How to Win Coding Competitions: Secrets of Champions Week ... fileLecture 10: Introduction to binary search Maxim Buzdalov Saint Petersburg 2016. Binary search Recall the example from

How to Win Coding Competitions: Secrets of Champions

Week 3: Sorting and Search AlgorithmsLecture 10: Introduction to binary search

Maxim BuzdalovSaint Petersburg 2016

Binary search

Recall the example from Lecture 1: how to search in sorted data

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This is an example of binary search. We will have more in this video

2 / 6

Binary search

Recall the example from Lecture 1: how to search in sorted data

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s

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hi

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gew

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Sp

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ky

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stru

p

Su

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Ta

nen

ba

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Tu

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This is an example of binary search. We will have more in this video

2 / 6

Binary search

Recall the example from Lecture 1: how to search in sorted data

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n

Ah

o

Bro

ok

s

Cor

men

Dij

kst

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hi

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gew

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p

Su

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This is an example of binary search. We will have more in this video

2 / 6

Binary search

Recall the example from Lecture 1: how to search in sorted data

Ab

elso

n

Ah

o

Bro

ok

s

Dij

kst

ra

Eck

el

For

isek

Ga

mm

a

Go

del

Gra

ha

m

Ho

are

Ker

nig

ha

n

Kn

uth

La

mp

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McC

on

nel

l

Mey

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Nor

vig

Pa

pa

dim

itri

ou

Pa

tash

nik

Rit

chie

Ru

ssel

Set

hi

Sed

gew

ick

Sk

ien

a

Sp

ols

ky

Str

ou

stru

p

Su

ssm

an

Ta

nen

ba

um

Tu

rin

g

Wir

th

Cor

men

This is an example of binary search. We will have more in this video

2 / 6

Binary search

Recall the example from Lecture 1: how to search in sorted data

Ab

elso

n

Ah

o

Bro

ok

s

Cor

men

Dij

kst

ra

Eck

el

For

isek

Ga

mm

a

Go

del

Gra

ha

m

Ho

are

Ker

nig

ha

n

Kn

uth

La

mp

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McC

on

nel

l

Mey

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Nor

vig

Pa

pa

dim

itri

ou

Pa

tash

nik

Rit

chie

Ru

ssel

Set

hi

Sed

gew

ick

Sk

ien

a

Sp

ols

ky

Str

ou

stru

p

Su

ssm

an

Ta

nen

ba

um

Tu

rin

g

Wir

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This is an example of binary search. We will have more in this video

2 / 6

Binary search – Problem to solve

A very general form of binary search

I Given a function F : D → C , such that:

I There exists a totally ordered set S equipped with an average-of-two operation

I We will write it Avg(a, b) for arguments a and bI Avg(a, b) should be between a and b and should not be equal to neither a nor b,

unless there is no element of S between a and b

I D is a bounded subset of S : D = {s | s ∈ S ,Dmin ≤ S ,S ≤ Dmax}I Simply speaking, a piece of S between Dmin and Dmax

I C = {−1, 0,+1} with the following meanings:I −1: “too early”I 0: “just in time”I +1: “too late”

I Monotonicity: Whenever a < b, F (a) ≤ F (b)

I Need to find x ∈ D such that F (x) = 0

I Or, if impossible, find x and y as near as possible, such that F (x) = −1, F (y) = 1

3 / 6

Binary search – Problem to solve

A very general form of binary searchI Given a function F : D → C , such that:

I There exists a totally ordered set S equipped with an average-of-two operation

I We will write it Avg(a, b) for arguments a and bI Avg(a, b) should be between a and b and should not be equal to neither a nor b,

unless there is no element of S between a and b

I D is a bounded subset of S : D = {s | s ∈ S ,Dmin ≤ S ,S ≤ Dmax}I Simply speaking, a piece of S between Dmin and Dmax

I C = {−1, 0,+1} with the following meanings:I −1: “too early”I 0: “just in time”I +1: “too late”

I Monotonicity: Whenever a < b, F (a) ≤ F (b)

I Need to find x ∈ D such that F (x) = 0

I Or, if impossible, find x and y as near as possible, such that F (x) = −1, F (y) = 1

