## The Coin in Three Fountains

```Draw three intersecting circles and place a coin, heads up, in each
of the seven enclosed regions.  Two operations are allowed: a FLIP
turns over all four of the coins within a single circle, whereas a
CLEAR turns all four coins in a single circle heads up.  Is it
possible to reach a condition in which the coin in the central
region (contained within all three of the circles) is tails while

The answer is no.  In fact, if, at any stage, all the coins in at
least one circle are heads, it's impossible to proceed to the desired
final configuration by means of FLIPS and CLEARS.  This not only
covers a wide range of initial conditions, it also implies that for
arbitrary initial configurations no sequence of operations containing
a CLEAR can yield the desired final configuration.

PROOF:  Assume that at some stage (possibly the initial config-
uration) all the coins in at least one circle are heads. Therefore,
any sequence of FLIPS and CLEARS can be assumed to contain at least
one CLEAR.  Let A denote the circle that is the subject of the last
CLEAR in the sequence, and let B and C denote the other two circles.
Every operation following the last CLEAR is a FLIP of either A, B,
or C.  Let nA, nB, and nC denote the number of these FLIPS respect-
ively.  Since the coin in the triple-region was turned heads up by
the last CLEAR, the sum (nA+nB+nC) must be odd for the triple-region
to end up tails.  The double-regions of AB and AC were also turned
heads up by the last CLEAR, so the sums (nA+nB) and (nA+nC) must
both be even to leave these regions heads up.  It follows that nA,
nB, and nC must each be odd.

However, the oddness of nA implies that the single-region of A, which
was turned heads up by the last CLEAR, ends up tails.  Thus, the
desired final configuration cannot be reached.
```