1. CS1010E Programming Methodology Tutorial 3 Control Structures and Data Files C14,A15,D11,C08,C11,A02

2. Question 1 (a) (b) 1.What is output ? 2.Correct the code to print [0, input)! 1. #include 2. int main(void) { 3. int i, input; 4. scanf(“%d”, &input) 5. while(i < input) 6.printf("%d\n", i); 7.i++; 8. return 0; 9. } 1. #include 2. int main(void) { 3. int i = 0, input; 4. scanf(“%d”, &input); 5. while(i < input) { 6.printf("%d\n", i); 7. i++; 8. } 9. return 0; 10.}

3. Question 1 (c) Change to Do-while Loop Which one is correct ? 1. #include 2. int main(void) { 3. int i = 0, input; 4. scanf(“%d”, &input); 5. while(i < input) { 6.printf("%d\n", i); 7. i++; 8. } 9. return 0; 10. } int i = 0, input; scanf(“%d”, &input); do { printf("%d\n", i); i++; } while(i < input); int i = 0, input; scanf(“%d”, &input); do { printf("%d\n", i); } while(i++ < input); int i = 0, input; scanf(“%d”, &input); if ( i < input) { do { printf("%d\n", i); i++; } while(i < input); }

4. Question 1 (d) Rewrite using for-loop 1. #include 2. int main(void) { 3. int i = 0, input; 4. scanf(“%d”, &input); 5. while(i < input) { 6.printf("%d\n", i); 7. i++; 8. } 9. return 0; 10. } for(i = 0;i < input; i++) { printf("%d\n", i); }

5. Question 1 (d) Three loops are equivalent !! Why we have three loops instead of one ?!! For convenience While - used when you know stopping condition For - used when you know the number of iterations while(i < input){ printf("%d\n", i); i++; } if ( i < input){ do{ printf("%d\n", i); i++; } while(i < input); } for(i = 0;i < input; i++){ printf("%d\n", i); }

6. Question 1 (e) Print Out all the odd number from [0, input) How to test parity of a number ? num % 2 != 0 for(i = 0;i < input; i++) { printf("%d\n", i); } for(i = 0;i < input; i++){ if(i%2 != 0) { printf("%d\n", i); } Can you use num%2 == 1 for odd number test? What is wrong here?

7. Question 1 (f) Print Out all the odd number divisible by 5 from [0, input) How to test such condition ? num is a not multiple of 2 AND num is multiple of 5 for(i = 0;i < input; i++) { printf("%d\n", i); } for(i = 0;i < input; i++){ if(i%2 != 0 && i%5 == 0){ printf("%d\n", i); } Note the difference between “&&” and “&”

8. Question 2 Iterate through a table (matrix) Standard trick is to use a Nested-For-Loop “i” indicate row number “j” indicate column number i =1 j =1j =3 i =2 j =2 for(i=1; i<=input; i++){ for(j=1; j<=input; j++){ do sth… } (1,1,)->(1,2)->(1,3)-> … -> (2,1) -> (2,2) ->(2,3)-> ….-> …. Column First Order

9. Question 2 (a) Write a program to take in an integer from 1 to 10 and print a multiplication table of that number. Analyze: 1.For every cell, cell value = row number * column number for(i=1; i<=input; i++){ for(j=1; j<=input; j++){ printf("%4d", i * j); } printf(“\n”); } for(i=1; i<=input; i++){ for(j=1; j<=input; j++){ do sth… } “i” indicate row number “j” indicate column number End of each row, we need a newline

10. Question 2 (b) Only the bottom triangle is obtained. Analyze: 1.For every cell, cell value = row number * column number 2.Bottom triangle ≡ row number >= column number for(i=1; i<=input; i++){ for(j=1; j<=i; j++){ printf("%4d", i * j); } printf(“\n”); } for(i=1; i<=input; i++){ for(j=1; j<=input; j++){ do sth… } “i” indicate row number “j” indicate column number Indicate row number >= column number How to print out Top triangle ? How to print out Pascal's triangle ?

11. Question 2 (c) Only the bottom triangle is obtained. Square Number -> ‘-’ Prime Number -> ‘p’ Analyze: 1.For every cell, cell value = row number * column number 2.Bottom triangle ≡ row number >= column number 3.We need to test the cell value: 1.Square number 2.Prime number for(i=1; i<=input; i++){ for(j=1; j<=i; j++){ cell = i *j; if(isPrime) { printf(“p”); } else if (isSquare) { printf(“-”); } else { printf("%4d", cell); } printf(“\n”); }

12. Question 2 (c) Test cell for prime 1.If cell is 1, it is not prime 2.Else If cell is divided by any number in [2, itself-1], it is not prime 3.Else it is prime int isPrime = 1; if(cell == 1) isPrime = 0; else{ for(k=2; k<= cell - 1; k++){ if(cell % k == 0) isPrime = 0; } Assume cell is prime Find all cases where cell is not prime

13. Question 2 (c) Test cell for square 1.If exists a number k in [1, cell), s.t. k * k == cell then cell is square isSquare = 0; for(k=1; k<=cell; k++) { if(k * k == cell) isSquare = 1; } Assume cell is not Square Try to find a k

14. Question 2 (c) Put all these together. int i, j, k, cell, input, isSquare, isPrime; scanf(“%d”, &input) for(i=1; i<=input; i++){ for(j=1; j<=i; j++){ cell = i * j; /* Square number check */ isSquare = 0; for(k=1; k<=cell; k++){ if(k * k == cell) isSquare = 1; } /* Prime number check */ isPrime = 1; if(cell == 1) isPrime = 0; for(k=2; k<cell; k++){ if(cell % k == 0) isPrimeNumber = 0; } /* Choose what to print */ if(isPrime == 1) printf(“ p”); else if(isSquare == 1) printf(“ -”); else printf(“%4d”, i*j); } printf(“\n”); } Can you improve your code to run faster ?

15. Question 3 Solution to Diophantine Equation In mathematics, we build a linear system and solve it using Guassian’s Elimination. In Computer Science, we can do brute force using power of computers. Given a set of equations of X,Y. Test all the possible combinations of to find a solution Analyze: 1.Read input from file 2.Test all possible values of 3.Output to file

16. Question 3 Read Input From File: #define NOSOLUTION -99999 FILE *fin, *fout; fin = fopen(“input.txt”, “r”); int x, y, solutionX, solutionY, a1, a2, b1, b2, c1, c2; fscanf(fin, “%d %d %d”, &a1, &b1, &c1); fscanf(fin, “%d %d %d”, &a2, &b2, &c2); Will a1 b1 c1 has the same value of a2 b2 c2? Other opening mode : w write a append r+ read & write … “w” will erase all the content of original file Be careful when read and write on the same file!!!

17. Question 3 Test all combinations of possible X, Y’s value Since we have two variable, we use nested loop solutionX = NOSOLUTION; solutionY = NOSOLUTION; for(x=-100; x<=100; x++) { for(y=-100; y<=100; y++) { if(a1*x + b1*y == c1 && a2*x + b2 * y == c2) { solutionX = x; solutionY = y; } Can we insert a break here ? solutionX = x; solutionY = y; break;

18. Question 3 Output to File Accordingly: fout = fopen(“ouptut.txt”, “w”); if(solutionX != NOSOLUTION) fprintf(fout,“X = %d, Y = %d”, solutionX, solutionY); else fprintf(fout,“Integer solution not found in range [-100,100]”); fclose(fout); fclose(in); return 0; If not closed, the result is not guaranteed to be written into file !

19. Thank You ! See you next week!