programming

Programming

Loops & Simulation

2

Some General Tips on Programming

Use Debugger to follow your execution flow and find what went wrong

Understanding is good but not enough - you must practice!

Example 1 - Fibonacci series

Write a program that prints Fibonacci series elements which are smaller than 5.

Example 1 - Fibonacci seriesF(0)=0F(1)=1F(n)=F(n-1)+F(n-2)

http://www.dorbanot.com/3757

http://www.dorbanot.com/3757

Example 1 - Fibonacci series

0 1 1 2 3 5 8 13 21 34 …

0 10 1 1 2 3 5 8 13 210 1 1 20 1 1+ +0 1 1 2 3 5 8 13 21 34+

F(0)=0F(1)=1F(n)=F(n-1)+F(n-2)

int fib1 = 0, fib2 = 1, fib_next = 0, lim = 5;

...printf("%d ", fib1);

while (fib2 < lim){

printf("%d ", fib2);fib_next = fib1 +

fib2;fib1 = fib2;fib2 = fib_next;

}

printf("\n");

Fibonacci – step by step

0

screen:

5lim

0fib1

1fib2

0fib_next

Fibonacci – step by stepint fib1 = 0, fib2 = 1, fib_next = 0, lim = 5;


while (fib2 < lim){



}

printf("\n");

0

screen:

5lim

0fib1

1fib2

0fib_next



while (fib2 < lim){



}

printf("\n");

0 1

screen:

5lim

0fib1

1fib2

0fib_next



while (fib2 < lim){



}

printf("\n");

0 1

screen:

5lim

0fib1

1fib2

1fib_next



while (fib2 < lim){



}

printf("\n");

0 1

screen:

5lim

1fib1

1fib2

1fib_next



while (fib2 < lim){



}

printf("\n");

0 1 1

screen:

5lim

1fib1

1fib2

1fib_next



while (fib2 < lim){



}

printf("\n");

0 1 1

screen:

5lim

1fib1

1fib2

2fib_next



while (fib2 < lim){



}

printf("\n");

0 1 1

screen:

5lim

1fib1

2fib2

2fib_next



while (fib2 < lim){



}

printf("\n");

0 1 1 2

screen:

5lim

1fib1

2fib2

2fib_next



while (fib2 < lim){



}

printf("\n");

0 1 1 2

screen:

5lim

1fib1

2fib2

3fib_next



while (fib2 < lim){



}

printf("\n");

0 1 1 2

screen:

5lim

2fib1

2fib2

3fib_next



while (fib2 < lim){



}

printf("\n");

0 1 1 2

screen:

5lim

2fib1

3fib2

3fib_next



while (fib2 < lim){



}

printf("\n");

0 1 1 2 3

screen:

5lim

2fib1

3fib2

3fib_next



while (fib2 < lim){



}

printf("\n");

0 1 1 2 3 3

screen:

5lim

2fib1

3fib2

5fib_next



while (fib2 < lim){



}

printf("\n");

0 1 1 2 3 3

screen:

5lim

3fib1

3fib2

5fib_next



while (fib2 < lim){



}

printf("\n");

0 1 1 2 3 3

screen:

5lim

3fib1

5fib2

5fib_next

We first perform the mathematical operation with the variable (=) and then evaluate it (++).

We first evaluate the variable (++) and then perform the mathematical operation with it (=).

Shorter lines

e = e + 1;e;++

f = f + 1;++f;

j=5i=6

i = 5;j = i;++

i=j=6i = 5;j = ++i;

We first perform the mathematical operation with the variable (=) and then evaluate it (--).

We first evaluate the variable (--) and then perform the mathematical operation with it (=).

Shorter lines

e = e - 1;e;--

f = f - 1;--f;

j=5i=4

i = 5;j = i;--

i=j=4i = 5;j = --i;

Now your turn!Class exercise: print a series Write a program that prints the first 10

elements in the following series : 1 -2 3 -4 5 -6 7 -8 9 -10……

Example 2: increasing series Write a program that gets positive integers

from the user and prints an increasing series (without duplications):

Every number is compared to its previous as follows:

Number > previous print it. Number = previous ignore it. Number < previous exit.

Solution 2: increasing series#include <stdio.h>int main() {

int x, prev = -1;printf ("please enter a series of positive integers :\n");do {

scanf("%d", &x );if (x < prev)

break;if (x == prev)

continue;printf("%d ", x);prev = x;

} while (x >= prev);return 0;

}

Example 3: Prime Numbers

Write a program that prints all the prime numbers smaller than or equal to N.

N is given by the user (input)

How do we go over all the numbers <= N? How do we know if a number is prime?

Solution 3: Prime Numbersint main(){

int candidate = 0, divisor = 0, last = 0;printf("Enter a number\n");scanf("%d", &last);for (candidate = 2; candidate <= last; ++candidate){

for (divisor = 2; divisor < candidate; ++divisor) if (candidate % divisor == 0)

break;

if (divisor == candidate) printf("the number %d is prime\n",

candidate); }return 0;

}

getchar

getchar() gets a single character from the user.

Requires including stdio.h Returns a non-positive number on failure. Similar to scanf.

char c;c = getchar();

char c;scanf("%c", &c);

====

putchar

putchar(char c) prints out the character inside the brackets.

Requires including stdio.h Similar to printf.

char c;putchar(c);

char c;printf("%c", c);

====

Example 4: Copy

int main(){ int c = 0;

c = getchar(); while (c != '\n') { putchar(c); c = getchar(); }

return 0;}

Example 5: Counting Letters

Write a program that read input from the user until end-of-line (‘\n’) is encountered.

The program counts the number of letters before the first space (or until the end-of-line , if space is not found).

Example 5: Counting Lettersint main() {

int counter = 0, c = 0;printf("Enter a line of text:\n");c = getchar();while (c != '\n') {

if (c == ' ') /* found the first space - exit the loop */break;

++counter;c = getchar();

}

if (c == ' ') /* we found a space */printf("There are %d letters before the first space.\n",

counter);else /* we didn’t find a space */

printf("There are no spaces in the input line.\n");return 0;

}

ctype library The <ctype.h> library contains

character classification functions. Useful functions:

int isalpha (int c) test for alphabetic character

int isdigit(int c)test for digit

int tolower(int c)convert character to lowercase

int toupper(int c)convert character to uppercase

and more… see ctype.h library

http://www.utas.edu.au/infosys/info/documentation/C/CStdLib.html

Exercise 6: Low to Up

Read input from the user until end-of-line (‘\n’) is encountered.

Convert lowercase letters to uppercase ones.

Solution 6: Low to Up#include <stdio.h>#include <ctype.h> // needed for the function toupper

int main(){

int c = 0;printf("Please, enter a string (end with an enter):\n");c = getchar();

while (c != '\n'){

putchar((c >= 'a' && c <= 'z') ? toupper(c) : c);c = getchar();

}putchar('\n');return 0;

}

Simulation: estimating pi Throw darts to compute the area of a unit

circle inside a square Pi can be inferred from the ratio between the

area of a square (total hits) to the area of a circle (hits inside the circle) resides in it:

151.39007094

44 2

2

square

circle

square

circle

AA

rr

AA

r =1

Random numbers We can generate (psuedo)-random

variables using rand() rand() returns a random integer between 0

and MAX_INT (MAX_INT may vary between implementations but it is at least 32767).

We need to include <stdlib.h> in the program.

Example#include <stdio.h>#include <stdlib.h>

int main(){

int number=0;number=rand();printf ("Random number is %d\n",number);return 0;

}

However, if we run it many times, it will always generate the same “random” number…

This is because the computer uses a formula to generate the “random” number.

What do we do?

Not so random…

The algorithm is based on seed, an initial state that will always produce the same sequence of numbers.

If we change the seed, the sequence will be different, resulting in different random number.

We initialize the seed with srand(seed). It is common to initialize with computer time

srand(time(NULL)). We need to include <time.h>.

Change the seed

Example#include <stdio.h>#include <time.h>#include <stdlib.h>

int main(){

int number=0;srand(time(NULL));number=rand();printf ("Random number is %d\n",number);return 0;

}

Random numbers within a range

We can choose a random variable in a range using %.0 up to 100: rand()%100;50 to up to 75: rand()%25+50;double between 0 and 1:

rand()/(double)RAND_MAX;

#include <stdio.h>#include <stdlib.h>#include <time.h>int main(){

long countAs=0, countAc=0,numDarts=1000000;double x, y;

srand(time(NULL)); // Used to initialize the random number generatorfor(countAs = 0; countAs < numDarts; countAs++){

// generate random number between -1 and 1x = -1 + 2 * ((double)rand()) / RAND_MAX;// generate random number between -1 and 1y = -1 + 2 * ((double)rand()) / RAND_MAX;// count if it hits the circle i.e. smaller than the radiusif (x*x + y*y <= 1.0)

countAc++;}printf("The estimation of pi is: %g\n", (4.0 * countAc) / countAs);return 0;

}

Class ex. solution (using while)/* Prints the first 10 elements in the following series: * 1 -2 3 -4 5 -6 7 -8 9 -10 ... */ #include <stdio.h>

int main(){

int counter = 1, sign = 1;

while (counter <= 10){

printf("%d ", sign*counter);sign*=-1;counter++;

}return 0;

}

Class ex. solution (using for)/* Prints the first 10 elements in the following series: * 1 -2 3 -4 5 -6 7 -8 9 -10 ... */ #include <stdio.h> int main(){

int counter, sign;

for(counter = 1,sign = 1; counter <= 10; counter++, sign*=-1)printf("%d ", sign*counter);

return 0;

}

programming

Documents

fib1 fib2fib1

fib2 lim

stepint fib1

nextfibonacci step

printfnfibonacci step

fibonacci serieswrite

fibonacci seriesf0

fibonacci series0