The Magic of Logic (15)
Finding the three mistakes in a sentence; and how to identify a poisoned cask of wine
Today was Enders-and-Hennes day. We spent it at the local lake, where an unusual number of families had gathered, drawn there by the sun and the easing of Covid restrictions. After months of abstinence, the boys were glad to exuberantly run around the water-side playground, full of boundless energy. But they were also ready for some new logical puzzles — even asked for one themselves.
So this is what I gave them, a German version of the classic: “Their are three misteaks in this sentence” — “Diser Satz hat drei Feler”, scratched out in the sand.
They spent a minute pondering, examining each word, but could only find two that were misspelled (“Diser” and “Feler”). And then Enders had it: the third mistake was that it had only two, not three mistakes! Easy-peasy, he said.
But Hennes had a nagging discomfort. It’s a paradox, he said. It is a word he loves to use, without being able to precisely define it. What he meant was: there is definitely something wrong with the puzzle. Enders agreed, and with a little help we worked it out:
If the third error in the sentence is that it contains only two, not three mistakes, then “three” is actually correct, which means it contains only two mistakes, which means “three” is wrong, etc. It was quite a remarkable discussion, one that for a while had my own head spinning.
Enders wanted something harder, and this he got:
There is this king who has invited guests over for a feast, and has received four casks of wine for the celebration. His spies have told him that one of the casks contains poison. The king fetches his two food tasters, who are willing to die for him, and asks them to determine which of the casks is poisoned. He wants to serve wine from the other three casks to his guests.
Unfortunately, you cannot taste the poison, and it only shows symptoms — sudden death — 24 hours later. But the guests will be arriving on the following day. How can he determine which of the casks are safe, before they arrive?
Enders and his grandmother (it was communal solving) hit upon the following method: the tasters get a glass of wine, one from cask A and one from cask B. If one taster is dead the next day, we know which cask is poisonous. If both are well, we give the tasters each a glass from C and from D. The one who survives reveals which of the two casks is safe.
Another method: the first taster must drink two glasses of wine, one from cask A and one from B; and the second taster two glasses from cask C and D. The one who is dead the next day tells us which pair of casks includes the one with the poison. But which one is it, specifically? Well, the surviving taster gets a glassful to drink from one of the two casks his companion had tried. If he now too perishes, then that is the poisoned cask; if he survives, it’s the other one.
But: both methods required two days. And the guests would be arriving before then. Enders suggested they might be content to drink only from two safe casks?! Maybe, but there is a way to find out, reliably in one day, which three casks are safe.
A friend came up with an interesting try: make one taster drink a glass from cask A and the other from cask B. Four hours later they drink from casks C and D. Now if the first taster dies after 24 hours, then the poison is in cask A, if he dies after 28 hours it is in cask B, if the second taster dies after 24 hours… etc. Very nice try, Divya, but it can also be done in 24 hours. [Addendum: in the meantime Divya and a 16-year-old chess grandmaster have found the correct solution].
The boys at the lake shore could not find a solution, but Enders is determined to work it out and has promised to call when he has found an answer. I urge you, dear readers, to think about it while we wait. I will post the solution here after I hear from Enders.
Solution
So Enders found the solution and sent me the following picture:
Taster one gets a glass from cask A and one from cask B. Taster two gets one from B and one from C. If taster one is dead the next day, and taster two alive, then cask A is poisoned. If both are dead it is cask B. If only taster 2 is dead, then it is cask C. If both are alive it is D.
