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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
[xfut -> lugnet.off-topic.geek]
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Message has 6 Replies:
 | | Re: How to maximize 90-piece bags?
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| (...) Sounds like they've set themselves up to give a pretty good bargain since a best case (for TLC) would be only 22 bags, so being able to get 26 bags is getting an extra 18.2% (or a 15% discount). Of course they may budget the thing for 25 bags (...) (24 years ago, 8-Nov-00, to lugnet.off-topic.geek)
| | | Re: How to maximize 90-piece bags?
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| (...) 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 (24 years ago, 8-Nov-00, to lugnet.off-topic.geek)
| | | Re: Lego Mosaic Customer Service call
|
| Jake, I just wanted to take a moment to say two things to you: 1. Congratulations on your new position with Lego Direct. :-) 2. I hope that we continue to hear from you regarding our questions in such a timely fashion. It is nice to be hearing (...) (24 years ago, 9-Nov-00, to lugnet.lego.direct)
| | | Re: How to maximize 90-piece bags?
|
| (...) Actually, subtract ($9.99 + $2.95 shipping = $12.94) the X-large baseplate, and (7.99 + $2.95 shipping = $10.94) the 100 pack of black slopes and a ($1.69 + $2.95 shipping) brick separator from the ($29.99 + $4.95 shipping = $39.94) Mosaic (...) (22 years ago, 1-Sep-02, to lugnet.off-topic.geek)
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Message is in Reply To:
| | Re: So did you wonder about me?
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| (...) Really! Cool. I was a bit worried about the pliers comment in the faq. (...) 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 (...) (24 years ago, 7-Nov-00, to lugnet.lego.direct)
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