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Telecooperation/RBG Technische Universität Darmstadt Copyrighted material; for TUD student use only Introduction to Computer Science I Topic 7: Complexity of Algorithms Prof. Dr. Max Mühlhäuser Dr. Guido Rößling

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Algorithm Selection Time complexity –How long does the execution take? Space complexity –How much memory is required for execution? Required network bandwidth 2 We need to consider non-functional properties of algorithm: Two algorithms solve the same problem. Example: merge sort/insertion sort. Which one is the "better" one? In the following, we will concentrate on "time". Analog treatment of "space"

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 How to measure the cost of computing? Timing the application of a program to specific arguments can help to understand its behavior in one situation –However, applying the same program to other inputs may require a radically different amount of time… Timing programs for specific inputs has the same status as testing programs for specific examples: –Just like testing may reveal bugs, timing may reveal anomalies of the execution behavior for specific inputs –It does not, however, provide a foundation for general statements about the behavior of a program This lecture provides a first glimpse at tools for making general statements about the cost of programs –ICS 2 provides more in-depth considerations 3

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Outline Abstract time for complexity analysis O-Notation and other notations for asymptotic growth Techniques for measuring algorithm complexity Vectors in Scheme 4

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Concrete Time, Abstract Time Real time" execution depends on many factors –speed of computer –type of computer –programming language –quality of compiler, … 5 To allow a reasonable comparison between different algorithms for the same problem, we need a way of measuring time complexity that is independent of all such factors Home computer Desktop Computer Minicomputer Mainframe computer Supercomputer 51.915 11.508 2.382 0.431 0.087 computer type Time Real time (milliseconds) for computing some f

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Measuring Complexity Idea: Describe resource consumption as a mathematical function Domain of the cost function: Size of the input n – depends on the problem being studied: –e.g., for sorting a list, n = number of list items –e.g., for matrix multiplication n = number of rows and m = number of columns –e.g., for graph algorithms, n = number of nodes, e = number of edges Range of the cost functions: Required time T(n) Number of natural recursions is a good measure of the size of an evaluation sequence. 6

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Illustrating Abstract Time Let us study the behavior of length, a function that we understand well: –It takes a list of arbitrary data and computes how many items the list contains 7 (define (length a-list) (cond [(empty? a-list) 0] [else (+ (length (rest a-list)) 1)])) length (list 'a 'b 'c)) = (+ (length (list 'b 'c)) 1) = (+ (+ (length (list 'c)) 1) 1) = (+ (+ (+ (length empty) 1) 1) 1) = 3 natural recursion steps

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Illustrating Abstract Time Only steps that are natural recursions are relevant for the complexity judgment. The steps between the natural recursions differ only as far as the substitution for a-list is concerned. 8 (length (list 'a 'b 'c)) = (cond [(empty? (list 'a 'b 'c)) 0] [else (+ (length (rest (list 'a 'b 'c))) 1)]) = (cond [false 0] [else (+ (length (rest (list 'a 'b 'c))) 1)]) = (cond [else (+ (length (rest (list 'a 'b 'c))) 1)]) = (+ (length (rest (list 'a 'b 'c))) 1) = (+ (length (list 'b 'c)) 1)

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Illustrating Abstract Time The example suggests two things: –The number of evaluation steps depends on input size –The number of natural recursions is a good measure of the size of an evaluation sequence. After all, we can reconstruct the actual number of steps from this measure and the function definition The abstract running time measures the performance of a program as a relationship between the size of the input and the number of recursion steps in an evaluation –abstract means: the measure ignores the details of how much time primitive steps take and how much real time the overall evaluation takes. 9

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 (define (contains-doll? alos) (cond [(empty? alos) false] [(symbol=? (first alos) 'doll) true] [else (contains-doll? (rest alos))])) Abstract Time and the Input Shape Let us walk through a second example: a recursive function 10 The following application requires no recursion step In contrast, the evaluation below requires as many recursion steps as there are items in the list In the best case, the function finds an answer immediately In the worst case, the function must search the entire input list (contains-doll? (list 'doll 'robot 'ball 'game-boy)) (contains-doll? (list 'robot 'ball 'game-boy 'doll))

