algorithm will produce an optimal flow with integer (hence 1 or 0) flows on each edge. If the given condition on cuts holds, the optimal flow has value at least k, and we can decompose it into at least k edge-disjoint paths by a greedy algorithm.

  • (b, 10) Prove that for any positive integer k, there are k vertex-disjoint paths from a to b if and only if every set of vertices, whose removal disconnects a from b, has at least k vertices.

    If the k vertex-disjoint paths exist and we delete fewer than k vertices, at least one of the paths will still be intact and vertex a will not be disconnected from vertex b.

    So it remains to prove that if there is no such set of k vertices, the paths will exist. To show this, I will make a new directed graph G' from our graph G, with each vertex v (except a and b) replaced by two vertices v_in and v_out, and directed edges from u_out to v_in and from v_out to u_in for each edge (u, v) in G. We also add vertices a_out and b_in, and directed edges in only one direction for edges in G involving a or b.

    I claim that the minimum cut in this graph has at least k edges, because if it had fewer than k I could disconnect G by deleting one endpoint of each of those edges. By the argument of part (a), there are then k edge-disjoint paths from a to b in G'. But these must correspond to vertex-disjoint paths in G, because any path in G' using a vertex v_in must use the edge to v_out.

• Question 3 (20): Once again we are distributing identical treats to distinct dogs. Let f(r) be the number of ways to distribute r treats to Cardie, Duncan, and Whistle, with the conditions that (1) each dog gets at least one treat, (2) Cardie gets strictly more treats than Duncan, and (3) Whistle gets no more than three treats.
  • (a, 10) Find f(11) by any valid method (but explain your reasoning).

    We need triples of positive integers (C, D, W) with C > D and W ≤ 3. We can find these exhaustively: if W = 1 we have 9-1-1, 8-2-1, 7-3-1, and 6-4-1; if W = 2 we have 8-1-2, 7-2-2, 6-3-2, and 5-4-2, and if W = 3 we have 7-1-3, 6-2-3, and 5-3-3. There are eleven combinations in all.

    We could also use the generating function from (b), but that strikes me as more complicated.

  • (b, 10) Find the ordinary generating function for the function f, and express this function as a ratio of polynomials (with no "..." or infinite sums). How would you use this generating function to answer part (a)?

    The trick here is to count the excess of Cardie's treats over Duncan's, rather than just counting Cardie's. If we let C' = C - D, we are now looking for a triple of positive numbers (C', D, W) with C' + 2D + W = r and W ≤ 3. A generating function for this is (x + x² + x³ + ...)(x² + x⁴ + x⁶ + ...)(x + x² + x³). We can rewrite this as x/(1-x) times x²/(1-x² times (x + x² + x³), which multiplies out to a ratio of polynomials.

    The answer to part a is the x¹¹ coefficient of this power series. (We've now finished part (b), but I'll continue the GF solution to get the answer of (a).)

    We can rewrite this, factoring the (1-x²) in the denominator, as x⁴(1-x)^-2 times (1+x+x²)/(1+x). This latter term is 1 + x²/(1+x). So to get the x¹¹ coefficient of the whole power series, we need the x⁷ coefficient of 1/(1-x)² which is C(8, 7) = 8, plus the x⁷ coefficient of 1/(1-x)² times x²/(1+x). This last is the x⁷ coefficient of (1+2x+3²+4x^{3+5x4+6x5+...) times (x2 - x3 + x4 - x5 + x6 - x7 + ...) which is 6-5+4-3+2-1 = 3. The total coefficient is 8+3 = 11 as we found above.}

• Question 4 (20): For any non-negative integer r, let g(r) be the number of binary strings of length r that do not contain 000 or 111 as a substring, i.e., that never have the same bit three times in a row.
  • (a, 10) Explain why, for r ≥ 3, the function g satisfies the recurrence g(n) = g(n-1) + g(n-2). Give sufficient initial condititions for this recurrence and use it to calculate g(6).

    If n ≥ 2 I can take any valid string w of length n-1 and make a valid string of length n by appending the bit that does not equal the last bit of w. If n ≥ 3 I can take any valid string w of length n-2 and make a valid string of length n by appending two copies of the bit that does equal the last bit of w. And any valid string is made in this way: if the last two bits are equal it must come from the second method, and if they are equal it must come from the first.

    All strings of length 1 and 2 are valid, so g(1) = 2 and g(2) = 4. The recurrence gives us g(3) = 6, g(4) = 10, g(5) = 16, and g(6) = 26. (Note that each value of g is twice a Fibonacci number.)

  • (b, 10) Represent an arbitrary binary string w as w₁w₂...w_r. For any positive integer i with i ≤ r-2, let A_i be the set of strings of length r such that w_i = w_i+1 = w_i+2. Use Inclusion/Exclusion on these sets to calculate g(6), and verify that this matches your answer from part (a).

