Subject:
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Re: How to maximize 90-piece bags?
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Newsgroups:
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lugnet.off-topic.geek
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Date:
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Wed, 8 Nov 2000 00:28:23 GMT
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Viewed:
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1658 times
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In lugnet.lego.direct, Todd Lehman writes:
> In lugnet.lego.direct, Larry Pieniazek writes:
> > > - Some number of 90 count bag(s) of 1x1 Plates - White
> > > - Some number of 90 count bag(s) of 1x1 Plates - Black
> > > - Some number of 90 count bag(s) of 1x1 Plates - Grey
> > > - Some number of 90 count bag(s) of 1x1 Plates - Dark Grey
> > > - Some number of 90 count bag(s) of 1x1 Plates - Light Grey
> >
> > Ho ho!!!
> >
> > Clearly we need a design that uses 92 (or 182) of each thing so that they
> > have to supply 88 extra of each color but one. (91 wouldn't work, they
> > always give one extra.)
>
> Hmm. I bet they're *really* 90-count bags and the extras are for if you go
> over 88, not 90. That is, I'll speculate that if you need 87 or 88, you get
> one bag. But if you need 89, 90, 91, 92, or any number up to 176, you get
> two bags. (Just a guess.)
>
> Anyway, ya, this is an interesting partitioning problem. The goal would be
> to get as many "free" elements as possible. I wonder if a straightforward
> greedy 1/5 partition is optimal for this?
>
> Let's see...first, there are 44 x 44 = 1936 squares to cover. Note that 88
> (which is 90 minus 2 -- a coincidence?) evenly divides 1936 (22 x 88 = 1936).
> So it should be easy to "tip the scales" toward extra bags.
>
> A first-order crude approximation on the theoretical upper bound of an optimal
> solution is 1936 + 90 x 5, or 2386. Of course, it's likely that no solution
> exists which yields this high quantity, but you can probably get pretty close
> to it.
>
> Well, here's one solution -- I haven't proven that it's optimal, but I
> conjecture that it is optimal given certain assumptions. Assuming the
> threshold for each bag is 88 and you get actual 90-ct bags, then
>
> 353 Black => 5 bags of 90 Black
> 353 DGray => 5 bags of 90 DGray
> 353 MGray => 5 bags of 90 MGray
> 353 LGray => 5 bags of 90 LGray
> 353 White => 5 bags of 90 White
>
> would result in 2250 elements. (353 is 88 x 4 + 1.)
>
> Now, since 353 x 5 = 1765 is much less than 1936, it makes sense, not knowing
> the precise partitioning rules, to "pad" the required quantities. Thus
>
> 387 Black => 5 bags of 90 Black
> 387 DGray => 5 bags of 90 DGray
> 387 MGray => 5 bags of 90 MGray
> 387 LGray => 5 bags of 90 LGray
> 388 White => 5 bags of 90 White
>
> adds up perfectly to requiring precisely 1936 elements, yet receiving 2250
> elements.
>
> But, if we assume that we know the precise partitioning rules, are there
> better (less greedy) ways to "spend" the "padding"? Look how much is wasted:
> (387 - 353) x 5 is 170 -- almost two whole bags!
>
> 353 Black => 5 bags of 90 Black
> 353 DGray => 5 bags of 90 DGray
> 353 MGray => 5 bags of 90 MGray
> 353 LGray => 5 bags of 90 LGray
> 524 White => 6 bags of 90 White
>
> adds up to requiring 1936 elements, yet receiving now 2340 elements! But is
> this the best we can do? No, we're still wasting padding here, because it
> only takes 441 elements to result in 6 bags, not the full 524 we have left
> over. Unfortunately, the excess 83 (524 - 441) isn't enough to push another
> bag up to the next integral quantity.
>
> Therefore, it makes more sense to distribute the 83 evenly across all colors
> as follows:
>
> 369 Black => 5 bags of 90 Black
> 369 DGray => 5 bags of 90 DGray
> 369 MGray => 5 bags of 90 MGray
> 369 LGray => 5 bags of 90 LGray
> 460 White => 6 bags of 90 White
>
> I believe this is an optimal solution (proof left as an exercise to the
> reader) with sufficient padding to ensure a total of 26 bags of 90 elements,
> for a total of 2340 elements, whether the threshold is 88->90 or 89->91 or
> 90->92. This gives approximately 10% padding on each color in the worst case
> scenario of 90->92.
>
> Now the question is, what would I do with several thousand 1x1 plates?! :-)
>
> And if I subtract the cost of a 48x48 gray baseplate (normal US$10??) and
> sloped bricks around the boder from the US$29.99 price of the mosaic, is it
> worth $.007 (that's 0.7 cents) per 1x1 plate? Seems like an excellent deal
> if I need 1x1 plates! :-s
>
> --Todd
FWIW, my first order resulted in:
1 Black => 1 bag of 90 Black
1 DGray => 1 bag of 90 DGray
1 MGray => 1 bag of 90 MGray
1932 LGray => 22 bags of 90 LGray
1 White => 1 bag of 90 White
What did everyone else order?
-Jon
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Message has 2 Replies:
 | | Re: How to maximize 90-piece bags?
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| (...) <snip> (...) I want to order exactly what you ordered if it is indeed the *new* gray, and they (TLC) don't tweek your image. Rich -- Have Fun! C-Ya! Legoman34 ***** Legoman34 (Richard W. Schamus)... (My views do not necessarily express the (...) (24 years ago, 8-Nov-00, to lugnet.off-topic.geek)
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Message is in Reply To:
| | How to maximize 90-piece bags?
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| (...) Hmm. I bet they're *really* 90-count bags and the extras are for if you go over 88, not 90. That is, I'll speculate that if you need 87 or 88, you get one bag. But if you need 89, 90, 91, 92, or any number up to 176, you get two bags. (Just a (...) (24 years ago, 7-Nov-00, to lugnet.lego.direct, lugnet.off-topic.geek)
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