我正在寻找确定一个long值是否是完美平方的最快方法(即它的平方根是另一个整数):
long
Math.sqrt()
这是我现在正在做的非常简单直接的方法:
public final static boolean isPerfectSquare(long n) { if (n < 0) return false; long tst = (long)(Math.sqrt(n) + 0.5); return tst*tst == n; }
注意:我在很多Project Euler问题中都使用了这个函数。因此,没有其他人将不得不维护此代码。这种微优化实际上可以产生影响,因为部分挑战是在不到一分钟的时间内完成每个算法,并且在某些问题中需要调用数百万次这个函数。
我尝试了不同的解决方案来解决这个问题:
0.5
or
switch
or``switch
if(lookup[(int)(n&0x3F)]) { test } else return false;
我想出了一个比你的 6bits+Carmack+sqrt 代码快 35% 的方法,至少在我的 CPU (x86) 和编程语言 (C/C++) 上是这样。您的结果可能会有所不同,尤其是因为我不知道 Java 因素将如何发挥作用。
我的方法有三个:
int64 x
。)
java if( x < 0 || (x&2) || ((x & 7) == 5) || ((x & 11) == 8) ) return false; if( x == 0 ) return true;
java int64 y = x; y = (y & 4294967295LL) + (y >> 32); y = (y & 65535) + (y >> 16); y = (y & 255) + ((y >> 8) & 255) + (y >> 16); // At this point, y is between 0 and 511. More code can reduce it farther.
为了实际检查余数是否为正方形,我在预先计算的表格中查找答案。
java if( bad255[y] ) return false; // However, I just use a table of size 512
Hensel 引理
的方法计算平方根。(我不认为它直接适用,但它可以通过一些修改来工作。)在此之前,我用二分搜索除以 2 的所有幂:
java if((x & 4294967295LL) == 0) x >>= 32; if((x & 65535) == 0) x >>= 16; if((x & 255) == 0) x >>= 8; if((x & 15) == 0) x >>= 4; if((x & 3) == 0) x >>= 2;
此时,要使我们的数字成为正方形,它必须是 1 mod 8。
java if((x & 7) != 1) return false;
亨塞尔引理的基本结构如下。(注意:未经测试的代码;如果不起作用,请尝试 t=2 或 8。)
java int64 t = 4, r = 1; t <<= 1; r += ((x - r * r) & t) >> 1; t <<= 1; r += ((x - r * r) & t) >> 1; t <<= 1; r += ((x - r * r) & t) >> 1; // Repeat until t is 2^33 or so. Use a loop if you want.
这个想法是,在每次迭代中,您将一位添加到 r 上,即 x 的“当前”平方根;每个平方根都精确模数越来越大的 2 次方,即 t/2。最后,r 和 t/2-r 将是 x 模 t/2 的平方根。(请注意,如果 r 是 x 的平方根,那么 -r 也是如此。偶数模数也是如此,但请注意,对某些数模数,事物的平方根甚至可能超过 2 个;值得注意的是,这包括 2 的幂。 ) 因为我们的实际平方根小于 2^32,此时我们实际上可以检查 r 或 t/2-r 是否是真正的平方根。在我的实际代码中,我使用了以下修改后的循环:
java int64 r, t, z; r = start[(x >> 3) & 1023]; do { z = x - r * r; if( z == 0 ) return true; if( z < 0 ) return false; t = z & (-z); r += (z & t) >> 1; if( r > (t >> 1) ) r = t - r; } while( t <= (1LL << 33) );
这里的加速是通过三种方式获得的:预先计算的起始值(相当于循环的约 10 次迭代)、循环的提前退出和跳过一些 t 值。对于最后一部分,我看一下
z = r - x * x
,并将 t 设置为 2 除 z 的最大幂,有点技巧。这使我可以跳过不会影响 r 值的 t 个值。在我的例子中,预先计算的起始值选择了“最小正”平方根模 8192。
即使这段代码对你来说运行得更快,我希望你喜欢它包含的一些想法。完整的、经过测试的代码如下,包括预先计算的表。
typedef signed long long int int64; int start[1024] = {1,3,1769,5,1937,1741,7,1451,479,157,9,91,945,659,1817,11, 1983,707,1321,1211,1071,13,1479,405,415,1501,1609,741,15,339,1703,203, 129,1411,873,1669,17,1715,1145,1835,351,1251,887,1573,975,19,1127,395, 1855,1981,425,453,1105,653,327,21,287,93,713,1691,1935,301,551,587, 257,1277,23,763,1903,1075,1799,1877,223,1437,1783,859,1201,621,25,779, 1727,573,471,1979,815,1293,825,363,159,1315,183,27,241,941,601,971, 385,131,919,901,273,435,647,1493,95,29,1417,805,719,1261,1177,1163, 1599,835,1367,315,1361,1933,1977,747,31,1373,1079,1637,1679,1581,1753,1355, 513,1539,1815,1531,1647,205,505,1109,33,1379,521,1627,1457,1901,1767,1547, 1471,1853,1833,1349,559,1523,967,1131,97,35,1975,795,497,1875,1191,1739, 