From The Sunday Times, 6th June 1971 [link]
There were six. Messrs. Butcher, Carpenter, Draper, Farmer, Grocer and Miller, who shared the fire-watching on Friday nights — three this week, three the next. By occupation they were (not necessarily respectively) a butcher, a carpenter, a draper, a farmer, a grocer and a miller.
Incidents were few and far between, until that Saturday morning when they found the “log” book signed by the Acting Deputy Assistant something or other, as follows: “All three present and fast asleep”.
Something had to be done about it: they decided to watch, to play whist, and to keep awake in the future. It was arranged thus: four did duty this week, next week two stood down and two others came in, and so on. Each did two turns in three weeks.
On the first Friday the carpenter, the draper, the farmer and the miller watched. Next week Mr Carpenter, Mr Draper, Mr Farmer and Mr Grocer played. On the third occasion Mr Butcher played against Mr Grocer, Mr Farmer against the butcher, and the miller against the draper. Each night the four cut for partners and kept them till morning.
If Mr Carpenter’s occupation is the same as the name of the one whose occupation is the same as the name of the one whose occupation is the same as the one whose occupation is the same as the name of the miller:
What is Mr Miller’s occupation?
As presented this puzzle has no solutions. In the comments I give a revised version that does have a unique answer.
[teaser521]
Like this:
Like Loading...
GeoffR and 
John Crabtree and 
Jim Randell 9:56 am on 12 July 2020 Permalink |
I used the [[
DominoGrid()]] solver from the enigma.py library (see: Teaser 1590). We pair up one of the edge squares with a 0 to make a domino, then attempt to fit the rest into the remainder of the grid. The following program runs in 88ms.Run: [ @repl.it ]
from enigma import DominoGrid, update, printf # the grid (X, Y, grid) = (5, 11, [ 6, 2, 6, 2, 0, 6, 3, 1, 2, 0, 3, 5, 1, 2, 0, 3, 5, 3, 3, 1, 6, 6, 5, 3, 0, 4, 3, 0, 3, 2, 2, 5, 4, 4, 4, 4, 2, 4, 6, 6, 5, 5, 4, 1, 1, 2, 4, 1, 5, 0, 1, 1, 0, 5, 6, ]) # remove one of the border squares and solve for the rest for (i, b) in enumerate(grid): (y, x) = divmod(i, X) if x == 0 or x == X - 1 or y == 0 or y == Y - 1: # place the 0-b domino and solve for the remaining dominoes p = DominoGrid(X, Y, update(grid, [(i, None)])) for s in p.solve(used=[(0, b)]): printf("[0-{b} placed @ {x},{y}]\n") p.output_solution(s, prefix=" ")Solution: The layout of the dominoes is shown below:
So the missing square is part of 0-2 domino.
LikeLike