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Abstract Time and the Input Shape 11 Programmers cannot safely assume that inputs are always of the best possible type –They also cannot just hope that the inputs will not be of the worst possible type Instead, they must analyze how much time their functions take on average For example, contains-doll? may - on average - find 'doll somewhere in the middle of the list –If the input contains n items, the abstract running time of contains-doll? is (in average) n/2 On average, there are half as many recursion steps as there are elements in the list

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 On the order of… Because we measure the running time of a function in an abstract manner, we can ignore the division by 2. More precisely, –We assume that each basic step takes K units of time –If we use K/2 as the constant, then: –This shows that we can ignore constant factors. To indicate that we are hiding such constants we say that contains-doll? takes "on the order of n steps (~ n)'' to find 'doll in a list of n items. 12

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Complexity Classes 13 T n 1000 0 F ~ on the order of n G ~ on the order of n 2 F G n1105050010005000 F (1000 · n)1000100005000050000010000005000000 G (n · n)11002500250000100000025000000

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 (insertion-sort (list 3 1 2)) = (insert 3 (insertion-sort (list 1 2))) = (insert 3 (insert 1 (insertion-sort (list 2)))) = (insert 3 (insert 1 (insert 2 (insertion-sort empty)))) = (insert 3 (insert 1 (insert 2 empty))) = (insert 3 (insert 1 (list 2))) = (insert 3 (cons 2 (insert 1 empty))) = (insert 3 (list 2 1)) = (list 3 2 1) Analysis: insertion-sort 14 ;; insertion-sort : list-of-numbers -> list-of-numbers ;; creates a sorted list of numbers from numbers in alon (define (insertion-sort alon) (cond [(empty? alon) empty] [else (insert (first alon) (insertion-sort (rest alon)))]))

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Analysis: insert-sort The evaluation consists of two phases: 1.The recursions for sort set up as many applications of insert as there are items in the list 2.Each application of insert traverses a list of 1, 2,...n - 1 elements, where n is the number of items in the original list Inserting an item is similar to finding an item : –Applications of insert to a list of n items may trigger on the average, n/2 natural recursions, i.e., ~ n. –Because there are n applications of insert, we have an average of ~ n 2 natural recursions of insert In summary, if lst contains n items, –evaluating (insert-sort lst) requires n natural recursions of insert-sort and –on the order of n 2 natural recursions of insert. –Taken together: n 2 + n, i.e. ~ n 2 15

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 On the order of… We know that insert-sort takes time roughly equal to c 1 n 2 to sort n items (proportional to n 2 ) –c 1 is a constant that does not depend on n We now analyze the efficiency of merge-sort –It takes time roughly equal to c 2 n log n for n items c 2 is a constant that does not depend on n, c 2 > c 1 –No matter how much smaller c 1 is than c 2, there will always be a crossover point in the input data (n big enough) beyond which merge-sort is faster –Hence, we can safely ignore constants 16

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 On the order of… Imagine that we used a very fast computer A for running insert-sort and a slow computer B for merge-sort –A is 100 times faster than B in raw computing power A executes one billion instructions per second B executes only ten million instructions per second In addition, we assume that –The best programmer in the world codes insert-sort in machine language for A The resulting code requires 2n 2 (c 1 = 2) instructions –An average programmer codes merge-sort in a high-level language with an inefficient compiler resulting code requires 50n log n (c 2 = 50) instructions 17

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 On the order of… To sort a list of one million numbers (n = 10 6 ) … 18 2 x (10 6 ) 2 instructions 10 9 instructions/sec = 2000 sec A (insert-sort) 50 x 10 6 x log 10 6 instructions 10 7 instructions/sec 100 sec B (merge-sort) By using an algorithm whose running time grows more slowly, even with a poor compiler, B runs 20 times faster than A! In general, as the problem size increases, so does the relative advantage of merge-sort.

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Summary: abstract running time Our abstract description is always a claim about the relationship between two quantities: –A (mathematical) function that maps an abstract size measure of input to an abstract measure of running time (number of natural recursions evaluated) The exact number of executed operations is less important It is important to know the complexity class to which the algorithm belongs –e.g., linear or logarithmic When we compare on the order of properties of procedures, such as n, n 2, 2 n,…, we mean to compare corresponding functions that use n and produce the above results 19

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Outline Abstract time for complexity analysis O-Notation and other notations for asymptotic growth Techniques for measuring algorithm complexity Vectors in Scheme 20

