Competitive programming

Competitive_Programming/C++/Maths & Number Theory/Euler_toitent_function.ipynb

@@ -0,0 +1,141 @@
+{
+  "nbformat": 4,
+  "nbformat_minor": 0,
+  "metadata": {
+    "colab": {
+      "name": "Euler_toitent_function.md",
+      "provenance": []
+    },
+    "kernelspec": {
+      "name": "python3",
+      "display_name": "Python 3"
+    },
+    "language_info": {
+      "name": "python"
+    }
+  },
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "dP9LH9ODdmir"
+      },
+      "source": [
+        "# **EULER TOITENT FUNCTION:**"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "UCesKC4Jd6yA"
+      },
+      "source": [
+        "Euler’s Totient function Φ (n) for an input n is the count of numbers belonging to set {1,2,3,....n} such that the number is coprime with n, i.e gcd(m,n)=1 such that 1<=m<n."
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "45B_hNXcefjl"
+      },
+      "source": [
+        "**PROPERTIES:**\n",
+        "\n",
+        "\n",
+        "1.   Φ (n)= n*(1-1/p1)*(1-1/p2)....*(1-1/pk),  where p1,p2 .....,pk are prime factors of n.\n",
+        "\n",
+        "2.   Φ (p)=p-1, where p=prime number\n",
+        "\n",
+        "3.   Φ (ab)=Φ (a)*Φ (b)\n",
+        "\n",
+        "4.   Φ (p^k)=(p^k)-(p^(k-1)) ,where p=prime number\n",
+        "\n",
+        "\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "XquPBAxYgZ3z"
+      },
+      "source": [
+        "**PROOF:**\n",
+        "\n",
+        "> Let n=(p1^a)*(p2^b)......*(pk^k) ,where p1,p2 .....,pk are prime factors of n and a,b,....k are powers of prime factors of n.\n",
+        "\n",
+        "\n",
+        "> Then, Φ (n)=Φ ((p1^a)*(p2^b)......*(pk^k))\n",
+        "\n",
+        "> Using The Properties we get: Φ (n)= Φ ((p1^a))*Φ ((p2^b))*.....Φ ((pk^k)),\n",
+        "\n",
+        "\n",
+        "> Since,p1,p2,...pk are prime numbers Therefore,Using the Properties Φ (pi^x)=(pi^x)-(pi^(x-1))\n",
+        "\n",
+        "\n",
+        "> => Φ (pi^x) = (pi^x)*(1-1/pi)\n",
+        "\n",
+        "> => Φ (n)= ((p1^a)*(1-1/p1))*((p2^b)*(1-1/p2))*........((pk^k)*(1-1/pk))\n",
+        "\n",
+        "\n",
+        "> => Φ (n) = ((p1^a)*(p2^b)......*(pk^k))*(1-1/p1)*(1-1/p2)*.....(1-1/pk)\n",
+        "\n",
+        "\n",
+        "> Therefore, Φ (n) = n*(1-1/p1)*(1-1/p2)*.....(1-1/pk)\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n",
+        "\n"
+      ]
+    },
+    {
+      "cell_type": "markdown",
+      "metadata": {
+        "id": "cc--D9uuj10W"
+      },
+      "source": [
+        "**FUNCTION TO CALCULATE EULER TOTIENT FUNCTION FOR A NATURAL NUMBER N:**"
+      ]
+    },
+    {
+      "cell_type": "code",
+      "metadata": {
+        "id": "Ku3SKXSEkFZN"
+      },
+      "source": [
+        "int euler_totient_function(int n){\n",
+        "    int phi[n+1];\n",
+        "    for(int i=1;i<=n;i++){\n",
+        "        phi[i]=i;\n",
+        "    }\n",
+        "    for(int i=2;i<=n;i++){\n",
+        "        if(phi[i]==i){\n",
+        "            phi[i]=i-1;\n",
+        "            for(int j=2*i;j<=n;j+=i){\n",
+        "                phi[j]=(phi[j]*(i-1))/i;\n",
+        "            }\n",
+        "        }\n",
+        "    }\n",
+        "    return phi[n];\n",
+        "}"
+      ],
+      "execution_count": null,
+      "outputs": []
+    }
+  ]
+}