Enigma 1460

Posted on: Tuesday 11/27, 2007;  8:19 PM

This problem is designed to make brute force attack difficult because the set of candidate solutions starts off with elements, i.e. it is the Cartesian product of 2 sets each with 9! elements. However, a methodical approach can be used to prune this set down until there is only one case left, i.e. the solution. The main trick is to prune each of the 9! element sets as much as possible before forming the Cartesian product of the residual sets.

Create a list of digits 1,...,9 and a list of all permutations of this.

Create a list of the 9 colours and a list of all permutations of this.

Without loss of generality, assume that "grey" is at circular position 1, and select all digit permutations where each circular position 2, ... , 9 (i.e. not including the "grey" position) has an odd and even digit on its neighbours.

Select the colour permutations where "hazel", "indigo", "jade" and "khaki" are in consecutive clockwise circular positions. There is no need to look for wrapped-round cases because "grey" is locked in circular position 1.

Build a Cartesian product list of all remaining digit permutations and colour permutations.

Select the cases where the "hazel", "indigo" and "jade" digits add up to give the "khaki" digit.

Select the colour permutations where "lemon", "mauve" and "navy" are in consecutive clockwise circular positions. There is no need to look for wrapped-round cases because "grey" is locked in circular position 1.

Select the cases where the "lemon" and "mauve" digits add up to give 2 times the "navy" digit.

Select the cases where the "hazel" digit is the same as the number of times you can find a digit equal to the sum of the digits on its neighbours. This leaves just one possibility which is the required solution.

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