Algorithms R OBERT S EDGEWICK | K EVIN W AYNE 1.4 A NALYSIS A LGORITHMS OF ‣ introduction ‣ observations Algorithms F O U R T H E D I T I O N R OBERT S EDGEWICK | K EVIN W AYNE ‣ mathematical models ‣ order-of-growth classifications ‣ memory http://algs4.cs.princeton.edu Last updated on Sep 21, 2015, 5:44 PM Running time “ As soon as an Analytical Engine exists, it will necessarily guide the future course of the science. Whenever any result is sought by its aid, the question will then arise—By what course of calculation can these results be arrived at by the machine in the shortest time? ” — Charles Babbage (1864) how many times do you have to turn the crank? 2 Reasons to analyze algorithms Predict performance. Compare algorithms. this course (COS 226) Provide guarantees. theory of algorithms (COS 423) Understand theoretical basis. Primary practical reason: avoid performance bugs. 3 An algorithmic success story N-body simulation. ・Simulate gravitational interactions among N bodies. ・Applications: cosmology, fluid dynamics, semiconductors, ... ・Brute force: N steps. ・Barnes-Hut algorithm: N log N steps, enables new research. 2 Andrew Appel PU '81 time quadratic 64T 32T limit on available time 16T linearithmic 8T size linear 1K 2K 4K 8K 4 Another algorithmic success story Discrete Fourier transform. ・Express signal as weighted sum of sines and cosines. ・Applications: DVD, JPEG, MRI, astrophysics, …. ・Brute force: N steps. ・FFT algorithm: N log N steps, enables new technology. 2 James Cooley John Tukey time quadratic 64T 32T limit on available time 16T linearithmic 8T size linear 1K 2K 4K 8K 5 1.4 A NALYSIS OF A LGORITHMS ‣ introduction ‣ observations Algorithms R OBERT S EDGEWICK | K EVIN W AYNE http://algs4.cs.princeton.edu ‣ mathematical models ‣ order-of-growth classifications ‣ memory The challenge Q. Will my program be able to solve a large practical input? Why is my program so slow ? Why does it run out of memory ? Insight. [Knuth 1970s] Use scientific method to understand performance. 7 Scientific method applied to the analysis of algorithms A framework for predicting performance and comparing algorithms. Scientific method. ・Observe some feature of the natural world. ・Hypothesize a model that is consistent with the observations. ・Predict events using the hypothesis. ・Verify the predictions by making further observations. ・Validate by repeating until the hypothesis and observations agree. Principles. ・Experiments must be reproducible. ・Hypotheses must be falsifiable. Feature of the natural world. Computer itself. Francis Bacon René Descartes John Stuart Mills 8 Example: 3-SUM 3-SUM. Given N distinct integers, how many triples sum to exactly zero? Context. Deeply related to problems in computational geometry. % more 8ints.txt 8 30 -40 -20 -10 40 0 10 5 % java ThreeSum 8ints.txt 4 a[i] a[j] a[k] sum 1 30 -40 10 0 2 30 -20 -10 0 3 -40 40 0 0 4 -10 0 10 0 9 3-SUM: brute-force algorithm public class ThreeSum { public static int count(int[] a) { int N = a.length; int count = 0; for (int i = 0; i < N; i++) for (int j = i+1; j < N; j++) for (int k = j+1; k < N; k++) if (a[i] + a[j] + a[k] == 0) count++; return count; } check each triple for simplicity, ignore integer overflow public static void main(String[] args) { In in = new In(args[0]); int[] a = in.readAllInts(); StdOut.println(count(a)); } } 10 Measuring the running time Q. How to time a program? % java ThreeSum 1Kints.txt A. Manual. tick tick tick 70 % java ThreeSum 2Kints.txt tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick 528 % java ThreeSum 4Kints.txt 4039 tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick tick Observing the running time of a