The Critical Line: Volume 6

In the spirit of the recent Rio Olympics, Jevon Fulbrook and Chris Ebbs have an Olympic-themed puzzle for you; the sixth installment of the Critical Line.

The first step is to arrange these five words in the Olympic rings. Each word goes inside one larger coloured ring, with one letter in each of the smaller circles. All words can be placed with the letters going clockwise. Where the rings intersect, the same letters must be used for both words.

  • Accomplish
  • Asymmetric
  • Haemolysis
  • Plicamycin
  • Sculptural


When you’ve worked out how the letters are arranged in the circles, the letters in the crossover sections (8 in total) can be used to complete the following puzzle.

Taking the 8 letters, research how many countries competing at this year’s Olympics begin with that letter. Once you have those numbers, write them in roman numerals. Each letter is represented by the number of letters used in the roman numeral. For example, there are 13 competing countries beginning with the letter A; this is XIII in roman numerals and uses 4 characters, hence A=4. Once you have converted each of the letters to a number, complete the Olympic rings below with the 8 numbers used so the sum of the numbers within each ring is the same.


Send your answer to for your chance to win a $50 Dymocks voucher!

Critical Line Volume 5 – Solution
By Dan Mayoh (

Congratulations to Andrew Parker and Oliver Chambers for both sending in correct solutions. The prize is awarded to Andrew, congratulations!

The answers to the Mastermind challenges in Critical Line Volume 5 are as follows:

Probabilities of winning:
Part 1: 14/1296
Part 2: 106/1296
Part 3: 132/1296

Optimal strategies:
Part 1: Choose any code that contains at least 3 different digits (e.g. 1123, 5465, 1652 etc). Any of these will give you a 14 in 1296 chance of winning the game within your first two guesses. 1080 of the 1296 possible initial guesses fit this criterion (720 of which contain exactly 3 different digits, and 360 of which contain 4 different digits).

Part 2: There are many combinations of guesses that are equally optimal. The first of these in numerical order is the guesses 1123 and 2453. Another optimal combination is 1234 and 1355. In all there are 17,280 equally optimal combinations. All of these involve a code with exactly 3 distinct digits and a code with 4 distinct digits, where the repeated digit in one code is not present in the other code. There are other criteria to the pairs of codes too, but that gives you a flavour.

Part 3: Here, the optimal strategy is to first pick any code with exactly 3 distinct digits (e.g. 1123), which will then give one of 14 possible scores of black and white pegs. For each of these 14 outcomes, there is one or more optimal second guesses. The table below presents the lowest-number optimal second guess for each of the scoring outcomes, given a first guess of 1123. The format of the score is given as “B-W” where B is the number of black pegs, and W is the number of white pegs.

First Score

Second Guess






























Any guess


What I find interesting about this is that some of the second guesses are codes that have already been excluded as being correct, yet making that guess is still the way to maximise your chances. For example, if the score is 3-0 (3 blacks and 0 whites after guessing 1123, the code cannot possibly be 2413 (or else the guess 1123 would have scored 1 black 2 white), yet 2413 is an optimal second guess.

Method for solving:
Parts 2 and 3 are solved by brute force searching, with some shortcuts to eliminate combinations that are symmetrical to combinations we’ve already considered. Doing this on a computer first requires writing a formula or block of code to determine the peg score given a specified code and specified guess. That task alone is an interesting little Excel challenge.

Part 1 can be solved by brute force as well, but can also be solved without a computer. The approach is described here. Please excuse the lazy formatting.

Noting that the peg outcome 3 Black 1 White is always impossible, then for any nominated first guess you will get one of at most 14 different peg-outcomes depending on the code, with frequencies F1, F2, F3, …, F14 that add up to 1296 (since there are 1296 equally likely codes).
Given peg-outcome j, of which there is an Fj / 1296 chance of achieving, you have a 1 / Fj chance of your second guess being correct.

The overall probability of your second guess being correct given your nominated first guess is then:
Sigma(j=1 to 14) [ Fj/1296 * 1/Fj ] for all non-zero Fj’s.
That simplifies down to Sigma(j=1 to 14) [ 1/1296 ] without even needing to find the Fj’s, provided the Fj’s are non-zero.
This sums to 14 / 1296 provided our first guess has at least 3 distinct digits. (A first guess with only 1 or 2 distinct digits will mean that some of the 14 peg-outcomes can never be achieved, giving us less than 14 values of Fj that are non-zero.)

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