Hello guzzlers.
Lewis Carroll - whose birthday was on Friday - was a keen deviser of mathematical puzzles, such as this one:
If these three sentences are true, what can you conclude?
No experienced person is incompetent.
Jenkins is always blundering.
No competent person is always blundering.
The problem is from Carroll’s 1896 book Symbolic Logic, a lesser known work than Alice in Wonderland. Carroll was a maths don at Oxford and Symbolic Logic was not aimed at children...
You can solve the Jenkins riddle by taking each sentence as it comes. If Jenkins is always blundering, then by sentence 3 he is incompetent, and if he is incompetent then by sentence 1 he cannot be experienced. So, we can conclude that Jenkins is inexperienced.
If Carroll was alive today, he may have come up with the following puzzle, which I have adapted from another example in Symbolic Logic. On the assumption that the following ten sentences are true, what can you conclude?
The only people in the cereal cafe are from Stoke.
Every person would make a great Uber driver, if he or she is not allergic to gluten.
When I love someone, I avoid them.
No one is a werewolf, unless they have orange skin and blond hair.
No one from Stoke fails to Instagram their breakfast.
No one ever asks me whether I prefer Wills to Harry, except the people in the cereal cafe.
People from Thanet wouldn’t make great Uber drivers.
None but werewolves Instagram their breakfast.
The people I love are the ones who do not ask me whether I prefer Wills to Harry.
People with orange skin and blond hair are not allergic to gluten.
I’ll be back at 5pm with the solution.
You may, however, be curious as to why these puzzles appeared in a book called Symbolic Logic. It’s because you can rewrite each statement in the form ‘If A is true, then B is true’, where A and B are statements, which we write in symbolic logic as A > B. (Usually the symbol used is an arrow, but since the Guardian doesn’t have a horizontal arrow, I’m using the ‘>’.) Sometimes this might be the easiest way to solve the problem. Let’s try it for the opener. First, extract the statements from the sentences. We have:
A = the person is experienced
B = the person is competent
C = the person is Jenkins
D = the person is always blundering
The three sentences are
A > B,
C > D,
B > not D (meaning that ‘not D’ is true, i.e D is false)
[Note that each sentence of the form ‘X > Y’ can be replaced by its contrapositive ‘not Y > not X’, since if all Xs are Ys, then everything that is not a Y is not an X]
We can string the three statements together to get C > D > not B > not A. So C > not A, or Jenkins is inexperienced.
Now back to the cereal cafe.
And NO SPOILERS!
I set a puzzle here every two weeks on a Monday. I’m always on the look-out for great puzzles. If you would like to suggest one, email me.
My most recent book is Can You Solve My Problems, A Casebook of Ingenious, Perplexing and Totally Satisfying Puzzles. My children’s book Football School: Where Football Explains The World, co-written with Ben Lyttleton, was recently shortlisted for the Blue Peter Book Award 2017.
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