This month, Jevon takes us a step away from the enigma theme from previous columns, and towards a more spatial /word puzzle.

To solve the puzzle you must place the tiles into the grid below in order to form words reading both vertically and horizontally, as given in the example:

**Sample:**

**Sample solution:**

**Puzzle:**

**For your chance to win a $50 book voucher, send your solutions to ActuariesMag@actuaries.asn.au**

**Solution to Critical Line 20 – by Dan Mayoh**

The snooker themed problem was a popular puzzle, with eight people submitting answers that were all in agreement with me. They were (in reverse date of submission):

Ean Chan, Kelvin Duong, Preetham Arvind, Bruce Dawkins, Grant Lian, Corey Plover, Andrew Parker, Min Chen.

The winner of the book voucher is Grant Lian. **Congratulations to all of you.**

**Solution:**

So it turns out the number of ways to make a break of exactly 100 is a lot. In the tens of billions in fact. To be precise, the answer we all arrived at is 93,194,837,248. The break score that has the most ways of achieving it is 88, with 133,835,290,090 ways being the consensus. The graph of partitions (P) vs score (n) looks quite interesting when plotted on a log scale. The approximately straight slope on the left hand side of the log scale graph means that the ratio of P(n)/P(n-1) is almost constant (up until P = about 60).

Most solutions were arrived at using some kind of recursive algorithm, hence making the problem of finding the number of ways for all x from x = 1 to 100 no harder than finding the number of ways for only x = 100. And I’m pretty sure everyone used a computer. I did it a slightly different way, based around brute force counting (with the help of some efficient VBA code) all the partitions of scoring P points from N coloured balls only, for P in the range of 2n to 7n and n = 0 to 15. From there I adjusted the partition results as necessary to take into account the reds between each colour, and the run out of the final six colours. The workings that came out of this gave rise to some interesting graphs too.

Now the observant among you may notice my choice of language so far has refrained from saying ‘the answer **is** 93,194,837,248’. Here’s why. Several people who answered gave a table showing the number of partitions for all break values from 1 to 147, including a count of 74,095,894,242 ways to make a break of 105, which agreed with my count. Having confidence in this figure, I then had a ‘Oh. Oops’ moment watching Frame 16 of the 2018 world championship second round match between Ronnie O’Sullivan and Ali Carter, when Ronnie made a break of 105 in a way that **was not** on my 74-billion-plus-change list. As it turns out, I forgot to consider the situation where a player, when shooting at a red ball, pots more than one red ball from a single shot. This is perfectly legal, and scores the player 1 point per red, at which point they then shoot at a coloured ball, and then back to normal. When defining the problem I forgot to make specific mention of this situation and whether to include or exclude it. Kelvin made a brief reference to it, saying his answer excluded these cases, but no one else (myself included) thought to consider it.

Ronnie schooled me on that one. Ronnie O’Sullivan 1, Actuaries 0.

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