USAGE

 combig.pl [OPTIONS] BIGRAM

INPUT PARAMETERS

• BIGRAM

  Specify a file of bigram counts created by NSP programs count.pl. The entries in BIGRAM will be formatted as follows:

          word1<>word2<>n11 n1p np1
      

  Here, word1 is followed by word2 n11 times. word1 occurs as the 1st word in total n1p bigrams and word2 occurs as the 2nd word in np1 bigrams.

• OPTIONS

   --help
      

  Displays this message.

   --version
      

  Displays the version information.

OUTPUT

combig.pl produces a count of the number of times two words make up a bigram in either order, whereas count.pl produces counts for a single fixed ordering. In other words, combig.pl combines the counts of bigrams that are composed of the same words but in reverse order. While the BIGRAM shows pairs of words forming bigrams, output of combig will show the pairs of words that are co-occurrences or that co-occur irrespective of their order.

e.g. if bigrams

        word1<>word2<>n11 n1p np1 
and 
        word2<>word1<>m11 m1p mp1

are found in BIGRAM file, then combig.pl treats these as a single unordered bigram

        word1<>word2<>n11+m11 n1p+mp1 np1+m1p

where the new bigram will show a combined contingency table in which the order of words doesn't matter.

                        word2           ~word2
        ___________________________________________________________
         word1  |       n11+m11         n12+m21     | n1p+mp1 
                |                                   |
        ~word1  |       n21+m12         n22+m22-n   | n2p+mp2-n
                ___________________________________________________
                        np1+m1p         np2+m2p-n   |  n
here the entry

• (word1,word2)=n11+m11

  shows the number of bigrams having both word1 and word2 in either order

  i.e. word1<>word2 + word2<>word1

• (word1,~word2)=n12+m21

  shows the number of bigrams having word1 but not word2 at either position

  i.e. word1<>~word2 + ~word2<>word1

• (~word1,word2)=n21+m12

  shows the number of bigrams having word2 but not word1 at either position

  i.e. ~word1<>word2 + word2<>~word1

• (~word1,~word2)=n22+m22

  shows the number of bigrams not having word1 nor word2 at either position

  i.e. ~word1<>~word2 + ~word2<>~word1 - n

  where n=total number of bigrams

  The mathematical proof of how the cell counts in the above contingency table are counted is explained in section Proof.

When a bigram appears in only one order i.e.

word1<>word2<>n11 n1p np1

appears but

word2<>word1<>m11 m1p mp1

does not, then the combined bigram will be same as the original bigram that appears. Or in other words,

word1<>word2<>n11 n1p np1

is displayed as it is.

PROOF OF CORRECTNESS

                  word2         ~word2
                --------------------------------------
        word1   n11     |       n12     |       n1p     
                        |               |
        ~word1  n21     |       n22     |       n2p
                --------------------------------------
                np1     |       np2     |       n

while a bigram word2<>word1<>m11 m1p mp1 represents a contingency table

                  word1         ~word1
                --------------------------------------  
        word2   m11     |       m12     |       m1p
                        |               |
        ~word2  m21     |       m22     |       m2p
                --------------------------------------
                mp1     |       mp2     |       n

Here,

 n11+n12+n21+n22 = n

Also,

 m11+m12+m21+m22 = n

combig.pl combines bigram counts into a single order independant word pair

 word1<>word2<>n11+m11 n12+m21 n21+m12

And the corresponding contingency table will be shown as

                        word2           ~word2  
                -----------------------------------------
        word1   n11+m11   |     n12+m21   |     n1p+mp1
                          |               |
        ~word1  n21+m12   |     n22+m22-n |     n2p+mp2 
                -----------------------------------------
                np1+m1p   |     np2+m2p   |     n

The first cell (n11+m11) shows the #bigrams having word1 and word2 (irrespective of their positions) i.e. word1<>word2 or word2<>word1 which is n11+m11.

The second cell (n12+m21) shows the #bigrams having word1 but not word2 at any position i.e. word1<>~word2 or ~word2<>word1 which is n12+m21.

The third cell (n21+m12) shows the #bigrams having word2 but not word1 at any position i.e. ~word1<>word2 or word2<>~word1 which is n21+m12.

The fourth cell (m22+n22-n) shows the #bigrams not having word1 nor word2 at any position which

 = n - (n11+m11) - (n12+m21) - (n21+m12)

 = n - (n11+n12+n21) - (m11+m12+m21)

 = n - (n-n22) - (n-m22)

 = n22 + m22 - n

Alternative proof -

 n22 = m11 + m12 + m21 + X      (a)

 m22 = n11 + n12 + n21 + X      (b)

where X = #bigrams not having either word1 or word2.

as both n22 and m22 have some terms in common which show the bigrams not having either word1 or word2. But,

 m11+m12+m21 = n - m22

Substituting this in eqn (a)

 n22 = n - m22 + X

Or

 X = n22 + m22 - n

Or add (a) and (b) to get

 n22+m22 = (n11+m11) + (n12+m21) + (n21+m12) + 2X

rearranging terms,

 n22+m22 = (n11+n12+n21) + (m11+m12+m21) + 2X

but

 n11+n12+n21 = n - n22 and 

 m11+m12+m21 = n - m22

Hence,

 n22+m22 = (n-n22) + (n-m22) + 2X

 2(n22+m22-n) = 2X

Or

 (n22+m22-n) = X

which is the fourth cell count.

Viewing Bigrams as Graphs

In this case,

n11, which is the number of times bigram word1<>word2 occurs, becomes the weight of the directed edge word1->word2.

n1p, which is the number of bigrams having word1 at 1st position, becomes the out degree of vertex word1

and

np1, which is the number of bigrams having word2 at 2nd position, becomes the in degree of vertex word2

combig.pl creates a new list of word pairs from these bigrams such that the order of words can be ignored. Viewed another way, it converts the directed graph of given bigrams to an undirected graph showing new word pairs.

A pair say

        word1<>word2<>n11 n1p np1

shown in the output of combig can be viewed as an undirected edge joining word1 and word2 having weight n11. If we count the degree of vertex word1 it will be n1p and degree of vertex word2 will be np1.