try all 2512 possible p's, but an astronomical number. In practice ..... n2000 is a nice time bound, or that the differences among n and 2n and n2 are...

0 downloads 0 Views 959KB Size

Winter 2011

Larry Ruzzo

Thanks to Paul Beame, James Lee, Kevin Wayne for some slides

1

goals

Design of Algorithms – a taste

design methods

common or important types of problems

analysis of algorithms - efficiency

2

goals

Complexity & intractability – a taste

solving problems in principle is not enough

algorithms must be efficient

some problems have no efficient solution

NP-complete problems

important & useful class of problems whose solutions (seemingly) cannot be found efficiently

3

complexity example Cryptography (e.g. RSA, SSL in browsers)

Secret: p,q prime, say 512 bits each

Public: n which equals p x q, 1024 bits

In principle

there is an algorithm that given n will find p and q: try all 2512 possible p’s, but an astronomical number

In practice

no fast algorithm known for this problem (on non-quantum computers)

security of RSA depends on this fact

(and research in “quantum computing” is strongly driven by the possibility of changing this)

4

algorithms versus machines

Moore’s Law and the exponential improvements in hardware...

Ex: sparse linear equations over 25 years

10 orders of magnitude improvement!

5

algorithms or hardware?

Seconds

25 years progress solving sparse linear systems

Hardware alone: 4 orders of magnitude

107

G.E. / CDC 3600

G.E. = Gaussian Elimination

CDC 6600

106

CDC 7600

Cray 1

105

Cray 2

104

Cray 3 (Est.)

103

102

101

100

1960

!

Source: Sandia, via M. Schultz 1970

1980

1990

6

2000

algorithms or hardware?

Seconds

25 years progress solving sparse linear systems

Hardware alone: 4 orders of magnitude

107

G.E. / CDC 3600

106

CDC 7600

Cray 1

105

Cray 2

104

103

Cray 3 (Est.)

Sparse G.E.

Gauss-Seidel

102

Software alone: 6 orders of magnitude

G.E. = Gaussian Elimination

SOR = Successive OverRelaxation

CG = Conjugate Gradient

CDC 6600

101

100

1960

SOR

CG

!

Source: Sandia, via M. Schultz 1970

1980

1990

7

2000

algorithms or hardware? The N-Body Problem:

in 30 years 107 hardware 1010 software

!

Source: T.Quinn

8

algorithms: a definition

Procedure to accomplish a task or solve a well-specified problem

Well-specified: know what all possible inputs look like and what output looks like given them

“accomplish” via simple, well-defined steps

Ex: sorting names (via comparison)

Ex: checking for primality (via +, -, *, /, ≤)

9

algorithms: a sample problem

Printed circuit-board company has a robot arm that solders components to the board

Time: proportional to total distance the arm must move from initial rest position around the board and back to the initial position

For each board design, find best order to do the soldering

10

printed circuit board

11

printed circuit board

12

more precise problem definition

Input: Given a set S of n points in the plane

Output: The shortest cycle tour that visits each point in the set S.

Better known as “TSP”

How might you solve it?

13

nearest neighbor heuristic Start at some point p0

Walk first to its nearest neighbor p1

Walk to the nearest unvisited neighbor p2, then nearest unvisited p3, … until all points have been visited

Then walk back to p0

heuristic: A rule of thumb, simplification, or educated guess that reduces or limits the search for solutions in domains that are difficult and poorly understood. May be good, but usually not guaranteed to give the best or fastest solution.

14

nearest neighbor heuristic

p 0!

p 1!

p 6!

15

an input where nn works badly

length ~ 84

16!

4!

1! .9! 2! p 0!

8! 16

an input where nn works badly

optimal soln for this example length ~ 64

16!

4!

1! .9! 2! p 0!

8! 17

revised heuristic – closest pairs first

Repeatedly join the closest pair of points

(such that result can still be part of a single loop in the end. I.e., join endpoints, but not points in middle, of path segments already created.)

?

How does this work on our bad example?

16!

