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Steven Traver

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All the King's Wine

You may cover this contingency when you write "unfortunately not as strong as the binary solution", but your solution won't identify the poisoned bottle in all cases. To illustrate, reduce the number of bottles to nine, and the number of servants to three (the square root of the number of bottles, like in the original problem). Arrange the bottles in a 3x3x3 grid, and let's say the poisoned bottle is in the middle (x-axis) front (z-axis) of the middle (y-axis) layer. Let's call the servants S, K and M. On day 1 of the first week, the servants each drink from all nine bottles in their assigned z-axis layer, S from the back layer, K from the middle layer, and M from the front. So M is poisoned and will die first. Midweek, they each drink from all nine bottles in their x-axis, S from the left, K from the middle, M from the right. So K is poisoned and will die second. At the end of the week, each drinks from their z-axis, S from the top, K from the middle, M from the bottom. K is poisoned again, S is unscathed. When M and K die, it can only be said that the poison is in the middle (x-axis) front (z-axis) of either the middle or bottom layer along the y-axis.

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