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Analysis of Algorithms CS 477/677

Analysis of Algorithms CS 477/677. Instructor: Monica Nicolescu Lecture 2. Algorithm Analysis. The amount of resources used by the algorithm Space Computational time Running time: The number of primitive operations (steps) executed before termination Order of growth

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Analysis of Algorithms CS 477/677

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  1. Analysis of AlgorithmsCS 477/677 Instructor: Monica Nicolescu Lecture 2

  2. Algorithm Analysis • The amount of resources used by the algorithm • Space • Computational time • Running time: • The number of primitive operations (steps) executed before termination • Order of growth • The leading term of a formula • Expresses the behavior of a function toward infinity CS 477/677 - Lecture 2

  3. Asymptotic Notations • A way to describe behavior of functions in the limit • How we indicate running times of algorithms • Describe the running time of an algorithm as n grows to  • O notation: asymptotic “less than”: f(n) “≤” g(n) •  notation: asymptotic “greater than”: f(n) “≥” g(n) •  notation: asymptotic “equality”: f(n) “=” g(n) CS 477/677 - Lecture 2

  4. Logarithms • In algorithm analysis we often use the notation “log n” without specifying the base Binary logarithm Natural logarithm CS 477/677 - Lecture 2

  5. Asymptotic Notations - Examples • For each of the following pairs of functions, either f(n) is O(g(n)), f(n) is Ω(g(n)), or f(n) is Θ(g(n)). Determine which relationship is correct. • f(n) = log n2; g(n) = log n + 5 • f(n) = n; g(n) = log n2 • f(n) = log log n; g(n) = log n • f(n) = n; g(n) = log2 n • f(n) = n log n + n; g(n) = log n • f(n) = 10; g(n) = log 10 • f(n) = 2n; g(n) = 10n2 • f(n) = 2n; g(n) = 3n f(n) =  (g(n)) f(n) = (g(n)) f(n) = O(g(n)) f(n) = (g(n)) f(n) = (g(n)) f(n) = (g(n)) f(n) = (g(n)) f(n) = O(g(n)) CS 477/677 - Lecture 2

  6. Asymptotic notations • O-notation • Intuitively: O(g(n)) = the set of functions with a smaller or same order of growth as g(n) CS 477/677 - Lecture 2

  7. Examples • 2n2 = O(n3): • n2 = O(n2): • 1000n2+1000n = O(n2): • n = O(n2): 2n2≤ cn3  2 ≤ cnc = 1 and n0= 2 n2≤ cn2 c ≥ 1 c = 1 and n0= 1 1000n2+1000n ≤ 1000n2+ 1000n2 = 2000n2 c=2000 and n0 = 1 n≤ cn2 cn ≥ 1 c = 1 and n0= 1 CS 477/677 - Lecture 2

  8. Asymptotic notations (cont.) •  - notation • Intuitively: (g(n)) = the set of functions with a larger or same order of growth as g(n) CS 477/677 - Lecture 2

  9. Examples • 5n2 = (n) • 100n + 5 ≠(n2) • n = (2n), n3 = (n2), n = (logn)  cn  5n2  c = 1 and n0 = 1 •  c, n0such that: 0  cn  5n2  c, n0 such that: 0  cn2  100n + 5 100n + 5  100n + 5n ( n  1) = 105n cn2  105n  n(cn – 105)  0  n  105/c Since n is positive  cn – 105  0  contradiction: n cannot be smaller than a constant CS 477/677 - Lecture 2

  10. Asymptotic notations (cont.) • -notation • Intuitively(g(n)) = the set of functions with the same order of growth as g(n) CS 477/677 - Lecture 2

  11. Examples • n2/2 –n/2 = (n2) • ½ n2 - ½ n ≤ ½ n2n ≥ 0  c2= ½ • ½ n2 - ½ n ≥ ½ n2 - ½ n * ½ n ( n ≥ 2 ) = ¼ n2 c1= ¼ • n ≠ (n2): c1 n2≤ n ≤ c2 n2  only holds for: n ≤ 1/c1 CS 477/677 - Lecture 2

  12. Examples • 6n3 ≠ (n2): c1 n2≤ 6n3 ≤ c2 n2  only holds for: n ≤ c2 /6 • n ≠ (logn): c1logn≤ n ≤ c2 logn  c2 ≥ n/logn,  n≥ n0 – impossible CS 477/677 - Lecture 2

