Relevant Mathematics Jeff Edmonds York University COSC 3101 Lecture skipped. Lecture skipped. (Done when needed) Logic Quantifiers The Time Complexity.

Relevant Mathematics Jeff Edmonds York University COSC 3101 Lecture skipped. Lecture skipped. (Done when needed) Logic Quantifiers The Time Complexity of an Algorithm Classifying Functions Adding Made Easy Recurrence Relations In More Details

Some Math Time Complexity t(n) =  (n 2 ) Logic Quantifiers  g  b Loves(b,g)  b  g Loves(b,g) In Iterative Algs (GCD) See logic slides

Some Math Logs and Exps 2 a × 2 b = 2 a+b 2 log n = n You are on your own. Input Size Time Classifying Functions f(i) = n  (n) See logic slides

Some Math Recurrence Relations T(n) = a T(n/b) + f(n) Adding Made Easy ∑ i=1 f(i). In Iterative Algs (Insertion Sort) In Recursive Algs (Multiplying) In Recursive Algs (Multiplying, Towers of Hanoi, Merge Sort)

The Time Complexity of an Algorithm Specifies how the running time depends on the size of the input.

Purpose?

Purpose To estimate how long a program will run. To estimate the largest input that can reasonably be given to the program. To compare the efficiency of different algorithms. To help focus on the parts of code that are executed the largest number of times. To choose an algorithm for an application.

Time Complexity Is a Function Specifies how the running time depends on the size of the input. A function mapping “size” of input “time” T(n) executed.

Definition of Time?

Definition of Time # of seconds (machine dependent). # lines of code executed. # of times a specific operation is performed (e.g., addition). Which?

Definition of Time # of seconds (machine dependent). # lines of code executed. # of times a specific operation is performed (e.g., addition). These are all reasonable definitions of time, because they are within a constant factor of each other.

Size of Input Instance? 83920

Size of Input Instance Size of paper # of bits # of digits Value - n = 2 in 2 - n = 17 bits - n = 5 digits - n = 83920 2’’ 83920 5 1’’ Which are reasonable?

Size of Input Instance Size of paper # of bits # of digits Value - n = 2 in 2 - n = 17 bits - n = 5 digits - n = 83920 Intuitive 2’’ 83920 5 1’’

Size of Input Instance Size of paper # of bits # of digits Value - n = 2 in 2 - n = 17 bits - n = 5 digits - n = 83920 Intuitive Formal 2’’ 83920 5 1’’

Size of Input Instance Size of paper # of bits # of digits Value - n = 2 in 2 - n = 17 bits - n = 5 digits - n = 83920 Intuitive Formal Reasonable # of bits = 3.32 * # of digits 2’’ 83920 5 1’’

Size of Input Instance Size of paper # of bits # of digits Value - n = 2 in 2 - n = 17 bits - n = 5 digits - n = 83920 Intuitive Formal Reasonable Unreasonable # of bits = log 2 (Value) Value = 2 # of bits 2’’ 83920 5 1’’

Two Example Algorithms Sum N array entries: A(1) + A(2) + A(3) + … Factor value N: Input N=5917 & Output N=97*61. Algorithm N/2, N/3, N/4, …. ? Time? Time = N Is this reasonable? No! One is considered fast and the other slow!

Two Example Algorithms Sum N array entries: A(1) + A(2) + A(3) + … Factor value N: Input N=5917 & Output N=97*61. Algorithm N/2, N/3, N/4, …. ? Standard input size? Time = N N = hard drive = 60G < 2 36 N = crypto key = 2 100

Two Example Algorithms Sum N array entries: A(1) + A(2) + A(3) + … Factor value N: Input N=5917 & Output N=97*61. Algorithm N/2, N/3, N/4, …. ? Size of Input Instance? Time = N size n = N size n = log N

Two Example Algorithms Sum N array entries: A(1) + A(2) + A(3) + … Factor value N: Input N=5917 & Output N=97*61. Algorithm N/2, N/3, N/4, …. ? size n = N size n = log N Time as function of input size? Time for you to solve problem instance as a function of time for me to give you the problem instance. = n = 2 n Time = N

Two Example Algorithms Sum N array entries: A(1) + A(2) + A(3) + … Factor value N: Input N=5917 & Output N=97*61. Algorithm N/2, N/3, N/4, …. ? Linear vs Exponential Time! size n = N size n = log N = n = 2 n Time = N

Size of Input Instance? 14,23,25,30,31,52,62,79,88,98

Size of Input Instance # of elements = n = 10 elements 14,23,25,30,31,52,62,79,88,98 10

Size of Input Instance # of elements = n = 10 elements 14,23,25,30,31,52,62,79,88,98 10 Is this reasonable?

Size of Input Instance # of elements = n = 10 elements Reasonable 14,23,25,30,31,52,62,79,88,98 10 If each element has size c # of bits = c * # of elements

Size of Input Instance # of elements = n = 10 elements Reasonable 14,23,25,30,31,52,62,79,88,98 10 If each element is in [1..n] each element has size log n # of bits = n log n ≈ n ~

Time Complexity Is a Function Specifies how the running time depends on the size of the input. A function mapping # of bits n needed to represent the input # of operations T(n) executed.

Which Input of size n? There are 2 n inputs of size n. Which do we consider for the time T(n)?

Which Input of size n? Typical Input Average Case Worst Case

Which Input of size n? Typical InputBut what is typical? Average CaseFor what distribution? Worst CaseTime for all inputs is bound. Easiest to compute

What is the height of tallest person in the class? Bigger than this? Need to look at only one person Need to look at every person Smaller than this?

Time Complexity of Algorithm O(n 2 ): Prove that for every input of size n, the algorithm takes no more than cn 2 time. Ω(n 2 ): Find one input of size n, for which the algorithm takes at least this much time. θ (n 2 ): Do both. The time complexity of an algorithm is the largest time required on any input of size n.

Time Complexity of Problem O(n 2 ): Provide an algorithm that solves the problem in no more than this time. Ω(n 2 ): Prove that no algorithm can solve it faster. θ (n 2 ): Do both. The time complexity of a problem is the time complexity of the fastest algorithm that solves the problem.

Classifying Functions T(n)T(n) 101001,00010,000 log n 36913amoeba n 1/2 31031100bird 101001,00010,000human n log n 306009,000130,000my father n2n2 10010,00010 6 10 8 elephant n3n3 1,00010 6 10 9 10 12 dinosaur 2n2n 1,02410 30 10 300 10 3000 the universe Note: The universe contains approximately 10 50 particles. n

Classifying Functions Functions Poly Logarithmic Polynomial Exponential ExpDouble Exp Constant (log n) 5 n5n5 2 5n 5 2 n5n5 2 << (log n) θ(1) n θ(1) 2 θ(n) θ(1)2 n θ(1) 2 θ(n) 2

Classifying Functions LinearQuadratic Cubic ? θ(n 2 ) θ(n) θ(n 3 ) Polynomial = n θ(1) θ(n 4 ) Others θ(n 3 log 7 (n)) log(n) not absorbed because not Mult-constant

Classifying Functions Functions Poly. Exp. 8 · 2 4n / n 100 θ(2 4n / n 100 ) 8 · 2 4n + n 3 n θ(1) 2 θ(n) θ(2 4n ) 8·n28·n2 θ(n 2 ) θ(n 3 log(n)) 7 · n 3 log(n) We consider two (of many) levels in this hierarchy Individuals Ignore Mult-constant Ignore Power-constant

BigOh and Theta? 5n 2 + 8n + 2log n =  (n 2 ) Drop low-order terms. Drop multiplicative constant. 5n 2 log n + 8n + 2log n =  (n 2 log n)

Which Functions are Polynomials? n c n 0.0001 n 10000 5n 2 + 8n + 2log n 5n 2 log n 5n 2.5 Drop low-order terms. Drop constant, log, (and poly) factors. Ignore power constant. Write n θ(1)

Which Functions are Exponential? 2 n 2 0.0001 n 2 10000 n 8 n 2 n / n 100 2 n · n 100 Drop low-order terms. Drop constant, log, and poly factors. Ignore power constant. Write 2 θ(n)

Notations Theta f(n) = θ(g(n))f(n) ≈ c g(n) BigOh f(n) = O(g(n))f(n) ≤ c g(n) Omega f(n) = Ω(g(n))f(n) ≥ c g(n) Little Oh f(n) = o(g(n))f(n) << c g(n) Little Omega f(n) = ω(g(n))f(n) >> c g(n)

Definition of Theta f(n) = θ(g(n))

Definition of Theta f(n) is sandwiched between c 1 g(n) and c 2 g(n) f(n) = θ(g(n))

Definition of Theta f(n) is sandwiched between c 1 g(n) and c 2 g(n) for some sufficiently small c 1 (= 0.0001) for some sufficiently large c 2 (= 1000) f(n) = θ(g(n))

Definition of Theta For all sufficiently large n f(n) = θ(g(n))

Definition of Theta For all sufficiently large n For some definition of “sufficiently large” f(n) = θ(g(n))

Definition of Theta 3n 2 + 7n + 8 = θ(n 2 )

Order of Quantifiers No! It cannot be a different c 1 and c 2 for each n. f(n) = θ(g(n))

Gauss ∑ i=1..n i = 1 + 2 + 3 +... + n = ? Arithmetic Sum

1 2........ n = number of white dots 1+2+3+...+n-1+n=S Adding Made Easy

1+2+3+...+n-1+n=S n+ +n-2+... +2+1=S 1 2........ n = number of yellow dots n....... 2 1 = number of white dots Adding Made Easy

1+2+3+...+n-1+n=S n+ +n-2+... +2+1=S = number of white dots = number of yellow dots n+1 n+1 n+1 n+1 n+1 n n n n n n = n(n+1) dots in the grid 2S dots S = n (n+1) 2 Adding Made Easy

Gauss ∑ i=1..n i = 1 + 2 + 3 +... + n =  (# of terms · last term) Arithmetic Sum True when ever terms increase slowly

∑ i=0..n 2 i = 1 + 2 + 4 + 8 +... + 2 n = ? Geometric Sum

∑ i=0..n 2 i = 1 + 2 + 4 + 8 +... + 2 n = 2 · last term - 1 Geometric Sum

∑ i=0..n r i = r 0 + r 1 + r 2 +... + r n = ? Geometric Sum

S Srrrr...rr S 1r S r1 r1 23nn1 n1 n1  1rrr... 23 r n          Geometric Sum

∑ i=0..n r i r1 r1 n1     Geometric Sum When r>1?

