Photo:
THE MIRROR
Bert is an architect and has fulfilled many strange customer requests.
However, when it comes to the new seven-story office building, he begins to ponder.
The client would like to install elevators, but not ordinary ones.
So that the elevators don't stop constantly, as she knows from her old building early in the morning, the elevators shouldn't serve every floor.
All lifts start on the ground floor and go up to the top floor - the seventh floor.
On floors one to six, only three intermediate stops should be allowed per elevator.
At the same time, the client wants people from every floor to be able to reach every other floor with a lift without having to change trains along the way.
What is the smallest possible number of elevators that Bert can use to fulfill the client's wishes?
Six elevators
must
be installed.
We look at the connections between floors one through six.
If they fulfill the client's wishes, this also applies to the ground floor and the seventh floor, because that's where all the lifts stop.
Between the first and sixth floors there are a total of 6*5/2 = 15 different combinations of two floors.
All of these must be possible with the elevators.
Each elevator has three stops on floors one through six.
An elevator thus covers three different connections.
For example, if the lift stops on floors one, three and six, these are the following three connections:
1 <-> 3
3 <-> 6
1 <-> 6
Since there are exactly 15 different connections between two floors with six floors and one elevator covers three different connections, five elevators should theoretically be enough.
In this case, however, no connection may occur twice, because otherwise the five lifts would not be able to serve all 15 connections.
In fact, five lifts are not enough.
It is unavoidable that a connection appears at two different lifts.
We see this in the example of the first floor.
From there, the five floors above must be reached without changing trains.
Since a lift stopping on the first floor is only allowed to make two more stops on the five floors above, we need at least three lifts - for example (1,2,3), (1,4,5) and (1 ,5,6).
With two lifts we can only cover four of the five necessary stops.
So there must be at least three lifts, and then at least one connection is double (three lifts = six stops, but only five different ones are possible).
The minimum number of lifts is therefore six.
And it actually works with six elevators, as the following breakdown shows:
(1,2,3)
(1,4,5)
(1,2,6)
(2,4,5)
(3,4,6)
(3,5,6)
I discovered this puzzle in the book "Garden of the Sphinx" by Pierre Berloquin.
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