green_t 提問：
實現如下轉換的最佳演算法是什麼？
0010 0000 => 0000 0100
具體的轉換是從MSB->LSB 到 LSB->MSB，所有的位都必須反轉，那意味著，這並不是位元組順序的交換。
LSB(Least Significant Bit)，意為最低有效位；MSB(Most Significant Bit)，意為最高有效位。
最佳答案（來自Matt J）
注意：下面的演算法都用C實現，但應該可以遷移到其它語言(只是不那麼快的時候可別找我)。
可選方案
記憶體佔用少(32位int，32位機器)(來源於這裡)
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unsigned int reverse(register unsigned int x) { x = (((x & 0xaaaaaaaa) >> 1) | ((x & 0x55555555) << 1)); x = (((x & 0xcccccccc) >> 2) | ((x & 0x33333333) << 2)); x = (((x & 0xf0f0f0f0) >> 4) | ((x & 0x0f0f0f0f) << 4)); x = (((x & 0xff00ff00) >> 8) | ((x & 0x00ff00ff) << 8)); return((x >> 16) | (x << 16)); } |
最快(查詢表)
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static const unsigned char BitReverseTable256[] = { 0x00, 0x80, 0x40, 0xC0, 0x20, 0xA0, 0x60, 0xE0, 0x10, 0x90, 0x50, 0xD0, 0x30, 0xB0, 0x70, 0xF0, 0x08, 0x88, 0x48, 0xC8, 0x28, 0xA8, 0x68, 0xE8, 0x18, 0x98, 0x58, 0xD8, 0x38, 0xB8, 0x78, 0xF8, 0x04, 0x84, 0x44, 0xC4, 0x24, 0xA4, 0x64, 0xE4, 0x14, 0x94, 0x54, 0xD4, 0x34, 0xB4, 0x74, 0xF4, 0x0C, 0x8C, 0x4C, 0xCC, 0x2C, 0xAC, 0x6C, 0xEC, 0x1C, 0x9C, 0x5C, 0xDC, 0x3C, 0xBC, 0x7C, 0xFC, 0x02, 0x82, 0x42, 0xC2, 0x22, 0xA2, 0x62, 0xE2, 0x12, 0x92, 0x52, 0xD2, 0x32, 0xB2, 0x72, 0xF2, 0x0A, 0x8A, 0x4A, 0xCA, 0x2A, 0xAA, 0x6A, 0xEA, 0x1A, 0x9A, 0x5A, 0xDA, 0x3A, 0xBA, 0x7A, 0xFA, 0x06, 0x86, 0x46, 0xC6, 0x26, 0xA6, 0x66, 0xE6, 0x16, 0x96, 0x56, 0xD6, 0x36, 0xB6, 0x76, 0xF6, 0x0E, 0x8E, 0x4E, 0xCE, 0x2E, 0xAE, 0x6E, 0xEE, 0x1E, 0x9E, 0x5E, 0xDE, 0x3E, 0xBE, 0x7E, 0xFE, 0x01, 0x81, 0x41, 0xC1, 0x21, 0xA1, 0x61, 0xE1, 0x11, 0x91, 0x51, 0xD1, 0x31, 0xB1, 0x71, 0xF1, 0x09, 0x89, 0x49, 0xC9, 0x29, 0xA9, 0x69, 0xE9, 0x19, 0x99, 0x59, 0xD9, 0x39, 0xB9, 0x79, 0xF9, 0x05, 0x85, 0x45, 0xC5, 0x25, 0xA5, 0x65, 0xE5, 0x15, 0x95, 0x55, 0xD5, 0x35, 0xB5, 0x75, 0xF5, 0x0D, 0x8D, 0x4D, 0xCD, 0x2D, 0xAD, 0x6D, 0xED, 0x1D, 0x9D, 0x5D, 0xDD, 0x3D, 0xBD, 0x7D, 0xFD, 0x03, 0x83, 0x43, 0xC3, 0x23, 0xA3, 0x63, 0xE3, 0x13, 0x93, 0x53, 0xD3, 0x33, 0xB3, 0x73, 0xF3, 0x0B, 0x8B, 0x4B, 0xCB, 0x2B, 0xAB, 0x6B, 0xEB, 0x1B, 0x9B, 0x5B, 0xDB, 0x3B, 0xBB, 0x7B, 0xFB, 0x07, 0x87, 0x47, 0xC7, 0x27, 0xA7, 0x67, 0xE7, 0x17, 0x97, 0x57, 0xD7, 0x37, 0xB7, 0x77, 0xF7, 0x0F, 0x8F, 0x4F, 0xCF, 0x2F, 0xAF, 0x6F, 0xEF, 0x1F, 0x9F, 0x5F, 0xDF, 0x3F, 0xBF, 0x7F, 0xFF }; unsigned int v; // reverse 32-bit value, 8 bits at time unsigned int c; // c will get v reversed // Option 1: c = (BitReverseTable256[v & 0xff] << 24) | (BitReverseTable256[(v >> 8) & 0xff] << 16) | (BitReverseTable256[(v >> 16) & 0xff] << 8) | (BitReverseTable256[(v >> 24) & 0xff]); // Option 2: unsigned char * p = (unsigned char *) &v; unsigned char * q = (unsigned char *) &c; q[3] = BitReverseTable256[p[0]]; q[2] = BitReverseTable256[p[1]]; q[1] = BitReverseTable256[p[2]]; q[0] = BitReverseTable256[p[3]]; |
來自於著名的Bit Twiddling Hacks page:
你可以擴充套件這個演算法到64位int的場景，或者為了更快的速度而犧牲多一些的記憶體(假設你的L1資料快取足夠大)，有一個64K的查詢表且每次反轉16位。
其它方案
簡單
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unsigned int v; // input bits to be reversed unsigned int r = v; // r will be reversed bits of v; first get LSB of v int s = sizeof(v) * CHAR_BIT - 1; // extra shift needed at end for (v >>= 1; v; v >>= 1) { r <<= 1; r |= v & 1; s--; } r <<= s; // shift when v's highest bits are zero |
更快(32位處理器)
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unsigned char b = x; b = ((b * 0x0802LU & 0x22110LU) | (b * 0x8020LU & 0x88440LU)) * 0x10101LU >> 16; |
更快(64位處理器)
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unsigned char b; // reverse this (8-bit) byte b = (b * 0x0202020202ULL & 0x010884422010ULL) % 1023; |
如果你想在32位int環境這樣做，那麼只需要把每一個byte反轉，然後再顛倒byte的次序即可。如下:
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unsigned int toReverse; unsigned int reversed; unsigned char inByte0 = (toReverse & 0xFF); unsigned char inByte1 = (toReverse & 0xFF00) >> 8; unsigned char inByte2 = (toReverse & 0xFF0000) >> 16; unsigned char inByte3 = (toReverse & 0xFF000000) >> 24; reversed = (reverseBits(inByte0) << 24) | (reverseBits(inByte1) << 16) | (reverseBits(inByte2) << 8) | (reverseBits(inByte3); |
結果
我測試了兩個最有效的方案，查詢表和按位與(第一個方案)。測試機器為一臺膝上型電腦，配置為4G DDR2記憶體，2.4GHz的
雙核T7500處理器，4MB的L2快取。我使用的是gcc 4.3.2，64位Linux。OpenMP(外加GCC繫結)被用來提高計時器的排程能力。
reverse.c
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#include <stdlib.h> #include <stdio.h> #include <omp.h> unsigned int reverse(register unsigned int x) { x = (((x & 0xaaaaaaaa) >> 1) | ((x & 0x55555555) << 1)); x = (((x & 0xcccccccc) >> 2) | ((x & 0x33333333) << 2)); x = (((x & 0xf0f0f0f0) >> 4) | ((x & 0x0f0f0f0f) << 4)); x = (((x & 0xff00ff00) >> 8) | ((x & 0x00ff00ff) << 8)); return((x >> 16) | (x << 16)); } int main() { unsigned int *ints = malloc(100000000*sizeof(unsigned int)); unsigned int *ints2 = malloc(100000000*sizeof(unsigned int)); for(unsigned int i = 0; i < 100000000; i++) ints[i] = rand(); unsigned int *inptr = ints; unsigned int *outptr = ints2; unsigned int *endptr = ints + 100000000; // Starting the time measurement double start = omp_get_wtime(); // Computations to be measured while(inptr != endptr) { (*outptr) = reverse(*inptr); inptr++; outptr++; } // Measuring the elapsed time double end = omp_get_wtime(); // Time calculation (in seconds) printf("Time: %f seconds\n", end-start); free(ints); free(ints2); return 0; } |
reverse_lookup.c
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#include <stdlib.h> #include <stdio.h> #include <omp.h> static const unsigned char BitReverseTable256[] = { 0x00, 0x80, 0x40, 0xC0, 0x20, 0xA0, 0x60, 0xE0, 0x10, 0x90, 0x50, 0xD0, 0x30, 0xB0, 0x70, 0xF0, 0x08, 0x88, 0x48, 0xC8, 0x28, 0xA8, 0x68, 0xE8, 0x18, 0x98, 0x58, 0xD8, 0x38, 0xB8, 0x78, 0xF8, 0x04, 0x84, 0x44, 0xC4, 0x24, 0xA4, 0x64, 0xE4, 0x14, 0x94, 0x54, 0xD4, 0x34, 0xB4, 0x74, 0xF4, 0x0C, 0x8C, 0x4C, 0xCC, 0x2C, 0xAC, 0x6C, 0xEC, 0x1C, 0x9C, 0x5C, 0xDC, 0x3C, 0xBC, 0x7C, 0xFC, 0x02, 0x82, 0x42, 0xC2, 0x22, 0xA2, 0x62, 0xE2, 0x12, 0x92, 0x52, 0xD2, 0x32, 0xB2, 0x72, 0xF2, 0x0A, 0x8A, 0x4A, 0xCA, 0x2A, 0xAA, 0x6A, 0xEA, 0x1A, 0x9A, 0x5A, 0xDA, 0x3A, 0xBA, 0x7A, 0xFA, 0x06, 0x86, 0x46, 0xC6, 0x26, 0xA6, 0x66, 0xE6, 0x16, 0x96, 0x56, 0xD6, 0x36, 0xB6, 0x76, 0xF6, 0x0E, 0x8E, 0x4E, 0xCE, 0x2E, 0xAE, 0x6E, 0xEE, 0x1E, 0x9E, 0x5E, 0xDE, 0x3E, 0xBE, 0x7E, 0xFE, 0x01, 0x81, 0x41, 0xC1, 0x21, 0xA1, 0x61, 0xE1, 0x11, 0x91, 0x51, 0xD1, 0x31, 0xB1, 0x71, 0xF1, 0x09, 0x89, 0x49, 0xC9, 0x29, 0xA9, 0x69, 0xE9, 0x19, 0x99, 0x59, 0xD9, 0x39, 0xB9, 0x79, 0xF9, 0x05, 0x85, 0x45, 0xC5, 0x25, 0xA5, 0x65, 0xE5, 0x15, 0x95, 0x55, 0xD5, 0x35, 0xB5, 0x75, 0xF5, 0x0D, 0x8D, 0x4D, 0xCD, 0x2D, 0xAD, 0x6D, 0xED, 0x1D, 0x9D, 0x5D, 0xDD, 0x3D, 0xBD, 0x7D, 0xFD, 0x03, 0x83, 0x43, 0xC3, 0x23, 0xA3, 0x63, 0xE3, 0x13, 0x93, 0x53, 0xD3, 0x33, 0xB3, 0x73, 0xF3, 0x0B, 0x8B, 0x4B, 0xCB, 0x2B, 0xAB, 0x6B, 0xEB, 0x1B, 0x9B, 0x5B, 0xDB, 0x3B, 0xBB, 0x7B, 0xFB, 0x07, 0x87, 0x47, 0xC7, 0x27, 0xA7, 0x67, 0xE7, 0x17, 0x97, 0x57, 0xD7, 0x37, 0xB7, 0x77, 0xF7, 0x0F, 0x8F, 0x4F, 0xCF, 0x2F, 0xAF, 0x6F, 0xEF, 0x1F, 0x9F, 0x5F, 0xDF, 0x3F, 0xBF, 0x7F, 0xFF }; int main() { unsigned int *ints = malloc(100000000*sizeof(unsigned int)); unsigned int *ints2 = malloc(100000000*sizeof(unsigned int)); for(unsigned int i = 0; i < 100000000; i++) ints[i] = rand(); unsigned int *inptr = ints; unsigned int *outptr = ints2; unsigned int *endptr = ints + 100000000; // Starting the time measurement double start = omp_get_wtime(); // Computations to be measured while(inptr != endptr) { unsigned int in = *inptr; // Option 1: //*outptr = (BitReverseTable256[in & 0xff] << 24) | // (BitReverseTable256[(in >> 8) & 0xff] << 16) | // (BitReverseTable256[(in >> 16) & 0xff] << 8) | // (BitReverseTable256[(in >> 24) & 0xff]); // Option 2: unsigned char * p = (unsigned char *) &(*inptr); unsigned char * q = (unsigned char *) &(*outptr); q[3] = BitReverseTable256[p[0]]; q[2] = BitReverseTable256[p[1]]; q[1] = BitReverseTable256[p[2]]; q[0] = BitReverseTable256[p[3]]; inptr++; outptr++; } // Measuring the elapsed time double end = omp_get_wtime(); // Time calculation (in seconds) printf("Time: %f seconds\n", end-start); free(ints); free(ints2); return 0; } |
在不同的優化級別(Optimizations)，兩個方案我都嘗試了，每個級別跑3個案例，每個案例反轉
1億個隨機的無符號整數。對於查詢表方案，bitwise hacks page上面的兩種方法(Option 1 and Option 2)我都測試過。
結果如下:
按位與
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mrj10@mjlap:~/code$ gcc -fopenmp -std=c99 -o reverse reverse.c mrj10@mjlap:~/code$ ./reverse Time: 2.000593 seconds mrj10@mjlap:~/code$ ./reverse Time: 1.938893 seconds mrj10@mjlap:~/code$ ./reverse Time: 1.936365 seconds mrj10@mjlap:~/code$ gcc -fopenmp -std=c99 -O2 -o reverse reverse.c mrj10@mjlap:~/code$ ./reverse Time: 0.942709 seconds mrj10@mjlap:~/code$ ./reverse Time: 0.991104 seconds mrj10@mjlap:~/code$ ./reverse Time: 0.947203 seconds mrj10@mjlap:~/code$ gcc -fopenmp -std=c99 -O3 -o reverse reverse.c mrj10@mjlap:~/code$ ./reverse Time: 0.922639 seconds mrj10@mjlap:~/code$ ./reverse Time: 0.892372 seconds mrj10@mjlap:~/code$ ./reverse Time: 0.891688 seconds |
查詢表(Option 1)
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mrj10@mjlap:~/code$ gcc -fopenmp -std=c99 -o reverse_lookup reverse_lookup.c mrj10@mjlap:~/code$ ./reverse_lookup Time: 1.201127 seconds mrj10@mjlap:~/code$ ./reverse_lookup Time: 1.196129 seconds mrj10@mjlap:~/code$ ./reverse_lookup Time: 1.235972 seconds mrj10@mjlap:~/code$ gcc -fopenmp -std=c99 -O2 -o reverse_lookup reverse_lookup.c mrj10@mjlap:~/code$ ./reverse_lookup Time: 0.633042 seconds mrj10@mjlap:~/code$ ./reverse_lookup Time: 0.655880 seconds mrj10@mjlap:~/code$ ./reverse_lookup Time: 0.633390 seconds mrj10@mjlap:~/code$ gcc -fopenmp -std=c99 -O3 -o reverse_lookup reverse_lookup.c mrj10@mjlap:~/code$ ./reverse_lookup Time: 0.652322 seconds mrj10@mjlap:~/code$ ./reverse_lookup Time: 0.631739 seconds mrj10@mjlap:~/code$ ./reverse_lookup Time: 0.652431 seconds |
查詢表(Option 2)
