

What the U.S. needs is an 18-cent coin - donteatbark
http://radio-weblogs.com/0105910/2003/05/16.html

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platykurtic
An interesting wrinkle to this is that an 18 cent coin would make it much
harder to make change with the fewest coins for certain values.

How do we make change in everyday life? The simple algorithm everyone knows,
even if they don't know what an algorithm is, is to start with the largest
coin and move down, taking as many of each as you can. Thus to make change for
72 cents we: take 2 quarters, leaving us with 22 cents take 2 dimes, leaving
us with 2 cents take no nickels take 2 pennies, leaving us with 0 cents and
we're done

This algorithm as it turns out is only optimal so long as each denomination is
at least twice as much as the previous one. So what happens if we have an 18
cent coin? Let's make change for 37 cents. With the simple algorithm we end up
with {1 quarter, 1 dime, 2 pennies}. That's four coins. However you can do it
with three coins: {2 18 cent pieces, 1 penny}.

The algorithm for the case with arbitrary denominations isn't np-complete
(it's a fun algorithms question to figure out), but it's way too difficult to
be doing in your head all the time.

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eddington
I don't know why you're saying it isn't NP-complete. For an arbitrary set of
coins, it's definitely NP-complete. You could easily reduce 0-1 ILP to it, or
subset-sum.

If you don't believe me, believe <http://graal.ens-
lyon.fr/~abenoit/algo09/coins2.pdf>

"Optimally making change—representing a given value with the fewest coins from
a set of denominations—is in general NP-hard."

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__alexs
> Shallit assumed that every amount of change between 0 and 99 cents is
> equally likely.

Fail. He could at least have blagged some transaction data from a retailer
that actually does a bunch of small cash transactions. Just getting copies of
the receipt roles from his local bodega would be better than this.

~~~
jack-r-abbit
Why does it have to be small transactions? Every transaction, regardless of
size, has a chance to end in anything from 0 to 99. The more items on your
receipt the more chances to have different variations. Plus tax rate varies
from area to area. I don't think that his assumption is that far off. But yes,
he probably could have done something to get actual data.

~~~
bunderbunder
It should be small transactions because cash is being used less and less often
for medium-to-large transactions.

It's not safe to assume that all cent values from 0 to 99 are equally likely
because of the fact that cash transactions tend to be small. The number of
items being purchased will also tend to be small. So the fact that most prices
for individual items tend to end in 99, 95 or some other multiple of five
cents should have a huge impact on the distributions in real life, especially
for transactions to which sales tax does not apply.

For example in Oregon, a state with no sales tax, the fewest number of items
you'd be able to buy at $0.99 or $0.95 to get a $x.32 total is at least 20.
(That's a whole lot of "fun size" bags of potato chips.) The total would come
out to $19.32, at which point a lot of people would just pay with plastic,
anyway. Perhaps most people would.

~~~
jack-r-abbit
I suppose. I guess I use cash more than others. I would easily toss a $20 at a
$19.32 bill.

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btgeekboy
The assumption that every possible value from 0 to 99 is equally likely
doesn't seem like the best idea. In a country where the cost of items usually
ends in .99, .97, or the like, and the sales tax is added after the fact (and
usually from 0-10%), is it still the best choice? The post (and cited paper)
don't seem to mention this.

That said, I don't know where you'd acquire these sorts of statistics.

~~~
dfc
I also wondered about the random assumption. The thing that made me question
the assumption the most is Benford's law.

~~~
ShardPhoenix
I don't think Benford's law applies here: it's about the first digit of
numbers (that are distribute in a certain way), but in most cases the first
digit of a price will be denominating dollars, not cents.

~~~
dfc
The leading digit of change is tens of cents...

~~~
mooism2
The change may be more than a dollar, so tens of cents is not necessarily the
leading digit.

~~~
dfc
the amount of coin change is never more than a dollar.

~~~
mooism2
The amount of coin change is the remainder left over after dividing the total
change by a dollar. Thus although the total change may or may not obey
Benford's Law, the coin change certainly doesn't.

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xianshou
People can multiply 1, 5, 10, and 25 with zero difficulty. An average person
would likely resent being forced to count multiples of a less round number
such as 18. For that reason alone, such a change is unlikely to come about.

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mcmire
No. What the U.S. needs is NO coins.

(Seriously, does ANYONE use coins anymore? It's either bills or a credit card
for me.)

~~~
duncanbojangles
Yes, I still use coins. They're very handy for vending machines, parking
meters, highway tolls, stupid magic tricks, decision making, purchasing one
taco at taco bell, carnival food, making wishes, receiving change, and so on.
I don't think coins will go away as long as paper currency remains, unless we
start printing bills worth less than a whole dollar, or we become so rich that
we don't concern ourselves with values less than one dollar.

~~~
naudo
exactly. I've been in singapore the past couple of weeks and they have a S$ 1
coin and it's been awesome.

They also have <http://en.wikipedia.org/wiki/Polymer_banknote> which is great
in terms of durability.

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HairyMezican
Or, they could get rid of every coin smaller than a quarter. Assuming every
amount of change less than a dollar is equally likely, the average number of
coins needed to make change now becomes 1.5 - and it's not like anybody really
uses anything less than a quarter for anything other than coinstar anyways

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Zarathust
Whenever taxation changes it will shift the "average transaction value" by a
few cents and invalidate the added usefulness of already minted coins

