Hey Everyone,

Please post your solutions to the Oct 17 In-class Exercise to his thread.

Best,

Chris

ScoreBM25(q, d) = 5* log( 64) * (k1+1) * 1/( 1+ k1) = 5* 6 = 30

if l_d = 2 * l_average

ScoreBM25(q, d) = 5 * log (64) * (k1+1) * 1/ ( 1+ k1* (1+b)) = 30 * 2.2 / ( 1+ 1.2*1.75) = 21.3

for term1 of query = log(N/Nt) * (1*(k+1))/1+k1(1-b+b(ld/lavg)) = log(N/Nt) * (K+1)/(K1+1) = 6

for 5 docs = 6 * 5 = 30 Nt = N/64

second part = 5 * 6 * (k1+1)/(k1+1+k1b) = 300.75 = 21.3

IDF = log(64/1)=6

TF = 5*(k1+1)/(1+k1)=5

TF-IDF=30

TF= 5*(k1+1)/(1+k1*(1-b+2b))

TF-IDF=30(k1+1)/(k1+k1b+1) =30(1.2+1)/(1.2+1.20.75+1) = 21.29

N/N_t = 64, k = 1.2, b = 0.75

mbox{Score_BM25} (q, d) = log_2 64 * (5 * (1.2+1)/(1.2+1)) = 6 * 5 * (22)/(22) = 30

mbox{Score_BM25} (q, d_2) = log_2 64 * (5 * (1.2+1)/(1+1.75*1.2)) = 6 * 5 * (22)/(31) = 21.29

1. Score_BM25(q, d) = 5 * log(64) * (1 * (1.2+1)) / (1 + 1.2((1-0.75) + 0.75*1)) = 5 * 6 * 2.2/2.2 = 30
2. Score_BM25(q, d) = 5 * log(64) * (1 * (1.2+1)) / (1 + 1.2((1-0.75) + 0.75*2)) = 5 * 6 * 2.2/(1 + 1.2 * 1.75) = 30 * 2.2/(3.1) = 21.29

1.  log(N/Nt) = log(64) = 6
       ScoreBM25(q,d)  = 6*5 = 30

1. ScoreBM25(q,d) = (65((k1+1)/(1+k1(1+b))) = 6(5*(2.2/3.1)) = 21.29

a) IDF = log(N/(N/64)) = log 64 = 6; TF = ((1 * (k1 + 1))/ (1 + (k1 * (1-b + b( 1))))) = 1 score = 5 terms * (IDF*TF) = 5 * 6 *1 = 30

b) score = 5 * 6 * (2.2 */ (2.2+0.9)) = 5 * 6 * 0.71 = 21.29