3 / 6

Binary search – Problem to solve

A very general form of binary searchI Given a function F : D → C , such that:

I There exists a totally ordered set S equipped with an average-of-two operation

I We will write it Avg(a, b) for arguments a and bI Avg(a, b) should be between a and b and should not be equal to neither a nor b,

unless there is no element of S between a and bI D is a bounded subset of S : D = {s | s ∈ S ,Dmin ≤ S ,S ≤ Dmax}

I Simply speaking, a piece of S between Dmin and Dmax

I C = {−1, 0,+1} with the following meanings:I −1: “too early”I 0: “just in time”I +1: “too late”

I Monotonicity: Whenever a < b, F (a) ≤ F (b)

I Need to find x ∈ D such that F (x) = 0

I Or, if impossible, find x and y as near as possible, such that F (x) = −1, F (y) = 1

3 / 6

Binary search – Problem to solve

A very general form of binary searchI Given a function F : D → C , such that:

I There exists a totally ordered set S equipped with an average-of-two operationI We will write it Avg(a, b) for arguments a and bI Avg(a, b) should be between a and b and should not be equal to neither a nor b,

unless there is no element of S between a and b

I D is a bounded subset of S : D = {s | s ∈ S ,Dmin ≤ S ,S ≤ Dmax}I Simply speaking, a piece of S between Dmin and Dmax

I C = {−1, 0,+1} with the following meanings:I −1: “too early”I 0: “just in time”I +1: “too late”

I Monotonicity: Whenever a < b, F (a) ≤ F (b)

I Need to find x ∈ D such that F (x) = 0

I Or, if impossible, find x and y as near as possible, such that F (x) = −1, F (y) = 1

3 / 6

Binary search – Problem to solve

A very general form of binary searchI Given a function F : D → C , such that:

I There exists a totally ordered set S equipped with an average-of-two operationI We will write it Avg(a, b) for arguments a and bI Avg(a, b) should be between a and b and should not be equal to neither a nor b,

unless there is no element of S between a and bI D is a bounded subset of S : D = {s | s ∈ S ,Dmin ≤ S ,S ≤ Dmax}

I Simply speaking, a piece of S between Dmin and Dmax

I C = {−1, 0,+1} with the following meanings:I −1: “too early”I 0: “just in time”I +1: “too late”

I Monotonicity: Whenever a < b, F (a) ≤ F (b)

I Need to find x ∈ D such that F (x) = 0

I Or, if impossible, find x and y as near as possible, such that F (x) = −1, F (y) = 1

3 / 6

Binary search – Problem to solve

A very general form of binary searchI Given a function F : D → C , such that:

I There exists a totally ordered set S equipped with an average-of-two operationI We will write it Avg(a, b) for arguments a and bI Avg(a, b) should be between a and b and should not be equal to neither a nor b,

unless there is no element of S between a and bI D is a bounded subset of S : D = {s | s ∈ S ,Dmin ≤ S ,S ≤ Dmax}

I Simply speaking, a piece of S between Dmin and Dmax

I C = {−1, 0,+1} with the following meanings:I −1: “too early”I 0: “just in time”I +1: “too late”

I Monotonicity: Whenever a < b, F (a) ≤ F (b)

I Need to find x ∈ D such that F (x) = 0

I Or, if impossible, find x and y as near as possible, such that F (x) = −1, F (y) = 1

3 / 6

Binary search – Problem to solve

A very general form of binary searchI Given a function F : D → C , such that:

I There exists a totally ordered set S equipped with an average-of-two operationI We will write it Avg(a, b) for arguments a and bI Avg(a, b) should be between a and b and should not be equal to neither a nor b,

unless there is no element of S between a and bI D is a bounded subset of S : D = {s | s ∈ S ,Dmin ≤ S ,S ≤ Dmax}

I Simply speaking, a piece of S between Dmin and Dmax

I C = {−1, 0,+1} with the following meanings:I −1: “too early”I 0: “just in time”I +1: “too late”

I Monotonicity: Whenever a < b, F (a) ≤ F (b)

