Number of ways to make mobile lock pattern

A mobile pattern is a grid of 3X3 cell, where drawing a specific pattern (connecting specific sequence of cells in order) will unlock the mobile. In this problem, the task is to calculate number of ways of making the lock pattern with number of connections in given range. In general terms, we are given a range as min and max, we need to tell how many patterns can be made which use at least min connection cell and at most max connection cell.

Input  : min = 5, max = 7
Output : 106080
Below are some example patterns
pattern1

A simple solution is to do simple DFS by going through various connections from all starting points. We can optimize by seeing the symmetry between patterns, we can treat cell 1, 3, 7 and 9 as same. Similarly, we can treat 2, 4, 6 and 8 as same. Answer returned by a member of same group can be multiplied by 4 to get result by all members.
Next thing to note is that below patterns are not valid, because jumping some cell is not allowed as in below diagram cell 8 and cell 5, 6 are jumped.
pattern2

To take care of such invalid moves, a 2D jump array is taken in below code which stores possible jump cells in jump array. When we call recursively we impose an extra condition that if we are moving from one cell to another which involves some jumping cell, then this cell should be visited already to ignore invalid moves.
In below code we have made recursive calls from 1, 2 and 5 only because 1 will cover 3, 5 and 7 and 2 will cover 4, 6 and 8 due to symmetry.
See below code for better understanding :

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//  C/C++ program to find number of ways to lock the mobile
// pattern
#include <bits/stdc++.h>
using namespace std;
#define DOTS 10
  
//  method to find total pattern starting from current cell
int totalPatternFromCur(bool visit[DOTS], int jump[DOTS][DOTS],
                                          int cur, int toTouch)
{
    if (toTouch <= 0)
    {
        //  if last cell then return 1 way
        return (toTouch == 0)? 1 : 0;
    }
  
    int ways = 0;
  
    //  make this cell visited before going to next call
    visit[cur] = true;
  
    for (int i=1; i<DOTS; i++)
    {
       /*  if this cell is not visit AND
           either i and cur are adjacent (then jump[i][cur] = 0)
           or between cell must be visit already (
           then visit[jump[i][cur]] = 1)   */
       if (!visit[i] && (!jump[i][cur] || visit[jump[i][cur]]))
         ways += totalPatternFromCur(visit, jump, i, toTouch - 1);
    }
  
    //  make this cell not visited after returning from call
    visit[cur] = false;
  
    return ways;
}
  
//  method returns number of pattern with minimum m connection
// and maximum n connection
int waysOfConnection(int m, int n)
{
    int jump[DOTS][DOTS] = {0};
  
    //  2 lies between 1 and 3
    jump[1][3] = jump[3][1] = 2;
  
    //  8 lies between 7 and 9
    jump[7][9] = jump[9][7] = 8;
  
    //  4 lies between 1 and 7
    jump[1][7] = jump[7][1] = 4;
  
    //  6 lies between 3 and 9
    jump[3][9] = jump[9][3] = 6;
  
    //  5 lies between 1, 9  2, 8  3, 7 and 4, 6
    jump[1][9] = jump[9][1] = jump[2][8] = jump[8][2]
     = jump[3][7] = jump[7][3] = jump[4][6] = jump[6][4] = 5;
  
    bool visit[DOTS] = {0};
    int ways = 0;
    for (int i = m; i <= n; i++)
    {
        //  1, 3, 7, 9 are symmetric so multiplying by 4
        ways += 4 * totalPatternFromCur(visit, jump, 1, i - 1);
  
        //  2, 4, 6, 8 are symmetric so multiplying by 4
        ways += 4 * totalPatternFromCur(visit, jump, 2, i - 1);
  
        ways += totalPatternFromCur(visit, jump, 5, i - 1);
    }
  
    return ways;
}
  
//  Driver code to test above method
int main()
{
    int minConnect = 4;
    int maxConnect = 6;
  
    cout << waysOfConnection(minConnect, maxConnect);
  
    return 0;
}

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Output:

34792

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