program 11 Measuring the running time Q. How to time a program? A. Automatic. public class Stopwatch (part of stdlib.jar ) Stopwatch() double elapsedTime() create a new stopwatch time since creation (in seconds) public static void main(String[] args) { In in = new In(args[0]); int[] a = in.readAllInts(); Stopwatch stopwatch = new Stopwatch(); StdOut.println(ThreeSum.count(a)); client code double time = stopwatch.elapsedTime(); StdOut.println("elapsed time = " + time); } 12 Empirical analysis Run the program for various input sizes and measure running time. 13 Empirical analysis Run the program for various input sizes and measure running time. N time (seconds) 250 0.0 500 0.0 1,000 0.1 2,000 0.8 4,000 6.4 8,000 51.1 † † on a 2.8GHz Intel PU-226 with 64GB DDR E3 memory and 32MB L3 cache; running Oracle Java 1.7.0_45-b18 on Springdale Linux v. 6.5 14 Data analysis Standard plot. Plot running time T (N) vs. input size N. standard plot log-log plot 50 51.2 s 25.6 12.8 lg(T(N)) running time T(N) 40 30 20 6.4 3.2 1.6 .8 .4 10 .2 .1 1K 2K 4K 8K 1 problem size N Analysis of experimental data (the running time of ThreeSum) 15 Data analysis Log-log plot. Plot running time T (N) vs. input size N using log-log scale. log-log plot 51.2 25.6 straight line of slope 3 lg (T(N )) 12.8 lg(T (N)) = b lg N + c b = 2.999 c = -33.2103 6.4 3.2 1.6 T (N) = a N b, where a = 2 c .8 3 orders of magnitude .4 .2 .1 8K 1K 2K 4K 8K lgN ental data (the running time of ThreeSum) Regression. Fit straight line through data points: a N b. power law slope Hypothesis. The running time is about 1.006 × 10 –10 × N 2.999 seconds. 16 Prediction and validation Hypothesis. The running time is about 1.006 × 10 –10 × N 2.999 seconds. Predictions. ・51.0 seconds for N = 8,000. ・408.1 seconds for N = 16,000. Observations. N time (seconds) 8,000 51.1 8,000 51.0 8,000 51.1 16,000 410.8 † validates hypothesis! 17 Doubling hypothesis Doubling hypothesis. Quick way to estimate b in a power-law relationship. Run program, doubling the size of the input. N 250 time (seconds) † ratio 0.0 lg ratio – 500 0.0 4.8 2.3 1,000 0.1 6.9 2.8 2,000 0.8 7.7 2.9 4,000 6.4 8.0 3.0 8,000 51.1 8.0 3.0 T (N ) aN b = T (N/2) a(N/2)b = 2b lg (6.4 / 0.8) = 3.0 seems to converge to a constant b ≈ 3 Hypothesis. Running time is about a N b with b = lg ratio. Caveat. Cannot identify logarithmic factors with doubling hypothesis. 18 Doubling hypothesis Doubling hypothesis. Quick way to estimate b in a power-law relationship. Q. How to estimate a (assuming we know b) ? A. Run the program (for a sufficient large value of N) and solve for a. N 8,000 time (seconds) † 51.1 51.1 = a × 80003 8,000 51.0 8,000 51.1 ⇒ a = 0.998 × 10 –10 Hypothesis. Running time is about 0.998 × 10 –10 × N 3 seconds. almost identical hypothesis to one obtained via regression 19 Analysis of algorithms quiz 1 Estimate the running time to solve a problem of size N = 96,000. N A. 39 seconds. B. 52 seconds. C. time (seconds) 1000 0.02 2000 0.05 117 seconds. 4,000 0.20 D. 350 seconds. 8,000 0.81 E. I don't know. 16,000 3.25 32,000 13.00 † 20 Experimental algorithmics System independent effects. ・Algorithm. ・Input data. determines exponent b in power law a N b System dependent effects. ・Hardware: CPU, memory, cache, … ・Software: compiler, interpreter, garbage collector, … ・System: operating system, network, other apps, … determines constant a in power law a N b Bad news. Sometimes difficult to get precise measurements. Good news. Much easier and cheaper than other sciences. 21 An aside Algorithmic experiments are virtually free by comparison with other sciences. Chemistry (1 experiment) Biology (1 experiment) Physics (1 experiment) Computer Science (1 million experiments) Bottom line. No excuse for not running experiments to understand costs. 