确定整数的平方根是否为整数的最快方法 [英] Fastest way to determine if an integer's square root is an integer
问题描述
我正在寻找最快的方法来确定 long
值是否是一个完美的平方(即它的平方根是另一个整数):
- 我使用内置的
Math.sqrt()
功能,但我想知道是否有办法通过将自己限制在仅限整数的域中. - 维护一个查找表是不切实际的(因为有大约231.5 平方小于 263 的整数.
这是我现在使用的非常简单直接的方法:
public final static boolean isPerfectSquare(long n){如果 (n <0)返回假;long tst = (long)(Math.sqrt(n) + 0.5);返回 tst*tst == n;}
注意:我在许多 Project Euler 问题中使用了这个函数.所以没有其他人需要维护这个代码.这种微优化实际上可能会有所作为,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该函数数百万次.><小时>
我尝试了不同的解决方案:
- 经过详尽的测试,我发现在 Math.sqrt() 的结果中添加
0.5
是没有必要的,至少在我的机器上没有. - 快速逆平方根速度更快,但它给出了不正确的结果 n >= 410881. 然而,正如 BobbyShaftoe 所建议的,我们可以使用 FISR hack for n <410881.
- Newton 的方法比
Math.sqrt()
慢一点.这可能是因为Math.sqrt()
使用类似于牛顿法的东西,但在硬件中实现,所以它比在 Java 中快得多.此外,牛顿法仍然需要使用双打. - 一种修改后的牛顿方法,它使用了一些技巧,因此只涉及整数数学,需要一些技巧来避免溢出(我希望这个函数可以处理所有 64 位有符号正整数),但它仍然比
Math.sqrt()
. - 二进制切割甚至更慢.这是有道理的,因为二进制斩波平均需要 16 次才能找到 64 位数字的平方根.
- 根据 John 的测试,在 C++ 中使用
or
语句比使用switch
更快,但在 Java 和 C# 中,or 似乎没有区别
和switch
. - 我还尝试制作了一个查找表(作为 64 个布尔值的私有静态数组).然后,我会说
if(lookup[(int)(n&0x3F)]) { test } else return false;
,而不是 switch 或or
语句.令我惊讶的是,这(只是稍微)慢了一点.这是因为 在 Java 中检查数组边界.
我找到了一种比 6bits+Carmack+sqrt 代码快 35% 的方法,至少在我的 CPU (x86) 和编程语言 (C/C++).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:
- 首先,过滤掉明显的答案.这包括负数和查看最后 4 位.(我发现查看最后六个没有帮助.)我也回答是 0.(在阅读下面的代码时,请注意我的输入是
int64 x
.)if( x <0 || (x&2) || ((x & 7) == 5) || ((x & 11) == 8) )返回假;如果( x == 0 )return true;
- 接下来,检查它是否是模 255 = 3 * 5 * 17 的平方.因为这是三个不同素数的乘积,所以只有大约 1/8 的模 255 的余数是平方.但是,根据我的经验,调用模运算符 (%) 的成本高于获得的收益,因此我使用涉及 255 = 2^8-1 的小技巧来计算残差.(无论好坏,我没有使用从单词中读取单个字节的技巧,只是按位和和移位.)
int64 y = x;y = (y & 4294967295LL) + (y > > 32);y = (y & 65535) + (y > > 16);y = (y & 255) + ((y > > 8) & 255) + (y > > 16);//此时y在0到511之间,代码越多,可以进一步缩小.
为了实际检查残差是否为正方形,我在预先计算的表格中查找答案.
if( bad255[y] )返回假;//但是,我只使用了一个大小为 512 的表
- 最后,尝试使用类似于 Hensel 引理 的方法计算平方根.(我认为它不能直接适用,但经过一些修改后可以使用.)在此之前,我使用二分搜索来划分 2 的所有幂:
if((x & 4294967295LL) == 0)x>>=32;如果((x & 65535) == 0)x>>=16;如果((x & 255) == 0)x>>=8;如果((x & 15) == 0)x>>=4;如果((x & 3) == 0)x >>= 2;
此时,对于我们的数字是一个正方形,它必须是 1 mod 8.
if((x & 7) != 1)return false;
Hensel 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.)
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;//重复直到 t 为 2^33 左右.如果需要,请使用循环.
