# Program for a game similar to rock-paper-scissors

by sandy.bridge
Tags: game, program, rockpaperscissors, similar
 P: 778 1. The problem statement, all variables and given/known data I'm writing a program for a game similar to rock-paper-scissors. It is called proton-neutron-electron (P, N, E). There are two players. In the event of a tie, player 2 wins P1 denotes player 1, P2 denotes player 2, P = proton, N = neutron, E = electron. P1P means player 1 picks proton. There are 9 possibilities. Here they are: P1P + P2P = P2 wins P1P + P2N = P1 wins P1P + P2E = P1 wins P1N + P2P = P2 wins P1N + P2N = P2 wins P1N + P2E = P2 wins P1E + P2P = P1 wins P1E + P2N = P2 wins P1E + P2E = P2 wins 3. The attempt at a solution Here is what I have written in the compiler; it does not work. I am not looking for another solution, I would just like to know where I went wrong in my syntax. PHP Code:      char user1, user2;    bool A, B, C, D, E, F;    cout << "Player 1, please enter your move: ";        cin >> user1;        cout << "Player 2, please enter your move: ";        cin >> user2;    if (user2 = 'P')    {        A = true;    }    else    {        A = false;    }    if (user2 = 'N')    {        B = true;    }    else    {        B = false;    }    if (user2 = 'E')    {        C = true;    }    else    {        C = false;    }    if (user1 = 'P')    {        D = true;        }    else    {        D = false;    }    if (user1 = 'N')    {        E = true;    }    else    {        E = false;    }    if (user1 = 'E')    {        F = true;    }    else    {        F = false;    }    if ((B&&D)||(C&&D)||(A&&E)||(B&&F))        {            cout << "The winner is player 1!";        }    else        {            cout << "The winner is player 2!";        }    return 0;   The way that it is set up now always results in P1 winning and I can't see why. Thanks in advance!
 Sci Advisor HW Helper P: 4,300 You could start by replacing all = with == and see what happens?
 P: 778 That didn't work.
 P: 778 Program for a game similar to rock-paper-scissors I managed to just do as such: PHP Code:  #include  using namespace std; int main() {     char user1, user2;             cout << "What move P1? ";             cin >> user1;             cout << "What move P2? ";             cin >> user2;             if (((user1 == 'P')&&(user2 == 'N'))||((user1 == 'P')&&(user2 == 'E'))||((user1 == 'N')&&(user2 == 'P'))||((user1 == 'E')&&(user2 == 'N')))                 {                     cout << "Player 1 is the winner!";                 }             else                 {                     cout << "Player 2 is the winner!";                 }             return 0; }   I'd still like to know what I was doing wrong for the first part.
 Sci Advisor HW Helper P: 4,300 Can you post the changed code that you tried but didn't work?
 P: 778 PHP Code:      char user1, user2;    bool A, B, C, D, E, F;    cout << "Player 1, please enter your move: ";        cin >> user1;        cout << "Player 2, please enter your move: ";        cin >> user2;    if (user2 == 'P')    {        A == true;    }    else    {        A == false;    }    if (user2 == 'N')    {        B == true;    }    else    {        B == false;    }    if (user2 == 'E')    {        C == true;    }    else    {        C == false;    }    if (user1 == 'P')    {        D == true;        }    else    {        D == false;    }    if (user1 == 'N')    {        E == true;    }    else    {        E == false;    }    if (user1 == 'E')    {        F == true;    }    else    {        F == false;    }    if ((B&&D)||(C&&D)||(A&&E)||(B&&F))        {            cout << "The winner is player 1!";        }    else        {            cout << "The winner is player 2!";        }    return 0;   This didn't work ^^^^ PHP Code:  #include using namespace std;int main() {    char user1, user2;            cout << "What move P1? ";            cin >> user1;            cout << "What move P2? ";            cin >> user2;            if (((user1 == 'P')&&(user2 == 'N'))||((user1 == 'P')&&(user2 == 'E'))||((user1 == 'N')&&(user2 == 'P'))||((user1 == 'E')&&(user2 == 'N')))                {                    cout << "Player 1 is the winner!";                }            else                {                    cout << "Player 2 is the winner!";                }            return 0;}   This did work ^^^
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 6,265 Note: '==' is a logical operator which you use, for example, in an 'if' statement to test if a certain condition is true: if (user1 == 'P') {'some calculations'} If you wish to assign a certain value to a variable, then the following statement is used: { user1 = 'P'; }
 P: 778 Ah yes, that was where my issue was. Works now! Thanks! I will obviously just use the other one though because it is much shorter.
Mentor
P: 21,216
 Quote by sandy.bridge 1. The problem statement, all variables and given/known data I'm writing a program for a game similar to rock-paper-scissors. It is called proton-neutron-electron (P, N, E). There are two players. In the event of a tie, player 2 wins P1 denotes player 1, P2 denotes player 2, P = proton, N = neutron, E = electron. P1P means player 1 picks proton. There are 9 possibilities. Here they are: 1. P1P + P2P = P2 wins 2. P1P + P2N = P1 wins 3. P1P + P2E = P1 wins 4. P1N + P2P = P2 wins 5. P1N + P2N = P1 wins << P2 should be the winner here 6. P1N + P2E = P2 wins 7. P1E + P2P = P1 wins 8. P1E + P2N = P2 wins 9. P1E + P2E = P2 wins
I added numbers to the rules for easier identification.

