2 Glasses Puzzle

M K Townsend

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Apr 29, 1998, 3:00:00 AM4/29/98

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Does anyone know the difinitive 'solution' to the 2 glasses puzzle?

A glass of water and a glass of wine.

A spoonful of water is taken and put into the wine and mixed up.

A spoonful of the mixture is put into the glass of water.

Is there more water in the wine; more wine in the water or are they equal??

Jonathan Dushoff

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Apr 29, 1998, 3:00:00 AM4/29/98

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M K Townsend (michael....@bradford.gov.uk) wrote:
: Does anyone know the difinitive 'solution' to the 2 glasses puzzle?

Since the total amount in both glasses is unchanged, all the water that
left the water glass must be exactly replaced by wine, and vice versa.
The amounts are equal.

Jonathan

Joel Duncan

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Apr 29, 1998, 3:00:00 AM4/29/98

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M K Townsend <michael....@bradford.gov.uk> wrote in article
<01bd736e$2f92f980$6e016206@F_MICHAELT>...

| Does anyone know the difinitive 'solution' to the 2 glasses puzzle?
|
| A glass of water and a glass of wine.
|
| A spoonful of water is taken and put into the wine and mixed up.
|
| A spoonful of the mixture is put into the glass of water.
|
| Is there more water in the wine; more wine in the water or are they
equal??
|

equal

note:
x=y=# of volume units in each full glass
x is for water
y is for wine

1/x=1/y=1 spoonful of liquid
in which case each glass begins with x/x or y/y or 1 full glass

observe the contents of each glass in each step

(water glass) (wine glass)
BEGIN:
x/x y/y

PLACE ONE SPOONFUL WATER INTO WINE GLASS

(x-1)/x y/y + 1/x OR (xy+y)/xy

PLACE ONE SPOONFUL WINE/WATER MIX INTO WATER GLASS

water glass wine glass

(x-1)/x + 1/((xy+y)/xy) ((xy+y)/xy) - 1/((xy+y)/xy)
OR
(x-1)/x + xy/(xy+y) (xy+y)/xy - xy/(xy+y)
OR
((x^2)y+xy-xy-y+(x^2)y)/((x^2)y+xy)
((x^2)(y^2)+2x(y^2)+(y^2)-(x^2)(y^2))/((x^2)(y^2)+x(y^2))

OR

2(x^2)y - y 2x(y^2) + (y^2)
----------------- --------------------------
(x^2)y + xy (x^2)(y^2) + x(y^2)

If these concentrations are equal, then these fractions will be equal.
cross-multiply--

2(x^4)(y^3) + 2(x^3)(y^3) - (x^2)(y^3) - x(y^3) =
2(x^3)(y^3) + (x^2)(y^3) + 2(x^2)(y^3) + x(y^3)

the two [2(x^3)(y^3)] terms cancel leaving

2(x^4)(y^3) - (x^2)(y^3) - x(y^3) =
3(x^2)(y^3) + x(y^3)

factor out an xy from each term and divide by xy on both sides

2(x^3)(y^2) - x(y^2) - (y^2) = 3x(y^2) + (y^2)

subtract to set equal to zero

2(x^3)(y^2) - 4x(y^2) - 2(y^2) = 0

divide by (y^2)

2(x^3) - 4x -2 = 0

divide by 2

(x^3) - 2x -1 = 0

subract 1 from each side

(x^3) - 2x = 1

the only solution for x is 1

showing the concentrations to be equal

Isaac Kuo

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Apr 29, 1998, 3:00:00 AM4/29/98

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In article <6i7eni$2fd$1...@news1.sinica.edu.tw>,

Jonathan Dushoff <dus...@gate.sinica.edu.tw> wrote:
>M K Townsend (michael....@bradford.gov.uk) wrote:

>: Does anyone know the difinitive 'solution' to the 2 glasses puzzle?

>: A glass of water and a glass of wine.

>: A spoonful of water is taken and put into the wine and mixed up.

>: A spoonful of the mixture is put into the glass of water.

>: Is there more water in the wine; more wine in the water or are they equal??

>Since the total amount in both glasses is unchanged, all the water that

>left the water glass must be exactly replaced by wine, and vice versa.
>The amounts are equal.

I'm not sure this is true. For example, 1 gallon of pure water
mixed with 1 gallon of pure alcohol will take up _less_ than
2 gallons. It's something about how the molecules pack together
better.

