CMPSCI 250: Introduction to Computation

Solutions to Second Midterm Exam

David Mix Barrington

4 April 2011

Directions:

• Answer the problems on the exam pages.
• There are four problems for 100 total points plus 10 extra credit. Actual scale is A=100, C=68.
• If you need extra space use the back of a page.
• No books, notes, calculators, or collaboration.

Question text is in black, solutions are in blue.

Correction in orange added 1 April 2012.

  Q1: 25 points
  Q2: 15 points
  Q3: 35+10 points
  Q4: 25 points
Total: 100+10 points

• Question 1 deals with a function f(i, j), which takes two natural numbers as arguments and returns a natural number. If j < i, then f(i, j) is defined to be 0. Otherwise f(i, j) is the sum 2ⁱ + ... + 2^j. We can define this inductively by the rules f(i, i) = 2ⁱ and, if j > i, f(i, j) = f(i, j-1) + 2^j.

• Question 1 (25): Two induction proofs concerning this function:
  • (a, 10): Prove that for all natural numbers j, f(0, j) = 2^j+1 - 1. (Hint: Use the rules above and ordinary induction on j.)

    Base Case: j = 0, f(0, 0) is defined to be 2⁰ = 1, and this equals 2⁰⁺¹ - 1 = 1.

    Inductive Case: Assume that f(0, j) = 2^j+1 - 1. Then f(0, j+1) is defined to be f(0, j) + 2^j+1, which by the IH is equal to 2^j+1 - 1 + 2^j+1 = 2^j+2 - 1 which is 2^(j+1)+1 - 1 as required.

  • (b, 15) Prove that for any natural i and for any natural j with j ≥ i, f(i, j) = 2^j+1 - 2ⁱ. (Hint: Let i be arbitrary and use induction starting with j = i to prove the statement for all j with j ≥ i.)

    Let i be arbitrary and use induction on all j with j ≥ i.

    Base Case: i = j, f(i, i) is defined to be 2ⁱ, and this is equal to 2^j+1 - 2ⁱ.

    Inductive Case: Assume that f(i, j) = 2^j+1 - 2ⁱ. Then f(i, j+1) is defined to be f(i, j) + 2^j+1 which by the IH is 2^j+1 - 2ⁱ + 2^j+1 = 2^j+2 - 2ⁱ. This is 2^(j+1)+1 - 2ⁱ as required.

  Question 2 deals with the inductive definition of addition of natural numbers, and with an inductively defined predicate LE(x, y) that takes two natural numbers as arguments.

  Recall that x + 0 is defined to be x, and that x + succ(y) is defined to be succ(x + y), where "succ" is the successor function on naturals.

  We define LE(x, y) by three rules. LE(0, y) is defined to be true for any y. LE(succ(x), 0) is defined to be false for any x. Finally, LE(succ(x), succ(y)) is defined to be equal to LE(x, y) for any naturals x and y.

• Question 2 (15): Prove that for any three natural numbers x, y, and z, if LE(x, y) is true, then LE(x + z, y + z) is true. (Hint: Let x and y be arbitrary, assume LE(x, y), and use ordinary induction on z.)

  Let x and y be arbitrary, and use ordinary induction on z.

  Base Case: z = 0, we must prove LE(x, y) → LE(x + 0, y + 0), which is true because x + 0 = x and y + 0 = y.

  Inductive Case: Assume that LE(x, y) implies LE(x + z, y + z). We want to prove that LE(x, y) implies LE(x + succ(z), y + succ(z)). Assume that LE(x, y) is true. Then the IH implies that LE(x + z, y + z) is true. By the definition of addition, LE(x + succ(z), y + succ(z)) equals LE(succ(x + z), succ(y + z)), which by the definition of LE is equal to LE(x + z, y + z), which is true.

  Question 3 deals with a particular directed graph with nine vertices s, a₁, a₂, a₃, a₄, b₁, b₂, b₃, and b₄. There are arcs from s to a₁ and b₁. For i = 1, 2, or 3, the nodes a_i and b_i each have arcs to nodes a_i+1 and b_i+1. Those are the only arcs -- there are 14 in all. We will consider depth-first and breadth-first searches of this directed graph starting at s, both with and without a closed list. (With a closed list, we can recognize visited nodes, including nodes on the stack or in the queue -- without one we cannot.) When we use DFS, the children a_i and b_i of a node are placed on the stack with b_i placed first, so that a_i is taken off the stack first. When we use BFS, a_i is placed on the queue first, so that it comes off first.

