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Chapter 9

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.

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Chapter 9

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  1. Chapter 9 Efficiency of Algorithms

  2. 9.2 Big-O, Big-Omega & Big-Theta

  3. 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.

  4. Approximations of Real Functions • Approximations • 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 mostg, or f(x) is O(g(x)). • 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)). • 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)).

  5. Graph Examples

  6. Definition • Let f and g be real-valued functions defined on the same set of nonnegative real numbers. Then • 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 thatA|g(x)|≤ |f(x)| for real numsx>a • 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 numsx>b • 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 thatA|g(x)| ≤ |f(x)| ≤ B|g(x)| for real numsx > k

  7. 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 numsx>k • A=10, B=30, g(x)=x6f(x)=17x6 – 45x3 + 2x + 8 • 17x6 – 45x3 + 2x + 8 is Θ(x6) • Express in omega notation • 15|x1/2| ≤ • Recall: A|g(x)|≤ |f(x)| for real numsx>a • A=15, a=0, g(x)=x1/2 • is Ω(x1/2)

  8. Example • Express in Big-OO notation • ≤ 45 |x1/2|, x > 6 • Recall: |f(x)| ≤ B|g(x)| for real numsx>b • B=45, b=6, g(x)=x1/2 • is O(x1/2)

  9. Example • Justify the statement is Θ(x) given the previous two examples. • 15|x1/2| ≤ ≤ 45|x1/2| • Recall:A|g(x)| ≤ |f(x)| ≤ B|g(x)| for real numsx > 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 • A|x1/2| ≤ ≤ B|x1/2|

  10. 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. • f(x) is Ω(g(x)) and f(x) is O(g(x)) if, and only if, f(x) is Θ(g(x)). • f(x) is Ω(g(x)) if, and only if, g(x) is O(f(x)). • f(x) is O(f(x)), f(x) is Ω(g(x)), and f(x) is Θ(f(x)) • If f(x) is O(g(x)) and g(x) is O(h(x)), then f(x) is O(h(x)). • If f(x) is O(g(x)) and c is any nonzero real number, then cf(x) is O(g(x)). • 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. • 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)).

  11. 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)

  12. 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 < 2x3and 7 < 7x3 (from LS 3x3+ 2x + 7) • Add up inequalities: • 3x3 ≤ 3x3, 2x < 2x3 and 7 < 7x3 • 3x3 + 2x + 7 ≤ 3x3 + 2x3+ 7x3 = 12x3

  13. 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)

  14. 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).

  15. Example • Show that 3x3 – 1000x -200 is O(x3) by definition of O-notation. • 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)

  16. 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| ≤ Bxsfor all real numbers x > b. • Hence, by definition of O-notation, 3x3 – 1000x – 200 is O(xs) for s > 3 .

  17. Polynomial Orders • Theorem (9.2.2) On Polynomial Orders • Suppose a0, a1, a2, … , an are real numbers and an ≠ 0. • anxn + an-1xn-1 + … + a1x + a0 is O(xs) for all intss≥n. • anxn + an-1xn-1 + … + a1x + a0 isΩ(xr) for all intsr≤n. • anxn + an-1xn-1 + … + a1x + a0 isΘ(xn).

  18. 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) = , for all real numbers x • solution: (x-1)(x+1) = x2 – 1, g(x) is Θ(x2)

  19. 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)

  20. 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)

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