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% (the others are not needed to compute the AP) 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 Write A as product of 3 x 2 and 2 x 5 ... what does this say about rank(A)
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.
TASK 2 (Encodings)
c(1) = 1, c(2) = 01, c(3) = 001, c(4) = 0001, ...
2.1 Decode 110000100111001 + why prefix-free
Integer sequence encoded: 1, 1, 5, 3, 1, 1, 3
All valid codes end in 1. Any non-empty prefix of any of the codes ends with a zero. Therefore, no prefix of any of the code is a valid code.
2.2 Function decode
def decode(bits):
x = 0
result = []
for bit in bits:
x += 1
if bit == 1:
result.append(x)
x = 0
return result if x == 0 else []
2.3 Random code, 0 or 1 with probability 1/2 each until 1
The code for x has exactly x bits. The probability of generating exactly those x bits with the process described is 1/2 * ... * 1/2 (x times), that is, 2 to the -x.
2.4 Entropy-optimality for this probability distribution
Let L_i be the length of the code for integer i. Then L_i = i.
The probability p_i of generating that code is 2^-i (according to Task 2.3).
For entropy-optimality, we need L_i <= log_2 (1/p_i) + 1 = i + 1. This is true.
TASK 3 ()
$(document).ready(function() {
$("#query").keyup(function() {
var query = $("#query").val();
$.get("http://www.gugel.de:8888/?q="+ query, function(result) {
$("#result").html(result);
})
})
})
3.1 Simple HTML that makes sense for this JavaScript (without head part)
<html>
<body>
<input id="query"/>
<div id="result"></div>
</body>
</html>
3.2 First fifteen characters of GET requests when typing "hey" + to which machine
GET /?q=h HTTP/ GET /?q=he HTTP GET /?q=hey HTT
The requests are sent to the machine with the address www.gugel.de
3.3 Number of valid UTF-8 sequences of length 2
Valid UTF-8 sequences of length 2 look like this
110xxxxx 10yyyyyy
where the 11-bit codepoint xxxxxyyyyyy encodes an integer that must be > 127 (otherwise a one-byte UTF-8 must be used).
The number of such codepoints is 2048 - 128 = 1920.
3.4 UTF-8 code and URL-escaped code of ΓΌ
ISO-8859-1 code in binary: 11111100 With padding, so that 11 bits: 00011111100 As UTF-8 code: 11000011 10111100 (Byte 1: 110 + first five bits of codepoint, Byte 2: 10 + last five bits) In decimal: 195 188 In hexadecimal: C3 BC URL-escaped: %C3%BC
TASK 4 (Perceptrons)
Training set: {1, 2, 3, 4} (negative), {5, 6, 7, 8} (positive)
4.1 First round of Perceptron algorithm
x = 1: w * x - b = 0 --> update: w = 0 - 1 = -1, b = 0 + 1 = 1 x = 2: w * x - b = -3 --> no update x = 3: w * x - b = -4 --> no update x = 4: w * x - b = -5 --> no update x = 5: w * x - b = -6 --> update: w = -1 + 5 = 4, b = 1 - 1 = 0 x = 6: w * x - b = 24 --> no update x = 7: w * x - b = 28 --> no update x = 8: w * x - b = 32 --> no update
4.2 Function perceptron
def perceptron():
(w, b) = (0, 0)
for round in range(50):
for x in [1, 2, 3, 4, 5, 6, 7, 8]:
if x <= 4 and w * x - b >= 0:
(w, b) = (w - x, b + 1)
if x >= 5 and w * x - b <= 0:
(w, b) = (w + x, b - 1)
return (w, b)
4.3 Prove ww* = - w* and bb* = -b when training labels swapped
Let wi, bi and wwi, bbi as in the hint. Then we can prove by induction that wwi = -wi and bbi = -bi:
In the beginning, all values are zero, so correct.
Assume that wwi = -wi and bbi = -bi after iteration i and look at what happens in iteration i+1. There are four cases:
1. x negative in original, no update --> x positive in variant, no update 2. x positive in original, no update --> x negative in variant, no update 3. x negative in original, update (wi+1, bi+1) = (wi - x, bi + 1) --> x positive in variant, update (wwi+1, bbi+1) = (wwi + x, bbi - 1) 4. x positive in original, update (wi+1, bi+1) = (wi + x, bi - 1) --> x negative in variant, update (wwi+1, bbi+1) = (wwi - x, bbi + 1)