All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Intuitive Coins (Posted on 2005-03-29)
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.

 See The Solution Submitted by Federico Kereki Rating: 3.8000 (5 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re(2): There's more to it.... (solution? spoiler?) -- larger sets | Comment 9 of 14 |
(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 better-than-greedy algorithms can exist where the better choice does not involve equal denomination coins.

 Posted by Charlie on 2005-03-31 13:08:57

 Search: Search body:
Forums (0)