∑ i=0..n r i r1 r1 n1     Geometric Sum θ(rn)θ(rn) Biggest Term When r>1

∑ i=0..n r i = r 0 + r 1 + r 2 +... + r n =  (biggest term) Geometric Increasing True when ever terms increase quickly

∑ i=0..n r i  Geometric Sum When r<1? 1 r n1   

∑ i=0..n r i  Geometric Sum 1 r n1     θ(1) When r<1 Biggest Term

∑ i=0..n r i = r 0 + r 1 + r 2 +... + r n =  (1) Bounded Tail True when ever terms decrease quickly

∑ i=1..n 1 / i = 1 / 1 + 1 / 2 + 1 / 3 + 1 / 4 + 1 / 5 + …+ 1 / n = ? Harmonic Sum

f(i) = 1 ∑ i=1..n f(i) = n n Sum of Shrinking Function

f(i) = ? ∑ i=1..n f(i) = n 1/2 n Sum of Shrinking Function

f(i) = 1/2 i ∑ i=1..n f(i) = 2  Sum of Shrinking Function

f(i) = 1/i ∑ i=1..n f(i) = ? n Sum of Shrinking Function

Harmonic Sum

∑ i=1..n 1 / i = 1 / 1 + 1 / 2 + 1 / 3 + 1 / 4 + 1 / 5 + …+ 1 / n =  (log(n)) Harmonic Sum

Approximating Sum by Integrating The area under the curve approximates the sum ∑ i=1..n f(i) ≈ ∫ x=1..n f(x) dx

Adding Made Easy (For +, -, ×, , exp, log functions f(n))

Adding Made Easy

Geometric Like: If f(n)  2 Ω(n), then ∑ i=1..n f(i) = θ(f(n)). Arithmetic Like: If f(n) = n θ(1)-1, then ∑ i=1..n f(i) = θ(n · f(n)). Harmonic: If f(n) = 1 / n, then ∑ i=1..n f(i) = log e n + θ(1). Bounded Tail: If f(n)  n -1-Ω(1), then ∑ i=1..n f(i) = θ(1). ( For +, -, , , exp, log functions f(n) ) This may seem confusing, but it is really not. It should help you compute most sums easily. Adding Made Easy

Recurrence Relations T(1) = 1 T(n) = a T(n/b) + f(n)

Recurrence Relations  Time of Recursive Program With all this recursing and looping, how do we determine the running time of this recursive program? procedure Eg(I n ) n = |I n | if(n  1) then put “Hi” else loop i=1..f(n) put “Hi” loop i=1..a I n/b = I n cut in b pieces Eg(I n/b ) T(1) = 1 T(n) = a T(n/b) + f(n) Recurrence relations.

Recurrence Relations  Time of Recursive Program Let n be the “size” of our input. procedure Eg(I n ) n = |I n | if(n  1) then put “Hi” else loop i=1..f(n) put “Hi” loop i=1..a I n/b = I n cut in b pieces Eg(I n/b ) T(1) = 1 T(n) = a T(n/b) + f(n)

Recurrence Relations  Time of Recursive Program Let T(n) be the # of “Hi”s printed by me procedure Eg(I n ) n = |I n | if(n  1) then put “Hi” else loop i=1..f(n) put “Hi” loop i=1..a I n/b = I n cut in b pieces Eg(I n/b ) T(1) = 1 T(n) = a T(n/b) + f(n)

Recurrence Relations  Time of Recursive Program Let T(n) be the # of “Hi”s printed by me And by all of my friends. procedure Eg(I n ) n = |I n | if(n  1) then put “Hi” else loop i=1..f(n) put “Hi” loop i=1..a I n/b = I n cut in b pieces Eg(I n/b ) T(1) = 1 T(n) = a T(n/b) + f(n)

If my input is sufficiently small (according to my measure of size) then I print  (1) “Hi”s. Recurrence Relations  Time of Recursive Program procedure Eg(I n ) n = |I n | if(n  1) then put “Hi” else loop i=1..f(n) put “Hi” loop i=1..a I n/b = I n cut in b pieces Eg(I n/b ) T(1) = 1 T(n) = a T(n/b) + f(n)

I personally output f(n) “Hi”s, i.e. top level stack frame. Recurrence Relations  Time of Recursive Program procedure Eg(I n ) n = |I n | if(n  1) then put “Hi” else loop i=1..f(n) put “Hi” loop i=1..a I n/b = I n cut in b pieces Eg(I n/b ) T(1) = 1 T(n) = a T(n/b) + f(n)

I have a friends i.e. recurse a times. Recurrence Relations  Time of Recursive Program procedure Eg(I n ) n = |I n | if(n  1) then put “Hi” else loop i=1..f(n) put “Hi” loop i=1..a I n/b = I n cut in b pieces Eg(I n/b ) T(1) = 1 T(n) = a T(n/b) + f(n)

Let b be the factor by which I shrink the input before giving it to a friend. i.e. if my input has size n then my friends are of size n/b. Recurrence Relations  Time of Recursive Program procedure Eg(I n ) n = |I n | if(n  1) then put “Hi” else loop i=1..f(n) put “Hi” loop i=1..a I n/b = I n cut in b pieces Eg(I n/b ) T(1) = 1 T(n) = a T(n/b) + f(n)

By definition of T, the number of “Hi”s printed by one of my friends and all of his friends is T(n/b). Recurrence Relations  Time of Recursive Program procedure Eg(I n ) n = |I n | if(n  1) then put “Hi” else loop i=1..f(n) put “Hi” loop i=1..a I n/b = I n cut in b pieces Eg(I n/b ) T(1) = 1 T(n) = a T(n/b) + f(n)

Hence, the total number printed by me and my a friends is T(n) = a T(n/b) + f(n) Recurrence Relations  Time of Recursive Program procedure Eg(I n ) n = |I n | if(n  1) then put “Hi” else loop i=1..f(n) put “Hi” loop i=1..a I n/b = I n cut in b pieces Eg(I n/b ) T(1) = 1 T(n) = a T(n/b) + f(n)

Evaluating: T(n) = aT(n/b)+f(n) Level Instance size Work in stack frame # stack frames Work in Level 0n f(n)f(n) 1 1 · f(n) 1 n/bn/bf(n/b)f(n/b) a a · f(n/b)a · f(n/b) 2 n/b2n/b2 f(n/b2)f(n/b2) a2a2 a 2 · f(n/b 2 ) i n/bin/bi f(n/bi)f(n/bi) aiai a i · f(n/b i ) h = log n / log b n/bhn/bh T(1) n log a / log b n · T(1) log a / log b Dominated by Top Level or Base Cases

Evaluating: T(n) = aT(n/b)+f(n)

Merge Sort Sort Time: T(n) = =  (n log(n)) 2T(n/2) +  (n) Total work at any level =  (n) # levels =  (log(n))

Evaluating: T(n) = aT(n/b)+f(n) Level Instance size Work in stack frame # stack frames Work in Level 0n f(n)f(n) 1 1 · f(n) 1 n/bn/bf(n/b)f(n/b) a a · f(n/b) 2 n/b2n/b2 f(n/b2)f(n/b2) a2a2 a 2 · f(n/b 2 ) i n/bin/bi f(n/bi)f(n/bi) aiai a i · f(n/b i ) h = log n / log b n/bhn/bh T(1) n log a / log b n · T(1) log a / log b Total Work T(n) = ∑ i=0..h a i  f(n/b i ) 2T(n/2) +  (n) 2 i ·  (n/2 i ) =  (n) =  (n) · log(n) All levels basically the same.

Logs log 27 log 9 =

Logs log ? 27 log ? 9 = Which base?

Logs log 2 27 log 2 9 = It does not matter. log 2 c · log c 27 · log c 9 = log c 27 log c 9 Which is easiest?

Logs log 3 27 log 3 9 = log 3 3 3 log 3 3 2 = 3 2 Please no calculators in exams. And I wont ask log 25 log 9

Evaluating: T(n) = 4T(n/2)+ n 3 / log 5 n

Time for top level?: Evaluating: T(n) = 4T(n/2)+ n 3 / log 5 n

Time for top level: f(n) = n 3 / log 5 n Time for base cases?: Evaluating: T(n) = 4T(n/2)+ n 3 / log 5 n

Time for top level: f(n) = n 3 / log 5 n Time for base cases: Evaluating: T(n) = 4T(n/2)+ n 3 / log 5 n θ(n ) = log a / log b θ(n ) log ? / log ?

Time for top level: f(n) = n 3 / log 5 n Time for base cases: Evaluating: T(n) = 4T(n/2)+ n 3 / log 5 n θ(n ) = log a / log b θ(n ) = ? log 4 / log 2

Time for top level: f(n) = n 3 / log 5 n Time for base cases: Dominated?: c = ? Evaluating: T(n) = 4T(n/2)+ n 3 / log 5 n θ(n ) = log a / log b θ(n ) = θ(n 2 ) log 4 / log 2 log a / log b = ?

