I'm supposed to take two sentences each time and compute if they are similar. By similar I mean, both syntactically and semantically.
INPUT1: Obama signs the law. A new law is signed by Obama.
INPUT2: A Bus is stopped here. A vehicle stops here.
INPUT3: Fire in NY. NY is burnt down.
INPUT4: Fire in NY. 50 died in NY fire.
I don't want to use ontology tree as a soul. I wrote a code to compute Levenshtein distance (LD) between sentences and then decide if the 2nd sentence:
- can be ignored (INPUT1 and 2),
- should replace the first sentence (INPUT 3), or
- store along with the first sentence (INPUT4).
I'm not happy with the code as LD only computes syntactical level (what other methods ?). How can semantic be incorporated (like bus is sort of a vehicle?) .
The code goes here:
%# As the difference is computed, a decision is made on the new event
%# (string 2) to be ignored, to replace existing event (string 1) or to be
%# stored separately. The higher the LD metric, the higher the difference
%# between two strings. Of course, lower difference indices either identical
%# or similar events. However, the higher difference indicates the new event
%# as a fresh event.
%#.........................................................................
%# Calculating the LD between two strings of events.
%#.........................................................................
L1=length(str1)+1;
L2=length(str2)+1;
L=zeros(L1,L2); %# Initializing the new length.
g=+1; %# just constant
m=+0; %# match is cheaper, we seek to minimize
d=+1; %# not-a-match is more costly.
% do BC's
L(:,1)=([0:L1-1]*g)';
L(1,:)=[0:L2-1]*g;
m4=0; %# loop invariant
%# Calculating required edits.
for idx=2:L1;
for idy=2:L2
if(str1(idx-1)==str2(idy-1))
score=m;
else
score=d;
end
m1=L(idx-1,idy-1) + score;
m2=L(idx-1,idy) + g;
m3=L(idx,idy-1) + g;
L(idx,idy)=min(m1,min(m2,m3)); % only minimum edits allowed.
end
end
%# The LD between two strings.
D=L(L1,L2);
%#....................................................................
%# Making decision on what to do with the new event (string 2).
%#...................................................................
if (D<=4) %# Distance is so less that string 2 seems identical to string 1.
store=str1; %# Hence string 2 is ignored. String 1 remains stored.
elseif (D>=5 && D<=15) %# Distance is larger to be identical but not enough to
%# make string 2 an individual event.
store= str2; %# String 2 is somewhat similar to string 1.
%# So, string 1 is replaced with string 2 and stored.
else
%# For all other distances, string 2 is stored along with string 1.
store={str1; str2};
end
Any help is appreciated.