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Lab of CS103 Data Structure and C++

Lab of CS103 Data Structure and C++. by Han Young Ryoo, Joseph Gomes. Big-O Notation. Definition: T(n) is O(f(n)) iff (if and only if) there exist positive constants c and n 0 such that T(n) <= c f(n), for all n >= n 0. Examples. T(n) = 3n + 2 is O(n)

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Lab of CS103 Data Structure and C++

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  1. Lab of CS103 Data Structure and C++ by Han Young Ryoo, Joseph Gomes

  2. Big-O Notation • Definition: T(n) is O(f(n)) iff (if and only if) there exist positive constants c and n0 such that T(n) <= c f(n), for all n >= n0

  3. Examples • T(n) = 3n + 2 is O(n) since 3n + 2 <= 4n for all n >= 2 where c=4 and n0 = 2. • T(n) = 10n2 + 4n + 2 is O(n2) since 10n2 + 4n + 2 <= 11n2 for all n >= 5 where c=11 and n0 = 5.

  4. Why do we need Big O • To decide whether an algorithm is adequate, • For a better implementation * The algorithm will always be too slow on a big enough input. • Quicksort vs. Bubble sort (O(n log n) vs. O(n2) Quicksort running on a small desktop computer can beat bubble sort running on a super-computer if there are a lot of numbers to sort. To sort 1,000,000 numbers, the quicksort takes 20,000,000 steps on average, while the bubble sort takes 1,000,000,000,000 steps!

  5. How to determine • Sequence of statements: statement 1; statement 2; ... statement k; The total time is found by adding the times for all statements: total time = time(statement 1) + time(statement 2) + ... + time(statement k)

  6. Rules for using big-O • Ignoring constant factors O(c f(N)) = O(f(N)), where c is a constant Example) O(20 N3) = O(N3)

  7. Rules for using big-O • Ignoring smaller terms If a<b then O(a+b) = O(b) Example) O(N2+N) = O(N2)

  8. Rules for using big-O • Upper bound only If a<b then an O(a) algorithm is also an O(b) algorithm. Example) an O(N) algorithm is also an O(N2) algorithm (but not vice versa).

  9. Rules for using big-O • N and log N are "bigger" than any constant, from an asymptotic view (that means for large enough N). So if k is a constant, an O(N + k) algorithm is also O(N), by ignoring smaller terms. Similarly, an O(log N + k) algorithm is also O(log N).

  10. Rules for using big-O • An O(N log N + N) algorithm, which is O(N(log N + 1)), can be simplified to O(N log N).

  11. Simple Statements • If each statement is "simple" (only involves basic operations) then the time for each statement is constant and the total time is also constant: O(1).

  12. if-then-else statements • if (condition) { sequence of statements 1 } else { sequence of statements 2 } • Either sequence 1 will execute, or sequence 2 will execute. • The worst-case time: the slowest of the two possibilities: max(time(sequence 1), time(sequence 2)). • If sequence 1 is O(N) and sequence 2 is O(1), what is the worst-case time for the whole if-then-else statement?

  13. loops • for (i = 0; i < N; i++) { sequence of statements } • The loop executes N times, so the sequence of statements also executes N times. • If the statements are O(1), what is the total time for the for loop?

  14. Nested loops • for (i = 0; i < N; i++) { for (j = 0; j < M; j++) { sequence of statements } } • The complexity is O(N * M). • Need Generalization: In a common special case where the stopping condition of the inner loop is j < N instead of j < M (i.e., the inner loop also executes N times), the total complexity for the two loops is O(N2).

  15. Function/Method calls • When a statement involves a method call, the complexity of the statement includes the complexity of the method call. Assume that you know that method f takes constant time, and that method g takes time proportional to (linear in) the value of its parameter k. Then the statements below have the time complexities indicated. f(k); // O(1) g(k); // O(k) • Example: for (j = 0; j < N; j++) { g(N);} What is the complexity?

  16. A program that calculates execution time in milliseconds #include <sys/timeb.h> #include "stdio.h" #include "math.h" int main(int argc, char* argv[]) { struct _timeb time_start,time_end; _ftime( &time_start ); for (int i=1;i<=10000;i++) { int m = 1; for(int j=1;j<100000;j++) { m *= 2; } } _ftime( &time_end ); printf("elapsed time = %d milliseconds\n",time_end.millitm-time_start.millitm); return 0; }

  17. Exercises • Two loops in a row: for (i = 0; i < N; i++) { sequence of statements } for (j = 0; j < M; j++) { sequence of statements } How would the complexity change if the second loop went to N instead of M?

  18. Exercises • A nested loop followed by a non-nested loop: for (i = 0; i < N; i++) { for (j = 0; j < N; j++) { sequence of statements } } for (k = 0; k < N; k++) { sequence of statements }

  19. Exercises • A nested loop with dependent loop index: for (i = 0; i < N; i++) { for (j = i; j < N; j++) { sequence of statements } } • A nested loop in which the number of times the inner loop executes depends on the value of the outer loop index

  20. Quiz • for (i = 0; i < N; i++) { for (j = 0; j < N; j++) { C[i][j] = 0; for (k = 0; k < N; k++) { C[i][j] = C[i][j] + A[i][k] * B[k][j]; } } }

  21. Quiz • Sorting for (i = N-1; i > 0; i--) for (j = 0; j < i; i++) if (a[j] > a[j+1]) swap a[j] and a[j+1];

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