Fastest way to determine if an integer's square root is an integer

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心在旅途
心在旅途 2020-11-22 02:17

I\'m looking for the fastest way to determine if a long value is a perfect square (i.e. its square root is another integer):

  1. I\'ve done it the ea
30条回答
  •  旧巷少年郎
    2020-11-22 03:05

    I figured out a method that works ~35% faster than your 6bits+Carmack+sqrt code, at least with my CPU (x86) and programming language (C/C++). Your results may vary, especially because I don't know how the Java factor will play out.

    My approach is threefold:

    1. First, filter out obvious answers. This includes negative numbers and looking at the last 4 bits. (I found looking at the last six didn't help.) I also answer yes for 0. (In reading the code below, note that my input is int64 x.)
      if( x < 0 || (x&2) || ((x & 7) == 5) || ((x & 11) == 8) )
          return false;
      if( x == 0 )
          return true;
    2. Next, check if it's a square modulo 255 = 3 * 5 * 17. Because that's a product of three distinct primes, only about 1/8 of the residues mod 255 are squares. However, in my experience, calling the modulo operator (%) costs more than the benefit one gets, so I use bit tricks involving 255 = 2^8-1 to compute the residue. (For better or worse, I am not using the trick of reading individual bytes out of a word, only bitwise-and and shifts.)
      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.
      
      To actually check if the residue is a square, I look up the answer in a precomputed table.
      if( bad255[y] )
          return false;
      // However, I just use a table of size 512
      
    3. Finally, try to compute the square root using a method similar to Hensel's lemma. (I don't think it's applicable directly, but it works with some modifications.) Before doing that, I divide out all powers of 2 with a 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;
      At this point, for our number to be a square, it must be 1 mod 8.
      if((x & 7) != 1)
          return false;
      The basic structure of Hensel's lemma is the following. (Note: untested code; if it doesn't work, try t=2 or 8.)
      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.
      The idea is that at each iteration, you add one bit onto r, the "current" square root of x; each square root is accurate modulo a larger and larger power of 2, namely t/2. At the end, r and t/2-r will be square roots of x modulo t/2. (Note that if r is a square root of x, then so is -r. This is true even modulo numbers, but beware, modulo some numbers, things can have even more than 2 square roots; notably, this includes powers of 2.) Because our actual square root is less than 2^32, at that point we can actually just check if r or t/2-r are real square roots. In my actual code, I use the following modified loop:
      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) );
      The speedup here is obtained in three ways: precomputed start value (equivalent to ~10 iterations of the loop), earlier exit of the loop, and skipping some t values. For the last part, I look at z = r - x * x, and set t to be the largest power of 2 dividing z with a bit trick. This allows me to skip t values that wouldn't have affected the value of r anyway. The precomputed start value in my case picks out the "smallest positive" square root modulo 8192.

    Even if this code doesn't work faster for you, I hope you enjoy some of the ideas it contains. Complete, tested code follows, including the precomputed tables.

    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;
    }

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