Title goes here

Design and Analysis Algorithm

Drs. Achmad Ridok M.KomFitra A. Bachtiar, S.T., M. EngImam Cholissodin, S.Si., M.KomAryo Pinandito, MT

Pertemuan 04Contents2Asymptotic Notation

2CS3024-FAZ3ContentsAsymptotic Notations:O (big oh) (big omega) (big theta)Basic Efficiency Classes3CS3024-FAZ4In the following discussiont(n) & g(n): any nonnegative functions defined on the set of natural numbers

t(n) an algorithms running timeUsually indicated by its basic operation count C(n)g(n) some simple function to compare the count withCS3024-FAZ5O(g(n)): InformallyO(g(n)) is a set of all functions with a smaller or same order of growth as g(n)Examples:n O(n2); 100n + 5 O(n2) n (n-1) O(n2)n3 O(n2); 0.0001 n3 O(n2); n4+n+1 O(n2)CS3024-FAZ6(g(n)): Informally(g(n)) is a set of all functions with a larger or same order of growth as g(n)Examples:n3 (n2) n (n-1) (n2)100n + 5 (n2)CS3024-FAZ7(g(n)): Informally(g(n)) is a set of all functions with a same order of growth as g(n)Examples:an2+bn+c; a>0 (n2); n2+sin n (n2) n (n-1) (n2); n2+log n (n2)100n + 5 (n2); n3 (n2)=CS3024-FAZ8O-notation: FormallyDEF1: A function t(n) is said to be in O(g(n)), denoted t(n) O(g(n)), if t(n) is bounded above by some constant multiple of g(n) for all large n

i.e. there exist some positive constant c and some nonnegative integer n0, such thatt(n) cg(n) for all n n0CS3024-FAZ9t(n) O(g(n)): Illustration

CS3024-FAZ10Proving Example: 100n + 5 O(n2)Remember DEF1: find c and n0, such that t(n) cg(n) for all n n0

100n + 5 100n + n (for all n 5) = 101n 101n2 c=101, n0=5100n + 5 100n + 5n (for all n 1) = 105n 105n2 c=105, n0=1

CS3024-FAZ11-notation: FormallyDEF2: A function t(n) is said to be in (g(n)), denoted t(n) (g(n)), if t(n) is bounded below by some constant multiple of g(n) for all large n

i.e. there exist some positive constant c and some nonnegative integer n0, such thatt(n) cg(n) for all n n0CS3024-FAZ12t(n) (g(n)): Illustration

CS3024-FAZ13Proving Example: n3 (n2)Remember DEF2: find c and n0, such that t(n) cg(n) for all n n0

n3 n2 (for all n 0) c=1, n0=0

CS3024-FAZ14-notation: FormallyDEF3: A function t(n) is said to be in (g(n)), denoted t(n) (g(n)), if t(n) is bounded both above and below by some constant multiple of g(n) for all large n

i.e there exist some positive constant c1 and c2 and some nonnegative integer n0, such thatc2g(n) t(n) c1g(n) for all n n0CS3024-FAZ15t(n) (g(n)): Illustration

CS3024-FAZ16Proving Example: n(n-1) (n2)Remember DEF3: find c1 and c2 and some nonnegative integer n0, such that c2g(n) t(n) c1g(n) for all n n0

The upper bound: n(n-1) = n2 n n2 (for all n 0)The lower bound: n(n-1) = n2 n n2 - n n (for all n 2) = n2 c1 = , c2 = , n0 = 2CS3024-FAZ17Useful PropertyTheorem: If t1(n) O(g1(n)) and t2(n) O(g2(n)), then t1(n) + t2(n) O(max{g1(n), g2(n)})The analogous assertions are true for the and notations as wellCS3024-FAZ18ExampleAlg to check whether an array has identical elements:Sort the array Scan the sorted array to check its consecutive elements for equality(1) = n(n-1) comparison O(n2)(2) = n-1 comparison O(n)The efficiency of (1)+(2) = O(max{n2,n}) = O(n2)CS3024-FAZ19Using Limits for Comparing OoGA convenient method for comparing order of growth of two specific functionsThree principal cases:

The first two cases t(n) O(g(n)); the last two cases t(n) (g(n)); the second case alone t(n) (g(n))

CS3024-FAZ20Limit-based: why convenient?It can take advantage of the powerful calculus techniques developed for computing limits, such asLHopitals rule

Stirlings formula

CS3024-FAZ21Example (1)Compare OoG of n(n-1) and n2.

