I have set up a script to automatically test  your assignment 2
submissions. It only really tests the tmi and ll3 portions of your 
assignment of course, since you all used different measures for 
user2 and user3. 

You should get an automatically generated message in the near
future that shows the results of my script running your program on
my test cases. The output will show the test case, your answer,
and the proposed "correct" answer. Hopefully it makes sense, and
you can see how your programs did. If your program did not run 
according to specifications then you will see whatever error 
messages NSP might have generated. 

A few words on the test cases:

For the 2x2 case I chose the following (pictured as
a 2x2 table, not in NSP format)

     W2  -W2
 W1   2   1  | 3
-W1   2  95  |97
     ---------
      4  96   100

(TMI = .0742)
(LL = 10.2850)

The above is just a general purpose table that looks alot
like "typical" bigram data. What I mean by that is that it's
rather messy, W1 and W2 occur together, but they also occur
separately. The number of times they occur together is dwarfed
by the same size. In real data of course the sample size would
be much bigger, and the skewness of the contingency table
would be even more so.

  1   0  | 1
  0  99  |99
---------
  1  99   100

(TMI = .0808)
(LL = 11.2003)

 10   0  |10
  0  90  |90
---------
 10  90   100

(TMI = .4690)
(LL = 65.0166) 

These two cases are rather extreme, and generally speaking I
think we'd agree that the second case (where the bigram
occurs 10 times) would be a "stronger" bigram than the first
case (where it only occurs 1 time). You'll note that the TMI 
and LL scores reflect this. 

I will check and see how user2 (your proposed measure) 
differentiates between these two. Certainly they should be scored  
differently, and I think it would be hard to argue that the
first case (of the bigram occurring 1 time) would deserve
a higher score than the second. I am always willing to hear
arguments on these points of course. 

Here are the test cases for the 2x2x2 case. (I hope the tables 
make a bit of sense):

      W3  W3  -W3 -W3
      W2 -W2   W2 -W2
W1     1   0    0   0
-W1    0   0    0  99

(LL3 = 22.4006)

      W3  W3  -W3 -W3
      W2 -W2   W2 -W2
W1    10   0    0   0
-W1    0   0    0  90

(LL3 = 130.0332)

These are similar to the cases for 2x2, where the first
has a bit less evidence in favor of dependence, and note that
the ll3 scores reflect this point. Again, I will be looking
at how your user3 handles this same situation. 

Ted Pedersen
October 22, 2002