Algorithms Design & Analysis. Analysis of Algorithm
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1 Algorithms Design & Analysis Analysis of Algorithm
2 Review Internship Stable Matching Algorithm 2
3 Outline Time complexity Computation model Asymptotic notions Recurrence Master theorem 3
4 The problem of sorting Input: sequence <a 1, a 2,, a n > of numbers. Output: permutation <a 1, a 2,, a n > such that a 1 a 2 a n. Example: Input: Output:
5 Example of insertion sort
6 Insertion sort INSERTION-SORT (A, n) // A[1..n] for j 2 to n do key A[j] i j - 1 while i > 0 and A[i] > key do A[i+1] A[i] i i - 1 A[i+1] key 6
7 Correctness Loop invariant Some properties always keep in the lifecycle Initiation It is true prior to the first iteration of the loop Maintenance If it is true before an iteration, it remains true before the next iteration Termination 7
8 Running time The running time depends on the input: an already sorted sequence is easier to sort. Parameterize the running time by the size of the input n Seek upper bounds on the running time T(n) for the input size n, because everybody likes a guarantee. It also depends on computation models. 8
9 Input Size Integer n 1 word log 2 n bits Product of two n-by-n matrices The order of the matrix The total number of the elements Spell checking Character word 9
10 Random access machine Sequential Instruction is executed one by one Instructions Arithmetic, data movement, and control Constant time Data type Integer and floating point 10
11 Machine-independent time What is time cost of insertion sort? It depends on the speed of our computer: absolute speed (on the same machine), relative speed (on different machines). Big Idea: Ignore machine-dependent constants. Look at growth of T(n) as n. Asymptotic Analysis 11
12 Kinds of analyses Worst-case: (usually) T(n) = maximum time of algorithm on any input of size n. Average-case: (sometimes) T(n) = expected time of algorithm over all inputs of size n. Need assumption of statistical distribution of inputs. Best-case: (bogus) Cheat with a slow algorithm that works fast on some input. 12
13 Θ-notation Math: Θ(g(n)) = {f(n) : there exist positive constants c 1, c 2, and n 0 such that 0 c 1 g(n) f(n) c 2 g(n) for all n n 0 } Engineering: Drop low-order terms; ignore leading constants. Example: 3n 3 +90n 2-5n+6046= Θ(n 3 ) 13
14 Asymptotic performance When n gets large enough, a Θ(n 2 ) algorithm always beats a Θ(n 3 ) algorithm We shouldn t ignore asymptotically slower algorithms, however. Real-world design situations often call for a careful balancing of engineering objectives. Asymptotic analysis is a useful tool to help to structure our thinking. 14
15 Worst-Case Polynomial-Time Def. An algorithm is efficient if its running time is polynomial. Justification: It really works in practice! Although N 20 is technically poly-time, it would be useless in practice. In practice, the poly-time algorithms that people develop almost always have low constants and low exponents. Breaking through the exponential barrier of brute force typically exposes some crucial structure of the problem. Exceptions. Some poly-time algorithms do have high constants and/or exponents, and are useless in practice. Some exponential-time (or worse) algorithms are widely used because the worst-case instances seem to be rare. simplex method Unix grep 15