I Need to find x ∈ D such that F (x) = 0

I Or, if impossible, find x and y as near as possible, such that F (x) = −1, F (y) = 1

3 / 6

Binary search – Problem to solve

A very general form of binary searchI Given a function F : D → C , such that:

I There exists a totally ordered set S equipped with an average-of-two operationI We will write it Avg(a, b) for arguments a and bI Avg(a, b) should be between a and b and should not be equal to neither a nor b,

unless there is no element of S between a and bI D is a bounded subset of S : D = {s | s ∈ S ,Dmin ≤ S ,S ≤ Dmax}

I Simply speaking, a piece of S between Dmin and Dmax

I C = {−1, 0,+1} with the following meanings:I −1: “too early”I 0: “just in time”I +1: “too late”

I Monotonicity: Whenever a < b, F (a) ≤ F (b)

I Need to find x ∈ D such that F (x) = 0

I Or, if impossible, find x and y as near as possible, such that F (x) = −1, F (y) = 1

3 / 6

Binary search – Problem to solve

A very general form of binary searchI Given a function F : D → C , such that:

I There exists a totally ordered set S equipped with an average-of-two operationI We will write it Avg(a, b) for arguments a and bI Avg(a, b) should be between a and b and should not be equal to neither a nor b,

unless there is no element of S between a and bI D is a bounded subset of S : D = {s | s ∈ S ,Dmin ≤ S ,S ≤ Dmax}

I Simply speaking, a piece of S between Dmin and Dmax

I C = {−1, 0,+1} with the following meanings:I −1: “too early”I 0: “just in time”I +1: “too late”

I Monotonicity: Whenever a < b, F (a) ≤ F (b)

I Need to find x ∈ D such that F (x) = 0I Or, if impossible, find x and y as near as possible, such that F (x) = −1, F (y) = 1

3 / 6

Binary search – Problem examples

I Given a function F : D → C , such that:I There exists a totally ordered set S equipped with an Avg(a, b) operationI D is a bounded subset of S : D = {s | s ∈ S ,Dmin ≤ S ,S ≤ Dmax}I C = {−1 (“too early”), 0 (”just in time”),+1 (“too late”)}I Monotonicity: Whenever a < b, F (a) ≤ F (b)

I Need to find x ∈ D such that F (x) = 0I Or, if impossible, find x , y as near as possible, such that F (x) = −1, F (y) = 1

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Binary search – Problem examples

I Given a function F : D → C , such that:I There exists a totally ordered set S equipped with an Avg(a, b) operationI D is a bounded subset of S : D = {s | s ∈ S ,Dmin ≤ S ,S ≤ Dmax}I C = {−1 (“too early”), 0 (”just in time”),+1 (“too late”)}I Monotonicity: Whenever a < b, F (a) ≤ F (b)

I Need to find x ∈ D such that F (x) = 0I Or, if impossible, find x , y as near as possible, such that F (x) = −1, F (y) = 1

Example: Find where q is in a sorted array

I D = [1;N]: the set of array indices, ordered naturally

I Avg(a, b) = b(a + b)/2cI F (x): “0” if q = x , “−1” if x < q, “+1” if x > q

I If no such element, “0” is infeasible: then you will find where to insert q

4 / 6

Binary search – Problem examples

I Given a function F : D → C , such that:I There exists a totally ordered set S equipped with an Avg(a, b) operationI D is a bounded subset of S : D = {s | s ∈ S ,Dmin ≤ S ,S ≤ Dmax}I C = {−1 (“too early”), 0 (”just in time”),+1 (“too late”)}I Monotonicity: Whenever a < b, F (a) ≤ F (b)

I Need to find x ∈ D such that F (x) = 0I Or, if impossible, find x , y as near as possible, such that F (x) = −1, F (y) = 1

Example: Find where q is in a sorted array

I D = [1;N]: the set of array indices, ordered naturally

I Avg(a, b) = b(a + b)/2cI F (x): “0” if q = x , “−1” if x < q, “+1” if x > q

I If no such element, “0” is infeasible: then you will find where to insert q

4 / 6

Binary search – Problem examples

I Given a function F : D → C , such that:I There exists a totally ordered set S equipped with an Avg(a, b) operationI D is a bounded subset of S : D = {s | s ∈ S ,Dmin ≤ S ,S ≤ Dmax}I C = {−1 (“too early”), 0 (”just in time”),+1 (“too late”)}I Monotonicity: Whenever a < b, F (a) ≤ F (b)