If the problem used a glass of water and a glass of alcohol, I
suspect the answer is that there is more alcohol in the water
than water in the alcohol.

For example, assume both glasses have 100 spoonfuls.

After the first step, the water glass has 99 spoonfuls of water
while the other has 1 spoonful of water and 100 spoonfuls of
alcohol. However, the second glass has _less_ than 101 spoonfuls
of mixture.

The second step takes 1 spoonful of mixture from the second
glass, so it takes away slightly more than 1/101 of the total
amount of water in the mixture. That leaves slightly less
than 1-1/101, or 100/101 spoonfuls of water in the alcohol
glass.

The second step also took away slightly more than 1/101 of
the total amount of alcohol in the mixture. That means it
has slightly more than 100/101 spoonfuls of alcohol in it.

Thus, the water glass has slightly more than 100/101 spoonfuls
of alcohol in it, while the alcohol glass has slightly less
than 100/101 spoonfuls of water in it. There is more alcohol
in the water than water in the alcohol.

I don't know if this will also be the case with wine.
--
_____ Isaac Kuo k...@bit.csc.lsu.edu http://www.csc.lsu.edu/~kuo
__|_)o(_|__
/___________\ "Mari-san... Yokatta...
\=\)-----(/=/ ...Yokatta go-buji de..." - Karigari Hiroshi

Jonathan Dushoff

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Apr 30, 1998, 3:00:00 AM4/30/98

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Isaac Kuo (k...@bit.csc.lsu.edu) wrote:
: In article <6i7eni$2fd$1...@news1.sinica.edu.tw>,

: Jonathan Dushoff <dus...@gate.sinica.edu.tw> wrote:
: >M K Townsend (michael....@bradford.gov.uk) wrote:
: >: Does anyone know the difinitive 'solution' to the 2 glasses puzzle?

: >Since the total amount in both glasses is unchanged, all the water that

: >left the water glass must be exactly replaced by wine, and vice versa.
: >The amounts are equal.

: I'm not sure this is true. For example, 1 gallon of pure water
: mixed with 1 gallon of pure alcohol will take up _less_ than
: 2 gallons. It's something about how the molecules pack together
: better.

: ...

: I don't know if this will also be the case with wine.

I don't think so. Since wine is already ~90% water, the alcohol
molecules are probably already 'packed' in solution.

I like your anwer though. A nice new way of looking at an old problem.

Jonathan

Busted

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Apr 30, 1998, 3:00:00 AM4/30/98

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I know a simpler solution to this puzzle. Instead of taking a spoonful of
water and put it in wine and then a spoonful back. Dump the whole glass of
water in the glass of wine and then pour the same amount back. There's your
answer. Of course you'll need big enough glasses to do that. None of that
algebrac crap.

M K Townsend <michael....@bradford.gov.uk> wrote in article
<01bd736e$2f92f980$6e016206@F_MICHAELT>...

> Does anyone know the difinitive 'solution' to the 2 glasses puzzle?
>

Randall M! Gee

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Apr 30, 1998, 3:00:00 AM4/30/98

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In article <6i8li5$8o7$1...@news1.sinica.edu.tw>, dus...@gate.sinica.edu.tw

(Jonathan Dushoff) wrote:
> I don't think so. Since wine is already ~90% water, the alcohol
> molecules are probably already 'packed' in solution.

It works with wine too. Unlike oil and water, alcohol and water are
quite miscable. Any amount of water will mix with any amount of
alcohol and the result will take up less volume than the two did
seperately.

-- Randall M! Gee, Keeper of Gummi Wisdom
(g...@math.berkeley.edu)

Paul Lord

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Apr 30, 1998, 3:00:00 AM4/30/98

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> Does anyone know the difinitive 'solution' to the 2 glasses puzzle?
>
> A glass of water and a glass of wine.
>
> A spoonful of water is taken and put into the wine and mixed up.
>
> A spoonful of the mixture is put into the glass of water.
>
> Is there more water in the wine; more wine in the water or are they
equal??
>

Below is a solution for idiots like myself to understand!

[SPOILER]

Assumption: Glasses contain == 95ml of the stated liquid.

Starting condition.
WINE GLASS == 95ml Wine, 0ml Water.
WATER GLASS == 95ml Water, 0ml Wine.

