On Time Versus Input Size
Great Theoretical Ideas In Computer Science
S. RudichV. Adamchik
CS 15-251 Spring 2006
Lecture 17 March 21, 2006 Carnegie Mellon University
# of bits
time
How to add 2 n-bit numbers.
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“Grade school addition”
Time complexity of grade school addition
+T(n) = amount of time grade school
addition uses to add two n-bit numbers
What do you mean by “time”?
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Our Goal
We want to define “time” taken by the method of grade school addition without
depending on the implementation details.
Hold on!
But you agree that T(n) does depend on the implementation!
A given algorithm will take different amounts of time on the same inputs depending on such factors as:
Processor speedInstruction setDisk speedBrand of compiler
These objections are serious, but they are not undefeatable.
There is a very nice sense in which we can analyze grade
school addition without having to worry about implementation
details.
Here is how it works . . .
Grade school addition
On any reasonable computer M, adding 3 bits and writing down 2 bits (for short ) can be done in constant time c.
Total time to add two n-bit numbers using grade school addition: c*n[c time for each of n columns]
Grade school addition
On another computer M’, the time to perform may be c’.
Total time to add two n-bit numbers using grade school addition: c’*n[c’ time for each of n columns]
Different machines result in different slopes, but time grows linearly as input size increases.
# of bits in the numbers
time
Machine M
: cn
Machine M’: c’n
You measure “time” asthe number of elementary “steps” defined in any other way, provided each such “step” takes constant timein a reasonable implementation.
Constant: independent of the length n of the input.
Thus we arrive at an implementation independent
insight:
Grade School Addition is a linear time algorithm.
AbstractionAbstraction: : Abstract away the inessential Abstract away the inessential
features of a problem or solutionfeatures of a problem or solution
=
“Keep it simple, stupid!”
The process of abstracting away details and determining the rate of resource usagein terms of the problem size is one of the fundamental ideas in computer science.
How to multiply 2 n-bit numbers.
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How to multiply 2 nHow to multiply 2 n--bit numbers.bit numbers.
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The total time is bounded by c n2 (abstracting away
the implementation details).
Grade School Addition: Linear timeGrade School Multiplication: Quadratic
time
No matter how dramatic the difference in the constants, the quadratic curve will eventually
dominate the linear curve
# of bits in the numbers
time
Time vs. Input Size
For any algorithm, define n – the input size
and T(n) - the amount of time used
on inputs of size n
We will ask:
What is the growth rate of T(n) ?
Useful notation to discuss growth rates
For any monotonic functions f, g from the positive integers to the positive integers, we say
f(n) = O( g(n) )if
g(n) eventually dominates f(n)
[Formally: there exists a constant c such that for all sufficiently large n: f(n) ≤ c*g(n) ]
More useful notation: Ω
For any monotonic functions f, g from the positive integers to the positive integers, we say
f(n) = Ω( g(n) )if:
f(n) eventually dominates g(n)
[Formally: there exists a constant c such that for all sufficiently large n: f(n) ≥ c*g(n) ]
Yet more useful notation: Θ
For any monotonic functions f, g from the positive integers to the positive integers, we say
f(n) = Θ( g(n) )if:
f(n) = O( g(n) ) and f = Ω( g(n) )
# of bits in numbers
time
f = Θ(n) means that f can be sandwiched between two lines
from some point on.
Quickies
• n = O(n2) ?– YES
• n = O(√n) ?– NO
• n2 = Ω(n log n) ?– YES
• 3n2 + 4n + = Ω(n2) ?– YES
• 3n2 + 4n + = Θ(n2) ?– YES
• n2 log n = Θ(n2) ?– NO
Names For Some Growth Rates
Linear Time: T(n) = O(n)Quadratic Time: T(n) = O(n2)Cubic Time: T(n) = O(n3)
Polynomial Time: for some constant k, T(n) = O(nk).
Example: T(n) = 13 n5
Large Growth Rates
Exponential Time: for some constant k, T(n) = O(kn)
Example: T(n) = n*2n
Logarithmic Time: T(n) = O(logn) Example: T(n) = 15 log2(n)
Complexity of Songs
Suppose we want to sing a song of length n.
Since n can be large, we want to memorize songs which require only a small amount of brain space.
Let S(n) be the space complexity of a song.
Complexity of Songs
The amount of space S(n) can be measured in either characters of words.
What is asymptotic relation between # of characters and the # of words?
Complexity of Songs
The amount of space S(n) can be measured in either characters of words.
What is asymptotic relation between # of characters and the # of words?
# of characters = Θ (# of words)
Complexity of Songs
S(n) is the space complexity of a song.
What is the upper and lower bounds for S(n)?
Complexity of Songs
S(n) is the space complexity of a song.
What is the upper and lower bounds for S(n)?
S(n) = O(n) S(n) = Ω(1)
Complexity of Songs with a refrain
Let S be a song with m verses of length V and a refrain of length R.The refrain is first, last and between.
What is the space complexity of S?
Complexity of Songs with a refrain
Let S be a song with m verses of length V and a refrain of length R.