Time for top level: f(n) = n 3 / log 5 n Time for base cases: Dominated?: c = 3 > 2 = log a / log b Evaluating: T(n) = 4T(n/2)+ n 3 / log 5 n θ(n ) = log a / log b θ(n ) = θ(n 2 ) log 4 / log 2 Hence, T(n) = ?

Time for top level: f(n) = n 3 / log 5 n Time for base cases: Dominated?: c = 3 > 2 = log a / log b Evaluating: T(n) = 4T(n/2)+ n 3 / log 5 n θ(n ) = log a / log b θ(n ) = θ(n 2 ) log 4 / log 2 Hence, T(n) = θ(top) = θ(f(n)) = θ(n 3 / log 5 n ).

Time for top level: f(n) = 2 n Time for base cases: Dominated?: c = ? > 2 = log a / log b Evaluating: T(n) = 4T(n/2)+ 2 n θ(n ) = log a / log b θ(n ) = θ(n 2 ) log 4 / log 2 Hence, T(n) = θ(top) = θ(f(n)) = θ(2 n ). bigger >big The time is even more dominated by the top level of recursion

Time for top level: f(n) = n log 5 n Time for base cases: Dominated?: c = ? 2 = log a / log b Evaluating: T(n) = 4T(n/2)+ n log 5 n θ(n ) = log a / log b θ(n ) = θ(n 2 ) log 4 / log 2 Hence, T(n) = ? 1< = θ(base cases) = θ(n ) = θ(n 2 ). log a / log b

Time for top level: f(n) = log 5 n Time for base cases: Dominated?: c = ? 2 = log a / log b Evaluating: T(n) = 4T(n/2)+ log 5 n θ(n ) = log a / log b θ(n ) = θ(n 2 ) log 4 / log 2 Hence, T(n) = ? 0< = θ(base cases) = θ(n ) = θ(n 2 ). log a / log b

Time for top level: f(n) = n 2 Time for base cases: Dominated?: c = ? 2 = log a / log b Evaluating: T(n) = 4T(n/2)+ n 2 θ(n ) = log a / log b θ(n ) = θ(n 2 ) log 4 / log 2 Hence, T(n) = ? 2= = θ(f(n) log(n) ) = θ(n 2 log(n) ).

Time for top level: f(n) = n 2 log 5 n Time for base cases: Dominated?: c = 2 = 2 = log a / log b Evaluating: T(n) = 4T(n/2)+ n 2 log 5 n θ(n ) = log a / log b θ(n ) = θ(n 2 ) log 4 / log 2 d = ? 5 ? Hence, T(n) = ? > -1 = θ(f(n) log(n) ) = θ(n 2 log 6 (n) ).

Evaluating: T(n) = aT(n-b)+f(n) h = ? n-hb n-ib n-2b n-b T(0) f(n-ib) f(n-2b) f(n-b) f(n) n Work in Level # stack frames Work in stack frame Instance size a aiai a2a2 a 1 n/bn/b a · T(0) a i · f(n-ib) a 2 · f(n-2b) a · f(n-b) 1 · f(n) n/bn/b |base case| = 0 = n-hb h = n / b i 2 1 0 Level Likely dominated by base cases Exponential

Evaluating: T(n) = 1T(n-b)+f(n) h = ? Work in Level # stack frames Work in stack frame Instance size 1 1 1 1 1 f(0) f(n-ib) f(n-2b) f(n-b) f(n) n-b n n-hb n-ib n-2b T(0) f(n-ib) f(n-2b) f(n-b) f(n) h = n / b i 2 1 0 Level Total Work T(n) = ∑ i=0..h f(b·i) = θ(f(n)) or θ(n · f(n))

Relevant Mathematics In more detail. Jeff Edmonds York University COSC 3101 Lecture skipped Classifying Functions The Time Complexity of an Algorithm Adding Made Easy Recurrence Relations

Assumed Knowledge Read the section on Existential and Universal Quantifiers, Logarithms and Exponentials  g  b Loves(b,g)  b  g Loves(b,g)

Start With Some Math Input Size Time Classifying Functions f(i) = n  (n) Recurrence Relations T(n) = a T(n/b) + f(n) Adding Made Easy ∑ i=1 f(i). Time Complexity t(n) =  (n 2 )

Classifying Functions Giving an idea of how fast a function grows without going into too much detail.

Which are more alike?

Mammals

Which are more alike?

Classifying Animals Vertebrates Birds Mammals Reptiles Fish Dogs Giraffe We consider two (of many) levels in this hierarchy Class Genus Individuals

Classifying Functions T(n)T(n) 101001,00010,000 log n 36913amoeba n 1/2 31031100bird 101001,00010,000human n log n 306009,000130,000my father n2n2 10010,00010 6 10 8 elephant n3n3 1,00010 6 10 9 10 12 dinosaur 2n2n 1,02410 30 10 300 10 3000 the universe Note: The universe contains approximately 10 50 particles. n

Which are more alike? n 1000 n2n2 2n2n

Which are more alike? Polynomials n 1000 n2n2 2n2n

Which are more alike? 1000n 2 3n 2 2n 3 Polynomials

Which are more alike? Quadratic 1000n 2 3n 2 2n 3 Polynomials

Classifying Functions? Functions

Classifying Functions Functions Poly Logarithmic Polynomial Exponential ExpDouble Exp Constant (log n) 5 n5n5 2 5n 5 2 n5n5 2 Others 2 n log(n)

Classifying Functions? Polynomial

Classifying Functions Polynomial LinearQuadratic Cubic ? 5n 2 5n 5n 3 5n 4 Others 5n 3 log 7 (n)

Classifying Functions Functions Poly. Exp. 8 · 2 4n / n 100 θ(2 4n / n 100 ) 8 · 2 4n + n 3 n θ(1) 2 θ(n) θ(2 4n ) 8·n28·n2 θ(n 2 ) θ(n 3 log(n)) 7 · n 2 + n We consider two (of many) levels in this hierarchy Ignore Power-constant Ignore Mult-constant Individuals

Logarithmic log 10 n = # digits to write n log 2 n = # bits to write n = 3.32 log 10 n log(n 1000 ) = 1000 log(n) Differ only by a multiplicative constant.

Poly Logarithmic (log n) 5 = log 5 n

Logarithmic << Polynomial For sufficiently large n log 1000 n << n 0.001

Linear << Quadratic For sufficiently large n 10000 n << 0.0001 n 2

Polynomial << Exponential For sufficiently large n n 1000 << 2 0.001 n

Classifying Functions Functions Poly Logarithmic Polynomial Exponential ExpDouble Exp Constant (log n) 5 n5n5 2 5n 5 2 n5n5 2 <<

Which Functions are Constant? 5 1,000,000,000,000 0.0000000000001 -5 0 8 + sin(n) Yes No Yes

Which Functions are “Constant”? 5 1,000,000,000,000 0.0000000000001 -5 0 8 + sin(n) Yes No Yes Lie in between 7 9 The running time of the algorithm is a “Constant” It does not depend significantly on the size of the input. Write θ(1).

Which Functions are Quadratic? n 2 … ?

Which Functions are Quadratic? n 2 0.001 n 2 1000 n 2 Some constant times n 2.

Which Functions are Quadratic? n 2 0.001 n 2 1000 n 2 5n 2 + 3000n + 2log n

Which Functions are Quadratic? n 2 0.001 n 2 1000 n 2 5n 2 + 3000n + 2log n Lie in between

Which Functions are Quadratic? n 2 0.001 n 2 1000 n 2 5n 2 + 3000n + 2log n Ignore multiplicative constant. Low-order terms absorbed into constant. Ignore "small" values of n. Write θ(n 2 ).

Classifying Functions Functions Poly. Exp. 8 · 2 4n / n 100 θ(2 4n / n 100 ) 8 · 2 4n + n 3 n θ(1) 2 θ(n) θ(2 4n ) 8·n28·n2 θ(n 2 ) θ(n 3 log(n)) 7 · n 2 + n We consider two (of many) levels in this hierarchy Ignore Mult-constant Individuals Ignore Power-constant

Classifying Functions Functions Poly. Exp. 8 · 2 4n / n 100 θ(2 4n / n 100 ) 8 · 2 4n + n 3 n θ(1) 2 θ(n) θ(2 4n ) 8·n28·n2 θ(n 2 ) θ(n 3 log(n)) 7 · n 2 + n We consider two (of many) levels in this hierarchy Ignore Mult-constant Individuals Ignore Power-constant log(n) not absorbed because not Mult-constant

Which Functions are Polynomial? n 5 … ?

Which Functions are Polynomial? n c n 0.0001 n 10000 n to some constant power.

Which Functions are Polynomial? n c n 0.0001 n 10000 5n 2 + 8n + 2log n 5n 2 log n 5n 2.5

Which Functions are Polynomial? n c n 0.0001 n 10000 5n 2 + 8n + 2log n 5n 2 log n 5n 2.5 Lie in between

Which Functions are Polynomials? n c n 0.0001 n 10000 5n 2 + 8n + 2log n 5n 2 log n 5n 2.5 Ignore power constant. Low-order terms and log factors absorbed into constant. Write n θ(1)

If f(n) = c b an n d log e (n) c  ? a  ? b  ? d  ? e  ? > 0  0 or b  1 > 0 (- ,  ) Which Functions are Polynomials? then f(n)  n θ(1)

Which Functions are Exponential? 2 n … ?

Which Functions are Exponential? 2 n 2 0.0001 n 2 10000 n 2 raised to the power of n times some constant power

Which Functions are Exponential? 2 n 2 0.0001 n 2 10000 n 8 n 2 n / n 100 2 n · n 100 too small? too big?

Which Functions are Exponential? 2 n 2 0.0001 n 2 10000 n 8 n 2 n / n 100 2 n · n 100 = 2 3n > 2 0.5n < 2 2n Lie in between

Which Functions are Exponential? 2 n 2 0.0001 n 2 10000 n 8 n 2 n / n 100 2 n · n 100 = 2 3n > 2 0.5n < 2 2n 2 0.5n > n 100 2 n = 2 0.5n · 2 0.5n > n 100 · 2 0.5n 2 n / n 100 > n 0.5n

Which Functions are Exponential? Ignore power constant. Low-order terms, change in base, and polynomial factors absorbed into constant. Write 2 θ(n) 2 n 2 0.0001 n 2 10000 n 8 n 2 n / n 100 2 n · n 100

If f(n) = c b an n d log e (n) c  ? a  ? b  ? d  ? e  ? then f(n)  2 θ(n) > 0  0  1 (- ,  ) Which Functions are Exponential?

Classifying Functions Functions Poly Logarithmic Polynomial Exponential ExpDouble Exp Constant (log n) θ(1) n θ(1) 2 θ(n) θ(1) 2 n θ(1) 2 θ(n) 2 Ignore Power-constant

Classifying Functions LinearQuadratic Cubic ? θ(n 2 ) θ(n) θ(n 3 ) Polynomial = n θ(1) θ(n 4 ) Others θ(n 3 log 7 (n)) Ignore Mult-constant

Exponential 2 n vs 8 n Should these be classified together? 8n = 4 n · 2 n >> 1,000,000 · 2 n 8 n = 2 log(8) · n = 2 3n No: 8 n   (2 n ) Yes: 8 n = 2  (n) Ignore Power-constant Ignore Mult-constant

Classifying Functions Rank constants in significance. f(n) = 8·2 4n / n 100 + 5·n 3 Tell me the most significant thing about your function

Classifying Functions f(n) = 8·2 4n / n 100 + 5·n 3 f(n) = 9·2 4n / n 100 + 5·n 3 vs additive: multiplicative: 9/8 2 4n / n 100 Rank constants in significance.

Classifying Functions f(n) = 8·2 4n / n 100 + 5·n 3 f(n) = 8·2 4n / n 100 + 6·n 3 vs ≈ 1 n3n3 Rank constants in significance. additive: multiplicative:

Classifying Functions f(n) = 8·2 4n / n 100 + 5·n 3 f(n) = 8·2 4n / n 100 + 5·n 4 vs ≈ 1 ≈ 5· n 4 Rank constants in significance. additive: multiplicative:

Classifying Functions f(n) = 8·2 4n / n 100 + 5·n 3 f(n) = 8·2 5n / n 100 + 5·n 3 vs multiplicative: 2n2n Rank constants in significance.

Classifying Functions f(n) = 8·2 4n / n 100 + 5·n 3 f(n) = 8·2 4n / n 101 + 5·n 3 vs multiplicative: n Rank constants in significance.

Classifying Functions f(n) = 8·2 4n / n 100 + 5·n 3 Rank constants in significance. 1231’45

Classifying Functions f(n) = 8·2 4n / n 100 + 5·n 3 Rank constants in significance. Significant Less significant Irrelevant

Classifying Functions f(n) = 8·2 4n / n 100 + 5·n 3 Tell me the most significant thing about your function 2 θ(n) 2 3n << 2 4n f(n)because

Classifying Functions f(n) = 8·2 4n / n 100 + 5·n 3 Tell me the most significant thing about your function 2 θ(n) 2 4n /n θ(1) θ(2 4n /n 100 ) 8·2 4n / n 100 + θ(n 3 ) 8·2 4n / n 100 + n θ(1) 8·2 4n / n 100 + 5·n 3

Classifying Functions Functions Poly. Exp. 8 · 2 4n / n 100 + n 3, θ(2 4n / n 100 ) 8 · 2 4n + n 3 n θ(1) 2 θ(n) θ(2 4n ) Ignore Mult-constant Individuals Ignore Power-constant

Notations Theta f(n) = θ(g(n))f(n) ≈ c g(n) BigOh f(n) = O(g(n))f(n) ≤ c g(n) Omega f(n) = Ω(g(n))f(n) ≥ c g(n) Little Oh f(n) = o(g(n))f(n) << c g(n) Little Omega f(n) = ω(g(n))f(n) >> c g(n)

Definition of Theta f(n) = θ(g(n))

Definition of Theta f(n) is sandwiched between c 1 g(n) and c 2 g(n) f(n) = θ(g(n))

Definition of Theta f(n) is sandwiched between c 1 g(n) and c 2 g(n) for some sufficiently small c 1 (= 0.0001) for some sufficiently large c 2 (= 1000) f(n) = θ(g(n))

Definition of Theta For all sufficiently large n f(n) = θ(g(n))

Definition of Theta For all sufficiently large n For some definition of “sufficiently large” f(n) = θ(g(n))

Definition of Theta 3n 2 + 7n + 8 = θ(n 2 )

Definition of Theta 3n 2 + 7n + 8 = θ(n 2 ) c 1 ·n 2  3n 2 + 7n + 8  c 2 ·n 2 ??

Definition of Theta 3n 2 + 7n + 8 = θ(n 2 ) 3·n 2  3n 2 + 7n + 8  4·n 2 34

Definition of Theta 3n 2 + 7n + 8 = θ(n 2 ) 3·1 2  3·1 2 + 7·1 + 8  4·1 2 1 False

Definition of Theta 3n 2 + 7n + 8 = θ(n 2 ) 3·7 2  3·7 2 + 7·7 + 8  4·7 2 7 False

Definition of Theta 3n 2 + 7n + 8 = θ(n 2 ) 3·8 2  3·8 2 + 7·8 + 8  4·8 2 8 True

Definition of Theta 3n 2 + 7n + 8 = θ(n 2 ) 3·n 2  3n 2 + 7n + 8  4·n 2 ?

Definition of Theta 3n 2 + 7n + 8 = θ(n 2 ) 3·n 2  3n 2 + 7n + 8  4·n 2 n  8 8

Definition of Theta 3n 2 + 7n + 8 = θ(n 2 ) 3·n 2  3n 2 + 7n + 8  4·n 2 True n  8

Definition of Theta 3n 2 + 7n + 8 = θ(n 2 ) 3·n 2  3n 2 + 7n + 8  4·n 2 7n + 8  1 · n 2 7 + 8 / n  1 · n n  8 True

Definition of Theta 3n 2 + 7n + 8 = θ(n 2 ) 3·n 2  3n 2 + 7n + 8  4·n 2 348 n  8 True

Definition of Theta n 2 = θ(n 3 )

Definition of Theta n 2 = θ(n 3 ) c 1 · n 3  n 2  c 2 · n 3 ??

Definition of Theta n 2 = θ(n 3 ) 0 · n 3  n 2  c 2 · n 3 0 True, but ?

Definition of Theta n 2 = θ(n 3 ) 0 Constants c1 and c2 must be positive! False

Definition of Theta 3n 2 + 7n + 8 = θ(n)

Definition of Theta 3n 2 + 7n + 8 = θ(n) c 1 ·n  3n 2 + 7n + 8  c 2 ·n ??

Definition of Theta 3n 2 + 7n + 8 = θ(n) 3·n  3n 2 + 7n + 8  100·n 3 100

Definition of Theta 3n 2 + 7n + 8 = θ(n) 3·n  3n 2 + 7n + 8  100·n ?

Definition of Theta 3n 2 + 7n + 8 = θ(n) 3·100  3·100 2 + 7·100 + 8  100·100 100 False

Definition of Theta 3n 2 + 7n + 8 = θ(n) 3·n  3n 2 + 7n + 8  10,000·n 3 10,000

Definition of Theta 3n 2 + 7n + 8 = θ(n) 3·10,000  3·10,000 2 + 7·10,000 + 8  10,000·10,000 10,000 False

Definition of Theta 3n 2 + 7n + 8 = θ(n) What is the reverse statement?

Understand Quantifiers!!! SamMary BobBeth John Marilyn Monroe FredAnn SamMary BobBeth John Marilyn Monroe FredAnn  b,  Loves(b, MM)  b, Loves(b, MM)  b,  Loves(b, MM)]  b, Loves(b, MM)]

Definition of Theta 3n 2 + 7n + 8 = θ(n) The reverse statement

Definition of Theta 3n 2 + 7n + 8 = θ(n) 3n 2 + 7n + 8 > 100·n 3 100 ?

Definition of Theta 3n 2 + 7n + 8 = θ(n) 3 · 100 2 + 7 · 100 + 8 > 100 · 100 3 100 True 100

Definition of Theta 3n 2 + 7n + 8 = θ(n) 3n 2 + 7n + 8 > 10,000·n 3 10,000 ?

Definition of Theta 3n 2 + 7n + 8 = θ(n) 3 · 10,000 2 + 7 · 10,000 + 8 > 10,000 · 10,000 3 10,000 True 10,000

Definition of Theta 3n 2 + 7n + 8 = θ(n) 3n 2 + 7n + 8 > c 2 ·n c1c1 c2c2 ?

Definition of Theta 3n 2 + 7n + 8 = θ(n) 3 · c 2 2 + 7 · c 2 + 8 > c 2 · c 2 c1c1 c2c2 True c2c2

Definition of Theta 3n 2 + 7n + 8 = θ(n) 3n 2 + 7n + 8 > c 2 ·n c1c1 c2c2 ? n0n0

Definition of Theta 3n 2 + 7n + 8 = θ(n) 3 · c 2 2 + 7 · c 2 + 8 > c 2 · c 2 c1c1 max(c 2,n 0 ) True c2c2 n0n0

Definition of Theta 3n 2 + 7n + 8 = g(n) θ(1)  c 1, c 2, n 0,  n  n 0 g(n) c1  f(n)  g(n) c2 3n 2 + 7n + 8 = n θ(1)

Definition of Theta 3n 2 + 7n + 8 = n θ(1) n c1  3n 2 + 7n + 8  n c2  c 1, c 2, n 0,  n  n 0 g(n) c1  f(n)  g(n) c2 ??

Definition of Theta 3n 2 + 7n + 8 = n θ(1) n 2  3n 2 + 7n + 8  n 3  c 1, c 2, n 0,  n  n 0 g(n) c1  f(n)  g(n) c2 23

Definition of Theta 3n 2 + 7n + 8 = n θ(1) n 2  3n 2 + 7n + 8  n 3  c 1, c 2, n 0,  n  n 0 g(n) c1  f(n)  g(n) c2 23? n  ?

Definition of Theta 3n 2 + 7n + 8 = n θ(1) n 2  3n 2 + 7n + 8  n 3  c 1, c 2, n 0,  n  n 0 g(n) c1  f(n)  g(n) c2 235 n  5

Definition of Theta 3n 2 + 7n + 8 = n θ(1) n 2  3n 2 + 7n + 8  n 3 True  c 1, c 2, n 0,  n  n 0 g(n) c1  f(n)  g(n) c2 235 n  5

Order of Quantifiers f(n) = θ(g(n)) ?

Understand Quantifiers!!! One girl Could be a separate girl for each boy. SamMary BobBeth John Marilyn Monroe FredAnn SamMary BobBeth John Marilyn Monroe FredAnn

Order of Quantifiers No! It cannot be a different c 1 and c 2 for each n. f(n) = θ(g(n))

Other Notations Theta f(n) = θ(g(n))f(n) ≈ c g(n) BigOh f(n) = O(g(n))f(n) ≤ c g(n) Omega f(n) = Ω(g(n))f(n) ≥ c g(n) Little Oh f(n) = o(g(n))f(n) << c g(n) Little Omega f(n) = ω(g(n))f(n) >> c g(n)

BigOh Notation n 2 = O(n 3 ) n 3 = O(n 2 )

BigOh Notation O(n 2 ) = O(n 3 ) O(n 3 ) = O(n 2 ) Odd Notation f(n) = O(g(n)) Standard f(n) ≤ O(g(n)) Stresses one function dominating another. f(n)  O(g(n)) Stress function is member of class.

Adding The Classic Techniques Evaluating ∑ i=1 f(i). n

Gauss ∑ i=1..n i = 1 + 2 + 3 +... + n = ? Arithmetic Sum

1+2+3+...+ n-1+n=S n+n-1+n-2+... +2+1=S (n+1) + (n+1) + (n+1) +... + (n+1) + (n+1) = 2S n (n+1) = 2S

Let’s restate this argument using a geometric representation Algebraic argument

1 2........ n = number of white dots.

1 2........ n = number of white dots = number of yellow dots n....... 2 1

n+1 n+1 n+1 n+1 n+1 = number of white dots = number of yellow dots n n n n n n There are n(n+1) dots in the grid

n+1 n+1 n+1 n+1 n+1 = number of white dots = number of yellow dots n n n n n n Note =  (# of terms · last term))

Gauss ∑ i=1..n i = 1 + 2 + 3 +... + n =  (# of terms · last term) Arithmetic Sum True when ever terms increase slowly

∑ i=0..n 2 i = 1 + 2 + 4 + 8 +... + 2 n = ? Geometric Sum

∑ i=0..n 2 i = 1 + 2 + 4 + 8 +... + 2 n = 2 · last term - 1 Geometric Sum

∑ i=0..n r i = r 0 + r 1 + r 2 +... + r n = ? Geometric Sum

S Srrrr...rr S 1r S r1 r1 23nn1 n1 n1  1rrr... 23 r n          Geometric Sum

∑ i=0..n r i r1 r1 n1     Geometric Sum When r>1?

∑ i=0..n r i r1 r1 n1     Geometric Sum θ(rn)θ(rn) Biggest Term When r>1

∑ i=0..n r i = r 0 + r 1 + r 2 +... + r n =  (biggest term) Geometric Increasing True when ever terms increase quickly

∑ i=0..n r i  Geometric Sum When r<1? 1 r n1   

∑ i=0..n r i  Geometric Sum 1 r n1     θ(1) When r<1 Biggest Term

∑ i=0..n r i = r 0 + r 1 + r 2 +... + r n =  (1) Bounded Tail True when ever terms decrease quickly

∑ i=1..n 1 / i = 1 / 1 + 1 / 2 + 1 / 3 + 1 / 4 + 1 / 5 + …+ 1 / n = ? Harmonic Sum

f(i) = ? ∑ i=1..n f(i) = n 1/2 n Sum of Shrinking Function

f(i) = 1/2 i ∑ i=1..n f(i) = 2  Sum of Shrinking Function

f(i) = 1/i ∑ i=1..n f(i) = ? n Sum of Shrinking Function

Harmonic Sum

∑ i=1..n 1 / i = 1 / 1 + 1 / 2 + 1 / 3 + 1 / 4 + 1 / 5 + …+ 1 / n =  (log(n)) Harmonic Sum

Approximating Sum by Integrating The area under the curve approximates the sum ∑ i=1..n f(i) ≈ ∫ x=1..n f(x) dx

Approximating Sum by Integrating 1 / c+1 x c+1 d dx = x c ∫ x=1..n x c dx = 1 / c+1 n c+1 ∑ i=1..n i c ≈ Arithmetic Sums  θ(# of terms · last term) True when ever terms increase slowly = θ(n · n c )

Approximating Sum by Integrating 1 / ln(b) e ln(b)x d dx = e ln(b)x = b x ∫ x=1..n b x dx = 1 / ln(b) b x ∑ i=1..n b i ≈ Geometric Sums  θ(last term) True when ever terms increase quickly = θ(b n )

Harmonic Sum Approximating Sum by Integrating

Problem: Integrating may be hard too.

Outline II) Math proofs of sums V) Ability to know the answer fast I) What is the sum Eg. ∑ i=1..n f(i) = ∑ i=1..n 1 / i 0.9999 =θ(n 0.0001 ) III) Pattern of sums ∑ i=1..n f(i) = θ(n · f(n)) IV) Intuition why this is the sum I flash it up If you follow great. If not don’t panic.

Adding Made Easy We will now classify (most) functions f(i) into four classes: –Geometric Like –Arithmetic Like –Harmonic –Bounded Tail

Adding Made Easy We will now classify (most) functions f(i) into four classes For each class, we will give an easy rule for approximating it’s sum θ( ∑ i=1..n f(i) )

Classifying Animals Vertebrates Birds Mammals Reptiles Fish Dogs Giraffe Mammal  Complex social networks Dog  Man’s best friend

Classifying Functions Functions Poly. Exp. 2 θ(n)  ∑ i=1..n f(i) = θ(f(n)) 8 · 2 n / n 100 + n 3, θ(2 n / n 100 ) 8 · 2 n + n 3 n θ(1) 2 θ(n) θ(2n)θ(2n) f(n) = 8 · 2 n / n 100 + n 3 θ(2 n / n 100 )  ∑ i=1..n f(i) = θ(2 n / n 100 ) 2 0.5n << 2 n 8 · 2 n / n 100 + n 3 Significant Less significant Irrelevant

Adding Made Easy 11 1 2 3 4 Four Classes of Functions

Adding Made Easy

If the terms f(i) grow sufficiently quickly, then the sum will be dominated by the largest term. Silly example: 1+2+3+4+5+ 1,000,000,000 ≈ 1,000,000,000 f(i)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like:

If the terms f(i) grow sufficiently quickly, then the sum will be dominated by the largest term. Classic example: ? f(i)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like:

If the terms f(i) grow sufficiently quickly, then the sum will be dominated by the largest term. Classic example: ∑ i=1..n 2 i = 2 n+1 -1 ≈ 2 f(n) f(i)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like:

If the terms f(i) grow sufficiently quickly, then the sum will be dominated by the largest term. For which functions f(i) is this true? How fast and how slow can it grow? f(i)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like:

r1 r1 n1      θ( ) r n Last Term when r>1. ∑ i=1..n r i 1rrr... 23 r n    θ(f(n)) ∑ i=1..n (1.0001) i ≈ 10,000 (1.0001) n = 10,000 f(n) Lower Extreme: Upper Extreme: ∑ i=1..n (1000) i ≈ 1.001(1000) n = 1.001 f(n) Recall, when f(n) = r i = 2 θ(n) f(i)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like:

Because the constant is sooo big, the statement is just barely true. ∑ i=1..n (1.0001) i ≈ 10,000 (1.0001) n = 10,000 f(n) Lower Extreme: Upper Extreme: ∑ i=1..n (1000) i ≈ 1.001(1000) n = 1.001 f(n) f(i)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like:

∑ i=1..n (1.0001) i ≈ 10,000 (1.0001) n = 10,000 f(n) Lower Extreme: Upper Extreme: ∑ i=1..n (1000) i ≈ 1.001(1000) n = 1.001 f(n) Even bigger? f(i)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like:

2n2n 2i2i ∑ i=1..n 2 2 ≈ 2 2 = 1f(n) No Upper Extreme: Even bigger? ∑ i=1..n (1.0001) i ≈ 10,000 (1.0001) n = 10,000 f(n) Lower Extreme: f(i)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like:

2n2n 2i2i ∑ i=1..n 2 2 ≈ 2 2 = 1f(n) No Upper Extreme: Functions in between? ∑ i=1..n (1.0001) i ≈ 10,000 (1.0001) n = 10,000 f(n) Lower Extreme: f(i)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like:

