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PostPosted: Wed Jan 13, 2010 11:29 pm 
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Posts: 44
Can someone tell me how to explain my DD questions like this?

All bridges must have 2 section A's (in the start and end) and nowhere else. Length of individual sections is as follows:

Section A - 8.19m, B - 5.21m, C=2.91m, D=3.64m, E=7.52m, F=2.40m

Find the best combination of sections that will make the following bridges.

a) 44.18m
b) 65m
c) 94m

I know the following:

1) 2 section A's = 8.19 + 8.19 = 16.38
2) For (a) then the length required becomes 44.18 - 16.38 = 27.80

Now how to get the other sections add upto 27.8?


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PostPosted: Thu Jan 14, 2010 7:55 am 
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I think I'd suggest mine skipped it!


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PostPosted: Thu Jan 14, 2010 10:43 am 
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Joined: Mon Jun 18, 2007 2:32 pm
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Location: East Kent
could you repeat the question?

Quote:
All bridges must have 2 section A's and nowhere else



:?: :?:


possibly best way to get it is estimation, you need something to add up to approx 28

either that or trial and error


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PostPosted: Thu Jan 14, 2010 11:36 am 
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Location: Watford, Herts
This is an example of what's known in the computing literature as a Knapsack Problem. It's one of a large class of problems (the NP-complete problems, also including the Travelling Salesman Problem) that are known to be equally hard, but no-one knows whether there is an efficient way to solve them. That question is one of the Millenium Prize Problems. Most people think the answer is no, but no-one has proved it.

In short, I think BarnetDad's advice is sound.

But then the one thing I can't resist is temptation, so, getting a computer to try all the possibilities:
a) 5D + 4F (unique solution)
b) 3B + C + 4E (shortest solution)
c) 2B + 4D + 7E or 6C + 8E (shortest solutions)
(It's typical of NP-complete problems that it's easy to check solutions, but the number of potential solutions to check is very large.)


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PostPosted: Thu Jan 14, 2010 4:38 pm 
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Joined: Mon Jun 30, 2008 3:08 pm
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This kind of question should be kept until your DD finishes all others. Here is what I think from what you have done for a)

If you take a look at 27.80 m, you will see that you require the last digit as 0.

Now look at each section

B = 5.21m - you need 10B to make the last digit = 0 ... No (52.10m)
C = 2.91m - you need 10C to make the last digit = 0 ... No (29.10m)
D = 3.64m - you need 5D to make the last digit = 0 ... Yes - 3.64x5 = 18.20m
E = 7.52m - you need 5E to make the last digit = 0 ... No (37.6m)
F = 2.40m - you always have the last digit = 0 ... Yes

So 27.80 - 18.20 = 9.60m
Then 2.40m x 4 = 9.60m

Answer 5D + 4F

For b) and c) you have to start from the biggest section, E.


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PostPosted: Thu Jan 14, 2010 5:32 pm 
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Crazydad wrote:
If you take a look at 27.80 m, you will see that you require the last digit as 0.

It's a lot harder than that. You've assumed that two numbers ending in 0 were added to get the final 0, but it could have been one ending in 6 and another ending in 4, etc. And it could have been adding three, four or five numbers.


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PostPosted: Thu Jan 14, 2010 6:37 pm 
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I have a heuristic method that meant I got at least some of the answers without trying all possibilities...

Here's how it works.. Assume 2A

Take all the other lengths and sort in descending length... giving order.. 7.52, 5.21, 3.64, 2.91, 2.4

Now choose the most of the biggest length that fits the remaining gap and then successively allocate the remainder using the next size down, when there is a remainder reduce the larger size and use a multiple of a smaller size...

I'm not sure such an approach works but I have just deduced that
44.18 is 2x8.19 + 5x3.64 and 4x2.4 without enumerating all the possibilities....

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PostPosted: Thu Jan 14, 2010 6:39 pm 
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and...

65 = 2x8.19 + 4x7.52 + 3x5.21 and 2.91

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PostPosted: Thu Jan 14, 2010 6:42 pm 
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94 = 2x8.19 + 7x7.52 + 2x5.21+4x3.64

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PostPosted: Thu Jan 14, 2010 6:46 pm 
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The moral of this story is that when packing your rucksac... start with the biggest things first. :lol:

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