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Thanks Wakeling and Carpox1. It was lucky that I had C&D for rings 5&6. Now your value of V for ring 2 makes perfect sense. As do the other ring clues. But - and it was Murky who made the point originally - I'm not convinced about counting the hole for 13 and 14 only once.

meursault - given the preamble total of 16 we can't assume a "Robin Hood", alas. Points given for holes, not for clues with holes in them.

Points are not given for the holes. Please read the preamble. If people wish to persist in their belief that the shared hole should only count once, that's fine, but perhaps they could tell the rest of us which entry, 13 or 14 is denied a score, and therefore which entry is only five letters long. The score can only be denied on the premise that the contest is biased. There's no thematic inconsistency in assuming that two shots hit the target in exactly the same place.

Phi must be really chuckling at the arguments raging here.

I believe that Planks is correct regarding the values: I have a gift for creating unnecessary complications.

Demeter's total for Inward includes an extra 500 for the second shot which travelled through the existing hole for 13/14. This sort of thing happens from time to time in target shooting. The better the marksman, the more likely it becomes, in particular towards the centre of the target. The setter would be unwise to discount that possibility. Both the options should be marked correct.

Murky - we don't know how many missed the grid entirely so it might be a tad unfair to assume the contest was biased. But perhaps it was. We'll find out when they publish the official solution.

The rubric says that each radial entry gets a value according to the location of the interrupting letter, so a scorecard would have 20 values of which 17 have such a location, even though there are 16 locations. So the two entries that share a location should each get a value and I do not think they share the value.

As a matter of fact, I could make an argument justifying a solution where there are only three occurrences in the centre, with a shared occurrence from two co-linear entries, and two separate occurrences in the cell in ring 6. There would still be 16 locations. Clearly not what is intended but why would it be wrong?

Yes, I think buzzb is right, so my totals are correct.

Inward and Outward have 10 shots each. The shot at 13/14 has to be, as Meursault says, one in which the 2nd shooter fires through the hole made by the first, and so each are awarded points.

Suppose the shared location happened to be a result of one Inside entry and one Outside entry. If the location only counted once, which player would get the value?

This wasn’t the easiest nor the most satisfying Listener to complete. But it was certainly a bit of light relief after my first (hopefully successful!) attempt at the numbers last week. Thanks Demeter for your binary hints. Given the randomness of the number totals, it was reassuring to find that my totals matched yours. I still can’t parse Ring 1, clue 1 but I’m not going to lose sleep over it.