∑ i=1..n 2 i = 2 · 2 n –2 = 2 1 + 2 2 + 2 3 + 2 4 +…+ 2 n 8 · [ ] 8·8· 8·8· 8·8· 8·8· 8·8· 8·8· = θ( 2 n ) 1 100 2 100 3 100 4 100 n 100 i 100 n 100 + i 3 + 1 3 + 2 3 + 3 3 + 4 3 + n 3 + n 4 Dominated by the largest term. = θ(f(n)) f(n) = 2 n 8·8· n 100 + n 3 = θ( 2 n ) n 100 f(i)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like: n 100 ≈

f(i)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like: f(n) = 2 n 8·8· n 100 ∑ i=1..n f(i) f(n)  c · f(n) f(i)  · f(n) f(i)  · f(i +1 ) EasyHard  (1 + 1 / i ) 100 1 2 1.9 = 8 · 2 (i+1) (i +1 ) 100 f(i +1 ) f(i) 8·2i8·2i i 100 =  1/.51  · f(i +1 ) f(i) (i +1 ) 100 i 100 2 =

f(i)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like: f(n) = 2 n 8·8· n 100 ∑ i=1..n f(i) f(n)  c · f(n) f(i)  · f(n)   · f(i +2 )  · f(i +1 ) f(i)  · f(i +1 ) f(i)   · f(i +3 ) .51 (n-i) ·f(i +(n-i) ) .51 (n-i) ·f( n ) f(i) .51 (n-i) ·f( n ) Eg. i = n, f(n) .51 (n-n) ·f( n )  1·f( n ) i = 0, f(0) .51 n ·f( n )

f(i)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like: f(n) = 2 n 8·8· n 100 ∑ i=1..n f(i) f(n)  c · f(n) ∑ i=1..n f(i)  ∑ i=1..n.51 (n-i) ·f( n ) = θ(1) · f(n) f(i)  · f(n)  · f(i +1 ) f(i) f(i) .51 (n-i) ·f( n ) = [∑ i=1..n.51 (n-i) ] · f( n )

Functions Poly. Exp. 2 θ(n)  ∑ i=1..n f(i) = θ(f(n)) 8 · 2 n / n 100 + n 3 θ(2 n / n 100 ) 8 · 2 n + n 3 n θ(1) 2 θ(n) θ(2n)θ(2n) f(n) = 8 · 2 n / n 100 + n 3 θ(2 n / n 100 )  ∑ i=1..n f(i) = θ(2 n / n 100 ) f(i)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like:

If f(n) = c b an n d log e (n) c  ? a  ? b  ? d  ? e  ? f(n) = Ω, θ ? then ∑ i=1..n f(i) = θ(f(n)).  2 Ω(n) > 0  0  1 (- ,  ) f(i)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like:

f(i)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like:

All functions in 2 Ω(n) ? Maybe not. f(i)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like:

Functions that oscillate continually do not have the property. f(i)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like:

Functions expressed with +, -, , , exp, log do not oscillate continually. They are well behaved for sufficiently large n. These do have the property. f(i)  2 Ω(n)  ∑ i=1..n f(i) = θ(f(n)) Geometric Like:

Adding Made Easy Done

Adding Made Easy

If most of the terms f(i) have roughly the same value, then the sum is roughly the number of terms, n, times this value. Silly example: 1,001 + 1,002 + 1,003 + 1,004 + 1,005 ≈ 5 · 1,000 f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

If most of the terms f(i) have roughly the same value, then the sum is roughly the number of terms, n, times this value. Another silly example: ∑ i=1..n 1 = n · 1 f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

If most of the terms f(i) have roughly the same value, then the sum is roughly the number of terms, n, times this value. Is the statement true for this function? ∑ i=1..n i = 1 + 2 + 3 +... + n f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

If most of the terms f(i) have roughly the same value, then the sum is roughly the number of terms, n, times this value. Is the statement true for this function? ∑ i=1..n i = 1 + 2 + 3 +... + n The terms are not roughly the same. f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

But half the terms are roughly the same. ∑ i=1..n i = 1 +... + n/2 +... + n f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

But half the terms are roughly the same and the sum is roughly the number terms, n, times this value ∑ i=1..n i = 1 +... + n/2 +... + n ∑ i=1..n i = θ(n · n) f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

Is the statement true for this function? ∑ i=1..n i 2 = 1 2 + 2 2 + 3 2 +... + n 2 Even though, the terms are not roughly the same. f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

Again half the terms are roughly the same. ∑ i=1..n i = 1 2 +... + (n/2) 2 +... + n 2 1 / 4 n 2 f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

Again half the terms are roughly the same. ∑ i=1..n i = 1 2 +... + (n/2) 2 +... + n 2 1 / 4 n 2 ∑ i=1..n i 2 = θ(n · n 2 ) f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

∑ i=1..n f(i) ≈ area under curve f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

area of small square  ∑ i=1..n f(i) ≈ area under curve  area of big square f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

n / 2 · f( n / 2 ) = area of small square  ∑ i=1..n f(i) ≈ area under curve  area of big square = n · f(n) f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

θ(n · f(n)) = n / 2 · f( n / 2 ) = area of small square  ∑ i=1..n f(i) ≈ area under curve  area of big square = n · f(n) f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

θ(n · f(n)) = n / 2 · f( n / 2 ) = area of small square  ∑ i=1..n f(i) ≈ area under curve  area of big square = n · f(n) ? f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

∑ i=1..n i 2 = 1 2 + 2 2 + 3 2 +... + n 2 f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like: f(i) = n 2 f( n / 2 ) = θ(f(n)) The key property is = ?

The key property is f( n / 2 ) = (n/2) 2 = 1 / 4 n 2 = θ( n 2 ) = θ(f(n)) ∑ i=1..n i 2 = 1 2 + 2 2 + 3 2 +... + n 2 f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like: f(i) = n 2  = θ(n·f(n)) = θ(n · n 2 )

∑ i=1..n f(i) = ? f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like: f(i) = n θ(1)

∑ i=1..n i r = 1 r + 2 r + 3 r +... + n r f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like: f(i) = n θ(1) f( n / 2 ) = θ(f(n)) The key property is

f( n / 2 ) = (n/2) r = 1 / 2 r n r = θ( n r ) = θ(f(n)) ∑ i=1..n i r = 1 r + 2 r + 3 r +... + n r f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:  = θ(n·f(n)) = θ(n · n r ) f(i) = n θ(1)

∑ i=1..n 2 i = 2 1 + 2 2 + 2 3 +... + 2 n f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like: f(i) = n θ(1) f( n / 2 ) = θ(f(n)) The key property is

∑ i=1..n 2 i = 2 1 + 2 2 + 2 3 +... + 2 n f( n / 2 ) = 2 (n/2) = θ( 2 n ) = θ(f(n)) f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like: f(i) = n θ(1)  = θ(n·f(n)) = θ(n · 2 n )

∑ i=1..n 1 = n · 1 = n · f(n) Middle Extreme: Upper Extreme: ∑ i=1..n i 1000 = 1 / 1001 n 1001 = 1 / 1001 n · f(n) All functions in between. f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

Adding Made Easy Half done

f(i) = ? ∑ i=1..n f(i) = n 1/2 n Sum of Shrinking Function 1/i 1/2

f(i) = 1/i ∑ i=1..n f(i) = log n n Sum of Shrinking Function

f(i) = 1/2 i ∑ i=1..n f(i) = 2 Sum of Shrinking Function 

If most of the terms f(i) have roughly the same value, then the sum is roughly the number of terms, n, times this value. Does the statement hold for functions f(i) that shrink? f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

If most of the terms f(i) have roughly the same value, then the sum is roughly the number of terms, n, times this value. ∑ i=1..n 1 / i = 1 / 1 + 1 / 2 + 1 / 3 + 1 / 4 + 1 / 5 + … + 1 / n Does the statement hold for the Harmonic Sum? f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

Does the statement hold for the Harmonic Sum? ∑ i=1..n 1 / i = 1 / 1 + 1 / 2 + 1 / 3 + 1 / 4 + 1 / 5 + … + 1 / n ∑ i=1..n f(i) = ? θ(n · f(n)) f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

Does the statement hold for the Harmonic Sum? ∑ i=1..n 1 / i = 1 / 1 + 1 / 2 + 1 / 3 + 1 / 4 + 1 / 5 + … + 1 / n ∑ i=1..n f(i) θ(n · f(n)) = ∑ i=1..n 1 / i = ? f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

Does the statement hold for the Harmonic Sum? ∑ i=1..n 1 / i = 1 / 1 + 1 / 2 + 1 / 3 + 1 / 4 + 1 / 5 + … + 1 / n ∑ i=1..n f(i) θ(n · f(n)) = ∑ i=1..n 1 / i =θ(log n) f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

Does the statement hold for the Harmonic Sum? ∑ i=1..n 1 / i = 1 / 1 + 1 / 2 + 1 / 3 + 1 / 4 + 1 / 5 + … + 1 / n ∑ i=1..n f(i) = θ(n · f(n)) = ∑ i=1..n 1 / i =θ(log n) ? f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

Does the statement hold for the Harmonic Sum? ∑ i=1..n 1 / i = 1 / 1 + 1 / 2 + 1 / 3 + 1 / 4 + 1 / 5 + … + 1 / n ∑ i=1..n f(i) = θ(n · f(n)) = ∑ i=1..n 1 / i =θ(log n) θ(1) = θ(n · 1 / n ) f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

Does the statement hold for the Harmonic Sum? ∑ i=1..n 1 / i = 1 / 1 + 1 / 2 + 1 / 3 + 1 / 4 + 1 / 5 + … + 1 / n ∑ i=1..n f(i) = θ(n · f(n)) = ∑ i=1..n 1 / i =θ(log n) θ(1) = θ(n · 1 / n ) ≠ No the statement does not hold! f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

