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Repeat step three until there is a new row with one more number than the previous row (do step 3 until = +) The number on the left hand side of a given row is the Bell number for that row. (,) Here are the first five rows of the triangle constructed by these rules:
The subdomains together must cover the whole domain; often it is also required that they are pairwise disjoint, i.e. form a partition of the domain. [5] In order for the overall function to be called "piecewise", the subdomains are usually required to be intervals (some may be degenerated intervals, i.e. single points or unbounded intervals).
A partition of a set S is a set of non-empty, pairwise disjoint subsets of S, called "parts" or "blocks", whose union is all of S.Consider a finite set that is linearly ordered, or (equivalently, for purposes of this definition) arranged in a cyclic order like the vertices of a regular n-gon.
However, the unit interval [0, 1] and the set of rational numbers Q are not almost disjoint, because their intersection is infinite. This definition extends to any collection of sets. A collection of sets is pairwise almost disjoint or mutually almost disjoint if any two distinct sets in the collection are almost disjoint. Often the prefix ...
In mathematics, a function on the real numbers is called a step function if it can be written as a finite linear combination of indicator functions of intervals. Informally speaking, a step function is a piecewise constant function having only finitely many pieces. An example of step functions (the red graph).
Finding such an exact cover is an NP-complete problem, even in the special case in which the size of all sets is 3 (this special case is called exact 3 cover or X3C). However, if we create a singleton set for each element of S and add these to the list, the resulting problem is about as easy as set packing.
The vertex-connectivity statement of Menger's theorem is as follows: . Let G be a finite undirected graph and x and y two nonadjacent vertices. Then the size of the minimum vertex cut for x and y (the minimum number of vertices, distinct from x and y, whose removal disconnects x and y) is equal to the maximum number of pairwise internally disjoint paths from x to y.
I can see sense in speaking of a family of "pairwise disjoint" sets (as opposed to the weaker condition that the intersection of all the sets must be empty), and a family of "pairwise relatively prime" numbers (as opposed to the common gcd of all numbers being unity), or even a "pairwise linearly independent" set of vectors (but don't ask me ...