by Ron Gutman
Listing One
public class BitLinear { public static long reverse (long bits) { long rl = 0; for (int i = 0; i < 64; i++) { rl = (rl << 1) + (bits & 1); bits = bits >>> 1; } return rl; } public static int count (long bits) { int cnt = 0; while (bits != 0) { cnt += bits & 1; bits = bits >>> 1; } return cnt; } } Listing Two public class BitRecursive { // reverse leftmost n bits of V static long reversen (long V, int n) { if (n <= 1) return V; int n2 = n/2; // reverse rightmost n/2 bits long right = reversen( V & ((1L<<n2)-1), n2); // reverse lefttmost n/2 bits long left = reversen( V >>> n2, n2); // combine in reverse order return (right << n2) | left; } public static long reverse (long bits) { return reversen (bits, 64); } } Listing Three public class BitLogN { public static long reverse (long bits) { // >>> fills bits on the left with 0 (no sign extension) bits = ((bits&0x00000000ffffffffL) << 32) | ((bits&0xffffffff00000000L) >>> 32); bits = ((bits&0x0000ffff0000ffffL) << 16) | ((bits&0xffff0000ffff0000L) >>> 16); bits = ((bits&0x00ff00ff00ff00ffL) << 8) | ((bits&0xff00ff00ff00ff00L) >>> 8); bits = ((bits&0x0f0f0f0f0f0f0f0fL) << 4) | ((bits&0xf0f0f0f0f0f0f0f0L) >>> 4); bits = ((bits&0x3333333333333333L) << 2) | ((bits&0xccccccccccccccccL) >>> 2); bits = ((bits&0x5555555555555555L) << 1) | ((bits&0xaaaaaaaaaaaaaaaaL) >>> 1); return bits; } public static int count (long bits) { bits = (bits&0x5555555555555555L) + ((bits&0xaaaaaaaaaaaaaaaaL) >>> 1); bits = (bits&0x3333333333333333L) + ((bits&0xccccccccccccccccL) >>> 2); bits = (bits&0x0f0f0f0f0f0f0f0fL) + ((bits&0xf0f0f0f0f0f0f0f0L) >>> 4); bits = (bits&0x00ff00ff00ff00ffL) + ((bits&0xff00ff00ff00ff00L) >>> 8); bits = (bits&0x0000ffff0000ffffL) + ((bits&0xffff0000ffff0000L) >>> 16); bits = (bits&0x00000000ffffffffL) + ((bits&0xffffffff00000000L) >>> 32); return (int) bits; } public static long mortonKey (int x, int y) { /* In C++, the calls to spreadBits could be made in-line */ /* to avoid function call overhead. */ /* In C, make the function a macro (admittedly an ugly one) */ return (spreadBits(x) << 1) | spreadBits(y); } // For j = 1 to 31, shift bit j j positions to the left static long spreadBits (int i) { long bits = i; // shift bits 16 to 31 16 bits bits = (bits & 0x000000000000ffffL) | ((bits & 0x00000000ffff0000L) << 16); // shift originally odd-numbered bytes 8 bits bits = (bits & 0x000000ff000000ffL) | ((bits & 0x0000ff000000ff00L) << 8); // shift originally odd-numbered nibbles 4 bits bits = (bits & 0x000f000f000f000fL) | ((bits & 0x00f000f000f000f0L) << 4); // shift originally odd-numbered bit pairs 2 bits bits = (bits & 0x0303030303030303L) | ((bits & 0x0c0c0c0c0c0c0c0cL) << 2); // shift originally odd-numbered bit pairs 1 bits bits = (bits & 0x1111111111111111L) | ((bits & 0x2222222222222222L) << 1); return bits; } } Listing Four public class BitTable { short[] table = new short[256]; public BitTable() { BitLinear lin = new BitLinear(); for (int i = 0; i < 256; i++) { table[i] = (short) (lin.reverse(i) >>> 56); } } public long reverse (long bits) { long rl = 0; rl = table[(int)(bits & 255)]; bits = bits >>> 8; rl = (rl << 8) | table[(int)(bits & 255)]; bits = bits >>> 8; rl = (rl << 8) | table[(int)(bits & 255)]; bits = bits >>> 8; rl = (rl << 8) | table[(int)(bits & 255)]; bits = bits >>> 8; rl = (rl << 8) | table[(int)(bits & 255)]; bits = bits >>> 8; rl = (rl << 8) | table[(int)(bits & 255)]; bits = bits >>> 8; rl = (rl << 8) | table[(int)(bits & 255)]; bits = bits >>> 8; rl = (rl << 8) | table[(int)(bits & 255)]; return rl; } }
Example 1:
if n equals 1, return V set R = right most n/2 bits of V set L = left most n/2 bits of V R = reversen(R,n/2) L = reversen(L,n/2) set RL = n bit value whose left most n/2 bits equals R and whose right most n/2 bits equals L return RL
Example 2:
if n equals 1, return V set R = right most n/2 bits of V set L = left most n/2 bits of V return countn(L,n/2) + countn(R,n/2)
See also:
file: /Techref/method/aa0900.htm, 5KB, , updated: 2006/8/11 07:47, local time: 2024/12/24 23:36,
owner: JMN-EFP-786,
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