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mrj10@mjlap:~/code$ gcc -fopenmp -std=c99 -o reverse_lookup reverse_lookup.c mrj10@mjlap:~/code$ ./reverse_lookup Time: 1.671537 seconds mrj10@mjlap:~/code$ ./reverse_lookup Time: 1.688173 seconds mrj10@mjlap:~/code$ ./reverse_lookup Time: 1.664662 seconds mrj10@mjlap:~/code$ gcc -fopenmp -std=c99 -O2 -o reverse_lookup reverse_lookup.c mrj10@mjlap:~/code$ ./reverse_lookup Time: 1.049851 seconds mrj10@mjlap:~/code$ ./reverse_lookup Time: 1.048403 seconds mrj10@mjlap:~/code$ ./reverse_lookup Time: 1.085086 seconds mrj10@mjlap:~/code$ gcc -fopenmp -std=c99 -O3 -o reverse_lookup reverse_lookup.c mrj10@mjlap:~/code$ ./reverse_lookup Time: 1.082223 seconds mrj10@mjlap:~/code$ ./reverse_lookup Time: 1.053431 seconds mrj10@mjlap:~/code$ ./reverse_lookup Time: 1.081224 seconds |
結論
如果你比較在意效能，那麼使用查詢表Option 1(Byte的定址不出意外的慢)。如果你需要儘可能的利用完每一個Byte記憶體
(且你也在意bit反轉的效能)，那麼優化後的按位與方案也還不賴。
附加說明
我知道上面的程式碼只是一個粗略的版本，非常歡迎大家提供一些優化的建議。以下是我知道的幾點:
- 我沒有許可權訪問ICC，那可能更快些(如果你可以測試請在評論中回覆)。
- 在一些L1快取比較大的現代機器上面，64K的查詢表可能工作得更好。
-mtune=native對 -O2/-O3(發生符號重定義的錯誤)無效，所以我不相信產生的程式碼是為我的微架構而優化。- SSE環境下應該有一種方法處理得更快。我不知道怎麼做，但又更快的記憶體複製，批量的按位與，調整的指令集，
總是有一些手段的。 - 我知道僅僅x86的指令集是危險的，下面是GCC在-O3環境產生的程式碼，所以比我更厲害的大牛可以檢查一下。
32-bit
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.L3: movl (%r12,%rsi), %ecx movzbl %cl, %eax movzbl BitReverseTable256(%rax), %edx movl %ecx, %eax shrl $24, %eax mov %eax, %eax movzbl BitReverseTable256(%rax), %eax sall $24, %edx orl %eax, %edx movzbl %ch, %eax shrl $16, %ecx movzbl BitReverseTable256(%rax), %eax movzbl %cl, %ecx sall $16, %eax orl %eax, %edx movzbl BitReverseTable256(%rcx), %eax sall $8, %eax orl %eax, %edx movl %edx, (%r13,%rsi) addq $4, %rsi cmpq $400000000, %rsi jne .L3 |
更改: 我也嘗試在自己機器上使用uint64，看看是否效能有所提高。相對於32-bit效能大概提高了10%。
無論你是每次用64-bit型別去反轉2個32-bit的int，或者實際上看作64-bit並分兩次來反轉，效能都大致相當。
程式碼如下(對於前者，每次反轉2個32-bit的int):
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.L3: movq (%r12,%rsi), %rdx movq %rdx, %rax shrq $24, %rax andl $255, %eax movzbl BitReverseTable256(%rax), %ecx movzbq %dl,%rax movzbl BitReverseTable256(%rax), %eax salq $24, %rax orq %rax, %rcx movq %rdx, %rax shrq $56, %rax movzbl BitReverseTable256(%rax), %eax salq $32, %rax orq %rax, %rcx movzbl %dh, %eax shrq $16, %rdx movzbl BitReverseTable256(%rax), %eax salq $16, %rax orq %rax, %rcx movzbq %dl,%rax shrq $16, %rdx movzbl BitReverseTable256(%rax), %eax salq $8, %rax orq %rax, %rcx movzbq %dl,%rax shrq $8, %rdx movzbl BitReverseTable256(%rax), %eax salq $56, %rax orq %rax, %rcx movzbq %dl,%rax shrq $8, %rdx movzbl BitReverseTable256(%rax), %eax andl $255, %edx salq $48, %rax orq %rax, %rcx movzbl BitReverseTable256(%rdx), %eax salq $40, %rax orq %rax, %rcx movq %rcx, (%r13,%rsi) addq $8, %rsi cmpq $400000000, %rsi jne .L3 |