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Discrete Maths

Discrete Maths. 242-213 , Semester 2, 2013-2014. Objective to describe the Big-Oh notation for estimating the running time of programs . 10 . Running Time of Programs. Overview. Running Time Big-Oh and Approximate Running Time Big-Oh for Programs Analyzing Function Calls

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Discrete Maths

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  1. Discrete Maths 242-213, Semester 2,2013-2014 • Objective • to describe the Big-Oh notation for estimating the running time of programs 10. Running Time of Programs

  2. Overview • Running Time • Big-Oh and Approximate Running Time • Big-Oh for Programs • Analyzing Function Calls • Analyzing Recursive Functions • Further Information

  3. 1. Running Time • What is the running time of this program? void main(){ int i, n; scanf("%d", &n); for(i=0; i<n; i++) printf("%d"\n", i);} continued

  4. There is no single answer! • the running time depends on the size of the n value • Instead of a time answer in seconds, we want a time answer which is related to the size of the input. continued

  5. For example: • programTime(n) = constant * n • this means that as n gets bigger, so does the program time • running time is linearly related to the input running time constant * n size of n

  6. Running Time Theory • A program/algorithm has a running time T(n) • n is some measure of the input size • T(n) is the largest amount of time the program takes on any input of size n • Time units are left unspecified. continued

  7. A typical result is: • T(n) = c*n, where c is some constant • but often we just ignore c • this means the program has linear running time • T(n) values for different programs can be used to compare their relative running times • selection sort: T(n) = n2 • merge sort: T(n) = n log n • so, merge sort is “better” for larger n sizes

  8. 1.1. Different Kinds of Running Time • Usually T(n) is the worst-case running time • the maximum running time on any input of size n • Tavg(n) is the average running time of the program over all inputs of size n • more realistic • very hard to calculate • not considered by us

  9. 1.2. T(n) Example • Loop fragment for finding the index of the smallest value in A[] array of size n: (2) small = 0;(3) for(j = 1; j < n; j++)(4) if (A[j] < A[small])(5) small = j; • Count each assignment and test as 1 “time unit”.

  10. Calculation • The for loop executes n-1 times • each loop carries out (in the worse case) 4 ops • test of j < n, if test, small assign, j increment • total loop time = 4(n-1) • plus 3 ops at start and end • small assign (line 2), init of j (line 3), final j < n test • Total time T(n) = 4(n-1) + 3 = 4n -1 • running time is linear with the size of the array

  11. 1.3. Comparing Different T()’s 20000 Tb(n) = 2n2 • If input size < 50, program B is faster. • But for large n’s, which are more common in real code, program B gets worse and worse. 15000 T(n)value 10000 Ta(n) = 100n 5000 input size n 20 40 60 80 100

  12. 1.4. Common Growth Formulae & Names • Formula (n = input size) Name n linear n2 quadratic n3 cubic nm polynomial, e.g. n10 mn ( m >= 2) exponential, e.g. 5n n! factorial 1 constant log n logarithmic n log n log log n

  13. 1.5. Execution Times Assume 1 instruction takes 1 microsec (10-6 secs) to execute. How long will n instructions take? • 3 9 50 100 1000 106n 3 9 50 100 1ms 1secn2 9 81 2.5ms 10ms 1sec 12 daysn3 27 729 125ms 1sec 16.7 min 31,710yr2n 8 512 36yr 4*1016 yr 3*10287 yr 3*10301016yrlog n 2 3 6 7 10 20 n (no. of instructions) growth formula T() if n is 50, you will wait 36 years for an answer!

  14. Notes • Logarithmic running times are best. • Polynomial running times are acceptable, if the power isn’t too big • e.g. n2 is ok, n100 is terrible • Exponential times mean sloooooooow code. • some size problems may take longer to finish than the lifetime of the universe!

  15. 1.6. Why use T(n)? • T() can guide our choice of which algorithm to implement, or program to use • e.g. selection sort or merge sort? • T() helps us look for better algorithms in our own code, without expensive implementation, testing, and measurement.

