If you must pay an amount in coins, the "intuitive" algorithm is: pay as much as possible with the largest denomination coin, and then go on to pay the rest with the other coins. For example, if there are 25, 5 and 1 cent coins, to pay someone 32 cents, you'd first give him a 25 cents coin, then one 5 cent coin, and finally two 1 cent coins.)
However, this doesn't always end paying with as few coins as possible: if we had 25, 10 and 1 cent coins, paying 32 cents with the "intuitive" algorithm would use 8 coins, while three 10 cent coins and two 1 cent coins would be better.
We can call a set "intuitive", if the "intuitive algorithm" always pays out any amount with as few coins as possible.
The problem: give an algorithm that allows you to decide that {25,5,1} is an "intuitive" set, while {25,10,1} isn't.
(In reply to
re: There's more to it.... (solution? spoiler?)  larger sets by Charlie)
In looking over the literature on the subject, it seems that checking multiples of a single denomination is not sufficient in weeding out sets where the greedy algorithm does not work.
I see an example of {14, 12, 5, 1}. While it's true that the multiple test does show a flaw at a value of 2*12=24, where "greedy" chooses 3 coins(14+2*5), there's another flaw that would not show up in the multiple test: 17. The greedy algorithm chooses 4 coins (14+3*1) while 12+5 works with only 2 coins. So betterthangreedy algorithms can exist where the better choice does not involve equal denomination coins.

Posted by Charlie
on 20050331 13:08:57 