问题
How do you write your own function for finding the most accurate square root of an integer?
After googling it, I found this (archived from its original link), but first, I didn\'t get it completely, and second, it is approximate too.
Assume square root as nearest integer (to the actual root) or a float.
回答1:
The following computes floor(sqrt(N)) for N > 0:
x = 2^ceil(numbits(N)/2)
loop:
y = floor((x + floor(N/x))/2)
if y >= x
return x
x = y
This is a version of Newton's method given in Crandall & Pomerance, "Prime Numbers: A Computational Perspective". The reason you should use this version is that people who know what they're doing have proven that it converges exactly to the floor of the square root, and it's simple so the probability of making an implementation error is small. It's also fast (although it's possible to construct an even faster algorithm -- but doing that correctly is much more complex). A properly implemented binary search can be faster for very small N, but there you may as well use a lookup table.
To round to the nearest integer, just compute t = floor(sqrt(4N)) using the algorithm above. If the least significant bit of t is set, then choose x = (t+1)/2; otherwise choose t/2. Note that this rounds up on a tie; you could also round down (or round to even) by looking at whether the remainder is nonzero (i.e. whether t^2 == 4N).
Note that you don't need to use floating-point arithmetic. In fact, you shouldn't. This algorithm should be implemented entirely using integers (in particular, the floor() functions just indicate that regular integer division should be used).
回答2:
Depending on your needs, a simple divide-and-conquer strategy can be used. It won't converge as fast as some other methods but it may be a lot easier for a novice to understand. In addition, since it's an O(log n) algorithm (halving the search space each iteration), the worst case for a 32-bit float will be 32 iterations.
Let's say you want the square root of 62.104. You pick a value halfway between 0 and that, and square it. If the square is higher than your number, you need to concentrate on numbers less than the midpoint. If it's too low, concentrate on those higher.
With real math, you could keep dividing the search space in two forever (if it doesn't have a rational square root). In reality, computers will eventually run out of precision and you'll have your approximation. The following C program illustrates the point:
#include <stdio.h>
#include <stdlib.h>
int main (int argc, char *argv[]) {
float val, low, high, mid, oldmid, midsqr;
int step = 0;
// Get argument, force to non-negative.
if (argc < 2) {
printf ("Usage: sqrt <number>\n");
return 1;
}
val = fabs (atof (argv[1]));
// Set initial bounds and print heading.
low = 0;
high = mid = val;
oldmid = -1;
printf ("%4s %10s %10s %10s %10s %10s %s\n",
"Step", "Number", "Low", "High", "Mid", "Square", "Result");
// Keep going until accurate enough.
while (fabs(oldmid - mid) >= 0.00001) {
oldmid = mid;
// Get midpoint and see if we need lower or higher.
mid = (high + low) / 2;
midsqr = mid * mid;
printf ("%4d %10.4f %10.4f %10.4f %10.4f %10.4f ",
++step, val, low, high, mid, midsqr);
if (mid * mid > val) {
high = mid;
printf ("- too high\n");
} else {
low = mid;
printf ("- too low\n");
}
}
// Desired accuracy reached, print it.