• Question 3 (35+10): Here are several questions about these searches:
  • (a, 5): Explain why each of the four searches will eventually terminate, even if there is no goal node.

    Of course the closed list searches will terminate because there are only finitely many nodes and arcs in the graph, and those searches will only process each of these once.

    With an open list, we are only guaranteed termination because this particular directed graph has no cycles. A node is placed on the open list once for each path from the start node to it, and in this directed graph all paths are of length at most four and every node has at most two neighbors, so there are finitely many paths.

  • (b, 10): Give the DFS tree for the depth-first search with no goal node and with a closed list.

    The tree has root s, with children a₁ and b₁. Node a₁ has children a₂ and b₂ in the tree, a₂ has children a₃ and b₃, a₃ has children a₄ and b₄, and the other four nodes are leaves of the tree. The two arcs each out of b₁, b₂, and b₃ are the non-tree edges.

  • (c, 10): Give the BFS tree for the breadth-first search with no goal node and with a closed list.

    The BFS tree is actually identical to the DFS tree, although the nodes are visited in a different order.

  • (d, 5): Find a choice for a goal node t such that the DFS will find t after exploring fewer nodes that the BFS needs to find t, both with or without a closed list. (Here a node is said to have been "explored" when it comes off the open list.) Justify your answer.

    The node a₄, for example, is the fifth node explored by the DFS, after s, a₁, a₂, and a₃. But it is the seventh node explored by the BFS, followed only by b₄.

  • (e, 5): Find a choice for a goal node t such that the BFS will find t after exploring fewer nodes than the DFS needs to find t, both with or without a closed list. Justify your answer.

    The node b₁ is the third node explored by the BFS, after only s and a₁. But it is the last node explored by the DFS, as all the descendents of a₁ are explored before b₁ is considered.

  • (f, 5XC): Determine how many total visits the DFS, with no closed list and with no goal node, will make to the two childless nodes a₄ and b₄. Justify your answer.

    There are eight visits to each of a₄ and b₄, or sixteen total. There is one visit of a node for each path from s to the node. There is one path to each node on level 1, two paths to each on level 2, four to each node on level 3, and finally eight to each on level 4.

  • (g, 5XC): Determine how many total visits the BFS, with no closed list and with no goal node, will make to the two childless nodes a₄ and b₄. Justify your answer.

    The answer again is eight visits to each node, sixteen in all. When s is explored, a₁ and b₁ are put on the queue one time each. These two nodes each put their two children on the queue, so after they are explored there are four nodes on the queue. Similarly there are eight nodes (four copies each of a₃ and b₃) on the queue after the second level has been explored, and sixteen after the third level has been explored. These nodes each represent one visit to a childless node.

  Question 4 deals with a two-player game called Stones. In Stones, there is a single pile with some natural number of stones in it. On her turn, a player must remove either one or two stones from the pile. The first player who cannot move loses. (That is, the player who takes the last stone wins.) The game is somewhat similar to another game called Nim that you may have seen, but Stones is easier to analyze.

• Question 4 (25): Four questions concerning this game:
  • (a, 5) Give a winning strategy for the second player if the game starts with three stones.

    If the first player takes one, take two. If she takes two, take one. Either way you (the second player) get the last stone.

  • (b, 5) Give a winning strategy for the first player if the game starts with four stones.

    Take one stone. The second player is now in the position of the first player in part (a), and will lose if you (the first player) follow the strategy given their for the second player.

  • (c, 10) Prove by (ordinary) induction, for all natural numbers k, that the second player has a winning strategy for the game that starts with 3k stones.

    Base case: k = 0, by definition of the game the first player loses because she cannot take any stones.

    Inductive case: Assume that the second player has a winning strategy for the game with 3k stones. In the game with 3(k+1) stones, the second player should take two on her first move if the first player takes one, and should take one if the first player takes two. She (the second player) is now the second player in a game with 3k stones, and by the IH she has a winning strategy for the remainder of the game.

    Many students gave a correct strategy (always move to make the remaining number of stones divisible by 3) but did not express it as an induction. Others said to play the 3k game first, then the 3 game (or vice versa), but did not explain how the 3k+3 game could be decomposed into these two smaller games.

  • (d, 5) Using the result of part (c), prove that if the starting number of stones is not divisible by 3, then the first player has a winning strategy.

    If the original number of stones is not divible by 3, it must be either 3k+1 or 3k+2 for some natural number k. On her first move, the first player should take one stone in the first case and two stones in the second case, so that she becomes the second player in a game with 3k stones. By part (c), she has a winning strategy for this remaining game.

Last modified 1 April 2012