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 O-Notation Comparing functions on N is difficult: they are infinite –If a function f produces larger outputs than some function g for all n in N, then f is clearly larger than g –But what if comparison fails for a few inputs, e.g.,1000? To make approximate judgments, –we adapt the mathematical notion of comparing functions –up to some factor –and some finite number of exceptions 21 T n 1000 0 F G

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 O-Notation The O-Notation goes back to the number theoretician Edmund Landau (1877-1938); the "O" is also called the Landau Symbol We say that "f(n) grows at least as quickly as g(n)" (g is an upper bound for f) 22 ORDER-OF (BIG-O): Given a function g on the natural numbers, O(g) (pronounced: big-O of g) denotes a class of functions on natural numbers. A function f is in O(g) if there exist numbers c and bigEnough such that for all n > bigEnough, it is true that f(n) <= cg(n)

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 O-Notation 23 f(n) O(g(n)), if two positive constants, C and n 0 exist, where |f(n)| C|g(n)| for all n n 0 12525050010002000 0 1000 2000 3000 C g(n) f(n) n0n0 function g defines the complexity class function f asymptotically behaves like g; from n 0 on, it is always < C g(n) f is at most of order g (n) An asymptotic measure ( n ). It abstracts from irrelevant details for the complexity class

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 O-Notation: Examples For F and G above: f(n) = 1000*n, g(n)=n 2 –We can say that f is O(g), because for all n > 1000, f(n) <= g(n) (bigEnough = 1000, c = 1) The definition of big-O provides us with a shorthand for stating claims about a function's running time: –The running time of length is in O(n). n is the abbreviation of the (mathematical) function g(n) = n –In the worst case, the running time of insert-sort is O(n 2 ) –Here, n and n 2 are standard abbreviations for the (mathematical) functions f(n) = n and g(n) =n 2 24

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Complexity Classes polynomial growth (O (n x )) drastically reduces the useful input size exponential growth (O(e n )) even more so 25 O(n) O(log n) O(n log n) O(n 2 ) O(n 3 ) number of comparisons input size (n)

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Properties of the O-Notation 26 Comparison of "O" complexity are only useful for large input sizes -For small input sizes, an inefficient algorithm may be faster than an efficient one -E.g., a function in 2n 2 grows faster than a function in (184 log 2 n), but is better for small input sizes (n<20) Algorithms with linear or even weaker growth rate are sensitive against such factors -Comparing the complexity class may be insufficient in such cases

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Properties of the O-Notation 27 -The Notation neglects proportional factors, small input sizes, and smaller terms 2n 3 +n 2 -20 O(n 3 ) log 10 n O(log 2 n) n + 10000 O(n) n O(n 2 ) O(1) O(log n) O(n) O(n 2 ) O(n 3 ) O(2 n ) O(10 n ) Examples 10 125 250 500 1000 2000 0.121 2.8 11.0 43.4 172.9 690.5 f(n)an 2 0.017 2.7 10.8 43.1 172.4 689.6 n 2 - term in % of the whole f(n) = an 2 + bn +c with a = 0.0001724, b = 0.0004 and c = 0.1 14.2 94.7 98.2 99.3 99.7 99.9 n

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 O-Notation: Other Symbols Other symbols: - Asymptotic lower bound –f(n) (g(n)) if positive constants c and n 0 such that 0 c g(n) f(n) n n 0 –"f(n) grows at least as quickly as g(n) Θ - Asymptotic tight bound –f(n) (g(n)) if positive constants c 1, c 2, and n 0 such that c 1 g(n) f(n) c 2 g(n) n n 0 –f(n) (g(n)) if and only if f(n) O(g(n)) and f(n) (g(n)) –"f(n) grows as quickly as g(n)" 28

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 O-Notation: Other Symbols Schema for O, and : 29 n0n0 n0n0 n0n0 upper boundlower boundexact bound

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Other Asymptotic Notations A function f(n) is o(g(n)) if positive constants c and n 0 exist such that f(n) < c g(n) n n 0 A function f(n) is (g(n)) if positive constants c and n 0 exist such that c g(n) < f(n) n n 0 Intuitively, 30 –o() is like < –O() is like – () is like > – () is like – () is like =

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Outline Abstract time for complexity analysis O-Notation and other notations for asymptotic growth Techniques for measuring algorithm complexity Vectors in Scheme 31

