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   "source": [
    "**Example 4.40.** Classify emails represented by the binary term-to-document vectors as \"spam\" and \"not spam\". Consider the following dataset.\n",
    "\n",
    "|                     | \"and\" | \"offer\" | \"the\" | \"of\" | \"sale\" | $y_i$           |\n",
    "| -                   |  -    |       - | -     |  -   |  -     |  -              |\n",
    "|                     |       |         |       |      |        |                 |\n",
    "| $\\mathbf{\\hat x}_1$ | 1     | 1       | 0     | 1    | 1      | $+1$ (spam)     |\n",
    "| $\\mathbf{\\hat x}_2$ | 0     | 0       | 1     | 1    | 0      | $-1$ (not spam) |\n",
    "| $\\mathbf{\\hat x}_3$ | 0     | 1       | 1     | 0    | 0      | $+1$ (spam)     |\n",
    "| $\\mathbf{\\hat x}_4$ | 1     | 0       | 0     | 1    | 0      | $-1$ (not spam) |\n",
    "|                     |       |         |       |      |        |                 |  \n",
    "| $\\mathbf{\\hat x}_5$ | 1     | 0       | 1     | 0    | 1      | $+1$ (spam)     |\n",
    "| $\\mathbf{\\hat x}_6$ | 1     | 0       | 1     | 1    | 0      | $-1$ (not spam) |\n"
   ]
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  {
   "cell_type": "code",
   "execution_count": null,
   "id": "6e1dc389-7410-4593-8373-1a640224d1a9",
   "metadata": {},
   "outputs": [],
   "source": [
    "import numpy as np\n",
    "Xh = np.array([[ 1 ,  1 ,  0 , 1 , 1],   # x_1   spam\n",
    "               [ 0 ,  0 ,  1 , 1 , 0],   # x_2   not spam\n",
    "               [ 0 ,  1 ,  1 , 0 , 0],   # x_3   spam\n",
    "               [ 1 ,  0 ,  0 , 1 , 0],   # x_4   not spam\n",
    "               [ 1 ,  0 ,  1 , 0 , 1],   # x_5   spam\n",
    "               [ 1 ,  0 ,  1 , 1 , 0]    # x_6   not spam\n",
    "             ])\n",
    "y = np.array([1,-1,1,-1,1,-1])\n",
    "X = np.hstack((np.ones((Xh.shape[0],1)), Xh)) # Affine -> Homogeneous form"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "id": "0d68c106-6acc-43e7-aaa2-b8013adba42f",
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   "source": [
    "# Perceptron algorithm selecting ik without replacement\n",
    "def Perceptron(X,y, K=100):\n",
    "    theta = np.zeros(X.shape[1])\n",
    "    k = 0\n",
    "    while (k<K):\n",
    "        for i in range(y.size):\n",
    "            if y[i] * X[i] @ theta <= 0:\n",
    "                theta += y[i] * X[i]\n",
    "                k += 1\n",
    "                print(f\"iteration {k}\")\n",
    "        if np.all(y * (X @ theta) > 0):\n",
    "            break\n",
    "    return theta"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "id": "6a44173a",
   "metadata": {},
   "outputs": [],
   "source": [
    "# Perceptron algorithm selecting ik with replacement\n",
    "def PerceptronR(X,y, K=100):\n",
    "    theta = np.zeros(X.shape[1])\n",
    "    k = 0\n",
    "    while (k<K):\n",
    "        i = np.random.randint(y.size)\n",
    "        if y[i] * X[i] @ theta <= 0:\n",
    "            theta += y[i] * X[i]\n",
    "            k += 1\n",
    "            print(f\"iteration {k}\")\n",
    "        if np.all(y * (X @ theta) > 0):\n",
    "            break\n",
    "    return theta"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "id": "864f91f5-0286-4697-9832-f7cb204c1bfc",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "iteration 1\n",
      "iteration 2\n",
      "iteration 3\n",
      "iteration 4\n",
      "theta: [ 0.  0.  2.  0. -1.  1.]\n",
      "predicted labels: [ 1. -1.  1. -1.  1. -1.]\n"
     ]
    }
   ],
   "source": [
    "theta = Perceptron(X[:4], y[:4])\n",
    "print(\"theta: \" + str(theta))\n",
    "print(\"predicted labels: \" + str(np.sign(X @ theta)))"
   ]
  }
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