4!

1! .9! 2! p 0!

8! 18

a bad example for closest pair

1! 1.5!

1.5!

19

a bad example for closest pair

1! 1.5!

1.5!

6+√10 = 9.16 !

vs !

8!

20

something that works

“Brute Force Search”:

For each of the n! = n(n-1)(n-2)…1 orderings of the points, check the length of the cycle;

Keep the best one

21

two notes The two incorrect algorithms were greedy

Often very natural & tempting ideas

They make choices that look great “locally” (and never reconsider them)

When greed works, the algorithms are typically efficient

BUT: often does not work - you get boxed in

Our correct alg avoids this, but is incredibly slow

20! is so large that checking one billion per second would take 2.4 billion seconds (around 70 years!)

And growing: n! ~ √2 π n • (n/e)n ~ 2O(n log n)

22

the morals of the story

Algorithms are important

Many performance gains outstrip Moore’s law

Simple problems can be hard

Factoring, TSP, many others

Simple ideas don’t always work

Nearest neighbor, closest pair heuristics

Simple algorithms can be very slow

Brute-force factoring, TSP

A point we hope to make: for some problems, even the best algorithms are slow

23

my plan

A brief overview of the theory of algorithms

Efficiency & asymptotic analysis

Some scattered examples of simple problems where clever algorithms help

A brief overview of the theory of intractability

Especially NP-complete problems

“Basics every educated CSE student should know”

24

computational complexity

The complexity of an algorithm associates a number T(n), the worst-case time the algorithm takes, with each problem size n.

Mathematically,

T: N+ → R+

i.e.,T is a function mapping positive integers (problem sizes) to positive real numbers (number of steps).

26

computational complexity

Time!

T(n)!

Problem size ! 27

computational complexity: general goals Characterize growth rate of worst-case run time as a function of problem size, up to a constant factor

Why not try to be more precise?

Average-case, e.g., is hard to define, analyze

Technological variations (computer, compiler, OS, …) easily 10x or more

Being more precise is a ton of work

A key question is “scale up”: if I can afford this today, how much longer will it take when my business is 2x larger? (E.g. today: cn2, next year: c(2n)2 = 4cn2 : 4 x longer.)

28

computational complexity

T(n)!

Time!

2n log2n!

n log2n! Problem size ! 29

asymptotic analysis & big-O Given two functions f and g: N→R, f(n) is O(g(n)) iff

∃ constant c > 0 so that f(n) is eventually always ≤ c g(n)

Example:

10n2-16n+100 is O(n2)

(and also O(n3)…)

why?:

10n2-16n+100 ≤ 11n2 for all n ≥ 10

30

polynomial vs exponential

For all r > 1 (no matter how small) and all d > 0, (no matter how large) nd = O(rn).

1.01n

n100

In short, every exponential grows faster than every polynomial!

31

the complexity class P: polynomial time

P: Running time O(nd) for some constant d

(d is independent of the input size n)

Nice scaling property: there is a constant c s.t. doubling n, time increases only by a factor of c.

(E.g., c ~ 2d)

Contrast with exponential: For any constant c, there is a d such that n → n+d increases time by a factor of more than c.

(E.g., 2n vs 2n+1)

32

polynomial vs exponential growth

22n! 22n

2n/10!

2n/10

1000n2

1000n2!

33

why it matters

not only get very big, but do so abruptly, which likely yields erratic performance on small instances

34

another view of poly vs exp Next year's computer will be 2x faster. If I can solve problem of size n0 today, how large a problem can I solve in the same time next year?

Complexity

Increase

O(n) O(n2) O(n3) 2n /10 2n

n0 → 2n0 n0 → √2 n0 n0 → 3√2 n0 n0 → n0+10 n0 → n0 +1

E.g. T=1012 1012 106 104 400 40

→ → → → →

2 x 1012 1.4 x 106 1.25 x 104 410 41

35

complexity summary

Typical initial goal for algorithm analysis is to find an

asymptotic

upper bound on

worst case running time

as a function of problem size

This is rarely the last word, but often helps separate good algorithms from blatantly poor ones - concentrate on the good ones!

