Comments on: On ordering appetizers

By: Peter Makholm

Peter Makholm — Wed, 11 Jul 2007 07:06:58 +0000

Not just a Python version but a quite different solution. I don’t think it’is you data structure but an much better algorithm that makes the difference. Reimplementing it in perl with a dumb list gives a comparable speed:


#!/usr/bin/perl

$, = " ";

@p = (580, 420, 355, 335, 275, 215);
$target = 1505;

my @solution = ([0]);
while (($val,@data) = @{ shift @solution }) {
        next if $val > $target;
        do { print @data; last } if $val == $target;

        push @solution, [$val + $_, @data, $_] for @p;
}

For harder problems and for a general solution which is to be run again and again speed will be an issue. But for solving a specific problem I couldn’t care less if it takes a minute or two to run - it’s all about getting things done.

But my original code is crude, brute and ignorant and probally didn’t deserved to be made public.

By: Matthias Urlichs

Matthias Urlichs — Tue, 10 Jul 2007 18:01:41 +0000

Sorry about the indent lossage.

indent everything below while: once, everything below for: twice, and the print+break twice.

By: Matthias Urlichs

Matthias Urlichs — Tue, 10 Jul 2007 17:59:20 +0000

Well, here’s a Python version. It happens to be a whole lot faster than yours.
Using appropriate data structures helps, I would say. ;-)

It prefers the shortest solution because the pqueue module is a FIFO for equal amounts (aka priority values).

#!/usr/bin/python
import pqueue

p = (580, 420, 355, 335, 275, 215)
target = 1505

q = pqueue.PQueue()
q.insert(0,(0,)*len(p))

current = -1
while True:
val,data = q.pop()
if val == target:
print [a for a in zip(data,p) if a[0]]
break
if current == val: continue
current = val
data = list(data)
for f in range(len(p)):
val += p[f]
data[f] += 1
q.insert(val,tuple(data))
data[f] -= 1
val -= p[f]