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Given a polygon P with n + 2 sides and a triangulation, mark one of its sides as the base, and also orient one of its 2n + 1 total edges. There are (4n + 2)C n such marked triangulations for a given base. Given a polygon Q with n + 3 sides and a (different) triangulation, again mark one of its sides as the base. Mark one of the sides other than ...
The formula follows from considering the set {1, 2, 3, ..., n} and counting separately (a) the k-element groupings that include a particular set element, say "i", in every group (since "i" is already chosen to fill one spot in every group, we need only choose k − 1 from the remaining n − 1) and (b) all the k-groupings that don't include "i ...
The function calls itself recursively on a smaller version of the input (n - 1) and multiplies the result of the recursive call by n, until reaching the base case, analogously to the mathematical definition of factorial. Recursion in computer programming is exemplified when a function is defined in terms of simpler, often smaller versions of ...
In mathematics, a recurrence relation is an equation according to which the th term of a sequence of numbers is equal to some combination of the previous terms. Often, only previous terms of the sequence appear in the equation, for a parameter that is independent of ; this number is called the order of the relation.
The sequence 2, 1, 3, 4, 7, 11, ... of Lucas numbers satisfies the same recurrence as the Fibonacci sequence but with initial conditions = and =. More generally, every Lucas sequence is constant-recursive of order 2.
The Fermat numbers satisfy the following recurrence relations: = + = + for n ≥ 1, = + = for n ≥ 2.Each of these relations can be proved by mathematical induction.From the second equation, we can deduce Goldbach's theorem (named after Christian Goldbach): no two Fermat numbers share a common integer factor greater than 1.
A common algorithm design tactic is to divide a problem into sub-problems of the same type as the original, solve those sub-problems, and combine the results. This is often referred to as the divide-and-conquer method; when combined with a lookup table that stores the results of previously solved sub-problems (to avoid solving them repeatedly and incurring extra computation time), it can be ...
If an element n is in then n + 1 is in . is the smallest set satisfying (1) and (2). There are many sets that satisfy (1) and (2) – for example, the set {1, 1.649, 2, 2.649, 3, 3.649, …} satisfies the definition. However, condition (3) specifies the set of natural numbers by removing the sets with extraneous members.