641,1149,1385,133,529,845,1657,725,161,1309,375,37,463,1555,615,1931, 1343,445,937,1083,1617,883,185,1515,225,1443,1225,869,1423,1235,39,1973, 769,259,489,1797,1391,1485,1287,341,289,99,1271,1701,1713,915,537,1781, 1215,963,41,581,303,243,1337,1899,353,1245,329,1563,753,595,1113,1589, 897,1667,407,635,785,1971,135,43,417,1507,1929,731,207,275,1689,1397, 1087,1725,855,1851,1873,397,1607,1813,481,163,567,101,1167,45,1831,1205, 1025,1021,1303,1029,1135,1331,1017,427,545,1181,1033,933,1969,365,1255,1013, 959,317,1751,187,47,1037,455,1429,609,1571,1463,1765,1009,685,679,821, 1153,387,1897,1403,1041,691,1927,811,673,227,137,1499,49,1005,103,629, 831,1091,1449,1477,1967,1677,697,1045,737,1117,1737,667,911,1325,473,437, 1281,1795,1001,261,879,51,775,1195,801,1635,759,165,1871,1645,1049,245, 703,1597,553,955,209,1779,1849,661,865,291,841,997,1265,1965,1625,53, 1409,893,105,1925,1297,589,377,1579,929,1053,1655,1829,305,1811,1895,139, 575,189,343,709,1711,1139,1095,277,993,1699,55,1435,655,1491,1319,331, 1537,515,791,507,623,1229,1529,1963,1057,355,1545,603,1615,1171,743,523, 447,1219,1239,1723,465,499,57,107,1121,989,951,229,1521,851,167,715, 1665,1923,1687,1157,1553,1869,1415,1749,1185,1763,649,1061,561,531,409,907, 319,1469,1961,59,1455,141,1209,491,1249,419,1847,1893,399,211,985,1099, 1793,765,1513,1275,367,1587,263,1365,1313,925,247,1371,1359,109,1561,1291, 191,61,1065,1605,721,781,1735,875,1377,1827,1353,539,1777,429,1959,1483, 1921,643,617,389,1809,947,889,981,1441,483,1143,293,817,749,1383,1675, 63,1347,169,827,1199,1421,583,1259,1505,861,457,1125,143,1069,807,1867, 2047,2045,279,2043,111,307,2041,597,1569,1891,2039,1957,1103,1389,231,2037, 65,1341,727,837,977,2035,569,1643,1633,547,439,1307,2033,1709,345,1845, 1919,637,1175,379,2031,333,903,213,1697,797,1161,475,1073,2029,921,1653, 193,67,1623,1595,943,1395,1721,2027,1761,1955,1335,357,113,1747,1497,1461, 1791,771,2025,1285,145,973,249,171,1825,611,265,1189,847,1427,2023,1269, 321,1475,1577,69,1233,755,1223,1685,1889,733,1865,2021,1807,1107,1447,1077, 1663,1917,1129,1147,1775,1613,1401,555,1953,2019,631,1243,1329,787,871,885, 449,1213,681,1733,687,115,71,1301,2017,675,969,411,369,467,295,693, 1535,509,233,517,401,1843,1543,939,2015,669,1527,421,591,147,281,501, 577,195,215,699,1489,525,1081,917,1951,2013,73,1253,1551,173,857,309, 1407,899,663,1915,1519,1203,391,1323,1887,739,1673,2011,1585,493,1433,117, 705,1603,1111,965,431,1165,1863,533,1823,605,823,1179,625,813,2009,75, 1279,1789,1559,251,657,563,761,1707,1759,1949,777,347,335,1133,1511,267, 833,1085,2007,1467,1745,1805,711,149,1695,803,1719,485,1295,1453,935,459, 1151,381,1641,1413,1263,77,1913,2005,1631,541,119,1317,1841,1773,359,651, 961,323,1193,197,175,1651,441,235,1567,1885,1481,1947,881,2003,217,843, 1023,1027,745,1019,913,717,1031,1621,1503,867,1015,1115,79,1683,793,1035, 1089,1731,297,1861,2001,1011,1593,619,1439,477,585,283,1039,1363,1369,1227, 895,1661,151,645,1007,1357,121,1237,1375,1821,1911,549,1999,1043,1945,1419, 