The limit = c n(n-1) (n2 )

Compare OoG of log2n and n

The limit = 0 log2n has smaller order of n

CS3024-FAZ22Example (2)Compare OoG of n! and 2n.

The limit = n! (2n )

Asymptotic NotationThink of n as the number of records we wish to sort with an algorithm that takes f(n) to run. How long will it take to sort n records?What if n is big?We are interested in the range of a function as n gets large.Will f stay bounded?Will f grow linearly?Will f grow exponentially?Our goal is to find out just how fast f grows with respect to n.23Asymptotic NotationMisalkan:T(n) = 5n2 + 6n + 25T(n) proporsional untuk ordo n2 untuk data yang sangat besar.

24Memperkirakan formula untuk run-timeIndikasi kinerja algoritma (untuk jumlah data yang sangat besar)

Asymptotic NotationIndikator efisiensi algoritma bedasar pada OoG pada basic operation suatu algoritma. Penggunaan notasi sebagai pembanding urutan OoG:O (big oh) (big omega) (big theta)t(n) : algoritma running time (diindikasikan dengan basic operation count (C(n))g(n) : simple function to compare the count25Classifying functions by theirAsymptotic Growth Rates (1/2)asymptotic growth rate, asymptotic order, or order of functions Comparing and classifying functions that ignores constant factors and small inputs.

O(g(n)), Big-Oh of g of n, the Asymptotic Upper Bound;W(g(n)), Omega of g of n, the Asymptotic Lower Bound.Q(g(n)), Theta of g of n, the Asymptotic Tight Bound; and

ExampleExample: f(n) = n2 - 5n + 13.The constant 13 doesn't change as n grows, so it is not crucial. The low order term, -5n, doesn't have much effect on f compared to the quadratic term, n2.We will show that f(n) = Q(n2) .

Q: What does it mean to say f(n) = Q(g(n)) ?A: Intuitively, it means that function f is the same order of magnitude as g.Example (cont.)Q: What does it mean to say f1(n) = Q(1)?A: f1(n) = Q(1) means after a few n, f1 is bounded above & below by a constant.Q: What does it mean to say f2(n) = Q(n log n)?A: f2(n) = Q(n log n) means that after a few n, f2 is bounded above and below by a constant times n log n. In other words, f2 is the same order of magnitude as n log n.More generally, f(n) = Q(g(n)) means that f(n) is a member of Q(g(n)) where Q(g(n)) is a set of functions of the same order of magnitude. Big-OhThe O symbol was introduced in 1927 to indicate relative growth of two functions based on asymptotic behavior of the functions now used to classify functions and families of functions

Upper Bound NotationWe say Insertion Sorts run time is O(n2)Properly we should say run time is in O(n2)Read O as Big-O (youll also hear it as order)

In general a functionf(n) is O(g(n)) if positive constants c and n0 such that f(n) c g(n) n n0

e.g. if f(n)=1000n and g(n)=n2, n0 > 1000 and c = 1 then f(n0) < 1.g(n0) and we say that f(n) = O(g(n))

Asymptotic Upper Boundf(n)g(n)c g(n) f(n) c g(n) for all n n0 g(n) is called an asymptotic upper bound of f(n). We write f(n)=O(g(n)) It reads f(n) is big oh of g(n).

n031We use O-notation to give an upper bound of a function, to within a constant factor.Big-Oh, the Asymptotic Upper BoundThis is the most popular notation for run time since we're usually looking for worst case time. If Running Time of Algorithm X is O(n2) , then for any input the running time of algorithm X is at most a quadratic function, for sufficiently large n.

e.g. 2n2 = O(n3) .From the definition using c = 1 and n0 = 2. O(n2) is tighter than O(n3).6g(n)f(n)for all n>6, g(n) > 1 f(n). Thus the function f is in the big-O of g.that is, f(n) in O(g(n)).Example 1g(n)f(n)5There exists a n0=5 s.t. for all n>n0, f(n) < 1 g(n).Thus, f(n) is in O(g(n)).Example 2There exists a n0=5, c=3.5, s.t. for all n>n0, f(n) < c h(n).Thus, f(n) is in O(h(n)).5h(n)f(n)3.5 h(n)Example 3Example of Asymptotic Upper Boundf(n)=3n2+5g(n)=n24g(n)=4n24 g(n) = 4n2 = 3n2 + n2 3n2 + 9 for all n 3 > 3n2 + 5 = f(n)Thus, f(n)=O(g(n)).

