// SPDX-License-Identifier: GPL-2.0-only
/*
* Copyright 2023 Red Hat
*/
#include "radix-sort.h"
#include <linux/limits.h>
#include <linux/types.h>
#include "memory-alloc.h"
#include "string-utils.h"
/*
* This implementation allocates one large object to do the sorting, which can be reused as many
* times as desired. The amount of memory required is logarithmically proportional to the number of
* keys to be sorted.
*/
/* Piles smaller than this are handled with a simple insertion sort. */
#define INSERTION_SORT_THRESHOLD 12
/* Sort keys are pointers to immutable fixed-length arrays of bytes. */
typedef const u8 *sort_key_t;
/*
* The keys are separated into piles based on the byte in each keys at the current offset, so the
* number of keys with each byte must be counted.
*/
struct histogram {
/* The number of non-empty bins */
u16 used;
/* The index (key byte) of the first non-empty bin */
u16 first;
/* The index (key byte) of the last non-empty bin */
u16 last;
/* The number of occurrences of each specific byte */
u32 size[256];
};
/*
* Sub-tasks are manually managed on a stack, both for performance and to put a logarithmic bound
* on the stack space needed.
*/
struct task {
/* Pointer to the first key to sort. */
sort_key_t *first_key;
/* Pointer to the last key to sort. */
sort_key_t *last_key;
/* The offset into the key at which to continue sorting. */
u16 offset;
/* The number of bytes remaining in the sort keys. */
u16 length;
};
struct radix_sorter {
unsigned int count;
struct histogram bins;
sort_key_t *pile[256];
struct task *end_of_stack;
struct task insertion_list[256];
struct task stack[];
};
/* Compare a segment of two fixed-length keys starting at an offset. */
static inline int compare(sort_key_t key1, sort_key_t key2, u16 offset, u16 length)
{
return memcmp(&key1[offset], &key2[offset], length);
}
/* Insert the next unsorted key into an array of sorted keys. */
static inline void insert_key(const struct task task, sort_key_t *next)
{
/* Pull the unsorted key out, freeing up the array slot. */
sort_key_t unsorted = *next;
/* Compare the key to the preceding sorted entries, shifting down ones that are larger. */
while ((--next >= task.first_key) &&
(compare(unsorted, next[0], task.offset, task.length) < 0))
next[1] = next[0];
/* Insert the key into the last slot that was cleared, sorting it. */
next[1] = unsorted;
}
/*
* Sort a range of key segments using an insertion sort. This simple sort is faster than the
* 256-way radix sort when the number of keys to sort is small.
*/
static inline void insertion_sort(const struct task task)
{
sort_key_t *next;
for (next = task.first_key + 1; next <= task.last_key; next++)
insert_key(task, next);
}
/* Push a sorting task onto a task stack. */
static inline void push_task(struct task **stack_pointer, sort_key_t *first_key,
u32 count, u16 offset, u16 length)
{
struct task *task = (*stack_pointer)++;
task->first_key = first_key;
task->last_key = &first_key[count - 1];
task->offset = offset;
task->length = length;
}
static inline void swap_keys(sort_key_t *a, sort_key_t *b)
{
sort_key_t c = *a;
*a = *b;
*b = c;
}
/*
* Count the number of times each byte value appears in the arrays of keys to sort at the current
* offset, keeping track of the number of non-empty bins, and the index of the first and last
* non-empty bin.
*/
static inline void measure_bins(const struct task task, struct histogram *bins)
{
sort_key_t *key_ptr;
/*
* Subtle invariant: bins->used and bins->size[] are zero because the sorting code clears
* it all out as it goes. Even though this structure is re-used, we don't need to pay to
* zero it before starting a new tally.
*/
bins->first = U8_MAX;
bins->last = 0;
for (key_ptr = task.first_key; key_ptr <= task.last_key; key_ptr++) {
/* Increment the count for the byte in the key at the current offset. */
u8 bin = (*key_ptr)[task.offset];
u32 size = ++bins->size[bin];
/* Track non-empty bins. */
if (size == 1) {
bins->used += 1;
if (bin < bins->first)
bins->first = bin;
if (bin > bins->last)
bins->last = bin;
}
}
}
/*
* Convert the bin sizes to pointers to where each pile goes.
*
* pile[0] = first_key + bin->size[0],
* pile[1] = pile[0] + bin->size[1], etc.
*
* After the keys are moved to the appropriate pile, we'll need to sort each of the piles by the
* next radix position. A new task is put on the stack for each pile containing lots of keys, or a
* new task is put on the list for each pile containing few keys.
