CSC302 2011S, Class 10: Prolog (3)

Overview:
* Add to End example
* Thinking in Prolog.
* Lab.

Admin:
* Late or missing: JL, AT
* Are there questions on Assignment 4?
* EC for Thursday extra: Our own Aaron Todd!
* EC for Friday CS Table: Knuth on the History of Combinatorial Generation.
* Due to popular demand, we will do a fourth day of Prolog on Wednesday,
  completely screwing up my schedule.
  * There is no reading for Tuesday night.

Problem: Write add_to_end(L, X, M), which holds when M is the result of
adding X to the end of L.  (Alternately: Which adds X to the end of L.)

add_to_end([], X, [X]).
add_to_end([H|T], X, M) :- add_to_end(T, X, U), M=[H|U].

vs.

add_to_end([], X, [X]).
add_to_end([H|T], X, [H|U]) :- add_to_end(T, X, U).

--

Thinking in Prolog

* 1: Try to think in terms of patterns.
* 2. Worry about the evaluation strategy (agh!)
* 3: Worry about multiple paths to the same answer.
  * Add explicit rules to ensure that, for example, V<P, V=P, V>P
  * Side note: =< not <=
4. Conflict between two views:
   This predicate holds when ...
   Given X, this predicate computes Y.q

Example: Length

(See Examples/Day3)

--

Lab

For printing the puzzle:

Helpful to write:

(a) split(L, N, L1, L2), which holds when L1 contains the first N
characters of L and L2 contains the remaining characters.

(b) print_list(L), which prints a list separated by spaces