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A Bitonic Sequence is a sequence of numbers which is first strictly increasing then after a point strictly decreasing.

According to the definition of a bitonic sequence, we know that a graph of a bitonic sequence can have at most one "peak" and one "valley".

So, why is the sequence 1,3,2,4 not a bitonic sequence?

I have referred the book, "Parallel Computing", by M. J. Quinn.

Sagnik
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user40628
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  • One "valley"? If you define a valley as a number that is smaller than all of its neighbor(s), then there are two valleys in a bitonic sequence, which are the two numbers at both ends of the bitonic sequence. If you define a valley as a number that has two larger neighbors, then there is no valley in a bitonic sequence. Either way, you cannot have exactly one valley in a bitonic sequence. – John L. Aug 22 '18 at 04:51

2 Answers2

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In this sequence, lets consider 3 as the bitonic point. So 1,3 makes for a strictly increasing series.

However after we cross the bitonic point 3, 2 and 4 do not make a strictly decreasing series.

Another point in Bitonic Series' is that we have to take into account circular bitonic patterns. So a series like 1 2 5 0 -5 -2 is bitonic since we monotonically increase from 1 to 5, then monotonically decrease from 5 to -5 and again we monotonically increase from -2 to 5.

In the case of 1 *3* 2 *4* 1, there are 2 peaks occurring and hence its not even circularly bitonic.

Sagnik
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According to your definition, a sequence $x_1,\ldots,x_n$ is bitonic if there exists an index $i$ such that $x_1 < x_2 < \cdots < x_i$ and $x_i > x_{i+1} > \cdots > x_n$.

In your case, this means that one of the following must holds: $$ 1 > 3 > 2 > 4 \\ 1 < 3 > 2 > 4 \\ 1 < 3 < 2 > 4 \\ 1 < 3 < 2 < 4 $$ You can check that none of these options works.

Another way to see this is to notice that your sequence has an ebb, $x_3 = 2$, which cannot happen in a bitonic sequence.

A third way is to notice that a sequence is bitonic if the $<,>$ pattern is of the form $<^*>^*$ (in regular expression parlance). In your case, $1 < 3 > 2 < 4$, and so the pattern is $<><$, which is not of this form.

Yuval Filmus
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