15 Apr '04 13:03>
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
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