Big Oh Notation• Greek letter Omicron (Ο) is used to denote the

limit of asymptotic growth of an algorithm• If algorithm processing time grows linearly with

the input set n, then we say the algorithm is Order n, or O(n).

• This notation isolates an algorithm’s run-time from other factors:– Size of the problem set– Initialization time– Processor speed and instruction set

Big-Oh notation• Let b(x) be the bubble sort algorithm• We say b(x) is O(n2)

– This is read as “b(x) is big-oh n2”– This means that the input size increases, the running time

of the bubble sort will increase proportional to the square of the input size

• In other words, by some constant times n2

• Let l(x) be the linear (or sequential) search algorithm• We say l(x) is O(n)

– Meaning the running time of the linear search increases directly proportional to the input size

Big-Oh notation

• Consider: b(x) is O(n2)– That means that b(x)’s running time is less

than (or equal to) some constant times n2

• Consider: l(x) is O(n)– That means that l(x)’s running time is less

than (or equal to) some constant times n

Big-Oh proofs• Show that f(x) = x2 + 2x + 1 is O(x2)

– In other words, show that x2 + 2x + 1 ≤ c*x2

• Where c is some constant• For input size greater than some x

• We know that 2x2 ≥ 2x whenever x ≥ 1• And we know that x2 ≥ 1 whenever x ≥ 1• So we replace 2x+1 with 3x2

– We then end up with x2 + 3x2 = 4x2

– This yields 4x2 ≤ c*x2

• This, for input sizes 1 or greater, when the constant is 4 or greater, f(x) is O(x2)

• We could have chosen values for c and x that were different

Big-Oh proofs

Sample Big-Oh problems

• Show that f(x) = x2 + 1000 is O(x2)– In other words, show that x2 + 1000 ≤ c*x2

• We know that x2 > 1000 whenever x > 31– Thus, we replace 1000 with x2

– This yields 2x2 ≤ c*x2

• Thus, f(x) is O(x2) for all x > 31 when c ≥ 2

Sample Big-Oh problems

• Show that f(x) = 3x+7 is O(x)– In other words, show that 3x+7 ≤ c*x

• We know that x > 7 whenever x > 7– Uh huh….– So we replace 7 with x– This yields 4x ≤ c*x

• Thus, f(x) is O(x) for all x > 7 when c ≥ 4

A variant of the last question• Show that f(x) = 3x+7 is O(x2)

– In other words, show that 3x+7 ≤ c*x2

• We know that x > 7 whenever x > 7– Uh huh….– So we replace 7 with x– This yields 4x < c*x2

– This will also be true for x > 7 when c ≥ 1

• Thus, f(x) is O(x2) for all x > 7 when c ≥ 1

What that means

• If a function is O(x)– Then it is also O(x2)– And it is also O(x3)

• Meaning a O(x) function will grow at a slower or equal to the rate x, x2, x3, etc.

Function growth rates• For input size n = 1000

• O(1) 1• O(log n) ≈10• O(n) 103

• O(n log n) ≈104

• O(n2) 106

• O(n3) 109

• O(n4) 1012

• O(nc) 103*c c is a consant• 2n ≈10301

• n! ≈102568

• nn 103000

Many interesting problems fall into these categories

Function growth rates

Logarithmic scale!

Integer factorization

• Factoring a composite number into it’s component primes is O(2n)– Where n is the number of bits in the number

• This, if we choose 2048 bit numbers (as in RSA keys), it takes 22048 steps– That’s about 10617 steps!

