Representasi objek untuk pembelajaran mesin berbasis kisi

Ini adalah artikel keempat dalam sebuah seri (tautan ke yang pertama , kedua dan ketigaartikel), yang dikhususkan untuk sistem pembelajaran mesin berdasarkan teori kisi, yang disebut "sistem VKF". Program ini menggunakan algoritme berdasarkan rantai Markov untuk menghasilkan penyebab properti target dengan menghitung subset acak dari kesamaan antara beberapa kelompok objek pembelajaran. Artikel ini menjelaskan representasi objek melalui string bit untuk menghitung kesamaan dengan perkalian bitwise dari representasi yang sesuai. Objek dengan fitur diskrit memerlukan beberapa teknik dari Analisis Konsep Formal. Kasus objek dengan fitur kontinu menggunakan regresi logistik, membagi area perubahan menjadi sub-interval menggunakan teori informasi dan representasi yang sesuai dengan lambung cembung dari interval yang dibandingkan.

1 Tanda diskrit

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$$$\langle{L,\wedge,\vee}\rangle$ $$$G$ () $$$\wedge$- $$$M$ () $$$\vee$- . $$$gIm\Leftrightarrow{g\geq{m}}$ $$$(G,M,I)$ $$$L(G,M,I)$, $$$\langle{L,\wedge,\vee}\rangle$.

$$$x\in{L}$ $$$\langle{L,\wedge,\vee}\rangle$ $$$\vee$-, $$$x\neq\emptyset$ $$$y,z\in{L}$ $$$y<x$ $$$z<x$ $$$y\vee{z}<x$.

$$$x\in{L}$ $$$\langle{L,\wedge,\vee}\rangle$ $$$\wedge$-, $$$x\neq\textbf{T}$ $$$y,z\in{L}$ $$$x<y$ $$$x<z$ $$$x<y\wedge{z}$.

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2.1

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$$$E=G\cup{O}$ $$$G$ - $$$O$. $$$[a,b)\subseteq\textbf{R}$ $$$V:G\to\textbf{R}$ $$$G[a,b)=\lbrace{g\in{G}: a\leq{V(g)}<b}\rbrace,$ $$$O[a,b)=\lbrace{g\in{O}: a\leq{V(g)}<b}\rbrace$

$$$E[a,b)=\lbrace{g\in{E}: a\leq{V(g)}<b}\rbrace$.

$$$[a,b)\subseteq\textbf{R}$ $$$V:G\to\textbf{R}$

$$

$${\rm{ent}}[a,b)=-\frac{\vert{G[a,b)}\vert}{\vert{E[a,b)}\vert}\cdot\log_{2}\left(\frac{\vert{G[a,b)}\vert}{\vert{E[a,b)}\vert}\right)-\frac{\vert{O[a,b)}\vert}{\vert{E[a,b)}\vert}\cdot\log_{2}\left(\frac{\vert{O[a,b)}\vert}{\vert{E[a,b)}\vert}\right)$$

$$$a<r<b$ $$$[a,b)\subseteq\textbf{R}$ $$$V:G\to\textbf{R}$

$$

$${\rm{inf}}[a,r,b)=\frac{\vert{E[a,r)}\vert}{\vert{E[a,b)}\vert}\cdot{\rm{ent}}[a,r)+\frac{\vert{E[r,b)}\vert}{\vert{E[a,b)}\vert}\cdot{\rm{ent}}[r,b).$$

— $$$V=r$ .

$$$V:G\to\textbf{R}$ $$$a=\min\{V\}$ $$$v_{0}$, $$$v_{l+1}$ , $$$b=\max\{V\}$. $$$\lbrace{v_{1}<\ldots<v_{l}}\rbrace$ .

2.2

$$$2l$, $$$l$ — . ()

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$$\delta_{i}^{V}(g)=1 \Leftrightarrow V(g)\geq{v_{i}} \\ \sigma_{i}^{V}(g)=1 \Leftrightarrow V(g)<v_{i},$$

$$$1\leq{i}\leq{l}$.

$$$\delta_{1}^{V}(g)\ldots\delta_{l}^{V}(g)\sigma_{1}^{V}(g)\ldots\sigma_{l}^{V}(g)$ $$$V$ $$$g\in{E}$.

