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Nattee Niparnan

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Recall What is the measurement of algorithm? How to compare two algorithms? Definition of Asymptotic Notation

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Today Topic

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Interesting Topics of Upper Bound Rule of thumb! We neglect Lower order terms from addition E.g. n 3 +n 2 = O(n 3 ) Constant E.g. 3n 3 = O(n 3 ) Remember that we use = instead of (more correctly)

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Why Discard Constant? From the definition We can use any constant E.g. 3n = O(n) Because When we let c >= 3, the condition is satisfied

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Why Discard Lower Order Term? Consider f(n) = n 3 +n 2 g(n) = n 3 If f(n) = O(g(n)) Then, for some c and n 0 c * g(n)-f(n) > 0 Definitely, just use any c >1

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Why Discard Lower Order Term? Try c = 1.1 Does 0.1n 3 -n 2 > 0 ? It is when 0.1n > 1 E.g., n > 10 1.1 * g(n)-f(n) = 0.1n 3 -n 2 0.1n 3 -n 2 > 0 0.1n 3 > n 2 0.1n 3 /n 2 > 1 0.1n > 1

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Lower Order only? In fact, Its only the dominant term that count Which one is dominating term? The one that grow faster Why? Eventually, it is g*(n)/f*(n) If g(n) grows faster, g(n)/f*(n) > some constant E.g, lim g(n)/f*(n) infinity The non- dominant term The dominant term

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What dominating what? nana n b (a > b) n log nn n 2 log nn log 2 n cncn ncnc Log n1 nlog n Left side dominates

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Putting into Practice What is the asymptotic class of 0.5n 3 +N 4 -5(n-3)(n-5)+n 3 log 8 n+25+n 1.5 (n-5)(n 2 +3)+log(n 20 ) 20n 5 +58n 4 +15n 3.2 *3n 2

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Putting into Practice What is the asymptotic class of 0.5n 3 +N 4 -5(n-3)(n-5)+n 3 log 8 n+25+n 1.5 (n-5)(n 2 +3)+log(n 20 ) 20n 5 +58n 4 +15n 3.2 *3n 2 O(n 4 ) O(n 3 ) O(n 5.4 )

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Asymptotic Notation from Program Flow Sequence Conditions Loops Recursive Call

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Sequence Block A Block B f (n) g (n) f(n) + g(n) = O(max (f(n),g(n))

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Example Block A Block B O(n) O(n 2 )

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Example Block A Block B (n) O(n 2 )

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Example Block A Block B (n) (n 2 )

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Example Block A Block B O(n 2 ) (n 2 )

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Condition Block ABlock B f (n)g (n) O(max (f(n),g(n))

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Loops for (i = 1;i

Example : While loops While (n > 0) { n = n - 10; } While (n > 0) { n = n - 10; } (n/10) = (n)

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Example : While loops While (n > 0) { n = n / 2; } While (n > 0) { n = n / 2; } (log n)

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Example : Euclids GCD function gcd(a, b) { while (b > 0) { tmp = b b = a mod b a = tmp } return a } function gcd(a, b) { while (b > 0) { tmp = b b = a mod b a = tmp } return a }

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Example : Euclids GCD function gcd(a, b) { while (b > 0) { tmp = b b = a mod b a = tmp } return a } function gcd(a, b) { while (b > 0) { tmp = b b = a mod b a = tmp } return a } How many iteration? Compute mod and swap Until the modding one is zero

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Example : Euclids GCD a b

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a b a mod b If a > b a mod b < a / 2 If a > b a mod b < a / 2

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Case 1: b > a / 2 a b a mod b

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Case 1: b a / 2 a b a mod b We can always put another b

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Example : Euclids GCD function gcd(a, b) { while (b > 0) { tmp = b b = a mod b a = tmp } return a } function gcd(a, b) { while (b > 0) { tmp = b b = a mod b a = tmp } return a } B always reduces at least half O( log n)

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Theorem If a i