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Cost of Euclids GCD Complexity analysis of Euclids algorithm gcd is highly non-trivial Lamés Theorem: –If Euclids Algorithm requires k steps to compute the gcd of some pair, then the smaller number in the pair must be greater than or equal to the k-th Fibonacci number –Hence, if n is the smaller number in the pair, n Fib(k) k Order of growth is O(log (n)) 32 (define (gcd a b) (cond [(= b 0) a] [else (gcd b (remainder a b))]))

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Example: Exponentiation Input: Base b, positive integer exponent n Output: b n Idea: b n = b* b (n-1), b 0 = 1 Assume multiplication needs a constant time c Time: T(n) = cn = O(n) 33 (define (expt b n) (cond [(= n 0) 1] [else (* b (expt b (- n 1)))]))

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Example: Faster Exponentiation Idea: Use fewer steps with successive squaring –Example: instead of computing b 8 as b*b*b*b*b*b*b*b we can compute it as b 2 = b*b, b 4 = (b 2 ) 2, b 8 = (b 4 ) 2 –In general we can use the rule b n = (b n/2 ) 2 if n is even b n = b*b n-1 if n is odd What is the time complexity of this algorithm? 34 (define (fast-expt b n) (cond [(= n 0) 1] [(even? n) (sqr (fast-expt b (/ n 2)))) (else (* b (fast-expt b (- n 1))))))

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Analyzing divide-and-conquer-algorithms The running time of a recursive algorithm can often be described by a recurrence equation or recurrence –A recurrence describes overall running time in terms of the running time on smaller inputs –Example: –Mathematical tools help to solve recurrences to find bounds on the performance of the algorithm A recurrence for the running time, T(n), of a divide- and-conquer algorithm on a problem of size n is based on the three steps of the paradigm… 35

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Analyzing divide-and-conquer-algorithms Case 1: Problem size is small enough, say n <= c for some constant c, so that it can be solved trivially –Direct solution takes constant time (1) Case 2: Problem is non-trivial: –The division of the problem yields a sub-problems, each of which is 1/b of the size of the original problem Example: for merge-sort we have a = b = 2 –D(n) : time to divide the problem into sub-problems –C(n) : time to combine the solutions of sub-problems 36

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Towers of Hanoi A Buddhist monastery at Hanoi has 3 large posts with 64 golden discs –Monks, acting out the command of an ancient prophecy, have been moving these disks, according to the rules of a puzzle, once every day since the monastery was founded, over a 1000 years ago. –When the last move of the puzzle is completed, the world will end in a clap of thunder. 37 ! The objective is to transfer the entire tower to one of the other pegs, moving only one disk at a time and never a larger one onto a smaller.

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Towers of Hanoi: Algorithm 38 1 2 3 To move n discs from peg A to peg B: 1.move n1 discs from A to C. This leaves disc #n alone on peg A 2.move disc #n from A to B 3.move n1 discs from C to B so they sit on disc #n A B C Recursive algorithm: to carry out steps 1 and 3, apply the same algorithm again for n1. The entire procedure is a finite number of steps, since at some point the algorithm will be required for n = 1. This step, moving a single disc from peg A to peg B, is trivial. n

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Towers of Hanoi 39 How many disk movements are required for moving a stack of height n? (define (move T1 T2 T3 n) (cond [(= n 0) empty] [else (append (move T1 T3 T2 (- n 1)) (list (list T1 T2)) (move T3 T2 T1 (- n 1))) ] ) ) (list (list 'A 'B) (list 'A 'C) (list 'B 'C) (list 'A 'B) (list 'C 'A) (list 'C 'B) (list 'A 'B)) Result is the list of disk movements A B C (move 'A 'B 'C 3)

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Solving the Recurrence for Towers of Hanoi 40 = 2 n ×T(0) + 2 n -1 = 2 n -1 => exponential complexity! T(n)= T(n-1)+1+T(n-1) = 2×T(n-1)+1 = 2×(2×T(n-2)+1)+1 = 2×(2×(2×T(n-3)+1)+1)+1 = 2 i ×T(n-i) +, for i=n, n-i becomes 0 0 2k2k i-1 k How many disk movements are required for moving a stack of height n from pole 1 to pole 2? -for n<2 the answer is easy: T(0)=0, T(1)=1 -for n>1 the answer is recursively defined:

