TASK 1 (Ranking and evaluation)
1.1 P@2, P@R, AP
P@2 = 1/2 P@R = 1/3 P@R1 = 1/2, P@R2 = 1/2, P@R3 = 0, AP = (1/2 + 1/2 + 0) / 3 = 1/3
1.2 DCG@2, iDCG@2, nDCG@2
DCG@2 = 0 + 3 / 1 = 3 iDCG@2 = 3 + 2 / 1 = 5 nDCG@2 = 3/5 = 60%
1.3 Function dcg that computes DCG@k
1. def dcg(rels, k): 2. sum = rels[0] 3. for in in range(1, len(rels)): 4. sum += rels[i] / log2(i) 5. return sum
1.4 Function ndcg that computes nDCG@k
1. def ndcg(rels, k): 2. dcg = dcg(rels, k) 3. sorted_rels = sort(rels, DESC) 4. idcg = dcg(rels, k) 5. return dcg / idcg
TASK 2 (Encodings, Niklas)
2.1 Function unary_encode that encodes given sequence of integers
1. def unary_encode(input): 2. result = [] 3. for x in input: 4. for i in range(0, x-1): 5. result.add(0) 6. result.add(1) 7. return result
2.2 Expected number of bits in sequence of n integers i.i.d from {1, ..., n}
1. The code length for integer i is exactly i
2. For a random integer from {1, ..., n}, the expected code length is sum_i (i * 1/n) = 1/n * sum_i i = 1/n * 1/2 * n * (n+1) = (n+1) / 2
3. The expected number of bits in the whole sequence is hence n * (n+1) / 2
2.3 Is this encoding entropy-optimal for the distribution from task 2.2
1. The entropy H of the distribution from 2.2 is -sum_i 1/n * log_2 (1/n) = 1/n * log_2 n * sum_i 1 = log_2 n 2. The expected code length is (n+1)/2 3. For entropy-optimality, we should have expected code length = entropy + O(1) 4. So the answer is NO.
2.4 Name a prefix-free entropy-optimal encoding for the distribution from task 2.2
1. Just encode each integer with a fixed number of ceil(log_2 n) bits 2. This is prefix-free, because for two different codes of the same length, none can be the prefix of the other 3. Since ceil(log_2 n) <= log_2 + 1, this encoding is entropy-optimal
TASK 3 (Fuzzy search and UTF-8, Axel)
3.1 (5 points) ED(x, y), ED(y, x), PED(x, y), PED(y, x) for x = cute and y = computer
ED(cute, computer) = 4 ... insert omp and r ED(computer, cute) = 4 ... ED is symmetric PED(cute, computer) = ED(cute, compute) = 3 PED(computer, cute) = ED(computer, cute) = 4
3.2 (5 points) Find x and y with ED(x, y) = 2 and comm($$x$$, $$y$$) = max{|x|, |y|} - 4
1. max{|x|, |y|} - 4 is the lower bound from the lecture for q = 3 and ED(x, y) = 2
2. The lower bound is achieved when destroying the maximal number of q-grams with each edit distance operation, in this case 3
3. For example, x = ABC, y = DBE
3.3 (10 points) Function valid_utf8_char8 that checks if a given array of three bytes is valid UTF-8 for a single character
1. def valid_utf8_char(bytes):
# Check that first byte start with 1110 and the other two with 10
# 10000000b = 128, 11000000b = 192, 11100000b = 224, 11110000b = 240
3. if (bytes[0] & 240 != 224 OR bytes[1] & 192 != 128 OR bytes[2] & 192 != 192):
4. return False
# Check that the code point is >= 2048
# 11100000b = 224, , 10100000b = 160
5. if (bytes[0] > 224 OR bytes[1] >= 160):
6. return True
7. return False
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