Transcript ppt - People Server at UNCW

Chapter 9
Efficiency of Algorithms
9.2
Big-O, Big-Omega & Big-Theta
Approximations of Real Functions
• The O (big-O), Ω (big Omega), and Θ (big
theta) are methods for approximating real
functions.
• These approximations can be used to
determine the efficiency of computer
programs via approximation.
Approximations of Real Functions
• Approximations
1. For large values of x, the values of f(x) are less than
those of a multiple of g(x), then f is of order at most
g, or f(x) is O(g(x)).
2. For large values of x, the values of f(x) are greater
than those of a multiple of g(x), then f is of order at
least g, or f(x) is Ωg(x)).
3. For large values of x, the values of f(x) are bounded
both above and below by those of a multiple of g(x),
then f is of order g, or f(x) is Θ(g(x)).
Graph Examples
Definition
• Let f and g be real-valued functions defined on the
same set of nonnegative real numbers. Then
1. f is of order at least g, written f(x) is Ω(g(x)), if, and only
if, there exist a positive real number A and a nonnegative
real number a such that
A|g(x)|≤ |f(x)| for real nums x>a
2. f is of order at most g, written f(x) is O(g(x)), if, and only
if, there exist a positive real number B and a nonnegative
real number b such that
|f(x)| ≤ B|g(x)| for real nums x>b
3. f is of order g, written f(x) is Θ(g(x)), if, and only if, there
exist positive real numbers A and B, and a nonnegative
real number k such that
A|g(x)| ≤ |f(x)| ≤ B|g(x)| for real nums x > k
Example
• Express in theta notation
–
–
–
–
10|x6| ≤ 17x6 – 45x3 + 2x + 8 ≤ 30|x6| for all real x>2.
Recall: A|g(x)| ≤ |f(x)| ≤ B|g(x)| for real nums x>k
A=10, B=30, g(x)=x6 f(x)=17x6 – 45x3 + 2x + 8
17x6 – 45x3 + 2x + 8 is Θ(x6)
• Express in omega notation
– 15|x1/2| ≤
15
x (2 x 9)
x 1
,x 0
– Recall: A|g(x)|≤ |f(x)| for real nums x>a
1/2
– A=15,
 a=0, g(x)=x
– 15 x (2 x 9) is Ω(x1/2)
x 1
Example
• Express in Big-OO notation
–


