Re: sum of coins article in mathematica journal
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- Subject: [mg128415] Re: sum of coins article in mathematica journal
- From: daniel.lichtblau0 at gmail.com
- Date: Sun, 14 Oct 2012 00:15:22 -0400 (EDT)
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On Saturday, October 13, 2012 12:08:01 AM UTC-5, von dremel wrote:
> a long time ago I read a short text in mathematica journal.
>
> A neat little program that calculated the numer of coins needed to reach a certain sum. you put in the coin values e.g. 25 cent 50 cent etc and got the minimum anount of how many coins you need.
>
> I think Allan Hayes wrote it.
>
> does anyone have a pointer to it please?
>
>
> Peter W
You might try looking in library.wolfram.com. I believe (some? many?) back issues of TMJ are archived there.
If I'm not mistaken, the method involves looking at a coefficient of 1/((1-x[1]*t)*(1-x[5]^t^5)*(1-x[10]*t^10)*(1-x[25]*t^25)*(1-x[50]*t^50)) expanded at the origin. Specifically, to make j cents in total change, look at the jth coefficient. Find the summand that uses smallest total powers of x[k]s.
Here is a bit of code for this.
minimalChange[coins_, amount_] := Module[
{len = Length[coins], mu, mus, x, ratfun, coeff, totals, bestpos},
mus = Map[mu, coins];
ratfun = Times @@ (1/Table[1 - mus[[j]]*x^coins[[j]], {j, len}]);
coeff = (List @@
Expand[SeriesCoefficient[ratfun, {x, 0, amount}]]);
coeff = coeff /. Times -> Plus;
totals = Map[# /. mu[_]^j_. :> j &, coeff];
bestpos = Ordering[totals, 1];
coeff[[bestpos]] /. Plus -> List /. mu[k_]^j_. :> {k, j}
]
Example:
coins = {1, 5, 10, 25, 50};
In[339]:= minimalChange[coins, 247]
Out[339]= {{{1, 2}, {10, 2}, {25, 1}, {50, 4}}}
So 4 50 cent pieces, 1 quarter, 2 dimes, two pennies. No surprises here.
I think this is a bit slow. There may be a more efficient way to get at the=
relevant series term.
I should allso note that the "obvious" greedy method does not work on all p=
ossible coin sets.
In[341]:= minimalChange[{4, 7, 11, 23}, 247]
Out[341]= {{{7, 1}, {11, 3}, {23, 9}}}
Also one can have a tie.
In[353]:= minimalChange[{5, 7, 11, 26}, 377]
Out[353]= {{{7, 4}, {11, 1}, {26, 13}}}
Could instead have had {5,2}, {7,1}, {11,2}, {26,13} for that.
Daniel Lichtblau
Wolfram Research