my pi problem

poolet (6)
hello, I'm a new on c++ an i want your help... here is my problem
PI is an irrational number, i.e. it cannot be written as a fraction.
It's approximate value of PI is 3.141592653589793. Below are five
different series which can be used to approximate PI:

you can find it here::


the function ask which series he/she wants to use
and the precision (number of digits, up to 10). If after
10000 iterations the series doesn't find the value of PI, it
should display an error message.

//example::
Which series do you want?
1
How many digits of accuracy ?
4
\\example
output:: The series 1 it's approximate on 1000 iterations = 3.1415


It will display either an error message or the value of PI and
the number of iterations necessary to obtain that value.
In order to round a real number to n decimal places, you
can use the following algorithm 78375 divided by 10^3 = 78.375

here is my code:: (My problem is that I CAN"T find the way to round the pi1 right)


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void menu4() //menu4() - Find the number of iterations to approximate Pi.
{
int counter=0,flag=0,digits;
char series;

long double pi1=0;
long double result=0;
long double pi12=0;
long double ValuePi1=31;
long double ValuePi2=314;
long double ValuePi3=3141;
long double ValuePi4=31415;
long double ValuePi5=314159;
long double ValuePi6=3141592;
long double ValuePi7=31415926;
long double ValuePi8=314159265;
long double ValuePi9=3141592653;
long double ValuePi10=31415926535;

system("cls"); // clear screen 
cout <<setw(68)<< "|--------------------------------------------------|" << endl;
cout <<setw(68)<< "|                    Choice 3.                     |" << endl;
cout <<setw(68)<< "|--------------------------------------------------|" << endl;
cout <<setw(68)<< "|                                                  |" << endl;
cout <<setw(68)<< "|                                                  |" << endl;
cout <<setw(68)<< "| Find the number of iterations to approximate Pi. |" << endl;
cout <<setw(68)<< "|                                                  |" << endl;              
cout <<setw(68)<< "|                                                  |" << endl;
cout <<setw(68)<< "|--------------------------------------------------|" << endl;
cout <<setw(24)<< "|Series|"<<endl;
cout <<setw(24)<< "--------"<<endl;
cout << "1.  P= 4*(1-(1/3)+(1/5)-(1/7)+(1/9)-..."<<endl;
cout << "2.  P= sqrt(12*(1- (1/4)+(1/9)-(1/16)+(1/25)-...))"<<endl;
cout << "3.  P= sqrt(6*(1+ (1/4)+(1/9)+(1/16)+(1/25)+...)) "<<endl;// menu of the series 
cout << "4.  P= sqrt(8*(1+ (1/9)+(1/25)+(1/49)+...      )) "<<endl;
cout << "5.  P= sqrt(24*((1/4)+(1/16)+(1/36)+(1/64)+...  ))"<<endl;
cout << "--------------------------------------------------"<<endl;
cout << " Which series? (1-5): ";
cin>>series;     // ask the series 




if (series !='1' && series !='2' && series!='3' && series !='4' && series!='5') 
{  // if start
cout <<endl;
cout <<" Error Input....."<<endl;
cout <<" Their are Only 5 series...(1,2,3,4,or5)"<<endl;
cout <<" And you can add different digits number.. digits>=1 or digits <=10"<<endl;  
cout <<" The program will be terminated in 5 seconds..."<<endl;
Sleep(5000);    // stop the program in 5 seconds 
} // if end

else 

if (series=='1') 
{ // end of if SERIES 1 
cout << " How many digits of accuracy? (<10): ";
cin>>digits;  // ask the number of digits 

switch (digits)
{ //star of switch 
case 1:  result = ValuePi1/pow(10.0,digits);break;
case 2:  result = ValuePi2/pow(10.0,digits);break;
case 3:  result = ValuePi3/pow(10.0,digits);break;
case 4:  result = ValuePi4/pow(10.0,digits);break;
case 5:  result = ValuePi5/pow(10.0,digits);break;
case 6:  result = ValuePi6/pow(10.0,digits);break;
case 7:  result = ValuePi7/pow(10.0,digits);break;
case 8:  result = ValuePi8/pow(10.0,digits);break;
case 9:  result = ValuePi9/pow(10.0,digits);break;
case 10: result = ValuePi10/pow(10.0,digits);break;
} //end of switch 




cout<<endl;
cout<<"           ________        "<<endl;
cout<<"          |Series 1|       "<<endl;
cout<<" --------------------------"<<endl;
cout<<" Pi Aproximation is: "<<result<<" with "<<digits<<" digits accuracy"<<endl;
cout<<" --------------------------"<<endl;

for (int long n=1; n<=10000 && flag!=1; n++) 
{//star for loop
	
	   pi1+= (4*pow(-1.0, n+1))/(2*n - 1);
       pi12=pi1/pow(10.0, digits);
	   
	 if (pi12==result)
	 {
	    flag=1;
	 }	  
	 else
		 counter++;
		