这个想法是在每次迭代时,你在 r 上加一位,即 x 的当前"平方根;每个平方根都是精确的模 2 的越来越大的幂,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.即使是模数也是如此,但要注意,对某些数字进行模数,事物甚至可以有 2 个以上的平方根;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 是否为实数平方根.在我的实际代码中,我使用了以下修改后的循环:
int64 r, t, z;r = 开始[(x > > 3) & 1023];做 {z = x - r * r;如果( z == 0 )返回真;如果( z <0 )返回假;t = z & (-z);r += (z & t) >>1;如果(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)、更早退出循环和跳过一些 t 值.对于最后一部分,我查看
z = r - x * x
,并通过一些技巧将 t 设置为 2 除 z 的最大幂.这允许我跳过无论如何都不会影响 r 值的 t 值.在我的情况下,预先计算的起始值挑选出最小正"平方根模 8192.
即使此代码对您来说不能更快地工作,但我希望您喜欢其中包含的一些想法.完整的、经过测试的代码如下,包括预先计算的表格.
typedef signed long long int int64;int开始[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};内联布尔方块(int64 x){//快速失败if( x < 0 || (x&2) || ((x & 7) == 5) || ((x & 11) == 8) )返回假;如果( x == 0 )返回真;//检查 mod 255 = 3 * 5 * 17,为了好玩int64 y = x;y = (y & 4294967295LL) + (y > > 32);y = (y & 65535) + (y > > 16);y = (y & 255) + ((y > > 8) & 255) + (y > > 16);如果(坏255[y])返回假;//使用二分查找除以 4 的幂如果((x & 4294967295LL) == 0)x>>=32;如果((x & 65535) == 0)x>>=16;如果((x & 255) == 0)x>>=8;如果((x & 15) == 0)x>>=4;如果((x & 3) == 0)x>>=2;如果((x & 7) != 1)返回假;//使用类似 Hensel 引理的东西计算 sqrtint64 r, t, z;r = 开始[(x > > 3) & 1023];做 {z = x - r * r;如果( z == 0 )返回真;如果( z <0 )返回假;t = z & (-z);r += (z & t) >>1;如果(r>(t>>1))r = t - r;} while( t <= (1LL << 33) );返回假;}
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):
- I've done it the easy way, by using the built-in
Math.sqrt()
function, but I'm wondering if there is a way to do it faster by restricting yourself to integer-only domain. - Maintaining a lookup table is impractical (since there are about 231.5 integers whose square is less than 263).
Here is the very simple and straightforward way I'm doing it now:
public final static boolean isPerfectSquare(long n)
{
if (n < 0)
return false;
long tst = (long)(Math.sqrt(n) + 0.5);
return tst*tst == n;
}
Note: I'm using this function in many Project Euler problems. So no one else will ever have to maintain this code. And this kind of micro-optimization could actually make a difference, since part of the challenge is to do every algorithm in less than a minute, and this function will need to be called millions of times in some problems.
I've tried the different solutions to the problem:
- After exhaustive testing, I found that adding
0.5
to the result of Math.sqrt() is not necessary, at least not on my machine. - The fast inverse square root was faster, but it gave incorrect results for n >= 410881. However, as suggested by BobbyShaftoe, we can use the FISR hack for n < 410881.
- Newton's method was a good bit slower than
Math.sqrt()
. This is probably becauseMath.sqrt()
uses something similar to Newton's Method, but implemented in the hardware so it's much faster than in Java. Also, Newton's Method still required use of doubles. - A modified Newton's method, which used a few tricks so that only integer math was involved, required some hacks to avoid overflow (I want this function to work with all positive 64-bit signed integers), and it was still slower than
Math.sqrt()
. - Binary chop was even slower. This makes sense because the binary chop will on average require 16 passes to find the square root of a 64-bit number.
- According to John's tests, using
or
statements is faster in C++ than using aswitch
, but in Java and C# there appears to be no difference betweenor
andswitch
. - I also tried making a lookup table (as a private static array of 64 boolean values). Then instead of either switch or
or
statement, I would just sayif(lookup[(int)(n&0x3F)]) { test } else return false;
. To my surprise, this was (just slightly) slower. This is because array bounds are checked in Java.
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:
- 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;
- 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
- 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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