TuckerTensorTrain.has_numerically_minimal_ranks

t3toolbox.tucker_tensor_train.TuckerTensorTrain.has_numerically_minimal_ranks(rtol=1e-09)

def has_numerically_minimal_ranks(self, rtol: float = 1e-9) -> bool:

True if the ranks are numerically minimal: no stored rank is numerically redundant.

Distinct from has_minimal_ranks, which is structural (ranks equal the structural minimum). A tensor can be structurally minimal yet have a near-zero singular value at some rank boundary (numerically redundant); this catches that. Algorithm: the cheap structural check first (structural redundancy implies numerical redundancy), then – only if structurally minimal – an t3svd() at relative tolerance rtol and a comparison of the truncated ranks to the stored ranks. The t3svd makes this O(tensor) – a diagnostic, not a hot-path check.

For an orthonormal frame prefer T3Frame.has_numerically_minimal_ranks(), which needs no SVD (orthonormal cores are full-rank, so structurally-minimal => numerically-minimal).

Examples

>>> import numpy as np
>>> import t3toolbox.tucker_tensor_train as t3
>>> np.random.seed(0)
>>> x = t3.TuckerTensorTrain.randn((6, 7, 5), (2, 2, 2), (1, 2, 2, 1))
>>> print(x.has_minimal_ranks, x.has_numerically_minimal_ranks())   # full-rank random tensor
True True
>>> xbig = x.resize((6, 7, 5), (3, 2, 2), (1, 2, 2, 1))  # tucker_0 padded -> a redundant rank
>>> print(xbig.has_minimal_ranks, xbig.has_numerically_minimal_ranks())
False False

Parameters:

rtol (float)

Return type:

bool