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Ling 472 Lab, November 5, 2004
Revised pseudo code for the (non-probabilistic) CKY algorithm:
Create and clear chart [ #words , #words ]
for i 1 to #words
chart [ i , i ] { | inputi }
for span 2 to #words
for begin 1 to #words โ span + 1
end begin + span โ 1
for m begin to end โ 1
if (
(^1)
2 ^ P^
1 ^ chart [ begin , m ] ^
2 chart [ m + 1, end ] then
chart[begin, end] = chart[begin, end] { }
Step through the (non-probabilistic) CKY algorithm, using this grammar:
S NP VP S Aux S
VP V S VP V NP VP VP PP
NP Det N NP NP PP
NP Waikiki NP Oslo NP Kim NP snow
PP P NP PP P S
V adores VP snores
Aux does Aux can Aux is
P in P on P before
Det this Det these Det the
Use this sentence:
Snow in Oslo snores 1 2 3 4
First, start out with a chart with the appropriate cells. Each one corresponds to a substring of the input string:
The first loop:
for i 1 to #words chart [ i , i ] { | inputi }
This fills in the chart with pre-terminals.
So we go through i = 1 to i = 4; for each of these, we put an element in the corresponding cell in the chart for each preterminal that expands to that input. We end up with a chart that looks like this:
1 NP
2 P
3 NP
4 VP
In the next set of nested loops, we build new constituents out of existing ones. Each time we execute the innermost loop, we are looking at two potential daughters and seeing if they form a constituent. If they do, we add that constituent to the appropriate place in the chart. The loops have these variables:
In the final iteration, weโre building constituents of length 4, so span will be 4. begin can just be 1. end can only be 4. m can range from 1 to 3.
we look at 1,1 (NP) and 2,4 (PP) and add NP to 1, we look at 1,2 and donโt find anything we look at 1,3 (NP) and 4,4 (VP) and add S to 1,
We end up with this table:
1 NP NP NP, S
2 P PP PP
3 NP S
4 VP
