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Commit 727b15ac by Stalin Munoz
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algoritmos genéticos para TSP

parent 51b94bd1
Showing with 8912 additions and 8 deletions
curso-search/geneticos/Ruleta.svg 0 → 100644
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curso-search/geneticos/aptitud.svg 0 → 100644
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********************************************************
Package `algorithm2e' Release 5.1 -- october 19 2015 --
- algorithm2e-announce@lirmm.fr mailing list for announcement about releases
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- Author: Christophe Fiorio (christophe.fiorio@umontpellier.fr)
********************************************************
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curso-search/geneticos/ga-permutacion.tex 0 → 100644
\documentclass[xcolor=dvipsnames]{beamer}
\usepackage[spanish]{babel}
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\usepackage{amsmath}
\usepackage{pgfpages}
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% descomentar para generar handouts con notas
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\tikzstyle{every picture}+=[remember picture]
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\title{Algoritmos genéticos de permutación}
\author{Stalin Muñoz Gutiérrez}
\institute{
Centro de Ciencias de la Complejidad\\
Universidad Nacional Aut\'onoma de Mexico (UNAM)}
\date{}
\begin{document}
\parindent 2em
\frame
{
\titlepage
\note<1>{
Hoy platicaremos un poco acerca del tipo de procesos en los que se inspiran los algoritmos metaheurísticos y abordaremos un tipo de algoritmo muy interesante, el algortimo genético para problemas de permutación. Veremos como puede aplicarse este algoritmo para resolver el problema del agente viajero.
}
}
\begin{frame}
\frametitle{Algoritmos metaheurísticos}
\begin{block}{Inspirados en la física}
\begin{enumerate}
\item Templado Simulado (Simulated Annealing)
\item Gran Explosión Gran Colapso (Big Bang Big Crunch)
\item Gravitacional
\item Similar a Electromanetismo
\item Optimización de Fuerza Central
\item Gotas de Agua Inteligentes
\item Dinámica de Formación de Rios
\item Colisión de Particulas
\end{enumerate}
\end{block}
\note{
Existe un gran número de algoritmos metaheurísticos.
Se pueden clasificar de acuerdo a la fuente de inspiración de los mismos.
Aquí tenemos una lista de algunos de los algoritmos basados en procesos físicos.
Entre los elementos de la lista es el algoritmo de Templado Simulado, el cual ya hemos discutido.
}
\end{frame}
\begin{frame}
\frametitle{Algoritmos metaheurísticos}
\begin{block}{Inspirados en la biología}
\begin{enumerate}
\item Sistema Inmune Artificial
\item Algoritmos Genéticos
\item Colonia de Hormigas
\item Colonia de Abejas
\item Algoritmo de Luciérnagas
\item Algoritmo de Murciélagos
\item Búsqueda del Cuco
\item Búsqueda de Cardumen
\end{enumerate}
\end{block}
\note{
Aquí tenemos otra lista de algoritmos metaheurísticos inspirados en la biología y en diferentes organismos.
Un gran número de estas metáforas se basan en abstraer la forma en que poblaciones de organismos resuelven colectivamente un problema.
Son de particular importancia los algoritmos genéticos, los cuales se inspiran en la teoría de evolución de Darwin para encontrar soluciones a problemas.
}
\end{frame}
\begin{frame}
\frametitle{Algoritmos metaheurísticos}
\begin{block}{Otras fuentes de inspiración}
\begin{enumerate}
\item Algoritmos Culturales
\item Optimización de Partículas
\item Sociedad Anarquísta
\item Competitivo Imperialista
\item Búsqueda de Armonía
\item Rueda Chirriante
\item Optimización social cognitiva
\item Explosión de minas
\end{enumerate}
\end{block}
\note{
No hay límite a las posibles fuentes de inspiración.
Otros algoritmos se basan en temas culturales, políticos, artísticos, industriales, incluso en aforismos como "The squeaky wheel gets the grease" (La rueda chirriante obtiene la atención).
}
\end{frame}
\begin{frame}
\frametitle{Algoritmo Genético}
\begin{center}
\includegraphics[scale=0.5]{ga-intro-01.pdf}
\end{center}
\note{
Los algoritmos genéticos fueron propuestos por John Holland en los años 60s, y fueron popularizados por David Goldberg en los años 80s.
Los algoritmos genéticos se basan en metáforas de la evolución de organismos.
La información necesaria para describir la construcción de un organismo artificial esta codificada en su genotipo.
El genotipo se expresa o decodifica en un fenotipo el cual representa una solución candidata al problema a resolver.
Los problemas son vistos como un entorno o medio ambiente que recompensa a las organismos más aptos. Siendo los fenotipos de los organismos artificiales las soluciones candidatas al problema.
}
\end{frame}
\begin{frame}[<+->]
\only<1>{
\begin{center}
\includegraphics[scale=0.4]{ga-intro-02.pdf}
\end{center}
}
\note<1>{
La evolución artificial opera sobre una población de organismos.
}
\only<2>{
\begin{center}
\includegraphics[scale=0.4]{ga-intro-03.pdf}
\end{center}
}
\note<2>{
Cada individuo es evaluado en el entorno obteniendo una aptitud.
La aptitud es más alta para individuos mejor adaptados al entorno.
}
\only<3>{
\begin{center}
\includegraphics[scale=0.4]{ga-intro-04.pdf}
\end{center}
}
\note<3>{
Una solución con baja aptitud representa una solución candidata que no es buena.
}
\only<4>{
\begin{center}
\includegraphics[scale=0.4]{ga-intro-05.pdf}
\end{center}
}
\note<4>{
En el algoritmo, los organismos menos aptos tendrán menor probabilidad de reproducirse.
}
\only<5>{
\begin{center}
\includegraphics[scale=0.4]{ga-intro-06.pdf}
\end{center}
}
\note<5>{
La aptitud asignada a un organismo artificial debe ser informativa para el algortimo de selección.
}
\only<6>{
\begin{center}
\includegraphics[scale=0.4]{ga-intro-07.pdf}
\end{center}
}
\note<6>{
La selección artificial seleccionará a los individuos más aptos.
}
\only<7>{
\begin{center}
\includegraphics[scale=0.4]{ga-intro-08.pdf}
\end{center}
}
\note<7>{
Sin embargo, en el algoritmo genético canónico, no hay garantía de que el mejor organismo de la población será seleccionado, ya que la selección artificial depende de una simulación estocástica.
}
\end{frame}
\begin{frame}
\frametitle{Algoritmos genéticos para resolver el problema del agente viajero}
\begin{center}
\includegraphics[scale=0.18]{tsp-mapa.pdf}
\end{center}
\note{
Utilizaremos el ejemplo del problema del agente viajero para ilustrar el funcionamiento de un algoritmo genético para problemas de permutación.
}
\end{frame}
\begin{frame}
\frametitle{Genotipo}
\begin{center}
\includegraphics[scale=0.8]{genotipo.pdf}
\end{center}
\note{Los organismos artificiales cuentan con un genotipo.
En el algoritmo genético canónico se trata de una cadena binaria.
En un algoritmo genético de permutación la cadena esta formada por una permutación de un segmento inicial de números enteros.
}
\end{frame}
\begin{frame}
\frametitle{Fenotipos}
\begin{center}
\includegraphics[scale=0.8]{genotipo-tsp.pdf}
\end{center}
\note{
Análogamente a un organismo biológico la información genética contiene un código que se expresa en un fenotipo. En un organismo biológico el fenotipo son los atributos físicos del organismo, su anatomía, fisiología, etc.
En un algoritmo genético el fenotipo es una solución candidata del problema a resolver.
En el caso del agente viajero, cada número entero representa una ciudad a visitar.
La secuencia de numeros es la ruta a tomar.
}
\end{frame}
\begin{frame}
\frametitle{Aptitud}
\begin{center}
\includegraphics[scale=0.5]{aptitud.pdf}
\end{center}
\note{
La aptitud de una ruta puede definirse en función de la distancia total de la ruta o del tiempo estimado.
En este ejemplo la aptitud de la solución candidata A es menor que la aptitud de la solución candidata B, dado que el tiempo estimado de A es 54 horas mientras que el tiempo estimado de B es 40 horas.
Dependiendo del algoritmo de selección podemos requerir o no de una función de aptitud explícita. De ser así, una opción puede ser el inverso de la distancia o el tiempo.
}
\end{frame}
\begin{frame}
\frametitle{Selección Artificial}
\begin{center}
\includegraphics[scale=0.2]{seleccion.pdf}
\end{center}
\note{
El algoritmo de selección discrimina las mejores soluciones candidatas.
Estas soluciones serán sujetas a reproducción para generar nuevas soluciones
candidatas.
}
\end{frame}
\begin{frame}
\frametitle{Selección tipo ruleta}
\begin{center}
\includegraphics[scale=0.55]{Ruleta.pdf}
\end{center}
\note{
Un algoritmo muy utilizado para la selección es el de selección de ruleta.
La probabibilidad de seleccionar a un organismo es proporcional a su aptitud.
Para calcular esta probabilidad simplemente dividimos la aptitud del organismo entre la suma de las aptitudes de todos los organismos de la población.
Al simular la ruleta, los organismos más aptos tienen un área más grande y se seleccionan con mayor frecuencia.
En este ejemplo la aptitud del organismo A es mayor a la del organismo B.
La selección tipo ruleta tiene varios inconvenientes, uno de ellos es que si existe un organismo con una aptitud significativamente mejor a la de los demás organismos, la selección tipo ruleta tendrá un sesgo muy fuerte hacia este organismo, resultando en convergencia temprana a soluciones subóptimas.
}
\end{frame}
\begin{frame}
\frametitle{Selección tipo torneo}
\begin{center}
\includegraphics[scale=0.15]{selecciontorneo.pdf}
\end{center}
\note{
Por ello existen otras formas de seleccionar organismos. Una muy efectiva es la selección tipo torneo.
Se define un tamaño de torneo \emph{k}. Por ejemplo torneo binario o entre dos soluciones.
El torneo se forma tomando una muestra aleatoria de \emph{k} soluciones de la población.
Se define como ganador del torneo el organismo con la mayor aptitud de la muestra.
Este ejemplo ilustra dos torneos binarios.
Las soluciones candidatas seleccionadas son las rutas más cortas en cada caso.
Al ser una muestra aleatoria no hay garantia de que el mejor individuo se seleccione en un torneo.
}
\end{frame}
\begin{frame}
\frametitle{Reproducción}
\begin{center}
\includegraphics[scale=0.55]{cruza.pdf}
\end{center}
\note{Las soluciones candidatas seleccionadas se reproducen para generar nuevas soluciones.
Hay muchas formas de hacer la recombinación de los genotipos.
Aquí ilustramos una muy simple. La cruza requiere dos padres y genera dos hijos.
Seleccionamos una región aleatoria de la permutación.
Esta región es copiada a cada uno de los hijos.
Después, considerando que el cromosoma es circular, colocamos los números que vienen de posiciones subsecuentes del otro padre, siempre que no se repitan números.
En el ejemplo el primer hijo tiene la secuencia 7,1,3 del primer padre.
Despues tomamos la secuencia del segundo padre a partir de esa posición.
Copiamos el número 9, considerando el cromosoma circular le sigue el número 2,
luego el 3, pero este se repite, por lo que no lo copiamos, luego copiamos el 6, sigue el 1, pero se repite por lo que no lo copiamos, copiamos el cero, sigue el 7 pero se repite, copiamos 8, 4 y 5.
}
\end{frame}
\begin{frame}
\frametitle{Mutación}
\begin{center}
\includegraphics[scale=0.55]{mutacion.pdf}
\end{center}
\note{
La mutación es otro operador genético del algoritmo.
Un ejemplo de mutación es el intercambio aleatorio de ciudades.
En el ejemplo tras la mutación se han intercamibado las posiciones de las ciudades 6 y 7.
}
\end{frame}
\begin{frame}
\frametitle{Evolución}
En la evolución aplicamos estas operaciones en el orden siguiente:
\begin{enumerate}
\item Generar población inicial
\item Repetir cierto número de generaciones
\begin{enumerate}
\item Asignar aptitud de la población actual
\item Seleccionar mejores soluciones
\item Cruzar las soluciones generando nueva población
\item Mutar la nueva población
\item La población actual es la nueva población
\end{enumerate}
\end{enumerate}
\note{
Aquí un algoritmo genético muy simple.
Generamos la población inicial con cromosomas aleatorios.
Repetimos los pasos siguientes cierto número de generaciones.
Decodificamos y evaluamos a los organismos en el problema para obtener aptitud.
Seleccionamos a los mejores.
Reproducimos a los organismos seleccionados.
Mutamos a los organismos.
Hacemos la población actual igual a la nueva población.
}
\end{frame}
\begin{frame}
\frametitle{Comentario final.}
\begin{center}
\includegraphics[scale=0.5]{logoesp.jpeg}
\end{center}
Curso: Cómputo evolutivo,
Especialización: Introducción a la inteligencia artificial de la UNAM.
\note{
Los algoritmos genéticos son metaheurísticos muy poderosos.
Aquí solo presentamos una versión muy simple.
Hay muchas variaciones de estos algoritmos.
Si quieres saber más respecto de estos algoritmos te recomendamos tomar el curso:
Cómputo evolutivo,
dentro de la especialización Introducción a la inteligencia artificial de la UNAM.}
\end{frame}
\end{document}
\ No newline at end of file
curso-search/geneticos/ga-permutacion.tex.bak 0 → 100644
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\usepackage[spanish]{babel}
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\usepackage{pgfpages}
\usepackage[makeroom]{cancel}
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\usepackage{algpseudocode}
\usepackage{multicol}
% descomentar para generar handouts con notas
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\tikzstyle{every picture}+=[remember picture]
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\DeclareMathOperator*{\argmax}{argmax}
\spanishdecimal{.}
\title{Algoritmos genéticos de permutación}
\author{Stalin Muñoz Gutiérrez}
\institute{
Centro de Ciencias de la Complejidad\\
Universidad Nacional Aut\'onoma de Mexico (UNAM)}
\date{}
\begin{document}
\parindent 2em
\frame
{
\titlepage
\note<1>{
Hoy platicaremos un poco acerca del tipo de procesos en los que se inspiran los algoritmos metaheurísticos y abordaremos un tipo de algoritmo muy interesante, el algortimo genético para problemas de permutación. Veremos como puede aplicarse este algoritmo para resolver el problema del agente viajero.
}
}
\begin{frame}
\frametitle{Algoritmos metaheurísticos}
\begin{block}{Inspirados en la física}
\begin{enumerate}
\item Templado Simulado (Simulated Annealing)
\item Gran Explosión Gran Colapso (Big Bang Big Crunch)
\item Gravitacional
\item Similar a Electromanetismo
\item Optimización de Fuerza Central
\item Gotas de Agua Inteligentes
\item Dinámica de Formación de Rios
\item Colisión de Particulas
\end{enumerate}
\end{block}
\note{
Existe un gran número de algoritmos metaheurísticos.
