The Critical Line – Volume 24

Jevon serves up a new type of wordy puzzle for you to unscramble while Dan sifts through 11 responses to find our Volume 23 winner!

Your task is to fit all the given words into the grid. Each word starts in the square with the corresponding number, the second letter being in the given direction. The remaining letters in the words may follow in any direction, left, right, up or down, one letter per square. The black cell is blank. The yellow cells unscrambled will deliver the final solution.

Note that letters in the grid may be shared by multiple entries, though no letter may be reused within a single entry.

For your chance to win a $50 book voucher, send your solutions to

Critical Line Volume 23 Answer – by Dan Mayoh (

11 people sent in responses to this problem, of which nine got the problem essentially correct.  A few different methods were used, which is what I like to see.

Congratulations to Kelvin Duong, Jackie Lok, Benjamin Howe, Mike Fowlds, Hemant Rupani, Andrew Parker, Weixuan Li, Edward Peyton and Tim Yip!  (And well done to the five of you who also sent in a correct answer to the second part of the question.)

However only four of those nine gave the answer in exact fraction form as requested, and only two of those four gave the answer expressed to the relevant fraction of a second as requested, as opposed to fraction of a minute.  And so, by toss of a coin, the winner of the book voucher is Benjamin Howe! (Bad luck Andrew Parker who lost the coin toss.)

The answer to part A is 12:04:02 and 174/1953 seconds (which equates to 4.034818228 turns rolling dice).
The answer to part B is 12:06:32 and 199708/364819 seconds (which equates to 6.542456944 turns rolling dice).

For a discussion on solving the problem (well, the equivalent problem using apple trees that I adapted to using dice) using Markov chains, see

For a discussion on solving the problem using (i) a tidy approach that can be done entirely by hand, and (ii) a recursive approach that can also be done by hand, see

The above ‘tidy’ approach can also be found at solution 16 here, which is where the problem originally came from:

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