```Date: Apr 16, 2013 4:21 AM
Author: Bob Hanlon
Subject: Re: Turning a Sequence into a List?

Just put list brackets around the sequence.minCoin[sol_List]:=Module[{m=Min[(q+d+n+p)/.sol]},Select[sol,((q+d+n+p)/.#)==m&][[1]]]v=Range[52,53];sol={Reduce[25q+10d+5n+p==#&&0<=p<5&&0<=n<2&&0<=d<3&&0<=q<4,{q,d,n,p},Integers]//ToRules}&/@v;minCoin/@sol//Column{q->2,d->0,n->0,p->2}{q->2,d->0,n->0,p->3}Or use Solve rather than Reducesol=Solve[25q+10d+5n+p==#&&0<=p<5&&0<=n<2&&0<=d<3&&0<=q<4,{q,d,n,p},Integers]&/@v;minCoin/@sol//Column{q->2,d->0,n->0,p->2}{q->2,d->0,n->0,p->3}Bob HanlonOn Tue, Apr 16, 2013 at 12:35 AM, Rob <rob@piovere.com> wrote:> Hello, I'm playing with a problem with minimum coins to make change.> Here's a problem spot where I look at ways to make up 52 and 53 cents> (later I'll use v =  Range[1,99].>> v=Range[52,53];> sol=(Reduce[25q+10d+5n+p==# &&0<=p<5 &&0<=n<2 &&0<=d<3> &&0<=q<4,{q,d,n,p},Integers])& /@v>> (* which gives {(q == 1 && d == 2 && n == 1 && p == 2) || (q == 2 && d> == 0 &&>      n == 0 && p == 2), (q == 1 && d == 2 && n == 1 &&>      p == 3) || (q == 2 && d == 0 && n == 0 && p == 3)} *)>> There are two cases for each coin and I need to pick the one with the> smallest number of coins. But the only way I know to begin that is with> a list of rules. I can get a Sequence for the first element of the List> sol but I can't figure how to get the Sequence to a list.>> x = (sol[[1]]) // ToRules> (* Sequence[{q -> 1, d -> 2, n -> 1, p -> 2}, {q -> 2, d -> 0, n -> 0,> p -> 2}] *)>> I've always had problems figuring out what a Sequence is and the HELP> doesn't really help me. Can someone please suggest how I can use this> sequence to select the {q,d,n,p} set that has the minimum coin count?> It's got me stumped. Thanks.>> -->>>
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