Se pueden clasificar de acuerdo a la fuente de inspiración de los mismos.
Aquí tenemos una lista de algunos de los algoritmos basados en procesos físicos.
Entre los elementos de la lista es el algoritmo de Templado Simulado, el cual ya hemos discutido.
}
\end{frame}
\begin{frame}
\frametitle{Algoritmos metaheurísticos}
\begin{block}{Inspirados en la biología}
\begin{enumerate}
\item Sistema Inmune Artificial
\item Algoritmos Genéticos
\item Colonia de Hormigas
\item Colonia de Abejas
\item Algoritmo de Luciérnagas
\item Algoritmo de Murciélagos
\item Búsqueda del Cuco
\item Búsqueda de Cardumen
\end{enumerate}
\end{block}
\note{
Aquí tenemos otra lista de algoritmos metaheurísticos inspirados en la biología y en diferentes organismos.
Un gran número de estas metáforas se basan en abstraer la forma en que poblaciones de organismos resuelven colectivamente un problema.
Son de particular importancia los algoritmos genéticos, los cuales se inspiran en la teoría de evolución de Darwin para encontrar soluciones a problemas.
}
\end{frame}
\begin{frame}
\frametitle{Algoritmos metaheurísticos}
\begin{block}{Otras fuentes de inspiración}
\begin{enumerate}
\item Algoritmos Culturales
\item Optimización de Partículas
\item Sociedad Anarquísta
\item Competitivo Imperialista
\item Búsqueda de Armonía
\item Rueda Chirriante
\item Optimización social cognitiva
\item Explosión de minas
\end{enumerate}
\end{block}
\note{
No hay límite a las posibles fuentes de inspiración.
Otros algoritmos se basan en temas culturales, políticos, artísticos, industriales, incluso en aforismos como "The squeaky wheel gets the grease" (La rueda chirriante obtiene la atención).
}
\end{frame}
\begin{frame}
\frametitle{Algoritmo Genético}
\begin{center}
\includegraphics[scale=0.5]{ga-intro-01.pdf}
\end{center}
\note{
Los algoritmos genéticos fueron propuestos por John Holland en los años 60s, y fueron popularizados por David Goldberg en los años 80s.
Los algoritmos genéticos se basan en metáforas de la evolución de organismos.
La información necesaria para describir la construcción de un organismo artificial esta codificada en su genotipo.
El genotipo se expresa o decodifica en un fenotipo el cual representa una solución candidata al problema a resolver.
Los problemas son vistos como un entorno o medio ambiente que recompensa a las organismos más aptos. Siendo los fenotipos de los organismos artificiales las soluciones candidatas al problema.
}
\end{frame}
\begin{frame}[<+->]
\only<1>{
\begin{center}
\includegraphics[scale=0.4]{ga-intro-02.pdf}
\end{center}
}
\note<1>{
La evolución artificial opera sobre una población de organismos.
}
\only<2>{
\begin{center}
\includegraphics[scale=0.4]{ga-intro-03.pdf}
\end{center}
}
\note<2>{
Cada individuo es evaluado en el entorno obteniendo una aptitud.
La aptitud es más alta para individuos mejor adaptados al entorno.
}
\only<3>{
\begin{center}
\includegraphics[scale=0.4]{ga-intro-04.pdf}
\end{center}
}
\note<3>{
Una solución con baja aptitud representa una solución candidata que no es buena.
}
\only<4>{
\begin{center}
\includegraphics[scale=0.4]{ga-intro-05.pdf}
\end{center}
}
\note<4>{
En el algoritmo, los organismos menos aptos tendrán menor probabilidad de reproducirse.
}
\only<5>{
\begin{center}
\includegraphics[scale=0.4]{ga-intro-06.pdf}
\end{center}
}
\note<5>{
La aptitud asignada a un organismo artificial debe ser informativa para el algortimo de selección.
}
\only<6>{
\begin{center}
\includegraphics[scale=0.4]{ga-intro-07.pdf}
\end{center}
}
\note<6>{
La selección artificial seleccionará a los individuos más aptos.
}
\only<7>{
\begin{center}
\includegraphics[scale=0.4]{ga-intro-08.pdf}
\end{center}
}
\note<7>{
Sin embargo, en el algoritmo genético canónico, no hay garantía de que el mejor organismo de la población será seleccionado, ya que la selección artificial depende de una simulación estocástica.
}
\end{frame}
\begin{frame}
\frametitle{Algoritmos genéticos para resolver el problema del agente viajero}
\begin{center}
\includegraphics[scale=0.18]{tsp-mapa.pdf}
\end{center}
\note{
Utilizaremos el ejemplo del problema del agente viajero para ilustrar el funcionamiento de un algoritmo genético para problemas de permutación.
}
\end{frame}
\begin{frame}
\frametitle{Genotipo}
\begin{center}
\includegraphics[scale=0.8]{genotipo.pdf}
\end{center}
\note{Los organismos artificiales cuentan con un genotipo.
En el algoritmo genético canónico se trata de una cadena binaria.
En un algoritmo genético de permutación la cadena esta formada por una permutación de un segmento inicial de números enteros.
}
\end{frame}
\begin{frame}
\frametitle{Fenotipos}
\begin{center}
\includegraphics[scale=0.8]{genotipo-tsp.pdf}
\end{center}
\note{
Análogamente a un organismo biológico la información genética contiene un código que se expresa en un fenotipo. En un organismo biológico el fenotipo son los atributos físicos del organismo, su anatomía, fisiología, etc.
En un algoritmo genético el fenotipo es una solución candidata del problema a resolver.
En el caso del agente viajero, cada número entero representa una ciudad a visitar.
La secuencia de numeros es la ruta a tomar.
}
\end{frame}
\begin{frame}
\frametitle{Aptitud}
\begin{center}
\includegraphics[scale=0.5]{aptitud.pdf}
\end{center}
\note{
La aptitud de una ruta puede definirse en función de la distancia total de la ruta o del tiempo estimado.
En este ejemplo la aptitud de la solución candidata A es menor que la aptitud de la solución candidata B, dado que el tiempo estimado de A es 54 horas mientras que el tiempo estimado de B es 40 horas.
Dependiendo del algoritmo de selección podemos requerir o no de una función de aptitud explícita. De ser así, una opción puede ser el inverso de la distancia o el tiempo.
}
\end{frame}
\begin{frame}
\frametitle{Selección Artificial}
\begin{center}
\includegraphics[scale=0.2]{seleccion.pdf}
\end{center}
\note{
El algoritmo de selección discrimina las mejores soluciones candidatas.
Estas soluciones serán sujetas a reproducción para generar nuevas soluciones
candidatas.
}
\end{frame}
\begin{frame}
\frametitle{Selección tipo ruleta}
\begin{center}
\includegraphics[scale=0.55]{Ruleta.pdf}
\end{center}
\note{
Un algoritmo muy utilizado para la selección es el de selección de ruleta.
La probabibilidad de seleccionar a un organismo es proporcional a su aptitud.
Para calcular esta probabilidad simplemente dividimos la aptitud del organismo entre la suma de las aptitudes de todos los organismos de la población.
Al simular la ruleta, los organismos más aptos tienen un área más grande y se seleccionan con mayor frecuencia.
En este ejemplo la aptitud del organismo A es mayor a la del organismo B.
La selección tipo ruleta tiene varios inconvenientes, uno de ellos es que si existe un organismo con una aptitud significativamente mejor a la de los demás organismos, la selección tipo ruleta tendrá un sesgo muy fuerte hacia este organismo, resultando en convergencia temprana a soluciones subóptimas.
}
\end{frame}
\begin{frame}
\frametitle{Selección tipo torneo}
\begin{center}
\includegraphics[scale=0.15]{selecciontorneo.pdf}
\end{center}
\note{
Por ello existen otras formas de seleccionar organismos. Una muy efectiva es la selección tipo torneo.
Se define un tamaño de torneo \emph{k}. Por ejemplo torneo binario o entre dos soluciones.
El torneo se forma tomando una muestra aleatoria de \emph{k} soluciones de la población.
Se define como ganador del torneo el organismo con la mayor aptitud de la muestra.
Este ejemplo ilustra dos torneos binarios.
Las soluciones candidatas seleccionadas son las rutas más cortas en cada caso.
Al ser una muestra aleatoria no hay garantia de que el mejor individuo se seleccione en un torneo.
}
\end{frame}
\begin{frame}
\frametitle{Reproducción}
\begin{center}
\includegraphics[scale=0.55]{cruza.pdf}
\end{center}
\note{Las soluciones candidatas seleccionadas se reproducen para generar nuevas soluciones.
Hay muchas formas de hacer la recombinación de los genotipos.
Aquí ilustramos una muy simple. La cruza requiere dos padres y genera dos hijos.
Seleccionamos una región aleatoria de la permutación.
Esta región es copiada a cada uno de los hijos.
Después, considerando que el cromosoma es circular, colocamos los números que vienen de posiciones subsecuentes del otro padre, siempre que no se repitan números.
En el ejemplo el primer hijo tiene la secuencia 7,1,3 del primer padre.
Despues tomamos la secuencia del segundo padre a partir de esa posición.
Copiamos el número 9, considerando el cromosoma circular le sigue el número 2,
luego el 3, pero este se repite, por lo que no lo copiamos, luego copiamos el 6, sigue el 1, pero se repite por lo que no lo copiamos, copiamos el cero, sigue el 7 pero se repite, copiamos 8, 4 y 5.
}
\end{frame}
\begin{frame}
\frametitle{Mutación}
\begin{center}
\includegraphics[scale=0.55]{mutacion.pdf}
\end{center}
\note{
La mutación es otro operador genético del algoritmo.
Un ejemplo de mutación es el intercambio aleatorio de ciudades.
En el ejemplo tras la mutación se han intercamibado las posiciones de las ciudades 6 y 7.
}
\end{frame}
\begin{frame}
\frametitle{Evolución}
En la evolución aplicamos estas operaciones en el orden siguiente:
\begin{enumerate}
\item Generar población inicial
\item Repetir cierto número de generaciones
\begin{enumerate}
\item Asignar aptitud de la población actual
\item Seleccionar mejores soluciones
\item Cruzar las soluciones generando nueva población
\item Mutar la nueva población
\item La población actual es la nueva población
\end{enumerate}
\end{enumerate}
\note{
Aquí un algoritmo genético muy simple.
Generamos la población inicial con cromosomas aleatorios.
Repetimos los pasos siguientes cierto número de generaciones.
Decodificamos y evaluamos a los organismos en el problema para obtener aptitud.
Seleccionamos a los mejores.
Reproducimos a los organismos seleccionados.
Mutamos a los organismos.
Hacemos la población actual igual a la nueva población.
}
\end{frame}
\begin{frame}
\frametitle{Comentario final.}
Cómputo evolutivo,
dentro de la especialización Introducción a la inteligencia artificial de la UNAM.
\note{
Los algoritmos genéticos son metaheurísticos muy poderosos.
Aquí solo presentamos una versión muy simple.
Hay muchas variaciones de estos algoritmos.
Si quieres saber más respecto de estos algoritmos te recomendamos tomar el curso:
Cómputo evolutivo,
dentro de la especialización Introducción a la inteligencia artificial de la UNAM.}
\end{frame}
\end{document}
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curso-search/geneticos/poblacion.svg 0 → 100644
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curso-search/geneticos/población.svg 0 → 100644
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curso-search/geneticos/seleccion.svg 0 → 100644
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curso-search/geneticos/selecciontorneo.svg 0 → 100644
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mdps.log
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......@@ -243,7 +243,7 @@ En lo que trataremos asumimos que el agente actua en forma racional.
Para el estado $s$ tenemos que el valor esperado de utilidad es: \\
\begin{center}
$\sum\limits_{s'} {P(S^{t+1}{=}s' \mid S^{t+1}{=}s) V(s')}$
$\sum\limits_{s'} {P(S^{t+1}{=}s' \mid S^{t}{=}s) V(s')}$
\end{center}
\note{
Recordemos que para una loteria podemos calcular su utilidad sumando los productos de las utilidades de las salidas con su probabilidad respectiva.
......@@ -263,7 +263,7 @@ Aquí el valor de utilidad $V(s)$ no será la recompensa instantanea sino la utili
\begin{itemize}
\item La utilidad {\color{blue}\emph{horizonte infinito}} considera la recompensa instantanea y el valor esperado de recompensa para tiempos futuros con su factor de descuento:\\
\begin{center}
$V(s) = R(s) + \gamma \sum\limits_{s'} {P(S^{t+1}{=}s' \mid S^{t+1}{=}s) V(s')}$
$V(s) = R(s) + \gamma \sum\limits_{s'} {P(S^{t+1}{=}s' \mid S^{t}{=}s) V(s')}$
\end{center}
\item Matricialmente:\\
\begin{center}
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\documentclass[xcolor=dvipsnames]{beamer}
\usepackage[spanish]{babel}
\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{pgfpages}
% descomentar para generar handouts con notas
%\setbeameroption{show notes on second screen=right}
\usepackage{tikz}
\tikzstyle{every picture}+=[remember picture]
\usefonttheme[onlymath]{serif}
\usepackage{verbatim}
\usepackage{pifont}
\usetheme{Antibes}
\usecolortheme{lily}
\usepackage{graphicx}
\usepackage{hyperref}
\newcommand{\I}[1]{\mathit{#1}}
\newcommand{\cierto}{\I{cierto}}
\newcommand{\falso}{\I{falso}}
\DeclareMathOperator*{\argmax}{argmax}
\spanishdecimal{.}
\title{Procesos de Decisión de Markov}
\author{Stalin Muñoz y David A.\ Rosenblueth}
\institute{
Instituto de Investigaciones en Matem\'aticas Aplicadas y en Sistemas (IIMAS)\\
Universidad Nacional Aut\'onoma de Mexico (UNAM)}
\date{}
\begin{document}
\parindent 2em
\frame
{
\titlepage
}
\section{Contenido}
\begin{frame}
\frametitle{Contenido}
\begin{block}{Contenido}
\begin{enumerate}
\item Conceptos preliminares
\item Políticas de un agente
\item Estados con utilidad en Cadenas de Markov
\item Procesos de decisión de Markov
\item Algoritmo de iteración de valores
\end{enumerate}
\end{block}
\note{
Hoy presentaremos un formalismo matemático
para toma de decisiones en entornos con incertidumbre.
Los procesos de decisión de Markov.
