LPN: reorg, split notes from exocortex page.

The exocortex page was getting unwieldy, this makes more sense.

lpn/ch02/crossword.pl

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+%% Exercise 2.4: Here are six Italian words:
+%% 
+%%     - astante
+%%     - astoria
+%%     - baratto
+%%     - cobalto
+%%     - pistola
+%%     - statale
+%% 
+%% They are to be arranged, crossword puzzle fashion, in the following grid: 
+%% 
+%%      V1  V2  V3
+%%      .   .   .
+%% H1 . 1 . 4 . 7 .
+%%      .   .   .
+%% H2 . 2 . 5 . 8 .
+%%      .   .   .
+%% H3 . 3 . 6 . 9 .
+%%      .   .   .
+%% 
+%% The following knowledge base represents a lexicon containing these words: 
+
+word(astante, a,s,t,a,n,t,e).
+word(astoria, a,s,t,o,r,i,a).
+word(baratto, b,a,r,a,t,t,o).
+word(cobalto, c,o,b,a,l,t,o).
+word(pistola, p,i,s,t,o,l,a).
+word(statale, s,t,a,t,a,l,e).
+
+%% Write a predicate crossword/6 that tells us how to fill in the
+%% grid. The first three arguments should be the vertical words from
+%% left to right, and the last three arguments the horizontal words
+%% from top to bottom.
+
+crossword(V1, V2, V3, H1, H2, H3) :-
+    word(V1, _, _1, _, _2, _, _3, _),
+    word(V2, _, _4, _, _5, _, _6, _),
+    word(V3, _, _7, _, _8, _, _9, _),
+    word(H1, _, _1, _, _4, _, _7, _),
+    word(H2, _, _2, _, _5, _, _8, _),
+    word(H3, _, _3, _, _6, _, _9, _),
+    V1 \= V2, V1 \= V3, V1 \= H1, V1 \= H2, V1 \= H3,
+    V2 \= V3, V2 \= H1, V2 \= H2, V2 \= H3,
+    V3 \= H1, V3 \= H2, V3 \= H3,
+    H1 \= H2, H1 \= H3, H2 \= H3.
+
+%% NB: execute the following query to find out the correct arrangement.
+%% crossword(A, B, C, D, E, F).
+
+%% $ swipl -q                                                                                             
+%% 1 ?- [italiano].                 
+%% true.                            
+%% 
+%% 2 ?- crossword(A, B, C, D, E, F). 
+%% A = astante,
+%% B = cobalto,
+%% C = pistola,
+%% D = astoria,
+%% E = baratto,
+%% F = statale ;
+%% A = astoria,
+%% B = baratto,
+%% C = statale,
+%% D = astante,
+%% E = cobalto,
+%% F = pistola ;
+%% false.
+
+%% Note that there are two solutions; these are mirrors of each other
+%% with the vertical words substituted for the horizontal words.

lpn/ch02/notes.md

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+## Unification
+
+Two terms unify if they are
+1. The same term
+2. They contain variables that can be uniformly instantiated such that the
+   resulting terms are equal.
+
+Unification is foundational to Prolog.
+
+`=/2` checks whether its two arguments unify:
+
+```
+12 ?- =(a, a).
+true.
+13 ?- =(a, b).
+false.
+```
+
+*NB* variable assignment with `=`:
+
+```
+24 ?- vincent = X.
+X = vincent.
+25 ?- jealous(Y, X).
+Y = vincent,
+X = marsellus ;
+Y = marsellus,
+X = vincent ;
+false.
+```
+
+### Lines
+```
+/* lines.pl */
+vertical(line(point(X,Y),point(X,Z))).
+horizontal(line(point(X,Y),point(Z,Y))).
+```
+
+Check this out:
+
+```
+39 ?- vertical(line(point(1, 7), P)).
+P = point(1, _402).
+```
+
+The answer is structured: `_402` is a placeholder variable that means "any old
+value of Y will do." This answer was derived solely through unification, and
+didn't require logic (e.g. modus ponens) or any sort of mathematical
+relationships. "Moreover, when a program is written that makes heavy use of
+unification, it is likely to be extremely efficient."

lpn/ch02/proof22.pl

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+f(a).
+f(b).
+
+g(a).
+g(b).
+
+h(b).
+
+k(X) :- f(X), g(X), h(X).