}//loop end
        
	 if (flag==1)
	 { 
	   cout<<"Series "<<series<<" is approximate on "<<counter<<" iterations"<<" Series "<<series<<" is "<<pi1<<endl;
	   cin.get();
	 } // if end 
	 else 
		
		 cout<<"This series doesn't approximate on Value PI, Change series or digits of accuracy"<<endl;
	 cin.get();

}//end of if staytment series 1 
cin.get();
 }//end of void ()


Hammurabi (399)
Your problem is twofold. Firstly you do this:
pi12=pi1/pow(10.0, digits);
which is simply wrong. Don't do that.

Secondly, and perhaps more fundamentally, you cannot (generally) simply compare two floating point numbers for equality. Try something like this instead:

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const long REPS = 10000;
// ...
double rounder = pow( 10.0, -(digits+1) ) * 5;
long n;
for (n=1; n<=REPS; n++)
{
    pi1 += (4*pow(-1.0, n+1.0))/(2*n - 1);
    if( pi1 + rounder >= result && pi1 + rounder < result + rounder*2 )
        break;
}

if (n <= REPS)
{
    cout<<"Series "<<series<<" is approximate on "<<n
        <<" iterations"<<" Series "<<series<<" is "<<result<<endl;
    cin.get();
}
else
    cout<<"This series doesn't approximate on Value PI, "
          "Change series or digits of accuracy\n";


Additinally, as is painfully common in beginner programs, you have too much repetition:

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double ValuePi1  = 31.0;
double ValuePi2  = 314.0;
double ValuePi3  = 3141.0;
double ValuePi4  = 31415.0;
double ValuePi5  = 314159.0;
double ValuePi6  = 3141592.0;
double ValuePi7  = 31415926.0;
double ValuePi8  = 314159265.0;
double ValuePi9  = 3141592653.0;
double ValuePi10 = 31415926535.0;

// ...

switch (digits)
{ //star of switch
case 1:  result = ValuePi1/pow(10.0,digits);break;
case 2:  result = ValuePi2/pow(10.0,digits);break;
case 3:  result = ValuePi3/pow(10.0,digits);break;
case 4:  result = ValuePi4/pow(10.0,digits);break;
case 5:  result = ValuePi5/pow(10.0,digits);break;
case 6:  result = ValuePi6/pow(10.0,digits);break;
case 7:  result = ValuePi7/pow(10.0,digits);break;
case 8:  result = ValuePi8/pow(10.0,digits);break;
case 9:  result = ValuePi9/pow(10.0,digits);break;
case 10: result = ValuePi10/pow(10.0,digits);break;
} //end of switch


You can do the same thing without so much repetition:

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const double ValuePi = 31415926535.0;
// ...
modf( ValuePi / pow( 10.0, 10 - digits ), &result );
result /= pow( 10.0, digits );

Last edited on
poolet (6)
Thanks your advice my friend.........

i have another code but i don't know if it's correct... if you can check it....

here is the code:: (i post the code only for the first series, so it's the same thing on this others but with different's results....

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for (int long n=1; n<=10000 && flag!=1; n++)  //for loop for series 1 
{//star for loop series 1
	
	   pi1+= (4*pow(-1.0, n+1))/(2*n - 1);   //calculation series 1 
       pi12=ceil((pi1*pow( 10.0,digits ) )- 0.49  ) / pow( 10.0,digits ); //round the results 
	 if (pi12==result) //if check..... for the result
	 {//if check  start
	    flag=1;  // if pi12==result  flag== 1 
	 }//if check end 	  
	 else //else of the result
		 counter++; // if the program don't find the approximate series change iterations ( calculation of iteratios)
		
}//loop end series 1 
        if (flag==1) // if check = 1 
	 {  // start if 
	   cout<<" Series "<<series<<" is approximate on "<<counter<<" iterations"<<" Series "<<series<<" is "<<pi12<<endl; 
	   cin>>pi12; //cin the result to print 
	 } // if end 
	 else  // else of check
	
     cout<<" This series doesn't approximate on Value PI...."<<endl; //print error message....
	 cin>>pi;//cin the result to print 

}//end of if staytment series 1
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