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Programming Enigma Puzzles

21 July 2014

Posted by on **From New Scientist #1354, 21st April 1983** [link]

In a recent frame at a snooker match the were no penalty points and after each of the 15 reds was potted a colour was potted (each of the six colours following two or three of the reds). The surprising thing about the result was that the winner’s total score was twice that of the loser, and yet they had both potted the same total number of balls.

What was the loser’s total and how many reds, how many yellows, how many greens, how many browns, how many blues, how many pinks and how many blacks did he pot?

(In snooker the potting of a red is followed by the potting of one of the other colours, the red remaining down but the other colour returning to the table. After 15 such events the remaining six colours are potted in the order stated above, reds are worth 1 point and the rest 2-7 in the order stated above).

[enigma208]

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This Python program runs in 238ms.

Solution:The loser scored 36 points, and potted 8 reds, 4 yellows, 4 greens, 2 browns, and none of the other colours.Along with the 8 reds the loser potted 3 yellows, 3 greens and 2 browns, and then in the final clearing of the colours also potted the yellow and the green.

It follows that the winner scored 72 points, potting 7 reds and in the final clearing of the colours the brown, blue, pink and black, meaning that the colours potted with the 7 reds made a total of 43 points. One way to achieve this would be 6 pinks and 1 black. But as Brian points out below the actual colours potted is determined by the program – it is 2 blues, 2 pinks and 3 blacks.

My results are:

loser: 36 points, 8 reds, colours (4, 4, 2, 0, 0, 0).

winner: 72 points, 7 reds, colours (0, 0, 1, 3, 3, 4).

I don’t think there is a free choice of how the winners score is made up since any colours that the loser doesn’t pot are potted by the winner.

You’re right of course. The solution tells you the way the winner made his 43 points from the colours potted along with the 7 reds – 2 blues, 2 pinks and 3 blacks.

Jim, it seems the winner potted only 14 balls before the ‘clearing’. So is your solution valid?

Hi Hugh,

I’m not quite sure what your objection is to the solution I gave.

The solution that my program finds is as follows:

The loser pots 8 of the reds, and after each red one colour to make 3 yellows, 3 greens and 2 browns in total.

The winner pots 7 of the reds, and after each red one colour to make 2 blues, 2 pinks and 3 blacks in total.

So, in the clearing of the reds each colour is potted 2 or 3 times, as required.

Then in the final clearing of the colours the loser pots the yellow and the green, and the winner pots the remaining brown, blue, pink and black.

Each player has potted 18 balls, and the winner scores 72 points, the loser scores 36, as required.

I’m not a snooker fan, but I think that there could be many matches that correspond to this, for example:

1. The winner pots 7 reds and 7 colours.

2. The loser pots 8 red and 8 colours, then starts to clear the colours and pots the yellow and the green.

3. The winner pots the remaining four colours.

I was misled by the statement “After 15 such events the remaining six colours are potted …”. Susan Denham’s explanation was not very clear for those who have never watched a game of snooker. 7 reds and 7 colours appears to make only 14 “events”; I now see she meant 15 reds between the two players.