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Let $x_1,\ldots,x_n$ be a finite sequence of real numbers. Let $f(\{x_i\}_{i=1}^n)=f(\{x_i\})$ be the length of the largest non-decreasing subsequence, and let $g(\{x_i\})$ be the length of the largest non-increasing subsequence.

Define the function

$$m(n)=\min_{\{x_i\}}f(\{x_i\})g(\{x_i\})$$

Considering a strictly increasing sequence, we can see that $m(n)\le n$. Is it true that $m(n)=n$ for all $n\in\mathbb{N}$?

Similarly, define $f'$ ($g'$) as the length of the largest strictly increasing (decreasing) subsequence, and define

$$M(n)=\max_{\{x_i\}}f'(\{x_i\})g'(\{x_i\})$$

Considering a 'mountain', that is, a sequence that strictly increases through its first half, and then strictly decreases through its second half, we see that $$M(n)\ge \begin{cases}\frac{n(n+2)}{4}&n\text{ even}\\\frac{(n+1)^2}{4}&n\text{ odd}\end{cases}$$

Is this actually an equality for all $n\in\mathbb{N}$? This all checks out for small values of $n$, but as the answers to this question point out, we should be wary about trusting patterns for small values.

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  • $\begingroup$ I don't understand.. Isn't $M(n) = 1,\,\forall n$? Consider the sequence $\pi,\pi,\pi,\pi,...$? $\endgroup$ – Lord Soth Apr 12 '13 at 23:57
  • $\begingroup$ Oh the second one is a maximization, OK. $\endgroup$ – Lord Soth Apr 12 '13 at 23:59
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The statement $m(n)=n$ is (also) true, because of the (very slightly stronger) fact:

If for a sequence of numbers the lengths of the longest weakly increasing and of the longest strictly decreasing sequence are respectively $pq$, then $n\leq pq$.

Two proofs. One is to assign to each term $x_i$ of the sequence a pair $(k,l)$ where $k$ is the length of the longest weakly increasing subsequence ending with $x_i$, and $l$ the length of the longest strictly decreasing subsequence ending with $x_i$. Then $1\leq k\leq p$ and $1\leq l\leq q$ and for $i<j$ the pair $(k',l')$ attached to $x_j$ has either $k'>k$ (if $x_i\leq x_j$) or $l'>l$ (otherwise), so in particular $(k',l')\neq(k,l)$ always. As there are only $pq$ possible pairs for $n$ terms, one must have $n\leq pq$. See this note by E. W. Dijkstra (second example) and also this answer by @Ross Millikan.

The other proof is to apply Schensted insertion to the sequence, which results in a pair of tableaux of a shape with $q$ rows and $p$ columns; since the shape (a Young diagram) fits inside the rectangle of those dimensions, one gets $n\leq pq$.

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  • $\begingroup$ Wonderful. Thank you! $\endgroup$ – Jared Apr 15 '13 at 17:13
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For $M(n)$ you are correct. This follows from the AM-GM inequality. You have two numbers that sum to $n$ and want to maximize the product. This happens when you split it as evenly as possible.

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  • $\begingroup$ Sure. This is clear. Thank you. Any ideas for $m(n)$? $\endgroup$ – Jared Apr 13 '13 at 18:21

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