[Jack Simth]

Originally posted by Nocturnal:
Here's a nice one.

While vacationing, you discover that you have been robbed. Your wallet and all your cash have been stolen.

You cannot afford to pay for a hotel room, and since it's a local holiday all the banks are closed. You won't be able to send for any money for seven days.

You go to the hotel you reserved and explain the situation, but they refuse to extend credit.
However, you are wearing a gold bracelet with seven links, of the kind that closes with a clasp. Each link is worth enough to pay for a night in your hotel. The hotel is willing to accept this as collateral until you can pay in full.

Being a suspicious type yourself, you'll only let the hotel have as much collateral as they are due on any given day. Today you leave one link, tomorrow two, the next day three, and so on until the week is up when you'll leave all seven. By then you will be able to pay in full and retrieve your bracelet.

This agreed to, you prepare to cut the bracelet. Considering its high emotional-value, you want to do as little damage as possible so it may be easily repaired. What is the fewest number of cuts you can make so the hotel may hold onto all the collateral they need on any day, but no more?

It can be done with a single cut:
0: Undo clasp, so it is a line;
1: Cut link 3 and remove (one + two connected, four+five+six+seven connected)

Process:
First day: give single link. (you: 2+4, them: 1)
Second Day: retrieve single, give double (you: 1+4, them: 2)
Third day: give single (you: 4, them: 2+1)
Fourth day: retrieve all, give quad (you: 2+1, them: 4)
Fifth day: give single (you: 2, them: 4+1)
Sixth day: retrieve single, give double: (you: 1, them: 4+2)
Seventh (final) day: give single (you: none, them: 4+2+1)

It's disguised binary counting.

[ July 22, 2003, 09:46: Message edited by: Jack Simth ]