Comments (2)
Python/NumPy implementation:
def _transpose(x):
n, n_, _ = x.shape
assert n == n_
assert n == 1 << n.bit_length() - 1
if n == 1:
return
assert n & 1 == 0
_transpose(x[:n>>1, :n>>1])
_transpose_and_swap(x[:n>>1, n>>1:], x[n>>1:, :n>>1])
_transpose(x[n>>1:, n>>1:])
def _transpose_and_swap(x, y):
assert x.shape == y.shape
n, n_, _ = x.shape
assert n == n_
assert n == 1 << n.bit_length() - 1
if n == 1:
x[...], y[...] = y.copy(), x.copy()
return
_transpose_and_swap(x[:n>>1, :n>>1], y[:n>>1, :n>>1])
_transpose_and_swap(x[:n>>1, n>>1:], y[n>>1:, :n>>1])
_transpose_and_swap(x[n>>1:, :n>>1], y[:n>>1, n>>1:])
_transpose_and_swap(x[n>>1:, n>>1:], y[n>>1:, n>>1:])
def bit_reverse(x):
n, = x.shape
lb_n = n.bit_length() - 1
assert n == 1 << lb_n
if lb_n <= 1:
return
half_lb_n = lb_n >> 1
leftover = lb_n & 1
x = x.reshape(1 << half_lb_n, 1 << half_lb_n + leftover)
for row in x:
bit_reverse(row)
_transpose(x.reshape(1 << half_lb_n, 1 << half_lb_n, 1 << leftover))
for row in x:
bit_reverse(row)
Obviously, an efficient implementation needs n
≈ fast cache size as the base case, with that base case calling the iterative approach.
from plonky2.
Done in #442 (with a different algorithm).
from plonky2.
Related Issues (20)
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from plonky2.