Listen With Others

Are you sitting comfortably? Then we’ll begin

Listener No 4647, Roundabout Sums: A Setter’s Blog by Oyler

Posted by Listen With Others on 14 Mar 2021

The inspiration for this puzzle came from a problem that appeared in Mathematical Pie which is a Mathematical Association publication aimed at secondary school pupils. Issue 186 which appeared in the summer of 2012 contained a problem set by one of their regular contributors Wil Ransome. There were two concentric circles and the area in between the two circles had been divided into 15 parts. Solvers had to use all of the numbers 1 to 15 inclusive, placing each number into a different part such that the sum of the numbers in any two adjacent parts was a triangular number. To give solvers a start five of the numbers had already been entered. I ignored those and started from scratch and about 10 minutes later had it solved.

I found it a very nice puzzle and it reminded me of the puzzle where you had to place the numbers 1 to 23 in a line such that the sum of each adjacent pair was a square number. I used this in my Martin Gardner tribute puzzle in The Magpie and Sabre used it in a crossword also in The Magpie.

I reckoned that I could set a crossnumber puzzle that used this sequence of numbers. It was about 5 years later that I got around to it (see below) having let it ferment away in the back of my mind.

I decided to use a triangular grid as that would be in keeping with the theme. I had set one other puzzle that had a triangular grid and used triangular numbers called Triangular Torment and that appeared in issue 6 of Tough Crosswords. So, an equilateral triangle with 6 rows would have 15 perimeter cells which was perfect. I decided to have letter/number assignment clues as I wanted to have some message in the perimeter cells. Helpfully, the word triangle has 8 different letters that don’t appear in the word sum. I played around with the remaining letters and eventually decided on the word joy along with a letter z added to the end of sum to make sumz which is reasonably homophonic.

The nice thing about a triangular grid is that there are three directions and a cell can contribute to three entries as opposed to two. I barred off the grid and got to work. I assigned the letters to the numbers first and entered them in the relevant perimeter cells. After a few days the puzzle was set. However, there was something not quite right. I had another think then decided to have the corner cells empty and have solvers fill them in in order to get the full message.

This meant starting again from scratch and this time the barring pattern sadly resulted in the grid being split into three separate portions which is not ideal. There have been some crossnumber puzzles that have had distinct separate areas before but they are not as pretty in my view.

In the revamp I ensured that there would be some clues which were of an alcoholic nature in order to appease one blogger, so we had grog and rums. Once again it only took a couple of days to set and I did the cold solve. I sent it off to two test solvers both of whom solved and enjoyed it. One tester greatly improved on my preamble and the editors even managed to improve on that! So, thanks go to the editors Kea and Tiburon, test solvers Nod and Zag and of course Wil Ransome and Mathematical Pie.

 
 

 

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.

 
%d bloggers like this: