Enigmatic Code
Programming Enigma Puzzles
Enigma 830: Post mix
Posted by on 27 March 2023
From New Scientist #1985, 8th July 1995 [link] [link]
In an inebriated state I tried to address two letters to each of five friends. I have listed my efforts below. Each envelope has four items:
– first name
– surname
– house name
– townAll the correct items occurred somewhere on the ten envelopes. But on each envelope the four items which I wrote all turned out to be from different people.
1: David Davis, Rose Cottage, Ely
2: Brian Clark, The Meadow, Carlisle
3: Ed. Andrews, Riverside, Bradford
4: Alan Eyres, Waters’ Edge, Altrincham
5: Ed. Clark, Belle View, Carlisle
6: David Andrews, Waters’ Edge, Ely
7: Clive Brown, Belle View, Doncaster
8: Brian Eyres, Riverside, Doncaster
9: Clive Clark, Riverside, Ely
10: Ed. Davis, Rose Cottage, Carlisle.What is Brian’s name and address?
[enigma830]
Enigmatic Code 
Fortunately each of the components of the 5 addresses are unique, so we can use them to stand for themselves.
This Python program runs in 63ms. (Internal runtime is 3.1ms).
Run: [ @replit ]
from enigma import (cproduct, intersect, diff, unzip, uniq, printf) # the 10 envelopes envs = [ ("David", "Davis", "Rose Cottage", "Ely"), ("Brian", "Clark", "The Meadow", "Carlisle"), ("Ed.", "Andrews", "Riverside", "Bradford"), ("Alan", "Eyres", "Waters' Edge", "Altrincham"), ("Ed.", "Clark", "Belle View", "Carlisle"), ("David", "Andrews", "Waters' Edge", "Ely"), ("Clive", "Brown", "Belle View", "Doncaster"), ("Brian", "Eyres", "Riverside", "Doncaster"), ("Clive", "Clark", "Riverside", "Ely"), ("Ed.", "Davis", "Rose Cottage", "Carlisle"), ] # solve envs for the remaining groups gs def solve(envs, gs, ss=[]): # are we done? if not gs[0]: yield ss else: # find the first component n = gs[0][0] # choose the remaining components for rs in cproduct(gs[1:]): cs = (n,) + rs # check each envelope contains at most one of these components if all(len(intersect([cs, xs])) < 2 for xs in envs): # solve for the remaining components yield from solve(envs, list(diff(xs, {x}) for (xs, x) in zip(gs, cs)), ss + [cs]) # group the components together gs = list(list(uniq(xs)) for xs in unzip(envs)) # find solutions for ss in solve(envs, gs): for (i, (f, s, h, t)) in enumerate(ss, start=1): printf("{i}: {f} {s}, {h}, {t}") printf()Solution: Brian Andrews, Belle View, Altrincham.
The correct addresses are: