STL: map with string as key - access speedup

The idea is quite simple: we do not have a single stl::map<string, something>, but vector of maps, indexed with O(1) time - each map store keys sharing certain properties. Drawback: additional memory.

I've tested following grouping schemes:

1. length of string,
2. first letter of string (one level trie),
3. both length and first letter.

Third is the fastest - around 60% faster then plain std::map from gcc (red-black tree).

Tests: my program read plain text (I've used The Illiad from http://gutenberg.org), text is splitted into words (~190000) and then each words is inserted into dictonary (~28000 distinct words); then the same words are searched in dictionaries. Table below summarizes results on my computer (gcc 4.3.4 from cygwin).

+-------------+--------------------------------+--------------------------------+
|             |          running time [ms]     |            speedup [%]         |
| data struct +----------+----------+----------+----------+----------+----------+
|             |    min   |   avg    |   max    |   min    |   avg    |   max    |
+-------------+----------+----------+----------+----------+----------+----------+
|                                          inserting                            |
+-------------+----------+----------+----------+----------+----------+----------+
|  std::map   |    269   |   287    |   355    |   100    |   100    |   100    |
+-------------+----------+----------+----------+----------+----------+----------+
| first char  |    218   |   241    |   395    |    81    |    84    |   111    |
+-------------+----------+----------+----------+----------+----------+----------+
|   length    |    218   |   240    |   345    |    81    |    84    |    97    |
+-------------+----------+----------+----------+----------+----------+----------+
| len./char   |    165   |   172    |   207    |    61    |    60    |    58    |
+-------------+----------+----------+----------+----------+----------+----------+
|                                            finding                            |
+-------------+----------+----------+----------+----------+----------+----------+
|  std::map   |    295   |   322    |   483    |   100    |   100    |   100    |
+-------------+----------+----------+----------+----------+----------+----------+
| first char  |    243   |   263    |   460    |    82    |    82    |    95    |
+-------------+----------+----------+----------+----------+----------+----------+
|   length    |    238   |   248    |   292    |    80    |    77    |    60    |
+-------------+----------+----------+----------+----------+----------+----------+
| len./char   |    184   |   190    |   241    |    62    |    60    |    50    |
+-------------+----------+----------+----------+----------+----------+----------+

Download test program.

Branchless signum

Problem: calculate value of sign(x):

• -1 when x < 0
• 0 when x = 0,
• +1 when x > 0.

My solution do not involve any hardware specific things like ALU flags nor special instructions - just plain AND, OR, shifts.

 ; input: eax = X

 movl %eax, %ebx
 sarl $31, %eax  // eax = -1 if X less then zero, 0 otherwise

 andl $0x7fffffff, %ebx
 addl $0x7fffffff, %ebx // MSB is set if any lower bits were set
 shrl $31, $ebx  // eax = +1 if X greater then zero, 0 otherwise

 orl %ebx, %eax  // eax = result

C99 implementation:

int32_t sign(int32_t x) {
 int32_t y;
 y = (x & 0x7fffffff) + 0x7fffffff;
 return (x >> 31) | ((uint32_t)y >> 31);
}

Fill word with selected bit

This is continuation of subproblem from previous post: we have a word (byte, dword, whatever) and want to fill it with selected bit.

1. The most general algorithm:

1. mask bit
   [10111010] => [00010000]
   
2. clone word
   a=[00010000], b=[00010000]
3. shift bit in first word to MSB, and to LSB in second word
   a=[10000000], b=[00000001]
4. subtract c = a - b
   c=[01111111]
5. add missing MSB c = c OR a
   c=[11111111]

2. If arithmetic shifts are supported:

1. shift bit to MSB
   a=[10000000]
2. arithmetic shift right
   a=[11111111]

3. On processor 386+ we can copy selected bit from reg to carry flag (CF) with bt reg, reg and then clone CF with sbb reg, reg.

