{
 "cells": [
  {
   "cell_type": "markdown",
   "id": "a35eeb9f-df70-4ab1-a243-2d2025888eb0",
   "metadata": {},
   "source": [
    "# Exercice 1\n",
    "1) Codez une fonction qui permet de chiffrer avec le chiffrement parfait (C = M ⊕ K). Cette\n",
    "fonction prend en entrée une clé et un message, tous deux sous forme de chaîne de caractères,\n",
    "et sa sortie est également une chaîne de caractères. Vous aurez peut-être l’usage de l’opérateur\n",
    "∧."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 59,
   "id": "57595af7-bd28-427e-b407-eb0bf49f483a",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "72"
      ]
     },
     "execution_count": 59,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "ord('H')"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "id": "06c3d56f-6872-4179-a1f8-e2a44ba117f3",
   "metadata": {},
   "outputs": [],
   "source": [
    "def bit(nombre, n):\n",
    "    resBin = [0]*n\n",
    "    i = 0\n",
    "    while nombre > 0:\n",
    "        resBin[n-1-i] = nombre%2\n",
    "        nombre = nombre // 2\n",
    "        i += 1\n",
    "    return resBin\n",
    "\n",
    "def sToBit(s):\n",
    "    return bit(ord(s),8)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "id": "277ebad9-936c-4e84-9f3c-b45bb69dd388",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "[0, 1, 0, 0, 1, 0, 0, 0]"
      ]
     },
     "execution_count": 2,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "sToBit(\"H\")"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "id": "c1376cfe-4eae-4323-8ec2-6fd5e3504f68",
   "metadata": {},
   "outputs": [],
   "source": [
    "def stringToBit(string):\n",
    "    resBin = []\n",
    "    for i in string:\n",
    "        resBin.append(sToBit(i))\n",
    "    return resBin"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "id": "6f9311ce-b1d9-458a-bdb7-ce9cd2dc8f32",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "[[0, 1, 0, 0, 1, 0, 0, 0],\n",
       " [0, 1, 1, 0, 0, 1, 0, 1],\n",
       " [0, 1, 1, 0, 1, 1, 0, 0],\n",
       " [0, 1, 1, 0, 1, 1, 0, 0],\n",
       " [0, 1, 1, 0, 1, 1, 1, 1]]"
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "stringToBit(\"Hello\")"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "id": "7b1e1f48-45e8-4b60-94c2-f3fa2a1ac7e8",
   "metadata": {},
   "outputs": [],
   "source": [
    "def chiffrement(cle, message):\n",
    "    cleBin = stringToBit(cle)\n",
    "    messageBin = stringToBit(message)\n",
    "    print(cleBin[0][0])\n",
    "    print(message)\n",
    "    res = \"\"\n",
    "    for i in range(len(message)):\n",
    "        for j in range(8):\n",
    "            res = res + str(cleBin[i][j] ^ messageBin[i][j])\n",
    "    return res"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "id": "8515d99f-d0e2-4da5-a186-3ed78e233b02",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "0\n",
      "Hello\n"
     ]
    },
    {
     "data": {
      "text/plain": [
       "'0000101100000000000111110000110100011101'"
      ]
     },
     "execution_count": 6,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "chiffrement(\"Cesar\", \"Hello\")"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "9211c83a-2fde-4614-b066-36a1b09d9ff3",
   "metadata": {},
   "source": [
    "# Exercice 2"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "id": "bfd0dfc8-0ebb-4103-a12d-0ccfcdcb15aa",
   "metadata": {},
   "outputs": [],
   "source": [
    "def LSFR(s, c, nbtours):\n",
    "    i = 0\n",
    "    resTot = s\n",
    "    t = len(s)\n",
    "    while i < nbtours:\n",
    "        res = s[t]\n",
    "        j = t\n",
    "        \n",
    "        for j in range(t):\n",
    "            print(c[j])\n",
    "            res = res ^ c[j]\n",
    "        for j in range(t):\n",
    "            if i == 1:\n",
    "                resTot[0] = res\n",
    "                break\n",
    "            resTot[t - 1] = resTot[t - 2]\n",
    "    return resTot"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "id": "a2bbdc83-30f3-4aca-9026-b2a2aba64171",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "4\n",
      "4\n",
      "4\n",
      "4\n"
     ]
    },
    {
     "ename": "IndexError",
     "evalue": "list index out of range",
     "output_type": "error",
     "traceback": [
      "\u001b[0;31m---------------------------------------------------------------------------\u001b[0m",
      "\u001b[0;31mIndexError\u001b[0m                                Traceback (most recent call last)",
      "Cell \u001b[0;32mIn[9], line 3\u001b[0m\n\u001b[1;32m      1\u001b[0m s \u001b[38;5;241m=\u001b[39m [\u001b[38;5;241m0\u001b[39m, \u001b[38;5;241m0\u001b[39m, \u001b[38;5;241m1\u001b[39m, \u001b[38;5;241m0\u001b[39m]\n\u001b[1;32m      2\u001b[0m c \u001b[38;5;241m=\u001b[39m [\u001b[38;5;241m1\u001b[39m, \u001b[38;5;241m0\u001b[39m, \u001b[38;5;241m1\u001b[39m, \u001b[38;5;241m0\u001b[39m]\n\u001b[0;32m----> 3\u001b[0m \u001b[38;5;28mprint\u001b[39m(\u001b[43mLSFR\u001b[49m\u001b[43m(\u001b[49m\u001b[43ms\u001b[49m\u001b[43m,\u001b[49m\u001b[43m \u001b[49m\u001b[43mc\u001b[49m\u001b[43m,\u001b[49m\u001b[43m \u001b[49m\u001b[38;5;241;43m10\u001b[39;49m\u001b[43m)\u001b[49m)\n",
      "Cell \u001b[0;32mIn[8], line 6\u001b[0m, in \u001b[0;36mLSFR\u001b[0;34m(s, c, nbtours)\u001b[0m\n\u001b[1;32m      4\u001b[0m t \u001b[38;5;241m=\u001b[39m \u001b[38;5;28mlen\u001b[39m(s)\n\u001b[1;32m      5\u001b[0m \u001b[38;5;28;01mwhile\u001b[39;00m i \u001b[38;5;241m<\u001b[39m nbtours:\n\u001b[0;32m----> 6\u001b[0m     res \u001b[38;5;241m=\u001b[39m \u001b[43ms\u001b[49m\u001b[43m[\u001b[49m\u001b[43mi\u001b[49m\u001b[43m]\u001b[49m\n\u001b[1;32m      7\u001b[0m     j \u001b[38;5;241m=\u001b[39m t\n\u001b[1;32m      8\u001b[0m     \u001b[38;5;28mprint\u001b[39m(j)\n",
      "\u001b[0;31mIndexError\u001b[0m: list index out of range"
     ]
    }
   ],
   "source": [
    "s = [0, 0, 1, 0]\n",
    "c = [1, 0, 1, 0]\n",
    "print(LSFR(s, c, 10))"
   ]
  }
 ],
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