1217,957,599,571,81,371,1351,1003,1311,931,311,1381,1137,723,1575,1611, 767,253,1047,1787,1169,1997,1273,853,1247,413,1289,1883,177,403,999,1803, 1345,451,1495,1093,1839,269,199,1387,1183,1757,1207,1051,783,83,423,1995, 639,1155,1943,123,751,1459,1671,469,1119,995,393,219,1743,237,153,1909, 1473,1859,1705,1339,337,909,953,1771,1055,349,1993,613,1393,557,729,1717, 511,1533,1257,1541,1425,819,519,85,991,1693,503,1445,433,877,1305,1525, 1601,829,809,325,1583,1549,1991,1941,927,1059,1097,1819,527,1197,1881,1333, 383,125,361,891,495,179,633,299,863,285,1399,987,1487,1517,1639,1141, 1729,579,87,1989,593,1907,839,1557,799,1629,201,155,1649,1837,1063,949, 255,1283,535,773,1681,461,1785,683,735,1123,1801,677,689,1939,487,757, 1857,1987,983,443,1327,1267,313,1173,671,221,695,1509,271,1619,89,565, 127,1405,1431,1659,239,1101,1159,1067,607,1565,905,1755,1231,1299,665,373, 1985,701,1879,1221,849,627,1465,789,543,1187,1591,923,1905,979,1241,181}; bool bad255[512] = {0,0,1,1,0,1,1,1,1,0,1,1,1,1,1,0,0,1,1,0,1,0,1,1,1,0,1,1,1,1,0,1, 1,1,0,1,0,1,1,1,1,1,1,1,1,1,1,1,1,0,1,0,1,1,1,0,1,1,1,1,0,1,1,1, 0,1,0,1,1,0,0,1,1,1,1,1,0,1,1,1,1,0,1,1,0,0,1,1,1,1,1,1,1,1,0,1, 1,1,1,1,0,1,1,1,1,1,0,1,1,1,1,0,1,1,1,0,1,1,1,1,0,0,1,1,1,1,1,1, 1,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,0,0,1,1,1,1,1,0,1,1,0,1,1,1,1,1, 1,1,1,1,1,1,0,1,1,0,1,0,1,1,0,1,1,1,1,1,1,1,1,1,1,1,0,1,1,0,1,1, 1,1,1,0,0,1,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,1,1,1, 1,0,1,1,1,0,1,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,1,1,1, 0,0,1,1,0,1,1,1,1,0,1,1,1,1,1,0,0,1,1,0,1,0,1,1,1,0,1,1,1,1,0,1, 1,1,0,1,0,1,1,1,1,1,1,1,1,1,1,1,1,0,1,0,1,1,1,0,1,1,1,1,0,1,1,1, 0,1,0,1,1,0,0,1,1,1,1,1,0,1,1,1,1,0,1,1,0,0,1,1,1,1,1,1,1,1,0,1, 1,1,1,1,0,1,1,1,1,1,0,1,1,1,1,0,1,1,1,0,1,1,1,1,0,0,1,1,1,1,1,1, 1,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,0,0,1,1,1,1,1,0,1,1,0,1,1,1,1,1, 1,1,1,1,1,1,0,1,1,0,1,0,1,1,0,1,1,1,1,1,1,1,1,1,1,1,0,1,1,0,1,1, 1,1,1,0,0,1,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,1,1,1, 1,0,1,1,1,0,1,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,1,1,1, 0,0}; inline bool square( int64 x ) { // Quickfail if( x < 0 || (x&2) || ((x & 7) == 5) || ((x & 11) == 8) ) return false; if( x == 0 ) return true; // Check mod 255 = 3 * 5 * 17, for fun int64 y = x; y = (y & 4294967295LL) + (y >> 32); y = (y & 65535) + (y >> 16); y = (y & 255) + ((y >> 8) & 255) + (y >> 16); if( bad255[y] ) return false; // Divide out powers of 4 using binary search if((x & 4294967295LL) == 0) x >>= 32; if((x & 65535) == 0) x >>= 16; if((x & 255) == 0) x >>= 8; if((x & 15) == 0) x >>= 4; if((x & 3) == 0) x >>= 2; if((x & 7) != 1) return false; // Compute sqrt using something like Hensel's lemma int64 r, t, z; r = start[(x >> 3) & 1023]; do { z = x - r * r; if( z == 0 ) return true; if( z < 0 ) return false; t = z & (-z); r += (z & t) >> 1; if( r > (t >> 1) ) r = t - r; } while( t <= (1LL << 33) ); return false; }