#include <stdlib.h>
#include <stdio.h>
#include <stdint.h>

uint32_t fill1(uint32_t a, int bit) {
uint32_t b;

b = a = a & (1 << bit);

a <<= 31 - bit;
b >>= bit;

return (a - b) | a;
}

uint32_t fill2(uint32_t a, int bit) {
a <<= 31 - bit;
return (int32_t)(a) >> 31;
}

uint32_t fill386(uint32_t a, int bit) {
uint32_t result;             
__asm__ __volatile__ (
"bt  %1, %0\n"
"sbb %0, %0\n"
: "=r" (result)
: "r" (bit), "0" (a)
);
return result;
}

int main(int argc, char* argv[]) {
uint32_t x, i;
if (argc > 1)
x = (unsigned long)strtol(argv[1], NULL, 0);

printf("input = %08x\n", x);
for (i=0; i < 32; i++)
printf("bit %2d: fill1 = %08x, fill2 = %08x, fill386 = %08x\n",
i,
fill1(x, i),
fill2(x, i),
fill386(x, i)
);

return EXIT_SUCCESS;
}

Transpose bits in byte using SIMD instructions

Method presented here allows to get any bit permutation, transposition is just one of possible operations. Lookup-based approach would be faster, but algorithm is worth to (re)show.

Algorithm outline for 8-byte vector (with SSE instruction it is possible to get 2 operations in parallel):

1. fill vector with given byte
   [11010001] =>
   [11010001|11010001|11010001|11010001|11010001|11010001|11010001|11010001]
   
2. leave one bit per byte
   [10000000|01000000|00000000|00010000|00000000|00000000|00000000|00000001]
   
3. perform desired transposition ("move" bits around)
   [00000001|000000010|00000000|00001000|00000000|00000000|00000000|10000000]
   
4. perform horizontal OR of all bytes
   [10001011]
   

Here is my old MMX code (polish text); below SSE/SSE5 implementation details.

Ad 1. Series of punpcklbw/punpcklwb/shufps or pshufb if CPU supports SSSE3.

# 1.1
movd      %eax, %xmm0
shufps    $0x00, %xmm0, %xmm0
punpcklbw %xmm0, %xmm0
punpcklwd %xmm0, %xmm0

# 1.2
pxor      %xmm1, %xmm1
movd       %eax, %xmm0
pshufb    %xmm1, %xmm0

Ad 2. Simple pand with mask packed_qword(0x8040201008040201).

pand  MASK1, %xmm0

Ad 3. If plain SSE instructions are supported this step require some work. First each bit is populated to fill whole byte (using pcmpeq - we get negated result), then mask bits on desired positons.

SSE5 has powerful instruction protb that can do perform rotation of each byte with independent amount - so in this case just one instruction is needed.

# SSE
pcmpeq  %xmm1, %xmm0
pandn   MASK2, %xmm0

# SSE5
protb    ROT, %xmm0, %xmm0

Ad 4. Since bits are placed on distinct positions, we can use instruction psadbw, that calculate horizontal sums of bytewide differences from two registers (separately for low and high registers half). If one register is full of zeros, we get sum of bytes from other register.

psadbw  %xmm1, %xmm0
movd    %xmm0, %eax

Depending on instruction set three (SSE) or two (SSE5) additional tables are needed.

Opis biblioteki pthreads na Wikibooks

W ramach głębszego poznawania biblioteki pthreads utworzyłem w serwisie wikibooks opis POSIX Threads, zawierający bardzo podstawowe informacje o większości funkcji tej biblioteki oraz nieco o rozszerzeniach Linuxowej implementacji. Integralną częścią są przykładowe programy.

(Wikibooks niestety nie zyskał takiej popularności, jak Wikipedia, a to całkiem wygodna platforma do tworzenia różnego rodzaju podręczników).

PostgreSQL: get selected rows with given order

Suppose that database stores some kind of dictionary and user picks some items, but wants to keep order. For example dictionary has entries with id=0..10, and user picked 9, 2, 4 and 0. This simple query does the job (query splitted):

foo = SELECT (ARRAY[9,2,4,0])[i] AS index, i AS ord FROM  generate_series(1, 4) AS i;
SELECT * FROM dictionary INNER JOIN (foo) ON dictionary.id=foo.index ORDER BY foo.ord

Słowa mistrza

Wczoraj późnym wieczorem czytałem jakiś wywiad z mistrzem Andrzejem Poniedzielskim, z którego wynotowałem następujące zdanie:

Dziś każdy pisze sobie muzykę i teksty, zapowiada, wykonuje, a telewizja sprawia wrażenie, że jeszcze ktoś oprócz niego samego się tym bawi.

Przed chwilą natomiast przypadkiem oglądnąłem Kabaretowy Klub Dwójki prowadzony przez liderów dwóch najpopularniejszych formacji... i właśnie tak mi się ten cytat z Poniedzielskiego skojarzył.