Rolling a dice in a spiral

Question

Given a grid size of nxn, a dice is put on the top-left field (1,1) with the number 6 facing down, 5 facing (1,2) and 4 facing (2,1). The dice will roll in a spiral (clockwise) to fill every field with a number (only once). Calculate the total sum of the numbers printed. Visual representation of the moves of the dice and the numbers printed when n=5 (result = 81)

01 02 03 04 05
16 17 18 19 06
15 24 25 20 07
14 23 22 21 08
13 12 11 10 09

6 5 1 2 6
4 5 3 2 4
1 1 3 1 1
3 2 3 5 3
6 5 1 2 6

This is a homework question, but I can't figure out how to do this efficiently without going through all possible cases. If someone could give me a solution and explanation, that would be amazing (no code needed, I want to do it myself).

Answer 1

A possible clean way t do this is by define a class Dice as mentioned below.

class Dice
{
public:
    Dice();
    int face_down();
    void roll_west();
    void roll_east();
    void roll_north();
    void roll_south();
    void set_initial_config(int face_down,int east,int north,int west,int south);
    ~Dice();
private:
    /// variables for your state etc
};

If you implement Dice successfully then rest of work will be simple simulation of the rolling of the dice.

int roll(int n,int m){ // grid dimensions

    std::vector<std::vector<bool> > visited(n,std::vector<bool>(m,0));

    int i=0,j=-1;
    int dir=0;
    bool dir_changed;
    int dir_change_count=0;
    int sum=0;
    Dice d;

    while(dir_change_count<4){

        dir_changed=0;
        switch(dir){
            case 0:
                if(j+1<m and !visited[i][j+1]){
                    j++;
                    visited[i][j+1]=1;
                    d.roll_east();
                }else{
                    dir_changed=1;
                    dir++;
                }
                break;
            case 1:
                if(i+1<n and !visited[i+1][j]){
                    i++;
                    visited[i+1][j]=1;
                    d.roll_south();
                }else{
                    dir_changed=1;
                    dir++;
                }
                break;
            case 2:
                if(j-1>=0 and !visited[i][j-1]){
                    j--;
                    visited[i][j-1]=1;
                    d.roll_west();
                }else{
                    dir_changed=1;
                    dir++;
                }
                break;
            case 3:
                if(i-1>=0 and !visited[i-1][j]){
                    i--;
                    visited[i-1][j]=1;
                    d.roll_north();
                }else{
                    dir_changed=1;
                    dir++;
                }
                break;
        }
        if(!dir_changed){
            sum+=d.face_down();
            dir_change_count=0;
        }else{
            dir_change_count++;
        }
    }

    return sum;
}