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# Shortest path to reach one prime to other by changing single digit at a time

Given two four digit prime numbers, suppose 1033 and 8179, we need to find the shortest path from 1033 to 8179 by altering only single digit at a time such that every number that we get after changing a digit is prime. For example a solution is 1033, 1733, 3733, 3739, 3779, 8779, 8179

Examples:

Input : 1033 8179
Output :6

Input : 1373 8017
Output : 7

Input  :  1033 1033
Output : 0

The question can be solved by BFS and it is a pretty interesting to solve as a starting problem for beginners. We first find out all 4 digit prime numbers till 9999 using technique of Sieve of Eratosthenes. And then using those numbers formed the graph using adjacency list. After forming the adjacency list, we used simple BFS to solve the problem.

Implementation:

## C++

 // CPP program to reach a prime number from// another by changing single digits and// using only prime numbers.#include  using namespace std; class graph {    int V;    list* l;public:    graph(int V)    {        this->V = V;        l = new list[V];    }    void addedge(int V1, int V2)    {        l[V1].push_back(V2);        l[V2].push_back(V1);    }    int bfs(int in1, int in2);}; // Finding all 4 digit prime numbersvoid SieveOfEratosthenes(vector& v){    // Create a boolean array "prime[0..n]" and initialize    // all entries it as true. A value in prime[i] will    // finally be false if i is Not a prime, else true.    int n = 9999;    bool prime[n + 1];    memset(prime, true, sizeof(prime));     for (int p = 2; p * p <= n; p++) {         // If prime[p] is not changed, then it is a prime        if (prime[p] == true) {             // Update all multiples of p            for (int i = p * p; i <= n; i += p)                prime[i] = false;        }    }     // Forming a vector of prime numbers    for (int p = 1000; p <= n; p++)        if (prime[p])            v.push_back(p);} // in1 and in2 are two vertices of graph which are// actually indexes in pset[]int graph::bfs(int in1, int in2){    int visited[V];    memset(visited, 0, sizeof(visited));    queue que;    visited[in1] = 1;    que.push(in1);    list::iterator i;    while (!que.empty()) {        int p = que.front();        que.pop();        for (i = l[p].begin(); i != l[p].end(); i++) {            if (!visited[*i]) {                visited[*i] = visited[p] + 1;                que.push(*i);            }            if (*i == in2) {                return visited[*i] - 1;            }        }    }} // Returns true if num1 and num2 differ// by single digit.bool compare(int num1, int num2){    // To compare the digits    string s1 = to_string(num1);    string s2 = to_string(num2);    int c = 0;    if (s1[0] != s2[0])        c++;    if (s1[1] != s2[1])        c++;    if (s1[2] != s2[2])        c++;    if (s1[3] != s2[3])        c++;     // If the numbers differ only by a single    // digit return true else false    return (c == 1);} int shortestPath(int num1, int num2){    // Generate all 4 digit    vector pset;    SieveOfEratosthenes(pset);      // Create a graph where node numbers are indexes    // in pset[] and there is an edge between two    // nodes only if they differ by single digit.    graph g(pset.size());    for (int i = 0; i < pset.size(); i++)        for (int j = i + 1; j < pset.size(); j++)            if (compare(pset[i], pset[j]))                g.addedge(i, j);              // Since graph nodes represent indexes of numbers    // in pset[], we find indexes of num1 and num2.    int in1, in2;    for (int j = 0; j < pset.size(); j++)        if (pset[j] == num1)            in1 = j;    for (int j = 0; j < pset.size(); j++)        if (pset[j] == num2)            in2 = j;     return g.bfs(in1, in2);} // Driver codeint main(){    int num1 = 1033, num2 = 8179;    cout << shortestPath(num1, num2);    return 0;}

