1. COSC 2P03 Week 11 Reasons to study Data Structures & Algorithms Provide abstraction Handling of real-world data storage Programmer’s tools Modeling of real-world objects and concepts Programs more readable, understandable, maintainable If you have a good feeling for them, you are a better programmer! –You can solve the problem faster –Your program works better, with fewer bugs –You will know when you have been given an impossible task

2. COSC 2P03 Week 12 Considerations Speed Memory usage Complexity of structure Examples of tradeoffs: Insertion speed vs. search speed vs. deletion speed Structure vs. speed Your task as a programmer: think of the tradeoffs and select the best option

3. Complexity Expresses the efficiency of an algorithm Running time expressed as function of problem size Problem size: number of inputs Some complexity classes: –Logarithmic –Linear –Quadratic –Exponential COSC 2P03 Week 13

4. “Big O” notation Function g(n) is O(f(n)) if there are positive constants c and n 0 such that g(n) ≤ c*f(n) for all n ≥ n 0 Upper bound on its rate of growth Determines algorithm’s “cost” in terms of time to run Note: other measurements exist for lower bound etc COSC 2P03 Week 14

5. 5 Complexity Examples 1 and 2 for(i = 0; i < n; i++) { b[i] = a[i]+1; c[i] = a[i]+b[i]; } for(i = 0; i < n; i++) { for(j = 7; j < n; j=j+3) { b[i] = a[i]+1; } c[i] = a[i]+b[i]; }

6. COSC 2P03 Week 16 Complexity Example 3 for(i = 0; i < n; i++) { if(a[i] > 0) { for(j = 0; j < n; j++) c[i] = c[i] + b[j]; } else c[i] = b[i]; } Question: what if the outer loop were changed to: for(i = 1; i < n; i=i*2) ?

7. COSC 2P03 Week 17 int xyz(int n) { if(n <= 1) return 1; else return xyz(n/2) + n; } Question: what if the last statement were changed to return xyz(n-2) + n; ? Complexity Example 4

8. COSC 2P03 Week 18 Complexity Example 5 int bc(int n) { int c; c = 0; while(n > 0) { c = c + (n % 2); n = n / 2; } return c; }

9. COSC 2P03 Week 19 Complexity Example 6 int pc(int n) { int i, c; if(n == 1) { return 1; } c = 0; for(i = 0; i < n; i++) { c = c + pc(n-1); } return c; }

10. COSC 2P03 Week 110 Table of Computational Complexity If a step takes 0.001 ms and the problem size is N: N=10N=100N=1000 StepsTimeStepsTimeStepsTime f(N) = N100.01ms1000.1ms10001ms f(N) = N 2 1000.1ms10 4 10ms10 6 1s f(N) = N 3 10001.0ms10 6 1s10 9 16.67min f(N) = 2 N 10241.02ms1.27x10 30 4.0x10 16 yrs 1.07x10 301 3.4x10 287 yrs

11. COSC 2P03 Week 111 Comparison: O(n 2 ) vs O(n log n) Assume a machine is running at 200 000 MIPs Assume 1 instruction per iteration. Time to sort 1 billion numbers: Bubble sort O(n 2 ) (10 9 ) 2 steps / (2x10 11 steps/sec) ≈ 2 months. Quick sort O(n log n) 10 9 * log 2 (10 9 ) steps / (2x10 11 steps/sec) ≈ 10 9 * 30 steps / (2x10 11 steps/sec) ≈ 0.15 seconds.

12. COSC 2P03 Week 112 The Rules of Recursion (Weiss, section 1.3) 1.There must be base cases that can be solved without recursion 2.Recursive calls must always make progress towards a base case 3.Assume that all recursive calls work 4.Never duplicate work by solving the same instance of a problem in separate recursive calls

13. COSC 2P03 Week 113 Recursive method printOut (Weiss, section 1.3) public static void printOut(int n) { if(n >= 10) printOut(n/10); printDigit(n % 10); }

14. COSC 2P03 Week 1 14 printOut(5034) { if(5034 >= 10) // true printOut(5034 / 10); printDigit(5034 % 10); } printOut(503) { if(503 >= 10) // true printOut(503 / 10); printDigit(503 % 10); } printOut(50) { if(50 >= 10) // true printOut(50 / 10); printDigit(50 % 10); } printOut(5) { if(5 >= 10) // false printDigit(5 % 10); } Output 4 3 0 5

15. COSC 2P03 Week 115 Sequence of calls to determine F 5 fib(5) fib(4)fib(3) fib(2)fib(2) fib(3)fib(2) fib(2) fib(1) fib(2) fib(2) fib(1) fib(1) fib(0) fib(1)fib(0) fib(1) fib(0)