Inaccurate floats in the self-referential Tupper formula

Question

Inaccurate floats in the self-referential Tupper formula

I am trying to create a Python program that displays the Tupper self-reference formula and I am facing some problems.

At first, my program couldn't handle the big floats it had to handle, so I decided to use BigFloats

it to sort out my problems. It worked. My problem is that I have a 540 digit number that needs to be multiplied by BigFloat

, and when I do that, it rounds the number, making it inaccurate, which causes problems later. I tried raising precision

to 1000 and it still keeps rounding the variable.

The point is that some numbers can be handled without BigFloat

to the exact value, but some numbers cannot be handled normally and with BigFloat

right now.

Here is one of the calculations below (this one can be processed without BigFloat

):

import bigfloat
y = 960939379918958884971672962127852754715004339660129306651505519271702802395266424689642842174350718121267153782770623355993237280874144307891325963941337723487857735749823926629715517173716995165232890538221612403238855866184013235585136048828693337902491454229288667081096184496091705183454067827731551705405381627380967602565625016981482083418783163849115590225610003652351370343874461848378737238198224849863465033159410054974700593138339226497249461751545728366702369745461014655997933798537483143786841806593422227898388722980000748404719
x = 0
with bigfloat.precision(1000):
    (y//17 * bigfloat.pow(2,0)) % 2

This code should return 1, but it returns 0 instead.

Is there a way to make it BigFloat

more precise so that I can use it in my program?

+3

python-3.x

Alexander Simko Apr 22 15 at 17:44

source to share

1 answer

halex · Accepted Answer · 2015-04-22T18:45:33+0000

You don't need floating point math for Tupper's formula. The trick is that 2 ^{-x is the} same as 1/2 ^x so you only need to work with integers, because you don't need to calculate the floating point number 2 ^-x and multiply it by an integer, but instead calculate the integer 2 ^x and divide another integer by that integer. Applied by Tupper's formula, part 2 ^{-17 * int (x) - int (y)% 17} becomes 1/2 ^{17 * int (x) + int (y)% 17} .

See the following version of the Tupper function, which works entirely on the integer domain (if you don't know what it is **

, it's the Python Power Operator ):

def tuppers_formula(x, y):
    """Return True if point (x, y) (x and y both start at 0) is to be drawn black, False otherwise
    """
    k = 960939379918958884971672962127852754715004339660129306651505519271702802395266424689642842174350718121267153782770623355993237280874144307891325963941337723487857735749823926629715517173716995165232890538221612403238855866184013235585136048828693337902491454229288667081096184496091705183454067827731551705405381627380967602565625016981482083418783163849115590225610003652351370343874461848378737238198224849863465033159410054974700593138339226497249461751545728366702369745461014655997933798537483143786841806593422227898388722980000748404719
    return ((k + y)//17//2**(17*int(x) + int(y)%17))%2 > 0.5

You can test this function with the following code, which "draws" the result of Tupper's formula into a text file.

import codecs
import os
with codecs.open("tupper.txt", "w", "utf-8") as f:
    values = [[tuppers_formula(x, y) for x in range(106)] for y in range(17)]
    for row in values:
        for value in row[::-1]: # x = 0 starts at the left so reverse the whole row
            if value:
                f.write("\u2588") # Write a block
            else:
                f.write(" ")
        f.write(os.linesep)

The result is a file tupper.txt

with the content:

        █                   █                █ ██ █     █                █  █ █     █    █ ██ █      █   █
        █                   █ █      █       █  █ █     █                █  █ █     █    █  █ █      █   █
██      █                  █  █      █    ██ █  █ █ █ █ █ ██ ████  ███ ███ █  █ █ █ █    █  █  █      █  █
 █      █                  █  █  █ █ █       █ █  █  █  █    █ █ █ █ █ █ █ █  █ █ █ █    █ █   █      █  █
 █      █                  █  █  █ █ █       █ █  █ █ █ █    █ █ █ ███ ███ █  █  █  █    █ █   █      █  █
 █      █               █ █   █   █  █  ██        █     █                  █  █ █   █  █       █   ██  █ █
███   █ █               █ █   █  █   █ █  █       █     █                   █ █     █  █      █   █  █ █ █
     █  █ ██ █   ██   ███ █   █      █   █        ███ ███                   █ ███ ███ █       █     █  █ █
███ █   █ █ █ █ █  █ █  █ █   █ ████ █  █                                                          █   █ █
     █  █ █ █ █ █  █ █  █ █   █      █ █                                                          █    █ █
██    █ █ █ █ █  ██   ███ █   █ █ ██ █ ████                                                       ████ █ █
  █     █                 █   █ █  █ █                                                          █      █ █
 █      █                  █  █ █  █ █                                                          █     █  █
█       █                  █  █ █ █  █                                                         █      █  █
███     █                  █  █ █ █  █                                                                █  █
        █                   █ █      █                                                               █   █
        ███                 █ ███  ███                                                               █ ███

Inaccurate floats in the self-referential Tupper formula

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