🔒 Overflow error

[inner prawn]

x is a perfect square
How do i remove this error

[coral vortex]

@inner prawn

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[round rampart]

it's simply impossible with that data type, try using the decimal module instead

!d decimal.Decimal.sqrt

[coral vortex]

decimal.Decimal.sqrt

sqrt(context=None)```
Return the square root of the argument to full precision.

[chilly turtle]

if you know that it's a perfect square, try math.isqrt

[inner prawn]

why does it work only for perfect squares?

[chilly turtle]

well it'd also work for non-perfect squares (isqrt is just 'integer sqrt')
you'd just get some numerical imprecision

[glacial wyvern]

Al you have to do is clarify it as a double not an int

You are getting error because int or float is too long

[novel oracle]

math.sqrt works on floats and so it'd not work on ints this big. Use math.isqrt.

[novel oracle]

That's precisely the problem, actually - this integer is too big to be represented as a float, so the conversion fails.

(the biggest non-infinite float is about 1e+308, whereas this int has something like 10k digits.)

[glacial wyvern]

Correct

[chilly turtle]

ints can pretty much grow arbitrarily, but floats have a max that you can check with

>>> import sys
>>> sys.float_info
sys.float_info(max=1.7976931348623157e+308, max_exp=1024, max_10_exp=308, min=2.2250738585072014e-308, min_exp=-1021, min_10_exp=-307, dig=15, mant_dig=53, epsilon=2.220446049250313e-16, radix=2, rounds=1)
>>> sys.float_info.max
1.7976931348623157e+308
>>>

[glacial wyvern]

You are just bottle necking by the variable type you are using

[inner prawn]

So if it's capped at that level
is there no other way around to print a req set of decimal points if x is not a square

[glacial wyvern]

I got it!!!

[novel oracle]

print a req set of decimal points if x is not a square
You'd need to use variable-precision arithmetic - so, the decimal module.

[glacial wyvern]

If math.sqrt does convert it to a value that too large to be stored you need to use a different module

[glacial wyvern]

Yess

decimal module

Correct

from decimal import Decimal, getcontext

# Set the precision for the decimal calculations
getcontext().prec = 50  # You can adjust the precision as needed

# Define a large double (float) value using Decimal
large_value = Decimal('1.2345678901234567890123456789012345678901234567890e+50')

# Calculate the square root of the large double value
square_root = large_value.sqrt()

# Print the result
print(f"The square root of {large_value} is {square_root}") ```

[novel oracle]

large double (float) value
decimals are not floating-point at all, that's kind of the point of them

[glacial wyvern]

Does this answer your question @inner prawn

[inner prawn]

Yesss

Thanks

[glacial wyvern]

No problem, happy we could help!

[coral vortex]

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