printf ("sqrt(%.4f) = %.4f\n", val, mid);
return 0;
}
Here's a couple of runs so you hopefully get an idea how it works. For 77:
pax> sqrt 77
Step Number Low High Mid Square Result
1 77.0000 0.0000 77.0000 38.5000 1482.2500 - too high
2 77.0000 0.0000 38.5000 19.2500 370.5625 - too high
3 77.0000 0.0000 19.2500 9.6250 92.6406 - too high
4 77.0000 0.0000 9.6250 4.8125 23.1602 - too low
5 77.0000 4.8125 9.6250 7.2188 52.1104 - too low
6 77.0000 7.2188 9.6250 8.4219 70.9280 - too low
7 77.0000 8.4219 9.6250 9.0234 81.4224 - too high
8 77.0000 8.4219 9.0234 8.7227 76.0847 - too low
9 77.0000 8.7227 9.0234 8.8730 78.7310 - too high
10 77.0000 8.7227 8.8730 8.7979 77.4022 - too high
11 77.0000 8.7227 8.7979 8.7603 76.7421 - too low
12 77.0000 8.7603 8.7979 8.7791 77.0718 - too high
13 77.0000 8.7603 8.7791 8.7697 76.9068 - too low
14 77.0000 8.7697 8.7791 8.7744 76.9893 - too low
15 77.0000 8.7744 8.7791 8.7767 77.0305 - too high
16 77.0000 8.7744 8.7767 8.7755 77.0099 - too high
17 77.0000 8.7744 8.7755 8.7749 76.9996 - too low
18 77.0000 8.7749 8.7755 8.7752 77.0047 - too high
19 77.0000 8.7749 8.7752 8.7751 77.0022 - too high
20 77.0000 8.7749 8.7751 8.7750 77.0009 - too high
21 77.0000 8.7749 8.7750 8.7750 77.0002 - too high
22 77.0000 8.7749 8.7750 8.7750 76.9999 - too low
23 77.0000 8.7750 8.7750 8.7750 77.0000 - too low
sqrt(77.0000) = 8.7750
For 62.104:
pax> sqrt 62.104
Step Number Low High Mid Square Result
1 62.1040 0.0000 62.1040 31.0520 964.2267 - too high
2 62.1040 0.0000 31.0520 15.5260 241.0567 - too high
3 62.1040 0.0000 15.5260 7.7630 60.2642 - too low
4 62.1040 7.7630 15.5260 11.6445 135.5944 - too high
5 62.1040 7.7630 11.6445 9.7037 94.1628 - too high
6 62.1040 7.7630 9.7037 8.7334 76.2718 - too high
7 62.1040 7.7630 8.7334 8.2482 68.0326 - too high
8 62.1040 7.7630 8.2482 8.0056 64.0895 - too high
9 62.1040 7.7630 8.0056 7.8843 62.1621 - too high
10 62.1040 7.7630 7.8843 7.8236 61.2095 - too low
11 62.1040 7.8236 7.8843 7.8540 61.6849 - too low
12 62.1040 7.8540 7.8843 7.8691 61.9233 - too low
13 62.1040 7.8691 7.8843 7.8767 62.0426 - too low
14 62.1040 7.8767 7.8843 7.8805 62.1024 - too low
15 62.1040 7.8805 7.8843 7.8824 62.1323 - too high
16 62.1040 7.8805 7.8824 7.8815 62.1173 - too high
17 62.1040 7.8805 7.8815 7.8810 62.1098 - too high
18 62.1040 7.8805 7.8810 7.8807 62.1061 - too high
19 62.1040 7.8805 7.8807 7.8806 62.1042 - too high
20 62.1040 7.8805 7.8806 7.8806 62.1033 - too low
21 62.1040 7.8806 7.8806 7.8806 62.1038 - too low
22 62.1040 7.8806 7.8806 7.8806 62.1040 - too high
23 62.1040 7.8806 7.8806 7.8806 62.1039 - too high
sqrt(62.1040) = 7.8806
For 49:
pax> sqrt 49
Step Number Low High Mid Square Result
1 49.0000 0.0000 49.0000 24.5000 600.2500 - too high
2 49.0000 0.0000 24.5000 12.2500 150.0625 - too high
3 49.0000 0.0000 12.2500 6.1250 37.5156 - too low
4 49.0000 6.1250 12.2500 9.1875 84.4102 - too high
5 49.0000 6.1250 9.1875 7.6562 58.6182 - too high
6 49.0000 6.1250 7.6562 6.8906 47.4807 - too low
7 49.0000 6.8906 7.6562 7.2734 52.9029 - too high
8 49.0000 6.8906 7.2734 7.0820 50.1552 - too high
9 49.0000 6.8906 7.0820 6.9863 48.8088 - too low
10 49.0000 6.9863 7.0820 7.0342 49.4797 - too high
11 49.0000 6.9863 7.0342 7.0103 49.1437 - too high
12 49.0000 6.9863 7.0103 6.9983 48.9761 - too low
13 49.0000 6.9983 7.0103 7.0043 49.0598 - too high
14 49.0000 6.9983 7.0043 7.0013 49.0179 - too high
15 49.0000 6.9983 7.0013 6.9998 48.9970 - too low
16 49.0000 6.9998 7.0013 7.0005 49.0075 - too high
17 49.0000 6.9998 7.0005 7.0002 49.0022 - too high
18 49.0000 6.9998 7.0002 7.0000 48.9996 - too low
19 49.0000 7.0000 7.0002 7.0001 49.0009 - too high
20 49.0000 7.0000 7.0001 7.0000 49.0003 - too high
21 49.0000 7.0000 7.0000 7.0000 49.0000 - too low
22 49.0000 7.0000 7.0000 7.0000 49.0001 - too high
23 49.0000 7.0000 7.0000 7.0000 49.0000 - too high
sqrt(49.0000) = 7.0000
回答3:
A simple (but not very fast) method to calculate the square root of X:
squareroot(x)
if x<0 then Error
a = 1
b = x
while (abs(a-b)>ErrorMargin)
a = (a+b)/2
b = x/a
endwhile
return a;
Example: squareroot(70000)
a b
1 70000
35001 2
17502 4
8753 8
4381 16
2199 32
1116 63
590 119
355 197
276 254
265 264
As you can see it defines an upper and a lower boundary for the square root and narrows the boundary until its size is acceptable.
