# Red, White, Black Beans

talzamir
Posers and Puzzles 30 Oct '11 07:00
1. talzamir
Art, not a Toil
30 Oct '11 07:00
It is fairly well known that if you have two sacks of beans, red and white, with the same number of beans in each, and take a cupful of red beans from the red to the white, mix perfectly, and then move as many beans from the mix of mainly white and some red into the red sack, the number of red beans in one sack equals the number of white beans in the other.

What if there are three sacks and three kinds of beans?
From red to white, and mix
From mainly white to black, and mix
From mainly black to red

Are the proportions of different kinds of beans the same in each sack, though of course with colors changed?
2. 30 Oct '11 22:03
No,

In the white sack there are red and white beans.
In the black sack there are red, white and black beans.
Similarly with the red sack there are beans of every colour.
3. talzamir
Art, not a Toil
30 Oct '11 23:13
Indeed so..

so.. if someone moves a bucketful from black to red, and you can only move to black to red to white to black.. is the only way to get the proportions to be the same in a finite time to put all the beans together and then dividing by three?
4. 31 Oct '11 12:07
Originally posted by talzamir
so.. if someone moves a bucketful from black to red, and you can only move to black to red to white to black.. is the only way to get the proportions to be the same in a finite time to put all the beans together and then dividing by three?
Even that is only a guarantee if you can mix perfectly. Contrast the case of two bags, where it's a guarantee no matter how badly you mix, as long as the two amounts are the same.

Richard
5. talzamir
Art, not a Toil
31 Oct '11 13:49
Of course. Entirely possible to have two sacks with a million beans in each, take a bucketful of beans from white bean sack to black bean sack, mix, and take a bucketful back - and by a stroke of luck not only is the number of beans in both bucketfuls the same even though beans are of different sizes and take more or less space depending on how they arrange themselves - but that all the white beans are in the original sack again and no black bean ever moves between sacks. Just extremely unlikely. For all I know, the solution on various problems can be affected by precision. It is also true that very few things are equal. And who knows, maybe the gravity of the moon messes up some problems too.

I would hope that this of all places - a puzzle section inside a virtual chess club - is an environment for people who simply enjoy the mental challenge of puzzles. Apparently not. Sadly, some of the other ways take the fun out of this for me.
6. 31 Oct '11 16:561 edit
Maybe if you use two cups and simulataneously transfer a cup to each of the two other sacs?
7. 31 Oct '11 18:18
Nope, that doesn't work, seems tricky to do for 3
8. 31 Oct '11 19:04
Originally posted by talzamir

I would hope that this of all places - a puzzle section inside a virtual chess club - is an environment for people who simply enjoy the mental challenge of puzzles. Apparently not. Sadly, some of the other ways take the fun out of this for me.
Don't give up hope, I enjoy them!
9. 01 Nov '11 08:14
Same. I read more than I post, and I'm sure there are plenty others.
10. 01 Nov '11 09:01
Originally posted by talzamir
Indeed so..

so.. if someone moves a bucketful from black to red, and you can only move to black to red to white to black.. is the only way to get the proportions to be the same in a finite time to put all the beans together and then dividing by three?
I haven't done the math but so long as there is a probability you could end up with a particular arrangement, then my guess is that yes it is entirely possible you could end up with the same proportions. The probability will be minuscule but it is possible.

Consider the following example:

Container#1 contains 15 White, Container#2 contains 15 Red & Container#3 contains 15 Black

move 1: move 10W from C#1 to C#2
C#1 contains 5W, C#2 contains 10W & 15R, C#3 contains 15B

move 2: move 10R from C#2 to C#3
C#1 contains 5W, C#2 contains 10W & 5R, C#3 contains 10R & 15B

move 3: move 10B from C#3 to C#1
C#1 contains 5W & 10B, Bag#2 contains 10W & 5R, C#3 contains 10R & 5B

move 4: move 5W & 5B from C#1 to C#2
C#1 contains 5B, C#2 contains 15W & 5R & 5B, C#3 contains 10R & 5B

move 5: move 10W from C#2 to C#3
C#1 contains 5B, C#2 contains 5W & 5R & 5B, C#3 contains 10W & 10R & 5B

move 6: move 5W & 5R from C#3 to C#1
C#1 contains 5W & 5R & 5B, C#2 contains 5W & 5R & 5B, C#3 contains 5W & 5R & 5B

Solved. Absolutely minuscule probability and I'm guessing we need to use a lottery probability calculator for this. So this now comes down to defining a "finite time". If that was defined as while the sun was still burning, no problem. If however, this must be completed before the beans rot, then it is highly unlikely.

Yet another interesting problem.

Thanks!
11. talzamir
Art, not a Toil
01 Nov '11 09:291 edit
Glad you like. ðŸ™‚ In a concise format,

initial orders: 15W + 15R + 15B
After move 1: 05W + 10W15R + 15B
After move 2: 05W + 10W05R + 10W15B
After move 3: 05W10B + 10W05R + 10W05R
..
After move 6: 05W05B05R + 05W05B05R + 05W05B05R

Actually move 3 suffices for what I first had in mind but apparently was careless in putting in words, as after three moves each sack has beans in ratios 0:1:2, in order of least to most. The second round makes it far prettier though, with 1:1:1 in each, which is a more stringent way to interpret what I was saying - or the only solution if the ratio is in order W:R:B rather than least:middle:most.

However.. if one requires that the sack is perfectly mixed each time.. (yes, I know that in the real world it is not doable) it would not be possible to move 10 white beans. from a sack that has beans of many colors. The above, taking 2/3 of the contents each time.. I'll use 81 beans of each kind:

Initial orders: 81W + 81R + 81B
After move 1: 27W + 54W81R + 81B
After move 2: 27W + 18W27R + 36W54R81B
After move 3: 51W36R54B + 18W27R + 36W54R81B

..and that's a dilemma. It's possible to make it so that each sack has 81 beans again, but the ratios of least : middle : most may or may not be possible to make the same without making one or more of the sacks empty. So.. is it possible to get identical mixtures that are NOT 1:1:1 , such as 1W10R70B + 1R10B70W + 1B10W70R, by mixing fully between transfers?

With two sacks it is of course trivially easy. Move any number of beans from the first sack to the second, and the same number back (yes I know in real world that can't be done either), and the ratios in less : more are always the same. But with three sacks it is not that clear.
12. 01 Nov '11 09:44
Regarding "fully mixed"......when you pull a number of beans from the bag, are you assuming you are removing beans in the same ratio that exists within that bag?
13. talzamir
Art, not a Toil
01 Nov '11 10:17
Precisely. If I take 20% of the contents of a sack with 60 black, 30 white and 10 red, then I would get exactly 12 black, 6 white, and 2 red. Perhaps it would be more accurate to be mixing fluids than beans, but whether those mix perfectly either..
14. 03 Nov '11 13:10
Originally posted by talzamir
I would hope that this of all places - a puzzle section inside a virtual chess club - is an environment for people who simply enjoy the mental challenge of puzzles. Apparently not. Sadly, some of the other ways take the fun out of this for me.
I, by contrast, find it fascinating that for two colours, you can be a sloppy as you like and the exact correspondence is guaranteed, while for any higher number - even for three - the solution can only be asymptotically correct.

Richard
15. talzamir
Art, not a Toil
03 Nov '11 14:43
I agree on that, the case of two sacks is beautiful. Of course, it requires that you move exactly as much from B to A as you move from A to B. One bean more or less and the symmetry is ruined.