blob: 1bc32e5f6d9dbf5c82ed1fae02863399da3c5f44 [file] [log] [blame]
/*
* Copyright © 2021 Intel Corporation
*
* Permission is hereby granted, free of charge, to any person obtaining a
* copy of this software and associated documentation files (the "Software"),
* to deal in the Software without restriction, including without limitation
* the rights to use, copy, modify, merge, publish, distribute, sublicense,
* and/or sell copies of the Software, and to permit persons to whom the
* Software is furnished to do so, subject to the following conditions:
*
* The above copyright notice and this permission notice (including the next
* paragraph) shall be included in all copies or substantial portions of the
* Software.
*
* THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
* IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
* FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL
* THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
* LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING
* FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS
* IN THE SOFTWARE.
*/
#ifndef INTEL_PIXEL_HASH_H
#define INTEL_PIXEL_HASH_H
/**
* Compute an \p n x \p m pixel hashing table usable as slice, subslice or
* pixel pipe hashing table. The resulting table is the cyclic repetition of
* a fixed pattern with periodicity equal to \p period.
*
* If \p index is specified to be equal to \p period, a 2-way hashing table
* will be generated such that indices 0 and 1 are returned for the following
* fractions of entries respectively:
*
* p_0 = ceil(period / 2) / period
* p_1 = floor(period / 2) / period
*
* If \p index is even and less than \p period, a 3-way hashing table will be
* generated such that indices 0, 1 and 2 are returned for the following
* fractions of entries:
*
* p_0 = (ceil(period / 2) - 1) / period
* p_1 = floor(period / 2) / period
* p_2 = 1 / period
*
* The equations above apply if \p flip is equal to 0, if it is equal to 1 p_0
* and p_1 will be swapped for the result. Note that in the context of pixel
* pipe hashing this can be always 0 on Gfx12 platforms, since the hardware
* transparently remaps logical indices found on the table to physical pixel
* pipe indices from the highest to lowest EU count.
*/
UNUSED static void
intel_compute_pixel_hash_table_3way(unsigned n, unsigned m,
unsigned period, unsigned index, bool flip,
uint32_t *p)
{
for (unsigned i = 0; i < n; i++) {
for (unsigned j = 0; j < m; j++) {
const unsigned k = (i + j) % period;
p[j + m * i] = (k == index ? 2 : (k & 1) ^ flip);
}
}
}
/**
* Compute an \p n x \p m pixel hashing table usable as slice,
* subslice or pixel pipe hashing table. This generalizes the
* previous 3-way hash table function to an arbitrary number of ways
* given by the number of bits set in the expression "mask1 | mask2".
* If a way is only set in one of the two mask arguments it will
* appear on the table with half the frequency as a way set on both
* masks.
*/
UNUSED static void
intel_compute_pixel_hash_table_nway(unsigned n, unsigned m,
uint32_t mask1, uint32_t mask2,
uint32_t *p)
{
/* If both masks are equal all ways are expected to show up with
* the same frequency on the final table, so we can zero out one of
* the masks in order to halve the number of IDs we need to handle.
*/
if (mask1 == mask2)
mask2 = 0;
/* Construct a table mapping consecutive indices to the physical
* indices given by the bits set on the mask arguments. Ways
* enabled on both masks will appear twice on the mapping, so
* they'll show up with twice the frequency on the final table.
*/
unsigned phys_ids[(sizeof(mask1) + sizeof(mask2)) * CHAR_BIT];
unsigned num_ids = 0;
for (unsigned i = 0; i < sizeof(mask1) * CHAR_BIT; i++) {
if (mask1 & (1u << i))
phys_ids[num_ids++] = i;
if (mask2 & (1u << i))
phys_ids[num_ids++] = i;
}
assert(num_ids > 0);
/* Compute a permutation of the above indices that assigns indices
* as far as possible to adjacent entries. This permutation is
* designed to be equivalent to the bit reversal of each index in
* cases where num_ids is a power of two, but doesn't actually
* require it to be a power of two in order to satisfy the required
* properties (which is necessary to handle configurations with
* arbitrary non-power of two fusing). By construction, flipping
* bit l of its input will lead to a change in its result of the
* order of num_ids/2^(l+1) (see variable t below). The
* bijectivity of this permutation can be verified easily by
* induction. This permutation is applied cyclically to the
* vertical indices of the hashing table constructed below.
