[racket] HtDP Exercise 32.3.2

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: racketnoob at hotmail.com
To: matthias at ccs.neu.edu
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: matthias at ccs.neu.edu
Date: Mon, 15 Aug 2011 15:21:08 -0400
CC: users at racket-lang.org
To: racketnoob at hotmail.com




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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