I Need to find x ∈ D such that F (x) = 0I Or, if impossible, find x , y as near as possible, such that F (x) = −1, F (y) = 1

Example: Find where q is in a sorted array

I D = [1;N]: the set of array indices, ordered naturally

I Avg(a, b) = b(a + b)/2cI F (x): “0” if q = x , “−1” if x < q, “+1” if x > q

I If no such element, “0” is infeasible: then you will find where to insert q

4 / 6

Binary search – Problem examples

I Given a function F : D → C , such that:I There exists a totally ordered set S equipped with an Avg(a, b) operationI D is a bounded subset of S : D = {s | s ∈ S ,Dmin ≤ S ,S ≤ Dmax}I C = {−1 (“too early”), 0 (”just in time”),+1 (“too late”)}I Monotonicity: Whenever a < b, F (a) ≤ F (b)

I Need to find x ∈ D such that F (x) = 0I Or, if impossible, find x , y as near as possible, such that F (x) = −1, F (y) = 1

Example: Find first occurence of q in a sorted array

I D = [1;N]: the set of array indices, ordered naturally

I Avg(a, b) = b(a + b)/2cI F (x): “0” never, “−1” if x < q, “+1” if x ≥ q

I But first check if F (y) = q. . .

4 / 6

Binary search – Problem examples

I Given a function F : D → C , such that:I There exists a totally ordered set S equipped with an Avg(a, b) operationI D is a bounded subset of S : D = {s | s ∈ S ,Dmin ≤ S ,S ≤ Dmax}I C = {−1 (“too early”), 0 (”just in time”),+1 (“too late”)}I Monotonicity: Whenever a < b, F (a) ≤ F (b)

I Need to find x ∈ D such that F (x) = 0I Or, if impossible, find x , y as near as possible, such that F (x) = −1, F (y) = 1

Example: Find first occurence of q in a sorted array

I D = [1;N]: the set of array indices, ordered naturally

I Avg(a, b) = b(a + b)/2cI F (x): “0” never, “−1” if x < q, “+1” if x ≥ q

I But first check if F (y) = q. . .

4 / 6

Binary search – Problem examples

I Given a function F : D → C , such that:I There exists a totally ordered set S equipped with an Avg(a, b) operationI D is a bounded subset of S : D = {s | s ∈ S ,Dmin ≤ S ,S ≤ Dmax}I C = {−1 (“too early”), 0 (”just in time”),+1 (“too late”)}I Monotonicity: Whenever a < b, F (a) ≤ F (b)

I Need to find x ∈ D such that F (x) = 0I Or, if impossible, find x , y as near as possible, such that F (x) = −1, F (y) = 1

Example: Find first occurence of q in a sorted array

I D = [1;N]: the set of array indices, ordered naturally

I Avg(a, b) = b(a + b)/2cI F (x): “0” never, “−1” if x < q, “+1” if x ≥ q

I But first check if F (y) = q. . .

4 / 6

Binary search – Problem examples

I Given a function F : D → C , such that:I There exists a totally ordered set S equipped with an Avg(a, b) operationI D is a bounded subset of S : D = {s | s ∈ S ,Dmin ≤ S ,S ≤ Dmax}I C = {−1 (“too early”), 0 (”just in time”),+1 (“too late”)}I Monotonicity: Whenever a < b, F (a) ≤ F (b)

I Need to find x ∈ D such that F (x) = 0I Or, if impossible, find x , y as near as possible, such that F (x) = −1, F (y) = 1

Example: Find a root of a monotonically growing function f

I D = [min; max]: the segment of R which we are interested in

I Avg(a, b) = (a + b)/2

I F (x): “0” if f (x) = 0, “−1” if f (x) < 0, “+1” if f (x) > 0

I Warning: you are unlikely to find the root exactly. . .

