#acl Björn Buchhold:read,write Elmar Haussmann:read,write Claudius Korzen:read,write Florian Bäurle:read,write Jonas Sternisko:read,write All:read

== TASK 1 (Ranking and evaluation) ==

{{{
bla   1  1  2  2  3
bli   1  0  1  2  2
blu   0  1  1  0  1
}}}

=== 1.1 AP for query "bla bli" ===

{{{
The vector for "bla blu" is simply (1, 0, 1)
Dot product similarities for D1, ..., D5: 1, 2, 3, 2, 4
Documents ordered by these similarities: D5, D3, D2 and D4, D1   (D2 and D4 have the same score)
Bit array of relevances: 1, 0, 0, 0, 1
Precisions: P@1 = 100%, P@5 = 40%
Average precision (AP): 70%
}}}

=== 1.2 Function compute_ap ===

{{{
def compute_ap(relevances):
    prec_sum = 0
    num_rel = 0
    for i in range(len(relevances)):
        if relevances[i] == 1:
            num_rel += 1
            prec = num_rel / (i + 1)
            prec_sum += prec
    return prec_sum / num_rel
}}}

=== 1.3 Prove that AP = 100% if and only if all 1s at the front ===

If all 1s at the front, then each prec in the code above is 100%, and so will be the average.

If AP = 100%, then the last prec must be 100%, which can only happen if there is no 0 before the last 1.

=== 1.4 

{{{
1  1  2  2  3       1  1
1  0  1  2  2   =   1  0  *  1  0  1  2  2 
0  1  1  0  1       0  1     0  1  1  0  1
}}}

This shows that each column of A can be written as a linear combination of the two columns of the 3 x 2 matrix on the right hand side. This proves that rank(A) <= 2.