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Fundamentals Integer functions Floor x = greatest integer less than or equal to x. Ceiling x = smallest integer greater than or equal to x. Summation Powers n n (n 1) i i 1 2 n n (n 1) (2n 1) i i 1 2 6 Geometric n 1 r n 1 ri i 0 1 r , r 1 Probability Suppose X is a random variable with values x1 , x 2 ,..., x n and corresponding probabilities p1 , p 2 ,..., p n . Then the expected value of the random variable X is n E ( X ) xi p i i 1 Growth rates One of the fundamental goals of a course is to predict the amount of resources an algorithm will use. Often it is the running time that we want to estimate. More precisely, we want to know the relationship between input size and running time. Many of the algorithms in computer science have running times with the functional forms such as those listed in the table below. Assuming one calculation per microsecond, complete the following table for the following running times (first column) and input sizes (first row). T(n) 10 20 50 100 1000 1000000 1 1 1 1 1 1 1 us us us us us us 3.32 4.32 5.64 6.64 9.97 19.9 log n us us us us us us 10 20 50 100 1 1 n us us us us msec second 33.2 86.4 282 664 9.97 19.9 n log n us us us us msec seconds 100 400 2.5 10 1 11.57 n2 us us msec msec second days 100 400 2.5 10 16.7 31.7 1000 n2 msec msec seconds seconds minutes years 1 8 125 1 16.7 317 n3 msec msec msec second minutes centuries 1.02 1.05 18.8 4 * 108 0.000001 * 2n ns msec minutes centuries n 1.02 1.05 35.7 4 * 1014 2 msec seconds years centuries 3.63 771 9 * 1050 n! seconds centuries centuries Clearly, algorithms with exponential or factorial running times are not practical on even moderately sized inputs. In this class we will be classifying the running times of algorithms into one of the categories shown below. The categories are organized from the fastest running times to the slowest. O(1) O(log n) O(n) O(n log n) O(n^2) O(n^3) O(2^n) O(n!) O(n^3) The growth rates may be compared graphically using a graphing calculator or Maple. For example in the graph below we compare n log n and n2 > plot({n*log[2](n),n^2},n=1..20); Asymptotics The formulas for the running time of an algorithm could get quite complex. The time would depend on the architecture of the particular machine. Asymptotic analysis makes two simplifying assumptions that make it easier to classify growth rates. 1) Large input sizes: Suppose the input size is n. We are primarily concerned with how the running time increases for large n. For small n almost any algorithm will do and the real differences in efficiency between algorithms are not apparent. 2) Ignore constant factors: The actual running time of a program depends on how the algorithm is coded, compiler optimizations, hardware speed and other factors. Hence, we attempt only to find a function f(n) that is proportional to the actual running speed. Asymptotics makes use of a special notation that simplifies the comparison process. Let f(n) and g(n) be two functions. If the functions grow at approximately the same rate, than in terms of limits we have f ( n) lim c n g ( n) where c is a nonzero constant not equal to 0 or . We can express this as f(n) (g(n)) meaning that the two functions are asymptotically equivalent. If f(n) grows at the same rate or perhaps more slowly than g(n), then c [0, ) and we write f(n) O(g(n)). The table below provides a summary of the commonly used notations and their meanings. f(n) (g(n)) 0<c< asymptotically equal f(n) O(g(n)) 0c< asymptotically less than or equal to f(n) (g(n)) c>0 asymptotically greater than or equal to f(n) o(g(n)) c=0 asymptotically less than f(n) (g(n)) c= asymptotically greater than Example: Compare f (n) n 3 2n 2 and g (n) n 3 . f ( n) Since lim 1 we have f(n) ( n 3 ), f(n) O( n 3 ) and f(n) ( n 3 ). n g ( n) It should be noted that each of the above classes represents a family of functions. Example: 3n 2 2n 5 O(n 2 ) , 7n O(n 2 ) , sin n O (n 2 ) Note although all the above statements are true, the last two statements do not represent useful bounds. Tighter bounds would be 7n O(n) , sin n O(1) Example: Printing an array of length n. for (i=0; i<n; i++) cout << a[i] Assume each element prints in constant time and then the time to print n elements is O(n) . Example: Multiplying square matrices with dimensions nn. for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { double sum = 0; for (int k = 0; k < n; k++) { sum = sum + a(i, k) * b(k, j); } c(i, j) = sum; } } Assuming each assignment takes constant time, the time to multiply the matrices is given by t n n (n 2) n n 3 2n 2 . Hence, the running time of this matrix multiplication is O(n 3 ) . Recurrences A recurrence relation defines a function in terms of smaller instances if itself. For example n! = n(n-1)! with 1! = 1. Recurrence relations are often used to model the running times of recursive functions. In this section we look at a couple of methods for solving recurrence relations. 