16 Insertion sort analysis INSERTION-SORT (A, n) // A[1..n] for j 2 to n do key A[j] i j - 1 while i > 0 and A[i] > key do A[i+1] A[i] i i - 1 A[i+1] key cost c 0 c 1 c 2 c 3 c 4 c 5 c 6 loop n j-1 16
17 Insertion sort analysis Worst case: Input reverse sorted. n T(n) = Θ( j) = Θ(n 2 ) j=2 Average case: All permutations equally likely. n T(n) = Θ( j / 2) = Θ(n 2 ) j=2 Is insertion sort a fast sorting algorithm? Moderately so, for small n. Not at all, for large n. 17
18 Merge sort Merge-Sort A[1..n] 1. If n = 1, done. 2. Recursively sort A[1..n/2] and A[n/2+1..n] 3. Merge the 2 sorted lists. Key subroutine: Merge 18
19 Merging two sorted arrays Time = Θ(n) to merge a total of n elements (linear time). 19
20 Analyzing merge sort T(n) Θ(1) 2T(n/2) Abuse Θ(n) Merge-Sort A[1..n] 1. If n = 1, done. 2. Recursively sort A[1..n/2] and A[n/2+1..n] 3. Merge the 2 sorted lists. Sloppiness: Should be T ( n / 2 ) + T( n / turns out not to matter asymptotically. 2 ), but it 20
21 Recurrence for merge sort T ( n) = 2T ( n Θ(1) / 2) + Θ( n) if n = 1; if n > 1. We shall usually omit stating the base case when T(n) = Θ(1) for sufficiently small n, but only when it has no effect on the asymptotic solution to the recurrence. CLRS provides several ways to find good upper bound on T(n). 21
22 Recursion tree Solve T(n)=2T(n/2)+cn, where c>0 is constant. 22
23 Comparison Θ(nlg k n) grows more slowly than Θ(n 2 ). Therefore, merge sort asymptotically beats insertion sort in the worst case. In practice, merge sort beats insertion sort for n > 30 or so. Try test. 23
24 Asymptotic notation O-notation ( big-oh, upper bounds) f(n) = O(g(n)) if const c, n 0 such that 0 f(n) cg(n), n n 0. e.g. 2n 2 =O(n 3 ) (c=1, n 0 =2) Think of O(g(n)) as a set of functions: O(g(n))={f(n): const c, n 0 such that 0 f(n) cg(n), n n 0 } Then 2n 2 O(n 3 ) 24
25 Example f(n) = n 2 + O(n) means f(n) = n 2 + h(n) for some h(n) O(n). Note: It makes no sense to say f(n) is at least O(n). Why? 25
26 Ω-notation (lower bounds) Again, think of Ω(g(n)) as a set of functions: Ω(g(n))={f(n): const c, n 0 such that 0 cg(n) f(n), n n 0 } e.g. n ½ = Ω(lgn) (c=1, n 0 =16) 26
27 Θ-notation (tight bounds) Θ(g(n))={f(n): const c 1, c 2, n 0 such that 0 c 1 g(n) f(n) c 2 g(n), n n 0 } Example: n + 2n = Θ( n Pick c 1 = ¼, c 2 = ¾ (educated guess), then find satisfying n 0 =8. ) 27
28 Useful Theorems Theorem: Leading constants and lowerorder terms don t matter. Theorem: (O and Ω) Θ. Two more asymptotic notations: Little-Oh notation f(n) = o(g(n)) Little-omega notation f(n) = ω(g(n)) 28
29 Properties Transitivity. If f = O(g) and g = O(h) then f = O(h). If f = Ω(g) and g = Ω(h) then f = Ω(h). If f = Θ(g) and g = Θ(h) then f = Θ(h). Additivity. If f = O(h) and g = O(h) then f + g = O(h). If f = Ω(h) and g = Ω(h) then f + g = Ω(h). If f = Θ(h) and g = O(h) then f + g = Θ(h). 29
30 Basic efficiency classes Class Name Comments 1 constant Rare case. And we need ignore log n logarithmic the input n linear Scan the list of the whole input nlog n n-log-n Many divide-and-conquer alg n 2 quadratic Algorithm with two embedded loops 2 n exponential Generate all subsets of a set 30
31 Why better algorithm 31
