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I came across the numberdrum problem in the Evening Standard, where the objective is to obtain a number in the centre using each of the numbers in the outer ring exactly once, along with the four basic arithmetic operations.

Is there a way to cast this problem mathematically? I was thinking about an optimisation problem, but here we need to optimise over the operators in a way, which leads to a clumsy optimisation. Is there a neater method to solve this problem and find if its infeasible or not? Or find the closest value to the central number you can reach by using basic arithmetic operations on the outer numbers?


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Aren't there eight numbers in the outer ring? Or is the solver supposed to choose a subset of either size 5 or size 7 from the outer ring, to combine and obtain the center number? – coffeemath Feb 28 '13 at 22:36
@coffeemath: Sorry for the gaffe. No subset selection; all numbers in the outer ring must be used once. – Bravo Feb 28 '13 at 22:39
Can you combine arbitrary numbers or must they be adjacent? Say, can you start with $367-171$? – Hagen von Eitzen Feb 28 '13 at 22:44
@HagenvonEitzen: Any combination will do; but need to use a number just once. And only +/-/*/÷... – Bravo Feb 28 '13 at 22:48
Well, it's a finite problem ... :) – Hagen von Eitzen Feb 28 '13 at 22:52

Going around from the 155: $$155+7-2-367+15+42+171+56=77.$$ A program to check only sums of $\pm a$ would be only size $2^8=256.$

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That's true here, but in general how do you solve for $N$ numbers in the outer ring? Gerry says it's NP-complete and that was what I wanted to know :) – Bravo Mar 1 '13 at 10:25
Actually I was surprised no multiplication or division was required for this particular puzzle. I agree it's not very efficient for general $n$ ring size to check $2^n$ cases, so exponential in the ring size. And the size of the check would be even larger if multiplication/division are thrown in, since then also permutations of the ring numbers would have to be involved. And if parentheses are allowed it gets worse. – coffeemath Mar 1 '13 at 11:48

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