Most of Countdown's 3.5 million viewers must have suspected long before Matthew Parris's outing of the show that the glib skills of many visitors to Dictionary Dell owed much to invisible helpers. The alternative hypothesis that all celebrity guests were stunning wordsmiths was a non-starter.
Why can't the whisperers behind the scenes be seen openly dishing out the verb sap? Because then the celebrities would look bad. So the celebrities find themselves instead in a false position. Oh that a screen would topple revealing a blinking huddle of nerds clustered around a wordfinder!
What is interesting is the idea that such a programme needs celebrities in the first place. If a celebrity has a skill beyond the facility to become famous, it's unlikely to overlap with anagrams and sums. Indeed the latter skill is so rare nowadays it has made Carol Vorderman the best- paid woman on TV. She also shows that one may shine at mental tasks and yet not become an anorak - a great first step in undermining the social forces that discourage numeracy. The bumbling Richard Whiteley also has an important role, extending a right to watch to all who see only four- letter words in FLLXEOORK or cannot see the nine-letter anagram in ROYALMINT. If he didn't exist we'd have to invent him.
But the function of falsely intelligent celebrities runs counter to all this. It diminishes the achievement of the true stars - the contestants - and perpetuates the myth that there is something out of reach about celebrities. Life is confusing enough without false signals being beamed at us.
Points to ponder
1 What is the longest word findable in the letters FLLXEOORK?
2 Of what is ROYALMINT an anagram?
3 A classic one: can the modified chessboard be covered with 31 dominoes [See top diagram]?
Answers next week.
Last week's puzzle
Ben Slocock was the first out of the electronic hat with a solution to Rosie Forth's star puzzle (see bottom diagram). He says that any path visiting each star exactly once must trace the same number of "arms" of the network as there are stars - ie 25. But any closed loop must trace an even number of arms. So it is impossible to find a tour that starts out at one star, visits each of them in turn and returns to its starting point.
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