500 CIS Software Foundations Homework Assignment 3
Induction; Operational Semantics Due: Monday, September 22, 2002, by noon Submission instructions: You may submit your solutions either electronically (in ascii, postscript, or PDF format) or in handwritten hardcopy form. Hand-written solutions should be submitted to Jennifer Finley in Levine 302. Electronic solutions should be submitted following the same instructions as last time; these can be found at http://www.seas.upenn.edu/cis500/homework.html.
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1 Exercise Explain the aw in the following proof. Theorem(?!): All horses are the same color. Proof: Let P (n) be the predicate in all non-empty collections of n horses, all the horses are the same color. We show that P (n) holds for all n by induction on n (using 1 as the base case). Base case: Clearly, P (1) holds. Induction case: Given P (n), we must show P (n + 1). Consider an arbitrary collection of n + 1 horses. Remove one horse temporarily. Now we have n horses and hence, by the induction hypothesis, these n horses are all the same color. Now call the exiled horse back and send a dierent horse away. Again, we have a collection of n horses, which, by the induction hypothesis, are all the same color. Moreover, these n horses are the same color as the rst collection. Thus, the horse we brought back was the same color as the second horse we sent away, and all the n + 1 horses are the same color. 2 Exercise How about this one? Theorem(?!): n2 + n is odd for every n 1. Proof: By induction on n (again starting from 1). For the base case, observe that 1 is odd by denition. For the induction step, assume that n2 + n is odd; we then show that (n + 1)2 + (n + 1) is odd as follows. (n + 1)2 + (n + 1) = n2 + 2n + 1 + n + 1 = (n2 + n) + (2n + 2). But n2 + n is odd by the induction hypothesis, and 2n + 2 is clearly even. Thus, (n2 + n) + (2n + 2) is the sum of an odd number and an even number, hence odd. 3 Exercise [Optional] Make up your own false proof in which an incorrect use of the induction hypothesis leads to a surprising conclusion. 4 Exercise 3.5.13 in TAPL. 5 Exercise 3.5.17 in TAPL. 6 Exercise 4.2.2 in TAPL. To complete this exercise, you will need to download your own copy of the les for the arith implementation. These les can be found on the TAPL web site http://www.cis.upenn.edu/bcpierce/tapl/checkers/arith.tar.gz (as a single bundle) http://www.cis.upenn.edu/bcpierce/tapl/checkers/arith separate (as les) Instructions for compiling these les into a running implementation can be found here: http://www.cis.upenn.edu/bcpierce/tapl/resources.html#checkers Chapter 10 of Jason Hickeys notes has more information on OCaml compilation. The only changes you will need to make for this problem are to the eval function in the le core.ml. Please include your version of just this le in the solutions you hand in.
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7 Debrieng 1. Are you in a study group? 2. If so, who else is in it? 3. How many hours did you spend on this assignment? 4. Would you rate it as easy, moderate, or dicult? 5. Did you work on it mostly alone, or mostly with your study group? 6. How deeply do you feel you understand the material it covers (0%100%)? If you have any other comments, we would like to hear them; please send them directly to [e-mail] to make sure they dont get lost in a pile of homeworks!
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Solutions
The solutions not given here can be found in the back of the book. 1. The problem with this proof is that the inductive step does not work for all n, namely for n = 1. According to our proof, to show that n + 1 = 2 horses are the same color, we consider horse #1 with color C1 and then we separately consider horse #2 with color C2 . The two groups of horses have no overlapping members, and thus we have no reason to assert that these two horses have the same ...