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Solving the Recurrence for Towers of Hanoi 41 => exponential complexity! The monks are nowhere close to completion – Even if the monks were able to move discs at a rate of one per second, using the smallest number of moves, it would take 2 64 1 seconds or ~ 585 billion years! – The universe is currently about 13.7 billion years old

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Analysis of Merge Sort Simplification: We assume for the analysis that the original problem size is a power of 2 –Each divide step then yields two sub-sequences of exactly n/2 –There are proofs that such an assumption does not affect the order of growth of the solution to the recurrence Worst case: n > 1 –Divide: Extracting the elements of the two sub-lists takes on the order of n time each D(n) = 2n ~ (n) –Conquer: To recursively solve two sub-problems, each of size n/2 requires 2T(n/2) –Combine: merging two lists takes also on the order of (n) 42 worst-case running time for merge sort

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Solving Recurrences Question: How does one solve recurrences such as the one we constructed for the running time of merge sort? Answer: Mathematical methods help us with this: –Substitution method –Recursion tree method –Master method based on master theorem A recipe" to solve recurrences of the form T(n) = aT(n/b)+f(n), a 1, b>1, f(n) asymptotically positive function It can be used to show that T(n) of merge sort is (n log n) We neglect technical details: –ignore ceilings and floors (all absorbed in the O or notation) –We assume integer arguments to functions –Boundary conditions 43

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 The Master Theorem 44 If for some constant > 0 then If then If for some constant > 0 and if a f(n/b) <= c f(n) for some constant c < 1 and all n sufficiently large, then T(n) = (f (n)) if n < c if n > 1 Consider where a >= 1 and b >= 1

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 The Recursion Tree Method In a recursion tree, each node represents the cost of a single sub-problem somewhere in the set of recursive function invocations –Sum costs within each level to get per-level costs –Sum all per-level costs to determine the total cost Particularly useful method for recurrences that describe running time of divide-and-conquer algorithms Often used to make a good guess that is then verified by other methods –When built carefully, can also be used to as a direct proof of a solution to a recurrence 45

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Recursion Tree for Merge Sort Let us write the recurrence as: 46 T(n/2) cn T(n/2) Node for 1st recursive call T(n/4) cn cn/2 T(n/4) Tree for two recursive steps

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Recursion Tree for Merge Sort 47 cn cn/2 cn/4... c cc cc cc c n log n + 1 levels (by induction) Total: cn (log n + 1) cn... 2 i c(n/2 i )... Level i 2 i nodes

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 The Substitution Method A.k.a. the making a good guess method Guess the form of the answer, then use induction to find the constants and show that solution works Examples: T(n) = 2T(n/2) + (n) T(n) = (n log n) T(n) = 2T( n/2 ) + n ??? T(n) = 2T( n/2 + 17) + n ??? 48 T(n) = 2T ( n/2 ) + n T(n) = (n log n) T(n) = 2T ( n/2 + 17) + n (n log n)

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 The Substitution Method 49 We guess that the solution is We need to prove that for an appropriate choice of the constant c > 0 We start by assuming that this bound holds for that is: Solution: we need to prove the guess for n>n 0

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Performance of Quick Sort: Best Case Running time of quick-sort depends on quality of partitioning, i.e., on the elements used as pivots 50 master theorem Best case: partition produces in each step two sub- problems of size n/2

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Performance of Quick Sort: Worst Case 51 Worst case: partitioning produces one sub-problem with n-1 and one with 0 elements in each recursive step Worst case occurs when the list is already sorted! n n - 1 n - 2 n - 3 2 1 1 1 1 1 1 n n n n - 1 n - 2 3 2 (n 2 ) T(n) = T(n-1)+T(0)+ (n) = T(n-1)+ (n) -Intuitively, summing per-level costs yields an arithmetic series, which evaluates to (n 2 ) Let partition time be (n), recursive call on empty list T(0) = (1)

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Performance of Quick Sort: Balanced Partitioning Average case of quick-sort is much closer to the best case Θ (nlogn) comparisons to sort n items. To understand why, we need to see how the partitioning balance is reflected in the recurrence for the running time Balanced partitioning: Partitioning always produces a constant split 52

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Reason: splits of constant proportionality yield recursive trees of depth (log n) with the cost at each level O(n) Performance of Quick Sort: Balanced Partitioning E.g.: 9-to-1 proportional split seems quite unbalanced T(n) <= T(9n/10) + T(n/10) + n 53 (log n) Even a 99:1 split yields O(n log n)

