Just for fun: Countdown numbers game solver

dg.google.groups at thesamovar.net dg.google.groups at thesamovar.net
Wed Jan 23 21:54:20 CET 2008

Well I tried the NumPy array thing that I was talking about, to
parallelise the problem, and there were some difficulties with it.
Firstly, the pruning makes a really big difference to the speed, and
you don't get that if you're trying to parallelise the problem because
what is an equivalent calculation for one set of numbers is obviously
not for another set. I think this problem disappears when you consider
very large sets of numbers though (where the cost of doing equivalent
computations vanishes in comparison to the alternative cost of
starting up the whole recursive computation from scratch many times).
The second problem is that you can't weed out division by zero and
intermediate fractions. I haven't looked at the internals of how NumPy
deals with these though, so this might be fixable or it might not.

For my part, I'd consider the following equivalences to be right for
defining equivalent expressions (where here a, b and c are any
subexpression, and 1 and 0 means any subexpression that evaluates to 1
and 0).


a*b <-> b*a
a+b <-> b+a


(a+b)+c <-> a+(b+c)
(a+b)-c <-> a+(b-c)
(a-b)+c <-> a-(b-c)
(a-b)-c <-> a-(b+c)
(a*b)*c <-> a*(b*c)
(a*b)/c <-> a*(b/c)
(a/b)*c <-> a/(b/c)
(a/b)/c <-> a/(b*c)

Units (1 is multiplicative unit, 0 is additive unit):

a*1 <-> a
a/1 <-> a
a+0 <-> a
a-0 <-> a

Substitution (swapping equal starting numbers is equivalent):

expr(a,b,c,...,z) <-> expr(s(a,b,c,...,z)) where a,b,c,...,z are the
original numbers given and s is a permutation of (a,b,c...,z) so that
(a,b,c,...z) evaluates to the same thing as s(a,b,c,...,z)

or equivalently, expr1 <-> expr2 if str(expr1)==str(expr2)

Then, any two expressions which can be transformed into one another by
the equivalences above are equivalent. Commutativity and units can be
easily implemented as you go and most of the programs suggested so far
do this, by for example only allowing a*b or a+b if a>=b, and not
allowing a*1 or 1*a, etc. Substitution can be implemented by just
taking set(map(str,expressions)) at the end. The associativity ones
are a little more tricky to implement, but Arnaud's idea of the fully
ordered binary tree seems to do it. Another way of saying it is that
any subexpression consisting only of + and - operations should be
reduced to a+b+c+...-z-y-x-.... where a>b>c>... and z>y>x>... (and
similarly for an expression involving only * and /).

Terry, I'd also be interested in a copy of your stack simulation code,

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