Notes on computational linguistics.pdf - UCLA Department of ...

Stabler - Lx 185/209 2003
a. d
e c
b. d
b d
e c
b
a
b
a
d
b
Consider, for example, how we could substitute the non-empty b of in 17a into the position of the
empty b, obtaining the tree in 17b. This involves two basic steps. First, we must replace the nonempty
subtree b/[a/[]] by an empty subtree b/[], and then we must replace the other empty subtree b/[]
by b/[a/[]]. Both steps involve modifying a tree just by replacing one of its subtrees by something
else. We formalize this basic step first, as an operation on our special notation for nodes, with the
predicate replace_node.
(18) We define replace_node(A,DescA,B,DescB) to hold just in case nodes A and B are subtree isomorphic
except that where the subtree of the former has descendant DescA, the latter has descendant DescB:
replace_node(A, A, B, B). % A replaced by B
replace_node(A, DescendantA, B, DescendantB) :- % DescA repl by DescB
label(A, Label),
label(B, Label),
children(A, ChildrenA),
children(B, ChildrenB),
replace_nodes(ChildrenA, DescendantA, ChildrenB, DescendantB).
The first clause, in effect, just replaces the current node A by B, while the second clause uses the relation
replace_nodes to do the replacement in exactly one of the children of the current node.
(19) We extend the previous relation to node sequences:
replace_nodes([A|As], DA, [B|Bs], DB) :replace_node(A,
DA, B, DB),
iso_subtrees(As, Bs).
replace_nodes([A|As], DA, [B|Bs], DB) :iso_subtree(A,
B),
replace_nodes(As, DA, Bs, DB).
With these axioms, we can establish some basic relations among trees. For example, with two basic
replacement steps, we can transform the tree in Figure 17a into the tree in Figure 17b. The first step
replaces the subtree b/[a/[]] by b/[], and the second step replaces the original b/[] by b/[a/[]].
Consider the first step, and since we are working with our special node representations, let’s focus our
attention just on the subtrees dominated by c, where the action is. Taking just this subtree, we establish
a relation between the root nodes:
A=n(root,c/[b/[],d/[b/[a/[]]]],none),
B=n(root,c/[b/[],d/[b/[]]],none).
What we do to obtain B is to replace DescA in A by DescB, where
DescA=n(1,b/[a/[]],n(2,d/[b/[a/[]]],
n(root,c/[b/[],d/[b/[a/[]]]],none))),
DescB=n(1,b/[],n(2,d/[b/[]],
n(root,c/[b/[],d/[b/[]]],none))).
We can deduce that these elements stand in the relation
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