## Python3

 # Python3 program to reach a prime number# from another by changing single digits # and using only prime numbers.import queue class Graph:         def __init__(self, V):        self.V = V;        self.l = [[] for i in range(V)]             def addedge(self, V1, V2):        self.l[V1].append(V2);        self.l[V2].append(V1);     # in1 and in2 are two vertices of graph     # which are actually indexes in pset[]    def bfs(self, in1, in2):        visited = [0] * self.V        que = queue.Queue()        visited[in1] = 1        que.put(in1)        while (not que.empty()):            p = que.queue[0]            que.get()            i = 0            while i < len(self.l[p]):                if (not visited[self.l[p][i]]):                    visited[self.l[p][i]] = visited[p] + 1                    que.put(self.l[p][i])                if (self.l[p][i] == in2):                    return visited[self.l[p][i]] - 1                i += 1         # Returns true if num1 and num2    # differ by single digit.     # Finding all 4 digit prime numbersdef SieveOfEratosthenes(v):         # Create a boolean array "prime[0..n]" and     # initialize all entries it as true. A value    # in prime[i] will finally be false if i is    # Not a prime, else true.    n = 9999    prime = [True] * (n + 1)     p = 2    while p * p <= n:         # If prime[p] is not changed,        # then it is a prime        if (prime[p] == True):             # Update all multiples of p            for i in range(p * p, n + 1, p):                prime[i] = False        p += 1     # Forming a vector of prime numbers    for p in range(1000, n + 1):        if (prime[p]):            v.append(p)     def compare(num1, num2):         # To compare the digits    s1 = str(num1)    s2 = str(num2)    c = 0    if (s1[0] != s2[0]):        c += 1    if (s1[1] != s2[1]):        c += 1    if (s1[2] != s2[2]):        c += 1    if (s1[3] != s2[3]):        c += 1     # If the numbers differ only by a single    # digit return true else false    return (c == 1)     def shortestPath(num1, num2):         # Generate all 4 digit    pset = []    SieveOfEratosthenes(pset)     # Create a graph where node numbers    # are indexes in pset[] and there is    # an edge between two nodes only if    # they differ by single digit.    g = Graph(len(pset))    for i in range(len(pset)):        for j in range(i + 1, len(pset)):            if (compare(pset[i], pset[j])):                g.addedge(i, j)         # Since graph nodes represent indexes    # of numbers in pset[], we find indexes    # of num1 and num2.    in1, in2 = None, None    for j in range(len(pset)):        if (pset[j] == num1):            in1 = j    for j in range(len(pset)):        if (pset[j] == num2):            in2 = j     return g.bfs(in1, in2) # Driver codeif __name__ == '__main__':     num1 = 1033    num2 = 8179    print(shortestPath(num1, num2)) # This code is contributed by PranchalK

## C#

 // C# program to reach a prime number from// another by changing single digits and// using only prime numbers.using System;using System.Collections.Generic; class Graph {    private int V;    private List[] l;     // Constructor    Graph(int v)    {        V = v;        l = new List[ v ];        for (int i = 0; i < v; i++)            l[i] = new List();    }     // Function to Add an edge into the graph    void Addedge(int V1, int V2)    {        l[V1].Add(V2);        l[V2].Add(V1);    }     // Finding all 4 digit prime numbers    static void SieveOfEratosthenes(List v)    {        // Create a bool array "prime[0..n]" and        // initialize all entries it as true. A value in        // prime[i] will finally be false if i is Not a        // prime, else true.        int n = 9999;        bool[] prime = new bool[n + 1];        Array.Fill(prime, true);         for (int p = 2; p * p <= n; p++) {             // If prime[p] is not changed, then it is a            // prime            if (prime[p] == true) {                 // Update all multiples of p                for (int i = p * p; i <= n; i += p)                    prime[i] = false;            }        }         // Forming a vector of prime numbers        for (int p = 1000; p <= n; p++) {            if (prime[p])                v.Add(p);        }    }     // in1 and in2 are two vertices of graph which are    // actually indexes in pset[]    int Bfs(int in1, int in2)    {        int[] visited = new int[V];        Array.Fill(visited, 0);        Queue que = new Queue();        visited[in1] = 1;        que.Enqueue(in1);         while (que.Count > 0) {            int p = que.Dequeue();            foreach(int i in l[p])            {                if (visited[i] == 0) {                    visited[i] = visited[p] + 1;                    que.Enqueue(i);                }                if (i == in2) {                    return visited[i] - 1;                }            }        }         return 0;    }     // Returns true if num1 and num2 differ    // by single digit.    static bool Compare(int num1, int num2)    {        // To compare the digits        string s1 = num1.ToString();        string s2 = num2.ToString();        int c = 0;        if (s1[0] != s2[0])            c++;        if (s1[1] != s2[1])            c++;        if (s1[2] != s2[2])            c++;        if (s1[3] != s2[3])            c++;         // If the numbers differ only by a single        // digit return true else false        return (c == 1);    }     static int ShortestPath(int num1, int num2)    {        // Generate all 4 digit        List pset = new List();        SieveOfEratosthenes(pset);         // Create a graph where node numbers are indexes        // in pset[] and there is an edge between two        // nodes only if they differ by single digit.        Graph g = new Graph(pset.Count);        for (int i = 0; i < pset.Count; i++) {            for (int j = i + 1; j < pset.Count; j++) {                if (Compare(pset[i], pset[j]))                    g.Addedge(i, j);            }        }         // Since graph nodes represent indexes of numbers        // in pset[], we find indexes of num1 and num2.        int in1 = 0, in2 = 0;        for (int j = 0; j < pset.Count; j++) {            if (pset[j] == num1)                in1 = j;        }         for (int j = 0; j < pset.Count; j++) {            if (pset[j] == num2)                in2 = j;        }         return g.Bfs(in1, in2);    }     // Driver code    static void Main(string[] args)    {        int num1 = 1033, num2 = 8179;        int ans = ShortestPath(num1, num2);        Console.WriteLine(ans);    }} // This code is contributed by cavi4762.

