Useful Formulae
Analysis of Algorithms
 for (i = 0; i <= n; i++) {
T(n) = O(n2)
j = i;
do j = j - 2 while (j > 0);
}
T(n) = O(n log n)
 for (i = 0; i <= n; i++) {
j = i;
do j = j div 2 while (j > 0);
}
Recurrence Relations
•
A sequence of numbers can be defined by a recurrence
relation and a set of initial conditions
•
Examples:
– an = an-1 + 2 ; a0 = 5 gives the sequence
a0 = 5, a1 = 7, a2 = 9, a3 = 11
–
an = an-1 + an-2 ; a0 = 1 a1 = 1 gives the sequence
a0 = 1, a1 = 1, a2 = 2, a3 = 3, a4 = 5, a5 = 8
How to solve a recurrence relation?
Ad Hoc Method
-- Calculate the first few values of the recurrence
-- Look for regularity
-- Guess a suitable general form
-- Prove by mathematical induction
e.g., T(n) = 2T(n - 1) + 1, T(1) = 1
Iterated Method
-- Converts the recurrence into a summation
-- Use bounded summation to solve the recurrence
e.g., T(n) = 3T(n/4) + n, T(1) = 1
 Characteristic Equation or Annihilator Method
Linear Recurrence Relations with Constant Coefficients
Superposition of Solutions
Obtaining a General Solution
• A general solution is of the form g(n)=an where a is
some constant. To determine a, substitute g(n)=an into
the recurrence.
c0an - c1an-1 - ... - ckan-k = 0,
divide by an-k to get
c0ak - c1ak-1 - ... - ck = 0,
• Solving this polynomial equation gives all possible a's.
(This is called the characteristic equation.)
Example
Multiple Roots
• If a is the root of the characteristic equation with
multiplicity m, there is a general solution
Getting a Particular Solution
• Mainly try and error. However, some very good
suggestions do exist:
• Sometimes your initial guess "degenerates" and gives
contradictory conditions. Then try a solution of higher
degree.
Examples
A Total Example
• Once the general and particular solutions have been
found, they are added together to give the total solution.
• The initial conditions are then used to determine the
constants of the general solution.
A Total Example
Pseudo Nonlinear Recurrence
• Range Transformations: on the values of the sequences
• Domain Transformations: on the value of indices
Example
A Practical Example
• Mergesort
– Split list in half
– Sort both halves
– Merge the two halves
• Analysis: The recurrence relation is
T(0) = 0
Example Continued
T(n) = 2T(n/2) + (n – 1)
Example Continued
Example Continued
(3) Impose the initial conditions on A2n + n  2n + 1
(n=0) A + 0 + 1 = U(0) = 0  A = -1
So U(n) = n  2n + 2n + 1 = T(2n)
Replace n by log n gives
T(n) = nlog n + n + 1 = Q(n  log n)
How to deal with this?
• What if the master theorem cannot be applied?
• Extend the recurrence and compute the sum of the terms.
Polygon Triangulation
• Given a polygon in the plane, add diagonals so that each
face is a triangle None of the diagonals are allowed to
cross.
• Triangulation is an important first step in many geometric
algorithms.
Polygon Triangulation
• The simplest algorithm might be to try each pair of points
and check if they see each other. If so, add the diagonal
and recur on both halves, for a total of O(n3).
• However, Chazelle gave an algorithm which runs in
T(n) =2T(n/2)+O( n ) time. Since n = O(n1-e) by case 1
of the Master Theorem, Chazelle's algorithm is linear, i.e.,
T(n) =O(n).
Application to Algorithm Design
b
T (n)  aT ( )  Q(n c )
n
• To improve the efficiency of an algorithm
– if logba > c, try to decrease the ratio of a/b (i.e., split
the problem into fewer subproblems with larger size.
– if logba ≤ c, try to find a better way for splitting/merge
subproblems.