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https://github.com/thoppe/python-hyperoperators

Python library for representing really, really, ridiculously large numbers.
https://github.com/thoppe/python-hyperoperators

hyperoperation python tetration

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Python library for representing really, really, ridiculously large numbers.

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# Hyperoperators

[![Build Status](https://travis-ci.org/thoppe/python-hyperoperators.svg?branch=master)](https://travis-ci.org/thoppe/python-hyperoperators)
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[![PyPI version](https://badge.fury.io/py/hyperop.svg)](https://badge.fury.io/py/hyperop)

`hyperop` is a small library for representing really, really, ridiculously large numbers in pure python. It does so using [hyperoperations](https://en.wikipedia.org/wiki/Hyperoperation).

+ Hyperoperation 0, `H0` is the [successor function](https://en.wikipedia.org/wiki/Successor_function), `H0(None, 4) = 5`
+ `H1` is [addition](https://en.wikipedia.org/wiki/Addition), `H1(2,4) = 2 + (1+1+1+1) = 6`
+ `H2` is [multiplication](https://en.wikipedia.org/wiki/Multiplication) (repeated addition), `H2(2,4) = 2+2+2+2 = 8`
+ `H3` is [exponentiation](https://en.wikipedia.org/wiki/Exponentiation) (repeated multiplication), `H3(2,4) = 2*2*2*2 = 16`
+ `H4` is [tetration](https://en.wikipedia.org/wiki/Tetration) (repeated exponentiation) `H4(2,4) = 2^(2^(2^(2))) = 65536`
+ ...
+ Hyperoperation n is repeated Hyperoperation (n-1)

Fundamentally, hyperop works recursively by applying a [fold-right](https://en.wikipedia.org/wiki/Fold_(higher-order_function)) operation:

H[n](x,y) = reduce(lambda x,y: H[n-1](y,x), [a,]*b)

### Installation

pip install hyperop

To install the latest version use:

pip install git+https://github.com/thoppe/python-hyperoperators

### Examples

``` python
from hyperop import hyperop

H1 = hyperop(1)
print(H1(2,3), H1(3,2), H1(5,4))
# >> 5, 5, 9

H3 = hyperop(3)
print(H3(2,3), H3(3,2), H3(5,4))
# >> 8, 9, 625

from math import log
H = hyperop(4)
print(H(2,5))
>>> 200352993040684646497....45587895905719156736

print(log(log(log(log(H(2,5),2.0),2.0),2.0),2.0) == 2)
>>> True
```

Approximate infinite tetration. Show that sqrt(2)^sqrt(2)^... where the tower continues an infinite amount of times is 2.

``` python
H4 = hyperop(4)
print(H4(2**0.5, 200))
# >> 2.0
```

Calculate the incomprehensibly large, but finite [Graham's number](https://en.wikipedia.org/wiki/Graham%27s_number):

``` python
def GrahamsNumber():
# This may take awhile...
g = 4
for n in range(1,64+1):
g = hyperop(g+2)(3,3)
return g
```

Plot the phase angle on the complex plane over tetrating four times `H4(z,4)`

``` python
from hyperop import hyperop
import mpmath

H = hyperop(4)
f = lambda z: H(z,4)
mpmath.cplot(f, verbose=True, points=100000)
```

![Complex tetration plot](figures/tetration_example.png)

### Bounded hyperoperators

Sometimes, especially in the case of small complex numbers, you only care about numbers that stay bounded during the calculation.
That is, you'd only like to keep the result for some bound z such that `H[n](a,b) <= z`.
The class `bounded_hyperop` does just that:

``` python
from hyperop import bounded_hyperop
Hb = bounded_hyperop(4, bound=1000)
print(Hb(2,3), Hb(2,4))
# >> 16 inf
```

### Caveats

Higher order hyperoperations (from tetration and above) are _not_ associative, thus tetration `H4(2,4) = 2^(2^(2^(2))) = 65536` is not `H4(2,4) != 2^(2*2*2) = 256`.

Since tetration is not defined for non-integral heights, the second argument of tetration and both arguments of pentation and above are restricted to non-negative integer values.

### Talks & Press

![python weekly](figures/python_weekly.png)

Hyperop was featured in issue #231 of [Python Weekly](http://us2.campaign-archive2.com/?u=e2e180baf855ac797ef407fc7&id=39cb63fa64)!

This library was first presented at DC's Hack && Tell (Feb. 2016). [Talk link.](http://thoppe.github.io/python-hyperoperators/HnT_pres.html#/)

# License

This work is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License.