3 Bears, 3 Lions, Oh My!
An old tale tells this story... There are 3 Bears and 3 Lions, together in the town of Babylon. You must transport all 6 animals to the nearby town of Sippar. You have the following constraints:
- When you travel to Sippar (or return to Babylon) you must travel with either 1 or 2 animal(s), it is a dangerous journey, and having a bear or lion is a useful ally.
- The bears and lions do not get along; they will fight to death if not under your supervision; thus you cannot leave more bears than lions (or more lions than bears) unsupervised, as if you do the animal with the numerical advantage will attack the other!
- You can leave bears and lions alone unsupervised – provided there are the same number of bears as lions, both the bears and lions will not fight if they see an equal amount of the other animal - and thus no advantage, they will growl and roar of course, but neither will attack and induce mutual destruction...
- You can leave only bears (no lions), or only lions (no bears) unsupervised…
- An animal cannot travel anywhere unsupervised, as it get lost or stolen on this treacherous trip...
What is the least amount of trips you must take from Babylon to Sippar in order to transport all six animals safely to Sippar?
The answer is 5.
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1 solution
It is very helpful to get some items to represent this... i.e. 3 blue markers as the bears and 3 red markers as the lions...
You need 5 trips from Babylon to Sippar (thus 4 trips from Sippar to babylon) to accomplish the task at hand... sequence is as follows:
You can switch every Bear for Lion, and Lion for Bear for the "other" solution...
Here is my paper solution - horizontal lines to right go to Sippar, diagonal lines to left are to Babylon - I do note with a Y where "You" are located, left side you are in Babylon, right side in Sippar (it is important to see the supervised and unsupervised situations...)