336There could be many pairs of c,n0 which leads to f(n)=O(g(n))Exercise on O-notationShow that 3n2+2n+5 = O(n2)

10 n2 = 3n2 + 2n2 + 5n2 3n2 + 2n + 5 for n 1

c = 10, n0 = 1Classification of Function : BIG O (1/2)A function f(n) is said to be of at most logarithmic growth if f(n) = O(log n)A function f(n) is said to be of at most quadratic growth if f(n) = O(n2)A function f(n) is said to be of at most polynomial growth if f(n) = O(nk), for some natural number k > 1A function f(n) is said to be of at most exponential growth if there is a constant c, such that f(n) = O(cn), and c > 1A function f(n) is said to be of at most factorial growth if f(n) = O(n!).Classification of Function : BIG O (2/2)A function f(n) is said to have constant running time if the size of the input n has no effect on the running time of the algorithm (e.g., assignment of a value to a variable). The equation for this algorithm is f(n) = cOther logarithmic classifications: f(n) = O(n log n)f(n) = O(log log n)Review Tugas n!Menghitung kompleksitas pada FaktorialFunction Faktorial (input n : integer) integer {menghasilkan nilai n!, n 0} Algoritma If n=0 then Return 1Else Return n*faktorial (n-1) EndifKompleksitas waktu : untuk kasus basis, tidak ada operasi perkalian (0) untuk kasus rekurens, kompleksitas waktu diukur dari jumlah perkalian (1) ditambah kompleksitas waktu untuk faktorial (n-1)

Review Tugas n! (Lanjutan)Kompleksitas waktu n! :T(n)=1+T(n-1)=T(n)=1+1+T(n-2)=2+T(n-2)=T(n)=2+1+T(n-3)=3+T(n-3)= = = n+T(0)= n + 0 Jadi T(n) = n T(n) O(n)

Lower Bound NotationWe say InsertionSorts run time is (n)In general a functionf(n) is (g(n)) if positive constants c and n0 such that 0 cg(n) f(n) n n0Proof:Suppose run time is an + bAssume a and b are positive (what if b is negative?)an an + bBig Asymptotic Lower Boundf(n)c g(n) f(n) c g(n) for all n n0 g(n) is called an asymptotic lower bound of f(n). We write f(n)=(g(n)) It reads f(n) is omega of g(n).

n043We use O-notation to give an upper bound of a function, to within a constant factor.Example of Asymptotic Lower Boundf(n)=n2/2-7c g(n)=n2/4g(n)=n2g(n)/4 = n2/4 = n2/2 n2/4 n2/2 9 for all n 6 < n2/2 7Thus, f(n)= (g(n)).

6g(n)=n244There could be many pairs of c,n0 which leads to f(n)=O(g(n))Example: Big OmegaExample: n 1/2 = W( log n) . Use the definition with c = 1 and n0 = 16.

Checks OK.Let n 16 : n 1/2 (1) log n if and only if n = ( log n )2 by squaring both sides.This is an example of polynomial vs. log.Big Theta NotationDefinition: Two functions f and g are said to be of equal growth, f = Big Theta(g) if and only if both f=Q(g) and g = Q(f).

Definition: f(n) = Q(g(n)) means positive constants c1, c2, and n0 such that c1 g(n) f(n) c2 g(n) n n0

If f(n) = O(g(n)) and f(n) = W(g(n)) then f(n) = Q(g(n)) (e.g. f(n) = n2 and g(n) = 2n2)Theta, the Asymptotic Tight BoundTheta means that f is bounded above and below by g; BigTheta implies the "best fit".f(n) does not have to be linear itself in order to be of linear growth; it just has to be between two linear functions,Asymptotically Tight Boundf(n)c1 g(n) f(n) = O(g(n)) and f(n) = (g(n)) g(n) is called an asymptotically tight bound of f(n). We write f(n)=(g(n)) It reads f(n) is theta of g(n).

n0c2 g(n)48We use O-notation to give an upper bound of a function, to within a constant factor.Other Asymptotic NotationsA function f(n) is o(g(n)) if positive constants c and n0 such that f(n) < c g(n) n n0A function f(n) is (g(n)) if positive constants c and n0 such that c g(n) < f(n) n n0Intuitively,o() is like < O() is like () is like > () is like () is like =Examples1. 2n3 + 3n2 + n = 2n3 + 3n2 + O(n) = 2n3 + O( n2 + n) = 2n3 + O( n2 ) = O(n3 ) = O(n4)

2. 2n3 + 3n2 + n = 2n3 + 3n2 + O(n) = 2n3 + Q(n2 + n) = 2n3 + Q(n2) = Q(n3)Example (cont.)n3 = 503 * 729 3n = 350 * 729n = n = log3 (729 * 350)n = n = log3(729) + log3 350n = 50 * 9 n = 6 + log3 350n = 50 * 9 = 450 n = 6 + 50 = 56

Improvement: problem size increased by 9 times for n3 algorithm but only a slight improvement in problem size (+6) for exponential algorithm.

