From: Neil Smith Date: Wed, 12 Mar 2014 12:33:21 +0000 (+0000) Subject: Added vector diagrams X-Git-Url: https://git.njae.me.uk/?a=commitdiff_plain;h=2271caf60ec3042cc7dcf337c834a1a52564947a;p=cipher-training.git Added vector diagrams --- diff --git a/slides/caesar-break.html b/slides/caesar-break.html index d552209..c6d4023 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; + } @@ -44,7 +47,7 @@ # Breaking caesar ciphers -![centre-aligned Caesar wheel](caesarwheel1.gif) +![center-aligned Caesar wheel](caesarwheel1.gif) --- @@ -67,6 +70,7 @@ What steps do we know how to do? # How close is it to English? What does English look like? + * We need a model of English. How do we define "closeness"? @@ -97,26 +101,56 @@ 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 + +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 -## FIXME! Diagram of vector subtraction here +.float-right[![right-aligned Vector subtraction](vector-subtraction.svg)] Several different distance measures (__metrics__, also called __norms__): -* L₂ norm (Euclidean distance): `$|\mathbf{x} - \mathbf{y}| = \sqrt{\sum_i (\mathbf{x}_i - \mathbf{y}_i)^2} $` +* L₂ norm (Euclidean distance): +`$|\mathbf{a} - \mathbf{b}| = \sqrt{\sum_i (\mathbf{a}_i - \mathbf{b}_i)^2} $` -* L₁ norm (Manhattan distance, taxicab distance): `$|\mathbf{x} - \mathbf{y}| = \sum_i |\mathbf{x}_i - \mathbf{y}_i| $` +* L₁ norm (Manhattan distance, taxicab distance): +`$|\mathbf{a} - \mathbf{b}| = \sum_i |\mathbf{a}_i - \mathbf{b}_i| $` -* L₃ norm: `$|\mathbf{x} - \mathbf{y}| = \sqrt[3]{\sum_i |\mathbf{x}_i - \mathbf{y}_i|^3} $` +* L₃ norm: +`$|\mathbf{a} - \mathbf{b}| = \sqrt[3]{\sum_i |\mathbf{a}_i - \mathbf{b}_i|^3} $` The higher the power used, the more weight is given to the largest differences in components. (Extends out to: -* L₀ norm (Hamming distance): `$|\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| $` +* L₀ 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_∞ norm: `$|\mathbf{x} - \mathbf{y}| = \max_i{(\mathbf{x}_i - \mathbf{y}_i)} $` +* L_∞ norm: +`$|\mathbf{a} - \mathbf{b}| = \max_i{(\mathbf{a}_i - \mathbf{b}_i)} $` neither of which will be that useful.) --- @@ -125,9 +159,9 @@ neither of which will be that useful.) Frequency distributions drawn from different sources will have different lengths. For a fair comparison we need to scale them. -* Eucliean scaling (vector with unit length): `$ \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): `$$ \hat{\mathbf{x}} = \frac{\mathbf{x}}{\| \mathbf{x} \|} = \frac{\mathbf{x}}{ \mathbf{x}_1 + \mathbf{x}_2 + \mathbf{x}_3 + \dots }$$` --- @@ -135,6 +169,8 @@ 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[![right-aligned Vector dot product](vector-dot-product.svg)] + 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} $` @@ -144,17 +180,47 @@ But, and `$ \| \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! Diagram of vector dot product