{
 "cells": [
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   "cell_type": "markdown",
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   "source": [
    "# Chiffrement de Hill"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Le chiffrement de Hill consiste à  coder un mot de deux lettres  selon la procédure suivante : \n",
    "\n",
    "### Etape 1\n",
    "Chaque lettre du mot est remplacée par un entier en utilisant le tableau ci-dessous :\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "correspondance = [(chr(i+65),i) for i in range(26)]\n",
    "print(correspondance)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**chr** retourne le caractère associé à l'entier **ord** (dans le code ASCII)\n",
    "\n",
    "'A' est le caractère associé à l'entier 65."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "[chr(i) for i in range(256)]"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "ord('A')"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "On obtient un couple d'entiers $(x_1,x_2)$ où $x_1$ correspond à la première lettre du mot et $x_2$ correspond à  la deuxième lettre du mot.\n",
    "\n",
    "### Etape 2\n",
    "$(x_1,x_2)$ est transformé en $(y_1,y_2)$ tel que :\n",
    "$$(S_1) \\quad \\left\\lbrace    \\begin{array}{l} y_1 \\equiv 11x_1 + 3x_2 \\quad (mod \\quad 26)\\\\ y_2 \\equiv 7x_1 + 4x_2 \\quad (mod \\quad 26) \\end{array} \\right. $$\n",
    "avec $0 \\leqslant y_1 \\leqslant 25$ et  $0 \\leqslant y_2 \\leqslant 25$.\n",
    "\n",
    "\n",
    "### Etape 3\n",
    "$(y_1,y_2)$ est transformé en un mot de deux lettres en utilisant le tableau de correspondance donné dans l'étape 1.\n",
    "\n",
    "\n",
    "Exemple\n",
    "\n",
    "Pour le mot en clair **TE** :"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# étape 1\n",
    "x1 = ord('T')-65\n",
    "x2 = ord('E')-65\n",
    "print(x1,x2)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# étape 2\n",
    "y1 = ( 11*x1 + 3*x2 ) % 26\n",
    "y2 = ( 7*x1 + 4*x2 ) % 26\n",
    "print(y1,y2)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "### étape 3\n",
    "print('Le mot codé est ', chr(y1+65)+chr(y2+65))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**1. ** Coder le mot ST."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# étape 1\n",
    "x1 = ord('S')-65\n",
    "x2 = ord('T')-65\n",
    "print(x1,x2)\n",
    "# étape 2\n",
    "y1 = ( 11*x1 + 3*x2 ) % 26\n",
    "y2 = ( 7*x1 + 4*x2 ) % 26\n",
    "print(y1,y2)\n",
    "### étape 3\n",
    "print('Le mot codé est ', chr(y1+65)+chr(y2+65))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**2.** On admet que le système $(S_1)$ est équivalent au système \n",
    "$$(S_2) \\quad \\left\\lbrace    \\begin{array}{l} x_1 \\equiv 16y_1 + y_2 \\quad (mod \\quad 26)\\\\ x_2 \\equiv 11y_1 + 5y_2 \\quad (mod \\quad 26) \\end{array} \\right. $$\n",
    "Décoder le mot YJ."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "y1 = ord('Y')-65\n",
    "y2 = ord('J')-65\n",
    "print(y1,y2)\n",
    "x1 = ( 16*y1 + y2 ) % 26\n",
    "x2 = ( 11*y1 + 5*y2 ) % 26\n",
    "print(x1,x2)\n",
    "print('Le mot en clair est ', chr(x1+65)+chr(x2+65))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Au lieu de retaper tout pour chaque mot, on peut définir des fonctions pour le chiffrement et le déchiffrement :"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "def chiffrementHill(motDeDeuxLettres):\n",
    "    # motDeDeuxLettres est une chaîne de caractères\n",
    "    x1 = ord(motDeDeuxLettres[0])-65\n",
    "    x2 = ord(motDeDeuxLettres[1])-65\n",
    "    y1 = ( 11*x1 + 3*x2 ) % 26\n",
    "    y2 = ( 7*x1 + 4*x2 ) % 26\n",
    "    return (chr(y1+65)+chr(y2+65))    "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "chiffrementHill('TE')"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "chiffrementHill('ST')"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "def deChiffrementHill(motDeDeuxLettres):\n",
    "    y1 = ord(motDeDeuxLettres[0])-65\n",
    "    y2 = ord(motDeDeuxLettres[1])-65\n",
    "    x1 = ( 16*y1 + y2 ) % 26\n",
    "    x2 = ( 11*y1 + 5*y2 ) % 26\n",
    "    return (chr(x1+65)+chr(x2+65))    "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "deChiffrementHill('NT')"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "deChiffrementHill('YJ')"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": []
  }
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