}
\end{frame}
\section{Conceptos preliminares}
\begin{frame}[<+->]
\frametitle{Conceptos preliminares}
\begin{enumerate}
\item Estado = conjunto de asignaciones de valores para un
conjunto de variables. Por ejemplo: \\
\{lloviendo = cierto, puerta = abierta, pasajeros = 237, . . . \}
\note<1>{
Un estado describe las variables relevantes del problema
}
\item Utilidad = función que asigna a un estado un valor real que representa de manera cuantitativa las preferencias del agente.
\note<2>{
Asumiremos que existe una función de utilidad definida para todo estado del problema.
Esta función asigna un número real a los estados del problema. Entre mayor sea la preferencia del agente, mayor es el número asignado por la función de utilidad.
}
\item Loteria = un experimento aleatorio en el que una variable
toma un valor particular en un espacio muestral.
\note<3>{
Nuestra loteria representado un evento probabilístico.
}
\item Utilidad de una loteria = Valor esperado de utilidad para un experimento aleatorio.
\note<4>{
La utilidad de la lotería entendida como el valor esperado de utilidad de la misma.
}
\item Cadena de markov = Modelo probabilistico de la evolución temporal de la distribución de probabilidad para los estados de un sistema que tiene la propiedad de Markov.
\note<5>{
Las cadenas de Markov nos servirán como base para definir los procesos de decisión de Markov.
}
\item Decision = Acción que toma un agente racional para maximizar su utilidad.
\note<6>{
En lo que trataremos asumimos que el agente actua en forma racional.
}
\end{enumerate}
\end{frame}
\section{Políticas de un agente}
\begin{frame}[<+->]
\frametitle{Política óptima}
\begin{itemize}
\item La {\color{blue}política} de un agente es una función $\pi: S \to A$ que mapea los estados del problema en acciones del agente.
\note<1>{
Nos interesa encontrar una política para nuestro agente artificial.
La política de un agente puede entenderse formalmente como una función que le indica al agente que acción tomar en cada situación.
}
\item La {\color{blue} política óptima} es aquella política que maximiza el valor esperado de utilidad.
\note<2>{
El agente racional tomará las decisiones que más le convienen.
}
\item En un problema de decisiones secuenciales, o iteradas, existen diferentes formas de definir la política óptima:
\note<3>{
Pero, ¿qué es lo que más le conviene al agente?
Aunque esto puede depender mucho del problema,
podemos decir que hay maneras estándar de tratar la optimalidad para la utilidad esperada de un agente en un entorno de decisiones secuenciales.
}
\begin{itemize}
\item {\color{blue}miope}: solo le interesa maximizar la utilidad esperada del tiempo siguiente,
\note<4>{
Una política miope se limitará a tomar la acción que consigue el mayor beneficio inmediato.
Esta estrategia puede ser insuficiente en entornos complejos.
}
\item {\color{blue}horizonte finito}: le interesa maximizar la utilidad en una ventana de tiempo definido, y
\note<5>{
Una política de horizonte finito tratará de maximizar la suma de las utilidades para una ventana de tiempo en el futuro.
Por ejemplo, si el agente toma la acción que maximiza la utilidad para los próximos 7 días y no solo maximiza la utilidad del día de mañana.
}
\item {\color{blue}horizonte infinito}: se maximiza la utilidad para todo tiempo futuro.
\note<6>{
La política horizonte infinito considera todos los tiempos futuros.
Esta política es la más conveniente para el análisis matemático. Y es la que vamos a utilizar.
}
\end{itemize}
\end{itemize}
\end{frame}
\section{Estados con utilidad en Cadenas de Markov}
\begin{frame}[<+->]
\frametitle{Utilidad esperada en cadenas de Markov}
\note<1>{
Utilizamos cadenas de Markov para modelar procesos secuenciales estocásticos.
Observamos que la probabilidad de llegar a un estado al tiempo siguiente solo depende del estado en el que nos encontramos en el tiempo actual.
Representamos la cadena de Markov con un grafo de transición de estados.
}
\note<2>{
Definimos una utilidad para cada estado posible.
En nuestro ejemplo podría ser el consumo promedio por unidad de tiempo en cada uno de los vagones.
Para el vagón comedor 27.
Para el bar 10.
En el dormitorio no hay consumo.
}
\note<3>{
En adelante usaremos la letra $R$ en lugar de la letra $U$.
La $R$ significará \emph{recompensa}.
}
\note<4>{
A la recompensa de un estado la denotaremos como recompensa instantanea.
Considerando que nos interesa encontrar una política horizonte infinito, plantearemos una utilidad que integra el valor esperado de utilidad para todos los estados futuros.
}
\note<5>{
El valor de la recompensa en tiempos futuros se verá afectado por un factor de depreciación o descuento.
Una recompensa vale más hoy que mañana, y vale más mañana que pasado mañana.
Aún no tomamos en cuenta las acciones del agente, pero esto lo haremos más adelante.
}
\begin{minipage}{0.5\textwidth}
\includegraphics<1>[scale=0.8]{mdp00.pdf}
\includegraphics<2>[scale=0.8]{mdp0.pdf}
\includegraphics<3->[scale=0.8]{mdp1.pdf}
\end{minipage}
\begin{minipage}{0.4\textwidth}
\only<4>{
\begin{itemize}
\item Nos interesa conocer la utilidad de cada estado pero considerando también tiempos futuros.
\end{itemize}
}
\only<5>{
\begin{itemize}
\item Nos interesa conocer la utilidad de cada estado pero considerando también tiempos futuros.
\item Vamos a utilizar un factor de descuento {\color{blue}$\gamma \in (0,1]$} para ponderar la utilidad obtenida en tiempos futuros.
\end{itemize}
}
\end{minipage}
\end{frame}
\begin{frame}[<+->]
\frametitle{Utilidad esperada en cadenas de Markov}
\begin{minipage}{0.5\textwidth}
\includegraphics[scale=0.8]{mdp1.pdf}
\end{minipage}
\begin{minipage}{0.4\textwidth}
Podemos tratar cada nodo del grafo como una loteria.
Para el estado $s$ tenemos que el valor esperado de utilidad es: \\
\begin{center}
$\sum\limits_{s'} {P(S^{t+1}{=}s' \mid S^{t}{=}s) V(s')}$
\end{center}
\note{
Recordemos que para una loteria podemos calcular su utilidad sumando los productos de las utilidades de las salidas con su probabilidad respectiva.
Aquí podemos plantear la utilidad de un estado de la misma manera.
Para obtener la utilidad de un estado $s$ sumamos los productos de la utilidad de cada estado $s'$ al que podemos llegar desde $s$ con su respectiva probabilidad.
Aquí el valor de utilidad $V(s)$ no será la recompensa instantanea sino la utilidad horizonte infinito.
}
\end{minipage}
\end{frame}
\begin{frame}[<+->]
\frametitle{Utilidad esperada en cadenas de Markov}
\begin{itemize}
\item La utilidad {\color{blue}\emph{horizonte infinito}} considera la recompensa instantanea y el valor esperado de recompensa para tiempos futuros con su factor de descuento:\\
\begin{center}
$V(s) = R(s) + \gamma \sum\limits_{s'} {P(S^{t+1}{=}s' \mid S^{t}{=}s) V(s')}$
\end{center}
\item Matricialmente:\\
\begin{center}
$V\left(s\right) = R(s) + \gamma T^\intercal V\left(s\right)$
\end{center}
\item El cual podemos resolver: \\
\begin{center}
$V\left(s\right) = \left(I - \gamma T^\intercal\right)^{-1} R\left(s\right)$
\end{center}
\end{itemize}
\note<1>{
La expresión para el valor horizonte infinito de utilidad nos lleva a una ecuación donde incorporamos tanto la recompensa instantanea como el valor esperado de utilidad para tiempos futuros.
Aplicamos el factor de descuento $\gamma$ para las recompensas esperadas en tiempos futuros.
}
\note<2>{
Observamos que la expresión del término que corresponde con la utilidad de estados futuros puede expresarse como un producto matricial.
El producto de la transpuesta de la matriz de probabilidades de transición $T$ con el vector de utilidades horizonte infinito $V(s)$
Aquí la matriz $T$ es la matriz de probabilidades de transición de una cadena de Markov.
}
\note<3>{
Esta ecuación lineal puede resolverse invirtiendo una matriz.
}
\end{frame}
%\begin{frame}[<+->]
%\frametitle{Utilidad esperada en cadenas de Markov}
%\only<1>{
%\begin{center}
%\includegraphics[scale=0.8]{markovu5.pdf}
%\end{center}
%}
%\note<1>{
%
%En nuestra cadena de Markov de ejemplo vamos a calcular las utilidades horizonte infinito.
%}
%\only<2>{
%\begin{center}
%\includegraphics[scale=0.8]{markovu4.pdf}
%\end{center}
%}
%\note<2>{
%
%Primero definimos un factor de descuento $\gamma$ de $0.9$.
%}
%\only<3>{
%\begin{center}
%\includegraphics[scale=0.8]{markovu3.pdf}
%\end{center}
%}
%\note<3>{
%
%Colocamos las recompensas instantaneas en el vector $R(s)$ y las probabilidades de transición como entradas de $T$.
%}
%\only<4>{
%\begin{center}
%\includegraphics[scale=0.8]{markovu2.pdf}
%\end{center}
%}
%\note<4>{
%
%Para resolver habrá que calcular la inversa de la diferencia entre la matriz identidad y el producto del factor de descuento con la matriz de probabilidades de transición.
%}
%\only<5>{
%\begin{center}
%\includegraphics[scale=0.8]{markovu1.pdf}
%\end{center}
%}
%\note<5>{
%
%Sustituimos los valores y resolvemos.
%}
%\only<6>{
%\begin{center}
%\includegraphics[scale=0.8]{markovu0.pdf}
%\end{center}
%}
%\note<6>{
%
%Obteniendo los valores de utilidad horizonte infinito.
%
%Esto nos dice que el estado con mayor utilidad es el del vagón comedor y el de menor utilidad es el del bar.
%}
%\end{frame}
%
\section{Procesos de decisión de Markov}
\begin{frame}[<+->]
\frametitle{Incorporando las acciones del agente}
\begin{itemize}
\item En los {\color{blue}procesos de decisión de Markov} las acciones del agente cambian las probabilidades de transición de estado.
\note<1>{
Ahora vamos a incorporar las acciones del agente en la ecuación.
Tenemos un proceso estocástico donde el agente no puede anticipar con certidumbre el efecto de sus acciones.
A pesar de esto, las acciones sí tienen un efecto probabilístico.
Es decir, las probabilidades de transición de estado no sólo dependerán del estado actual, sino también de la acción que se toma.
De no ser así, cualquier política sería igual de buena y no habría problema que resolver.
}
\item Asumiremos que el agente es {\color{blue}racional} y las acciones son aquellas que maximizan la utilidad horizonte infinito.
\note<2>{
En el proceso de decisión de Markov nuestro agente es racional.
Es decir, desea maximizar su recompensa horizonte infinito.
}
\item El proceso de decisión de Markov se modela con la {\color{blue}ecuación de Bellman}: \\
\begin{center}
$V(s) = R(s) + \gamma \max\limits_{{\color{blue}a}}{\sum\limits_{s'} P(s'\mid s,{\color{blue}a})V(s')}$
\end{center}
\note<3>{
El proceso de decisión de Markov queda descrito por la ecuación de Bellman.
Observamos que aparece un término no lineal $\max$ el cual modela el efecto de la acción racional del agente, aquí representada con la letra $a$.
Adicionalmente se observa que la probabilidad de transición no solo está condicionada al estado actual, sino también a la acción del agente.
}
\end{itemize}
\end{frame}
\begin{frame}[<+->]
\frametitle{Un grafo de decisiones}
\only<1>{
\begin{center}
\includegraphics[scale=0.4]{decisiongraph8.pdf}
\end{center}
}
\note<1>{
En teoría de decisiones encadenamos nodos de decisión con loterias formando árboles de decisión.
Los procesos de decisión de Markov pueden interpretarse de manera similar, pero dado que resultarían árboles infinitos con probabilidades de transición estacionarias es mejor tratarlos como grafos de decisión.
Aquí comenzamos con los estados los que representaremos con nodos de decisión.
}
\only<2>{
\begin{center}
\includegraphics[scale=0.4]{decisiongraph7.pdf}
\end{center}
}
\note<2>{
Por ejemplo, un pasajero se encuentra en el vagón comedor. Nuestro agente, es decir el tren inteligente, puede decidir entre un conjunto de acciones para tratar de cambiar la probabilidad de que el pasajero continúe o transite a otro vagón.
}
\only<3>{
\begin{center}
\includegraphics[scale=0.4]{decisiongraph6.pdf}
\end{center}
}
\note<3>{
Una acción posible es hacer un anuncio publicitario en el vagón. Esto no definirá la acción del pasajero, pero si tendrá un efecto en la probabilidad de transición de estado.
}
\only<4>{
\begin{center}
\includegraphics[scale=0.4]{decisiongraph5.pdf}
\end{center}
}
\note<4>{
La transición la representamos con un nodo de lotería.
}
\only<5>{
\begin{center}
\includegraphics[scale=0.4]{decisiongraph4.pdf}
\end{center}
}
\note<5>{
La lotería definirá el estado siguiente.
}
\only<6>{
\begin{center}
\includegraphics[scale=0.4]{decisiongraph3.pdf}
\end{center}
}
\note<6>{
En este caso podemos transitar a cualquiera de los vagones: quedarnos en el comedor, ir al dormitorio o al bar.
Cada arista de la lotería tiene anotada la probabilidad de transición.
Para no saturar la figura solo ilustramos la probabilidad de transitar al dormitorio.
Observamos que esta probabilidad esta condicionada a encontrarnos en el vagón comedor y que el tren haga el anuncio 1.
}
\only<7>{
\begin{center}
\includegraphics[scale=0.4]{decisiongraph2.pdf}
\end{center}
}
\note<7>{
También es posible que el tren no haga anuncio alguno.
En este caso tenemos otra loteria para transitar de estado.
}
\only<8>{
\begin{center}
\includegraphics[scale=0.4]{decisiongraph1.pdf}
\end{center}
}
\note<8>{
Así para el estado $b$ tenemos acciones similares y las loterias respectivas.
}
\only<9>{
\begin{center}
\includegraphics[scale=0.4]{decisiongraph0.pdf}
\end{center}
}
\note<9>{
Y para el estado $d$.
La pregunta relevante que hacer es ¿cuál es la política óptima?
Es decir, ¿que acción debe tomar el tren en cada vagón, tal que maximice la utilidad horizonte infinito?
}
\end{frame}
\begin{frame}
\frametitle{Política óptima}
\begin{center}
\includegraphics[scale=0.4]{decisiongraphPi.pdf}
\end{center}
\note{
De la ecuación de Bellman podemos ver que la política óptima $\Pi^*(s)$ se obtiene con la acción que maximiza el valor esperado de utilidad para cada estado.