, — .

$$$\delta_{1}^{(1)}\ldots\delta_{l}^{(1)}\sigma_{1}^{(1)}\ldots\sigma_{l}^{(1)}$ $$$v_{i}\leq{V(A_{1})}<v_{j}$ $$$\delta_{1}^{(2)}\ldots\delta_{l}^{(2)}\sigma_{1}^{(2)}\ldots\sigma_{l}^{(2)}$ $$$v_{n}\leq{V(A_{2})}<v_{m}$.

$$

$$(\delta_{1}^{(1)}\cdot\delta_{1}^{(2)})\ldots(\delta_{l}^{(1)}\cdot\delta_{l}^{(2)})(\sigma_{1}^{(1)}\cdot\sigma_{1}^{(2)})\ldots(\sigma_{l}^{(1)}\cdot\sigma_{l}^{(2)})$$

$$$\min\lbrace{v_{i},v_{n}}\rbrace\leq{V((A_{1}\cup{A_{2}})'')}<\max\lbrace{v_{j},v_{m}}\rbrace$.

, $$$0\ldots00\ldots0$ $$$\min\{V\}\leq{V((A_{1}\cup{A_{2}})'')}\leq\max\{V\}$.

2.3

. ( 1). . , .

$$$p_{i_{1}}\vee\ldots\vee{p_{i_{k}}}$ $$$p_{i_{1}}+\ldots+{p_{i_{k}}}>\sigma$ $$$0<\sigma<1$.

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— $$$c:$R$$$^{d}\to\lbrace{0,1}\rbrace$, $$$\textbf{R}^{d}$ — ( $$$d$ ) $$$\lbrace{0,1}\rbrace$ — .

, $$$\langle{\vec{X},K}\rangle\in\text{R}^{d}\times\lbrace{0,1}\rbrace$,

$$

$$p_{\vec{X},K}(\vec{x},k)=p_{\vec{X}}(\vec{x})\cdot{p_{K\mid\vec{X}}(k\mid\vec{x})},$$

$$$p_{\vec{X}}(\vec{x})$ — () , a $$$p_{K\mid\vec{X}}(k\mid\vec{x})$ — , .. $$$\vec{x}\in\text{R}^{d}$

$$

$$p_{K\mid\vec{X}}(k\mid\vec{x})=\textbf{P}\lbrace{K=k\mid\vec{X}=\vec{x}}\rbrace.$$

$$$c:\textbf{R}^{d}\to\lbrace{0,1}\rbrace$

$$

$$R(c)=\textbf{P}\left\lbrace{c(\vec{X})\neq{K}}\right\rbrace.$$

$$$b:\textbf{R}^{d}\to\lbrace{0,1}\rbrace$ $$$p_{K\mid\vec{X}}(k\mid\vec{x})$

$$

$$b(\vec{x})=1 \Leftrightarrow p_{K\mid\vec{X}}(1\mid\vec{x})>\frac{1}{2}>p_{K\mid\vec{X}}(0\mid\vec{x})$$

$$$b$ :

$$

$$\forall{c:\textbf{R}^{d}\to\lbrace{0,1}\rbrace}\left[R(b)=\textbf{P}\lbrace{b(\vec{X})\neq{K}}\rbrace\leq{R(c)}\right]$$

$$

$$p_{K\mid\vec{X}}(1\mid\vec{x})=\frac{p_{\vec{X}\mid{K}}(\vec{x}\mid{1})\cdot\textbf{P}\lbrace{K=1}\rbrace}{p_{\vec{X}\mid{K}}(\vec{x}\mid{1})\cdot\textbf{P}\lbrace{K=1}\rbrace+p_{\vec{X}\mid{K}}(\vec{x}\mid{0})\cdot\textbf{P}\lbrace{K=0}\rbrace}= \\ =\frac{1}{1+\frac{p_{\vec{X}\mid{K}}(\vec{x}\mid{0})\cdot\textbf{P}\lbrace{K=0}\rbrace}{p_{\vec{X}\mid{K}}(\vec{x}\mid{1})\cdot\textbf{P}\lbrace{K=1}\rbrace}}=\frac{1}{1+\exp\lbrace{-a(\vec{x})}\rbrace}=\sigma(a(\vec{x})),$$

$$$a(\vec{x})=\log\frac{p_{\vec{X}\mid{K}}(\vec{x}\mid{1})\cdot\textbf{P}\lbrace{K=1}\rbrace}{p_{\vec{X}\mid{K}}(\vec{x}\mid{0})\cdot\textbf{P}\lbrace{K=0}\rbrace}$ $$$\sigma(y)=\frac{1}{1+\exp\lbrace{-y}\rbrace}$ — .

