The optimization idea is that 1,3,7,9 are symmetric, 2,4,6,8 are also symmetric. Hence we only calculate one among each group and multiply by 4.

```
public class Solution {
// cur: the current position
// remain: the steps remaining
int DFS(boolean vis[], int[][] skip, int cur, int remain) {
if(remain < 0) return 0;
if(remain == 0) return 1;
vis[cur] = true;
int rst = 0;
for(int i = 1; i <= 9; ++i) {
// If vis[i] is not visited and (two numbers are adjacent or skip number is already visited)
if(!vis[i] && (skip[cur][i] == 0 || (vis[skip[cur][i]]))) {
rst += DFS(vis, skip, i, remain - 1);
}
}
vis[cur] = false;
return rst;
}
public int numberOfPatterns(int m, int n) {
// Skip array represents number to skip between two pairs
int skip[][] = new int[10][10];
skip[1][3] = skip[3][1] = 2;
skip[1][7] = skip[7][1] = 4;
skip[3][9] = skip[9][3] = 6;
skip[7][9] = skip[9][7] = 8;
skip[1][9] = skip[9][1] = skip[2][8] = skip[8][2] = skip[3][7] = skip[7][3] = skip[4][6] = skip[6][4] = 5;
boolean vis[] = new boolean[10];
int rst = 0;
// DFS search each length from m to n
for(int i = m; i <= n; ++i) {
rst += DFS(vis, skip, 1, i - 1) * 4; // 1, 3, 7, 9 are symmetric
rst += DFS(vis, skip, 2, i - 1) * 4; // 2, 4, 6, 8 are symmetric
rst += DFS(vis, skip, 5, i - 1); // 5
}
return rst;
}
}
```