Oooh, ooh, this might be better... if impractical...
Assign each colour a number 1-50.
Leprechaun 1 looks at the all the hats in front of him and adds the numbers together.
He takes away 50 from this number until he has something in the interval 1-50.
He then calls out the colour corresponding to this number.
If that happens to be the colour of his hat he is very lucky and gets to go free, but he's probably gonna be sticking around.
Leprechaun 2 can see the all the hats L1 could except his own.
He repeats the calculation with the 998 hats in front of him.
The difference between his result and that called out by L1 will tell him the colour of his hat. Which he calls out.
Leprechaun 3 will hear this colour and can take its value from the colour(/number) that L1 called out. He then uses the same method as L2 to deduce his/her own colour.
Each subsequent Leprechaun can then copy the procedure followed by L3.
Of course if anybody makes a mistake then it's not gonna go too well for them!
- Mike
- Joined
- 26 Apr 03
- Moves
- 26771
Originally posted by mikenayCongratulations! That's it! With the best strategy 999 leprechauns go free. The one remaining one can't even breed more on her own. The king (who thought only half would escape) is desolated at the loss of his leprechauns who were almost irreplacable staff in the nation's banks and paint shops due to their infallible color sense and instant mental arithmetic.
Oooh, ooh, this might be better... if impractical...
Assign each colour a number 1-50.
Leprechaun 1 looks at the all the hats in front of him and adds the numbers together.
He takes away 50 from this number until he has something in th ...[text shortened]... kes a mistake then it's not gonna go too well for them!
- Mike
Anyone want to suggest a puzzle that lets the last poor lonely leprechaun escape to her friends?