36

why “polynomial”? Point is not that n2000 is a nice time bound, or that the differences among n and 2n and n2 are negligible.

Rather, simple theoretical tools may not easily capture such differences, whereas exponentials are qualitatively different from polynomials, so more amenable to theoretical analysis.

“My problem is in P” is a starting point for a more detailed analysis

“My problem is not in P” may suggest that you need to shift to a more tractable variant, or otherwise readjust expectations

37

algorithm design techniques

39

algorithm design techniques

We will survey two:

Later: Dynamic programming

Orderly solution of many smaller sub-problems, typically non-disjoint

Can give exponential speedups compared to more bruteforce approaches

Today: Divide & Conquer

Reduce problem to one or more sub-problems of the same type, typically disjoint

Often gives significant, usually polynomial, speedup

40

algorithm design techniques

Divide & Conquer

Reduce problem to one or more sub-problems of the same type

Each sub-problem’s size a fraction of the original

Subproblem’s typically disjoint

Often gives significant, usually polynomial, speedup

Examples:

Mergesort, Binary Search, Strassen’s Algorithm, Quicksort (roughly)

41

divide & conquer – the key idea

Suppose we've already invented DumbSort, taking time n2

Try Just One Level of divide & conquer:

DumbSort(first n/2 elements)

DumbSort(last n/2 elements)

Merge results

Time: 2 (n/2)2 + n = n2/2 + n << n2

D&C in a

nutshell

Almost twice as fast!

42

d&c approach, cont. Moral 1: “two halves are better than a whole”

Two problems of half size are better than one full-size problem, even given O(n) overhead of recombining, since the base algorithm has super-linear complexity.

Moral 2: “If a little's good, then more's better”

Two levels of D&C would be almost 4 times faster, 3 levels almost 8, etc., even though overhead is growing.

In the limit: you’ve just rediscovered mergesort.

43

mergesort (review)

T(n) = 2T(n/2)+cn, n≥2

T(1) = 0

Solution: O(n log n)

Log n levels!

Mergesort: (recursively) sort 2 half-lists, then merge results.

O(n) work per level!

44

A Divide & Conquer Example: Closest Pair of Points

46

closest pair of points: non-geometric version Given n points and arbitrary distances between them, find the closest pair. (E.g., think of distance as airfare – definitely not Euclidean distance!)

n) edges…)

(… and all the rest of the (2

Must look at all n choose 2 pairwise distances, else any one you didn’t check might be the shortest.

Also true for Euclidean distance in 1-2 dimensions?

47

closest pair of points: 1 dimensional version Given n points on the real line, find the closest pair

Closest pair is adjacent in ordered list

Time O(n log n) to sort, if needed

Plus O(n) to scan adjacent pairs

Key point: do not need to calc distances between all pairs: exploit geometry + ordering

48

closest pair of points. 2d, Euclidean distance: 1st try

Divide. Sub-divide region into 4 quadrants.

L

50

closest pair of points: 1st try

Divide. Sub-divide region into 4 quadrants.

Obstacle. Impossible to ensure n/4 points in each piece.

L

51

closest pair of points

Algorithm.

Divide: draw vertical line L with ≈ n/2 points on each side.

L

52

closest pair of points

Algorithm.

Divide: draw vertical line L with ≈ n/2 points on each side.

Conquer: find closest pair on each side, recursively.

L

21

12

53

closest pair of points

Algorithm.

Divide: draw vertical line L with ≈ n/2 points on each side.

Conquer: find closest pair on each side, recursively.

Combine: find closest pair with one point in each side.

seems Return best of 3 solutions.

like Θ(n2) ?

L

8

21

12

54

closest pair of points

Find closest pair with one point in each side, assuming that distance < δ.

L

21

12

δ = min(12, 21)

55

closest pair of points

Find closest pair with one point in each side, assuming that distance < δ.

Observation: suffices to consider points within δ of line L.