22 1.4 A NALYSIS OF A LGORITHMS ‣ introduction ‣ observations Algorithms R OBERT S EDGEWICK | K EVIN W AYNE http://algs4.cs.princeton.edu ‣ mathematical models ‣ order-of-growth classifications ‣ memory Mathematical models for running time Total running time: sum of cost × frequency for all operations. ・Need to analyze program to determine set of operations. ・Cost depends on machine, compiler. ・Frequency depends on algorithm, input data. Donald Knuth 1974 Turing Award In principle, accurate mathematical models are available. 24 Example: 1-SUM Q. How many instructions as a function of input size N ? int count = 0; for (int i = 0; i < N; i++) if (a[i] == 0) count++; N array accesses operation cost (ns) † frequency variable declaration 2/5 2 assignment statement 1/5 2 less than compare 1/5 N+1 equal to compare 1/10 N array access 1/10 N increment 1/10 N to 2 N † representative estimates (with some poetic license) 25 Example: 2-SUM Q. How many instructions as a function of input size N ? int count = 0; for (int i = 0; i < N; i++) for (int j = i+1; j < N; j++) if (a[i] + a[j] == 0) count++; 0 + 1 + 2 + . . . + (N 1) = = Pf. [ Gauss ] + T(N) = 0 + 1 + … + (N – 2) + (N – 1) T(N) = (N – 1) + (N – 2) + … + 1 + 0 2 T(N) = (N – 1) + (N – 1) + … + (N – 1) + (N – 1) ⇒ T(N) = N (N – 1) / 2 1 N (N 2 ⇥ N 2 1) 26 Example: 2-SUM Q. How many instructions as a function of input size N ? int count = 0; for (int i = 0; i < N; i++) for (int j = i+1; j < N; j++) if (a[i] + a[j] == 0) count++; 0 + 1 + 2 + . . . + (N operation cost (ns) frequency variable declaration 2/5 N+2 assignment statement 1/5 N+2 less than compare 1/5 ½ (N + 1) (N + 2) equal to compare 1/10 ½ N (N − 1) array access 1/10 N (N − 1) increment 1/10 ½ N (N + 1) to N 2 1) = = 1 N (N 2 ⇥ N 2 1) 1/4 N2 + 13/20 N + 13/10 ns to 3/10 N2 + 3/5 N + 13/10 ns (tedious to count exactly) 27 Simplifying the calculations “ It is convenient to have a measure of the amount of work involved in a computing process, even though it be a very crude one. We may count up the number of times that various elementary operations are applied in the whole process and then given them various weights. We might, for instance, count the number of additions, subtractions, multiplications, divisions, recording of numbers, and extractions of figures from tables. In the case of computing with matrices most of the work consists of multiplications and writing down numbers, and we shall therefore only attempt to count the number of multiplications and recordings. ” — Alan Turing ROUNDING-OFF ERRORS IN MATRIX PROCESSES By A. M. TURING {National Physical Laboratory, Teddington, Middlesex) [Received 4 November 1947] THIS paper contains descriptions of a number of methods for solving sets Downlo SUMMARY A number of methods of solving sets of linear equations and inverting matrices are discussed. The theory of the rounding-off errors involved is investigated for some of the methods. In all cases examined, including the well-known 'Gauss elimination process', it is found that the errors are normally quite moderate: no exponential build-up need occur. Included amongst the methods considered is a generalization of Choleski's method which appears to have advantages over other known methods both as regards accuracy and convenience. This method may also be regarded as a rearrangement of the elimination process. 