}
\end{frame}
\begin{frame}[<+->]
\frametitle{Política óptima: ¿cómo encontrar $\Pi^*(s)$?}
\begin{itemize}
\item Deseamos resolver $\Pi^{*}:S \to A$:\\
\begin{center}
$\Pi^*(s) = \argmax\limits_{a}{\sum\limits_{s'}{P(s' \mid s,a)}V(s')}$
\end{center}
\note<1>{
El problema de diseño del agente racional consiste en encontrar la política óptima $\Pi^{*}$. Dada un estado $s$ que acción debe tomar el agente.
}
\item Algunos algoritmos para encontrar $\Pi^*(s)$ son:
\note<2>{
Existen varias formas de resolver la política óptima.
}
\begin{itemize}
\item {\color{blue} Iteración de valores}: Primero resolvemos $V(s)$ en la ecuación de Bellman. Usamos $V(s)$ para encontrar $\Pi^*(s)$.
\note<3>{
El algoritmo de iteración de valores encuentra primero $V(s)$ iterando la ecuación de Bellman hasta que los valores convergen.
}
\item {\color{blue} Iteración de políticas}: Se fija la política y se resuelve un sistema de ecuaciones para encontrar $V(s)$, repitiendo hasta que la política no cambie.
\note<4>{
El algoritmo de iteración de políticas itera la política hasta que converge sin preocuparse de encontrar los valores de $V(s)$.
}
\end{itemize}
\end{itemize}
\end{frame}
\begin{frame}[<+->]
\frametitle{Algoritmo de iteración de políticas}
\only<1>{
\begin{center}
\includegraphics[scale=0.9]{mdpejemplo0.pdf}
\end{center}
}
\note<1>{
El algoritmo de iteración de políticas esta basado en el hecho de que dada una política conocida $\pi$, el proceso de decisión de Markov puede tratarse como una cadena de Markov.
Regresamos al ejemplo del tren.
Ahora hemos anota las aristas con tuplas de probabilidades.
Cada elemento de la tupla corresponde con una acción distinta.
En este ejemplo el primer elemento de la tupla corresponde con no anunciar,
el segundo elemento corresponde con la acción \emph{Anuncio}.
}
\only<2>{
\begin{center}
\includegraphics[scale=0.9]{mdpejemplo1.pdf}
\end{center}
}
\note<2>{
La política es una función.
Para cada estado nos dice que acción tomar.
Esta función puede representarse como una lista de acciones.
La posición de la acción es la misma que la del estado correspondiente.
}
\only<3>{
\begin{center}
\includegraphics[scale=0.9]{mdpejemplo2.pdf}
\end{center}
}
\note<3>{
Podemos pensar en un arreglo $T$ que tiene como entradas las tuplas.
La política selecciona de cada tupla un elemento particular.
En este caso si la política para todo estado es no anunciar, entonces nuestra matriz de probabilidades de transición es la misma que en el ejemplo usado en el tema de cadenas de Markov.
}
\only<4>{
\begin{center}
\includegraphics[scale=0.9]{mdpejemplo3.pdf}
\end{center}
}
\note<4>{
Encontraremos la utilidad horizonte infinitio para cada estado.
}
\only<5>{
\begin{center}
\includegraphics[scale=0.9]{mdpejemplo4.pdf}
\end{center}
}
\note<5>{
Observese que en la ecuación de Bellman matricial,
la matriz de transición de la cadena de Markov
aparece en forma transpuesta.
}
\only<6>{
\begin{center}
\includegraphics[scale=0.9]{mdpejemplo5.pdf}
\end{center}
}
\note<6>{
Sustituimos...
}
\only<7>{
\begin{center}
\includegraphics[scale=0.9]{mdpejemplo6.pdf}
\end{center}
}
\note<7>{
Obtenemos los valores horizonte infinito.
}
\only<8>{
\begin{center}
\includegraphics[scale=0.9]{mdpejemplo7.pdf}
\end{center}
}
\note<8>{
En este paso vamos a recalcular la política óptima $\pi_1(s)$
}
\only<9>{
\begin{center}
\includegraphics[scale=0.9]{mdpejemplo8.pdf}
\end{center}
}
\note<9>{
Para el estado $d$ encontraremos la acción que maximiza la utilidad.
Así quedan las sumatorias.
}
\only<10>{
\begin{center}
\includegraphics[scale=0.9]{mdpejemplo9.pdf}
\end{center}
}
\note<10>{
Sustituimos valores...
}
\only<11>{
\begin{center}
\includegraphics[scale=0.9]{mdpejemplo10.pdf}
\end{center}
}
\note<11>{
Tomaremos la acción que da el valor máximo.
}
\only<12>{
\begin{center}
\includegraphics[scale=0.9]{mdpejemplo11.pdf}
\end{center}
}
\note<12>{
En este caso hacer el anuncio resulta en una mayor utilidad.
}
\only<13>{
\begin{center}
\includegraphics[scale=0.9]{mdpejemplo12.pdf}
\end{center}
}
\note<13>{
$\pi_1$ queda definida como hacer el anuncio si se esta en el dormitorio o el comedor pero no hacerlo si se está en el bar.
Observamos que $\pi_0$ es diferente de $\pi_1$ por lo que el algoritmo continuará iterando.
Calculamos las probabilidades de transición para $\pi_1$ y resolvemos el sistema de ecuaciones obteniendo los nuevos valores de utilidad horizonte infinito $V_1(s)$.
}
\only<14>{
\begin{center}
\includegraphics[scale=0.9]{mdpejemplo13.pdf}
\end{center}
}
\note<14>{
Recalculamos la política óptima $\pi_2$ observando que el algoritmo convergió.
El algoritmo termina.
}
\end{frame}
\end{document}
python/.ipynb_checkpoints/dpll-checkpoint.ipynb 0 → 100644
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Algoritmo DPLL"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"El algoritmo DPLL..."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Literales"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {},
"outputs": [],
"source": [
"import re\n",
"class Literal:\n",
" \n",
" # regular expression for a valid variable name\n",
" ID_REGEXP = re.compile(r\"^[^\\d\\W]\\w*\\Z\", re.UNICODE)\n",
" \n",
" \"\"\"\n",
" Propositional sentence tha is a variable or a negated variable\n",
" \"\"\"\n",
" \n",
" def __init__(self,variable,positive=True):\n",
" \"\"\"\n",
" Creates the literal for the given variable\n",
" :param variable: the variable of the literal\n",
" :param positive: positive literal if True, negative literal if false\n",
"\n",
" \"\"\"\n",
" if re.match(Literal.ID_REGEXP,variable):\n",
" self.variable = variable\n",
" else:\n",
" raise SyntaxError(\"Invalid variable name: '\"+variable+\"'\")\n",
" self.positive = positive\n",
" \n",
" @staticmethod\n",
" def parse(s):\n",
" s = s.replace(\" \",\"\")\n",
" positive = not s.startswith(\"~\")\n",
" return Literal(s if positive else s[1:],positive)\n",
"\n",
" def __repr__(self):\n",
" return self.__str__()\n",
" \n",
" def __str__(self):\n",
" return (\"\" if self.positive else \"~\")+self.variable\n",
" \n",
" def __hash__(self):\n",
" return hash(self.__str__())\n",
" \n",
" def __eq__(self,other):\n",
" return self.__str__()==other.__str__()\n",
" \n",
" def __invert__(self):\n",
" return Literal(self.variable,not self.positive)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Ejemplo para crear las literales $L_1=A$ y $L_2 = \\neg B$"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"A ~B C ~C\n"
]
}
],
"source": [
"l1 = Literal(\"A\")\n",
"l2 = Literal(\"B\",False)\n",
"l3 = Literal(\"C\")\n",
"l4 = ~l3\n",
"print(l1,l2,l3,l4)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Cláusulas"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {},
"outputs": [],
"source": [
"class Clause:\n",
" \"\"\"\n",
" A propositional clause\n",
" This class keeps a singleton pattern of the clause with frozen hash\n",
" so that any simplification can be propagated to sets and dictionaries\n",
" in a single operation\n",
" \"\"\"\n",
" \n",
" def __init__(self, literals=None,frozen_hash=True):\n",
" \"\"\"\n",
" cretates the clause with the provided set of literals\n",
" :param literals: the set of literals\n",
" :param frozen_hash: if true the hash is not based on content\n",
" \"\"\"\n",
" self.frozen_hash = frozen_hash\n",
" if literals:\n",
" self.literals = frozenset(literals) if frozen_hash else literals\n",
" else:\n",
" self.literals = set()\n",
" \n",
" def __hash__(self):\n",
" \"\"\"\n",
" the hash number is the memory location when the flag\n",
" frozen_hash is True, otherwise the hash is the same as the \n",
" hash of the internal frozenset containing the literals\n",
" :returns: the hash number of the clause\n",
" \"\"\"\n",
" return id(self) if self.frozen_hash else hash(self.literals)\n",
" \n",
" def __eq__(self,other):\n",
" \"\"\"\n",
" Equality between different instances of the same clause is only checked\n",
" when frozen_hash is False. If frozen_hash is True, the method will\n",
" return True, only for the same instance of the class\n",
" \"\"\"\n",
" if self.__hash__()==hash(other):\n",
" if self.frozen_hash:\n",
" return True\n",
" else:\n",
" return self.literals == other.literals\n",
" else:\n",
" return False\n",
" \n",
" def __iadd__(self,literal):\n",
" \"\"\"\n",
" adds the literal to the clause\n",
" :param literal: the literal to add\n",
" \"\"\"\n",
" if isinstance(literal,Literal):\n",
" self.literals = self.literals.union({literal})\n",
" return self\n",
" else:\n",
" raise TypeError(\"An argument of type Literal was expected\")\n",
" \n",
" def __add__(self,literal):\n",
" \"\"\"\n",
" adds the literal to the clause\n",
" :param literal: the literal to add\n",
" \"\"\"\n",
" if isinstance(literal,Literal):\n",
" return Clause(self.literals.union({literal}),self.frozen_hash)\n",
" else:\n",
" raise TypeError(\"An argument of type Literal was expected\")\n",
" \n",
" def __isub__(self,literal):\n",
" \"\"\"\n",
" deletes a literal from the clause\n",
" :param literal: the literal to substract\n",
" \"\"\"\n",
" if isinstance(literal, Literal):\n",
" self.literals = self.literals.difference({literal})\n",
" return self\n",
" else:\n",
" raise TypeError(\"A literal of type string was expected\")\n",
"\n",
" def __sub__(self,literal):\n",
" \"\"\"\n",
" deletes a literal from the clause\n",
" :param literal: the literal to substract\n",
" \"\"\"\n",
" if isinstance(literal, Literal):\n",
" return Clause(self.literals.difference({literal}),self.frozen_hash)\n",
" else:\n",
" raise TypeError(\"A literal of type string was expected\")\n",
" \n",
" def __str__(self):\n",
" return \"( \"+\" | \".join(map(str,self.literals))+\" )\"\n",
" \n",
" def __repr__(self):\n",
" return self.__str__()\n",
" \n",
" def __iter__(self):\n",
" return (i for i in self.literals)\n",
" \n",
" def __len__(self):\n",
" return len(self.literals)\n",
" \n",
" def copy(self):\n",
" return Clause(self.literals.copy(),self.frozen_hash)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Ejemplo creamos una cláusula $c_1$ formada por las literales $l_1$ y $l_2$"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"( A | ~B )\n"
]
}
],
"source": [
"c1 = Clause({l1,l2})\n",
"print(c1)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Fórmula en Forma Normal Conjuntiva"
]
},
{
"cell_type": "raw",
"metadata": {},
"source": [
"Explicar forma normal."