Adding Made Easy not included

Does the statement hold for the almost Harmonic Sum? ∑ i=1..n 1 / i 0.9999 Shrinks slightly slower than harmonic. f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

Does the statement hold for the almost Harmonic Sum? ∑ i=1..n 1 / i 0.9999 ∑ i=1..n f(i) = θ(n · f(n)) = ∑ i=1..n 1 / i 0.9999 = ? f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

Does the statement hold for the almost Harmonic Sum? ∑ i=1..n 1 / i 0.9999 ∑ i=1..n f(i) = θ(n · f(n)) = ∑ i=1..n 1 / i 0.9999 =θ(n 0.0001 ) f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

Does the statement hold for the almost Harmonic Sum? ∑ i=1..n 1 / i 0.9999 ∑ i=1..n f(i) = θ(n · f(n)) = ∑ i=1..n 1 / i 0.9999 =θ(n 0.0001 ) ? f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

Does the statement hold for the almost Harmonic Sum? ∑ i=1..n 1 / i 0.9999 ∑ i=1..n f(i) = θ(n · f(n)) = ∑ i=1..n 1 / i 0.9999 =θ(n 0.0001 ) θ(n 0.0001 ) = θ(n · 1 / n 0.9999 ) f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

Does the statement hold for the almost Harmonic Sum? ∑ i=1..n 1 / i 0.9999 ∑ i=1..n f(i) = θ(n · f(n)) = ∑ i=1..n 1 / i 0.9999 =θ(n 0.0001 ) θ(n 0.0001 ) = θ(n · 1 / n 0.9999 ) = The statement does hold! f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

∑ i=1..n 1 = n · 1 = n · f(n) Middle Extreme: Lower Extreme: ∑ i=1..n 1 / i 0.9999 = θ(n 0.0001 ) = θ(n · f(n)) All functions in between. f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

∑ i=1..n 1 = n · 1 = n · f(n) Middle Extreme: Lower Extreme: ∑ i=1..n 1 / i 0.9999 = θ(n 0.0001 ) = θ(n · f(n)) Upper Extreme: ∑ i=1..n i 1000 = 1 / 1001 n 1001 = 1 / 1001 n · f(n) f(i) = n θ(1)-1  ∑ i=1..n f(i) = θ(n·f(n)) Arithmetic Like:

Arithmetic Like If f(n) = c b an n d log e (n) c  ? a  ? b  ? d  ? e  ? f(n) = Ω, θ ? then ∑ i=1..n f(i) = θ(n · f(n)). n -1+θ(1) > 0  0 or b  1 > -1 (- ,  ) (For +, -, , , exp, log functions f(n)) Conclusion

Adding Made Easy Harmonic

Harmonic Sum ∑ i=1..n 1 / i = 1 / 1 + 1 / 2 + 1 / 3 + 1 / 4 + 1 / 5 + …+ 1 / n =  (log(n))

Adding Made Easy

If the terms f(i) decrease towards zero sufficiently quickly, then the sum will be a constant. The classic example ∑ i=0..n 1 / 2 i = 1 + 1 / 2 + 1 / 4 + 1 / 8 + … = 2. f(n)  n -1-Ω(1)  ∑ i=1..n f(i) = θ(1) Bounded Tail:

If f(i) decays even faster, then the tail of the sum is more bounded. 2i2i ∑ i=1..n 2 2 = θ(1). 1 f(n)  n -1-Ω(1)  ∑ i=1..n f(i) = θ(1) Bounded Tail:

Upper Extreme: ∑ i=1..n 1 / i 1.0001 = θ(1) f(n)  n -1-Ω(1)  ∑ i=1..n f(i) = θ(1) Bounded Tail:

Upper Extreme: ∑ i=1..n 1 / i 1.0001 = θ(1) No Lower Extreme: 2i2i ∑ i=1..n 2 2 = θ(1). 1 All functions in between. f(n)  n -1-Ω(1)  ∑ i=1..n f(i) = θ(1) Bounded Tail:

Bounded Tail If f(n) = c b an n d log e (n) c  ? a  ? b  ? d  ? e  ? f(n) = Ω, θ ? then ∑ i=1..n f(i) = θ(1). > 0  (- ,  ) Conclusion or b  1 c n d log e (n) c  a  b  b an

Bounded Tail If f(n) = c n d log e (n) c  ? a  b = 1 d  ? e  ? f(n) = Ω, θ ? then ∑ i=1..n f(i) = θ(1).  n -1-Ω(1) > 0 < -1 (- ,  ) Conclusion

Adding Made Easy Missing Functions 1 / nlogn logn / n

Adding Made Easy Geometric Like: If f(n)  2 Ω(n), then ∑ i=1..n f(i) = θ(f(n)). Arithmetic Like: If f(n) = n θ(1)-1, then ∑ i=1..n f(i) = θ(n · f(n)). Harmonic: If f(n) = 1 / n, then ∑ i=1..n f(i) = log e n + θ(1). Bounded Tail: If f(n)  n -1-Ω(1), then ∑ i=1..n f(i) = θ(1). ( For +, -, , , exp, log functions f(n) ) This may seem confusing, but it is really not. It should help you compute most sums easily.

Recurrence Relations T(1) = 1 T(n) = a T(n/b) + f(n)

Recurrence Relations  Time of Recursive Program Recurrence relations arise from the timing of recursive programs. Let T(n) be the # of “Hi”s on an input of “size” n. procedure Eg(I n ) n = |I n | if(n  1) then put “Hi” else loop i=1..f(n) put “Hi” loop i=1..a I n/b = I n cut in b pieces Eg(I n/b ) ?

Given size 1, the program outputs T(1)=1 Hi’s. Recurrence Relations  Time of Recursive Program procedure Eg(I n ) n = |I n | if(n  1) then put “Hi” else loop i=1..f(n) put “Hi” loop i=1..a I n/b = I n cut in b pieces Eg(I n/b )

Given size n, the stackframe outputs f(n) Hi’s. Recurrence Relations  Time of Recursive Program procedure Eg(I n ) n = |I n | if(n  1) then put “Hi” else loop i=1..f(n) put “Hi” loop i=1..a I n/b = I n cut in b pieces Eg(I n/b )

Recursing on an instances of size n/b generates T(n/b) “Hi”s. Recurrence Relations  Time of Recursive Program procedure Eg(I n ) n = |I n | if(n  1) then put “Hi” else loop i=1..f(n) put “Hi” loop i=1..a I n/b = I n cut in b pieces Eg(I n/b )

Recursing a times generates a·T(n/b) “Hi”s. Recurrence Relations  Time of Recursive Program procedure Eg(I n ) n = |I n | if(n  1) then put “Hi” else loop i=1..f(n) put “Hi” loop i=1..a I n/b = I n cut in b pieces Eg(I n/b )

For a total of T(1) = 1 T(n) = a·T(n/b) + f(n) “Hi”s. Recurrence Relations  Time of Recursive Program procedure Eg(I n ) n = |I n | if(n  1) then put “Hi” else loop i=1..f(n) put “Hi” loop i=1..a I n/b = I n cut in b pieces Eg(I n/b )

Solving Technique 1 Guess and Verify Recurrence Relation: T(1) = 1 & T(n) = 4T(n/2) + n Guess: G(n) = 2n 2 – n Verify: Left Hand SideRight Hand Side T(1) = 2(1) 2 – 1 T(n) = 2n 2 – n 1 4T(n/2) + n = 4 [2( n / 2 ) 2 – ( n / 2 )] + n = 2n 2 – n

Recurrence Relation: T(1) = 1 & T(n) = 4T(n/2) + n Guess: G(n) = an 2 + bn + c Verify: Left Hand SideRight Hand Side T(1) = a+b+c T(n) = an 2 +bn+c 1 4T(n/2) + n = 4 [a ( n / 2 ) 2 + b ( n / 2 ) +c] + n = an 2 +(2b+1)n + 4c Solving Technique 2 Guess Form and Calculate Coefficients

Recurrence Relation: T(1) = 1 & T(n) = 4T(n/2) + n Guess: G(n) = an 2 + bn + 0 Verify: Left Hand SideRight Hand Side T(1) = a+b+c T(n) = an 2 +bn+c 1 4T(n/2) + n = 4 [a ( n / 2 ) 2 + b ( n / 2 ) +c] + n = an 2 +(2b+1)n + 4c Solving Technique 2 Guess Form and Calculate Coefficients c=4c c=0

Recurrence Relation: T(1) = 1 & T(n) = 4T(n/2) + n Guess: G(n) = an 2 - 1n + 0 Verify: Left Hand SideRight Hand Side T(1) = a+b+c T(n) = an 2 +bn+c 1 4T(n/2) + n = 4 [a ( n / 2 ) 2 + b ( n / 2 ) +c] + n = an 2 +(2b+1)n + 4c Solving Technique 2 Guess Form and Calculate Coefficients b=2b+1 b=-1

Recurrence Relation: T(1) = 1 & T(n) = 4T(n/2) + n Guess: G(n) = an 2 - 1n + 0 Verify: Left Hand SideRight Hand Side T(1) = a+b+c T(n) = an 2 +bn+c 1 4T(n/2) + n = 4 [a ( n / 2 ) 2 + b ( n / 2 ) +c] + n = an 2 +(2b+1)n + 4c Solving Technique 2 Guess Form and Calculate Coefficients a=a

Recurrence Relation: T(1) = 1 & T(n) = 4T(n/2) + n Guess: G(n) = 2n 2 - 1n + 0 Verify: Left Hand SideRight Hand Side T(1) = a+b+c T(n) = an 2 +bn+c 1 4T(n/2) + n = 4 [a ( n / 2 ) 2 + b ( n / 2 ) +c] + n = an 2 +(2b+1)n + 4c Solving Technique 2 Guess Form and Calculate Coefficients a+b+c=1 a-1+0=1 a=2

Recurrence Relation: T(1) = 1 & T(n) = aT(n/b) + f(n) Solving Technique 3 Approximate Form and Calculate Exponent which is bigger? Guess

Recurrence Relation: T(1) = 1 & T(n) = aT(n/b) + f(n) Guess: aT(n/b) << f(n) Simplify: T(n)  f(n) Solving Technique 3 Calculate Exponent In this case, the answer is easy. T(n) =  (f(n))

Recurrence Relation: T(1) = 1 & T(n) = aT(n/b) + f(n) Guess: aT(n/b) >> f(n) Simplify: T(n)  aT(n/b) Solving Technique 3 Calculate Exponent In this case, the answer is harder.

