From Newsgroup: rec.puzzles

The UK's National Cipher Challenge post a puzzle each Sunday, during the competition. This week's puzzle is below:

Four people come to a river in the night. There is a narrow bridge, and it
can only hold two people at a time. They have one torch and, because itrCOs night, the torch has to be used when crossing the bridge. Person A can
cross the bridge in 1 minute, B in 2 minutes, C in 5 minutes, and D in 8 minutes. When two people cross the bridge together, they must move at the slower personrCOs pace. The question is, can they all get across the bridge
if the torch lasts only 15 minutes?

The full version of this puzzle and other challenges are available here:

https://www.cipherchallenge.org/better-late-than-never-2/
--
David Entwistle
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From Newsgroup: rec.puzzles

On 08/12/2025 07:29, David Entwistle wrote:

The UK's National Cipher Challenge post a puzzle each Sunday, during the competition. This week's puzzle is below:

Four people come to a river in the night. There is a narrow bridge, and it can only hold two people at a time. They have one torch and, because itrCOs night, the torch has to be used when crossing the bridge. Person A can
cross the bridge in 1 minute, B in 2 minutes, C in 5 minutes, and D in 8 minutes. When two people cross the bridge together, they must move at the slower personrCOs pace. The question is, can they all get across the bridge if the torch lasts only 15 minutes?

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except this one:

For wxyz to cross without loss of generality, we have:

wx |______yz->________| yz
wxy |______<-y_________| z
w |______xy->________| xyz
wy |______<-y_________| xz
|______wy->________| wxyz

So we have two return trips (min 1 minute each) and three outward
trips in the remaining 13 minutes. If C and D cross separately,
their outward trips will take 13 minutes, leaving nothing for the
fourth traveller.

C and D must, then, cross together. But if they cross *first*, C
must return with the torch, which clogs up the works because
he'll end up making three of the five trips in 15 minutes,
leaving no leeway for the other two trips.

So A and B must cross first.

CD |_____ ab-> _______| AB (0+2=2)
ACD|_____<-a _______| B (2+1=3)
A |_____ cd->_______| BCD (3+8=11)
AB |_____<- b _______| CD (11+2=13)
AB |_____<- b _______| CD (11+2=13)
|_____ ab-> _______| ABCD (13+2=15)
--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within
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From Newsgroup: rec.puzzles


Richard Heathfield <rjh@cpax.org.uk> posted:

On 08/12/2025 07:29, David Entwistle wrote:

The UK's National Cipher Challenge post a puzzle each Sunday, during the competition. This week's puzzle is below:

Four people come to a river in the night. There is a narrow bridge, and it can only hold two people at a time. They have one torch and, because itrCOs

C and D must, then, cross together....

So A and B must cross first.

What is the more general solution, for five people?
Label the people A,B,C,D,E in increasing order of slowness.
The obvious approach (A always brings the torch back)
uses 3A+B+C+D+E time while Richard's uses 2A+3B+C+E; Richard's is better when the 2nd fastest time is less than the average of best and 2nd-worst time.

With a 6th person (F is the slow-poke) Richard's trick applies TWICE;
EF cross together and so do CD. This is better than the obvious when
3B < A + C + E.

MAYBE. Check my work? I give myself a 22% chance of being correct.

Cheers,
James
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