Formal Big-Oh definition

• Let f and g be functions. We say that f(x) is O(g(x)) if there are constants c and k such that

|f(x)| ≤ C |g(x)|whenever x > k

Formal Big-Oh definition

Big Omega (Ω) and Big Theta (Θ)

• If Big-Oh a less-than relationship:– then Big Omega is greater-than

•  | f(x) | > C | g(x) | when x > n0

– and Big Theta is equals• if f(x) is O(g(x)) and Ω(g(x)), then it is Θ(g(x))

• x2 is Θ(x2)• x is O(x2)• x2 is Ω(x)

A useful recursive algorithm

• Merge sort

procedureprocedure mergesortmergesort ( (LL = = aa11,…a,…ann))if if nn>1 >1 thenthen

mm := := floorfloor(n/2)(n/2)LL11 := := aa11,a,a22,…,a,…,amm

LL22 := := aam+1m+1,a,am+2m+2,…,a,…,ann

LL := := mergemerge((mergesortmergesort((LL11),),mergesortmergesort((LL22))))

{L is now sorted into elements of increasing order}{L is now sorted into elements of increasing order}

mergesort needs mergeprocedureprocedure mergemerge((LL11, L, L22: lists): lists)LL := empty list := empty listwhile while LL11 and and LL22 are both nonempty are both nonemptybeginbegin

remove smaller of first element of remove smaller of first element of LL11 and and LL22 from the list it is in and put it at the left end of from the list it is in and put it at the left end of LL

if if removal of this element make one list removal of this element make one list emptyempty

thenthenremove all elements from the other list and remove all elements from the other list and

append them to append them to LLendendreturnreturn LL

{L is the merged list with elements in increasing order}{L is the merged list with elements in increasing order}

Time complexity

• First: how many recursive calls are there for n inputs?

Satisfiability• Consider a Boolean expression of the form:• (x1 x2 x3) (x2 x3 x4) (x1 x4 x5)

– This is a conjunction of disjunctions• Is such an equation satisfiable?

– In other words, can you assign truth values to all the xi’s such that the equation is true?

– The above problem is easy (only 3 clauses of 3 variables) – set x1, x2, and x4 to true

• There are other possibilities: set x1, x2, and x5 to true, etc.– But consider an expression with 1000 variables and

thousands of clauses

Satisfiability• If given a solution, it is easy to check if such a solution

works– Plug in the values – this can be done quickly, even by hand

• However, there is no known efficient way to find such a solution– The only definitive way to do so is to try all possible values for the

n Boolean variables– That means this is O(2n)!– Thus it is not a polynomial time function

• NP stands for “Not Polynomial”

• Cook’s theorem (1971) states that SAT is NP-complete– There still may be an efficient way to solve it, though!

NP Completeness

• There are hundreds of NP complete problems– It has been shown that if you can solve one of them

efficiently, then you can solve them all– Example: the traveling salesman problem

• Given a number of cities and the costs of traveling from any city to any other city, what is the cheapest round-trip route that visits each city once and then returns to the starting city?

• Not all algorithms that are O(2n) are NP complete– In particular, integer factorization (also O(2n)) is not

thought to be NP complete

NP Completeness• It is “widely believed” that there is no efficient solution to

NP complete problems– In other words, everybody has that belief

• If you could solve an NP complete problem in polynomial time, you would be showing that P = NP– And you’d get a million dollar prize (and lots of fame!)

• If this were possible, it would be like proving that Newton’s or Einstein’s laws of physics were wrong

• In summary: – NP complete problems are very difficult to solve, but easy to

check the solutions of– It is believed that there is no efficient way to solve them

Reserve

An aside: inequalities• If you have a inequality you need to show:

x < y• You can replace the lesser side with something greater:

x+1 < y• If you can still show this to be true, then the original

inequality is true

• Consider showing that 15 < 20– You can replace 15 with 16, and then show that 16 < 20.

Because 15 < 16, and 16 < 20, then 15 < 20

An aside: inequalities• If you have a inequality you need to show:

x < y• You can replace the greater side with something lesser:

x < y-1• If you can still show this to be true, then the original

inequality is true

• Consider showing that 15 < 20– You can replace 20 with 19, and then show that 15 < 19.

Because 15 < 19, and 19 < 20, then 15 < 20

An aside: inequalities• What if you do such a replacement and can’t

show anything?– Then you can’t say anything about the original

inequality

• Consider showing that 15 < 20– You can replace 20 with 10– But you can’t show that 15 < 10– So you can’t say anything one way or the other about

the original inequality