precision in python upto 50 decimal places

Question

I am trying to code to find out the radius of the inner soddy circles, the code is running fine but i am not getting the required precision, i want the answer to be truncated to 50 digits Please suggest how can i get more precise calculations How should i obtain the precision of 50 digits

import math
t=input()
while t>0:
    t-=1
    r1,r2,r3=map(int,raw_input().split())
    k1=1.0/r1
    k2=1.0/r2
    k3=1.0/r3
    k4=k1+k2+k3+2*math.sqrt(k1*k2+k2*k3+k3*k1)
    r4=1.0/k4
    print r4

http://docs.python.org/2/library/decimal.html – ndpu Dec 01 '13 at 16:28 — ndpu, Dec 01 '13 at 16:28

aIKid · Answer 1 · 2013-12-01T17:19:12.973

3

Use the decimal module. Store all the variables you're using as decimal.Decimal objects.

Updated code:

from decimal import *
import math
context = Context(prec=1000)
setcontext(context)
t=input()
while t>0:
    t-=1
    r1,r2,r3=map(Decimal,raw_input().split())
    k1=Decimal(1.0)/Decimal(r1)
    k2=Decimal(1.0)/Decimal(r2)
    k3=Decimal(1.0)/Decimal(r3)
    k4=k1+k2+k3+2*(k1*k2+k2*k3+k3*k1).sqrt()
    r4=Decimal(1.0)/Decimal(k4)
    print r4

edited Dec 01 '13 at 17:19

answered Dec 01 '13 at 16:29

aIKid

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why is this giving a runtime error http://ideone.com/NzwcbT – user103260 Dec 01 '13 at 16:48
Almost right. I'll update your code in a minute. – aIKid Dec 01 '13 at 16:51
Ah, the only thing wrong was that you need to add `import decimal` at the beginning of your function, or remove `decimal.` in line 3 and 4. It should work now. – aIKid Dec 01 '13 at 16:53
can u please add the updated code here, thanks – user103260 Dec 01 '13 at 16:55
It gives runtime error http://ideone.com/NzwcbT – user103260 Dec 01 '13 at 16:58
Instead of using `math.sqrt` (which would toss away all the extra precision anyhow), use `(k1*k2+k2*k3+k3*k1).sqrt()`. – DSM Dec 01 '13 at 17:12
@user103260 What? It looks fine even in ideone... Can't you see the stdout? – aIKid Dec 01 '13 at 17:21
its showing 5.999999... though it should show 6.00000 why so and what can be done to correct it, plus i want total of 50 digits after decimal, there can be as many digits to the left of the decimal, what shud be done in that case, thanks – user103260 Dec 01 '13 at 17:26
What do you expect? You specified `prec=1000`.. set `prec=50` if you only want 50 decimal places.. – aIKid Dec 01 '13 at 17:41
you don't need to wrap every single argument in `Decimal()`. `Decimal()` is infectious; [it would spread by itself](http://stackoverflow.com/a/20315255/4279) – jfs Dec 01 '13 at 17:50

score 2 · Answer 2 · edited May 23 '17 at 12:05

If more iterations should yield better results then you could ignore input t parameter and iterate until result converges within current precision:

import decimal

#NOTE: separate user input from the algorithm
#      no input inside `radius()` function
def radius(r1, r2, r3):
    with decimal.localcontext() as ctx:
        ctx.prec += 2 # increase precision for intermediate calculations
        prev = None # previous value
        k1, k2, k3 = [1 / decimal.Decimal(r) for r in [r1, r2, r3]]
        while True: 
            # change some parameters to simulate converging algorithm
            #NOTE: you don't need to wrap anything using `Decimal()`
            k1 = k1 + k2 + k3 + 2*(k1*k2 + k2*k3 + k3*k1).sqrt()
            r = 1 / k1
            if prev == r: # compare using enhanced precision
                break
            prev = r # save previous value
    return +r # `+` applies current precision

decimal.getcontext().prec = 50 # set desired precision
print(radius(*map(int, raw_input().split())))

For a working example that demonstrates this technique, see Gauss-Legendre Algorithm in python that calculate Pi upto 100 digits.

precision in python upto 50 decimal places

2 Answers2