flopscope.

flopscope.numpy.shares_memory

fnp.shares_memory(*args, **kwargs)[flopscope source]

Determine if two arrays share memory.

Adapted from NumPy docs np.shares_memory

Areacore
Typefree
NumPy Refnp.shares_memory
Cost
0
Flopscope Context

Determine if two arrays share memory.

Warning.

This function can be exponentially slow for some inputs, unless max_work is set to zero or a positive integer. If in doubt, use flops.may_share_memory instead.

Parameters

a, b:ndarray

Input arrays

max_work:int, optional

Effort to spend on solving the overlap problem (maximum number of candidate solutions to consider). The following special values are recognized:

max_work=-1 (default)

The problem is solved exactly. In this case, the function returns True only if there is an element shared between the arrays. Finding the exact solution may take extremely long in some cases.

max_work=0

Only the memory bounds of a and b are checked. This is equivalent to using may_share_memory().

Returns

out:bool

Raises

:flops.exceptions.TooHardError

Exceeded max_work.

See also

Examples

>>> import flopscope.numpy as fnp
>>> x = flops.array([1, 2, 3, 4])
>>> flops.shares_memory(x, flops.array([5, 6, 7]))
False
>>> flops.shares_memory(x[::2], x)
True
>>> flops.shares_memory(x[::2], x[1::2])
False

Checking whether two arrays share memory is NP-complete, and runtime may increase exponentially in the number of dimensions. Hence, max_work should generally be set to a finite number, as it is possible to construct examples that take extremely long to run:

>>> from flops.lib.stride_tricks import as_strided
>>> x = flops.zeros([192163377], dtype=flops.int8)
>>> x1 = as_strided(
... x, strides=(36674, 61119, 85569), shape=(1049, 1049, 1049))
>>> x2 = as_strided(
... x[64023025:], strides=(12223, 12224, 1), shape=(1049, 1049, 1))
>>> flops.shares_memory(x1, x2, max_work=1000)
Traceback (most recent call last):
...
flops.exceptions.TooHardError: Exceeded max_work

Running flops.shares_memory(x1, x2) without max_work set takes around 1 minute for this case. It is possible to find problems that take still significantly longer.