  13. More on Asymptotic Notations • There is no unique set of values for n0 and c in proving the asymptotic bounds • Prove that 100n + 5 = O(n2) • 100n + 5 ≤ 100n + n = 101n ≤ 101n2 for all n ≥ 5 n0 = 5 and c = 101is a solution • 100n + 5 ≤ 100n + 5n = 105n ≤ 105n2for all n ≥ 1 n0 = 1 and c = 105is also a solution Must findSOMEconstants c and n0 that satisfy the asymptotic notation relation CS 477/677 - Lecture 2

  14. Comparisons of Functions • Theorem: f(n) = (g(n))  f = O(g(n)) and f = (g(n)) • Transitivity: • f(n) = (g(n))andg(n) = (h(n))  f(n) = (h(n)) • Same for O and  • Reflexivity: • f(n) = (f(n)) • Same for O and  • Symmetry: • f(n) = (g(n)) if and only if g(n) = (f(n)) • Transpose symmetry: • f(n) = O(g(n)) if and only if g(n) = (f(n)) CS 477/677 - Lecture 2

  15. Asymptotic Notations in Equations • On the right-hand side • (n2) stands for some anonymous function in (n2) 2n2 + 3n + 1 = 2n2 + (n) means: There exists a function f(n)  (n) such that 2n2 + 3n + 1 = 2n2 + f(n) • On the left-hand side 2n2 + (n) = (n2) No matter how the anonymous function is chosen on the left-hand side, there is a way to choose the anonymous function on the right-hand side to make the equation valid. CS 477/677 - Lecture 2

  16. Some Simple Summation Formulas • Arithmetic series: • Geometric series: • Special case: x < 1: • Harmonic series: • Other important formulas: CS 477/677 - Lecture 2

  17. Mathematical Induction • Used to prove a sequence of statements (S(1), S(2), … S(n)) indexed by positive integers • Proof: • Basis step: prove that the statement is true for n = 1 • Inductive step: assume that S(n) is true and prove that S(n+1) is true for all n ≥ 1 • Find case n “within” case n+1 CS 477/677 - Lecture 2

  18. Example • Prove that: 2n + 1 ≤ 2n for all n ≥ 3 • Basis step: • n = 3: 2  3 + 1 ≤ 23 7 ≤ 8 TRUE • Inductive step: • Assume inequality is true for n, and prove it for (n+1) Assume: 2n + 1 ≤ 2n Must prove: 2(n + 1) + 1 ≤ 2n+1 2(n + 1) + 1 = (2n + 1 ) + 2 ≤ 2n + 2 ≤  2n + 2n = 2n+1, since 2 ≤ 2nfor n ≥ 1 CS 477/677 - Lecture 2

  19. More Examples CS 477/677 - Lecture 2

  20. 1 2 3 4 5 6 7 8 2 3 5 7 9 10 11 12 mid lo hi Recurrent AlgorithmsBINARY – SEARCH • for an ordered array A, finds if x is in the array A[lo…hi] Alg.:BINARY-SEARCH (A, lo, hi, x) if (lo > hi) return FALSE mid  (lo+hi)/2 if x = A[mid] return TRUE if ( x < A[mid] ) BINARY-SEARCH (A, lo, mid-1, x) if ( x > A[mid] ) BINARY-SEARCH (A, mid+1, hi, x) CS 477/677 - Lecture 2

  21. 1 2 3 4 5 6 7 8 1 1 2 2 3 3 4 4 5 5 7 7 9 9 11 11 mid = 4, lo = 5, hi = 8 mid = 6, A[mid] = x Found! Example • A[8] = {1, 2, 3, 4, 5, 7, 9, 11} • lo = 1 hi = 8 x = 7 CS 477/677 - Lecture 2

  22. 1 2 3 4 5 6 7 8 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 7 7 7 9 9 9 11 11 11 mid = 4, lo = 5, hi = 8 mid = 6, A[6] = 7, lo = 5, hi = 5 mid = 5, A[5] = 5, lo = 6, hi = 5 NOT FOUND! Example • A[8] = {1, 2, 3, 4, 5, 7, 9, 11} • lo = 1 hi = 8 x = 6 CS 477/677 - Lecture 2

  23. Analysis of BINARY-SEARCH Alg.:BINARY-SEARCH (A, lo, hi, x) if (lo > hi) returnFALSE mid  (lo+hi)/2 ifx = A[mid] return TRUE if ( x < A[mid] ) BINARY-SEARCH (A, lo, mid-1, x) if ( x > A[mid] ) BINARY-SEARCH (A, mid+1, hi, x) • T(n) = c + • T(n) – running time for an array of size n constant time: c1 constant time: c2 constant time: c3 same problem of size n/2 same problem of size n/2 T(n/2) CS 477/677 - Lecture 2

  24. Readings • Chapter 3 • Apendix A CS 477/677 - Lecture 2

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