  16. 2. Big-Oh and Approximate Running Time • Big-Oh mathematical notation simplifies the process of estimating the running time of programs • it uses T(n), but ignores constant factors which depend on compiler/machine behaviour continued

  17. The Big-Oh value specifies running time independent of: • machine architecture • e.g. don’t consider the running speed of individual machine operations • machine load (usage) • e.g. time delays due to other users • compiler design effects • e.g. gcc versus Borland C

  18. Example • In the code fragment example on slide 9, we assumed that assigment and testing takes 1 “time unit”. This means: T(n) = 4n -1 • The Big-Oh value, O(), uses the T(n) value but ignores constants (which will actually vary from machine to machine). This means: T(n) is O(n) we say "T(n) is order n"

  19. More Examples • T(n) valueBig Oh value: O() • 10n2+ 50n+100 O(n2) • (n+1)2 O(n2) • n10 O(2n) • 5n3 + 1 O(n3) • These simplifications have a mathematical reason, which is explained in section 2.2. hard to understand

  20. 2.1. Is Big-Oh Useful? • O() ignores constant factors, which means it is a more reliable measure across platforms/compilers. • It can be compared with Big-Oh values for other algorithms. • i.e. linear is better than polynomial and exponential, but worse than logarithmic

  21. 2.2. Definition of Big-Oh • The connection between T() and O() is: • when T(n) is O( f(n) ), it means that f(n) is the most important thing in T() when n is large • More formally, for some integer n0 and constant c > 0 • T(n) is O( f(n) ),if for all integers n >= n0, T(n) <= c*f(n) • n0 and c are called witnesses to the relationship: T(n) is O( f(n) )

  22. Informally, the n2 part is the most important thing in the T() function Example 1 • T(n) = 10n2 + 50n + 100 • which allows that T(n) is O(n2) • Why? • Witnesses: n0 = 1, c = 160 • then T(n) <= c*f(n), n >= 1so 10n2 + 50n + 100 <= 160 n2since 10n2 + 50n + 100 <= 10n2 + 50n2 + 100n2 <= 160 n2

  23. Example 2 • T(n) = (n+1)2 • which allows that T(n) is O(n2) • Why? • Witnesses: n0 = 1, c = 4 • then T(n) <= c*f(n), n >= 1so (n+1)2 <= 4n2since n2 + 2n + 1 <= n2 + 2n2 + n2 <= 4n2

  24. Example 3 • T(n) = n10 • which allows that T(n) is O(2n) • Why? • Witnesses: n0 = 64, c = 1 • then T(n) <= c*f(n), n >= 64so n10 <= 2nsince 10*log2 n <= n (by taking log2s) which is true when n >= 64 (10*log2 64 == 10*6; 60 <= 64)

  25. 2.4. Some Observations about O() • When choosing an O() approximation to T(), remember that: • constant factors do not matter • e.g. T(n) = (n+1)2 is O(n2) • low-order terms do not matter • e.g. T(n) = 10n2 + 50n + 100 is O(n2) • there are many possible witnesses

  26. 3. Big-Oh for Programs • First decide on a size measure for the data in the program. This will become the n. • Data Type Possible Size Measure integer its valuestring its lengtharray its length

  27. 3.1. Building a Big-Oh Result • The Big-Oh value for a program is built up inductively by: • 1) Calculate the Big-Oh’s for all the simple statements in the program • e.g. assignment, arithmetic • 2) Then use those value to obtain the Big-Oh’s for the complex statements • e.g. blocks, for loops, if-statements

  28. Simple Statements (in C) • We assume that simple statements always take a constant amount of time to execute • written as O(1) • Kinds of simple statements: • assignment, break, continue, return, all library functions (e.g. putchar(),scanf()), arithmetic, boolean tests, array indexing

  29. Complex Statements • The Big-Oh value for a complex statement is a combination of the Big-Oh values of its component simple statements. • Kinds of complex statements: • blocks { ... } • conditionals: if-then-else, switch • loops: for, while, do-while continued

  30. 3.2. Structure Trees • The easiest way to see how complex statement timings are based on simple statements (and other complex statements) is by drawing a structure tree for the program.