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Performance of Quick Sort: Average Case Intuition for the average case –All permutations of the input numbers are equally likely –On a random input list, we expect a mix of well-balanced and unbalanced splits –Good and bad splits are randomly distributed across the tree 54 Alternate of a bad and a good split Nearly well- balanced split n n - 1 1 (n – 1)/2 n (n – 1)/2 + 1 Running time of quick-sort when levels alternate between good and bad splits is O(n log n) combined cost: 2n-1 = (n) combined cost: n = (n)

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Performance of Quick Sort Typically, quick-sort is faster in practice than other Θ (n log n) algorithms –Its inner loop can be efficiently implemented on most architectures –In most real-world data, it is possible to make design choices which minimize the possibility of requiring quadratic time 55

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Summary of Cost of Computations Algorithms can be classified according to their complexity O-Notation –only relevant for large input sizes "Measurements" are machine independent –counting operations in terms of the input size –worst-, average-, best-case analysis Algorithms strongly vary in their efficiency –Good programming one or two complexity classes less –Some problems are inherently complex 56

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Outline Abstract time for complexity analysis O-Notation and other notations for asymptotic growth Techniques for measuring algorithm complexity Vectors in Scheme 57

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 (define (find-route origin destination G) (cond [(symbol=? origin destination) (list destination)] [else (local ((define possible-route (find-route/list (neighbors origin G) destination G))) (cond [(boolean? possible-route) false] [else (cons origin possible-route)]))])) Cost of Lookup in List Recall the example of finding a route in a graph: 58

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Cost of Lookup in List neighbors is similar to contains-doll? i.e. neighbors is O(n) 59 ;; neighbors : node graph -> (listof node) ;; to lookup the node in graph (define (neighbors node graph) (cond [(empty? graph) (error no neighbors")] [(symbol=? (first (first graph)) node) (second (first graph))] [else (neighbors node (rest graph))]))

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Cost of Lookup in List 60 The complexity of neighbors is O(n) neighbors is used at every stage of find-route, i.e. n times in case of maximal route the algorithm requires O(n 2 ) steps in neighbors neighbors can be the bottleneck of find-route ! We need a data structure that allows the access to the neighbors of a given node by name within constant time Vectors are data structures in which allow element access in constant time.

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Operations on Vectors 61 vector creates a new vector from given values: build-vector is essentially the same as build-list : (vector V-0... V-n) (build-vector n f) = (vector (f 0)... (f (- n 1))) vector-ref extracts a value from the vector vector-length returns the number of elements: vector? is the vector predicate: (vector-length (vector V-0... V-n)) = (+ n 1) (vector? (vector V-0... V-n)) = true (vector? U) = false (vector-ref (vector V-0... V-n) i) = V-i, 0 i n

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Graphs as Vectors 62 (define Graph-as-list '((A (B E)) (B (E F)) (C (D)) (D ()) (E (C F)) (F (D G)) (G ()))) (define Graph-as-vector (vector (list 1 4) (list 4 5) (list 3) empty (list 2 5) (list 3 6) empty)) ABCDEFG 0123456 A node is a natural number between 0 and n – 1, n is the number of nodes List-based representation:Vector-based representation: The vector's i-th field contains the list of neighbors of the i-th node A graph is a vector of nodes: (vectorof (listof node))

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Graphs as Vectors Now neighbors of a node is a constant-time operation => We can ignore it when we study the abstract running time of find-route. 63 ;; neighbors : node graph -> (listof node) ;; to lookup the node in graph (define (neighbors node graph) (vector-ref graph node))

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Processing Vectors Example: The function vector-sum-for-3 uses vectors of three numbers and returns their sum: 64 ;; (vector number number number) -> number (define (vector-sum-for-3 v) (+ (vector-ref v 0) (vector-ref v 1) (vector-ref v 2))) Programming with vectors means programming with vector-ref – thinking about vectors and indices into them

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Processing Vectors Consider the more general function vector-sum that works with vectors of arbitrary size: 65 ;; vector-sum : (vectorof number) -> number ;; to sum up the numbers in v (define (vector-sum v)...) (= 0 (vector-sum (vector -1 3/4 1/4))) (= 1 (vector-sum (vector.1.1.1.1.1.1.1.1.1.1 )) (= 0 (vector-sum (vector))) Here are some examples:

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Processing Vectors vector-sum does not receive the length of vector – the number of elements to be processed. We must define an auxiliary function with such an argument: Then vector-sum is as follows: 66 (define (vector-sum v) (vector-sum-aux v (vector-length v))) ;; vector-sum-aux:(vectorof number) n -> number ;; to sum up the numbers in v relative to i (define (vector-sum-aux v i)...)

