Not so trivial, is it?

I found this problem quite interesting.

Sharky has a word of 2015 letters made of S and K letters only (e.g. SKSSS, KSKSK). A palindrome is a word which can be read the same as looking from left to right or right to left.

Sharky decides to cut up his word into sub-words such that each sub-word is a palindrome. Given that each sub-word must contain a natural number of letters (No cutting up a letter in half), find the minimum number of sub-words that can be cut out for any word of Sharky's.

#Combinatorics

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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`\boxed{123}`	$\boxed{123}$

Comments

Got an entry on OEIS. The formula there says 672.

Ivan Koswara - 5 years, 1 month ago

Damn, that looks neat. Anyone got a proof for that? The best I got was 674.

Sharky Kesa - 5 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$