The song length: n = R + (V+R)*mThe space: s = R + V*mEliminate m to get s = V*n/(V+R) +constTherefore, S(n) = O(n)
100 Bottles of Beer
n bottles of beer on the wall, n bottles of beer. You take one down and pass it around.n-1 bottles of beer on the wall.
What is the space complexity of S?
100 Bottles of Beer
n bottles of beer on the wall, n bottles of beer. You take one down and pass it around.n-1 bottles of beer on the wall.
What is the space complexity of S?
We have to remember one verse (template) and the current value of n. S = Θ(1) + O(log n)
The k Days of Christmas
On the first day of Christmas, my true love sent to me
A partridge in a pear tree.
The k Days of Christmas
On the k day of Christmas, my true love sent to me
gift1 ,…, giftk
What is the space complexity of this song?
The k Days of Christmas
What is the space complexity of this song?
The song length: n = template + k*gift1+(k-1)*gift2+…+ giftk
Let all gifts be the same length G.n = Θ(1) + G*(1+2+…+k) = O(k2)
The space: S = Θ(1) + k*G = O(k) = O(n)
•The worst time complexity •The best time complexity •The average time complexity•The amortized time complexity •The randomized time complexity
Time complexities
In sorting and searching (array) algorithms, the input size is the number of items.
In graph algorithm the input size is presented by V+E
In number algorithms, the input size is the number of bits.
The input size
Primality
boolean prime(int n) int k = 2; while( k++ <= Math.sqrt(n) ) if(n%k == 0) return false; return true;
What is the complexity of this algorithm?
For a given algorithm, the input size is defined as the number of characters it takes to write (or encode) the input.
The input size
Primality
The number of passes through the loop is n.
If we use base 10 encoding, it takes O( log10 n ) = input_sizecharacters (digits) to encode n.
Therefore, T(n) = 10^(input_size/2)The time complexity is not polynomial!
Fibonacci Numbers
int fib( int n ) int [] f = new int[n+2]; f[0]=0; f[1]=1; for( int k = 2; k<=n; k++)
f[k] = f[k-1] +f[k-2]; return f[n];
What is the complexity of this algorithm?
Fibonacci Numbers
public static int fib( int n )int [] f = new int[n+2];f[0]=0; f[1]=1;for( int k = 2; k<=n; k++)
f[k] = f[k-1] +f[k-2];return f[n];
input_size = log n, then n = 2^(input_size)The algorithm is exponential in terms of size.
0-1 Knapsack problem
A thief breaks into a jewelry store carrying a knapsack. Each item in the store has a value and a weight. The knapsack will break if the total weight exceeds W. The thief’s dilemma is to maximize the total value
What is the time complexity using the dynamic programming approach?
0-1 Knapsack problem
We create a table of size n*W,where n is the number of items andW is the maximum weight allowed
T = Θ(n*W) But the input size of W is log W, therefore
T = Θ( n*2^(input size of W) ) If you change W by one bit, the amount of work will double.
Grade School Multiplication:Quadratic time
c(log n)2 time to square the number n
# of bits in numbers
time
Some Big Ones
Doubly Exponential Time means that for some constant k
Triply Exponential
And so forth.
T(n) = 22k n
T(n) = 222k n
Exponential Times
2STACK(0) = 1
2STACK(n) = 22STACK(n-1)
2STACK(1) = 22STACK(2) = 42STACK(3) = 162STACK(4) = 65536
2STACK(5) ¸ 1080 = atoms in universe
2222: : :2
“tower of n 2’s”
2STACK(n) =
And the inverse of 2STACK: log*
2STACK(0) = 1 log*(1) = 0
2STACK(n) = 22STACK(n-1)
2STACK(1) = 2 log*(2) = 12STACK(2) = 4 log*(4) = 22STACK(3) = 16 log*(16) = 32STACK(4) = 65536 log*(65536) = 4
2STACK(5) ¸ 1080 log*(atoms) = 5 = atoms in universe
log*(n) = # of times you have to apply the log function to n to make it ≤ 1
So an algorithm that can be shown to run
in O(n log*n) Time
isLinear Time for all
practical purposes!!
Ackermann’s Function
A(0, n) = n + 1 for n ≥ 0 A(m, 0) = A(m - 1, 1) for m ≥ 1 A(m, n) = A(m - 1, A(m, n - 1)) for m, n ≥ 1
A(4,2) > # of particles in universeA(5,2) can’t be written out in this universe
Inverse Ackermann function
A(0, n) = n + 1 for n ≥ 0 A(m, 0) = A(m - 1, 1) for m ≥ 1 A(m, n) = A(m - 1, A(m, n - 1)) for m, n ≥ 1
Define: A’(k) = A(k,k) Inverse Ackerman α(n) is the inverse of A’
Practically speaking: n × α(n) ≤ 4n
Inverse Ackermann function
It grows slower than log* n
The amortized time of Union-Find takes
O(a(V, E))Finding a MST O(E a(V, E))
Time complexity of grade school addition
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T(n) = The amount of time grade school addition uses to add two n-bit numbers
We saw that T(n) was linear.
T(n) = Θ(n)
Time complexity of grade school multiplication
T(n) = The amount of time grade school multiplication uses
to add two n-bit numbers
We saw that T(n) was quadratic.
T(n) = Θ(n2)
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