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- Subject
- Posted on
- Diff for words in string
- 06-21-2006
- Csaba Gabor
June 21, 2006, 9:56 am
quite different or quite similar. In the case where they are similar,
I would like to display the more recent one and say something like:
Word 2 added [before word 2 in original]: "Jack be nimble"
Words 10-11 changed to: "the quick brown fox"
[from words 9-11 in original]: "the brown fast quick fox"
Words [20-22 in original] before word 20 removed: "sat in a corner on"
One way to do this is to replace all spacing chars with \n, write the
strings to two files, and then run diff (FC on my Win XP Pro = file
compare), collecting the output. Does anyone happen to have PHP code
for this where I don't have to write files? Note that the diff is on
words of the strings and not characters.
In particular, the normal algorithms for this (longest common
subsequence) only produce a number, and don't note the differences.
Also, I wanted to give a threshhold (about 10, say) and if the longest
common subsequence differs from the shorter string (strings are on the
order of 100 words) by at least this amount, then simply fail (since
the difference between the two strings would be deemed to great). This
should make the algorithm far more efficient. The corresponding
argument in FC would be /LB10.
Thanks,
Csaba Gabor from Vienna
Ref: http://www.ics.uci.edu/~dan/class/161/notes/6/Dynamic.html
http://en.wikipedia.org/wiki/Longest-common_subsequence_problem
Note that this problem is distinct from the longest common substring
problem.
Re: Diff for words in string
http://pear.php.net/package/Text_Diff/
The code in that PEAR library may give you a starting point to
implementing it in PHP rather than calling fc.
Cheers,
Andy
--
Andy Jeffries MBCS CITP ZCE | gPHPEdit Lead Developer
http://www.gphpedit.org | PHP editor for Gnome 2
http://www.andyjeffries.co.uk | Personal site and photos
Re: Diff for words in string
Thanks both Andy and Jeff for your suggestions regarding diff / longest
common subsequence. It got my creative juices flowing and I rolled my
own, which I didn't see in the literature. There may be some overlap
with Ukkonen (see
http://www.csse.monash.edu.au/~lloyd/tildeAlgDS/Dynamic/Edit /) since
the complexity appears to have similar overtones. The running time for
sequences which are highly similar should be linear in the length of
the strings. The one thing that is missing is an efficient
dissimilarity test so that one may bail out of the routine early.
Step 1: For each word in the first string, have an array of where it
is in the first string
Step 2: From this it is possible to construct what is known as a
permutation graph:
Start $ctr at 0; for each word in string 2 find the array of indeces
in string 1 and (iterating backwards through that array), unshift
++$ctr onto an array at that index point.
The array of arrays built up in this step should now be collapsed,
forming a permuation of the integers from 1 ... $ctr. The
corresponding indeces are assumed to be numbered consecutively starting
at 1 for string 2. This constitutes a permutation graph (a permutation
graph is one where the corresponding integers are connected, which
connections correspond to vertices; and vertices are adjacent iff their
corresponding segments intersect. This happens iff the order of the
two integers corresponding to the two segments differs above vs.
below).
The relevance to the problem is that any segment from the graph
indicates a common word in the original sequences. Two segments may be
in a common subsequence as long as they don't intersect.
Step 3:
Thus, the problem is reduced to finding the longest increasing
monotonic subsequence of the integers constructed in Step 2. I do this
by dynamic programming, working backwards through the sequence of
integers. For each position, I keep track of the longest monotonic
increasing subsequence (rightwards) from that point. I also keep track
of the maximum longest monotonic increasing subsequence (rightwards)
from that point as a speedup convenience.
Step 4:
Working from the left, reconstruct the sequence (longest increasing
monotonic subsequence).
Step 5:
Recover the mapping that the sequence from Step 4 represents (along
with the indexes into the original strings for each word). This is the
longest common subsequence between the two original strings.
Step 6.
Using Step 5, we can also determine which words from string 1 didn't
make it and which words from string 2 were not included, so we can
easily construct the diff where the words from string one that didn't
make it will have a strikethrough and the words from string 2 will be
bolded.
Csaba Gabor from Vienna
common subsequence. It got my creative juices flowing and I rolled my
own, which I didn't see in the literature. There may be some overlap
with Ukkonen (see
http://www.csse.monash.edu.au/~lloyd/tildeAlgDS/Dynamic/Edit /) since
the complexity appears to have similar overtones. The running time for
sequences which are highly similar should be linear in the length of
the strings. The one thing that is missing is an efficient
dissimilarity test so that one may bail out of the routine early.
Step 1: For each word in the first string, have an array of where it
is in the first string
Step 2: From this it is possible to construct what is known as a
permutation graph:
Start $ctr at 0; for each word in string 2 find the array of indeces
in string 1 and (iterating backwards through that array), unshift
++$ctr onto an array at that index point.
The array of arrays built up in this step should now be collapsed,
forming a permuation of the integers from 1 ... $ctr. The
corresponding indeces are assumed to be numbered consecutively starting
at 1 for string 2. This constitutes a permutation graph (a permutation
graph is one where the corresponding integers are connected, which
connections correspond to vertices; and vertices are adjacent iff their
corresponding segments intersect. This happens iff the order of the
two integers corresponding to the two segments differs above vs.
below).
The relevance to the problem is that any segment from the graph
indicates a common word in the original sequences. Two segments may be
in a common subsequence as long as they don't intersect.
Step 3:
Thus, the problem is reduced to finding the longest increasing
monotonic subsequence of the integers constructed in Step 2. I do this
by dynamic programming, working backwards through the sequence of
integers. For each position, I keep track of the longest monotonic
increasing subsequence (rightwards) from that point. I also keep track
of the maximum longest monotonic increasing subsequence (rightwards)
from that point as a speedup convenience.
Step 4:
Working from the left, reconstruct the sequence (longest increasing
monotonic subsequence).
Step 5:
Recover the mapping that the sequence from Step 4 represents (along
with the indexes into the original strings for each word). This is the
longest common subsequence between the two original strings.
Step 6.
Using Step 5, we can also determine which words from string 1 didn't
make it and which words from string 2 were not included, so we can
easily construct the diff where the words from string one that didn't
make it will have a strikethrough and the words from string 2 will be
bolded.
Csaba Gabor from Vienna
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