Recurrence Relation: T(1) = 1 & T(n) = aT(n/b) Guess: G(n) = cn  Verify: Left Hand SideRight Hand Side T(n) = cn  aT(n/b) = a [c ( n / b )  ] = c a b  n  Solving Technique 3 Calculate Exponent ( log a / log b ) = cn 1 = a b -  b  = a  log b = log a  = log a / log b

Recurrence Relation: T(1) = 1 & T(n) = 4T(n/2) Guess: G(n) = cn  Verify: Left Hand SideRight Hand Side T(n) = cn  aT(n/b) + f(n) = c [a ( n / b )  ] = c a b  n  Solving Technique 3 Calculate Exponent 1 = a b -  b  = a  log b = log a  = log a / log b ( log a / log b ) = cn ( log 4 / log 2 ) = cn 2

Logs log 27 log 9 =

Logs log ? 27 log ? 9 = Which base?

Logs log 2 27 log 2 9 = It does not matter. log 2 c · log c 27 · log c 9 = log c 27 log c 9 Which is easiest?

Logs log 3 27 log 3 9 = log 3 3 3 log 3 3 2 = 3 2 Please no calculators in exams. And I wont ask log 25 log 9

Recurrence Relation: Solving Technique 3 Calculate Exponent If bigger then T(n) =  (f(n)) If bigger then ( log a / log b ) T(n) =  (n ) And if aT(n/b)  f(n) what is T(n) then? T(1) = 1 & T(n) = aT(n/b) + f(n)

4: Decorate The Tree T(n) = a T(n/b) + f(n) T(1) = 1 & T(n) = aT(n/b) + f(n) f(n) T(n/b) T(n) = T(1) T(1) = 1 1 = a

f(n) T(n/b) T(n) = a

f(n) T(n/b) T(n) = a f(n/b) T(n/b 2 ) a

f(n) T(n) = a f(n/b) T(n/b 2 ) a f(n/b) T(n/b 2 ) a f(n/b) T(n/b 2 ) a f(n/b) T(n/b 2 ) a

11111111111111111111111111111111...... 111111111111111111111111111111111 f(n) T(n) = a f(n/b) T(n/b 2 ) a f(n/b) T(n/b 2 ) a f(n/b) T(n/b 2 ) a f(n/b) T(n/b 2 ) a

Evaluating: T(n) = aT(n/b)+f(n) Level 0 1 2 i h

Evaluating: T(n) = aT(n/b)+f(n) Level Instance size 0 1 2 i h

Evaluating: T(n) = aT(n/b)+f(n) Level Instance size 0n 1 2 i h

Evaluating: T(n) = aT(n/b)+f(n) Level Instance size 0n 1 n/b 2 i h

Evaluating: T(n) = aT(n/b)+f(n) Level Instance size 0n 1 n/b 2 n/b 2 i h

Evaluating: T(n) = aT(n/b)+f(n) Level Instance size 0n 1 n/b 2 n/b 2 i n/b i h n/b h

Evaluating: T(n) = aT(n/b)+f(n) Level Instance size 0n 1 n/b 2 n/b 2 i n/b i h n/b h = 1 base case

Evaluating: T(n) = aT(n/b)+f(n) Level Instance size 0n 1 n/b 2 n/b 2 i n/b i h n/b h = 1

Evaluating: T(n) = aT(n/b)+f(n) Level Instance size 0n 1 n/b 2 n/b 2 i n/b i h = log n / log b n/b h = 1 b h = n h log b = log n h = log n / log b

Evaluating: T(n) = aT(n/b)+f(n) Level Instance size Work in stack frame 0n 1 n/b 2 n/b 2 i n/b i h = log n / log b 1

Evaluating: T(n) = aT(n/b)+f(n) Level Instance size Work in stack frame 0n f(n) 1 n/bf(n/b) 2 n/b 2 f(n/b 2 ) i n/b i f(n/b i ) h = log n / log b 1T(1)

Evaluating: T(n) = aT(n/b)+f(n) Level Instance size Work in stack frame # stack frames 0n f(n) 1 n/bf(n/b) 2 n/b 2 f(n/b 2 ) i n/b i f(n/b i ) h = log n / log b n/b h T(1)

Evaluating: T(n) = aT(n/b)+f(n) Level Instance size Work in stack frame # stack frames 0n f(n) 1 1 n/bf(n/b) a 2 n/b 2 f(n/b 2 ) a2a2 i n/b i f(n/b i ) aiai h = log n / log b n/b h T(1) ahah

Evaluating: T(n) = aT(n/b)+f(n) Level Instance size Work in stack frame # stack frames 0n f(n) 1 1 n/bf(n/b) a 2 n/b 2 f(n/b 2 ) a2a2 i n/b i f(n/b i ) aiai h = log n / log b n/b h T(1) ahah a h = a = n log n / log b log a / log b

Evaluating: T(n) = aT(n/b)+f(n) Level Instance size Work in stack frame # stack frames Work in Level 0n f(n) 1 1 n/bf(n/b) a 2 n/b 2 f(n/b 2 ) a2a2 i n/b i f(n/b i ) aiai h = log n / log b n/b h T(1) n log a / log b

Evaluating: T(n) = aT(n/b)+f(n) Level Instance size Work in stack frame # stack frames Work in Level 0n f(n) 1 1 · f(n) 1 n/bf(n/b) a a · f(n/b) 2 n/b 2 f(n/b 2 ) a2a2 a 2 · f(n/b 2 ) i n/b i f(n/b i ) aiai a i · f(n/b i ) h = log n / log b n/b h T(1) n log a / log b n · T(1) log a / log b Total Work T(n) = ∑ i=0..h a i  f(n/b i )

Evaluating: T(n) = aT(n/b)+f(n) = ∑ i=0..h a i  f(n/b i ) If a Geometric Sum ∑ i=0..n x i  θ(max(first term, last term))

Evaluating: T(n) = aT(n/b)+f(n) Level Instance size Work in stack frame # stack frames Work in Level 0n f(n) 1 1 · f(n) 1 n/bf(n/b) a a · f(n/b) 2 n/b 2 f(n/b 2 ) a2a2 a 2 · f(n/b 2 ) i n/b i f(n/b i ) aiai a i · f(n/b i ) h = log n / log b n/b h T(1) n log a / log b n · T(1) log a / log b Dominated by Top Level or Base Cases

Evaluating: T(n) = aT(n/b)+f(n) Is the sum Geometric? Simplify by letting f(n) = n c log k n T(n) = ∑ i=0..h a i  f(n/b i ) = ∑ i=0..h a i  (n/b i ) c  log k (n/b i ) = n c ∑ i=0..h 2 [log a - c log b] · i  log k (n/b i ) Geometrically IncreasingArithmetic SumGeometrically Decreasing Sum = θ(last term) = θ( ) Sum = θ(any term  # terms) = θ(f(n)  log n) Sum = θ(first term) = θ(f(n)). log a - c log b > 0log a - c log b = 0log a - c log b < 0 c < log a / log b c = log a / log b & k<-1 c = log a / log b & k>-1 c > log a / log b n log a / log b (n/..) c = n c a i = 2 [log a]·i (.. /b i ) c = 2 [-c log b]·i = n c ∑ i=0..h 2 -d i  log k (n/b i )

Evaluating: T(n) = aT(n/b)+f(n) Is the sum Geometric? Simplify by letting f(n) = n c log k n T(n) = ∑ i=0..h a i  f(n/b i ) = ∑ i=0..h a i  (n/b i ) c  log k (n/b i ) = n c ∑ i=0..h 2 [log a - c log b] · i  log k (n/b i ) Geometrically IncreasingArithmetic SumGeometrically Decreasing Sum = θ(last term) = θ( ) Sum = θ(any term  # terms) = θ(f(n)  log n) Sum = θ(first term) = θ(f(n)). log a - c log b > 0log a - c log b = 0log a - c log b < 0 c < log a / log b c = log a / log b & k<-1 c = log a / log b & k>-1 c > log a / log b n log a / log b k =-1  log k ( n /b i ) = 1 / (logn - i  logb)  1 / i (backwards) = n c ∑ i=0..h 2 0 i  log k (n/b i ) (back wards) de

Evaluating: T(n) = aT(n/b)+f(n) Is the sum Geometric? Simplify by letting f(n) = n c log k n T(n) = ∑ i=0..h a i  f(n/b i ) = ∑ i=0..h a i  (n/b i ) c  log k (n/b i ) = n c ∑ i=0..h 2 [log a - c log b] · i  log k (n/b i ) Geometrically IncreasingArithmetic SumGeometrically Decreasing Sum = θ(last term) = θ( ) Sum = θ(any term  # terms) = θ(f(n)  log n) Sum = θ(first term) = θ(f(n)). log a - c log b > 0log a - c log b = 0log a - c log b < 0 c < log a / log b c = log a / log b & k<-1 c = log a / log b & k>-1 c > log a / log b n log a / log b = n c ∑ i=0..h 2 +d i  log k (n/b i )