L

21

δ = min(12, 21)

12

56

δ

closest pair of points

Find closest pair with one point in each side, assuming that distance < δ.

Observation: suffices to consider points within δ of line L.

Almost the one-D problem again: Sort points in 2δ-strip by their y coordinate.

L

7

5

4

6

21

δ = min(12, 21)

12

3

2

1

57

δ

closest pair of points

Find closest pair with one point in each side, assuming that distance < δ.

Observation: suffices to consider points within δ of line L.

Almost the one-D problem again: Sort points in 2δ-strip by their y coordinate. Only check pts within 8 in sorted list!

L

7

5

4

6

21

δ = min(12, 21)

12

3

2

1

58

δ

closest pair of points

Def. Let si be the point in the 2δ-strip, with the ith smallest ycoordinate.

j

39

31

Claim. If |i – j| > 8, then the distance between si and sj is > δ.

i Pf: No two points lie in same ½δ-by-½δ box; only 8 boxes within δ

30

½δ

29

28

27

26

25

59

δ

δ

½δ

closest pair of points: analysis

Number of pairwise distance calculations:

⎧ 0 D(n) ≤ ⎨2D n /2 + 7n ) ⎩ (

n = 1⎫ ⎬ n > 1⎭ ⇒ D(n) = O(n log n)

€ (A

mostly superfluous detail: straightforward implementation gives a running time that is a factor of log n larger, due to sorting in the various subproblems. Run time can be reduced to O(n log n) also, roughly by the trick of sorting by x at the top level, and returning/merging y-sorted lists from the subcalls.

Regardless of this nuance, the big picture is the same: divideand-conquer allows sharp speed gain over a naive n2 method.)

61

Integer Multiplication

63

integer arithmetic

Add. Given two n-digit integers a and b, compute a + b.

Add

O(n) bit operations.

1

1

1

1

1

1

0

1

1

1

0

1

0

1

0

1

+

0

1

1

1

1

1

0

1

1

0

1

0

1

0

0

1

0

1

1

0

1

0

1

0

1

Multiply. Given two n-digit integers a and b, compute a × b. The “grade school” method: Θ(n2) bit operations.

*

0

1

1

1

1

1

0

1

1

1

0

1

0

1

0

1

Multiply

0

0

0

0

0

0

0

0

1

1

0

1

0

1

0

1

1

1

0

1

0

1

0

1

1

1

0

1

0

1

0

1

1

1

0

1

0

1

0

1

1

1

0

1

0

1

0

1

0

0

0

0

0

0

0

0

0

1

1

0

1

0

0

0

0

0

0

0

0

064

1

0

integer arithmetic

Add. Given two n-digit integers a and b, compute a + b.

Add

O(n) bit operations.

1

1

1

1

1

1

0

1

1

1

0

1

0

1

0

1

+

0

1

1

1

1

1

0

1

1

0

1

0

1

0

0

1

0

1

1

0

1

0

1

0

1

Multiply. Given two n-digit integers a and b, compute a × b. The “grade school” method: Θ(n2) bit operations.

*

0

1

1

1

1

1

0

1

1

1

0

1

0

1

0

1

Multiply

0

0

0

0

0

0

0

0

1

1

0

1

0

1

0

1

1

1

0

1

0

1

0

1

1

1

0

1

0

1

0

1

1

1

0

1

0

1

0

1

1

1

0

1

0

1

0

1

0

0

0

0

0

0

0

0

0

1

1

0

1

0

0

0

0

0

0

0

0

065

1

0

divide-and-conquer multiplication: warmup

To multiply two 2-digit integers:

Multiply four 1-digit integers.

