1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
|
section
h1 Why understanding of basic math is imporant for computer programing
p September 21, 2018, rev2
a(href="https://twitter.com/mitjafelicijan") @mitjafelicijan
section.center
q We Cannot Solve Our Problems With The Same Thinking We Used When We Created Them.
footer — Albert Einstein
section
h2 How we usually find solutions and why this is problematic?
ul
li We search for code example instead of algorithms.
li We copy and paste and do testing on trial&error principle.
li We don't take enough time to properly understand problem we a re trying to solve.
li Brute force solutions we make are usually not optimized
section
h2 Levenshtein distance
p The Levenshtein distance is a string metric for measuring difference between two sequences. Informally, the Levenshtein distance between two words is the minimum number of single-character edits (i.e. insertions, deletions or substitutions) required to change one word into the other.
hr
div.center
img(src="levenshtein-distance.svg")
hr
pre
code.language-python
| def levenshtein(seq1, seq2):
| oneago = None
| thisrow = range(1, len(seq2) + 1) + [0]
| for x in xrange(len(seq1)):
| twoago, oneago, thisrow = oneago, thisrow, [0] * len(seq2) + [x + 1]
| for y in xrange(len(seq2)):
| delcost = oneago[y] + 1
| addcost = thisrow[y - 1] + 1
| subcost = oneago[y - 1] + (seq1[x] != seq2[y])
| thisrow[y] = min(delcost, addcost, subcost)
| return thisrow[len(seq2) - 1]
hr
h4 Going further
ol
li https://en.wikibooks.org/wiki/Algorithm_Implementation/Strings/Levenshtein_distance
li https://en.wikipedia.org/wiki/Levenshtein_distance
li https://rosettacode.org/wiki/Levenshtein_distance
section
h2 Basic linear algebra
pre
code.language-css
| body {
| background: black;
| }
pre
code.language-javascript
| $(document).ready(function() {
| $('pre code').each(function(i, block) {
| hljs.highlightBlock(block);
| });
| });
hr
figcaption.right Step 1: Finding nearest point
div.center
img(src="drawing1.svg")
hr
figcaption.right Step 1: Finding nearest point
$$ \large{ \mathbb{R}^2 ∈ \vec{a} \bar{a} } $$
$$ \large{ e^{i\pi} + 1 = 0 } $$
$$ \large{ x = {-b \pm \sqrt{b^2-4ac} \over 2a} } $$
hr
figcaption.right Step 2: Finding nearest point
pre
code.language-python
| fruits = ["apple", "banana", "cherry"]
| for x in fruits:
| if x == "banana":
| break
| print(x)
hr
figcaption.right Finding nearest point
$$ \large{ \mathbb{R}^2 ∈ \vec{a} \bar{a} } $$
$$ \large{ e^{i\pi} + 1 = 0 } $$
$$ \large{ x = {-b \pm \sqrt{b^2-4ac} \over 2a} } $$
hr
figcaption.right Finding nearest point
pre
code.language-c
| #include <stdio.h>
| int main () {
| for( ; ; ) {
| printf("This loop will run forever.\n");
| }
| return 0;
| }
hr
figcaption.right Finding nearest point
pre
code.language-python
| fruits = ["apple", "banana", "cherry"]
| for x in fruits:
| if x == "banana":
| break
| print(x)
hr
figcaption.right Finding nearest point
pre
code.language-sql
| SELECT `CustomerName`, `City` FROM `Customers`;
hr
figcaption.right Finding nearest point
pre
code.language-go
| package main
| import "fmt"
| func main() {
| sum := 0
| for i := 0; i < 10; i++ {
| sum += i
| }
| fmt.Println(sum)
| }
hr
figcaption.right Finding nearest point
pre
code.language-javascript
| $(document).ready(function() {
| $('pre code').each(function(i, block) {
| hljs.highlightBlock(block);
| });
| });
hr
figcaption.right Finding nearest point
pre
code.language-css
| body {
| background: black;
| }
section
h3 Grid example
div.grid.col-1-1
div Lipsum
div Lipsum
div.grid.col-2-1
div Lipsum
div Lipsum
div.grid.col-1-2
div Lipsum
div Lipsum
|
