# Blending Colors

Three one-gallon buckets of red, blue, and yellow paint are each two-thirds full. Without the ability to measure, is it possible to equally mix all of the paint through a finite sequence of pours from one bucket to another?

If a solution exists, then the intermediate stages all consist of a full bucket, a two-thirds full bucket, and a one-third full bucket. Also, all pours, except for the first pour maybe the final pour, must be from a full bucket. This is because an empty bucket is never helpful.

Under these constraints, there are only two choices for any intermediate pour, so the problem is fairly restrictive in nature, but I'm having a hard time keeping track of all of the proportions.

Edit: Considering Ross's observation below, it seems if a solution exists, the final state must consist of two full buckets and one empty bucket.

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I don't understand the question. You have given us three buckets that are each two thirds full. Are we pouring these into different buckets or are we given some starting configuration and asked to arrive at the three buckets? –  John Douma May 4 '13 at 0:51
@user69810: The starting configuration is three buckets of red, blue, and yellow paint, each two-thirds full. There are no other buckets. Since we can't measure, any pour from one bucket to another can only stop when the bucket being poured into is full, so that we preserve thirds. Can we obtain a uniform mixture in all three buckets after finitely many pours? –  Jared May 4 '13 at 0:56
Thanks. I got it. We need equal amounts of each color in each bucket and each bucket must be two thirds full at the end. It's like an evil towers of Hanoi puzzle. –  John Douma May 4 '13 at 1:00
@user69810: It is not required (and in fact impossible) that each bucket in the end be two thirds full. We only require that there be equal amounts of each color in each bucket. –  Jared May 4 '13 at 1:01
If there is a solution it is clear that you can never empty a bucket because you would lose the ability to measure –  Dale M May 4 '13 at 8:50

It is not possible. Assume it were. Then just before the last pour, you would have two buckets with equal mixes and one with an unequal mix. This violates the fact that the total quantities of each of the three colors are equal. Also, pouring from a bucket that has an equal mix into one that has an unequal mix can never fix the receiving bucket.

Even with the final move allowed to join two partial buckets, it is not possible. We will focus on the powers of $3$ in the denominators of the amount of the first color poured, say red. The first pour moves $\frac 13$ gallon of red to another bucket. Then each subsequent pour (except the last) must be made from the bucket that received the previous pour. It will move either $\frac 13$ or $\frac 23$ of the red that is in the donor bucket to the receiver bucket. The $n^{\text{th}}$ pour will move an amount that has a denominator of $3^n$. This means that after $n$ pours, the donor and receiver buckets will have an amount of red that has a denominator of $3^n$ and the third bucket will have an amount with lower denominator. The greatest denominators keep growing. But the move before the last must leave us with a full bucket containing $\frac 13$ gallon of red, which we can never achieve.

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Why couldn't we have a situation in which a full bucket has equal mixing, and a two-thirds bucket and one-third bucket have unequal mixing? Then pouring the one-third bucket into the two-thirds bucket will create two equally mixed full buckets, and an empty bucket. The final pour is allowed to empty a bucket if desired. Sorry if that wasn't clear. –  Jared May 4 '13 at 15:08
@Jared: Good point. I still believe it impossible, but don't have a proof (yet). The most likely approach I can see is to make a vector for each bucket with the fraction of a gallon of each color within. Then prove that the maximum power of $\frac 13$ never decreases with a pour. A dark horse would be to find some quantity that is conserved on pouring that the final state violates, but I don't have any good ideas. –  Ross Millikan May 5 '13 at 3:49

Disclaimer: I just noticed that move $3$ requires a measurement, which is not allowed. So this answer does not completely answer the question. I will leave it here, as it may prompt other ideas.

I am going to assume that after each pour, the colours are sufficiently mixed so that they are then poured out in equal proportions.

Each value indicates the volume of colour in a bucket, not the percentage.

Initial Setup: $$\begin{array}{cccc} &R&Y&B\\ \text{Bucket 1}&2/3&0&0\\ \text{Bucket 2}&0&2/3&0\\ \text{Bucket 3}&0&0&2/3 \end{array}$$

Moves $1$ and $2$: pour half any bucket into each of the other two buckets (here I choose bucket $3$): $$\begin{array}{cccc} &R&Y&B\\ \text{Bucket 1}&2/3&0&1/3\\ \text{Bucket 2}&0&2/3&1/3\\ \text{Bucket 3}&0&0&0 \end{array}$$

Moves $3$ and $4$: pour half of each of the other two buckets (in this case $1$ and $2$) into the empty bucket (here $3$): $$\begin{array}{cccc} &R&Y&B\\ \text{Bucket 1}&1/3&0&1/6\\ \text{Bucket 2}&0&1/3&1/6\\ \text{Bucket 3}&1/3&1/3&1/3 \end{array}$$

Moves $5$ and $6$: pour the two half empty buckets into each other to make one full bucket (in this case $2$ into $1$): Initial: $$\begin{array}{cccc} &R&Y&B\\ \text{Bucket 1}&1/3&1/3&1/3\\ \text{Bucket 2}&0&0&0\\ \text{Bucket 3}&1/3&1/3&1/3 \end{array}$$

Conclusion: After $6$ moves, you can have two buckets equally mixed with one empty bucket. To obtain each with $2/3$ of paint, pour $1/3$ of each of the equally mixed buckets into the empty bucket, making $8$ moves in total.

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Your disclaimer is correct. In fact, any time after the first pour we cannot pour half of a bucket without making a measurement. –  Jared May 4 '13 at 22:46

Absolutely possible. Just keep pouring. After a sufficient number of pours (let's say infinite) composition/complexion of paint in each bucket would be the same.

P.S. Remember, nobody asked in the question how many pours are needed.

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The OP clearly stated that the number of pours was finite. –  Dale M May 7 '13 at 1:36
Oh, just saw. My bad. –  OC2PS May 7 '13 at 1:48