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {},
"outputs": [],
"source": [
"import logging as log\n",
"class FormulaCNF:\n",
" \"\"\"\n",
" Formula in Conjunctive Normal Form\n",
" :param formula: CNF formula (string to parse)\n",
" example: '(A|!B)&(!A|B|C)'\n",
" \"\"\"\n",
" def __init__(self,formula=None,assignment=None):\n",
" #removes all white spaces\n",
" if formula:\n",
" formula = ''.join(formula.split())\n",
" (self.variables,self.clauses) = self.parse(formula)\n",
" self.build_dicts() \n",
" if assignment:\n",
" self.assignment = assignment.copy()\n",
" else:\n",
" self.assignment = set()\n",
" \n",
" if log.getLogger().isEnabledFor(log.DEBUG):\n",
" log.debug(self.string_internals())\n",
" \n",
" def __getitem__(self,literal):\n",
" \"\"\"\n",
" short notation for the simplify method\n",
" :returns: the simplified formula resulting from assuming the literal\n",
" is true\n",
" \"\"\"\n",
" return self.simplify(literal)\n",
" \n",
" def empty_sentence(self):\n",
" \"\"\"\n",
" :returns: true if the formula has no clauses\n",
" \"\"\"\n",
" return not self.clauses\n",
" \n",
" def empty_clause(self):\n",
" \"\"\"\n",
" :returns: true if there is an empty clause\n",
" \"\"\"\n",
" return 0 in self.n_to_c\n",
" \n",
" def get_unit_clause_literal(self):\n",
" \"\"\"\n",
" Gets a unit clause if there is one, None otherwise\n",
" \"\"\"\n",
" clause = next(iter(self.n_to_c[1])) \\\n",
" if 1 in self.n_to_c else {}\n",
" return next(iter(clause)) if clause else None\n",
"\n",
" def get_pure_literal(self):\n",
" \"\"\"\n",
" Gest a pure literal\n",
" \"\"\"\n",
" return next(iter(self.P)) if self.P else None\n",
" \n",
" def get_variable_literal(self):\n",
" \"\"\"\n",
" Obtiene una variable de la formula\n",
" \"\"\"\n",
" return Literal(next(iter(self.variables))) if self.variables else None\n",
"\n",
" def simplify(self,literal):\n",
" \"\"\"\n",
" Simplifies the current sentence assming the provided literal\n",
" :param literal: the literal to assume as true\n",
" \"\"\"\n",
" log.debug(\"simplifying literal: \"+str(literal))\n",
" if not literal:\n",
" raise ValueError(\n",
" \"Invalid literal provided as argument: \"+ str(literal))\n",
" \n",
" # stores the assignment\n",
" self.assignment.add(literal)\n",
" # deletes the variable of the literal\n",
" self.variables = self.variables - {literal.variable}\n",
" # updates the list of literals\n",
" self.L = self.L -{literal}\n",
" # updates the list of pure literals\n",
" if self.isPureLiteral(literal):\n",
" self.P = self.P - {literal}\n",
" # deals with the clauses to delete\n",
" if literal in self.l_to_c:\n",
" for clause in self.l_to_c[literal].copy():\n",
" self.remove_clause(clause)\n",
" #TODO remover las cláusulas de otras entradas en l_to_c\n",
" # update data structures for other literals\n",
" for l in clause.copy():\n",
" self.decrease_literal_count(l)\n",
" self.del_from_dictionary_of_sets(\n",
" self.l_to_c,l,clause,True)\n",
" #removes the entry from dicionary literal to clauses\n",
" #del self.l_to_c[literal]\n",
" \n",
" # deals with the literals to delete negated literal\n",
" neg_literal = ~literal \n",
" # updates list of literals\n",
" self.L = self.L - {neg_literal}\n",
" # deletes all negated literals from clauses\n",
" if neg_literal in self.l_to_c:\n",
" for clause in self.l_to_c[neg_literal]:\n",
" self.remove_literal_from_clause(neg_literal,clause)\n",
" # delete entry for negated literal from dictionary literal to clauses\n",
" if neg_literal in self.l_to_c:\n",
" del self.l_to_c[neg_literal]\n",
" # updates list of negated literals\n",
" if self.isPureLiteral(neg_literal):\n",
" self.P = self.P - {neg_literal}\n",
"\n",
" log.debug(\"simplified formula:\")\n",
" if log.getLogger().isEnabledFor(log.DEBUG):\n",
" log.debug(self.string_internals())\n",
" return self\n",
" \n",
" def remove_clause(self,clause):\n",
" \"\"\"\n",
" Removes the clause from data structures\n",
" :parama clause: the clause to delete\n",
" \"\"\"\n",
" # deletes clause from inverse map counts\n",
" self.clauses.remove(clause)\n",
" n = len(clause)\n",
" self.del_from_dictionary_of_sets(self.n_to_c,n,clause,True)\n",
" \n",
" def remove_literal_from_clause(self,literal,clause):\n",
" # frozen_hash allows for unique clause \n",
" # all clauses are updated (they are the same clause) \n",
" m = len(clause)\n",
" # remove clause from map of counts to clauses\n",
" self.del_from_dictionary_of_sets(self.n_to_c,m,clause,True)\n",
" # obtain simplified clause\n",
" clause -= literal\n",
" # update map of count to clauses\n",
" self.add_to_dictionary_of_sets(self.n_to_c,m-1,clause)\n",
" # update other counts related to the literal\n",
" self.decrease_literal_count(literal)\n",
" \n",
" \n",
" def decrease_literal_count(self,literal):\n",
" # decrease number to literal counts\n",
" # delete literal from previous count\n",
" n = self.l_to_n[literal]\n",
" self.del_from_dictionary_of_sets(self.n_to_l,n,literal,True)\n",
" # update the literal count if it is not zero\n",
" self.add_to_dictionary_of_sets(self.n_to_l,n-1,literal)\n",
" \n",
" # update counts for the literal\n",
" if literal in self.l_to_n:\n",
" self.l_to_n[literal] -= 1\n",
" else:\n",
" raise ValueError(\"Inconsistent counts for literal: \"+str(literal))\n",
" \n",
" def isPureLiteral(self,literal):\n",
" \"\"\"\n",
" :returns: true if the literal is pure\n",
" \"\"\"\n",
" return literal in self.P\n",
" \n",
" \n",
" def parse(self,formula):\n",
" \"\"\"\n",
" Parses the CNF formula string\n",
" \"\"\"\n",
" variables = set()\n",
" clauses = set()\n",
" clstr = re.findall(\"[^&]+\",formula)\n",
" for c in clstr:\n",
" literals = re.findall(\"[^\\(\\)\\|]+\",c)\n",
" clause = Clause()\n",
" for l in literals:\n",
" literal = Literal.parse(l)\n",
" variables.add(literal.variable)\n",
" clause += literal\n",
" clauses.add(clause)\n",
" return (variables,clauses)\n",
" \n",
" \n",
" def build_dicts(self):\n",
" \"\"\"\n",
" Builds dictionaries for efficient access\n",
" to clauses and literals\n",
" l_to_c: dictionary, keys are literals, values are clauses\n",
" n_to_c: dictionary, keys are sizes, values are clauses\n",
" l_to_n: dictionary, keys are literals, values are counts\n",
" n_to_c: dictionary, keys are counts, values are clauses\n",
" L: set, all literals in the formula\n",
" P: set, all pure literals in the formula\n",
" \"\"\"\n",
" # W gives fast access to clauses by literals\n",
" self.l_to_c = {}\n",
" # n gives fast access to clauses by size\n",
" self.n_to_c = {}\n",
" # literal to counts\n",
" self.l_to_n = {}\n",
" # counts to literals\n",
" self.n_to_l = {}\n",
" \n",
" for c in self.clauses:\n",
" # build dictionary indexed by size\n",
" self.add_to_dictionary_of_sets(self.n_to_c,len(c),c)\n",
" for l in c:\n",
" #builds dictionary indexed by literal\n",
" self.add_to_dictionary_of_sets(self.l_to_c,l,c)\n",
" #compute frecuency of literals\n",
" self.l_to_n[l] = self.l_to_n[l] + 1 if l in self.l_to_n else 1\n",
" \n",
" #reverse literal counts\n",
" self.L = set()\n",
" for k,v in self.l_to_n.items():\n",
" self.add_to_dictionary_of_sets(self.n_to_l,v,k)\n",
" self.L.add(k)\n",
" self.P = {v for v in self.L if ~v not in self.L}\n",
" \n",
" def add_to_dictionary_of_sets(self,d,k,v):\n",
" \"\"\"\n",
" Adds a value to a dictionary of sets\n",
" :param d: the dictionary\n",
" :param k: the key\n",
" :param v: the value to add\n",
" :param by_len: True if the key is the length of the set to add\n",
" \"\"\"\n",
" singleton = {v}\n",
" d[k] = d[k].union(singleton) if k in d else singleton\n",
" \n",
" def del_from_dictionary_of_sets(self,d,k,m,del_key):\n",
" \"\"\"\n",
" Deletes a set member from a dictionary of sets\n",
" :param d: the dictionary\n",
" :param k: the key\n",
" :parma m: the member of the set to delete\n",
" :del_key: if this flag is True the entry is removed when set is empty\n",
" \"\"\"\n",
" # removes with set difference\n",
" d[k] = d[k] - {m}\n",
" if del_key and not d[k]:\n",
" del d[k]\n",
" \n",
" def copy(self):\n",
" \"\"\"\n",
" Creates a shallow copy of the CNF formula\n",
" \"\"\"\n",
" log.debug(\"Creating a copy of the current CNF formula\")\n",
" formula = FormulaCNF(str(self),self.assignment)\n",
" return formula\n",
" \n",
" def string_internals(self):\n",
" return (\"clauses:\"+str(self.clauses)+\"\\n\"+\n",
" \"variables:\"+str(self.variables)+\"\\n\"+\n",
" \"assignment:\"+str(self.assignment)+\"\\n\"+\n",
" \"l_to_c:\"+str(self.l_to_c)+\"\\n\"+\n",
" \"n_to_c:\"+str(self.n_to_c)+\"\\n\"+\n",
" \"l_to_n:\"+str(self.l_to_n)+\"\\n\"+\n",
" \"n_to_l:\"+str(self.n_to_l)+\"\\n\"+\n",
" \"L:\"+str(self.L)+\"\\n\"+\n",
" \"P:\"+str(self.P))\n",
" \n",
" def __str__(self):\n",
" return \" & \".join(map(str,self.clauses))\n",
" \n",
" def __repr__(self):\n",
" return self.__str__()"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Ejemplo de fórmula $\\phi$ en FNC.\n"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"( A | ~B ) & ( ~A | C ) & ( ~B | ~C ) & ( C )\n"
]
}
],
"source": [
"phi = FormulaCNF(\"(A|~B)&(~A|C)&(~B|~C)&(C)\")\n",
"print(phi)"
]
},
{
"cell_type": "raw",
"metadata": {},
"source": [
"Ejemplos del uso de métodos"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"False"
]
},
"execution_count": 7,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#verificar si la formula es vacía\n",
"phi.empty_sentence()"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"False"
]
},
"execution_count": 8,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#¿tiene la formula una cláusula vacía?\n",
"phi.empty_clause()"
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"C\n"
]
}
],
"source": [
"#De existir una clúsula unitaria, obtiene la literal de la cláusula. Regresa None si no existe\n",
"l = phi.get_unit_clause_literal()\n",
"print(l)"
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"~B\n"
]
}
],
"source": [
"#Obtiene una literal pura de existir, None si no existe\n",
"l = phi.get_pure_literal()\n",
"print(l)"
]
},
{
"cell_type": "code",
"execution_count": 11,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"A\n"
]
}
],
"source": [
"#Obtiene una literal con una variable arbitraria de la formula phi\n",
"v = phi.get_variable_literal()\n",
"print(v)"
]
},
{
"cell_type": "code",
"execution_count": 12,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"( A | ~B ) & ( C ) & ( ~B | ~C ) & ( ~A | C )\n"
]
}
],
"source": [
"#Obtiene una copia de la formula\n",
"psi = phi.copy()\n",
"print(psi)"
]
},
{
"cell_type": "code",
"execution_count": 13,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"( ~A | C ) & ( C )\n"
]
}
],
"source": [
"#Simplifica la formula asumiendo que la literal l es verdadera\n",
"psi = phi[l]\n",
"print(psi)"
]
},
{
"cell_type": "code",
"execution_count": 15,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"{~B}\n"
]
}
],
"source": [
"#Obtiene las asignaciones aplicadas a una fórmula\n",
"a = phi.assignment\n",
"print(a)"
]
},
{
"cell_type": "raw",
"metadata": {},
"source": [
"La clase DLPP implementa..."
]
},
{
"cell_type": "code",
"execution_count": 14,
"metadata": {},
"outputs": [],
"source": [
"class DPLL:\n",
" \"\"\"\n",
" DPLL algorithm\n",
" \"\"\"\n",
" \n",
" @staticmethod\n",
" def satisfiable(phi):\n",
" \"\"\"\n",
" determines if phi is satisfiable\n",
" :param phi: a CNF formula\n",
" :returs: una tupla cuyo primer elemento indica\n",
" si la fórmula es satisfactible o no, y el segundo\n",
" la asignación que logró hacer la fórmula verdadera\n",
" \"\"\"\n",
" log.info(\"phi: \"+str(phi))\n",
" if phi.empty_sentence():\n",
" log.info(\"expresión vacía\")\n",
" return (True,phi.assignment)\n",
" \n",
" #inserta tu código aquí\n",
" "
]
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 3",
"language": "python",
"name": "python3"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 3
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.6.2"
}
},
"nbformat": 4,
"nbformat_minor": 2
}
python/.ipynb_checkpoints/markov-Copy1-checkpoint.ipynb 0 → 100644
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Cadenas de Markov"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Descripción.\n",
"![Cadena de Markov](\"markov.png\")"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": []
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 3",
"language": "python",
"name": "python3"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 3
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.6.2"
}
},
"nbformat": 4,
"nbformat_minor": 2
}
python/.ipynb_checkpoints/markov-checkpoint.ipynb 0 → 100644
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Cadenas de Markov"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Descripción.\n",
"![Cadena de Markov](markov.png)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Matriz de probabilidades de transición"
]
},
{
"cell_type": "code",
"execution_count": 13,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"[[ 0.4 0.1]\n",
" [ 0.6 0.9]]\n"
]
}
],
"source": [
"import numpy as np\n",
"\n",
"# define la matriz de probabilidades de transición\n",
"# utilizando el paquete de cómputo científico de Python numpy\n",
"\n",
"#A = | p(d|d), p(b|d), p(c|d) |\n",
"# | p(b|d), p(b|b), p(c|b) |\n",
"# | p(c|d), p(b|c), p(c|c) |\n",
"\n",
"# por ejemplo la matriz\n",
"#A = | 0.4, 0.1|\n",
"# | 0.6, 0.9|\n",
"# puede definirse en python como\n",
"A = np.matrix([[0.4,0.1],[0.6,0.9]])\n",
"\n",
"print(A)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Encontrar la distribución de probabilidad estacionaria elevando a un exponente grande"
]
},
{
"cell_type": "code",
"execution_count": 22,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"[[ 0.14285714 0.14285714]\n",
" [ 0.85714286 0.85714286]]\n",
"[[ 0.14285714 0.14285714]\n",
" [ 0.85714286 0.85714286]]\n"
]
}
],
"source": [
"B = np.linalg.matrix_power(A,50)\n",
"print(B)\n",
"B = np.linalg.matrix_power(A,100)\n",
"print(B)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Encontrar la misma distribución calculando los eigenvectores"
]
},
{
"cell_type": "code",
"execution_count": 23,
"metadata": {},
"outputs": [],
"source": [
"w,v = np.linalg.eig(A)"
]
},
{
"cell_type": "code",
"execution_count": 24,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"[ 0.3 1. ]\n",
"[[-0.70710678 -0.16439899]\n",
" [ 0.70710678 -0.98639392]]\n"
]
}
],
"source": [
"print(w)\n",
"print(v)"
]
},
{
"cell_type": "code",
"execution_count": 25,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"matrix([[ 0.14285714],\n",
" [ 0.85714286]])"
]
},
"execution_count": 25,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"v[:,1]/sum(v[:,1])"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": []
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 3",
"language": "python",
"name": "python3"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 3
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.6.2"
}
},
"nbformat": 4,
"nbformat_minor": 2
}
python/dpll.ipynb 0 → 100644
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Algoritmo DPLL"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"El algoritmo DPLL..."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Literales"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {},
"outputs": [],
"source": [
"import re\n",
"class Literal:\n",
" \n",
" # regular expression for a valid variable name\n",
" ID_REGEXP = re.compile(r\"^[^\\d\\W]\\w*\\Z\", re.UNICODE)\n",
" \n",
" \"\"\"\n",
" Propositional sentence tha is a variable or a negated variable\n",
" \"\"\"\n",
" \n",
" def __init__(self,variable,positive=True):\n",
" \"\"\"\n",
" Creates the literal for the given variable\n",
" :param variable: the variable of the literal\n",
" :param positive: positive literal if True, negative literal if false\n",
"\n",
" \"\"\"\n",
" if re.match(Literal.ID_REGEXP,variable):\n",
