From the setters’ perspective

First of all apologies for this being a bitty late but I’ve been away on holiday.

Can You Guess Where I was?  Of Course You Can!!

Jealous? You bet I am!!

A bit of a giveaway

The best bit about Can You Do Division? M100 was that it had that rare occurrence of a 100% success rate.  There were other bits of the puzzle that I liked – the parity part of odd numbered clues being odd and even numbered clues being even and the being divisible by their clue number for the down clues.  The unsatisfactory parts were the need to give the digit sums for the rows and columns and the clumsy divisibility property for some of the across clues.

I decided to see if I could make use of Piccadilly’s idea of palindromes that were exactly divisible by the number of digits and expand on this whilst still keeping bits of the previous puzzle.  I felt that the parity had to go as it was quite restrictive in obtaining a grid but I kept the divisible by their clue number and the digit sum part could be improved on.  Then came the size of grid and the barring pattern.  I felt that 7×7 was just about right so kept that.  I couldn’t have any 2 digit entries for the acrosses of course so the grid maybe looked a tad strange.  As it was a 7×7 grid I decided that the unchecked cells would sum to 49.

I started by writing a program to calculate the palindromes.  I knew full well that some solvers would just do the same thing and fig jig in their answers which they would probably find to be a fairly unsatisfying solve.  But hey – that’s their choice!!  If they don’t want to solve the puzzle in the spirit in which it was set and the manner it was intended then so be it!  Of course if they applied some logic first as AJ did in his published solution they would find that it would circumvent the need to do lots of laborious calculations and be somewhat more satisfying.

I started setting in the bottom half of the grid.  The reason for this is that there are fewer possible entries for the high numbered down clues.  I had hoped to have all the down entries exactly divisible by their digit sum as well but this proved to be a major stumbling block so relaxed that to just 3 symmetrically located pairs.  I worked steadily upwards in the grid until I encountered a problem at 10ac/2dn and had to give the information about there being an odd number of even digits in the grid. Damn! – ‘The best laid schemes o’ Mice an’ Men gang aft agley’ as we often say up here – or not as the case may be!!

Just in case you’re wondering

‘I see you went to Oxford‘

‘Yes‘

‘What did you do?‘

‘I bought a tie!‘

Thanks Spike.

Posted on Monday, July 9th, 2012 at 4:59 pm under Setting the scene.
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