There are three buckets, each of which contains an integral amount of liquid.
The goal is to empty one of the buckets! The rules are:
A: The capacity of each bucket in unlimited.
B. The liquid in a given bucket may be increased only doubling its
volume with liquid from another bucket.
C. The liquid in a given bucket may be decreased only by some of it into
another bucket.
Obviously one bucket may be emptied when,
1. Two buckets have the same volume of liquid, or
2. One bucket contains half of the total liquid.
If this is to be a viable problem, the initial amount of liquid and its
allocation among the three buckets should not permit an immediate solution.
There are certain interim positions which usually precede a solution. These are
1. A = 2^M; B = 2^N. where A and B are the volumes of liquid in any two buckets
M & N are integers and M <= N. If M < N, the liquid from C may be added to A to
make A = B
2. A + B = 2^M + 2^N, but A <> M(N) & B <> N(M). By alternately pouring liquid from
B to A and then from A into B, it may be possible to achieve position 1. EXAMPLE:
A B
8 26 M = 1, N = 5
16 18
32 2 QED The roles of A & B have been reversed but this does not invalidate
the result. To complete example, we must consider all possible initial positions.
1 33
2 32
3 31
6 28
12 22
24 10
14 20
28 6 Failure
4 30
32 2 QED
5 29
10 24 Failure
7 27
14 20 Failure
8 26
16 18
32 2 QED
9 25
18 16 QED
11 23
22 12
10 24 Failure
13 21
26 8 QED
15 19
30 4
2 32 QED
16 18
32 2 QED
A "seed" is any value for A that is not divisible by 2, Why do some seeds lead to a
successful partitioning into A & B, while others do not? Before we address this
problem, let's consider a third advantageous position.
3. A + B = |2^N - 2^M| ! If A & B can be partitioned into 2^M and 2^N, then
2^M can be taken from bucket C with the result that A + B = 2^N
7 27
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