" self.variable = variable\n",
" else:\n",
" raise SyntaxError(\"Invalid variable name: '\"+variable+\"'\")\n",
" self.positive = positive\n",
" \n",
" @staticmethod\n",
" def parse(s):\n",
" s = s.replace(\" \",\"\")\n",
" positive = not s.startswith(\"~\")\n",
" return Literal(s if positive else s[1:],positive)\n",
"\n",
" def __repr__(self):\n",
" return self.__str__()\n",
" \n",
" def __str__(self):\n",
" return (\"\" if self.positive else \"~\")+self.variable\n",
" \n",
" def __hash__(self):\n",
" return hash(self.__str__())\n",
" \n",
" def __eq__(self,other):\n",
" return self.__str__()==other.__str__()\n",
" \n",
" def __invert__(self):\n",
" return Literal(self.variable,not self.positive)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Ejemplo para crear las literales $L_1=A$ y $L_2 = \\neg B$"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"A ~B C ~C\n"
]
}
],
"source": [
"l1 = Literal(\"A\")\n",
"l2 = Literal(\"B\",False)\n",
"l3 = Literal(\"C\")\n",
"l4 = ~l3\n",
"print(l1,l2,l3,l4)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Cláusulas"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {},
"outputs": [],
"source": [
"class Clause:\n",
" \"\"\"\n",
" A propositional clause\n",
" This class keeps a singleton pattern of the clause with frozen hash\n",
" so that any simplification can be propagated to sets and dictionaries\n",
" in a single operation\n",
" \"\"\"\n",
" \n",
" def __init__(self, literals=None,frozen_hash=True):\n",
" \"\"\"\n",
" cretates the clause with the provided set of literals\n",
" :param literals: the set of literals\n",
" :param frozen_hash: if true the hash is not based on content\n",
" \"\"\"\n",
" self.frozen_hash = frozen_hash\n",
" if literals:\n",
" self.literals = frozenset(literals) if frozen_hash else literals\n",
" else:\n",
" self.literals = set()\n",
" \n",
" def __hash__(self):\n",
" \"\"\"\n",
" the hash number is the memory location when the flag\n",
" frozen_hash is True, otherwise the hash is the same as the \n",
" hash of the internal frozenset containing the literals\n",
" :returns: the hash number of the clause\n",
" \"\"\"\n",
" return id(self) if self.frozen_hash else hash(self.literals)\n",
" \n",
" def __eq__(self,other):\n",
" \"\"\"\n",
" Equality between different instances of the same clause is only checked\n",
" when frozen_hash is False. If frozen_hash is True, the method will\n",
" return True, only for the same instance of the class\n",
" \"\"\"\n",
" if self.__hash__()==hash(other):\n",
" if self.frozen_hash:\n",
" return True\n",
" else:\n",
" return self.literals == other.literals\n",
" else:\n",
" return False\n",
" \n",
" def __iadd__(self,literal):\n",
" \"\"\"\n",
" adds the literal to the clause\n",
" :param literal: the literal to add\n",
" \"\"\"\n",
" if isinstance(literal,Literal):\n",
" self.literals = self.literals.union({literal})\n",
" return self\n",
" else:\n",
" raise TypeError(\"An argument of type Literal was expected\")\n",
" \n",
" def __add__(self,literal):\n",
" \"\"\"\n",
" adds the literal to the clause\n",
" :param literal: the literal to add\n",
" \"\"\"\n",
" if isinstance(literal,Literal):\n",
" return Clause(self.literals.union({literal}),self.frozen_hash)\n",
" else:\n",
" raise TypeError(\"An argument of type Literal was expected\")\n",
" \n",
" def __isub__(self,literal):\n",
" \"\"\"\n",
" deletes a literal from the clause\n",
" :param literal: the literal to substract\n",
" \"\"\"\n",
" if isinstance(literal, Literal):\n",
" self.literals = self.literals.difference({literal})\n",
" return self\n",
" else:\n",
" raise TypeError(\"A literal of type string was expected\")\n",
"\n",
" def __sub__(self,literal):\n",
" \"\"\"\n",
" deletes a literal from the clause\n",
" :param literal: the literal to substract\n",
" \"\"\"\n",
" if isinstance(literal, Literal):\n",
" return Clause(self.literals.difference({literal}),self.frozen_hash)\n",
" else:\n",
" raise TypeError(\"A literal of type string was expected\")\n",
" \n",
" def __str__(self):\n",
" return \"( \"+\" | \".join(map(str,self.literals))+\" )\"\n",
" \n",
" def __repr__(self):\n",
" return self.__str__()\n",
" \n",
" def __iter__(self):\n",
" return (i for i in self.literals)\n",
" \n",
" def __len__(self):\n",
" return len(self.literals)\n",
" \n",
" def copy(self):\n",
" return Clause(self.literals.copy(),self.frozen_hash)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Ejemplo creamos una cláusula $c_1$ formada por las literales $l_1$ y $l_2$"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"( A | ~B )\n"
]
}
],
"source": [
"c1 = Clause({l1,l2})\n",
"print(c1)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Fórmula en Forma Normal Conjuntiva"
]
},
{
"cell_type": "raw",
"metadata": {},
"source": [
"Explicar forma normal."
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {},
"outputs": [],
"source": [
"import logging as log\n",
"class FormulaCNF:\n",
" \"\"\"\n",
" Formula in Conjunctive Normal Form\n",
" :param formula: CNF formula (string to parse)\n",
" example: '(A|!B)&(!A|B|C)'\n",
" \"\"\"\n",
" def __init__(self,formula=None,assignment=None):\n",
" #removes all white spaces\n",
" if formula:\n",
" formula = ''.join(formula.split())\n",
" (self.variables,self.clauses) = self.parse(formula)\n",
" self.build_dicts() \n",
" if assignment:\n",
" self.assignment = assignment.copy()\n",
" else:\n",
" self.assignment = set()\n",
" \n",
" if log.getLogger().isEnabledFor(log.DEBUG):\n",
" log.debug(self.string_internals())\n",
" \n",
" def __getitem__(self,literal):\n",
" \"\"\"\n",
" short notation for the simplify method\n",
" :returns: the simplified formula resulting from assuming the literal\n",
" is true\n",
" \"\"\"\n",
" return self.simplify(literal)\n",
" \n",
" def empty_sentence(self):\n",
" \"\"\"\n",
" :returns: true if the formula has no clauses\n",
" \"\"\"\n",
" return not self.clauses\n",
" \n",
" def empty_clause(self):\n",
" \"\"\"\n",
" :returns: true if there is an empty clause\n",
" \"\"\"\n",
" return 0 in self.n_to_c\n",
" \n",
" def get_unit_clause_literal(self):\n",
" \"\"\"\n",
" Gets a unit clause if there is one, None otherwise\n",
" \"\"\"\n",
" clause = next(iter(self.n_to_c[1])) \\\n",
" if 1 in self.n_to_c else {}\n",
" return next(iter(clause)) if clause else None\n",
"\n",
" def get_pure_literal(self):\n",
" \"\"\"\n",
" Gest a pure literal\n",
" \"\"\"\n",
" return next(iter(self.P)) if self.P else None\n",
" \n",
" def get_variable_literal(self):\n",
" \"\"\"\n",
" Obtiene una variable de la formula\n",
" \"\"\"\n",
" return Literal(next(iter(self.variables))) if self.variables else None\n",
"\n",
" def simplify(self,literal):\n",
" \"\"\"\n",
" Simplifies the current sentence assming the provided literal\n",
" :param literal: the literal to assume as true\n",
" \"\"\"\n",
" log.debug(\"simplifying literal: \"+str(literal))\n",
" if not literal:\n",
" raise ValueError(\n",
" \"Invalid literal provided as argument: \"+ str(literal))\n",
" \n",
" # stores the assignment\n",
" self.assignment.add(literal)\n",
" # deletes the variable of the literal\n",
" self.variables = self.variables - {literal.variable}\n",
" # updates the list of literals\n",
" self.L = self.L -{literal}\n",
" # updates the list of pure literals\n",
" if self.isPureLiteral(literal):\n",
" self.P = self.P - {literal}\n",
" # deals with the clauses to delete\n",
" if literal in self.l_to_c:\n",
" for clause in self.l_to_c[literal].copy():\n",
" self.remove_clause(clause)\n",
" #TODO remover las cláusulas de otras entradas en l_to_c\n",
" # update data structures for other literals\n",
" for l in clause.copy():\n",
" self.decrease_literal_count(l)\n",
" self.del_from_dictionary_of_sets(\n",
" self.l_to_c,l,clause,True)\n",
" #removes the entry from dicionary literal to clauses\n",
" #del self.l_to_c[literal]\n",
" \n",
" # deals with the literals to delete negated literal\n",
" neg_literal = ~literal \n",
" # updates list of literals\n",
" self.L = self.L - {neg_literal}\n",
" # deletes all negated literals from clauses\n",
" if neg_literal in self.l_to_c:\n",
" for clause in self.l_to_c[neg_literal]:\n",
" self.remove_literal_from_clause(neg_literal,clause)\n",
" # delete entry for negated literal from dictionary literal to clauses\n",
" if neg_literal in self.l_to_c:\n",
" del self.l_to_c[neg_literal]\n",
" # updates list of negated literals\n",
" if self.isPureLiteral(neg_literal):\n",
" self.P = self.P - {neg_literal}\n",
"\n",
" log.debug(\"simplified formula:\")\n",
" if log.getLogger().isEnabledFor(log.DEBUG):\n",
" log.debug(self.string_internals())\n",
" return self\n",
" \n",
" def remove_clause(self,clause):\n",
" \"\"\"\n",
" Removes the clause from data structures\n",
" :parama clause: the clause to delete\n",
" \"\"\"\n",
" # deletes clause from inverse map counts\n",
" self.clauses.remove(clause)\n",
" n = len(clause)\n",
" self.del_from_dictionary_of_sets(self.n_to_c,n,clause,True)\n",
" \n",
" def remove_literal_from_clause(self,literal,clause):\n",
" # frozen_hash allows for unique clause \n",
" # all clauses are updated (they are the same clause) \n",
" m = len(clause)\n",
" # remove clause from map of counts to clauses\n",
" self.del_from_dictionary_of_sets(self.n_to_c,m,clause,True)\n",
" # obtain simplified clause\n",
" clause -= literal\n",
" # update map of count to clauses\n",
" self.add_to_dictionary_of_sets(self.n_to_c,m-1,clause)\n",
" # update other counts related to the literal\n",
" self.decrease_literal_count(literal)\n",
" \n",
" \n",
" def decrease_literal_count(self,literal):\n",
" # decrease number to literal counts\n",
" # delete literal from previous count\n",
" n = self.l_to_n[literal]\n",
" self.del_from_dictionary_of_sets(self.n_to_l,n,literal,True)\n",
" # update the literal count if it is not zero\n",
" self.add_to_dictionary_of_sets(self.n_to_l,n-1,literal)\n",
" \n",
" # update counts for the literal\n",
" if literal in self.l_to_n:\n",
" self.l_to_n[literal] -= 1\n",
" else:\n",
" raise ValueError(\"Inconsistent counts for literal: \"+str(literal))\n",
" \n",
" def isPureLiteral(self,literal):\n",
" \"\"\"\n",
" :returns: true if the literal is pure\n",
" \"\"\"\n",
" return literal in self.P\n",
" \n",
" \n",
" def parse(self,formula):\n",
" \"\"\"\n",
" Parses the CNF formula string\n",
" \"\"\"\n",
" variables = set()\n",
" clauses = set()\n",
" clstr = re.findall(\"[^&]+\",formula)\n",
" for c in clstr:\n",
" literals = re.findall(\"[^\\(\\)\\|]+\",c)\n",
" clause = Clause()\n",
" for l in literals:\n",
" literal = Literal.parse(l)\n",
" variables.add(literal.variable)\n",
" clause += literal\n",
" clauses.add(clause)\n",
" return (variables,clauses)\n",
" \n",
" \n",
" def build_dicts(self):\n",
" \"\"\"\n",
" Builds dictionaries for efficient access\n",
" to clauses and literals\n",
" l_to_c: dictionary, keys are literals, values are clauses\n",
" n_to_c: dictionary, keys are sizes, values are clauses\n",
" l_to_n: dictionary, keys are literals, values are counts\n",
" n_to_c: dictionary, keys are counts, values are clauses\n",
" L: set, all literals in the formula\n",
" P: set, all pure literals in the formula\n",
" \"\"\"\n",
" # W gives fast access to clauses by literals\n",
" self.l_to_c = {}\n",
" # n gives fast access to clauses by size\n",
" self.n_to_c = {}\n",
" # literal to counts\n",
" self.l_to_n = {}\n",
" # counts to literals\n",
" self.n_to_l = {}\n",
" \n",
" for c in self.clauses:\n",
" # build dictionary indexed by size\n",
" self.add_to_dictionary_of_sets(self.n_to_c,len(c),c)\n",
" for l in c:\n",
" #builds dictionary indexed by literal\n",
" self.add_to_dictionary_of_sets(self.l_to_c,l,c)\n",
" #compute frecuency of literals\n",
" self.l_to_n[l] = self.l_to_n[l] + 1 if l in self.l_to_n else 1\n",
" \n",
" #reverse literal counts\n",
" self.L = set()\n",
" for k,v in self.l_to_n.items():\n",
" self.add_to_dictionary_of_sets(self.n_to_l,v,k)\n",
" self.L.add(k)\n",
" self.P = {v for v in self.L if ~v not in self.L}\n",
" \n",
" def add_to_dictionary_of_sets(self,d,k,v):\n",
" \"\"\"\n",
" Adds a value to a dictionary of sets\n",
" :param d: the dictionary\n",
" :param k: the key\n",
" :param v: the value to add\n",
" :param by_len: True if the key is the length of the set to add\n",
" \"\"\"\n",
" singleton = {v}\n",
" d[k] = d[k].union(singleton) if k in d else singleton\n",
" \n",
" def del_from_dictionary_of_sets(self,d,k,m,del_key):\n",
" \"\"\"\n",
" Deletes a set member from a dictionary of sets\n",
" :param d: the dictionary\n",
" :param k: the key\n",
" :parma m: the member of the set to delete\n",
" :del_key: if this flag is True the entry is removed when set is empty\n",
" \"\"\"\n",
" # removes with set difference\n",
" d[k] = d[k] - {m}\n",
" if del_key and not d[k]:\n",
" del d[k]\n",
" \n",
" def copy(self):\n",
" \"\"\"\n",
" Creates a shallow copy of the CNF formula\n",
" \"\"\"\n",
" log.debug(\"Creating a copy of the current CNF formula\")\n",
" formula = FormulaCNF(str(self),self.assignment)\n",
" return formula\n",
" \n",
" def string_internals(self):\n",
" return (\"clauses:\"+str(self.clauses)+\"\\n\"+\n",
" \"variables:\"+str(self.variables)+\"\\n\"+\n",
" \"assignment:\"+str(self.assignment)+\"\\n\"+\n",
" \"l_to_c:\"+str(self.l_to_c)+\"\\n\"+\n",
" \"n_to_c:\"+str(self.n_to_c)+\"\\n\"+\n",
" \"l_to_n:\"+str(self.l_to_n)+\"\\n\"+\n",
" \"n_to_l:\"+str(self.n_to_l)+\"\\n\"+\n",
" \"L:\"+str(self.L)+\"\\n\"+\n",
" \"P:\"+str(self.P))\n",
" \n",
" def __str__(self):\n",
" return \" & \".join(map(str,self.clauses))\n",
" \n",
" def __repr__(self):\n",
" return self.__str__()"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Ejemplo de fórmula $\\phi$ en FNC.\n"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"( A | ~B ) & ( ~A | C ) & ( ~B | ~C ) & ( C )\n"
]
}
],
"source": [
"phi = FormulaCNF(\"(A|~B)&(~A|C)&(~B|~C)&(C)\")\n",
"print(phi)"
]
},
{
"cell_type": "raw",
"metadata": {},
"source": [
"Ejemplos del uso de métodos"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"False"
]
},
"execution_count": 7,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#verificar si la formula es vacía\n",
"phi.empty_sentence()"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"False"
]
},
"execution_count": 8,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#¿tiene la formula una cláusula vacía?\n",
"phi.empty_clause()"
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"C\n"
]
}
],
"source": [
"#De existir una clúsula unitaria, obtiene la literal de la cláusula. Regresa None si no existe\n",
"l = phi.get_unit_clause_literal()\n",
"print(l)"
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"~B\n"
]
}
],
"source": [
"#Obtiene una literal pura de existir, None si no existe\n",
"l = phi.get_pure_literal()\n",
"print(l)"
]
},
{
"cell_type": "code",
"execution_count": 11,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"A\n"
]
}
],
"source": [
"#Obtiene una literal con una variable arbitraria de la formula phi\n",
"v = phi.get_variable_literal()\n",
"print(v)"
]
},
{
"cell_type": "code",
"execution_count": 12,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"( A | ~B ) & ( C ) & ( ~B | ~C ) & ( ~A | C )\n"
]
}
],
"source": [
"#Obtiene una copia de la formula\n",
"psi = phi.copy()\n",
"print(psi)"
]
},
{
"cell_type": "code",
"execution_count": 13,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"( ~A | C ) & ( C )\n"
]
}
],
"source": [
"#Simplifica la formula asumiendo que la literal l es verdadera\n",
"psi = phi[l]\n",
"print(psi)"
]
},
{
"cell_type": "code",
"execution_count": 15,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"{~B}\n"
]
}
],
"source": [
"#Obtiene las asignaciones aplicadas a una fórmula\n",
"a = phi.assignment\n",
"print(a)"
]
},
{
"cell_type": "raw",
"metadata": {},
"source": [
"La clase DLPP implementa..."
