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# Pouring water V.

17 replies to this topic

### #11 hoopshank

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Posted 23 September 2007 - 01:34 AM

Scenario 1: 4 and 5 litre bowls....

Fill the 5[-litre bowl], pour 4 litres into the 4[-litre bowl]; 1 litre remains in the 5.
Empty the 4 and put the 1 litre into the 4.
Refill the 5, pour into the 4 until full. 2 litres remain in the 5.

Scenario 2: 4 and 3 litre bowls....

Fill the 4, pour 3 litres into the 3; 1 litre remains in the 4.
Empty the 3, pour the 1 litre into the 3.
Refill the 4, pour into the 3 until full. 2 litres remain in the 4.

So exactly the same method, basically, but if I abstracted it with variables for smaller and larger bowls, it would probably be less clear.

Somewhat fewer than 1000 words....
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### #12 jhedges3

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Posted 13 December 2007 - 09:51 PM

The question is poorly worded. Among other things, it does not make clear that you only have each of the two differently sized bowls.
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### #13 joaopsr

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Posted 29 January 2008 - 01:56 PM

Er, most buckets are tapered, aren't they?, and so this would not work at all. Or have I misunderstood your suggestion?
You haven't misunderstood; you're just under the assumption that skbrown's solution won't work with a tapered bucket. It will, as long as it's symmetrical.

This is not true! If it were a cylinder it would work. If it's tapered, it will not. Consider the extreme example of a bucket tapered to a point - you get a cone. How do you know when to stop pouring? Besides, we have BOWLS not buckets, which suggests an entirely different shape to me. Symmetrical or not, this method does not work with standard bowl or bucket shapes.

ok... maths says he's right... but i sure liked the answer...
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### #14 zach

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Posted 15 April 2008 - 06:24 AM

You can solve both sets of buckets from both small-to-big and big-to-small. The green water is your solution, and the blue water is the alternate solution. My first solutions were the green solution for the 5/4 set and the blue solution for the 4/3 set. In both cases, your solution is faster, but I thought it interesting to note that you can do it either way.

really nice solution by fosley
i got the solution but i appreciate more the alternate ones.
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### #15 gnassy

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Posted 02 May 2008 - 10:22 PM

fill the 5 liter up and pour it in the 3 liter. The over flow is 2 liters.
There are many ways to solve this one.

---------------------------------------------------------------------------------------------------------------------------------------------- jus stop when the 3 liter bowl filled
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### #16 firepumpguy

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Posted 04 May 2008 - 02:17 AM

4 & 5 capacity, resolve to 2

A=4 size
B=5 size

A - B
is
0 - 0

Fill B(+5)
0 - 5

Pour B(-4) in to A(+4)
4 - 1

Dump A(-4)
0 - 1

Pour B(-1) in to A(+1)
1 - 0

Fill B(+5)
1 - 5

Pour B(-3) in to A(+3)
4 - 2

Edited by firepumpguy, 04 May 2008 - 02:23 AM.

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### #17 firepumpguy

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Posted 04 May 2008 - 02:21 AM

3 & 4 capacity, resolve to 2

A=3 size
B=4 size

A - B
is
0 - 0

Fill B(+4)
0 - 4

Pour B(-3) in to A(+3)
3 - 1

Dump A(-3)
0 - 1

Pour B(-1) in to A(+1)
1 - 0

Fill B(+4)
1 - 4

Pour B(-2) in to A(+2)
3 - 2

Very interesting that the steps with sizes 3 & 4 are the same as those of sizes 4 & 5.

Edited by firepumpguy, 04 May 2008 - 02:22 AM.

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### #18 bwmeekin

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Posted 03 July 2008 - 07:33 PM

1. fill the 4, put in the 5, fill four put as much as u can in five leaveing 3litres in the 4 then empty the 5 and put it in the 3 litres from the 4 in the 5 then fill the 4 and pour into 5 leaving 2 in the 4

2.fill the 3 pour into 4 fill 3 pour into 4 leaving 2 in the 3
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