2.4

$$$a(\vec{x})=\log\frac{p_{\vec{X}\mid{K}}(\vec{x}\mid{1})\cdot\textbf{P}\lbrace{K=1}\rbrace}{p_{\vec{X}\mid{K}}(\vec{x}\mid{0})\cdot\textbf{P}\lbrace{K=0}\rbrace}$ $$$\vec{w}^{T}\cdot\varphi(\vec{x})$ $$$\varphi_{i}:\textbf{R}^{d}\to\textbf{R}$ ($$$i=1,\ldots,m$) $$$\vec{w}\in\textbf{R}^{m}$.

$$$\langle\vec{x}_{1},k_{1}\rangle,\dots,\langle\vec{x}_{n},k_{n}\rangle$ $$$t_{j}=2k_{j}-1$.

$$

$$\log\lbrace{p(t_{1},\dots,t_{n}\mid\vec{x}_{1},\ldots,\vec{x}_{n},\vec{w})}\rbrace=-\sum_{j=1}^{n}\log\left[1+\exp\lbrace{-t_{j}\sum_{i=1}^{m}w_{i}\varphi_{i}(\vec{x}_{j})}\rbrace\right].$$

,

$$

$$L(w_{1},\ldots,w_{m})=-\sum_{j=1}^{n}\log\left[1+\exp\lbrace{-t_{j}\sum_{i=1}^{m}w_{i}\varphi_{i}(\vec{x}_{j})}\rbrace\right]\to\max$$

.

-

$$

$$\vec{w}_{t+1}=\vec{w}_{t}-(\nabla_{\vec{w}^{T}}\nabla_{\vec{w}}L(\vec{w}_{t}))^{-1}\cdot\nabla_{\vec{w}}L(\vec{w}_{t}).$$

$$$s_{j}=\frac{1}{1+\exp\lbrace{t_{j}\cdot{(w^{T}\cdot\Phi(x_{j}))}}\rbrace}$

$$

$$\nabla{L(\vec{w})}=-\Phi^{T}{\rm{diag}}(t_{1},\ldots,t_{n})\vec{s}, \nabla\nabla{L(\vec{w})}=\Phi^{T}R\Phi,$$

$$$R={\rm{diag}}(s_{1}(1-s_{1}), s_{2}(1-s_{2}), \ldots, s_{n}(1-s_{n}))$ —

$$$s_{1}(1-s_{1}), s_{2}(1-s_{2}), \ldots, s_{n}(1-s_{n})$ $$${\rm{diag}}(t_{1},\ldots,t_{n})\vec{s}$ — $$$t_{1}s_{1}, t_{2}s_{2}, \ldots, t_{n}s_{n}$.

$$

$$\vec{w}_{t+1}=\vec{w}_{t}+\left(\Phi^{T}R\Phi\right)^{-1}\Phi^{T}{\rm{diag}}(t)\vec{s}= (\Phi^{T}R\Phi)^{-1}\Phi^{T}R\vec{z},$$

$$$\vec{z}=\Phi\vec{w}_{t}+R^{-1}{\rm{diag}}(t_{1},\ldots,t_{n})\vec{s}$ — .

, - -

$$

$$\vec{w}_{t+1}=(\Phi^{T}R\Phi+\lambda\cdot{I})^{-1}\cdot(\Phi^{T}R\vec{z}).$$

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- $$$V_{k}$ ,

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$$R^{2}=1-\exp\lbrace{2(L(w_{0},\ldots, w_{k-1})-L(w_{0},\ldots,w_{k-1},w_{k}))/n}\rbrace\geq\sigma$$

$$$V_{k}$ ,

$$

$$1-\frac{L(w_{0},\ldots,w_{k-1},w_{k})}{L(w_{0},\ldots,w_{k-1})}\geq\sigma$$

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Tetapi situasi dengan pasangan ("pH", "alkohol") sangat berbeda. Bobot "alkohol" positif, sedangkan bobot "pH" negatif. Tetapi dengan bantuan transformasi logis yang jelas, kami mendapatkan implikasinya ("pH"$$$\Rightarrow$ "alkohol").

Penulis mengucapkan terima kasih kepada rekan-rekan dan mahasiswanya atas dukungan dan insentifnya.