Filed under: Frivolity, Games, Mathematics, Programming, python | Tags: Countdown, countdown numbers game, countdown numbers game solver

One of the things about being ill is that you have to spend a lot of time in bed with nothing much to do. Having watched the whole first series of the Sopranos, I had to find something else. So here’s the result. I revisited an old program I wrote many years ago to solve the Countdown numbers game.

In this game, you’re given six numbers between 1 and 100 and a target number between 100 and 999. You’re given 30 seconds to try to make the target using the six numbers and the operations plus, minus, times and divide.

I originally wrote a program to solve this many years ago (when I was about 14 I think), but the algorithm I used was pretty horrible. I worked out by hand all the possible arrangements of brackets you could have for six numbers, and then tried each operator and number in each appropriate slot. It worked, but it was ugly programming.

Recently I’ve been learning Python for an academic project, and so I thought I may as well try rewriting it in Python. I think the solution I’ve come up with is nicer than any of the solutions I’ve found on the internet (mostly written in Java or C), although having written it I found this paper which uses a very similar solution to mine (but in Haskell rather than Python).

Python programmers might get something from the minimal code below (all comments and docs stripped out), or you can take a look at the full source code here, including detailed comments and docs explaining the code and algorithm.

My ideal (as always with Python) was to write a program you could just look at and understand the source code without comments, but I don’t think I achieved that. I’d be interested if a more experienced Python programmer could do so. Let me know.

This version is incomplete, from the slower version, and is supposed to be understandable without explanations (takes about 40 seconds to find all solutions, too slow for Countdown):

def ValidExpressions(sources,operators=standard_operators,minimal_remaining_sources=0): for value, i in zip(sources,range(len(sources))): yield TerminalExpression(value=value, remaining_sources=sources[:i]+sources[i+1:]) if len(sources)>=2+minimal_remaining_sources: for lhs in ValidExpressions(sources,operators,minimal_remaining_sources+1): for rhs in ValidExpressions(lhs.remaining_sources, operators, minimal_remaining_sources): for f in operators: try: yield BranchedExpression(operator=f, lhs=lhs, rhs=rhs, remaining_sources=rhs.remaining_sources) except InvalidExpressionError: pass def TargetExpressions(target,sources,operators=standard_operators): for expression in ValidExpressions(sources,operators): if expression.value==target: yield expression

This version is actually complete, from the faster version which needs the comments to explain (takes about 15 seconds to run, good enough to win Countdown):

sub = lambda x,y: x-y def add(x,y): if x<=y: return x+y raise ValueError def mul(x,y): if x=2+minremsources: for e1, rs1, v1 in expressions(sources,ops,minremsources+1): for e2, rs2, v2 in expressions(rs1,ops,minremsources): for o in ops: try: yield ([o,e1,e2],rs2,o(v1,v2)) except ValueError: pass def findfirsttarget(target,sources,ops=standard_ops): for e,s,v in expressions(sources,ops): if v==target: return e return []