*
* @stack: pointer the top of the stack
* @end_of_stack: the end of the stack
* @list: pointer the head of the list
* @pile: array for pointers to the end of each pile
* @bins: the histogram of the sizes of each pile
* @first_key: the first key of the stack
* @offset: the next radix position to sort by
* @length: the number of bytes remaining in the sort keys
*
* Return: UDS_SUCCESS or an error code
*/
static inline int push_bins(struct task **stack, struct task *end_of_stack,
struct task **list, sort_key_t *pile[],
struct histogram *bins, sort_key_t *first_key,
u16 offset, u16 length)
{
sort_key_t *pile_start = first_key;
int bin;
for (bin = bins->first; ; bin++) {
u32 size = bins->size[bin];
/* Skip empty piles. */
if (size == 0)
continue;
/* There's no need to sort empty keys. */
if (length > 0) {
if (size > INSERTION_SORT_THRESHOLD) {
if (*stack >= end_of_stack)
return UDS_BAD_STATE;
push_task(stack, pile_start, size, offset, length);
} else if (size > 1) {
push_task(list, pile_start, size, offset, length);
}
}
pile_start += size;
pile[bin] = pile_start;
if (--bins->used == 0)
break;
}
return UDS_SUCCESS;
}
int uds_make_radix_sorter(unsigned int count, struct radix_sorter **sorter)
{
int result;
unsigned int stack_size = count / INSERTION_SORT_THRESHOLD;
struct radix_sorter *radix_sorter;
result = vdo_allocate_extended(struct radix_sorter, stack_size, struct task,
__func__, &radix_sorter);
if (result != VDO_SUCCESS)
return result;
radix_sorter->count = count;
radix_sorter->end_of_stack = radix_sorter->stack + stack_size;
*sorter = radix_sorter;
return UDS_SUCCESS;
}
void uds_free_radix_sorter(struct radix_sorter *sorter)
{
vdo_free(sorter);
}
/*
* Sort pointers to fixed-length keys (arrays of bytes) using a radix sort. The sort implementation
* is unstable, so the relative ordering of equal keys is not preserved.
*/
int uds_radix_sort(struct radix_sorter *sorter, const unsigned char *keys[],
unsigned int count, unsigned short length)
{
struct task start;
struct histogram *bins = &sorter->bins;
sort_key_t **pile = sorter->pile;
struct task *task_stack = sorter->stack;
/* All zero-length keys are identical and therefore already sorted. */
if ((count == 0) || (length == 0))
return UDS_SUCCESS;
/* The initial task is to sort the entire length of all the keys. */
start = (struct task) {
.first_key = keys,
.last_key = &keys[count - 1],
.offset = 0,
.length = length,
};
if (count <= INSERTION_SORT_THRESHOLD) {
insertion_sort(start);
return UDS_SUCCESS;
}
if (count > sorter->count)
return UDS_INVALID_ARGUMENT;
/*
* Repeatedly consume a sorting task from the stack and process it, pushing new sub-tasks
* onto the stack for each radix-sorted pile. When all tasks and sub-tasks have been
* processed, the stack will be empty and all the keys in the starting task will be fully
* sorted.
*/
for (*task_stack = start; task_stack >= sorter->stack; task_stack--) {
const struct task task = *task_stack;
struct task *insertion_task_list;
int result;
sort_key_t *fence;
sort_key_t *end;
measure_bins(task, bins);
/*
* Now that we know how large each bin is, generate pointers for each of the piles
* and push a new task to sort each pile by the next radix byte.
*/
insertion_task_list = sorter->insertion_list;
result = push_bins(&task_stack, sorter->end_of_stack,
&insertion_task_list, pile, bins, task.first_key,
task.offset + 1, task.length - 1);
if (result != UDS_SUCCESS) {
memset(bins, 0, sizeof(*bins));
return result;
}
/* Now bins->used is zero again. */
/*
* Don't bother processing the last pile: when piles 0..N-1 are all in place, then
* pile N must also be in place.
*/
end = task.last_key - bins->size[bins->last];
bins->size[bins->last] = 0;
for (fence = task.first_key; fence <= end; ) {
u8 bin;
sort_key_t key = *fence;
/*
* The radix byte of the key tells us which pile it belongs in. Swap it for
* an unprocessed item just below that pile, and repeat.
*/
while (--pile[bin = key[task.offset]] > fence)
swap_keys(pile[bin], &key);
/*
* The pile reached the fence. Put the key at the bottom of that pile,
* completing it, and advance the fence to the next pile.
*/
*fence = key;
fence += bins->size[bin];
bins->size[bin] = 0;
}
/* Now bins->size[] is all zero again. */
/*
* When the number of keys in a task gets small enough, it is faster to use an
* insertion sort than to keep subdividing into tiny piles.
*/
while (--insertion_task_list >= sorter->insertion_list)
insertion_sort(*insertion_task_list);
}
return UDS_SUCCESS;
}
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