    The total number of strings is 2⁶. The sets A₁, A₂, A₃, and A₄ each have size 2⁴. Of the six two-way intersections, three have size 2³ and three have size 2². Of the four three-way intersections, two have size 2² and two have size 2¹. The four-way intersection has size 2¹.

    We have 64 - 4(16) + 3(8) + 3(4) - 2(4) - 2(2) + 2 = 26.

• Question 5 (30): Consider the graph H with vertices {a, b, c, d} and edge set {(a, b), (a, c), (a, d)}.
  • (a, 10) An automorphism of a graph is a graph isomorphism (a bijection h of the vertices such that (u, v) is an edge if and only if (h(u), h(v)) is an edge) from a graph to itself. Prove that H has exactly six automorphisms.

    Vertex a must go to itself as it is the only vertex of degree 3. Any permutation sending a to itself is an automorphism because the three other vertices each have one edge to a and no other edges. There are 3! = 6 permutations of the set {b, c, d}.

  • (b, 10) Determine the number of 2-colorings of the vertices of H, where two colorings are considered equivalent if one can be taken to the other by an automorphism of H. Give the pattern inventory for these colorings. Stray text here removed 20 December 2016.

    The group of six automorphisms has cycle index polynomial (1/6)(x₁⁴ + 2x₁x₃ + 3x₁²x₂), as the group S₃ has the identity, two 3-cycles, and three two-cycles.

    We get the number of 2-colorings by substituting 2 for each variable in the cycle index polynomial to get (1/6)(2⁴ + 2×2² + 3×2³ = (16 + 8 + 24)/6 = 8. We could also use Burnside's Theorem directly by counting the colorings fixed by each of the permutations.

    The pattern inventory, by Polya's Enumeration Theorem, is just (1/6)((b+w)⁴ + 2(b+w)(b³+w³) + 3(b+w)²(b²+w²) = b⁴ + 2b³w + 2b²w² + 2bw³ + w⁴. We could obtain this directly by noting that all colorings with four black or four white vertices are isomorphic, and that the other three color-count classes each have two colorings up to isomorphism because vertex a can be black or white.

  • (c, 10) Repeat part (b) for 3-colorings of H under the same conditions. You need not simplify the pattern inventory.

    The number of classes is just (1/6)(3⁴ + 2×3² + 3×3^{3) = (81 + 18 + 81)/6 = 30.}

    The pattern inventory is (1/6)(b+w+r)⁴ + 3(b+w+r)²(b²+w²+r² + 2(b+w+r)(b³+w³+r³)) which evaluates to a sum of fifteen monomials, one for each possible choice of how many nodes of each color. The terms b⁴, w⁴, and r⁴ have coefficient 1, b²wr and the two terms like it have coefficient 3, and the other nine terms have coefficient 2.

• Question 6 (15): Here is a combinatorial game defined on rectangular arrays of stones. Given an n by k array of stones, the player whose move it is may delete either one row of stones, leaving a (k-1) by n array, or delete one column of stones, leaving a k by (n-1) array. As usual, the player taking the last stone wins.
  • (a, 5) Determine the winner, under optimal play, if we start with a 2 by 2 array. You may use any valid method, but justify your answer.

    The second player has a winning strategy. All of the first player's moves leave a 1 by 2 or 2 by 1 array, and the second player can just remove all the remaining stones on his first move.

  • (b, 10) Find the kernel of this game, or equivalently, find the exact set of start positions for which the second player has a winning strategy.

    Of course a 0 by k array is a zero game and is in the kernel.

    A 1 by k array is a losing position for any positive k, since the first player will take all the stones. So none of these positions are in the kernel.

    We saw in part (a) that the 2 by 2 game is in the kernel. The 2 by 3 game thus is not, since the first player has a winning move going to 2 by 2. For the 2 by 4 game, the first player has a choice of 2 by 3 or 1 by 4, both of which are losing, so 2 by 4 is winning and is in the kernel. Similarly 2 by 2k is in the kernal for any k ≥ 2. Then 3 by 3 can be seen to be in the kernel, 3 by 4 is out, 3 by 5 is in, and so forth.

    The kernel is all j by k arrays where either (1) j or k is 0, or (2) j ≥ 2, k ≥ 2, and j+k is even. To verify this we must show that no move takes one such position to another, and that every non-kernel position has a move to a kernel position. There are no moves if (1) holds. If (2) holds, any move makes j = 1, k = 1, or j, k ≥ 2 with j+k odd. So no move from our alleged kernel leads to another kernal position. If we are not in the kernel, either j or k is 1 and there is a move making (1) hold, or both are ≥ 2 and there is a move making (2) hold.

Last modified 20 December 2016