]
},
{
"cell_type": "code",
"execution_count": 14,
"metadata": {},
"outputs": [],
"source": [
"class DPLL:\n",
" \"\"\"\n",
" DPLL algorithm\n",
" \"\"\"\n",
" \n",
" @staticmethod\n",
" def satisfiable(phi):\n",
" \"\"\"\n",
" determines if phi is satisfiable\n",
" :param phi: a CNF formula\n",
" :returs: una tupla cuyo primer elemento indica\n",
" si la fórmula es satisfactible o no, y el segundo\n",
" la asignación que logró hacer la fórmula verdadera\n",
" \"\"\"\n",
" log.info(\"phi: \"+str(phi))\n",
" if phi.empty_sentence():\n",
" log.info(\"fórmula vacía\")\n",
" return (True,phi.assignment)\n",
" \n",
" #inserta tu código aquí\n",
" "
]
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 3",
"language": "python",
"name": "python3"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 3
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.6.2"
}
},
"nbformat": 4,
"nbformat_minor": 2
}
python/formulas.txt
......@@ -4,5 +4,5 @@ Satisfiable:
(p | q | r | s) & (~p | q | ~r) & (~q | ~r | s) & (p | ~q | r | s) & (q | ~r | ~s) & (~p | ~r | s) & (~p | ~s) & (p | ~q)
Unsatisfiable:
(__Z__21|__R__NEG__v1__v2)&(__Z__21|(__Z__22|v2__0)&(__Z__21|(__Z__22|~v2__2)&(__Z__21|(~__Z__22|v2__1)&(__Z__21|(~__Z__22|~v2__3)&(~__Z__21|~__R__NEG__v1__v2)&(~__Z__21|(~v2__0|v2__2)&(~__Z__21|(~v2__1|v2__3)&(__Z__23|__R__POS__v1__v2)&(__Z__23|(__Z__24|~v2__0)&(__Z__23|(__Z__24|v2__2)&(__Z__23|(~__Z__24|~v2__1)&(__Z__23|(~__Z__24|v2__3)&(~__Z__23|~__R__POS__v1__v2)&(~__Z__23|(v2__0|~v2__2)&(~__Z__23|(v2__1|~v2__3)&(__R__POS__v1__v2&~__R__NEG__v1__v2)&(__Z__25|__R__NEG__v1__v3)&(__Z__25|(__Z__26|v3__0)&(__Z__25|(__Z__26|~v3__2)&(__Z__25|(~__Z__26|v3__1)&(__Z__25|(~__Z__26|~v3__3)&(~__Z__25|~__R__NEG__v1__v3)&(~__Z__25|(~v3__0|v3__2)&(~__Z__25|(~v3__1|v3__3)&(__Z__27|__R__POS__v1__v3)&(__Z__27|(__Z__28|~v3__0)&(__Z__27|(__Z__28|v3__2)&(__Z__27|(~__Z__28|~v3__1)&(__Z__27|(~__Z__28|v3__3)&(~__Z__27|~__R__POS__v1__v3)&(~__Z__27|(v3__0|~v3__2)&(~__Z__27|(v3__1|~v3__3)&(__R__NEG__v1__v3&~__R__POS__v1__v3)&(__Z__29|__R__NEG__v2__v1)&(__Z__29|v1__0)&(__Z__29|~v1__1)&(~__Z__29|~__R__NEG__v2__v1)&(~__Z__29|(~v1__0|v1__1)&(__Z__30|__R__POS__v2__v1)&(__Z__30|~v1__0)&(__Z__30|v1__1)&(~__Z__30|~__R__POS__v2__v1)&(~__Z__30|(v1__0|~v1__1)&(__R__POS__v2__v1&~__R__NEG__v2__v1)&(__Z__31|__R__NEG__v3__v2)&(__Z__31|(__Z__32|v2__0)&(__Z__31|(__Z__32|~v2__1)&(__Z__31|(~__Z__32|v2__2)&(__Z__31|(~__Z__32|~v2__3)&(~__Z__31|~__R__NEG__v3__v2)&(~__Z__31|(~v2__0|v2__1)&(~__Z__31|(~v2__2|v2__3)&(__Z__33|__R__POS__v3__v2)&(__Z__33|(__Z__34|~v2__0)&(__Z__33|(__Z__34|v2__1)&(__Z__33|(~__Z__34|~v2__2)&(__Z__33|(~__Z__34|v2__3)&(~__Z__33|~__R__POS__v3__v2)&(~__Z__33|(v2__0|~v2__1)&(~__Z__33|(v2__2|~v2__3)&(__R__POS__v3__v2&~__R__NEG__v3__v2)&(__Z__35|__R__NEG__v3__v3)&(__Z__35|(__Z__36|v3__0)&(__Z__35|(__Z__36|~v3__1)&(__Z__35|(~__Z__36|v3__2)&(__Z__35|(~__Z__36|~v3__3)&(~__Z__35|~__R__NEG__v3__v3)&(~__Z__35|(~v3__0|v3__1)&(~__Z__35|(~v3__2|v3__3)&(__Z__37|__R__POS__v3__v3)&(__Z__37|(__Z__38|~v3__0)&(__Z__37|(__Z__38|v3__1)&(__Z__37|(~__Z__38|~v3__2)&(__Z__37|(~__Z__38|v3__3)&(~__Z__37|~__R__POS__v3__v3)&(~__Z__37|(v3__0|~v3__1)&(~__Z__37|(v3__2|~v3__3)&~__R__NEG__v3__v3)&(v3__0|(v1__0|v2__0)&(~v3__1|(v1__0|v2__1)&(v3__0|(v1__1|~v2__0)&(~v3__1|(v1__1|~v2__1)&(v3__2|(~v1__0|v2__2)&(~v3__3|(~v1__0|v2__3)&(v3__2|(~v1__1|~v2__2)&(~v3__3|(~v1__1|~v2__3)
Satisfiable:
(__Z__17|__R__NEG__v1__v2)&(__Z__17|(__Z__18|v2__0)&(__Z__17|(__Z__18|~v2__2)&(__Z__17|(~__Z__18|v2__1)&(__Z__17|(~__Z__18|~v2__3)&(~__Z__17|~__R__NEG__v1__v2)&(~__Z__17|(~v2__0|v2__2)&(~__Z__17|(~v2__1|v2__3)&(__Z__19|__R__POS__v1__v2)&(__Z__19|(__Z__20|~v2__0)&(__Z__19|(__Z__20|v2__2)&(__Z__19|(~__Z__20|~v2__1)&(__Z__19|(~__Z__20|v2__3)&(~__Z__19|~__R__POS__v1__v2)&(~__Z__19|(v2__0|~v2__2)&(~__Z__19|(v2__1|~v2__3)&(__R__POS__v1__v2&~__R__NEG__v1__v2)&(__Z__21|__R__NEG__v1__v3)&(__Z__21|v3__0)&(__Z__21|~v3__1)&(~__Z__21|~__R__NEG__v1__v3)&(~__Z__21|(~v3__0|v3__1)&(__Z__22|__R__POS__v1__v3)&(__Z__22|~v3__0)&(__Z__22|v3__1)&(~__Z__22|~__R__POS__v1__v3)&(~__Z__22|(v3__0|~v3__1)&(__R__NEG__v1__v3&~__R__POS__v1__v3)&(__Z__23|__R__NEG__v2__v1)&(__Z__23|v1__0)&(__Z__23|~v1__1)&(~__Z__23|~__R__NEG__v2__v1)&(~__Z__23|(~v1__0|v1__1)&(__Z__24|__R__POS__v2__v1)&(__Z__24|~v1__0)&(__Z__24|v1__1)&(~__Z__24|~__R__POS__v2__v1)&(~__Z__24|(v1__0|~v1__1)&(__R__POS__v2__v1&~__R__NEG__v2__v1)&(__Z__25|__R__NEG__v3__v2)&(__Z__25|(__Z__26|v2__0)&(__Z__25|(__Z__26|~v2__1)&(__Z__25|(~__Z__26|v2__2)&(__Z__25|(~__Z__26|~v2__3)&(~__Z__25|~__R__NEG__v3__v2)&(~__Z__25|(~v2__0|v2__1)&(~__Z__25|(~v2__2|v2__3)&(__Z__27|__R__POS__v3__v2)&(__Z__27|(__Z__28|~v2__0)&(__Z__27|(__Z__28|v2__1)&(__Z__27|(~__Z__28|~v2__2)&(__Z__27|(~__Z__28|v2__3)&(~__Z__27|~__R__POS__v3__v2)&(~__Z__27|(v2__0|~v2__1)&(~__Z__27|(v2__2|~v2__3)&(__R__POS__v3__v2&~__R__NEG__v3__v2)
Satisfiable: LA FORMULA NO ESTA CORRECTA
(__Z__17|__R__NEG__v1__v2)&(__Z__17|(__Z__18|v2__0)&(__Z__17|(__Z__18|~v2__2)&(__Z__17|(~__Z__18|v2__1)&(__Z__17|(~__Z__18|~v2__3)&(~__Z__17|~__R__NEG__v1__v2)&(~__Z__17|(~v2__0|v2__2)&(~__Z__17|(~v2__1|v2__3)&(__Z__19|__R__POS__v1__v2)&(__Z__19|(__Z__20|~v2__0)&(__Z__19|(__Z__20|v2__2)&(__Z__19|(~__Z__20|~v2__1)&(__Z__19|(~__Z__20|v2__3)&(~__Z__19|~__R__POS__v1__v2)&(~__Z__19|(v2__0|~v2__2)&(~__Z__19|(v2__1|~v2__3)&(__R__POS__v1__v2)&(~__R__NEG__v1__v2)&(__Z__21|__R__NEG__v1__v3)&(__Z__21|v3__0)&(__Z__21|~v3__1)&(~__Z__21|~__R__NEG__v1__v3)&(~__Z__21|(~v3__0|v3__1)&(__Z__22|__R__POS__v1__v3)&(__Z__22|~v3__0)&(__Z__22|v3__1)&(~__Z__22|~__R__POS__v1__v3)&(~__Z__22|(v3__0|~v3__1)&(__R__NEG__v1__v3&~__R__POS__v1__v3)&(__Z__23|__R__NEG__v2__v1)&(__Z__23|v1__0)&(__Z__23|~v1__1)&(~__Z__23|~__R__NEG__v2__v1)&(~__Z__23|(~v1__0|v1__1)&(__Z__24|__R__POS__v2__v1)&(__Z__24|~v1__0)&(__Z__24|v1__1)&(~__Z__24|~__R__POS__v2__v1)&(~__Z__24|(v1__0|~v1__1)&(__R__POS__v2__v1)&(~__R__NEG__v2__v1)&(__Z__25|__R__NEG__v3__v2)&(__Z__25|(__Z__26|v2__0)&(__Z__25|(__Z__26|~v2__1)&(__Z__25|(~__Z__26|v2__2)&(__Z__25|(~__Z__26|~v2__3)&(~__Z__25|~__R__NEG__v3__v2)&(~__Z__25|(~v2__0|v2__1)&(~__Z__25|(~v2__2|v2__3)&(__Z__27|__R__POS__v3__v2)&(__Z__27|(__Z__28|~v2__0)&(__Z__27|(__Z__28|v2__1)&(__Z__27|(~__Z__28|~v2__2)&(__Z__27|(~__Z__28|v2__3)&(~__Z__27|~__R__POS__v3__v2)&(~__Z__27|(v2__0|~v2__1)&(~__Z__27|(v2__2|~v2__3)&(__R__POS__v3__v2&~__R__NEG__v3__v2)
python/markov-Copy1.ipynb 0 → 100644
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Cadenas de Markov"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Descripción.\n",
"![Cadena de Markov](\"markov.png\")"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": []
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 3",
"language": "python",
"name": "python3"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 3
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.6.2"
}
},
"nbformat": 4,
"nbformat_minor": 2
}
python/markov.ipynb 0 → 100644
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Cadenas de Markov"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Descripción.\n",
"![Cadena de Markov](markov.png)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Matriz de probabilidades de transición"
]
},
{
"cell_type": "code",
"execution_count": 13,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"[[ 0.4 0.1]\n",
" [ 0.6 0.9]]\n"
]
}
],
"source": [
"import numpy as np\n",
"\n",
"# define la matriz de probabilidades de transición\n",
"# utilizando el paquete de cómputo científico de Python numpy\n",
"\n",
"#A = | p(d|d), p(b|d), p(c|d) |\n",
"# | p(b|d), p(b|b), p(c|b) |\n",
"# | p(c|d), p(b|c), p(c|c) |\n",
"\n",
"# por ejemplo la matriz\n",
"#A = | 0.4, 0.1|\n",
"# | 0.6, 0.9|\n",
"# puede definirse en python como\n",
"A = np.matrix([[0.4,0.1],[0.6,0.9]])\n",
"\n",
"print(A)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Encontrar la distribución de probabilidad estacionaria elevando a un exponente grande"
]
},
{
"cell_type": "code",
"execution_count": 22,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"[[ 0.14285714 0.14285714]\n",
" [ 0.85714286 0.85714286]]\n",
"[[ 0.14285714 0.14285714]\n",
" [ 0.85714286 0.85714286]]\n"
]
}
],
"source": [
"B = np.linalg.matrix_power(A,50)\n",
"print(B)\n",
"B = np.linalg.matrix_power(A,100)\n",
"print(B)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Encontrar la misma distribución calculando los eigenvectores"
]
},
{
"cell_type": "code",
"execution_count": 23,
"metadata": {},
"outputs": [],
"source": [
"w,v = np.linalg.eig(A)"
]
},
{
"cell_type": "code",
"execution_count": 24,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"[ 0.3 1. ]\n",
"[[-0.70710678 -0.16439899]\n",
" [ 0.70710678 -0.98639392]]\n"
]
}
],
"source": [
"print(w)\n",
"print(v)"
]
},
{
"cell_type": "code",
"execution_count": 25,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"matrix([[ 0.14285714],\n",
" [ 0.85714286]])"
]
},
"execution_count": 25,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"v[:,1]/sum(v[:,1])"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": []
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 3",
"language": "python",
"name": "python3"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 3
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.6.2"
}
},
"nbformat": 4,
"nbformat_minor": 2
}
python/tsp.py 0 → 100644
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Fri Nov 16 12:09:07 2018
@author: stan
"""
#%%
import simplejson, urllib
orig_coord = 19.3071075,-99.1843678
dest_coord = 20.6665935,-103.3530723
urlstr = "http://maps.googleapis.com/maps/api/distancematrix/json?origins={0}&destinations={1}&mode=driving&language=en-EN&sensor=false".format(str(orig_coord),str(dest_coord))
print(urlstr)
with urllib.request.urlopen(urlstr) as url:
result= simplejson.load(urllib.read(url))
driving_time = result['rows'][0]['elements'][0]['duration']['value']
#%%
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.basemap import Basemap
#plt.figure(num=None, figsize=(8, 6), dpi=300, facecolor='w', edgecolor='k')
m = Basemap(projection='lcc',
lon_0=-121,lat_0=16,\
llcrnrlat=16,urcrnrlat=30,\
llcrnrlon=-121,urcrnrlon=-84,rsphere=6371200.,resolution='h')