Add, shift some 2-digit integers to obtain result.

x y xy

= 10⋅ x1 + x 0 = 10⋅ y1 + y 0 = (10⋅ x1 + x 0 ) (10⋅ y1 + y 0 ) = 100 ⋅ x1 y1 + 10⋅ ( x1 y 0 + x 0 y1 ) + x 0 y 0

Same idea works for long integers –

can split them into 4 half-sized ints

4

5

y1 y0

3

2

x1 x0

1

0

x0⋅y0

0

8

x0⋅y1

1

5

x1⋅y0

1

2

x1⋅y1

1

4

4

0

66

divide-and-conquer multiplication: warmup

To multiply two n-digit integers:

Multiply four n/2-digit integers.

Add, shift some n/2-digit integers to obtain result.

x = 2 n / 2 ⋅ x1 + y = 2 n / 2 ⋅ y1 + xy = (2 n / 2 ⋅ x1 + = 2 n ⋅ x1 y1 +

x0 y0 x 0 ) (2 n / 2 ⋅ y1 + y 0 ) 2 n / 2 ⋅ ( x1 y 0 + x 0 y1 ) + x 0 y 0

T(n) = 4T (n /2 ) + Θ(n) ⇒ T(n) = Θ(n 2 ) recursive calls

€

add, shift

1

1

0

1

0

1

0

1

y1 y0

*

0

1

1

1

1

1

0

1

x1 x0

0

1

0

0

0

0

0

1

x0⋅y0

1

0

1

0

1

0

0

1

x0⋅y1

0

0

1

0

0

0

1

1

x1⋅y0

0

1

0

1

1

0

1

1

0

1

1

0

1

0

0

0

0

0

0

0

x1⋅y1

0

0

0

1

assumes n is a power of 2

67

key trick: 2 multiplies for the price of 1 x y xy

= 2 n / 2 ⋅ x1 + x 0 = 2 n / 2 ⋅ y1 + y 0 = (2 n / 2 ⋅ x1 + x 0 ) (2 n / 2 ⋅ y1 + y 0 )

Well, ok, 4 for 3 is more accurate…

= 2 n ⋅ x1 y1 + 2 n / 2 ⋅ ( x1 y 0 + x 0 y1 ) + x 0 y 0

€

α β αβ

( x1 y 0 + x 0 y1 )

= = = = =

x1 + x 0 y1 + y 0 ( x1 + x 0 ) ( y1 + y 0 ) x1 y1 + ( x1 y 0 + x 0 y1 ) + x 0 y 0 αβ − x1 y1 − x 0 y 0 68

€

Karatsuba multiplication To multiply two n-digit integers:

Add two ½n digit integers.

Multiply three ½n-digit integers.

Add, subtract, and shift ½n-digit integers to obtain result.

x = 2 n / 2 ⋅ x1 y = 2 n / 2 ⋅ y1 xy = 2 n ⋅ x1 y1 = 2 n ⋅ x1 y1

+ + + +

x0 y0 2 n / 2 ⋅ ( x1 y0 + x0 y1 ) + x0 y0 2 n / 2 ⋅ ( (x1 + x0 ) (y1 + y0 ) − x1 y1 − x0 y0 ) + x0 y0

A

B

A

C

C

Theorem. [Karatsuba-Ofman, 1962] Can multiply two n€ digit integers in O(n1.585) bit ops.

T(n) ≤ 3 T ( n /2) + O(n) recursive calls

⇒ T(n) = O(n

add, subtract, shift

log 2 3

) = O(n1.585 )

69

multiplication – the bottom line

Naïve:

Θ(n2)

Karatsuba:

Θ(n1.59…)

Amusing exercise: generalize Karatsuba to do 5 size n/3 subproblems → Θ(n1.46…)

Best known:

Θ(n log n loglog n)

"Fast Fourier Transform"

but mostly unused in practice (unless you need really big numbers - a billion digits of π, say)

High precision arithmetic IS important for crypto

71

d & c summary

Idea:

“Two halves are better than a whole”

if the base algorithm has super-linear complexity.

“If a little's good, then more's better”

repeat above, recursively

Applications: Many.

Binary Search, Merge Sort, (Quicksort), Closest points, Integer multiply,…

73

Our partners will collect data and use cookies for ad personalization and measurement. Learn how we and our ad partner Google, collect and use data. Agree & close