Add finished ch03 exercises.

lpn/ch03/notes.md

@@ -140,6 +140,14 @@ yes
 no
 ```
 
+Answer (in `succ.pl`):
+
+```
+greater_than(succ(succ(X)), succ(0)).
+greater_than(succ(X), succ(Y)) :-
+    greater_than(X, Y).
+```
+
 #### Exercise 3.5
 
 Binary trees are trees where all internal nodes have exactly two children. The
@@ -159,4 +167,6 @@ tree that is its first argument. For example:
 ?-  swap(tree(tree(leaf(1),  leaf(2)),  leaf(4)),T).
 T  =  tree(leaf(4),  tree(leaf(2),  leaf(1))).
 yes
-```
+```
+
+This one took a lot of thinking, but the solution is in `tree.pl`.

lpn/ch03/tree.pl

@@ -0,0 +1,42 @@
+%% Exercise 3.5
+%%
+%% Binary trees are trees where all internal nodes have exactly two
+%% children. The smallest binary trees consist of only one leaf node. We
+%% will represent leaf nodes as `leaf(Label)`. For instance, `leaf(3)`
+%% and `leaf(7)` are leaf nodes, and therefore small binary trees. Given
+%% two binary trees B1 and B2 we can combine them into one binary tree
+%% using the functor `tree/2` as follows: `tree(B1,B2)`. So, from the
+%% leaves `leaf(1)` and `leaf(2)` we can build the binary tree
+%% `tree(leaf(1),leaf(2))` . And from the binary trees
+%% `tree(leaf(1),leaf(2))` and `leaf(4)` we can build the binary tree
+%% `tree(tree(leaf(1), leaf(2)),leaf(4))`.
+%%
+%% Now, define a predicate `swap/2`, which produces the mirror image of
+%% the binary tree that is its first argument. For example:
+%%
+%% ```
+%% ?-  swap(tree(tree(leaf(1),  leaf(2)),  leaf(4)),T).
+%% T  =  tree(leaf(4),  tree(leaf(2),  leaf(1))).
+%% yes
+%% ```
+
+
+%% Let's start with the base case: swapping a leaf is the identity.
+swap(leaf(X), leaf(X)).
+
+%% What does it mean to swap a tree? Given a tree
+%%     tree(X, Y)
+%% it should have an equivalent
+%%     tree(swap(Y), swap(X))
+%% How do we express this?
+%% Let's start with a tree made of two leaves.
+%% swap(tree(X, leaf(Y)), tree(leaf(Y), Z).
+
+%% But this isn't quite right, is it? We need to express that
+%% the X node needs to be swapped. So we need to express this
+%% as a rule.
+swap(tree(X, leaf(Y)), tree(leaf(Y), Z)) :- swap(X, Z).
+
+%% And there it is:
+%% ?- swap(tree(tree(leaf(1),  leaf(2)),  leaf(4)),T).
+%% T = tree(leaf(4), tree(leaf(2), leaf(1))).