  31. Example: binary conversion void main() { int i;(1) scanf(“%d”, &i);(2) while (i > 0) {(3) putchar(‘0’ + i%2);(4) i = i/2; }(5) putchar(‘\n’); }

  32. Structure Tree for Example block1-5 1 5 while2-4 the time for this is the time for (3) + (4) block3-4 3 4

  33. 3.3. Details for Complex Statements • Blocks: Running time bound = summation of the bounds of its parts. • The summation rule means that only the largest Big-Oh value is considered. "summation" means 'add'

  34. Block Calculation Graphically O( f1(n) ) summation rule O( f2(n) ) O( f1(n) + f2(n) + ... + fk(n)) In other words: O( largest fi(n) ) O( fk(n) )

  35. Block Summation Rule Example • First block's time T1(n) = O(n2) • Second block's time T2(n) = O(n) • Total running time = O(n2 + n) = O(n2) the largest part

  36. Conditionals e.g. if statements, switches • Conditionals: Running time bound = the cost of the if-test + larger of the bounds for the if- and else- parts • When the if-test is a simple statement (a boolean test), it is O(1).

  37. Conditional Graphically O(1) Test O( max( f1(n), f2(n)) +1 ) which is the same as O( max( f1(n), f2(n)) ) IfPart ElsePart O( f1(n) ) O( f2(n) )

  38. If Example • Code fragment: if (x < y) // O(1) foo(x); // O(n)else bar(y); // O(n2) • Total running time = O( max(n,n2) + 1) = O(n2 + 1) = O(n2)

  39. Loops • Loops: Running time bound is usually = the max. number of times round the loop * the time to execute the loop body once • But we must include O(1) for the increment and test each time around the loop. • Must also include the initialization and final test costs (both O(1)).

  40. While Graphically Altogether this is:O( g(n)*(f(n)+1) + 1 ) which can be simplified to: O( g(n)*f(n) ) Test O(1) At mostg(n) timesaround O( g(n)*f(n) ) Body O( f(n) )

  41. While Loop Example • Code fragment: x = 0;while (x < n) { // O(1) for test foo(x, n); // O(n2) x++; // O(1)} • Total running time of loop: = O( n*( 1 + n2 + 1) + 1 ) = O(n3 + 2n + 1) = O(n3)

  42. For-loop Graphically O(1) Initialize Test O( g(n)*(f(n)+1+1) + 1) which can be simplified to: O( g(n)*f(n) ) O(1) At mostg(n) timesaround Body O( f(n) ) Increment O(1)

  43. For Loop Example • Code Fragment: for (i=0; i < n; i++) foo(i, n); // O(n2) • It helps to rewrite this as a while loop: i=0; // O(1)while (i < n) { // O(1) for test foo(i, n); // O(n2) i++; // O(1)} continued

  44. Running time for the for loop: = O( 1 + n*( 1 + n2 + 1) + 1 ) = O( 2 + n3 + 2n ) = O(n3)

  45. 3.4.1. Example: nested loops (1) for(i=0; i < n; i++)_ (2) for (j = 0; j < n; j++)(3) A[i][j] = 0; • line (3) is a simple op - takes O(1) • line (2) is a loop carried out n times • takes O(n *1) = O(n) • line (1) is a loop carried out n times • takes O(n * n) = O(n2)

  46. 3.4.2. Example: if statement (1) if (A[0][0] == 0) {(2) for(i=0; i < n; i++)_ (3) for (j = 0; j < n; j++)(4) A[i][j] = 0; }(5) else {(6) for (i=0; i < n; i++)(7) A[i][i] = 1; } continued

  47. The if-test takes O(1);the if block takes O(n2);the else block takes O(n). • Total running time: = O(1) + O( max(n2, n) ) = O(1) + O(n2) = O(n2) // using the summation rule

  48. 3.4.3. Time for a Binary Conversion void main() { int i;(1) scanf(“%d”, &i);(2) while (i > 0) {(3) putchar(‘0’ + i%2);(4) i = i/2; }(5) putchar(‘\n’); } continued

  49. Lines 1, 2, 3, 4, 5: each O(1) • Block of 3-4 is O(1) + O(1) = O(1) • While of 2-4 loops at most (log2 i)+1 times • total running time = O(1 * log2 i+1) = O(log2 i) • Block of 1-5: = O(1) + O(log2 i) + O(1) = O(log2 i) why?

  50. Why (log2 i)+1 ? • Assume i = 2k • Start 1st iteration, i = 2kStart 2nd iteration, i = 2k-1Start 3rd iteration, i = 2k-2Start kth iteration, i = 2k-(k-1) = 21 = 2Start k+1th iteration, i = 2k-k = 20 = 1 • the while will terminate after this iteration • Since 2k = i, so k = log2 i • So k+1, the no. of iterations, = (log2 i)+1

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