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I am looking for guidance on why there are an even number of solutions to this problem:

An eccentric has his own personal collection of wild animals that include a bear, a lion, a tiger, an elephant, a wolverine, a rhino and a dingo. These 7 animals constitute his dangerous animal collection and are housed in 7 animal pens that are arranged in a row, the leftmost pen being number 1, and the rightmost pen number 7.

Some of these animals agitate one another and therefore the man decides that he must separate the following pairs of animals:

-the bear and the wolverine -the rhino and the dingo -the lion and the tiger -the bear and the elephant -the elephant and the dingo -the elephant and the bear

Two animals are separate if they are non-adjacent, that is, if there is another pen between them. All animals must be placed in different pens.

I'd like to know why there will always be an even number of solutions to this problem? From what I understand constraint problems gain efficiency with symmetry - that is you can easily swap animals in the pens as long as it done symmetrically to get different solutions. But I don't quite understand this if anyone can provide a clearer explanation it would be very helpful. Thank you in advance.

false
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CompilerSaysNo
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2 Answers2

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You need to define the 'number of solutions' more precisely. Say, you have N solutions which assigns the bear to the leftmost (or first) pen. By reversing left and right (assigning the bear to the last pen) you get twice as many solutions. And twice as many is always going to be divisible by two.

user1666959
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Because you provide twice only one of the pairs (the elephant and the bear + the bear and the elephant ). That said either you have exclude other pairs in code or you add an extra constraint unpredictable. Either provide all of them or write right code that will not exclude the opposite in every pair.

ilias iliadis
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