*/
const unsigned bits = util_logbase2_ceil(num_ids);
unsigned swzy[ARRAY_SIZE(phys_ids)];
for (unsigned k = 0; k < num_ids; k++) {
unsigned t = num_ids;
unsigned s = 0;
for (unsigned l = 0; l < bits; l++) {
if (k & (1u << l)) {
s += (t + 1) >> 1;
t >>= 1;
} else {
t = (t + 1) >> 1;
}
}
swzy[k] = s;
}
/* Compute a second permutation applied cyclically to the
* horizontal indices of the hashing table. In cases where a
* single mask is present (which means that all ways are expected
* to have the same frequency) this permutation will be the
* identity and will have no effect.
*
* In cases where some ways have twice the frequency of the others,
* use a similar iterative halving of the range of the permutation
* as in the the swzy[] permutation defined above, but instead of
* scanning the bits of its argument (the "k" variable above) in
* the opposite order (from LSB to MSB), proceed by halving the
* domain of the permutation in the same order as its range, which
* would lead to an identity permutation if it wasn't because the
* LSB of its range is adjusted as early as possible instead of at
* the last iteration.
*
* The reason for the special casing of the LSB is that we want to
* avoid assigning adjacent IDs to adjacent elements of the table,
* since ways that appear duplicated in the phys_ids mapping above
* would then appear duplicated in adjacent positions of the final
* table, which would lead to poor utilization for small primitives
* that only cover a small contiguous portion of the hashing table
* and would have twice as much work as necessary submitted to the
* same way instead of spreading its processing over a larger
* number of ways.
*/
unsigned swzx[ARRAY_SIZE(phys_ids)];
if (mask1 && mask2) {
for (unsigned k = 0; k < num_ids; k++) {
unsigned l = k;
unsigned t = num_ids;
unsigned s = 0;
bool in_range = false;
while (t > 1) {
const bool first_in_range = t <= m && !in_range;
in_range |= first_in_range;
if (l >= (t + 1) >> 1) {
/* Apply the s++ increment (which could only occur in
* the last t == 2 iteration if we were constructing an
* identity permutation) as soon as the domain of the
* permutation has been decomposed into a chunk smaller
* than the width of the hashing table \p m (which
* causes in_range to be first set to true), since
* doing it earlier would prevent any alternation
* between even and odd indices in the first \p m
* elements of swzx[], which are the only ones actually
* used.
*
* Subsequent (in_range == true) increments of s need
* to be doubled since they are selecting between
* indices of the same parity.
*/
if (!in_range)
s += (t + 1) >> 1;
else if (first_in_range)
s++;
else
s += (t + 1) >> 1 << 1;
l -= (t + 1) >> 1;
t >>= 1;
} else {
t = (t + 1) >> 1;
}
}
swzx[k] = s;
}
} else {
for (unsigned k = 0; k < num_ids; k++)
swzx[k] = k;
}
/* Initialize the table with the cyclic repetition of a
* num_ids-periodic pattern.
*
* Note that the horizontal and vertical permutations (swzx and
* swzy respectively) are different, and the former is either an
* identity permutation or close to the identity. This asymmetry
* is intentional in order to minimize the size of the contiguous
* area that needs to be rendered in parallel in order to utilize
* the whole GPU: In cases where swzx is the identity a rendering
* rectangle of width W will need to be at least H blocks high,
* where H is bounded by 2^ceil(log2(num_ids/W)) thanks to the
* above definition of the swzy permutation.
*/
for (unsigned i = 0; i < n; i++) {
const unsigned k = i % num_ids;
for (unsigned j = 0; j < m; j++) {
const unsigned l = j % num_ids;
p[j + m * i] = phys_ids[(swzx[l] + swzy[k]) % num_ids];
}
}
}
#endif