15
x (2 x 9)
x 1
≤ 45 |x1/2|, x > 6
– Recall: |f(x)| ≤ B|g(x)| for real nums x>b
– B=45, b=6, g(x)=x1/2
– 15 xx (2 1x 9) is O(x1/2)
Example
• Justify the statement 15 xx (2 1x 9) is Θ(x) given the
previous two examples.
• 15|x1/2| ≤ 15 xx (2 1x 9) ≤ 45|x1/2|
– Recall:A|g(x)| ≤ |f(x)| ≤ B|g(x)| for real nums x > k
– k must be the largest real number that satisfies
the conditions, hence, x>0 and x>6 will require
k=6.
– A=15, B=45
15 x (2 x 9)
1/2|
– A|x1/2| ≤
≤
B|x
x 1
Properties of O, Ω, Θ-Notations
• Theorem 9.2.1
– Let f and g be real-valued functions defined on the same set of
nonnegative real numbers.
1. f(x) is Ω(g(x)) and f(x) is O(g(x)) if, and only if, f(x) is Θ(g(x)).
2. f(x) is Ω(g(x)) if, and only if, g(x) is O(f(x)).
3. f(x) is O(f(x)), f(x) is Ω(g(x)), and f(x) is Θ(f(x))
4. If f(x) is O(g(x)) and g(x) is O(h(x)), then f(x) is O(h(x)).
5. If f(x) is O(g(x)) and c is any nonzero real number, then cf(x) is
O(g(x)).
6. If f(x) is O(h(x)) and g(x) is O(k(x)), then f(x) + g(x) is O(G(x))
where G(x) = max(|h(x)|, |k(x)|) for each x in the domain of
the functions.
7. If f(x) is O(h(x)) and g(x) is O(k(x)), then f(x)g(x) is O(h(x)k(x)).
Orders of Power Functions
• If 1 < x, then x < x2, x2 <
x3 ….
• So, if 1 < x < x2 < x3 …
• For any rational
numbers r and s
– if 1 < x, and r < s, then xr
< xs (9.2.1)
– if r < s, then xr is O(xs)
Example
• Show that for any real number x, if x > 1, then
3x3 + 2x + 7 ≤ 12x3
– From previous slide, x is real number and x > 1.
– Then x < x3 and 1 < x3
– So 2x < 2x3 and 7 < 7x3 (from LS 3x3 + 2x + 7)
– Add up inequalities:
• 3x3 ≤ 3x3, 2x < 2x3 and 7 < 7x3
• 3x3 + 2x + 7 ≤ 3x3 + 2x3 + 7x3 = 12x3
Example
• Show that 2x4 + 3x3 + 5 is Θ(x4) by using the
def’s of big-Omega, big-O, and big theta.
• Solution
– f(x) = 2x4 + 3x3 + 5, g(x) = x4
– for x > 0, 2x4 ≤ 2x4 + 3x3 + 5 b/c 3x3 + 5 > 0
– 2|x4| ≤ |2x4 + 3x3 + 5|, A = 2, and a = 0
– A|x4| ≤ |2x4 + 3x3 + 5|, for all x > a
– by definition of Omega-notation, 2x4 + 3x3 + 5 is
Θ(x4)
Example (cont)
• Previous can be expanded for O-notation
• Solution
–
–
–
–
–
x > 1 and by 9.2.1
2x4 + 3x3 + 5 ≤ 2x4 + 3x4 + 5x4
2x4 + 3x3 + 5 ≤ 10x4
|2x4 + 3x3 + 5| ≤ 10|x4|
Let B = 10 and b = 1, |2x4 + 3x3 + 5| ≤ B|x4| for all x >
b.
– by definition of O-notation 2x4 + 3x3 + 5 is O(x4)
– Since 2x4 + 3x3 + 5 is both Ω(x4) and Θ(x4) then it is by
definition O(x4).
Example
• Show that 3x3 – 1000x -200 is O(x3) by definition of Onotation.
• Solution
–
–
–
–
–
–
–
|a + b| ≤ |a| + |b| for all real numbers (triangle inequality)
further, |a - b| ≤ |a + (-b)| ≤ |a| + |-b| ≤ |a| + |b|
for x>1, |3x3 – 1000x -200| ≤ |3x3| + |-1000x|+ |-200|
|3x3 – 1000x -200| ≤ 3x3 + 1000x + 200
|3x3 – 1000x -200| ≤ 3x3 + 1000x3 + 200x3|
|3x3 – 1000x -200| ≤ 1203x3
Let B = 1203 and b=1 then,
|3x3 – 1000x -200| ≤ Bx3 for all real numbers x > b
– So, by definition |3x3 – 1000x -200| is O(x3)
Example
• Show that 3x3 – 1000x – 200 is O(xs) for s > 3
Recall that O(g(x)) bounds (top) f(x)
• Solution
– s > 3, then x3 < xs for all real numbers x>1
– So, B|x3| < B|xs| for all real numbers x > 1 (b=1)
– |3x3 – 1000x -200| ≤ Bxs for all real numbers x > b.
– Hence, by definition of O-notation, 3x3 – 1000x –
200 is O(xs) for s > 3 .
Polynomial Orders
• Theorem (9.2.2) On Polynomial Orders
– Suppose a0, a1, a2, … , an are real numbers and an
≠ 0.
1. anxn + an-1xn-1 + … + a1x + a0 is O(xs) for all ints s≥n.
2. anxn + an-1xn-1 + … + a1x + a0 is Ω(xr) for all ints r≤n.
3. anxn + an-1xn-1 + … + a1x + a0 is Θ(xn).
Example
• Find the orders for the functions
– f(x) = 7x5 + 5x3 – x + 4, for all real numbers x
• solution: f(x) is Θ(x5)
– g(x) = ( x  1)( x  1) , for all real numbers x
4
• solution: (x-1)(x+1) = x2 – 1, g(x) is Θ(x2)

Limitation of Orders of Polynomial
• Theorem: Limitation on Orders of Polynomial
Functions (9.2.3)
– Let n be a positive integer, and let a0, a1, a2, … , an
be real numbers with an ≠ 0. if m is any integer
with m < n, then
anxn + an-1xn-1 + … + a1x + a0 is not O(xm), and
anxn + an-1xn-1 + … + a1x + a0 is not Θ(xm)
Example
• Show that 1 + 2 + 3 + … + n is Θ(n2)
• Solution
– 1 + 2 + 3 + … + n = n(n + 1)/2
– n(n + 1)/2 = n2/2 + n/2
– Hence, n2/2 + n/2 is Θ(n2)
– Therefore, 1 + 2 + 3 + … + n is Θ(n2)