28 Simplification 1: cost model Cost model. Use some basic operation as a proxy for running time. int count = 0; for (int i = 0; i < N; i++) for (int j = i+1; j < N; j++) if (a[i] + a[j] == 0) count++; 0 + 1 + 2 + . . . + (N operation cost (ns) frequency variable declaration 2/5 N+2 assignment statement 1/5 N+2 less than compare 1/5 ½ (N + 1) (N + 2) equal to compare 1/10 ½ N (N − 1) array access 1/10 N (N − 1) increment 1/10 ½ N (N + 1) to N 2 1) = = 1 N (N 2 ⇥ N 2 1) cost model = array accesses (we assume compiler/JVM do not optimize any array accesses away!) 29 Simplification 2: tilde notation ・Estimate running time (or memory) as a function of input size N. ・Ignore lower order terms. – when N is large, terms are negligible – when N is small, we don't care N 3/6 Ex 1. ⅙ N 3 + 20 N + 16 ~ ⅙N3 Ex 2. ⅙N ~ ⅙N Ex 3. ⅙N3 - ½N 3 + 100 N 2 4/3 + 56 + ⅓ N N 3/6 ! N 2/2 + N /3 166,666,667 3 ~ ⅙N3 166,167,000 N discard lower-order terms (e.g., N = 1000: 166.67 million vs. 166.17 million) Technical definition. f(N) ~ g(N) means 1,000 Leading-term approximation f (N) = 1 ∞ g(N) lim N→ 30 Simplification 2: tilde notation ・Estimate running time (or memory) as a function of input size N. ・Ignore lower order terms. – when N is large, terms are negligible – when N is small, we don't care operation frequency tilde notation variable declaration N+2 ~N assignment statement N+2 ~N less than compare ½ (N + 1) (N + 2) ~½N2 equal to compare ½ N (N − 1) ~½N2 array access N (N − 1) ~N2 increment ½ N (N + 1) to N 2 ~ ½ N 2 to ~ N 2 31 Example: 2-SUM Q. Approximately how many array accesses as a function of input size N ? int count = 0; for (int i = 0; i < N; i++) for (int j = i+1; j < N; j++) if (a[i] + a[j] == 0) count++; "inner loop" 0 + 1 + 2 + . . . + (N 1) = = 1 N (N 2 ⇥ N 2 1) A. ~ N 2 array accesses. Bottom line. Use cost model and tilde notation to simplify counts. 32 Example: 3-SUM Q. Approximately how many array accesses as a function of input size N ? int count = 0; for (int i = 0; i < N; i++) for (int j = i+1; j < N; j++) for (int k = j+1; k < N; k++) if (a[i] + a[j] + a[k] == 0) count++; "inner loop" N 3 A. ~ ½ N 3 array accesses. ⇥ = ⇥ N (N 1)(N 3! 2) 1 3 N 6 Bottom line. Use cost model and tilde notation to simplify counts. 33 Estimating a discrete sum Q. How to estimate a discrete sum? A1. Take a discrete mathematics course (COS 340). 34 Estimating a discrete sum Q. How to estimate a discrete sum? A2. Replace the sum with an integral, and use calculus! ⇥ N Ex 1. 1 + 2 + … + N. i Ex 2. 1 + 1/2 + 1/3 + … + 1/N. i=1 N N ⇥ 1 i ⇥ 1 Ex 4. 1 + ½ + ¼ + ⅛ + … x=0 x N x=1 i=1 j=i k=j 1 2 1 dx = ln 2 i=0 N x=1 N Ex 3. 3-sum triple loop. x dx x=1 i=1 N 1 2 N 2 N 1 2 ⇥ 1 dx = ln N x N y=x ⇥ N z=y dz dy dx 1 3 N 6 1.4427 i = 2 integral trick doesn't always work! 35 Mathematical models for running time In principle, accurate mathematical models are available. In practice, formulas can be complicated. costs (depend on machine, compiler) TN = c1 A + c2 B + c3 C + c4 D + c5 E A= B= C= D= E= array access integer add integer compare increment variable assignment frequencies (depend on algorithm, input) Bottom line. We use approximate models in this course: T(N) ~ c N 3. 36 Analysis of algorithms quiz 2 How many array accesses does the following code fragment make as a function of N ? int count = 0; for (int i = 0; i < N; i++) for (int j = i+1; j < N; j++) for (int k = 1; k < N; k = k*2) if (a[i] + a[j] >= a[k]) count++; A. ~ N 2 lg N B. ~ 3/2 N 2 lg N C. ~ 1/2 N 3 D. ~ 3/2 N 3 E. I don't know. 37 1.4 A NALYSIS OF A LGORITHMS ‣ introduction ‣ observations Algorithms R OBERT S EDGEWICK | K EVIN W AYNE http://algs4.cs.princeton.edu ‣ mathematical models ‣ order-of-growth classifications ‣ memory Common order-of-growth classifications Definition. If f (N) ~ c g(N) for some constant c > 0, then the order of growth of f (N) is g(N). ・Ignores leading coefficient. ・Ignores lower-order terms. Ex. The order of growth of the running time of this code is N 3. int count = 0; for (int i = 0; i < N; i++) for (int j = i+1; j < N; j++) for (int k = j+1; k < N; k++) if (a[i] + a[j] + a[k] == 0) count++; Typical usage. Mathematical analysis of running times. where leading coefficient depends on machine, compiler, JVM, ... 