There are more efficient methods but this one illustrates the process and is easy to understand.
Just beware to set the Errormargin to 1 if using integers else you have an endless loop.
回答4:
Let me point out an extremely interesting method of calculating an inverse square root 1/sqrt(x) which is a legend in the world of game design because it is mind-boggingly fast. Or wait, read the following post:
http://betterexplained.com/articles/understanding-quakes-fast-inverse-square-root/
PS: I know you just want the square root but the elegance of quake overcame all resistance on my part :)
By the way, the above mentioned article also talks about the boring Newton-Raphson approximation somewhere.
回答5:
Of course it's approximate; that is how math with floating-point numbers work.
Anyway, the standard way is with Newton's method. This is about the same as using Taylor's series, the other way that comes to mind immediately.
回答6:
Calculate square root with arbitrary precision in Python
#!/usr/bin/env python
import decimal
def sqrt(n):
assert n > 0
with decimal.localcontext() as ctx:
ctx.prec += 2 # increase precision to minimize round off error
x, prior = decimal.Decimal(n), None
while x != prior:
prior = x
x = (x + n/x) / 2 # quadratic convergence
return +x # round in a global context
decimal.getcontext().prec = 80 # desirable precision
r = sqrt(12345)
print r
print r == decimal.Decimal(12345).sqrt()
Output:
111.10805551354051124500443874307524148991137745969772997648567316178259031751676
True
回答7:
It's a common interview question asked by Facebook etc. I don't think it's a good idea to use the Newton's method in an interview. What if the interviewer ask you the mechanism of the Newton's method when you don't really understand it?
I provided a binary search based solution in Java which I believe everyone can understand.
public int sqrt(int x) {
if(x < 0) return -1;
if(x == 0 || x == 1) return x;
int lowerbound = 1;
int upperbound = x;
int root = lowerbound + (upperbound - lowerbound)/2;
while(root > x/root || root+1 <= x/(root+1)){
if(root > x/root){
upperbound = root;
} else {
lowerbound = root;
}
root = lowerbound + (upperbound - lowerbound)/2;
}
return root;
}
You can test my code here: leetcode: sqrt(x)
回答8:
Found a great article about Integer Square Roots.
This is a slightly improved version that it presents there:
unsigned long sqrt(unsigned long a){
int i;
unsigned long rem = 0;
unsigned long root = 0;
for (i = 0; i < 16; i++){
root <<= 1;
rem = (rem << 2) | (a >> 30);
a <<= 2;
if(root < rem){
root++;
rem -= root;
root++;
}
}
return root >> 1;
}
回答9:
Here's a way of obtaining a square root using trigonometry. It's not the fastest algorithm by a longshot, but it is precise. Code is in javascript:
var n = 5; //number to get the square root of
var icr = ((n+1)/2); //intersecting circle radius
var sqrt = Math.cos(Math.asin((icr-1)/icr))*icr; //square root of n
alert(sqrt);
回答10:
There is an algorithm that I studied in school that you can use to compute exact square roots (or of arbitrarily large precision if the root is an irrational number). It is definitely slower than Newton's algorithms but it is exact. Lets say you want to compute the square root of 531.3025
First thing is you divide your number starting from the decimal point into groups of 2 digits:
{5}{31}.{30}{25}
Then:
1) Find the closest square root for first group that is smaller or equal to the actual square root of first group: sqrt({5}) >= 2. This square root is the first digit of your final answer. Lets denote the digits we have already found of our final square root as B. So at the moment B = 2.
