But, if I choose your second approach (function that picks and returns one of the feasible configurations), than such function cannot be used in next exercise 32.3.3 in which I have to construct function solitaire for solving a puzzle.
From: [email protected] To: [email protected] Subject: RE: [racket] HtDP Exercise 32.3.2 Date: Mon, 15 Aug 2011 22:04:13 +0200 But, if I choose your second approach (function that picks and returns one of the feasible configurations), than such function cannot be used in next exercise 32.3.3 in which I have to construct function solitaire for solving a puzzle. Subject: Re: [racket] HtDP Exercise 32.3.2 From: [email protected] Date: Mon, 15 Aug 2011 15:21:08 -0400 CC: [email protected] To: [email protected] On Aug 15, 2011, at 3:10 PM, Racket Noob wrote:I don't understand this exercise: Exercise 32.3.2. Develop a function that, given a board and the board position of a peg, determines whether or not the peg can jump. We call such a peg enabled.Develop a function that, given a board and the board position of an enabled peg, creates a board that represents the next configuration. But, it may be the case that an enabled peg can jump to more than one of free places. Thus, we can have more then one new configurations, no? Good catch. Now design a function that returns a list of next configurations for an 'enabled' peg. Alternatively, design a function that picks one of the feasible successor configurations.
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