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Processing Vectors First, we develop a template for the function. Implementation of vector-sum-for-3 suggests that i is the varying argument in the template. The template suggests that we should consider i as the number of items of v that vector-sum-aux must consider 67 ;; vector-sum-aux :(vectorof number) n -> number ;; to sum up the numbers in v with index in [0, i) (define (vector-sum-aux v i) (cond [(zero? i)...] [else... (vector-sum-aux v (pred i))...]))

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Processing Vectors To transform the template into a complete function definition, consider each clause of the cond : 1.If i is 0, there are no further items to be considered =>The result is 0. 2.Otherwise, –compute the sum of the numbers in v with indices less than i-1 : –and take the value of vector field with index i-1: => the result is their sum: 68 (vector-sum-aux v (pred i)) (vector-ref v (pred i)) (+ (vector-ref v (pred i)) (vector-sum-aux v (pred i))

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Processing Vectors 69 ;; vector-sum : (vectorof number) -> number ;; to compute the sum of the numbers in v (define (vector-sum v) (vector-sum-aux v (vector-length v))) ;; vector-sum-aux :(vectorof number) n -> number ;; to sum the numbers in v with index in [0, i) (define (vector-sum-aux v i) (cond [(zero? i) 0] [else (+ (vector-ref v (pred i)) (vector-sum-aux v (pred i)))])) vector-sum-aux extracts the numbers from the vector in a right-to-left order as i decreases to 0

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Processing Vectors The sum can also be computed from left to right: 70 ;; lr-vector-sum : (vectorof number) -> number ;; to sum up the numbers in v (define (lr-vector-sum v) (vector-sum-aux v 0)) ;; vector-sum : (vectorof number) -> number ;; to sum up the numbers in v with index in ;; [i, (vector-length v)) (define (vector-sum-aux v i) (cond [(= i (vector-length v)) 0] [else (+ (vector-ref v i) (vector-sum-aux v (succ i)))]))

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Creating Vectors On the next slides, we will briefly examine how to create vectors To illustrate this, we develop a function that moves the elements of a given vector at a given velocity. 71

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Building Vectors The velocity of an object can be represented by a vector: –(vector 1 2) – the velocity of an object on a plane that moves 1 unit to the right and 2 down in each time unit. –(vector -1 2 1) – velocity in space; -1 units in the x direction, 2 units in the y direction, and 1 units in the z direction. 72 Let us develop a function that computes the displacement for an object with some velocity v in t time units: ;; (vectorof number) number -> (vectorof number) ;; to compute the displacement of v and t (define (displacement v t)...)

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Building Vectors 73 (equal? (displacement (vector 1 2) 3) (vector 3 6)) (equal? (displacement (vector -1 2 1) 6) (vector -6 12 6)) (equal? (displacement (vector -1 -2) 2) (vector -2 -4)) To compute the result, we just multiply each dimension of the velocity vector with the number representing the time and return a new vector. Some examples:

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Building Vectors 74 (build-vector (vector-length v)...) ;; new-item : n -> number (define (new-item index)...) ;; (vectorof number) number -> (vectorof number) ;; to compute the displacement of v and t (define (displacement v t) (local ((define (new-item i) (* (vector-ref v i) t))) (build-vector (vector-length v) new-item))) We construct a vector of the same length as v: Now we need to replace... with a function that computes the 0-th, 1-st, and so on items of the new vector: Then we multiply (vector-ref v i) with t and that's it:

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Dr. G. Rößling Prof. Dr. M. Mühlhäuser RBG / Telekooperation © Introduction to Computer Science I: T7 Building Vectors Since the local function new-item is not recursive, we can replace it with a lambda expression: 75 ;; (vectorof number) number -> (vectorof number) ;; to compute the displacement of v and t (define (displacement v t) (build-vector (vector-length v) (lambda (i) (* (vector-ref v i) t)))) In mathematics, this function is called scalar product… This and many other mathematical operations with vectors can be naturally expressed in Scheme.

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