I return to the subject of crosswords and find that everything that I wrote in this piece, back in 2020, still applies. As it says, “I learnt how to do cryptic crosswords at my father’s knee, as it were”, but rarely look at them these days. My father continued to get a daily paper until shortly before he died, aged 89, in 2020. He was still doing the daily crossword well into his 80s. His paper of choice by then, mainly for the crossword, but also because it was heavily discounted, was The Times.
It’s possible that somewhere in a box I have puzzles completed by my father, in his own hand, but I can’t find any right now. Instead, I offer this example, filled in by me, to show what a completed crossword would look like. It’s last Friday’s Guardian Cryptic Crossword (12 May 2023). I got a few of the answers but for most of them I needed the printed Solution in the following day’s paper.
As you can see, only the interconnecting letters have been filled in. My father adopted this habit in the late 1970s. He was working on construction projects in Saudi Arabia and the ballpoint pens they used would dry up much quicker than in the UK. To conserve ink, and prolong the life of a pen, he would only fill in the letters shared between the down and across clues. He continued to fill in puzzles this way for the rest of his life.
In the last 20 years of his life the usual location for my father and me to attempt a puzzle would be a pub somewhere in West London. Often this would be before, during or after a hurling match. I recall helping out with a couple of clues one afternoon about 10 years ago. The answer to one of them (10, 6), spread across two different lines, was “Camberwell Carrot”, a term for a large spliff, popularized by the film “Withnail and I”. Over the years many people I know have expressed great affection for the film but it never did much for me. I saw it at a preview screening one morning a few months before it was released, and a couple of times later, including a 10th anniversary release in 1997. My opinion of the film never changed. I was pretty sure that my father never saw it, so he had no chance of getting that answer.
Similarly we once attempted a crossword that was built around a clue with a five-word answer (4, 4, 2, 3, 4). It turned out it was an album title, and a number of the answers were the titles of the tracks. Once I’d worked out the theme (“Dark Side of the Moon”) answers like “Eclipse”, “Brain Damage” and even (3, 5, 3, 2, 3, 3) (“The Great Gig in the Sky”) were straightforward enough. Not really fair on my father though.
Over the years I have written more than once about the “4-Ring Man”, beginning with this piece in July 2018. It deals with the rings that are left behind as you drink a pint of stout, and there are photographic examples in pieces like this one. In the original piece I finished with these words:
“You might chance upon a bar where they don’t clear away the tables as soon as someone has finished their drink. If you do, and you notice a few glasses, each with the remnants of 4 or 5 rings from the froth on a pint of Guinness, and a few folded-up crisp packets, then you will know that I was there, or maybe it was my brother, or someone like us.”
When my father was having a few beers he would leave evidence of a different kind. Once the crossword was finished he would leave it on the table, beside his last pint of the day. In the summer of 1998, when my brother was visiting from Spain, I returned from work to find him still feeling a little fragile from our session the night before. I, for some reason, had not been so badly affected. He had just come back from a late afternoon drink with my Dad at one of the nearby pubs, a not-very-successful hair of the dog. I ran down to try and catch him but he had just left. I could tell which table he had been sat at. The evidence was overwhelming: the empty pint glass, 4 rings made by the head of a pint of Guinness evenly spaced down the side, and his paper, folded neatly to display the day’s crossword puzzle, completed just like the example shown above.
One thought on “My father’s way of doing crosswords”
> To conserve ink, and prolong the life of a pen, he would only fill in the letters shared between the down and across clues.
Clever huh – us engineers ;-D
That’s such a logical thing to do, why write redundant letters down.
Thr’s ths clm tht mst vwls r ls rdndnt, n ny rsnbly lng sntnc, th cntxt llws y t ndrstnd th sns f t.
My incoming supply of promotional and trade pens ended 5 years ago when I changed jobs. I had a stock of about 50 ballpoints then, and I generally pick a pen randomly from my stockpile to use. It’s curious how the law of large numbers is kicking in now. When I take a diary away for a day or a weekend and a pen and three as spares, it’s uncanny how when one pen dries up, another will now very often too on the same day.
My Gr. 12 English master used cryptic crosswords for vocabulary enrichment, we did them in class once a week. My Gr. 10 son has it much easier, they do Wordle challenges in class, kind of the ice breaker for the day.
The style of the cryptic-crossword-creator is critical to working out the answers. We learned the style of my English master’s daily broadsheet. Years later I tried again with a different newspaper and couldn’t cut it anymore. I then switched to the “two-speed” where the same answers were described by cryptic and easy clues, so one could get hints.
Another variant is to get someone to extract the words from a previous solution and try to design a white/black crossword grid to hold them all. That can’t be done with ink on paper I suspect, needs pencil, eraser and paper. Maybe the actual solution grid is needed, else there are too many freedoms.
[My Gr. 10 son …] He mostly does them in 3 or 4 tries. He once got a hole in one, I suppose if you do them often enough and realise that there are only about 1200 five letter words it can pick as the mystery word, and you say have eliminated 100 you’ve seen, the probability you get a hole in one in the next 100 words is 9%. Much better odds than a golf hole in one :
@#ME to Bard: What is the probability of getting a hole in one on a par 3, par 4 and par 5 hole?
[https://bard.google.com/]: The probability of getting a hole in one on a par 3 hole is about 1 in 12,500. The probability of getting a hole in one on a par 4 hole is about 1 in 125,000. The probability of getting a hole in one on a par 5 hole is about 1 in 1,000,000.
These probabilities are based on the average distance of each type of hole and the average driving distance of a professional golfer. The probability of getting a hole in one also depends on a number of other factors, such as the wind conditions, the slope of the green, and the golfer’s skill level.
Well, I know that Bard has messed up questions that depend on reasoning, but the latter answer is purely factual. It messed up the estimate of a wordle hole in one – it actually calculated 0.63% and then overwrote this with 63%, that’s a damning indictment of its maths abilities.
@#ME to Bard: I’ve seen 100 Wordle answer words so can eliminate these from my first word try in a new game. What is the probability that I guess the wordle word of the day on my first guess at least once in my next 100 games?
[https://bard.google.com/]: The probability that you guess the Wordle word of the day on your first guess at least once in your next 100 games is about 63%.
This is calculated by taking the probability that you do not guess the word on your first guess, which is 99.4%, and subtracting it from 100%. This gives us a probability of 0.63%, or 63%, that you will guess the word on your first guess at least once in your next 100 games.
Here is the calculation:
Probability of guessing the word on first guess = 1 – Probability of not guessing the word on first guess
Probability of not guessing the word on first guess = 99.4%
Probability of guessing the word on first guess = 1 – 99.4% = 0.63% = 63%
It is important to note that this is just an estimate. The actual probability of guessing the word on your first guess can vary depending on a number of factors, such as the words that you have already seen and the words that you choose to guess.
I calculated the 9% from
1 -prod( [1100 : -1 : 1000] -1) /prod( [1100 : -1 : 1000]),;
ans = 0.091818