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+ .float-right {
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+ }
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# Breaking caesar ciphers
-
+
+
+---
+
+# Human vs Machine
+
+Slow but clever vs Dumb but fast
+
+## Human approach
+
+Ciphertext | Plaintext
+---|---
+ | 
---
-# Brute force
+# Human vs machine
+
+## Machine approach
+
+Brute force.
-How many keys to try?
+Try all keys.
+
+* How many keys to try?
## Basic idea
# How close is it to English?
What does English look like?
+
* We need a model of English.
How do we define "closeness"?
+---
+
+# What does English look like?
+
+## Abstraction: frequency of letter counts
+
+Letter | Count
+-------|------
+a | 489107
+b | 92647
+c | 140497
+d | 267381
+e | 756288
+. | .
+. | .
+. | .
+z | 3575
+
+One way of thinking about this is a 26-dimensional vector.
+
+Create a vector of our text, and one of idealised English.
+
+The distance between the vectors is how far from English the text is.
+
+---
+
+# Frequencies of English
+
+But before then how do we count the letters?
+
+* Read a file into a string
+```python
+open()
+read()
+```
+* Count them
+```python
+import collections
+```
+
+---
+
+# Canonical forms
+
+Counting letters in _War and Peace_ gives all manner of junk.
+
+* Convert the text in canonical form (lower case, accents removed, non-letters stripped) before counting
+
+```python
+[l.lower() for l in text if ...]
+```
+---
+
+
+# Accents
+
+```python
+>>> caesar_encipher_letter('é', 1)
+```
+What does it produce?
+
+What should it produce?
+
+## Unicode, combining codepoints, and normal forms
+
+Text encodings will bite you when you least expect it.
+
+* urlencoding is the other pain point.
+
+---
+
+# Five minutes on StackOverflow later...
+
+```python
+def unaccent(text):
+ """Remove all accents from letters.
+ It does this by converting the unicode string to decomposed compatibility
+ form, dropping all the combining accents, then re-encoding the bytes.
+
+ >>> unaccent('hello')
+ 'hello'
+ >>> unaccent('HELLO')
+ 'HELLO'
+ >>> unaccent('héllo')
+ 'hello'
+ >>> unaccent('héllö')
+ 'hello'
+ >>> unaccent('HÉLLÖ')
+ 'HELLO'
+ """
+ return unicodedata.normalize('NFKD', text).\
+ encode('ascii', 'ignore').\
+ decode('utf-8')
+```
+
+---
+
+# Vector distances
+
+.float-right[]
+
+Several different distance measures (__metrics__, also called __norms__):
+
+* L<sub>2</sub> norm (Euclidean distance):
+`\(\|\mathbf{a} - \mathbf{b}\| = \sqrt{\sum_i (\mathbf{a}_i - \mathbf{b}_i)^2} \)`
+
+* L<sub>1</sub> norm (Manhattan distance, taxicab distance):
+`\(\|\mathbf{a} - \mathbf{b}\| = \sum_i |\mathbf{a}_i - \mathbf{b}_i| \)`
+
+* L<sub>3</sub> norm:
+`\(\|\mathbf{a} - \mathbf{b}\| = \sqrt[3]{\sum_i |\mathbf{a}_i - \mathbf{b}_i|^3} \)`
+
+The higher the power used, the more weight is given to the largest differences in components.
+
+(Extends out to:
+
+* L<sub>0</sub> norm (Hamming distance):
+`$$\|\mathbf{a} - \mathbf{b}\| = \sum_i \left\{
+\begin{matrix} 1 &\mbox{if}\ \mathbf{a}_i \neq \mathbf{b}_i , \\
+ 0 &\mbox{if}\ \mathbf{a}_i = \mathbf{b}_i \end{matrix} \right. $$`
+
+* L<sub>∞</sub> norm:
+`\(\|\mathbf{a} - \mathbf{b}\| = \max_i{(\mathbf{a}_i - \mathbf{b}_i)} \)`
+
+neither of which will be that useful.)
+---
+
+# Normalisation of vectors
+
+Frequency distributions drawn from different sources will have different lengths. For a fair comparison we need to scale them.
+
+* Eucliean scaling (vector with unit length): `$$ \hat{\mathbf{x}} = \frac{\mathbf{x}}{\| \mathbf{x} \|} = \frac{\mathbf{x}}{ \sqrt{\mathbf{x}_1^2 + \mathbf{x}_2^2 + \mathbf{x}_3^2 + \dots } }$$`
+
+* Normalisation (components of vector sum to 1): `$$ \hat{\mathbf{x}} = \frac{\mathbf{x}}{\| \mathbf{x} \|} = \frac{\mathbf{x}}{ \mathbf{x}_1 + \mathbf{x}_2 + \mathbf{x}_3 + \dots }$$`
+
+---
+
+# Angle, not distance
+
+Rather than looking at the distance between the vectors, look at the angle between them.
+
+.float-right[]
+
+Vector dot product shows how much of one vector lies in the direction of another:
+`\( \mathbf{A} \bullet \mathbf{B} =
+\| \mathbf{A} \| \cdot \| \mathbf{B} \| \cos{\theta} \)`
+
+But,
+`\( \mathbf{A} \bullet \mathbf{B} = \sum_i \mathbf{A}_i \cdot \mathbf{B}_i \)`
+and `\( \| \mathbf{A} \| = \sum_i \mathbf{A}_i^2 \)`
+
+A bit of rearranging give the cosine simiarity:
+`$$ \cos{\theta} = \frac{ \mathbf{A} \bullet \mathbf{B} }{ \| \mathbf{A} \| \cdot \| \mathbf{B} \| } =
+\frac{\sum_i \mathbf{A}_i \cdot \mathbf{B}_i}{\sum_i \mathbf{A}_i^2 \times \sum_i \mathbf{B}_i^2} $$`
+
+This is independent of vector lengths!
+
+Cosine similarity is 1 if in parallel, 0 if perpendicular, -1 if antiparallel.
+
+---
+
+# An infinite number of monkeys
+
+What is the probability that this string of letters is a sample of English?
+
+Given 'th', 'e' is about six times more likely than 'a' or 'i'.
+
+## Naive Bayes, or the bag of letters
+
+Ignore letter order, just treat each letter individually.
+
+Probability of a text is `\( \prod_i p_i \)`
+
+(Implmentation issue: this can often underflow, so get in the habit of rephrasing it as `\( \sum_i \log p_i \)`)
+
+---
+
+# Which is best?
+
+ | Euclidean | Normalised
+---|-----------|------------
+L1 | x | x
+L2 | x | x
+L3 | x | x
+Cosine | x | x
+
+And the probability measure!
+
+* Nine different ways of measuring fitness.
+
+## Computing is an empircal science
+
+
</textarea>
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projects / cipher-training.git / blobdiff