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SITE SCIENTIFIQUE D'AZROU

Informatique - Computer science

C/C++

Lesson four 4.

In this lesson we will learn about condition instructions:

If (condition1) {instruction 1 ;}
Else if (cond2) {instruction 2;}
.
.
.
Else if (cond n-1) {instruction n-1;}
Else (cond n) {instruction n;}

This function allow us to run instruction 1 if the first condition is true, if not, the computer Will try the second condition and run the instruction2 if it's true, if not, it goes To the next and so on till the n instruction.
Now we'll write an algorithm on how to resolve a 2^nd degree equation, Then we'll translate it to a C language.

Algorithm :

2^nd degree equation has the following prototype :
a x²+bx+c=0 with a,b,c are real
Display a message on the screenthat will let us know the resolution's steps. Here is the messages :
" 2nd degree equation resolution : "
" a*x^2 + b*x + c = 0 "
Displaying messages to enter data a,b and c
" enter a " ; " enter b " ; " enter c "
delta = (b*b)-(4*a*c) real
Test delta value:
- delta >0 we have two solutions :
  - x1 = (-b+ root (delta))/(2*a) : real
  - x1 = (-b- root (delta))/(2*a) : real
- delta =0
  - we have only one solution
- delta <0
  - there is no solution in real group
  - but there is two complex solutions
    - x1= y1 +iy2
    - x2= y1 -iy2
y1,y2 real code :

#include<stdio.h>
    #include<conio.h>
    #include<math.h>
    main() 
      { 
       float a,b,c,delta,x1,x2,y1,y2;
       clrscr();
       printf(("2nd degree equation resolution:\n");
       printf("The equation is like : a x² +bx +c=0\n");
       printf("\n\n\n");

printf("Enter a :);scanf("%f",&a);
       printf("Enter b :);scanf("%f",&b);
       printf("Enter c :);scanf("%f",&c);
       
       delta=(b*b)-(4*a*c);

if (delta>0) 
           {
             x1=(-b-sqrt(delta))/(2*a);
             x2=(-b+sqrt(delta))/(2*a);
             printf("We have two real solutions\n"):
             printf("x1 = %f et x2 = %f",x1,x2);
           } 
       else if (delta==0) 
           {
             x=-b/(2*a);
             printf("We have only one solution :\n"); 
             printf("x=%f",x);
           } 
       else 
           {
             y1=-b/(2*a);
             y2=sqrt(delta)/(2*a);
             printf("There is no solution in |R\n");
             printf("But there is two complexs solutions:\n"); 
             printf("x1 =%f + i %f et x2 = %f -i %f",y1,y2,y1,y2 ); 
           }

getch();
       return 0;
     }

Before writing this code, I must explain the news:

The root is (sqrt), for example root(4) =sqrt(4)=2, sqrt is in the header " math.h ", so we must include it.
We use the bracket, because of this problem :
5+3*4= ? may be 5+3=8 then 8*4=32 or 3*4=12 then 12+5 = 17, In reality it makes 3*4 =12 then 12+5 = 17 because the multiplication has the priority than the addition, so to have the result 32 we must use brackets : (5+3)*4 = 32.
We used (= =) and not (=) because (=) is the affectation, for exampl x=-b/(2*a) mean that we affect the -b/(2*a) value to x so x and this value will be equal, but to test if x and this value are equal we'll write if (x= =-b/(2*a)), its meaning is: if x= this value, the sentence is true, if not it's false but if we write: if (x=-b/(2*a)) then we will affect this value and the sentence will be always true .

To use conditions into your code, you must have some knowledge on logic:
Lets A, B, (A and B), (A or B), (A xor B) and (not A) 6 sentences, T : true ; F :false

The (And) :
and A , B and (A et B) are three sentences

A	B	A et B
T	T	T
T	F	F
F	T	F
F	F	F

equivalence :

A	B	A eqv B
T	T	T
T	F	F
F	T	F
F	F	T

The (Not):
Lets A be a sentence proposition, T : true ; F :false

A	not A
T	F
F	T

The (Xor):

A	B	A xor B
T	T	F
T	F	T
F	T	T
F	F	F

The (Or):

A	B	A or B
T	T	T
T	F	T
F	T	T
F	F	F

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