## Monday, March 9, 2009

### Two more measuring problems

See the previous two measuring problems

You have two lengths of rope. Each one takes exactly one hour to burn. However, they are not necessarily identical in length or size. In fact, neither of them is uniform. That is, after half an hour of burning, there might not be exactly half of the original length remaining. Therefore, you cannot simply wait for the fire to reach the half-way point and assume that exactly 30 minutes have passed.

Use these two ropes to measure 45 minutes.

And now for the challenge problem! It's an old one that I had written, and now it's yours.

You have three unmarked, asymmetrical containers, and no others. The first holds 6 cups, the second holds 10 cups, and the third holds 15 cups. The 10-cup pitcher is filled with lemonade, but it's too strong. You need to dilute it to half of its current concentration. You have an unlimited source and sink for water, but please don't throw out any of the lemonade! Remember, all the lemonade must be diluted to half-concentration.

Solutions have been posted, or you can just read the comments below.

Anonymous said...

Hmmm…for the rope one I think I've got it. Start both ropes burning, the first rope just on one end, and the second rope burning from both ends. The second rope will burn completely in exactly 30 minutes (since it is effectively burning twice as fast), at the end of which you will have exactly 30 minutes of rope left on the first rope. Now you can burn the first rope from both ends and it will take exactly 15 minutes.

Baumann Eduard said...

Fill 6 from 10
fill the rest of 10 (4) in 15
fill 10 with water
from 10 to 15
fill 10 with water again

Now you have 10 lemonade
and 20 water in your system.
This is the wanted concentration.

You have only to mix this.
It is possible with infinitly doing the following:
from 6 to 15
from 10 to 6
from 15 to 10

The proces is converging to concentration 0.5 for all pitches.

miller said...

Half-concentration means there are equal amounts of water and lemonade. So you'll have 20 cups total in the end.

It is possible to do it in fewer than infinite moves.

Baumann Eduard said...

Second try for the right concentration.

Fill 6 from 10
fill the rest of 10 (4) in 15
fill 10 with water

Now you have 10 lemonade
and 10 water in your system.
This is the wanted concentration.

You have only to mix this.
It is possible with infinitly doing the following:
from 10 to 15
from 6 to 10
from 15 to 6

The proces is converging to concentration 1:1 for all pitches.
Ten times is practically sufficient.

There may be other solutions.

Eduard said...

My friend Constantino has found a beautifull 11 step solution.

......6..............10............15
H2O + Jus....H2O + Jus....H2O + Jus

... 0 + 0.........0 + 10.........0 + 0
... 0 + 0.........0 + 10.......15 + 0
... 6 + 0.........0 + 10.........9 + 0
... 6 + 0.........0 + 10........0 + 0
... 0 + 0.........0 + 10........6 + 0
... 0 + 6.........0 + 4.........6 + 0
... 0 + 0.........0 + 4.........6 + 6
drink
... 0 + 0........0 + 4.........0 + 0
... 0 + 0........0 + 0.........0 + 4
... 0 + 0......10 + 0.........0 + 4
... 6 + 0........4 + 0.........0 + 4
... 0 + 0........4 + 0.........0 + 4
... 0 + 0........4 + 4.........0 + 0
drink

miller said...

Clever, drinking the lemonade to make more container-space. I love it.

There exists a solution in which all lemonade is prepared before anyone takes a drink.

Eduard said...

The ultimate answer with 8 steps of my friend Costantino:

......6..............10............15
H2O + Jus....H2O + Jus....H2O + Jus

... 0 + 0.........0 + 10.........0 + 0
... 0 + 6.........0 + 4.......... 0 + 0
... 0 + 0.........0 + 4...........0 + 6
... 6 + 0.........0 + 4...........0 + 6
... 0 + 0.........0 + 4...........6 + 6
... 0 + 4.........0 + 0...........6 + 6
... 1 + 5.........0 + 0...........5 + 5
this is the crucial step
... 0 + 0.........1 + 5...........5 + 5
... 0 + 0.........5 + 5...........5 + 5

Remark:
the big recipent can vary between 12 and 19 !

miller said...

Wonderful! You've improved upon my previous solution, which used 9 moves.

amit said...

C = container, L = lemonade, W = water

15C 10C 6C
start 0 10 L 0
step1 0 4L 6L
step2 6L 4L 0
step3 6l 4l 6W
step4 6l6W 0 4l
step5 10(L+W) 0 6(l+W)
step6 10(l+w) 6(l+w) 0
step7 10(l+w) 10(l+w)