# create polar stereographic Basemap instance.
#m = Basemap(projection='stere',lon_0=-120,lat_0=16,\
#llcrnrlat=16,urcrnrlat=29,\
#llcrnrlon=-119,urcrnrlon=-84,\
#rsphere=6371200.,resolution='h') #c, l, i, h, f
#m.fillcontinents(color='white', alpha=0.6, zorder=0)
#m.fillcontinents(color='coral',lake_color='aqua')
#m.bluemarble()
#m.etopo()
m.drawstates()
m.drawcoastlines()
m.drawcountries()
m.fillcontinents(color='lightgray',zorder=0)
#m.drawrivers(color='b', linewidth=.1)
# draw parallels.
#parallels = np.arange(0.,90,1.)
#m.drawparallels(parallels,labels=[1,0,0,0],fontsize=10)
## draw meridians
#meridians = np.arange(180.,360.,1.)
#m.drawmeridians(meridians,labels=[0,0,0,1],fontsize=10)
ciudades = {
('Aguascalientes Aguascalientes', -102.305481910559, 21.8475186),
('Mexicali Baja California', -115.4440698, 32.6200699),
('La Paz Baja California', -110.4135535, 24.1017563),
('Tuxtla Gutierrez Chiapas', -93.1418086426027, 16.74521845),
('Chihuahua Chihuahua', -106.384451167241, 28.938764),
('Campeche Campeche', -90.25414928915, 19.59188245),
('Saltillo Tamaulipas', -100.9887188, 25.4466476),
('Colima Colima', -103.654240318428, 19.0986013),
('Durango Durango', -104.919810561772, 23.9112721),
('Guanajuato Guanajuato', -101.242782801753, 21.02405745),
('Chilpancingo Guerrero', -99.5048416, 17.5542149),
('Pachuca Hidalgo', -98.7413535, 20.1165413),
('Guadalajara Jalisco', -103.3469982, 20.676143),
('Toluca Mexico', -99.4890962, 19.2819217),
('Morelia Michoacan', -101.1923818, 19.7027116),
('Cuernavaca Morelos', -99.2342282, 18.9218274),
('Tepic Nayarit', -105.1423419, 21.7206915),
('Monterrey Nuevo Leon', -100.293101629066, 25.63978365),
('Oaxaca Oaxaca', -96.7253575, 17.0604663),
('Puebla Puebla', -98.1984744, 19.0437227),
('Queretaro Queretaro', -100.3879904, 20.5878372),
('Chetumal Quintana Roo', -88.3006444, 18.5132414),
('San Luis Potosi San Luis Potosi', -100.926198413959, 22.3089625),
('Culiacan Sinaloa', -107.3936301, 24.8015732),
('Hermosillo Sonora', -110.9612378, 29.0894152),
('Villa Hermosa Tabasco', -91.5269144, 14.8411997),
('Ciudad Victoria Tamaulipas', -99.1419341, 23.7439573),
('Tlaxcala Tlaxcala', -98.2376413321235, 19.3173725),
('Xalapa Veracruz', -96.9238793, 19.5274085),
('Merida Yucatan', -89.6237402, 20.9670759),
('Zacatecas Zacatecas', -102.668598836424, 22.7293312)}
C,X,Y = zip(*ciudades)
X,Y = m(X,Y)
m.scatter(X, Y,color='red',s=50,marker='o',zorder=1)
plt.show()
#%%
from geopy.geocoders import Nominatim
capitales = ["Aguascalientes Aguascalientes",
"Mexicali Baja California",
"La Paz Baja California",
"Tuxtla Gutierrez Chiapas",
"Chihuahua Chihuahua",
"Campeche Campeche",
"Saltillo Tamaulipas",
"Colima Colima",
"Durango Durango",
"Guanajuato Guanajuato",
"Chilpancingo Guerrero",
"Pachuca Hidalgo",
"Guadalajara Jalisco",
"Toluca Mexico",
"Morelia Michoacan",
"Cuernavaca Morelos",
"Tepic Nayarit",
"Monterrey Nuevo Leon",
"Oaxaca Oaxaca",
"Puebla Puebla",
"Queretaro Queretaro",
"Chetumal Quintana Roo",
"San Luis Potosi San Luis Potosi",
"Culiacan Sinaloa",
"Hermosillo Sonora",
"Villa Hermosa Tabasco",
"Ciudad Victoria Tamaulipas",
"Tlaxcala Tlaxcala",
"Xalapa Veracruz",
"Merida Yucatan",
"Zacatecas Zacatecas"]
geolocator = Nominatim(user_agent='myapplication')
for city in capitales:
location = geolocator.geocode(city)
print((city,location.longitude,location.latitude))
\ No newline at end of file
python/tspmexico.py 0 → 100644
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Fri Nov 16 12:09:07 2018
@author: stan
"""
#%%
import simplejson, urllib
orig_coord = 19.3071075,-99.1843678
dest_coord = 20.6665935,-103.3530723
urlstr = "http://maps.googleapis.com/maps/api/distancematrix/json?origins={0}&destinations={1}&mode=driving&language=en-EN&sensor=false".format(str(orig_coord),str(dest_coord))
print(urlstr)
with urllib.request.urlopen(urlstr) as url:
result= simplejson.load(urllib.read(url))
driving_time = result['rows'][0]['elements'][0]['duration']['value']
#%%
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.basemap import Basemap
#
#fig = plt.figure()
#ax = fig.add_subplot(1,1,1)
#plt.figure(num=None, figsize=(8, 6), dpi=300, facecolor='w', edgecolor='k')
m = Basemap(projection='lcc',
lon_0=-121,lat_0=16,\
llcrnrlat=16,urcrnrlat=30,\
llcrnrlon=-121,urcrnrlon=-84,rsphere=6371200.,resolution='h')
# create polar stereographic Basemap instance.
#m = Basemap(projection='stere',lon_0=-120,lat_0=16,\
#llcrnrlat=16,urcrnrlat=29,\
#llcrnrlon=-119,urcrnrlon=-84,\
#rsphere=6371200.,resolution='h') #c, l, i, h, f
#m.fillcontinents(color='white', alpha=0.6, zorder=0)
#m.fillcontinents(color='coral',lake_color='aqua')
#m.bluemarble()
#m.etopo()
m.drawstates()
m.drawcoastlines()
m.drawcountries()
m.fillcontinents(color='lightgray',zorder=0)
#m.drawrivers(color='b', linewidth=.1)
# draw parallels.
#parallels = np.arange(0.,90,1.)
#m.drawparallels(parallels,labels=[1,0,0,0],fontsize=10)
## draw meridians
#meridians = np.arange(180.,360.,1.)
#m.drawmeridians(meridians,labels=[0,0,0,1],fontsize=10)
ciudades = {
('Aguascalientes Aguascalientes', -102.305481910559, 21.8475186),
('Mexicali Baja California', -115.4440698, 32.6200699),
('La Paz Baja California', -110.4135535, 24.1017563),
('Tuxtla Gutierrez Chiapas', -93.1418086426027, 16.74521845),
('Chihuahua Chihuahua', -106.384451167241, 28.938764),
('Campeche Campeche', -90.25414928915, 19.59188245),
('Saltillo Tamaulipas', -100.9887188, 25.4466476),
('Colima Colima', -103.654240318428, 19.0986013),
('Durango Durango', -104.919810561772, 23.9112721),
('Guanajuato Guanajuato', -101.242782801753, 21.02405745),
('Chilpancingo Guerrero', -99.5048416, 17.5542149),
('Pachuca Hidalgo', -98.7413535, 20.1165413),
('Guadalajara Jalisco', -103.3469982, 20.676143),
('Toluca Mexico', -99.4890962, 19.2819217),
('Morelia Michoacan', -101.1923818, 19.7027116),
('Cuernavaca Morelos', -99.2342282, 18.9218274),
('Tepic Nayarit', -105.1423419, 21.7206915),
('Monterrey Nuevo Leon', -100.293101629066, 25.63978365),
('Oaxaca Oaxaca', -96.7253575, 17.0604663),
('Puebla Puebla', -98.1984744, 19.0437227),
('Queretaro Queretaro', -100.3879904, 20.5878372),
('Chetumal Quintana Roo', -88.3006444, 18.5132414),
('San Luis Potosi San Luis Potosi', -100.926198413959, 22.3089625),
('Culiacan Sinaloa', -107.3936301, 24.8015732),
('Hermosillo Sonora', -110.9612378, 29.0894152),
('Villa Hermosa Tabasco', -92.9881407, 17.992608),
('Ciudad Victoria Tamaulipas', -99.1419341, 23.7439573),
('Tlaxcala Tlaxcala', -98.2376413321235, 19.3173725),
('Xalapa Veracruz', -96.9238793, 19.5274085),
('Merida Yucatan', -89.6237402, 20.9670759),
('Zacatecas Zacatecas', -102.668598836424, 22.7293312)}
C,X,Y = zip(*ciudades)
X,Y = m(X,Y)
m.scatter(X, Y,color='red',s=50,marker='o',zorder=1)
lines, = plt.plot(X,Y)
ax = plt.gca()
plt.pause(2)
ax.lines.remove(lines)
plt.draw()
plt.show()
#%%
np.array(list(zip(X,Y)))
#%%
from geopy.geocoders import Nominatim
capitales = ["Aguascalientes Aguascalientes",
"Mexicali Baja California",
"La Paz Baja California",
"Tuxtla Gutierrez Chiapas",
"Chihuahua Chihuahua",
"Campeche Campeche",
"Saltillo Tamaulipas",
"Colima Colima",
"Durango Durango",
"Guanajuato Guanajuato",
"Chilpancingo Guerrero",
"Pachuca Hidalgo",
"Guadalajara Jalisco",
"Toluca Mexico",
"Morelia Michoacan",
"Cuernavaca Morelos",
"Tepic Nayarit",
"Monterrey Nuevo Leon",
"Oaxaca Oaxaca",
"Puebla Puebla",
"Queretaro Queretaro",
"Chetumal Quintana Roo",
"San Luis Potosi San Luis Potosi",
"Culiacan Sinaloa",
"Hermosillo Sonora",
"Villa Hermosa Tabasco",
"Ciudad Victoria Tamaulipas",
"Tlaxcala Tlaxcala",
"Xalapa Veracruz",
"Merida Yucatan",
"Zacatecas Zacatecas"]
geolocator = Nominatim(user_agent='myapplication')
for city in capitales:
location = geolocator.geocode(city)
print((city,location.longitude,location.latitude))
\ No newline at end of file
regexpbayes.txt 0 → 100644
(?m)(.*$)*[^a-zA-Z]*P\(homicida\)\s*=\s*0.0(09[0-9]*|1[0-9]*)[^a-zA-Z]*$[^a-zA-Z]*P\(sangre\)\s*=\s*0.10[0-9]*[^a-zA-Z]*$[^a-zA-Z]*P\(cuchillo\)\s*=\s*0.25[0-9]*[^a-zA-Z]*$[^a-zA-Z]*P\(~cuchillo\)\s*=\s*0.74[0-9]*[^a-zA-Z]*$[^a-zA-Z]*P\(~cuchillo\|homicida\)\s*=\s*0.1[45][0-9]*[^a-zA-Z]*$[^a-zA-Z]*P\(homicida\|cuchillo,sangre\)\s*=\s*0.21[456][0-9]*[^a-zA-Z]*$(.*$)*
respuestabayes.txt 0 → 100644
P(homicida) = 0.01
P(sangre) = 0.10700000000000001
P(cuchillo) = 0.256
P(~cuchillo) = 0.744
P(~cuchillo|homicida) = 0.15000000000000002
P(homicida|cuchillo,sangre) = 0.21553090332805072
respuestabayes1.txt 0 → 100644
P(~cuchillo|homicida) = 0.15000000000000002
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