4 / 6

Binary search – Problem examples

I Given a function F : D → C , such that:I There exists a totally ordered set S equipped with an Avg(a, b) operationI D is a bounded subset of S : D = {s | s ∈ S ,Dmin ≤ S ,S ≤ Dmax}I C = {−1 (“too early”), 0 (”just in time”),+1 (“too late”)}I Monotonicity: Whenever a < b, F (a) ≤ F (b)

I Need to find x ∈ D such that F (x) = 0I Or, if impossible, find x , y as near as possible, such that F (x) = −1, F (y) = 1

Example: Find a root of a monotonically growing function f

I D = [min; max]: the segment of R which we are interested in

I Avg(a, b) = (a + b)/2

I F (x): “0” if f (x) = 0, “−1” if f (x) < 0, “+1” if f (x) > 0

I Warning: you are unlikely to find the root exactly. . .

4 / 6

Binary search – Problem examples

I Given a function F : D → C , such that:I There exists a totally ordered set S equipped with an Avg(a, b) operationI D is a bounded subset of S : D = {s | s ∈ S ,Dmin ≤ S ,S ≤ Dmax}I C = {−1 (“too early”), 0 (”just in time”),+1 (“too late”)}I Monotonicity: Whenever a < b, F (a) ≤ F (b)

I Need to find x ∈ D such that F (x) = 0I Or, if impossible, find x , y as near as possible, such that F (x) = −1, F (y) = 1

Example: Find a root of a monotonically growing function f

I D = [min; max]: the segment of R which we are interested in

I Avg(a, b) = (a + b)/2

I F (x): “0” if f (x) = 0, “−1” if f (x) < 0, “+1” if f (x) > 0

I Warning: you are unlikely to find the root exactly. . .

4 / 6

Binary search: Algorithm

function BinarySearch(F , Avg, Dmin, Dmax)L← Dmin, R ← Dmax, Vmin ← F (L), Vmax ← F (R)if Vmin = 1 then return 〈Null, Dmin〉 end ifif Vmax = −1 then return 〈Dmax,Null〉 end ifif Vmin = 0 then return 〈Dmin, Dmin〉 end ifif Vmax = 0 then return 〈Dmax, Dmax〉 end iffor ever do

M ← Avg(L,R)if M = L or M = R then return 〈L,R〉 end ifv ← F (M)if v = 0 then return 〈M,M〉 end ifif v = −1 then L← M else R ← M end if

end forend function

5 / 6

Binary search: Algorithm

function BinarySearch(F , Avg, Dmin, Dmax)L← Dmin, R ← Dmax, Vmin ← F (L), Vmax ← F (R) . First evaluate endpointsif Vmin = 1 then return 〈Null, Dmin〉 end ifif Vmax = −1 then return 〈Dmax,Null〉 end ifif Vmin = 0 then return 〈Dmin, Dmin〉 end ifif Vmax = 0 then return 〈Dmax, Dmax〉 end iffor ever do

M ← Avg(L,R)if M = L or M = R then return 〈L,R〉 end ifv ← F (M)if v = 0 then return 〈M,M〉 end ifif v = −1 then L← M else R ← M end if

end forend function

5 / 6

Binary search: Algorithm

function BinarySearch(F , Avg, Dmin, Dmax)L← Dmin, R ← Dmax, Vmin ← F (L), Vmax ← F (R) . First evaluate endpointsif Vmin = 1 then return 〈Null, Dmin〉 end if . If true, no zeros at allif Vmax = −1 then return 〈Dmax,Null〉 end ifif Vmin = 0 then return 〈Dmin, Dmin〉 end ifif Vmax = 0 then return 〈Dmax, Dmax〉 end iffor ever do

M ← Avg(L,R)if M = L or M = R then return 〈L,R〉 end ifv ← F (M)if v = 0 then return 〈M,M〉 end ifif v = −1 then L← M else R ← M end if

end forend function

5 / 6

Binary search: Algorithm

function BinarySearch(F , Avg, Dmin, Dmax)L← Dmin, R ← Dmax, Vmin ← F (L), Vmax ← F (R) . First evaluate endpointsif Vmin = 1 then return 〈Null, Dmin〉 end if . If true, no zeros at allif Vmax = −1 then return 〈Dmax,Null〉 end if . If true, no zeros at allif Vmin = 0 then return 〈Dmin, Dmin〉 end ifif Vmax = 0 then return 〈Dmax, Dmax〉 end iffor ever do

M ← Avg(L,R)if M = L or M = R then return 〈L,R〉 end ifv ← F (M)if v = 0 then return 〈M,M〉 end ifif v = −1 then L← M else R ← M end if

end forend function

5 / 6

Binary search: Algorithm

function BinarySearch(F , Avg, Dmin, Dmax)L← Dmin, R ← Dmax, Vmin ← F (L), Vmax ← F (R) . First evaluate endpointsif Vmin = 1 then return 〈Null, Dmin〉 end if . If true, no zeros at allif Vmax = −1 then return 〈Dmax,Null〉 end if . If true, no zeros at allif Vmin = 0 then return 〈Dmin, Dmin〉 end if . If true, zero is foundif Vmax = 0 then return 〈Dmax, Dmax〉 end iffor ever do

M ← Avg(L,R)if M = L or M = R then return 〈L,R〉 end ifv ← F (M)if v = 0 then return 〈M,M〉 end ifif v = −1 then L← M else R ← M end if

end forend function

5 / 6

Binary search: Algorithm

function BinarySearch(F , Avg, Dmin, Dmax)L← Dmin, R ← Dmax, Vmin ← F (L), Vmax ← F (R) . First evaluate endpointsif Vmin = 1 then return 〈Null, Dmin〉 end if . If true, no zeros at allif Vmax = −1 then return 〈Dmax,Null〉 end if . If true, no zeros at allif Vmin = 0 then return 〈Dmin, Dmin〉 end if . If true, zero is foundif Vmax = 0 then return 〈Dmax, Dmax〉 end if . If true, zero is foundfor ever do

M ← Avg(L,R)if M = L or M = R then return 〈L,R〉 end ifv ← F (M)if v = 0 then return 〈M,M〉 end ifif v = −1 then L← M else R ← M end if

end forend function

5 / 6

Binary search: Algorithm

function BinarySearch(F , Avg, Dmin, Dmax)L← Dmin, R ← Dmax, Vmin ← F (L), Vmax ← F (R) . First evaluate endpointsif Vmin = 1 then return 〈Null, Dmin〉 end if . If true, no zeros at allif Vmax = −1 then return 〈Dmax,Null〉 end if . If true, no zeros at allif Vmin = 0 then return 〈Dmin, Dmin〉 end if . If true, zero is foundif Vmax = 0 then return 〈Dmax, Dmax〉 end if . If true, zero is foundfor ever do . Invariant: F (L) = −1, F (R) = 1

M ← Avg(L,R)if M = L or M = R then return 〈L,R〉 end ifv ← F (M)if v = 0 then return 〈M,M〉 end ifif v = −1 then L← M else R ← M end if

end forend function

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Binary search: Algorithm

function BinarySearch(F , Avg, Dmin, Dmax)L← Dmin, R ← Dmax, Vmin ← F (L), Vmax ← F (R) . First evaluate endpointsif Vmin = 1 then return 〈Null, Dmin〉 end if . If true, no zeros at allif Vmax = −1 then return 〈Dmax,Null〉 end if . If true, no zeros at allif Vmin = 0 then return 〈Dmin, Dmin〉 end if . If true, zero is foundif Vmax = 0 then return 〈Dmax, Dmax〉 end if . If true, zero is foundfor ever do . Invariant: F (L) = −1, F (R) = 1

M ← Avg(L,R) . Getting new query pointif M = L or M = R then return 〈L,R〉 end ifv ← F (M)if v = 0 then return 〈M,M〉 end ifif v = −1 then L← M else R ← M end if

end forend function

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Binary search: Algorithm

function BinarySearch(F , Avg, Dmin, Dmax)L← Dmin, R ← Dmax, Vmin ← F (L), Vmax ← F (R) . First evaluate endpointsif Vmin = 1 then return 〈Null, Dmin〉 end if . If true, no zeros at allif Vmax = −1 then return 〈Dmax,Null〉 end if . If true, no zeros at allif Vmin = 0 then return 〈Dmin, Dmin〉 end if . If true, zero is foundif Vmax = 0 then return 〈Dmax, Dmax〉 end if . If true, zero is foundfor ever do . Invariant: F (L) = −1, F (R) = 1

M ← Avg(L,R) . Getting new query pointif M = L or M = R then return 〈L,R〉 end if . (L,R) empty → no zerosv ← F (M)if v = 0 then return 〈M,M〉 end ifif v = −1 then L← M else R ← M end if

end forend function

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Binary search: Algorithm

function BinarySearch(F , Avg, Dmin, Dmax)L← Dmin, R ← Dmax, Vmin ← F (L), Vmax ← F (R) . First evaluate endpointsif Vmin = 1 then return 〈Null, Dmin〉 end if . If true, no zeros at allif Vmax = −1 then return 〈Dmax,Null〉 end if . If true, no zeros at allif Vmin = 0 then return 〈Dmin, Dmin〉 end if . If true, zero is foundif Vmax = 0 then return 〈Dmax, Dmax〉 end if . If true, zero is foundfor ever do . Invariant: F (L) = −1, F (R) = 1

M ← Avg(L,R) . Getting new query pointif M = L or M = R then return 〈L,R〉 end if . (L,R) empty → no zerosv ← F (M) . Evaluating Mif v = 0 then return 〈M,M〉 end ifif v = −1 then L← M else R ← M end if

end forend function

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Binary search: Algorithm

function BinarySearch(F , Avg, Dmin, Dmax)L← Dmin, R ← Dmax, Vmin ← F (L), Vmax ← F (R) . First evaluate endpointsif Vmin = 1 then return 〈Null, Dmin〉 end if . If true, no zeros at allif Vmax = −1 then return 〈Dmax,Null〉 end if . If true, no zeros at allif Vmin = 0 then return 〈Dmin, Dmin〉 end if . If true, zero is foundif Vmax = 0 then return 〈Dmax, Dmax〉 end if . If true, zero is foundfor ever do . Invariant: F (L) = −1, F (R) = 1

M ← Avg(L,R) . Getting new query pointif M = L or M = R then return 〈L,R〉 end if . (L,R) empty → no zerosv ← F (M) . Evaluating Mif v = 0 then return 〈M,M〉 end if . Direct hit!if v = −1 then L← M else R ← M end if

end forend function

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Binary search: Algorithm

function BinarySearch(F , Avg, Dmin, Dmax)L← Dmin, R ← Dmax, Vmin ← F (L), Vmax ← F (R) . First evaluate endpointsif Vmin = 1 then return 〈Null, Dmin〉 end if . If true, no zeros at allif Vmax = −1 then return 〈Dmax,Null〉 end if . If true, no zeros at allif Vmin = 0 then return 〈Dmin, Dmin〉 end if . If true, zero is foundif Vmax = 0 then return 〈Dmax, Dmax〉 end if . If true, zero is foundfor ever do . Invariant: F (L) = −1, F (R) = 1

M ← Avg(L,R) . Getting new query pointif M = L or M = R then return 〈L,R〉 end if . (L,R) empty → no zerosv ← F (M) . Evaluating Mif v = 0 then return 〈M,M〉 end if . Direct hit!if v = −1 then L← M else R ← M end if . “Too early”: use right part

end forend function

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Binary search: Algorithm

function BinarySearch(F , Avg, Dmin, Dmax)L← Dmin, R ← Dmax, Vmin ← F (L), Vmax ← F (R) . First evaluate endpointsif Vmin = 1 then return 〈Null, Dmin〉 end if . If true, no zeros at allif Vmax = −1 then return 〈Dmax,Null〉 end if . If true, no zeros at allif Vmin = 0 then return 〈Dmin, Dmin〉 end if . If true, zero is foundif Vmax = 0 then return 〈Dmax, Dmax〉 end if . If true, zero is foundfor ever do . Invariant: F (L) = −1, F (R) = 1

