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Intro to Analysis of Algorithms

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Algorithm “A sequence of unambiguous instructions for solving a problem, i.e., for obtaining a required output for any legitimate input in a finite amount of time.” Named for Al Khwarizmi, who laid out basic procedures for arithmetic functions. (Read about him!)

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Analysis of Algorithms Correctness Generality Optimality Simplicity Time Efficiency Space Efficiency

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Measuring Efficiency What is basic unit to measure input size? (n) What is basic unit of resource? – Time: basic unit operation – Space: memory units Best, worst, or average case? Find its efficiency class

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Why do we care? Let’s look at Fibonacci numbers: 1, 1, 2, 3, 5, 8, 13, …

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Fibonacci Sequence We want to compute the nth number in the sequence. (F 3 = 2, for example.)

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This definition can be translated directly into code – a recursive method. How many additions does it take to compute F n ?

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Which is better? function fib2(n) create an array f[0...n] f[0] = 0, f[1] = 1 for i = 2...n: f[i] = f[i-1] + f[i-2] return f[n] function fib1(n) if n = 0: return 0 if n = 1: return 1 return fib1(n-1) + fib1(n-2)

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Consider calculating F 200. The fib1 method takes over 2 138 steps. Computers can do several billion instructions per second. Suppose we have a supercomputer that does 40 trillion instructions per second.

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Consider calculating F 200. The fib1 method takes over 2 138 steps. Computers can do several billion instructions per second. Suppose we have a supercomputer that does 40 trillion instructions per second. Even on this machine, fib1(200) takes at least 2 92 seconds, or 10 18 centuries, long after the expected end of our sun!!!

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Consider calculating F 200. The fib1 method takes over 2 138 steps. Computers can do several billion instructions per second. Suppose we have a supercomputer that does 40 trillion instructions per second. Even on this machine, fib1(200) takes at least 2 92 seconds, or 10 18 centuries, long after the expected end of our sun!!! But, fib2(200) would take less than a billionth of a second to compute!!!

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10 6 instructions/sec, runtimes N O(log N)O(N)O(N log N)O(N 2 ) 10 0.0000030.000010.0000330.0001 100 0.0000070.000100.0006640.1000 1,000 0.0000100.001000.0100001.0 10,000 0.0000130.010000.1329001.7 min 100,000 0.0000170.100001.6610002.78 hr 1,000,000 0.0000201.019.911.6 day 1,000,000,000 0.00003016.7 min18.3 hr318 centuries

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Some helpful mathematics 1 + 2 + 3 + 4 + … + N – N(N+1)/2 = N 2 /2 + N/2 is O(N 2 ) N + N + N + …. + N (total of N times) – N*N = N 2 which is O(N 2 ) 1 + 2 + 4 + … + 2 N – 2 N+1 – 1 = 2 x 2 N – 1 which is O(2 N )

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Basics of Efficiency Big-oh – upper bound on the efficiency class Efficiency classes don’t worry with constants A cubic is worse than a quadratic, quartic worse than cubic… Getting big-oh analysis of non-recursive code is pretty easy

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Maximum(A[1..n]) max <- A[0] for i<- 1 to n if A[i] > max max <- A[i] return max 1.What is input size? 2.What is unit of time? 3.What is big-oh analysis?

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Maximum(A[1..n]) max <- A[0] for i<- 1 to n if A[i] > max max <- A[i] return max 1 assignment n times: 1 compare maybe 1 assignment 1 addition to i 1 return ______________________ 1 + n(3) + 1 = O(3n+2)

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The algorithm is O(3n+2), which is O(n). We only care about the efficiency class. Why?

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The algorithm is O(3n+2), which is O(n). We only care about the efficiency class. Why? At some point, every parabola (n 2 ) overtakes any line (n). We only really care about large input.

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Efficiency classes So we really just care about the leading term, which determines the shape of the graph. This means for non-recursive algorithms, what’s important is the loops.

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Analyze this… AllUnique(A[1..n]) for i<-1 to n for j<- 1 to n if A[i] = A[j] and i ≠ j return false return true Best case? Worst case? Average case?

