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CMPS 256: Advanced Algorithms and Data Structures

CMPS 256: Advanced Algorithms and Data Structures. Growth of Functions. Credits. Slides based on: Monica Nicolescu @ UNR F. Abu Salem, J. Boulos, W. Keyrouz, & George Turkiyyah @ AUB. Algorithm Analysis. The amount of resources used by the algorithm Space Computational time Running time:

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CMPS 256: Advanced Algorithms and Data Structures

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  1. CMPS 256: Advanced Algorithms and Data Structures Growth of Functions

  2. Credits Slides based on: • Monica Nicolescu @ UNR • F. Abu Salem, J. Boulos, W. Keyrouz, & George Turkiyyah @ AUB

  3. 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

  4. Runtime Analysis • In this course, we will typically use T(n) to refer to worst-case running time, unless otherwise noted • It is difficult to determine T(n) experimentally • Too many possible inputs • Don’t know a priori which input leads to worst-case behavior • We will instead determine T(n) theoretically- by analyzing/counting number of primitive operations in algorithm pseudocode

  5. Order of growth • Alg.: MIN (a[1], …, a[n]) m ← a[1]; for i ← 2 to n if a[i] < m then m ← a[i]; • Running time: T(n) =1 [first step] + (n) [for loop] + (n-1) [if condition] + (n-1) [the assignment in then] = 3n - 1 • Order (rate) of growth: • The leading term of the formula • Gives a simple characterization of the algorithm’s efficiency • Expresses the asymptotic behavior of the algorithm • T(n) grows like n

  6. Analyzing Code • Conditional statements if C then S1 else S2 time  time(C) + Max( time(S1), time(S2)) • Loops for i  a1 to a2 do S time  time(Si) for all i

  7. Analyzing Code • Nested loops for i = 1 to n do for j = 1 to n do sum = sum + 1 Number_of_iterations * time_per_iteration

  8. Analyzing Code • Nested loops for i = 1 to n do for j = i to n do sum = sum + 1

  9. Exercise • A more complicated example for ( i=0; i<=n2; i++) for ( j=i; j>=1; j=j/5) sum = sum + 1 T(n) = ?

  10. Typical Running Time Functions • 1 (constant running time): • Instructions are executed once or a few times • logN (logarithmic) • A big problem is solved by cutting the original problem in smaller sizes, by a constant fraction at each step • N (linear) • A small amount of processing is done on each input element • N logN (log-linear)s • A problem is solved by dividing it into smaller problems, solving them independently and combining the solution

  11. Typical Running Time Functions • N(1+c) where c is a constant satisfying 0 < c < 1 (super-linear) • N2 (quadratic) • Typical for algorithms that process all pairs of data items (double nested loops) • N3(cubic) • Processing of triples of data (triple nested loops) • NK (polynomial) • kN (exponential) • Few exponential algorithms are appropriate for practical use

  12. Logarithms • In algorithm analysis we often use the notation “log n” without specifying the base Binary logarithm Natural logarithm

  13. Some Common Functions • Floors and ceilings • x : the greatest integer less than or equal to x • x : the least integer greater than or equal to x • x – 1 < x ≤ x ≤ x < x + 1

  14. Some Simple Summation Formulas • Arithmetic series: • Geometric series: • Special case: x < 1: • Harmonic series: • Other important formulae:

  15. 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 • The key to mathematical induction is to find case n “within” case n+1

  16. 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): 2n + 1 ≤ 2nmust prove: 2(n + 1) + 1 ≤ 2n+1 2(n + 1) + 1 = (2n + 1 ) + 2 ≤ 2n + 2 (by induction)  2n + 2n = 2n+1 (2 ≤ 2nfor n ≥ 1)

  17. Another Example • Prove that: • Basis step: • n = 1: • Inductive step: • Assume inequality is true for n, and prove it is true for (n+1): • Assume and prove:

  18. 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)

  19. Asymptotic notations • O-notation • Intuitively: O(g(n)) = the set of functions with a smaller or same order of growth as g(n) • An implicit assumption in order notation is that f(n) is an asymptotically nonnegative function, i.e. f(n)0  sufficiently large n

  20. Examples • 2n2 = O(n3): • n2 = O(n2): • 1000n2+1000n = O(n2): • n = O(n2): 2n2≤ cn3  2 ≤ cn  possibly c = 1 and n0= 2 n2≤ cn2  c ≥ 1  possibly c = 1 and n0= 1 1000n2+1000n ≤ cn2 ≤ cn2+ 1000n  possibly c=1001 and n0 = 1000 n ≤ cn2  cn ≥ 1  possibly c = 1 and n0= 1

  21. Examples • E.g.:prove that n2≠ O(n) • Assume  c & n0 such that:  n≥ n0: n2 ≤ cn • Choose n = max (n0, c) • n2 = n  n ≥ n  c  n2 ≥ cn contradiction!!!

  22. Comments • Our textbook uses “asymptotic notation” and “order notation” interchangeably • Order notation can be interpreted in terms of set membership, for e.g., • T(n) = O(f(n)) is equivalent to T(n) O(f(n)) • O(f(n)) is said to be “the set of all functions for which f(n) is an upper bound”

  23. Asymptotic notations (cont.) •  - notation • Intuitively: (g(n)) = the set of functions with a larger or same order of growth as g(n)

  24. Examples • 5n2 = (n) • 100n + 5 ≠(n2) • n = (2n), n3 = (n2), n = (logn)  cn  5n2 • Possibly c = 1 and n0 = 1 •  c, n0such that: 0  cn  5n2 Spse  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

  25. Asymptotic notations (cont.) • -notation • Intuitively(g(n)) = the set of functions with the same order of growth as g(n)

  26. 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 • 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

  27. 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

  28. Asymptotic Notations - Examples •  notation • n2/2 – n/2 • (6n3 + 1)lgn/(n + 1) • n vs. n2 •  notation • n vs. 2n • n3 vs. n2 • n vs. logn • n vs. n2 = (n2) = (n2lgn) n ≠ (n2) • O notation • 2n2 vs. n3 • n2 vs. n2 • n3 vs. nlogn n = (2n) 2n2 = O(n3) n2 = O(n2) n3 = (n2) n3 O(nlgn) n = (logn) n (n2)

  29. 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))

  30. 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) = 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))

  31. 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) with the desired properties 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.

  32. Limits and Comparisons of Functions Using limits for comparing orders of growth: • compare ½ n (n-1) and n2

  33. Limits and Comparisons of Functions L’Hopital rule: • compare lgn and

  34. Some Common Functions • Factorials Stirling’s approximation

  35. Exercises • Is this true: f(n)=O(g(n))  f(n)+g(n)=W(f(n)) If so, give a proof that uses formal definition of O, W, and Q • Order these functions asymptotically from slowest to fastest:

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