M ← Avg(L,R) . Getting new query pointif M = L or M = R then return 〈L,R〉 end if . (L,R) empty → no zerosv ← F (M) . Evaluating Mif v = 0 then return 〈M,M〉 end if . Direct hit!if v = −1 then L← M else R ← M end if . “Too early”: use right part

end for . “Too late”: use left partend function

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Binary search: Properties

Termination

I Guaranteed to terminate if D is finite

I Proof: [L,R] shrinks at least by one item on every iteration

Correctness

I Loop invariant: F (L) = −1, F (R) = 1

I Zeros are always between → will be found if exist

I Nonexisting zeros: will report a point where −1 switches to 1

Running time

I Strongly depends on properties of AvgI If Avg is a “real” average, the running time is O(logD)

I the size of [L;R) range is divided by two on every iteration

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Binary search: Properties

Termination

I Guaranteed to terminate if D is finite

I Proof: [L,R] shrinks at least by one item on every iteration

Correctness

I Loop invariant: F (L) = −1, F (R) = 1

I Zeros are always between → will be found if exist

I Nonexisting zeros: will report a point where −1 switches to 1

Running time

I Strongly depends on properties of AvgI If Avg is a “real” average, the running time is O(logD)

I the size of [L;R) range is divided by two on every iteration

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Binary search: Properties

Termination

I Guaranteed to terminate if D is finite

I Proof: [L,R] shrinks at least by one item on every iteration

Correctness

I Loop invariant: F (L) = −1, F (R) = 1

I Zeros are always between → will be found if exist

I Nonexisting zeros: will report a point where −1 switches to 1

Running time

I Strongly depends on properties of AvgI If Avg is a “real” average, the running time is O(logD)

I the size of [L;R) range is divided by two on every iteration

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Binary search: Properties

Termination

I Guaranteed to terminate if D is finite

I Proof: [L,R] shrinks at least by one item on every iteration

Correctness

I Loop invariant: F (L) = −1, F (R) = 1

I Zeros are always between → will be found if exist

I Nonexisting zeros: will report a point where −1 switches to 1

Running time

I Strongly depends on properties of AvgI If Avg is a “real” average, the running time is O(logD)

I the size of [L;R) range is divided by two on every iteration

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Binary search: Properties

Termination

I Guaranteed to terminate if D is finite

I Proof: [L,R] shrinks at least by one item on every iteration

Correctness

I Loop invariant: F (L) = −1, F (R) = 1

I Zeros are always between → will be found if exist

I Nonexisting zeros: will report a point where −1 switches to 1

Running time

I Strongly depends on properties of AvgI If Avg is a “real” average, the running time is O(logD)

I the size of [L;R) range is divided by two on every iteration

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Binary search: Properties

Termination

I Guaranteed to terminate if D is finite

I Proof: [L,R] shrinks at least by one item on every iteration

Correctness

I Loop invariant: F (L) = −1, F (R) = 1

I Zeros are always between → will be found if exist

I Nonexisting zeros: will report a point where −1 switches to 1

Running time

I Strongly depends on properties of AvgI If Avg is a “real” average, the running time is O(logD)

I the size of [L;R) range is divided by two on every iteration

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Binary search: Properties

Termination

I Guaranteed to terminate if D is finite

I Proof: [L,R] shrinks at least by one item on every iteration

Correctness

I Loop invariant: F (L) = −1, F (R) = 1

I Zeros are always between → will be found if exist

I Nonexisting zeros: will report a point where −1 switches to 1

Running time

I Strongly depends on properties of Avg

I If Avg is a “real” average, the running time is O(logD)I the size of [L;R) range is divided by two on every iteration

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Binary search: Properties

Termination

I Guaranteed to terminate if D is finite

I Proof: [L,R] shrinks at least by one item on every iteration

Correctness

I Loop invariant: F (L) = −1, F (R) = 1

I Zeros are always between → will be found if exist

I Nonexisting zeros: will report a point where −1 switches to 1

Running time

I Strongly depends on properties of AvgI If Avg is a “real” average, the running time is O(logD)

I the size of [L;R) range is divided by two on every iteration

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