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You have bricks which are 20 cm tall and 10 cm wide (and 10 cm deep). You have to build a hedge which is 20 cm tall, and 1 m long. How many different patterns can you make keeping the minimum number of bricks in use? What is special in this number?

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Seems long enough to keep the puzzle open! So here is the solution:

First, the minimum number of bricks needed to make the fence is 10 (Minimum volume of Hedge /Volume of 1 brick = (100 * 20 * 10) / (10*10*20) = 10)

Once that is fixed lets try to find the number of patterns that can be made using these bricks.

post-4406-1203495477_thumbjpg

From the image you can see f(1) = 1 & f(2) = 2 ; where f(n) is the number of patterns for n number of bricks used.

Note: There are only 2 ways in which you can start to have a hedge 20 cm high, either by keeping one brick standing (case green) or 2 bricks stacked on top of another (case blue). So a case f(n) consists of only 2 ways to make, by putting 2 bricks stacked in front of all f(n-2) pattern, or by keeping 1 brick stacked upright in front of all f(n-1) patter.

There fore, f(n) = f(n-1) +f(n-2) - given f(1)=1 and f(2)=2

Thus f(3)=3; f(4)=5; f(5)=8 ... f(10)=89!

This series is, by the way, also called as Fibonacci Series.

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Interesting...

It feels like the Fibonacci Series but it will take too long to explain how.

So i'll stick to my intuition.

1 1 2 3 5 8 13 21 34 55 89

The 11th number in the sequence!

Yup .. the sequence is Fibonacci series but why did you choose 89 as ur answer?!

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Yup .. the sequence is Fibonacci series but why did you choose 89 as ur answer?!

Well I simply started with the second 1 in the series with the first brick (1 possible answer)

The 2 bricks got me 2 easy...

3 got me 5 a little more complicated

so for 10 bricks 11th number not 10th!

This is because length = 2X width...

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Now what if the bricks were 30x10x10??

How can we still use the same sequence? (if possible)

Of course we consider the wall to be 30x120 cm

Umm it would be an offshoot of another Fibonacci I think.

For 1 brick, no. of patterns = f(1) = 1

For 2 bricks, no. of patterns = f(2) = 1

for 3 bricks, no. of patterns = f(3) = 2

Now, note that there are only 2 ways to start any pattern. Either with 1 brick stacked upright, or 3 bricks stacked on top of each other.

Therefore, f(n) = f(n-1)+f(n-3)

Thus f(4) = 3

f(5) = 4

So the series becomes, 1, 1, 2, 3, 4, 6, 9, 13, 19, 28, 41, 60....

Hence, f(12) = 60

For a general case, 10ax10x10 brick, making a 10ax10bx10 wall, total no. of patterns, f(b) can be given as,

f(b)=f(b-1)+f(b-a)

The problem will become more interesting if I would say that you have to make a 40x150x10 wall using minimum number of 20x10x10 brick. What are the no. of patterns? What are the no. of patterns for general case of this?

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Well I simply started with the second 1 in the series with the first brick (1 possible answer)

The 2 bricks got me 2 easy...

3 got me 5 a little more complicated

so for 10 bricks 11th number not 10th!

This is because length = 2X width...

Roolstar .. Actually 3 should have got you 3 only! 4 bricks would be needed t make 5 patterns .. but I figure, you got the point..

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The problem will become more interesting if I would say that you have to make a 40x150x10 wall using minimum number of 20x10x10 brick. What are the no. of patterns? What are the no. of patterns for general case of this?

I tried doing it for 40x150x10 wall, though I could not figure out the exact function for patterns. Basically, it seems to me, it is a convoluted function of combination of fibonacci series and combinatorics!

All I could figure for 40x100x10 wall it will have f(10) = 96 patterns (and I made them, though I might have missed a few)! Any takers for f(15)??

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