Your rule 5 violates what you said earlier about ties going to player 2.
Can you summarize the rules any better than in the table?

It looks like proton beats neutron (rules 2, 4) and electron beats neutron (rule 6) sometimes, but neutron beats electron sometimes (rule 8). Is that also a mistake?

In rock paper scissors, there is a simple set of rules: rock covers (beats) paper, scissors cut (beat) paper, and paper covers (beats) rock. Is there a similar, simple rule here that you didn't mention? If so, it would make your logic much simpler.
 P: 778 Yeah, I noticed that as well. There doesn't seem to be any logicality with it, but that is how the table was given in the assignment. At that point I just made sure the algorithm obeyed the table. PHP Code:               P1 plays P      P1 plays N             P1 plays E P2 plays P         P2                 P1                  P2 P2 plays N         P1                 P2                  P1 P2 plays E         P1                 P2                  P2  
Mentor
P: 21,216
I redid the indentation in your code so that the lines didn't scroll so far to the right.
 Quote by sandy.bridge #include using namespace std; int main() { char user1, user2; cout << "What move P1? "; cin >> user1; cout << "What move P2? "; cin >> user2; if (((user1 == 'P')&&(user2 == 'N')) || ((user1 == 'P')&&(user2 == 'E')) || ((user1 == 'N')&&(user2 == 'P')) || ((user1 == 'E')&&(user2 == 'N')) ) { cout << "Player 1 is the winner!"; } else { cout << "Player 2 is the winner!"; } return 0; } This did work ^^^
Mentor
P: 21,216
 Quote by sandy.bridge Yeah, I noticed that as well.
Then you have an error in post 1. According to the table below, if both players play a neutron, P2 wins.

 Quote by sandy.bridge There doesn't seem to be any logicality with it, but that is how the table was given in the assignment. At that point I just made sure the algorithm obeyed the table. PHP Code:               P1 plays P      P1 plays N             P1 plays E P2 plays P         P2                 P1                  P2 P2 plays N         P1                 P2                  P1 P2 plays E         P1                 P2                  P2  
 P: 778 Oops! I just edited that thanks.
 Mentor P: 21,216 Instead of one if block with a large condition, I would write three if blocks in which I check what player 1 has. I have implemented the first of these if blocks, and have added comments to suggest what the logic is. if (user1 == 'P') { if (user2 == 'P') { cout << "Player 2 is the winner!"; } else { cout << "Player 1 is the winner!"; } } else if (user1 == 'N') { // Player 2 wins only if player 1 has 'P' } else // user1 == 'E' { // Player 1 wins only if player 2 has 'N' }

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