## Teaser 2500

**From The Sunday Times, 22nd August 2010** [link]

A well-known puzzle asks:

“If among 12 snooker balls one is a different weight, how can the rogue ball be identified – together with deciding whether it is heavier or lighter – in three weighings on a balance?”

Recently I faced a very similar problem of finding a rogue ball among a consignment of 39 identical-looking balls – and deciding if it was heavier or lighter. I had at my disposal a two-pan balance.

How many weighings did I need?

[teaser2500]

## Jim Randell 9:50 am

on24 March 2020 Permalink |We dealt with a similar puzzle in

Enigma 1589(also set by Peter Harrison). For that puzzle I wrote a program that searches for a minimal decision tree.For an analytical solution the paper

Weighing Coins: Divide and Conquer to Detect a Counterfeitby Mario Martelli and Gerald Gannon [ link ] provides an elegant analysis to determine the maximum number of coins that can be distinguished using a certain number of weighings for four related problems.And this gives us the solution to this puzzle:

Solution:4 weighings can be used to find the rogue object in a set of 39.LikeLike

## Jim Randell 9:44 pm

on24 March 2020 Permalink |For

Enigma 1589I wrote the following code to generate the decision trees for the four types of problem described in the linked paper.It can be used to generate decision trees for either puzzle.

Run:[ @repl.it ]LikeLike