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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.

C++

// CPP program to reach a prime number from 
// another by changing single digits and 
// using only prime numbers.
#include <bits/stdc++.h>
  
using namespace std;
  
class graph {
    int V; 
    list<int>* l; 
public:
    graph(int V)
    {
        this->V = V;
        l = new list<int>[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 numbers
void SieveOfEratosthenes(vector<int>& 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<int> que;
    visited[in1] = 1;
    que.push(in1);
    list<int>::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<int> 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 code
int 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 numbers def 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 code if __name__ == '__main__': num1 = 1033 num2 = 8179 print(shortestPath(num1, num2)) # This code is contributed by PranchalK [tabbyending] Output :

6


This article is attributed to GeeksforGeeks.org

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