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.
This puzzle was originally published with no title.
[teaser521]
GeoffR and 
Frits, 
John Crabtree and 
Robert Brown are discussing. 
S2T2 
Jim Randell 8:11 am on 28 June 2020 Permalink |
We can use the [[
SubstitutedExpression()]] solver from the enigma.py library to solve this puzzle.The following run file executes in 96ms.
Solution: The numbers assigned to each line are shown below:
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Jim Randell 12:38 pm on 30 June 2020 Permalink |
The puzzle as originally published in The Sunday Times was worded as follows:
CD being the external edge of the 23 triangle (which is the value C in my solution).
We can adjust the run-file accordingly:
And we find that there are 24 ways to fill out the numbers, but the value on the required edge is always 9.
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GeoffR 4:09 pm on 29 June 2020 Permalink |
% A Solution in MiniZinc include "globals.mzn"; % using the vertex labels A,B,C,D and E (as original diagram in link) % plus F for the central point % outside edges of the pentagon var 0..9:AB; var 0..9:BC; var 0..9:CD; var 0..9:DE; var 0..9:EA; % internal lines to central point F var 0..9:AF; var 0..9:BF; var 0..9:CF; var 0..9:DF; var 0..9:EF; constraint all_different ([AB, BC, CD, DE, EA, AF, BF, CF, DF, EF]); % three given triangles are AEF, ABF and CDF constraint AF + EF + EA == 7; constraint AB + AF + BF == 9; constraint CF + DF + CD == 23; % the other two triangle sides (BCF and DEF) are the same total % and the two add together to give a perfect square constraint DE + EF + DF == BF + CF + BC; set of int: sq = {16, 25, 36, 49, 64, 81}; constraint (DE + EF + DF + BF + CF + BC) in sq; solve satisfy; % AB = 0; BC = 3; CD = 6; DE = 5; EA = 1; % AF = 2; BF = 7; CF = 8; DF = 9; EF = 4; % % time elapsed: 0.06 s % ---------- % ==========LikeLike
GeoffR 1:46 pm on 8 July 2020 Permalink |
My solution below is updated for the original Sunday Times description, the solution above being the solution being based on the teaser in the book Brainteasers (2002, edited by Victor Bryant).
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