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Unformatted text preview: 4. J UXTAPOSITIONS In the last section we studied one method, sums, to combine classes. In this lecture we study another such operation: concatenation, or juxtaposition. 4.1. A S IMPLE E XAMPLE Consider the class of all permutations which can be written as one increas ing segment followed by another, as shown in Figure 4.1. We call such permutations juxtapositions of two increasing sequences, and we denote the class by J Av 21 Av 21 . More formally, lies in J if can be written as the concatenation , where both and are increasing se quences (i.e., in the same relative order as a permutation in Av 21 ). As with the layered permutations in the last lecture, the basis question and the enumeration question immediately present themselves. Clearly 321 cannot be written as a juxtaposition of two increasing sequences but every permutation it contains can be, so 321 is a basis element for J . All other permutations of length 3 lie in J , but 2143 and 3142 do not lie in J and do not contain 321 , so 2143 and 3142 are also basis elements for J . These three permutations constitute the entire basis, as we show next. Proposition 4.1. The basis of the class J Av 21 Av 21 consists of 321 , 2143 , and 3142 . Proof. Since 321 , 2143 , and 3142 do not lie in J , J Av 321 , 2143 , 3142 , so it suffices to prove the reverse inclusion. Take any permutation Av 321 , 2143 , 3142 of length n and suppose that the longest initial increas ing segment of (every permutation has such a segment of length at least 1 ) is 1 2 i ; because this is the longest initial increasing segment, i i 1 . We need to show that i 1 i 2 n . Suppose, to the contrary, that j k for i j k . Note that we must have Figure 4.1: The permutation 235681479 can expressed as the concate nation of two increasing segments: 23568 and 1479 ....
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 Spring '11
 Vetter

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