From: Neil Smith <neil.git@njae.me.uk> Date: Wed, 12 Mar 2014 12:37:20 +0000 (+0000) Subject: Fixed conflicting equation typesetting X-Git-Url: https://git.njae.me.uk/?a=commitdiff_plain;h=5b51a469cc152b1035a5cf69f2c38d51f9d16eb8;hp=05677f98ecb9a499853a19cb299b1880c0359c9a;p=cipher-training.git Fixed conflicting equation typesetting --- diff --git a/slides/caesar-break.html b/slides/caesar-break.html index 6df8365..a47e236 100644 --- a/slides/caesar-break.html +++ b/slides/caesar-break.html @@ -37,6 +37,9 @@ text-shadow: 0 0 20px #333; padding: 2px 5px; } + .float-right { + float: right; + } </style> </head> <body> @@ -44,7 +47,7 @@ # Breaking caesar ciphers - + --- @@ -116,26 +119,56 @@ The distance between the vectors is how far from English the text is. --- -# Vector distances +# 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 -## FIXME! Diagram of vector subtraction here +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 + +--- + +# Vector distances + +.float-right[] Several different distance measures (__metrics__, also called __norms__): -* L<sub>2</sub> norm (Euclidean distance): `\(\|\mathbf{x} - \mathbf{y}\| = \sqrt{\sum_i (\mathbf{x}_i - \mathbf{y}_i)^2} \)` +* 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): `\(\|\mathbf{x} - \mathbf{y}\| = \sum_i |\mathbf{x}_i - \mathbf{y}_i| \)` +* 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: `\(\|\mathbf{x} - \mathbf{y}\| = \sqrt[3]{\sum_i |\mathbf{x}_i - \mathbf{y}_i|^3} \)` +* 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): `\(\|\mathbf{x} - \mathbf{y}\| = \sum_i \left\{\begin{matrix} 1 &\mbox{if}\ \mathbf{x}_i \neq \mathbf{y}_i , \\ 0 &\mbox{if}\ \mathbf{x}_i = \mathbf{y}_i \end{matrix} \right| \)` +* 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: `\(\|\mathbf{x} - \mathbf{y}\| = \max_i{(\mathbf{x}_i - \mathbf{y}_i)} \)` +* L<sub>∞</sub> norm: +`\(\|\mathbf{a} - \mathbf{b}\| = \max_i{(\mathbf{a}_i - \mathbf{b}_i)} \)` neither of which will be that useful.) --- @@ -144,9 +177,9 @@ neither of which will be that useful.) 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): `\( \hat{\mathbf{x}} = \frac{\mathbf{x}}{\| \mathbf{x} \|} = \frac{\mathbf{x}}{ \sqrt{\mathbf{x}_1^2 + \mathbf{x}_2^2 + \mathbf{x}_3^2 + \dots } }\)` +* 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): `\( \hat{\mathbf{x}} = \frac{\mathbf{x}}{\| \mathbf{x} \|} = \frac{\mathbf{x}}{ \mathbf{x}_1 + \mathbf{x}_2 + \mathbf{x}_3 + \dots }\)` +* Normalisation (components of vector sum to 1): `$$ \hat{\mathbf{x}} = \frac{\mathbf{x}}{\| \mathbf{x} \|} = \frac{\mathbf{x}}{ \mathbf{x}_1 + \mathbf{x}_2 + \mathbf{x}_3 + \dots }$$` --- @@ -154,6 +187,8 @@ Frequency distributions drawn from different sources will have different lengths 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: `\( \mathbf{A} \bullet \mathbf{B} = \| \mathbf{A} \| \cdot \| \mathbf{B} \| \cos{\theta} \)` @@ -163,17 +198,47 @@ But, and `\( \| \mathbf{A} \| = \sum_i \mathbf{A}_i^2 \)` A bit of rearranging give the cosine simiarity: -`\( \cos{\theta} = \frac{ \mathbf{A} \bullet \mathbf{B} }{ \| \mathbf{A} \| \cdot \| \mathbf{B} \| } = -\frac{\sum_i \mathbf{A}_i \cdot \mathbf{B}_i}{\sum_i \mathbf{A}_i^2 \times \sum_i \mathbf{B}_i^2} \)` +`$$ \cos{\theta} = \frac{ \mathbf{A} \bullet \mathbf{B} }{ \| \mathbf{A} \| \cdot \| \mathbf{B} \| } = +\frac{\sum_i \mathbf{A}_i \cdot \mathbf{B}_i}{\sum_i \mathbf{A}_i^2 \times \sum_i \mathbf{B}_i^2} $$` This is independent of vector lengths! -Cosine similarity is 1 if in same direction, 0 if at 90â°, -1 if antiparallel. +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. -## FIXME! Cosine distance bug: frequencies2 length not squared. +## Computing is an empircal science -## FIXME! 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