32 Solving Recurrences Like solving integrals, differential equations, etc. Three methods: Substitution, Iteration, Master Substitution Method (most general) Guess form of solution Verify by induction Solve for satisfying constant. 32
33 Example of substitution method proof Example: solve T(n) = 4T(n/2) + n Guess that T(n) O(n 3 ), i.e. that T of form cn 3. Assume T(k) ck 3 for k < n and Prove T(n) cn 3 by induction T(n)= 4T(n/2) + n 4c(n/2) 3 + n =cn 3 /2 + n = cn 3 - (cn 3 /2-n) cn 3 if c 2 and n 1 Thus T(n)=O(n 3 )! 33
34 Substitution Method (cont.) Notes: To prove inductive step, try to write expression as <answer you want> - <something> > 0 In example above: did not show that T(n)=Ω(n 3 ) or Θ(n 3 ) thus n 3 is not a tight bound! Can show O(n 2 )? 34
35 Substitution: Achieving Tighter Bounds Try to show T(n)=O(n 2 ) Assume T(k) ck 2 T(n)=4T(n/2)+n 4c(n/2) 2 +n =cn 2 +n cn 2 for no choice of c>0. Lose. What went wrong? 35
36 Fallacious argument The problem? We couldn t rewrite the equality as T(n) = cn 2 + n T(n) = cn 2 - <(something positive)> in order to show the inequality we wanted: T(n) cn 2 36
37 Corrected Proof Idea: Strength inductive hypothesis by subtracting lower-order term: Assume T(k) c 1 k 2 - c 2 k for k< n T(n)=4T(n/2)+n Given 4(c 1 (n/2) 2 -c 2 (n/2))+n Ind. Hyp. =c 1 n 2-2c 2 n+n simplify =c 1 n 2 -c 2 n-(c 2 n-n) Rearrange c 1 n 2 -c 2 n if c 2 1 Pick c 1 big enough to handle initial conditions. Thus T(n)=O(n 2 )! 37
38 Iterating recurrences Basic idea: expand, and convert to summation Example: T(n)=n+4T(n/2) =n+4(n/2+4t(n/4)) =n+4(n/2+4(n/4+4t(n/8))) =n+4(n/2+4(n/4+4(n/8+4t(n/16)))) =n+2n+4n+ +2 lgn T(1) =2 0 n+2 1 n+2 2 n+ +2 lgn T(1) lgn 1 = n 2 k= 0 k lgn Θ( n) = n ( n) Θ =Θ(n 2 )+Θ(n)=Θ(n 2 ) 38
39 To iterate recurrences Should know rules and have intuition for arithmetic and geometric series Math can be messy and hard. Often, use iterating recurrence to generate guess for substitution method. 39
40 Iteration method: Visualizing recursion tree Construction of recursion tree for T(n) = T(n/4) + T(n/2) + n 2 n 2 n 2 (n/4) 2 (n/2) 2 T(n/4) T(n/2) n 2 T(n/16) T(n/8) T(n/8) T(n/4) (n/4) 2 (n/2) 2 5n 2 /16 (n/16) 2 (n/8) 2 (n/8) 2 (n/4) 2 25n 2 /256 Θ(n 2 ) 40
41 Iteration method: Visualizing recursion tree Another recurrence: T(n) = T(n/3) + T(2n/3) + n Recursion tree: n n log 3/ 2 n n/3 2n/3 n n/9 2n/9 2n/9 4n/9 n Total: O(nlogn) 41
42 Master Theorem Let a 1 and b >1 be constants, let f(n) be a function, and let T(n) be defined on the nonnegative integers by the recurrence T(n) = at(n/b) + f(n) Then T(n) can be bounded asymptotically as follows. If f(n) = ( log b O n a ε ) for some constant ε > 0, then T( n) = Θ( n log a b If f(n) = Θ( n logb a ), then T(n) = ( log b a Θ n lgn) If f(n) = ( log b a Ω n + ε ) for some constant ε > 0, and if af ( n / b) cf ( n) for some constant c < 1 and all sufficiently large n, then T( n) = Θ( f ( n)) ) 42
43 Reference Books Knuth et al.. Concrete Mathematics Knuth. Selected Papers on Analysis of Algorithms Sedgewick and Flajolet. An Introduction to the Analysis of Algorithms 算法分析导论 - 塞奇威克 /dp/b0011bjvr0 Knuth. The Art of Computer Programming(TAOCP) Günter M. Ziegler and Martin Aigner. Proofs from THE BOOK 43
44 Next Class Sorting Order statistics Submit the hw1. 44
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