TASK 1 (Ranking and evaluation)
1.1 Term-document matrix
The idf for first and second is log_2(4/1) = 2, the idf for the other words is log_2 (4/2) = 1. Hence
first 2 0 0 0
second 0 2 0 0
nice 1 0 1 0
document 1 1 0 0
text 0 0 1 1
more 0 0 1 1
1.2 Function for conversion from tf to tf.idf
1. def tf_to_tf_idf(A, m, n): 2. for i in range(m): 3. df = 0 4. for j in range(n): 5. if A[i, j] > 0: 6. df += 1 7. for j in range(n): 8. A[i, j] *= log2(n/df)
1.3 Find query with P@2 and AP = 50%
Query: second nice document -> scores: D1 = 2, D2 = 3, D3 = 1, D4 = 0. Order: D2, D1, D3, D4 Relevant: D1 and D4 -> then P@2 and P@4 are 50%, and hence also AP
TASK 2 (Encodings)
2.1 Encoding for x = 1, ...,10
1 = 100. 2 = 101.0 3 = 101.1 4 = 110.00 5 = 110.01 6 = 110.10 7 = 110.11 8 = 111.000 9 = 111.001 10 = 111.010
2.2 Function for code for x < 16
1. def code(x): 2. result = [1] 3. length = floor(log2(x)) 4. result.extend([length // 2, length % 2]) 5. if x >= 2 and x < 4: result.append(x % 2) 7. if x >= 4 and x < 8: result.extend([(x & 2) >> 1, x % 2]) 8. if x >= 8 and x < 16: result.extend([(x & 4) >> 2, (x & 2) >> 1, x % 2]) 9. return result
2.3 Formula for code length for arbitrary x
Length of Golomb part: floor(floor(log_2 x) / 4) + 3 Length of binary part: floor(log_2 x) Sum of the two: floor(floor(log_2 x) / 4) + 3 + floor(log_2 x)
TASK 3 (Web applications and UTF-8)
3.1 Write HTML
<html> <head><script src="convert.js"></script></head> <body>Centimetres: <input id="cm"/>Inches: <input id="in"/></body> </html>
3.2 Write JavaScript
$(document).ready(function(){
$("#cm").keyup(function(){ $("#in").val($("#cm).val() / 2.54)); })
$("#in").keyup(function(){ $("#cm").val($("#in).val() * 2.54)); })
})
3.3 UTF-8 code for Euro sign
32 in binary is 0010.0000, 172 in binary is 1010.1100 (128 + 32 + 8 + 4) The code point in binary is hence: 0010.0000.1010.1100 We hence need a 3-byte code of the form 1110 xxxx 10 yyyyyy 10 zzzzzz (with a 16-bit code point xxxxyyyyyyzzzzzz) The 3-byte UTF-8 code is hence: 1110 0010 10 000010 10 101100
3.4 Function for counting #characters in UTF-8 sequence
1. def count_utf8_chars(bytes): 2. count = 0 3. for byte in bytes: 4. # Count all except the "follow bytes" of the form "10......". 5. if (byte & (128 + 64) != 128): 6. count += 1 7. return count
TASK 4 (Naive Bayes and k-means)
4.1 Steps of k-means
Points: P1 = (3, 1), P2 = (1, 2), P3 = (1, 4) Initial centroids: C1 = (3, 0), C2 = (0, 3) Step 1a: C1 <- P1, C2 <- P2, P3 Step 1b: C1 = (3, 1), C2 = (1, 3) Step 2a: C1 <- P1, C2 <- P2, P3 Step 2b: C1 = (3, 1), C2 = (1, 3) Centroid identical to those from previous round -> STOP
4.2 Compute centroids from A and P
1. def centroids(A, P): 2. return A.normalizecolumnsL1().transpose() * P
4.3 Determine w and b of Naive Bayes
Learned probabilities: p_1X = 3/4, p_2X = 1/4, p_1Y = 1/4, p_2Y = 3/4, p_X = 1/3, p_Y = 2/3 Linear classifier: w = (log_2 (p_1X/p_1Y), log_2 (p_2X/p_2Y)) = (log_2 3, -log_2 3) Linear classifier: b = - log_2 (p_X/p_Y)
4.4 Example such that Naive Bayes decides 2x > y
Solution for x > y:
Choose P1 = (2, 1) --> X and P2 = (1, 2) --> Y Then p_1X = 2/3, p_2X = 1/3, p_1Y = 1/3, p_2Y = 2/3 Then w = (1, -1) and b = 0 Then (w1, w2) * (x, y) - b > 0 <=> x > y.
TASK 5 (Latent Semantic Indexing)
5.1 Show that V is row-orthonormal
Product without factor 1/4 * sqrt(2) is: [[8, 0], [0, 8]] 1/4 * sqrt(2) * 1/4 * sqrt(2) = 1/16 * 2 = 1/8 This gives indeed the identity matrix
5.2 Compute missing S
A * At = [[40, 24], [24, 64]] A * At * [1, 1]t = [64, 64]t -> Eigenvalue 64 A * At * [1, -11]t = [16, -16]t -> Eigenvalue 16 The singular values are hence 8 and 4 S is hence diag(8, 4)
5.3 Rank of A and making it rank 3
A has two independent columns -> rank >= 2 A has two rows -> rank <= 2 Since the number of rows is an upper bound, adding another column cannot make the rank 3
5.4 Function for L2-normalization of a vector
1. def normalize(vector): 2. sum = 0 3. for x in vector: 4. sum += x * x 5. for x in vector: 6. x = x / sqrt(sum)
TASK 6 (Miscellaneous)
6.1 Total number of items in a 3-gram index
Each 3-gram of each word contributed one item. A word v has |v| + 2 3-grams. Hence the total number of items is sum_i |v_i| + 2 * n.
6.2 SQL query for persons who founded company in city they were born in
SELECT founded_by.person FROM founded_by, based_in, born_in WHERE founded_by.person = born_in.person AND founded_by.company = based_in.company AND based_in.city = born_in.city
6.3 Maximum entropy of distribution given by p1, ..., pn
Entropy is sum_i -p_i * log2 p_i Optimize sum_i -p_i * ln p_i (the constant factor log / ln does not change the position of the optimum) First derivative: -1 - ln p_i + lambda = 0 => all p_i equal and hence 1/n Second derivative: -1/p_i < 0 => MAX Resulting entropy: sum_i -1/n * log2 (1/n) = log2 n