From The Sunday Times, 21st January 1962 [link]
This unusual instrument is operated by selecting one of the four switch positions: A, B, C, D, and turning the power on. The effects are:
A: The pratching valve glows and the queech obulates;
B: The queech obulates and the urfer curls up, but the rumption does not get hot;
C: The sneeveling rod turns clockwise, the pratching valve glows and the queech fails to obulate;
D: The troglodyser gives off hydrogen but the urfer does not curl up.
Whenever the pratching valve glows, the rumption gets hot. Unless the sneeveling rod turns clockwise, the queech cannot obulate, but if the sneeveling rod is turning clockwise the troglodyser will not emit hydrogen. If the urfer does not curl up, you may be sure that the rumption is not getting hot.
In order to get milk chocolate from the machine, you must ensure:
(a) that the sneeveling rod is turning clockwise AND;
(b) that if the troglodyser is not emitting hydrogen, the queech is not obulating.
1. Which switch position would you select to get milk chocolate?
If, tiring of chocolate, you wish to receive the Third Programme, you must take care:
(a) that the rumption does not get hot AND;
(b) either that the urfer doesn’t curl and the queech doesn’t obulate or that the pratching valve glows and the troglodyser fails to emit hydrogen.
2. Which switch position gives you the Third Programme?
No setter was given for this puzzle.
This puzzle crops up in several places on the web. (Although maybe it’s just because it’s easy to search for: “the queech obulates” doesn’t show up in many unrelated pages).
And it is sometimes claimed it “appeared in a national newspaper in the 1930s” (although the BBC Third Programme was only broadcast from 1946 to 1967 (after which it became BBC Radio 3)), but the wording always seems to be the same as the wording in this puzzle, so it seems likely this is the original source (at least in this format).
“Omnibombulator” is also the title of a 1995 book by Dick King-Smith.
[teaser44]
Hugh Casement, 
John Crabtree, and 
Frits, 
Tony Brooke-Taylor, and 2 others are discussing. 
Geoff, 
GeoffR are discussing. 
Tony Brooke-Taylor, 
S2T2 
Jim Randell 10:26 am on 10 January 2021 Permalink |
I think the wording in this puzzle could have been made a bit clearer. (I have attempted to clarify the wording from the originally published text).
Evidently, if we write down the numbers of the houses that the postman delivered to, in the order he delivered, we are told that for every house he visited (after the first two houses), the number of that house is the sum of the numbers of the previous two houses delivered to.
So the numbers of the houses visited form a Fibonacci sequence. And we are told the sequence starts with two single digit numbers, and the number 1453 is in the sequence.
This Python program runs in 45ms.
from enigma import (irange, subsets, fib, printf) # collect solutions r = set() # choose a single digit number to start the sequence for (a, b) in subsets(irange(1, 9), size=2, select="P"): ns = list() for n in fib(a, b): if n > 1453: break ns.append(n) if n == 1453: r.add(ns[-2]) printf("[({a}, {b}) -> {ns}]") printf("answer = {r}")Solution: The postman had just come from house number 898.
There are in fact three possible starting pairs of house numbers:
but they all end up settling down to generating the same sequence of higher valued house numbers, which does eventually reach 1453:
And the number immediately preceding 1453 is 898.
As the ratio of consecutive terms in a Fibonacci sequence approaches the value of the Golden Ratio (𝛗 = (1 + √5) / 2), we can immediately calculate a good guess to the answer:
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Hugh Casement 2:02 pm on 11 January 2021 Permalink |
So for every third house visited, the postman crossed the road and back again.
That’s not the way our posties (male or female) work.
I think a better puzzle could have put Simpson in an even-numbered house,
all houses receiving post that day being on the one side of the street.
That was a good idea to divide by phi. Repeated division, each time rounding to the nearest integer, would get us most of the way to the start of the sequence, though it would be safer after a while to take S(n) = S(n+2) – S(n+1) .
Did we need to know that it starts with two single-digit terms? 1453/phi has a unique value!
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John Crabtree 4:26 pm on 11 January 2021 Permalink |
Good question, but I think that the answer is yes. Otherwise you can get shorter sequences which reach 1453, ie:
1453, 897, 556, 341, 215, 126, 89, 37, 52
1453, 899, 554, 345, 209, 136, 73, 63, 10
The ratio of successive terms oscillates about phi, and converges quite slowly.
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Hugh Casement 10:24 am on 12 January 2021 Permalink |
Thanks, John. The best approximations to phi are given by the original Fibonacci sequence.
So if we double each term we get numbers all on the same side of the street.
But perhaps the puzzle would then be too easy!
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