CMPSCI 250: Introduction to Computation

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.

  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.)

  • (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.)

  Question 2 deals with the inductive definition of addition of natural numbers, and with an inductively define 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.)

  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.

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

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

  • (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.

  • (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.

  • (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.

  • (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.

  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.

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

  • (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.

  • (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.

Last modified 9 April 2011