Error and Error Handling

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1 Error and Error Handling ˆ You can issue/throw an error if you do not allow the callee for some situations. 1 1 >> error('this is an error.'); ˆ Also, you can use a try-catch statement to handle errors. 1 try 2 % normal operations 3 catch 4 % handler operations 5 end 1 In other words, you can decide if the action can be performed. Zheng-Liang Lu 98 / 122

2 Example: Combinations ˆ For all nonnegative integers n k, ( n k) is given by ( ) n n! = k k!(n k)!. ˆ Note that factorial(n) returns n!. 1 clear; clc; 2 3 n = input('n =? '); 4 k = input('k =? '); 5 y = factorial(n) / (factorial(k) * factorial(n - k)) 6 disp('end of program.'); Zheng-Liang Lu 99 / 122

3 ˆ Try n = 2, k = 5. ˆ factorial( 3) is not allowed! ˆ The program is not designed to handle this error, so it is interrupted in Line 5 and does not reach the end of program. 1 try 2 y = factorial(n) / (factorial(k) *... factorial(n - k)) 3 catch e % capture the thrown exception 4 disp(['error: ' e.message]); % show the message 5 end 6 disp('end of program.'); Zheng-Liang Lu 100 / 122

4 Exercise: Divided by Zero ˆ Write a program which calculates x/y for any two real numbers x, y. ˆ The program restricts the user to do division by zero % assume x = 1 any y = 0 2 if y == 0 3 error('divided by zero!'); 4 else 5 ans = x / y; 6 end 7... Zheng-Liang Lu 101 / 122

5 Repetitions ˆ If some instructions are potentially repeated, you should wrap those in a loop. ˆ All loops can be done in the following three parts: ˆ find the repeated pattern for each iteration; ˆ warp them by a proper loop; ˆ set the continuation condition by defining a loop variable with some criterion. ˆ MATLAB has two types of loops: for loops and while loops. ˆ Use for loops if you know the number of iterations. ˆ Otherwise, use while loops. Zheng-Liang Lu 102 / 122

6 for Loops ˆ A for loop is the easiest choice when you know how many times you need to repeat the loop. 1 for loopvar = somearray 2 % body 3 end ˆ Particularly, we often use for loops to manipulate arrays (data)! Zheng-Liang Lu 103 / 122

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8 Examples 1 for i = 1 : 10 2 disp(i); 3 end ˆ How to show the odd integers from 1 to 9? 1 for student = {'Arthur', 'Alice'} 2 disp(student); 3 end ˆ Clearly, MATLAB has for-each loops, which is an enhanced one compared to the naive one in C. Zheng-Liang Lu 105 / 122

9 Example: Find Maximum (Revisited) 1 clear; clc; 2 3 data = [ ]; 4 result = data(1); 5 for item = data(2 : end) 6 if result < item 7 result = item; 8 end 9 end 10 result ˆ Can you find the location of the maximum element? ˆ Try to find the minimum element and its location. Zheng-Liang Lu 106 / 122

10 Exercise: Where is Maximum? ˆ Write a program which indicates where the maximum is. 1 clear; clc; 2 3 data = [ ]; 4 loc = 1; 5 for i = 2 : length(data) 6 if data(i) > data(loc) 7 loc = i; 8 end 9 end 10 loc ˆ Extension: maximum drawdown (mdd) in a trading strategy. Zheng-Liang Lu 107 / 122

11 Example: Running Sum ˆ Write a program which calculates the sum of data. ˆ We use randi to generate a random integer array as testing data. 1 clear; clc; 2 3 % data = 1 : 1 : 100; 4 data = randi(100, 1, 5) 5 s = 0; 6 for item = data 7 s = s + item; % running sum 8 end 9 s ˆ Of course, you could use sum for the same functionality. Zheng-Liang Lu 108 / 122

12 Example: Monte Carlo Simulation ˆ Let m be the number of sample points falling in the region of the quarter circle shown in the next page, n be the total number of sample points. ˆ Use rand to generate a value between 0 and 1 (exclusive). ˆ Write a program which estimates π by ˆπ = 4 m n. ˆ Note that ˆπ π as n by the law of large numbers (LLN). Zheng-Liang Lu 109 / 122

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14 1 clear; clc; 2 3 n = 1e5; 4 m = 0; 5 for i = 1 : n 6 x = rand(1); 7 y = rand(1); 8 if x ˆ 2 + y ˆ 2 < 1 9 m = m + 1; 10 end 11 end 12 pi estimator = 4 * m / n ˆ Try to vectorize this program. Zheng-Liang Lu 111 / 122

15 while Loops ˆ The while loops are preferred when you need to keep repeating the instructions until a continuation criterion is not satisfied. 1 while criterion 2 % body 3 end ˆ Be aware that if statement executes only once; you should use while statement if you want to repeat some actions. Zheng-Liang Lu 112 / 122

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17 Example: Compounding ˆ Let balance be the initial amount of some investment, and r be the annualized return rate. ˆ Write a program which calculates the holding years when this investment doubles it value. Zheng-Liang Lu 114 / 122

18 Solution ˆ In this case, we don t know how many iterations we need before the loop. 1 clear; clc; 2 3 balance = 100; 4 goal = 200; 5 r = 0.01; 6 years = 0; 7 8 while balance < goal 9 balance = balance * (1 + r); 10 years = years + 1; 11 end 12 years ˆ Note that the criterion is to continue the loop. Zheng-Liang Lu 115 / 122

19 Infinite Loops 1 while true 2 disp('press ctrl+c to stop me!!!'); 3 end ˆ Note that your program can terminate the program by pressing ctrl+c. Zheng-Liang Lu 116 / 122

20 More Exercises ˆ Let a > b be two any positive integers. ˆ Write a program which calculates the remainder of a divided by b. ˆ Do not use mod(a, b). ˆ Write a program which determines the greatest common divisor (GCD) of a and b. ˆ Do not use gcd(a, b). Zheng-Liang Lu 117 / 122

21 Numerical Example: Bisection Method for Root-finding Zheng-Liang Lu 118 / 122

22 Problem Formulation Input - Target function f (x) = x 3 x 2 - Initial search interval [a, b] = [1, 2] - Error tolerance ɛ = 1e 9 Output - ˆr Zheng-Liang Lu 119 / 122

23 Solution 1 clear; clc; 2 3 a = 0; b = 2; eps = 1e-9; 4 while b - a > eps 5 c = (a + b) / 2; 6 fa = a * a * a - a - 2; 7 fc = c * c * c - c - 2; 8 9 if fa * fc < 0 % a simplified case 10 b = c; 11 else 12 a = c; 13 end 14 end 15 c Zheng-Liang Lu 120 / 122

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25 Exercise: Square Roots 2 ˆ Using the aforesaid bisection algorithm, write a function which determines n for all n R +. ˆ It is easy to see that n = x is equivalent to x 2 n = 0. ˆ So n is the positive root of x 2 n = 0, which is the input of the bisection algorithm. ˆ Clearly, a = 0. ˆ Choose b = n if n > 1; b = 1, otherwise. 2 Also see Best-Square-Root-Method-Algorithm-Function-Precisi. Zheng-Liang Lu 122 / 122