## Javascript

 // JS Code class Graph {  constructor(V) {    this.V = V;    this.l = new Array(V);    for(let i=0;i 0) {      let p = que[0];      que.splice(0, 1);      for (let i = 0; i < this.l[p].length; i++) {        if (!visited[this.l[p][i]]) {          visited[this.l[p][i]] = visited[p] + 1;          que.push(this.l[p][i]);        }        if (this.l[p][i] == in2) {          return visited[this.l[p][i]] - 1;        }      }    }  }} // Finding all 4 digit prime numbersfunction SieveOfEratosthenes(v){   // Create a boolean array "prime[0..n]" and  // initialize all entries it as true. A value  // in prime[i] will finally be false if i is  // Not a prime, else true.  let n = 9999;  let prime = new Array(n + 1).fill(true);   let p = 2;  while (p * p <= n)  {       // If prime[p] is not changed,    // then it is a prime    if (prime[p] == true)    {           // Update all multiples of p      for (let i = p * p; i <= n; i += p) {        prime[i] = false;      }    }    p++;  }   // Forming a vector of prime numbers  for (let p = 1000; p <= n; p++) {    if (prime[p]) {      v.push(p);    }  }} function compare(num1, num2) {  // To compare the digits  let s1 = String(num1);  let s2 = String(num2);  let c = 0;  if (s1[0] != s2[0]) {    c += 1;  }  if (s1[1] != s2[1]) {    c += 1;  }  if (s1[2] != s2[2]) {    c += 1;  }  if (s1[3] != s2[3]) {    c += 1;  }   // If the numbers differ only by a single  // digit return true else false  return c == 1;} function shortestPath(num1, num2) {  // Generate all 4 digit  let pset = [];  SieveOfEratosthenes(pset);   // Create a graph where node numbers  // are indexes in pset[] and there is  // an edge between two nodes only if  // they differ by single digit.  let g = new Graph(pset.length);  for (let i = 0; i < pset.length; i++) {    for (let j = i + 1; j < pset.length; j++) {      if (compare(pset[i], pset[j])) {        g.addedge(i, j);      }    }  }   // Since graph nodes represent indexes  // of numbers in pset[], we find indexes  // of num1 and num2.  let in1, in2;  for (let j = 0; j < pset.length; j++) {    if (pset[j] == num1) {      in1 = j;    }  }  for (let j = 0; j < pset.length; j++) {    if (pset[j] == num2) {      in2 = j;    }  }   return g.bfs(in1, in2);} // Driver codeif (true) {  let num1 = 1033;  let num2 = 8179;  console.log(shortestPath(num1, num2));} // This code is contributed by ishankhandelwals.

Output :

6

Time Complexity : O(n*log(log(n)) + n^2 + n^2) = O(n^2).

Space complexity: O(n + V^2), which is equivalent to O(n).