Step 1:
Transfer 1 x 5ml spoonfull of the water to the wine glass.
Result of step 1:
WINE GLASS == 95ml Wine, 5ml water (MIXED EQUALLY).
WATER GLASS == 90ml water.

Step 2:
Transfer 1 x 5ml spoonfull of the mixture in the wine glass to the water
glass.
The "mixture" in the 5ml spoon will be 5% Water(0.25ml), 95% Wine(4.75ml).
Result of step 2.
WINE GLASS == 90.25ml Wine, 4.75ml Water.
WATER GLASS == 90.25ml Water, 4.75ml Wine.

Conclusion:
There is an equal amount of water in the wine glass as there is wine in the
water glass.

Hope this has helped.
Paul.

"In order to define recursion, we must first define recursion."

Jon Haugsand

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Apr 30, 1998, 3:00:00 AM4/30/98

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* Paul Lord

The point is that the conclusion is still valid even without a lot of
your assumptions. Assume you start with equal amount of wine and water
(separated) and the you take some amount of wine into the water you
mix it but not necessarily thoroughly. Then you take about the same
amount of the water/wine mixture into the wine. But, hups, you took
too much, so you take some of the wine (with a little water) into the
water, so you mix back and forth until you are sure that you have the
same volume in both glasses. Now, prove that you have the same amount
of wine in the water as water in the wine.

Proof: assume there are X ml of wine in the water. However, you have
the same volume of water as before the mixing so you have removed X ml
of water. Where is it? In the wine of course. QED.

Ob puzzle: if you have one litre wine and one litre water and you take
half the water into the wine, mix it, and then half a litre of
wine/water mixture back into the water, and then repeat this procedure
until you have exactly 1/2 litre water in the wine. How many times do
you have to repeat this process?

--
Jon Haugsand
Dept. of Informatics, Univ. of Oslo, Norway, mailto:jon...@ifi.uio.no
Just say NO to frames: http://www.ifi.uio.no/~jonhaug/say-no-to-frames.html

Paul Lord

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Apr 30, 1998, 3:00:00 AM4/30/98

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**************************************************
The point of my posting was to allow idiots like myself to understand the
solution.
As far as "a lot of your assumptions" go, I believe I only made one, which
was to define the volume of liquid in each glass we started with.

There is no need to go and get all technical on me!

Paul

Systems analyst to programmer: "You start coding. I'll go find out what they
want."
**************************************************

Chris Cole

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Apr 30, 1998, 3:00:00 AM4/30/98

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M K Townsend wrote:

> Does anyone know the difinitive 'solution' to the 2 glasses puzzle?
>
> A glass of water and a glass of wine.
>
> A spoonful of water is taken and put into the wine and mixed up.
>
> A spoonful of the mixture is put into the glass of water.
>
> Is there more water in the wine; more wine in the water or are they equal??

Welcome to rec.puzzles. You may be interested in this
entry from the rec.puzzles archive:

==> logic/mixing.p <==
Start with a half cup of tea and a half cup of coffee. Take one tablespoon
of the tea and mix it in with the coffee. Take one tablespoon of this mixture
and mix it back in with the tea. Which of the two cups contains more of its
original contents?

==> logic/mixing.s <==
Mixing Liquids

The two cups end up with the same volume of liquid they started with. The same
amount of tea was moved to the coffee cup as coffee to the teacup. Therefore
each cup contains the same amount of its original contents.
********************************************
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Abigail

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May 1, 1998, 3:00:00 AM5/1/98

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Paul Lord (PL...@DioneCorp.com) wrote on MDCCIII September MCMXCIII in
<URL: news:6i9rru$k64$1...@news.u-net.com>:
++
++
++ The point of my posting was to allow idiots like myself to understand the
++ solution.
++ As far as "a lot of your assumptions" go, I believe I only made one, which
++ was to define the volume of liquid in each glass we started with.

Uhm, no. You also made the assumption that the mixture was mixed
equally well.

You wrote:
>>> Result of step 1:
>>> WINE GLASS == 95ml Wine, 5ml water (MIXED EQUALLY).

>>> The "mixture" in the 5ml spoon will be 5% Water(0.25ml), 95% Wine(4.75ml).

Even without that assumption, there will as much wine in the water,
as there will be water in the wine.

Abigail