crotonite:
DIS LAND IS 4 SALE NOT.



here is my incomplete solution.

larisoft:
@OP... your questions are real brain heaters. I love them! But would you please provide the solutions after each of them has been unsolved for hours?. Many of us attempt them for hours and it will be nice to finally know the solution after trying so hard!

from decimal import *
import math

x = 0.7893

beginH = Decimal(1.0 )
beginB = Decimal(1.0 )

multi = 3
adder = 2

r = Decimal(0)
for i in range( 2, 5001 ) :
	r = r + Decimal( ( beginH/beginB ) * Decimal( x ** i ) )
	
	beginH = multi * beginH + adder
	beginB = multi * beginB
	multi +=1
	adder +=1
print(r)

crotonite:
DIS LAND IS 4 SALE NOT.



here is my incomplete solution.

Slyr0x:


I sure will.

Thanks

proxy20:

I have an answer with proxy20 infinite decimal equations . May aim was to solve it partially .

jacob05:

Hmmm. Interesting.

kosytedy:


@crotonite, your code contains some errors, I can't start typing all of them. But the fact() function is wrong as it should contain a loop.

Factorial of 4 (4!) is = 4 x 3 x 2 x 1 = 24

Programmatically, the loop will have to start right at 5000 (value passed into the function) and decrease by one with each decrement multiplied by the already gotten result till it stops looping, then the final answer is given. Please review your codes again. Nice try, keep it up.

Lyphiard:
Please do not answer this problem. The OP has taken this question from a competition and is attempting to cheat.

Slyr0x:


Calm your tits.

I solved the challenges before posting here, and the answer for this is 5.08xxxxxxxxxxxxxxxxxx.

I will release the write up later .

Cheers Kiddo,

R.

mojeed4:



Lol. I guess you have not heard of a recursion method. Try running the fact() function and you will see that it will run perfectly well. Although using a for loop will also do the job, but I believe the recursion method does the job finer and cleaner. (My opinion though).

Slyr0x:


Calm your tits.

I solved the challenges before posting here, and the answer for this is 5.08xxxxxxxxxxxxxxxxxx.

I will release the write up later .

Cheers Kiddo,

R.

olyjosh:
Please guys, do submit your computation time and memory usage for these solutions. Anybody could have solve it, but let's see how effective is your solution in term of space and time.
You can log ur codes here https://ideone.com/ or on your machine(please do state the machine processor and memory configuration).
Thanks

jacob05:


https://ideone.com/DSusUY

bcscale(25);
$itotal = "0";
$xt = "0.7893";$ii = 1;
$yyy = "1";while ($ii <= 5000)
{
    $y2 = intval(gmp_add($yyy,"1");
    $y3 = strval($y2);
    
    $factit = gmp_fact($y2);
    
    $at = bcdiv(gmp_mul("2", gmp_sub($factit,"1"), $factit);
    $xtpow = bcpow($xt,$y2);
    
    $anst = bcmul($at,$xtpow);
    
    $itotal = bcadd($itotal,$anst);
    $yyy = gmp_add($yyy,"1";
    $ii++;
    
}

echo "Solution: " .$itotal."<br />";

Slyr0x:
Great job Jacob05. Your solution is more efficient.

Solution below as solved by someone on my team

bcscale(25);
$itotal = "0";
$xt = "0.7893";$ii = 1;
$yyy = "1";while ($ii <= 5000)
{
    $y2 = intval(gmp_add($yyy,"1");
    $y3 = strval($y2);
    
    $factit = gmp_fact($y2);
    
    $at = bcdiv(gmp_mul("2", gmp_sub($factit,"1"), $factit);
    $xtpow = bcpow($xt,$y2);
    
    $anst = bcmul($at,$xtpow);
    
    $itotal = bcadd($itotal,$anst);
    $yyy = gmp_add($yyy,"1";
    $ii++;
    
}

echo "Solution: " .$itotal."<br />";

Slyr0x:
Great job Jacob05. Your solution is more efficient.

Solution below as solved by someone on my team

bcscale(25);
$itotal = "0";
$xt = "0.7893";$ii = 1;
$yyy = "1";while ($ii <= 5000)
{
    $y2 = intval(gmp_add($yyy,"1");
    $y3 = strval($y2);
    
    $factit = gmp_fact($y2);
    
    $at = bcdiv(gmp_mul("2", gmp_sub($factit,"1"), $factit);
    $xtpow = bcpow($xt,$y2);
    
    $anst = bcmul($at,$xtpow);
    
    $itotal = bcadd($itotal,$anst);
    $yyy = gmp_add($yyy,"1";
    $ii++;
    
}

echo "Solution: " .$itotal."<br />";

Slyr0x:
^^I will check that in a bit.

Updated


Lol, Jacob05 , you've got the wrong answer. I ran your code, saw the first 5.088459******* and then assumed you got it right.

import java.math.BigDecimal;
import java.math.BigInteger;
import java.math.RoundingMode;

/**
 *
 * @author Olyjosh
 */
public class SimpleFactorial {

    private static BigDecimal sum = new BigDecimal("0";
    private static final BigDecimal x=new BigDecimal(0.7893);
    private static final BigDecimal TWO=new BigDecimal(2);
    private static BigDecimal exp = x;
    private static BigInteger bi=BigInteger.ONE;

    
    private static BigDecimal factBig(int n){
        BigDecimal bd;
        BigDecimal div;
        for (int i = 2; i <= n; i++) {
            bi=bi.multiply(new BigInteger(""+i));
            //BigInteger factBig = factBig();
            bd = new BigDecimal(bi);
            div = bd.subtract(BigDecimal.ONE)
                    .divide(bd, 21, RoundingMode.HALF_UP);
            exp=exp.multiply(x);
            sum=sum.add(div.multiply(exp).multiply(TWO));
        }
        return sum.setScale(20, RoundingMode.HALF_UP);
    }
    
    
    public static void main(String[] args) {
        System.out.println(factBig(5000));
    }
}

jacob05:
@Slyr0x
An Attempt

 from decimal import *
 import math

 x = 0.7893

 beginH = Decimal(1.0 )
 beginB = Decimal(1.0 )

 multi = 3
 adder = 2

 r = Decimal(0)
 for i in range( 2, 5001 ) :
 	r = r + Decimal( ( beginH/beginB ) * Decimal( x ** i ) )
	
 	beginH = multi * beginH + adder
 	beginB = multi * beginB
 	multi +=1
 	adder +=1
 print(r)
 

. 
 from math import factorial
 from decimal import Decimal

 n = 5000
 x = Decimal('0.7893')
 y = Decimal('0')

 for i in range(1, n+1):
     y = y + 2 * (x**i) * (factorial(i)-1) / factorial(i)
 print y
 

dozies1:

here is my solution:

. 
 from math import factorial
 from decimal import Decimal

 n = 5000
 x = Decimal('0.7893')
 y = Decimal('0')

 for i in range(1, n+1):
     y = y + 2 * (x**i) * (factorial(i)-1) / factorial(i)
 print y
 

Please, would you mind telling me why mine is much more slower than yours.

jacob05:

Because you're finding and summing the factorial of numbers from 1 to 5000 : which is quite computation intensive.

dozies1:
Ok.
What is the principle behind making it go faster.

jacob05:
Don't use factorial.. Simple