Method of distinguished element

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"Distinguished element" redirects here. For sets with pre-defined distinguished elements, see Pointed set.

In enumerative combinatorial mathematics, identities are sometimes established by arguments that rely on singling out one "distinguished element" of a set.

Definition[edit]

Let \mathcal{A} be a family of subsets of the set A and let x \in A be a distinguished element of set A. Then suppose there is a predicate P(X,x) that relates a subset X\subseteq A to x. Denote \mathcal{A}(x) to be the set of subsets X from \mathcal{A} for which P(X,x) is true and \mathcal{A}-x to be the set of subsets X from \mathcal{A} for which P(X,x) is false, Then \mathcal{A}(x) and \mathcal{A}-x are disjoint sets, so by the method of summation, the cardinalities are additive[1]

|\mathcal{A}| = |\mathcal{A}(x)| + |\mathcal{A}-x|

Thus the distinguished element allows for a decomposition according to a predicate that is a simple form of a divide and conquer algorithm. In combinatorics, this allows for the construction of recurrence relations. Examples are in the next section.

Examples[edit]

  • The binomial coefficient {n \choose k} is the number of size-k subsets of a size-n set. A basic identity, one of whose consequences is that these are precisely the numbers appearing in Pascal's triangle, states that:
{n \choose k-1}+{n \choose k}={n+1 \choose k}.
Proof: In a size-(n + 1) set, choose one distinguished element. The set of all size-k subsets contains: (1) all size-k subsets that do contain the distinguished element, and (2) all size-k subsets that do not contain the distinguished element. If a size-k subset of a size-(n + 1) set does contain the distinguished element, then its other k − 1 elements are chosen from among the other n elements of our size-(n + 1) set. The number of ways to choose those is therefore {n \choose k-1}. If a size-k subset does not contain the distinguished element, then all of its k members are chosen from among the other n "non-distinguished" elements. The number of ways to choose those is therefore {n \choose k}.
  • The number of subsets of any size-n set is 2n.
Proof: We use mathematical induction. The basis for induction is the truth of this proposition in case n = 0. The empty set has 0 members and 1 subset, and 20 = 1. The induction hypothesis is the proposition in case n; we use it to prove case n + 1. In a size-(n + 1) set, choose a distinguished element. Each subset either contains the distinguished element or does not. If a subset contains the distinguished element, then its remaining elements are chosen from among the other n elements. By the induction hypothesis, the number of ways to do that is 2n. If a subset does not contain the distinguished element, then it is a subset of the set of all non-distinguished elements. By the induction hypothesis, the number of such subsets is 2n. Finally, the whole list of subsets of our size-(n + 1) set contains 2n + 2n = 2n+1 elements.
  • Let Bn be the nth Bell number, i.e., the number of partitions of a set of n members. Let Cn be the total number of "parts" (or "blocks", as combinatorialists often call them) among all partitions of that set. For example, the partitions of the size-3 set {abc} may be written thus:
\begin{matrix}abc \\  a/bc \\  b/ac \\  c/ab \\  a/b/c \end{matrix}
We see 5 partitions, containing 10 blocks, so B3 = 5 and C3 = 10. An identity states:
B_n+C_n=B_{n+1}.\,
Proof: In a size-(n + 1) set, choose a distinguished element. In each partition of our size-(n + 1) set, either the distinguished element is a "singleton", i.e., the set containing only the distinguished element is one of the blocks, or the distinguished element belongs to a larger block. If the distinguished element is a singleton, then deletion of the distinguished element leaves a partition of the set containing the n non-distinguished elements. There are Bn ways to do that. If the distinguished element belongs to a larger block, then its deletion leaves a block in a partition of the set containing the n non-distinguished elements. There are Cn such blocks.

See also[edit]

References[edit]

  1. ^ Petkovšek, Marko; Tomaž Pisanski (November 2002). "Combinatorial Interpretation of Unsigned Stirling and Lah Numbers". University of Ljubljana preprint series 40 (837): 1–6. Retrieved 12 July 2013.