Question of the Day  Move one matchstick to produce a square

Question of the Day  Move one matchstick to produce a square

“Anything that can go wrong…”  Big-Oh will calculate algorithm’s complexity  Worst-case  Worst-case analysis of algorithm performance  Usually reasonably correlated with execution time  Not always right to consider only worst-case  May be situation where worst-case is very rare  Solve for other cases similarly, but almost never done

Algorithmic Analysis

Primitive Statements O(1)  Basis of programming, take constant time: O(1)  Fastest possible big-Oh notation  Time to run sequence of primitive statements, too  But only if the input does not affect sequence Ignore constant multiplier 11 O(5) = O(5 * 1 ) = O( 1 )

Simple Loops for (int i = 0; i < n.length; i++){} -or- while (i < n) { i++; }  Each loop executed n times  Primitive statements only within body of loop O(1)  Big –oh complexity of single loop iteration: O(1) O(n)  Either loop runs O(n) iterations O(n)O(1)O(n)  So loop has O(n) * O(1) = O(n) complexity total

More Complicated Loops for (int i = 0; i < n; i += 2) { } i  0, 2, 4, 6,..., n  In above example, loop executes n / 2 iterations O(1)  Iterations takes O(1) time, so total complexity: O( n )O(1) = O( n / 2 ) * O(1) O(n ) = O(n * ½ * 1) O(n) = O(n)

Really Complicated Loops for (int i = 1; i < n; i *= 2) { } i  1, 2, 4, 8,..., n  In above code, loop executes log n iterations O(1)  Iterations takes O(1) time, so total complexity: O(log n)O(1) = O(log n) * O(1) O(log n) = O(log n * 1) O(log n) = O(log n)

Nested Loops for (int i = 0; i < n; i++){ for (int j = 0; j < n; j++) { } }  Program would execute outer loop n times  Inner loop run n times each iteration of outer loop  O(n)O(n)  O(n) iterations doing O(n) work each iteration O(n)O(n)O(n 2 )  So loop has O(n) * O(n) = O(n 2 ) complexity total  Loops complexity multiplies when nested

 Important to explain your answer  Saying O(n) not enough to make it O(n)  Methods using recursion especially hard to determine  Derive difficult answer using simple process Justifying an Answer

Proving Your Answer

 Important to explain your answer  Saying O(n) not enough to make it O(n)  Methods using recursion especially hard to determine  Derive difficult answer using simple process  May find that you can simplify big-Oh computation  Find smaller or larger big-Oh than imagined  Convincing others need not be very formal  Explaining your answer in clear way is critical, however Justifying an Answer

Algorithm sneaky(int n) total = 0 for i = 0 to n do for j = 0 to n do total += i * j return total end for end for  sneaky would take _____ time to execute  O(n) iterations for each loop in the method It’s About Time

Algorithm sneaky(int n) total = 0 for i = 0 to n do for j = 0 to n do total += i * j return total end for end for  sneaky would take O(1) time to execute  O(n) iterations for each loop in the method  But in first pass, method ends after return  Always executes same number of operations It’s About Time

Algorithm power(int a, int b ≥ 0) if a == 0 && b == 0 then return -1 endif exp = 1 repeat b times exp *= a end repeat return exp  power takes O(n) time in most cases  Would only take O(1) if a & b are 0  ____ algorithm overall Big-Oh == Murphy’s Law

Algorithm power(int a, int b ≥ 0) if a == 0 && b == 0 then return -1 endif exp = 1 repeat b times exp *= a end repeat return exp  power takes O(n) time in most cases  Would only take O(1) if a & b are 0 big-Oh uses worst-case  O(n) algorithm overall; big-Oh uses worst-case Big-Oh == Murphy’s Law

algorithm sum(int[][] a) total = 0 for i = 0 to a.length do for j = 0 to a[i].length do total += a[i][j] end for end for return total  Despite nested loops, this runs in O(n) time  Input is doubly-subscripted array for this method  For this method n is number entries in array How Big Am I?

Handling Method Calls  Method call is O(1) operation, …  … but then also need to add time running method  Big-Oh counts operations executed in total  Remember: there is no such thing as free lunch  Borrowing $5 to pay does not make your lunch free  Similarly, need to include all operations executed  In which method run DOES NOT MATTER

public static int sumOdds(int n) { int sum = 0; for (int i = 1; i <= n; i+=2) { sum+=i; } return sum; } public static void oddSeries(int n) { for (int i = 1; i < n; i++) { System.out.println(i + “ ” + sumOdds(n)); } }  oddSeries calls sumOdds n times  Each call does O(n) work, so takes O(n 2 ) total time! Methods Calling Methods

Your Turn  Get into your groups and complete activity

How to Prepare for Midterm DODON'T  Make cheat sheets for the test  Review how parts of Java work  Add post-its to important pages  Memorize  Drink case of 40s before test  Use post-its as clothing