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O -Notation. April 23, 2003 Prepared by Doug Hogan CSE 260. O -notation: The Idea. Big-O notation is a way of ranking about how much time it takes for an algorithm to execute How many operations will be done when the program is executed?

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o notation

O-Notation

April 23, 2003

Prepared by Doug Hogan

CSE 260

o notation the idea
O-notation: The Idea
  • Big-O notation is a way of ranking about how much time it takes for an algorithm to execute
  • How many operations will be done when the program is executed?
  • Find a bound on the running time, i.e. functions that are on the same order.
    • We care about what happens for large amounts of data  asymptotic order.
o notation the idea1
O-notation: The Idea
  • Use mathematical tools to find asymptotic order.
    • Real functions to approximate integer functions.
    • Depends on some variable, like n or X, which is usually the size of an array or how much data is going to be processed
o notation the complicated math behind it all
O-notationThe complicated math behind it all…
  • Given f and g, real functions of variable x…
  • First form:
    • g provides an upper bound for f ≡ graph of f lies closer to the x axis than g
  • More general form:
    • g provides an upper bound for f ≡ graph of f lies closer to the x axis than some positive multiple (M) of g after some minimum value of x(x0).
so what does closer to the x axis mean
So what does “closer to the x-axis” MEAN?
  • -M ∙ g(x) ≤ f(x) ≤ M ∙ g(x)
  • But that’s absolute value…
  • |f(x)| ≤ M ∙ |g(x)|
another graphical view
Another graphical view

y

M ∙g

f

g

After x0, |f(x)| ≤ M ∙ |g(x)|

Before x0, nothing claimed about f’s growth

x

x0

formal definition
Formal Definition
  • Let f and g be real-valued functions defined on the same set of reals.
  • f is of order g, written f(x) = O(g(x)), iff there exists
    • a positive real number M (multiple)
    • a real number x0 (starting point)

such that for all x in the domain of f and g, |f(x)| ≤ M ∙ |g(x)| when x > x0

example
Example
  • Use the definition of O-notationto express|17x6 – 3x3 + 2x + 8| ≤ 30|x6| for all x > 1
  • M = 30
  • x0 = 1
  • 17x6 – 3x3 + 2x + 8 is O(x6)
graphically
Graphically…

17x6 – 3x3 + 2x + 8 is O(x6); M = 30; x0 = 1

30x6

17x6 – 3x3 + 2x + 8

x6

problem
Problem
  • Use the definition of O-notationto express for all x > 6
  • M = 45
  • x0 = 6
graphically1
Graphically…

M = 45; x0 = 6

another graphical example
Another graphical example

f(x) = 7x3 - 2x + 3

12x3

x3

M = 12, x0 = 1 7x3 - 2x + 3 is O(x3)

using o notation
Using O-notation…
  • Order of Power Functions:
    • For any rational numbers r and s, if r < s,

xr is O(xs)

  • Order of Polynomial Functions:
    • If a0, a1,…, anare real numbers and an ≠ 0

anxn+an-1xn-1 +… + a1x + a0is O(xm) for all m ≥ n

examples
Examples
  • Example:
    • Find an order for
      • f(x) = 7x5 + 5x3 – x + 4 (all reals x)
    • O(x5)
    • Is that the only answer?
      • No…
      • But it’s the “best”
showing that a function is not big o of another
Showing that a function is NOT Big-O of another…
  • Show that x2 is not O(x).
  • [Arguing by contradiction.] Suppose not, that x2 is O(x).
  • By definition of O(…), then there exist
    • a positive real number M
    • a real number x0

such that |x2| ≤ M ∙ |x| for all x > x0 (1)

showing that a function is not big o of another ctd
Showing that a function is NOT Big-O of another…, ctd.
  • Let x be a positive real number greater than both M and x0, i.e. x>M and x>x0.
  • Then by multiplying both sides of x>M by x, x∙x>M∙x.
  • Since x is positive, |x2|>M∙|x|.
  • So there is a real number x>x0 s.t. |x2|>M∙|x|.
  • This contradicts (1) above. So, the supposition is false and thus x2 is not O(x). □
generalization
Generalization
  • If a0, a1,…, anare real numbers and an ≠ 0

anxn+an-1xn-1 +… + a1x + a0is NOTO(xm) for all m<n

best approximation
Best approximation

Definition

  • Suppose S is a set of functions from a subset of RtoR and fis a R->R function.
  • Function gis a best big-Oapproximation for f in Siff
    • f(x) is O(g(x))
    • for any h in S, if f(x) is O(h(x)), then g(x) is O(h(x)).
problem1
Problem
  • Find best big-O approx for f(x) = 5x3 – 2x + 1
  • By thm. on polynomial orders,
    • f(x) is O(xn) for all n ≥ 3
  • By previous property,
    • f(x) is NOT O(xm) for all m < 3
  • So O(x3) is the best approximation.
o arithmetic
O-Arithmetic

Let f and g be functions and k be a constant.

  • O(k*f) = O(f)
  • O(f *g) = O(f) * O(g)
  • O(f/g) = O(f) / O(g)
  • O(f) ≥ O(g) iff f dominates* g
  • O(f + g) = Max[O(f), O(g)]

(Headington 546)

dominance
Dominance

Let f and g be functions and a,b,m,n be constants.

  • xxdominates x!
  • x!dominates ax
  • axdominates bxif a > b
  • ax dominates xn if n > m
  • x dominates logax if a > 1
  • logax dominateslogbx if b > a > 1
  • logax dominates1 if a > 1

(Headington 547)

works cited
Works Cited

Epp, Susanna. Discrete Mathematics with Applications. 2nd Ed. Belmont, CA: Brooks, 1995.

Headington, Mark A., and David Riley. Data Abstraction and Structures using C++. Lexington, MA: Heath, 1994.

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