Question

I’m messing around with bitwise operations and wanted to implement the simple logic puzzle where you have a fox(wolf), chicken(goose), grain(corn) and a person trying to cross a river. I’m using the first 4 bits for each one to represent which side of the river they are on.

I’m having a little bit of trouble trying to implement the logic.

How can I check if the two bits are either 1 or 0 but not 1 and 0?

int game()
{
    int state = 0;
    int done = 0;
    while (!done)
    {
        int input = 0;
        input = move();

        /*
           0000 0000
                |||\_grain
                ||\__chicken
                |\___fox
                \____person
           0 = left side of the river
           1 = right side
        */
        if (input == 3)// Moves person and grain
        {
            // Move the grain if the person is on the same side.
            if (!(state & 1 << 3 ^ state & 1<< 0))
            {
                state ^= 1 << 3;
                state ^= 1 << 0;
            }
            else
            {
                // Always switch the person no matter what
                state ^= 1 << 3;
            }
        }
        else if (input == 2) // Moves person and chicken
        {
            // Move Chicken only if person is on the same side
            if (!(state & 1 << 3 ^ state & 1<< 1))
            {
                state ^= 1 << 3;
                state ^= 1 << 1;
            }
            else
            {
                // Always switch the person no matter what
                state ^= 1 << 3;
            }
        }
        else if (input == 1)// Moves person and fox
        {
            // Move the fox if the person is on the same side.
            if (!(state & 1 << 3 ^ state & 1<< 2))
            {
                state ^= 1 << 3;
                state ^= 1 << 2;
            }
            else
            {
                // Always switch the person no matter what
                state ^= 1 << 3;
            }
        }

        // Fox and Chicken on one side and person on the other = lost
        if ((state & 1 << 2 && state & 1 << 1) && ~(state & 1 << 3))
        {
            printf("Failed\n");
        }
    }
    return 1;
}

I’m guessing bitwise checking would be more elegant code but it seems like it’s more of a pain. I always end up doing this when I get sick of hitting my head against the wall with bitwise logic.

int game()
{
    int state = 0;
    int done = 0;
    while (!done)
    {
        int input = 0;
        input = move();

        /*
           0000 0000
                   | | | \_grain
                   | | \__chicken
                   | \___fox
                   \____person
           0 = left side of the river
           1 = right side
        */
        if (input == 3)// Moves person and grain
        {
            // Are they on the same side?
            if (state == 9 || state == 11 || state == 13 || state == 15 ||
                state == 0 || state == 2 || state == 4 || state == 6)
            {
                // Move the person and grain
                state ^= 1 << 3;
                state ^= 1 << 0;
            }
            else
            {
                state ^= 1 << 3;
            }
        }
        else if (input == 2) // Moves person and chicken
        {
            // Are they on the same side?
            if (state == 10 || state == 11 || state == 14 || state == 15 ||
                state == 0 || state == 1 || state == 4 || state == 5)
            {
                // Move the person and chicken
                state ^= 1 << 3;
                state ^= 1 << 1;
            }
            else
            {
                state ^= 1 << 3;
            }
        }
        else if (input == 1)// Moves person and fox
        {
            // Are they on the same side?
            if (state == 12 || state == 13 || state == 14 || state == 15 ||
                state == 0 || state == 1 || state == 2 || state == 3)
            {
                // Move the person and fox
                state ^= 1 << 3;
                state ^= 1 << 2;
            }
            else
            {
                state ^= 1 << 3;
            }
        }
        else
        {
            // Always switch the person no matter what
            state ^= 1 << 3;
        }

        //Check if you won or lost
        if (state == 3 || state == 6 || state == 8 || state == 9 || state == 12) // Lost
        {
            return 0;
        }
        if (state == 15) // Won
        {
            return 1;
        }
    }
    return 1;
}

Answer 1

How can I check if the two bits are either 1 or 0 but not 1 and 0?

How about comparing them for equality?

#define MAN_MASK   (1<<3)
#define GRAIN_MASK (1<<0)

if(  
     ((state & MAN_MASK) == 0)
            == 
     ((state & GRAIN_MASK) == 0)
  );