# pp4 - 4 J UXTAPOSITIONS In the last section we studied one...

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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, π lies in J if π can be written as the concatenation στ , where both σ and τ 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 π Av 321 , 2143 , 3142 of length n and suppose that the longest initial increas- ing segment of π (every permutation has such a segment of length at least 1 ) is π 1 π 2 π i ; because this is the longest initial increasing segment, π i π i 1 . We need to show that π i 1 π i 2 π n . Suppose, to the contrary, that π j π 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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pp4 - 4 J UXTAPOSITIONS In the last section we studied one...

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