More Examples0.5n2 - 5n + 2 = ( n2). Let c = 0.25 and n0 = 25.0.5 n2 - 5n + 2 = 0.25( n2) for all n = 25

(b) 0.5 n2 - 5n + 2 = O( n2). Let c = 0.5 and n0 = 1. 0.5( n2) = 0.5 n2 - 5n + 2 for all n = 1

(c) 0.5 n2 - 5n + 2 = ( n2) from (a) and (b) above. Use n0 = 25, c1 = 0.25, c2 = 0.5 in the definition.More Examples(d) 6 * 2n + n2 = O(2n). Let c = 7 and n0 = 4. Note that 2n = n2 for n = 4. Not a tight upper bound, but it's true.

(e) 10 n2 + 2 = O(n4). There's nothing wrong with this, but usually we try to get the closest g(n). Better is to use O(n2 ).Practical Complexity t < 250

Practical Complexity t < 500

Practical Complexity t < 1000

Practical Complexity t < 5000

CS3024-FAZ58Tugas (1)True or false: n(n+1)/2 O(n3)n(n+1)/2 O(n2)n(n+1)/2 (n3)n(n+1)/2 (n)Indicate the class (g(n)):(n2+1)10(10n2+7n+3) 2n log (n+2)2+(n+2)2 log (n/2)N^2/2 +n/2 < (n^2/2+n/2)n , 58Tugas 1 : O-notation3. Tentukan OoG dari masing-masing soalf1(n) = 10 n + 25 n2f2(n) = 20 n log n + 5 nf3(n) = 12 n log n + 0.05 n2f4(n) = n1/2 + 3 n log n

4.. True/false ?0.25n2 - 5n + 2 = ( n2). (b) 0.25n2 - 5n + 2 = O( n2). (c) 0.25n2 - 5n + 2 = ( n2).

O(n2)O(n log n)O(n2) O(n log n)

59For each function, find the simplest and g(n) such that f_i(n) = O(g(n))Tugas KelompokKerjakan soal di hal 29 no 2.2-1 sd. 2.2-4Tugas 2 s.d Tugas 6 di slide iniPengumpulan :Tulis dikertas folio bergarisDikumpulkan minggu depan di kelasKODE TUGAS : DAA_A_1_1 (MT DAA, kelas A, Kelompok1, tugas ke-1)DAA_D_5_1

Tugas 2

Tugas 3

Tugas 4

CS3024-FAZ64Tugas 55. Prove that every polynomial p(n) = aknk + ak-1nk-1 + + a0 with ak > 0belongs to (nk)

6. Prove that exponential functions an have different orders of growth for different values of base a > 0

Tugas 6: Examples (cont.)7. Suppose a program P is O(n3), and a program Q is O(3n), and that currently both can solve problems of size 50 in 1 hour. If the programs are run on another system that executes exactly 729 times as fast as the original system, what size problems will they be able to solve?Click to edit subtitle styleThank You !n

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f(n) = nf(n) = log(n)f(n) = n log(n)f(n) = n^2f(n) = n^3f(n) = 2^n

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f(n) = nf(n) = log(n)f(n) = n log(n)f(n) = n^2f(n) = n^3f(n) = 2^n

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f(n) = nf(n) = log(n)f(n) = n log(n)f(n) = n^2f(n) = n^3f(n) = 2^n

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Sheet3

Chart310011221248431.58496250074.7548875022927842816641652.321928094911.6096404744251253262.584962500715.5097750043362166472.807354922119.65148445444934312883246451225693.169925001428.52932501381729512103.321928094933.219280948910010001024113.459431618638.05374780512113312048123.584962500743.019550008714417284096133.700439718148.105716335816921978192143.807354922153.3029689088196274416384153.906890595658.603358934122533753276816464256409665536174.087462841369.48686830132894913131072184.169925001475.0586500263245832262144194.247927513480.71062275543616859524288204.321928094986.438561897740080001048576

f(n) = nf(n) = log(n)f(n) = n log(n)f(n) = n^2f(n) = n^3f(n) = 2^n

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