28 — Game: an opponent that knows the rule

An opponent worth playing needs a rule, not only luck. In this game you take 1, 2 or 3 stones from a pile, and whoever takes the last stone loses. The computer follows a rule that keeps it in a winning position, and the whole rule is one remainder.

#Steps

1. The rule first. If the pile you hand over is a multiple of four, then whatever your opponent takes, you can make it a multiple of four again. So the computer takes the remainder of the pile divided by four:

   set pile to 9
   set mine to the remainder of pile divided by 4
   show mine
   

   copyrun it →

   Nine divided by four leaves one, so it takes one and leaves eight.

2. When the remainder is nought there is no winning move, so it plays at random:

   set pile to 8
   set mine to the remainder of pile divided by 4
   if mine equals 0
       set mine to random number from 1 to 3
   end
   show mine
   

   copyrun it →

3. Your turn asks for a number, and asks again when it is not 1 to 3. skip abandons the rest of this turn and goes back to the top of the loop:

   repeat forever
       ask number taken How many? 1, 2 or 3
       if taken is less than 1
           show it has to be 1, 2 or 3
           skip
       end
       stop
   end
   

   copyrun it →

   skip only works inside a loop. Outside one, the compiler stops you.

4. Take the stones off the pile; at nought, whoever just took has lost:

   set pile to 12
   set taken to 3
   subtract taken from pile
   show pile
   

   copyrun it →

5. Here is the whole game. Put it in nim.nme and play:

   set pile to 12
   repeat forever
       show stones left
       show pile
       ask number taken How many? 1, 2 or 3
       if taken is less than 1
           show it has to be 1, 2 or 3
           skip
       end
       if taken is greater than 3
           show it has to be 1, 2 or 3
           skip
       end
       subtract taken from pile
       if pile equals 0
           show you took the last stone, you lose
           stop
       end
       set mine to the remainder of pile divided by 4
       if mine equals 0
           set mine to random number from 1 to 3
       end
       subtract mine from pile
       show the computer took
       show mine
       if pile equals 0
           show the computer took the last stone, you win
           stop
       end
   end
   

   copyrun it →

   Try 3, then 1, then 3. The computer keeps handing back a multiple of four, so from a pile of twelve you are the one who wins.

#Try it yourself

Change the starting pile to 15. Fifteen is not a multiple of four, so the computer takes the good position first. Play a few rounds and work out why. A name that counts your wins is a good next step.

#What you learned

• the remainder of … divided by … is what is left over, and it is this whole strategy.
• skip gives up the rest of this turn and starts the loop again.
• An opponent's "thinking" is one rule worked out from where the game is now.
• A random move where the rule cannot win keeps the opponent from collapsing.