39 Commonly-used notations in the theory of algorithms notation provides example shorthand for used to ½ N2 Big Theta asymptotic order of growth Θ(N2) 10 N 2 5 N 2 + 22 N log N + 3N classify algorithms ⋮ 10 N 2 Big O Θ(N2) and smaller O(N2) 100 N 22 N log N + 3 N develop upper bounds ⋮ ½N2 Big Omega Θ(N2) and larger Ω(N2) N5 N 3 + 22 N log N + 3 N ⋮ develop lower bounds time Common order-of-growth classifications 200T Good news. The set of functions 100T 1, log N, N, N log N, N 2, N 3, and 2N logarithmic constant suffices to describe the order of growth of most common algorithms. 100K 200K 500K size r ic ea hm lin lin ea qua r it dra tic cubi c 512T exponential log-log plot time 64T 8T 4T 2T logarithmic T constant 1K 2K 4K 8K size 512K Typical orders of growth 41 Common order-of-growth classifications order of growth name typical code framework description example T(2N) / T(N) 1 constant a = b + c; statement add two numbers 1 log N logarithmic while (N > 1) N = N/2; ... divide in half binary search N linear for (int i = 0; i < N; i++) { ... } single loop find the maximum 2 N log N linearithmic see mergesort lecture divide and conquer mergesort ~2 double loop check all pairs 4 triple loop check all triples 8 exhaustive search check all subsets 2N N2 N 3 2N quadratic cubic exponential { } for (int i = 0; i < N; i++) for (int j = 0; j < N; j++) { ... } for (int i = 0; i < N; i++) for (int j = 0; j < N; j++) for (int k = 0; k < N; k++) { ... } see combinatorial search lecture ~1 42 Binary search Goal. Given a sorted array and a key, find index of the key in the array? Binary search. Compare key against middle entry. ・Too small, go left. ・Too big, go right. ・Equal, found. 6 13 14 25 33 43 51 53 64 72 84 93 95 96 97 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 43 Binary search: implementation Trivial to implement? ・First binary search published in 1946. ・First bug-free one in 1962. ・Bug in Java's Arrays.binarySearch() discovered in 2006. http://googleresearch.blogspot.com/2006/06/extra-extra-read-all-about-it-nearly.html 44 Binary search: Java implementation Invariant. If key appears in array a[], then a[lo] ≤ key ≤ a[hi]. public static int binarySearch(int[] a, int key) { int lo = 0, hi = a.length - 1; while (lo <= hi) why not mid = (lo + hi) / 2 ? { int mid = lo + (hi - lo) / 2; if (key < a[mid]) hi = mid - 1; else if (key > a[mid]) lo = mid + 1; else return mid; } return -1; one "3-way compare" } 45 Binary search: mathematical analysis Proposition. Binary search uses at most 1 + lg N key compares to search in a sorted array of size N. Def. T (N) = # key compares to binary search a sorted subarray of size ≤ N. Binary search recurrence. T (N) ≤ T (N / 2) + 1 for N > 1, with T (1) = 1. left or right half (floored division) possible to implement with one 2-way compare (instead of 3-way) Pf sketch. [assume N is a power of 2] T (N) ≤ T (N / 2) + 1 [ given ] ≤ T (N / 4) + 1 + 1 [ apply recurrence to first term ] ≤ T (N / 8) + 1 + 1 + 1 [ apply recurrence to first term ] ≤ T (N / N) + 1 + 1 + … + 1 [ stop applying, T(1) = 1 ] = 1 + lg N ⋮ lg N 46 1.4 A NALYSIS OF A LGORITHMS ‣ introduction ‣ observations Algorithms R OBERT S EDGEWICK | K EVIN W AYNE http://algs4.cs.princeton.edu ‣ mathematical models ‣ order-of-growth classifications ‣ memory Basics Bit. 0 or 1. NIST most computer scientists Byte. 8 bits. Megabyte (MB). 