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POFind equivalent results under distributive and commutative laws
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copied!<p>I have a script that loops over 5 vectors of numbers and mix them with all the four operations searching for a combination that give a certain target number as a result.</p> <p>The script prints an output like:</p> <pre><code>312 / 130 x 350 - 122 + 282 = 1000.0 312 / 130 x 350 + 282 - 122 = 1000.0 312 - 282 x 372 / 15 + 256 = 1000.0 142 + 350 - 372 x 125 / 15 = 1000.0 142 + 350 - 372 / 15 x 125 = 1000.0 350 / 130 x 312 + 282 - 122 = 1000.0 350 + 142 - 372 x 125 / 15 = 1000.0 </code></pre> <p>Each line is formatted from a list of the numbers and a list of the operations.</p> <p>What I'd like to do, is to <strong>remove the equivalent results</strong>, i.e. having an output like:</p> <pre><code>312 / 130 x 350 - 122 + 282 = 1000.0 312 - 282 x 372 / 15 + 256 = 1000.0 142 + 350 - 372 x 125 / 15 = 1000.0 </code></pre> <p>As a solution, at first, I thought about "remembering" the numbers that already gave 1000 and skip those, but then I realized it might shadow new results, so I don't know what to do.</p> <p>How can I find the equivalent results under distributive and commutative laws?</p> <p><strong>Note:</strong> In the presented output the parentheses are NOT shown, but the order is reduce-like, meaning that, for example:</p> <pre><code>142 + 350 - 372 x 125 / 15 = 1000.0 </code></pre> <p>is calculated like:</p> <pre><code>(((142 + 350) - 372) x 125) / 15 = 1000.0 </code></pre> <hr> <p>This is the code I have so far:</p> <pre><code>import operator from itertools import permutations, product, count from functools import reduce vectors = [[87, 125, 209, 312], [29, 122, 254, 372], [15, 130, 277, 369], [142, 197, 282, 383], [64, 157, 256, 350]] OPER = {operator.add: '+', operator.sub: '-', operator.mul: 'x', operator.truediv: '/'} def format_result(nums, ops, res): s = ' '.join('{} {}'.format(n,OPER[op]) for n,op in zip(nums, ops)) s += ' {} = {}'.format(nums[-1], res) return s def calc(vectors, test=lambda x: x == 1000.): for vv in permutations(vectors): for indexes in product((0,1,2,3), repeat=5): numbers = tuple(v[i] for i,v in zip(indexes, vv)) for operations in permutations(OPER): res = reduce(lambda x,y,n=count(0): operations[next(n)](x,y), numbers) if test(res): print(format_result(numbers, operations, res)) calc(vectors) </code></pre>

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