1) Guess and check 2) Iteration 3) Characteristic equation 4) Generating functions 5) Recursion trees 6) Master theorem 7) Maple Example: Solve t n 2t n 1 1, t 0 0 using Guess and Check. Guess a solution to the recurrence and prove using mathematical induction. Since t 0 0, t1 1, t 2 3, t 3 7, t 4 15, t 5 31,... it seems reasonable to guess t n 2 n 1 . Prove by induction. Base case: t 0 2 0 1 0 . Assume t n 2 n 1 . Then t n 1 2t n 1 2(2 n 1) 1 2 n 1 1 as desired and hence t n 2 n 1 for all n 0 . Example: Solve t n 3t n 1 7, t 0 5 using iteration. t n 3t n 1 7 3 (3t n 2 7) 7 3 2 t n 2 3 7 7 3 3 t n 3 3 2 7 3 7 7 3 n t 0 3 n 1 7 3 n 2 7 3 2 7 3 7 7 3n 1 3 1 5 3n 7 7 5 3n 3n 1 2 Example: Solve t n 5t n 1 6t n 2 0, t 0 0, t1 1 using the characteristic polynomial. For homogeneous (right hand side equals zero), linear (no products), constant coefficient recurrences we can assume a solution of the form t n r n . Substituting this into the recurrence results in the characteristic polynomial r 2 5r 6 0 with solutions r 2,3 . Hence, the solution to the recurrence has the form t n c1 (2) n c 2 (3) n . Use the initial conditions to solve for the constants. Hence, t n (2) n (3) n . Example: Use Maple to solve t n 2t n 1 n > rsolve({t(n) = 2*t(n/2) + n,t(1)=1}, {t}); Towers of Hanoi At the creation of the universe, priests in the temple of Brahama were supposedly given three diamond needles, with one needle containing 64 golden disks. Each golden is slightly smaller than the one below it. The priests’ task is to move all 64 disks from the first needle to the third according to the following rules 1. Only one disk can be moved at a time. 2. The removed disk must be placed on one of the needles. 3. A larger disk cannot be placed on top of a smaller disk. Assuming the priests can move one disk per second and that they use the minimum number of moves, estimate the “life expectancy” of the universe. Three disk problem 1. Move disk 1 from needle 1 to needle 3. 2. Move disk 2 from needle 1 to needle 2. 3. Move disk 1 from needle 3 to needle 2. 4. Move disk 3 from needle 1 to needle 3. 5. Move disk 1 from needle 2 to needle 1. 6. Move disk 2 from needle 2 to needle 3. 7. Move disk 1 from needle 1 to needle 3. Recursion Suppose needle 1 contains n disks 1. Move n-1 disks from needle 1 to needle 2 using needle 3 as the intermediate needle. 2. Move disk n from needle 1 to needle 3. 3. Move the n-1 disks from needle 2 to needle 3 using needle 1 as the intermediate needle. Recurrence Suppose t n = number of times a disk is moved when solving the Towers of Hanoi problem with n disks. Then the work done in the n disk case is related to the work done in the n-1 disk case via t n t n 1 1 t n 1 2t n 1 1 with t1 1 . The equation t n 2t n 1 1, t1 1 is called a recurrence. Find the homogenous solution t n 2t n 1 0 by setting t n r n to obtain t n 2 n . Find the non-homogenous solution by setting t n A to obtain t n 1 . Combine the homogenous and non-homogenous solutions to obtain the complete solution t n 2 n 1 . Now compute the “life expectancy” of the universe. C++ code # include<iostream> using namespace std; void moveDisks(int,int,int,int); int main( ) { moveDisks(4,1,3,2); return 0; } void moveDisks(int count,int needle1,int needle3,int needle2) { if (count > 0) { moveDisks(count-1,needle1,needle2,needle3); cout << "Moving disk " << count << " from needle " << needle1 << " to needle " << needle3 << endl; moveDisks(count-1,needle2,needle3,needle1); } } Project 1 1) You have an O(n2) algorithm which takes 5 minutes to run on a problem of size n = 1000. How long will it take to solve a problem of size n = 10,000? What about n = 1,000,000? 2) Repeat problem one with an O(n log n) algorithm. 3) Chapter 1: 3a (sum of odd integers) 4) Chapter 1: 5b (number of assignment statements) 5) Give a simplified big-O bound for the following running times: Prove your result by computing an appropriate limit. a) 20 n2 b) 10 n3 + 6 n2 c) 5n log n + 30 n 6) Characterize the running time of the following code fragment using the big-oh notation. int i = n; while (i > 0) { cout << "howdy"; i = i/2; } 7) Fibonacci numbers: Suppose you have a pair of rabbits and suppose every month each pair bears a new pair that from the second month on becomes productive. How many pairs of rabbits will you have in a year? a) Write down the recurrence using the notation f(n) to represent the number of rabbits alive in month n. b) Solve the recurrence to show 1 1 5 1 5 n n f ( n) 5 2 2 c) There are three ways to compute the Fibonacci numbers (i) iteratively (ii) recursively and (iii) using the exact solution derived in part b). Implement these three methods as C++ functions. d) Characterize the running time of each method presented in part c) using the big-oh notation. Resources 1) http://www.mathsci.unco.edu/course/isaacson/ced509u01-complexity.html 2) http://www.cs.rutgers.edu/~djimenez/ut/utsa/cs3343/lecture3.html

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