1 million or 220 bytes. Gigabyte (GB). 1 billion or 230 bytes. 64-bit machine. We assume a 64-bit machine with 8-byte pointers. some JVMs "compress" ordinary object pointers to 4 bytes to avoid this cost 48 Typical memory usage for primitive types and arrays type bytes type bytes boolean 1 char[] 2 N + 24 byte 1 int[] 4 N + 24 char 2 double[] 8 N + 24 int 4 float 4 long 8 double one-dimensional arrays type bytes char[][] ~2MN int[][] ~4MN double[][] ~8MN 8 primitive types two-dimensional arrays 49 Typical memory usage for objects in Java Object overhead. 16 bytes. integer wrapper object Reference. 8 bytes. public class Integer Padding. { Each object uses a private int x; ... } Ex 1. A Date object uses 32 date object public class Date { private int day; private int month; private int year; ... } 24 bytes multiple of 8 bytes. object overhead int x value bytes padding of memory. 32 bytes object overhead 16 bytes (object overhead) day month year padding 4 bytes (int) int values 4 bytes (int) 4 bytes (int) 4 bytes (padding) 32 bytes counter object public class Counter 32 bytes 50 Typical memory usage summary Total memory usage for a data type value: ・Primitive type: 4 bytes for int, 8 bytes for double, … ・Object reference: 8 bytes. ・Array: 24 bytes + memory for each array entry. ・Object: 16 bytes + memory for each instance variable. ・Padding: round up to multiple of 8 bytes. + 8 extra bytes per inner class object (for reference to enclosing class) Note. Depending on application, we may want to count memory for any referenced objects (recursively). 51 Analysis of algorithms quiz 3 How much memory does a WeightedQuickUnionUF use as a function of N ? A. ~ 4 N bytes B. ~ 8 N bytes C. ~ 4 N 2 bytes D. ~ 8 N 2 bytes E. I don't know. public class WeightedQuickUnionUF { private int[] parent; private int[] size; private int count; public WeightedQuickUnionUF(int N) { parent = new int[N]; size = new int[N]; count = 0; for (int i = 0; i < N; i++) parent[i] = i; for (int i = 0; i < N; i++) size[i] = 1; } ... } 52 TECHNICAL INTERVIEW QUESTIONS 53 THE 3-SUM PROBLEM 3-SUM. Given N distinct integers, find three such that a + b + c = 0. Version 0. N 3 time, N space. Version 1. N 2 log N time, N space. Version 2. N 2 time, N space. Note. For full credit, running time should be worst case. *Fastest known algorithm (published in 2014): N 2 / (log N / log log N) 2/3 time 54 THE 3-SUM PROBLEM: AN N2 LOG N ALGORITHM input 30 -40 -20 -10 40 Algorithm. ・Step 1: ・Step 2: Sort the N (distinct) numbers. For each pair of numbers a[i] and a[j], binary search for -(a[i] + a[j]). -40 -20 -10 0 (-40, -10) 50 Analysis. Order of growth is N 2 log N. (-40, 0) 40 ・Step 1: (-40, 5) 35 (-40, 10) 30 (or N log N with mergesort). ・ Step 2: N 2 log N with binary search. 5 10 30 40 binary search 60 with insertion sort 5 sort (-40, -20) N2 0 10 ⋮ ⋮ (-20, -10) ⋮ (-10, 30 ⋮ 0) ⋮ 10 ⋮ ( 10, 30) -40 ( 10, 40) -50 ( 30, 40) -70 only count if a[i] < a[j] < a[k] to avoid double counting 55 Turning the crank: summary Empirical analysis. ・Execute program to perform experiments. ・Assume power law. ・Formulate a hypothesis for running time. ・Model enables us to make predictions. Mathematical analysis. ・Analyze algorithm to count frequency of operations. ・Use tilde notation to simplify analysis. ・Model enables us to explain behavior. ⌊lg N ⌋ ! h=0 ⌈N/2h+1 ⌉ h ∼ N Scientific method. ・Mathematical model is independent of a particular system; applies to machines not yet built. ・Empirical analysis is necessary to validate mathematical models and to make predictions. 56

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