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Analyze this… AllUnique(A[1..n]) for i<-1 to n for j<- 1 to n if A[i] = A[j] and i ≠ j return false return true Average case? Quit halfway through, O(n*n/2) Still O(n 2 ) Often, Average case = Worst case.

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AllUnique(A[1..n]) for i<-1 to n for j<- i to n if A[i] = A[j] return false return true

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AllUnique(A[1..n]) for i<-1 to n for j<- i to n if A[i] = A[j] return false return true n + (n-1) + (n-2) + … + 3 + 2 + 1 = n(n+1)/2

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MatrixMultiply(A[nxn], B[nxn]) Initialize empty C[nxn] for i<- 1 to n for j <- 1 to n C[i,j] <- 0 for k<- 1 to n C[i,j] <- C[i,j] + A[i,k]*B[k,j] return C

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MethodA(n) – answer <- 1 – for i<- 1 to n/2: answer <- answer + i for i<- 1 to n – answer <- answer + 1 – return answer MethodB(n) – answer <- 1 – for i <- 1 to lg n: answer = answer + MethodA(n) return answer

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Big-oh Definition Let f(n) and g(n) be functions from positive integers to positive reals. We say f = O(g) if there exists some constant c > 0 such that f(n) ≤ cg(n) for all n.

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Big-oh Definition Let f(n) and g(n) be functions from positive integers to positive reals. We say f = O(g) if there exists some constant c > 0 such that f(n) ≤ cg(n) for all n. Say what?

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Big-O f(n) = O(g(n), think: f(n) ≤ g(n) n = O(n 2 ) because I can find a big number to multiply the parabola by that makes it lie completely above the line. I can’t do that in reverse.

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Know the shapes constant logarithmic linear quadratic exponential

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Theta Notation Theta = tight bound Multiply constants by a function, results are always (asymptotically) above and below. What it means: If you have a parabola, you can always find parabolas that stay above and below the given parabola. Same for lines, and other curves. Just tells us the efficiency category.

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Practice – True or False? ½n(n-1) = Ө(n) if f(n) = Ө g(n), then g(n) = Ө(f(n)) -n – 100000 = Ө(n)

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O - notation f = O(g) if f grows slower than or the same as g That is, asymptotically, g is not below f g is an upper bound Think: O is “≤”

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Practice – True or False? n = O(n 2 ) n 3 = O(n 2 ).00000001n 3 = O(n 2 ) 100n + 5 = O(n 2 ) ½ n(n-1) = O(n 2 ) n 4 +n+1= O(n 2 )

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-Notation Opposite of big-oh f = (g) means f is always above or equal to g. It gives an asymptotic lower bound. Think: is “≥”

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Practice – True or False? n 3 = (n 2 ) ½ n(n-1) = (n 2 ) 100n + 5 = (n 2 )

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o and ω f = O(g) means g is above or equal to f f = o(g) means g is strictly above f Also means – there is a better, tighter bound ω is analogous for - there is a tighter bound available

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Function Growth Rates Say that f isMean that f isWrite “big oh of g”No faster than g, ≤ f = O(g) “theta g”About as fast as g, = f = Θ(g) “omega g”No slower than g, ≥ f = Ω(g)

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Polynomials are Easy – What about other functions? constant (1): Very few examples of algorithms that don’t grow with input size logarithmic (lg n): Usually result of cutting input size by a constant factor each time through the loop linear (n): Look at each input element a constant number of times nlgn: Divide and conquer quadratic (n 2 ): Two embedded loops cubic (n 3 ): Three embedded loops exponential (2 n ): Generate all subsets of the input elements factorial (n!): Generate all permutations of the input n n : Generate all permutations of input, allowing repetitions

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Other standard functions Polylog – log 3 n = (logn) 3 Any log grows slower than any polynomial: log a n = o(n b ), for any a,b > 0 Any polynomial grows slower than any exponential with base c > 1. n b = o(c n ) for any c > 1 n! = o(n n ) n! = ω(2 n ) lg(n!) = Ө(nlgn)

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Relative Growths –O, , or Ө? lg 2 n 2n n 4 +3n2 n √nn 2/3 4 n 4 n/2 n+2 n 3 n 100n + lgnn+(lgn) 2 lg(n 2 ) lg(n 3 )

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