2) Next compute the difference between {5} and B^2: 5 - 4 = 1.
3) For all subsequent 2 digit groups do the following:
Multiply the remainder by 100, then add it to the second group: 100 + 31 = 131.
Find X - next digit of your root, such that 131 >=((B*20) + X)*X. X = 3. 43 * 3 = 129 < 131. Now B = 23. Also because you have no more 2-digit groups to the left of decimal points, you have found all integer digits of your final root.
4)Repeat the same for {30} and {25}. So you have:
{30} : 131 - 129 = 2. 2 * 100 + 30 = 230 >= (23*2*10 + X) * X -> X = 0 -> B = 23.0
{25} : 230 - 0 = 230. 230 * 100 + 25 = 23025. 23025 >= (230 * 2 * 10 + X) * X -> X = 5 -> B = 23.05
Final result = 23.05.
The algorithm looks complicated this way but it is much simpler if you do it on paper using the same notation you use for "long division" you have studied in school, except that you don't do division but instead compute the square root.
回答11:
// Fastest way I found, an (extreme) C# unrolled version of:
// http://www.hackersdelight.org/hdcodetxt/isqrt.c.txt (isqrt4)
// It's quite a lot of code, basically a binary search (the "if" statements)
// followed by an unrolled loop (the labels).
// Most important: it's fast, twice as fast as "Math.Sqrt".
// On my pc: Math.Sqrt ~35 ns, sqrt <16 ns (mean <14 ns)
private static uint sqrt(uint x)
{
uint y, z;
if (x < 1u << 16)
{
if (x < 1u << 08)
{
if (x < 1u << 04) return x < 1u << 02 ? x + 3u >> 2 : x + 15u >> 3;
else
{
if (x < 1u << 06)
{ y = 1u << 03; x -= 1u << 04; if (x >= 5u << 02) { x -= 5u << 02; y |= 1u << 02; } goto L0; }
else
{ y = 1u << 05; x -= 1u << 06; if (x >= 5u << 04) { x -= 5u << 04; y |= 1u << 04; } goto L1; }
}
}
else // slower (on my pc): .... y = 3u << 04; } goto L1; }
{
if (x < 1u << 12)
{
if (x < 1u << 10)
{ y = 1u << 07; x -= 1u << 08; if (x >= 5u << 06) { x -= 5u << 06; y |= 1u << 06; } goto L2; }
else
{ y = 1u << 09; x -= 1u << 10; if (x >= 5u << 08) { x -= 5u << 08; y |= 1u << 08; } goto L3; }
}
else
{
if (x < 1u << 14)
{ y = 1u << 11; x -= 1u << 12; if (x >= 5u << 10) { x -= 5u << 10; y |= 1u << 10; } goto L4; }
else
{ y = 1u << 13; x -= 1u << 14; if (x >= 5u << 12) { x -= 5u << 12; y |= 1u << 12; } goto L5; }
}
}
}
else
{
if (x < 1u << 24)
{
if (x < 1u << 20)
{
if (x < 1u << 18)
{ y = 1u << 15; x -= 1u << 16; if (x >= 5u << 14) { x -= 5u << 14; y |= 1u << 14; } goto L6; }
else
{ y = 1u << 17; x -= 1u << 18; if (x >= 5u << 16) { x -= 5u << 16; y |= 1u << 16; } goto L7; }
}
else
{
if (x < 1u << 22)
{ y = 1u << 19; x -= 1u << 20; if (x >= 5u << 18) { x -= 5u << 18; y |= 1u << 18; } goto L8; }
else
{ y = 1u << 21; x -= 1u << 22; if (x >= 5u << 20) { x -= 5u << 20; y |= 1u << 20; } goto L9; }
}
}
else
{
if (x < 1u << 28)
{
if (x < 1u << 26)
{ y = 1u << 23; x -= 1u << 24; if (x >= 5u << 22) { x -= 5u << 22; y |= 1u << 22; } goto La; }
else
{ y = 1u << 25; x -= 1u << 26; if (x >= 5u << 24) { x -= 5u << 24; y |= 1u << 24; } goto Lb; }
}
else
{
if (x < 1u << 30)
{ y = 1u << 27; x -= 1u << 28; if (x >= 5u << 26) { x -= 5u << 26; y |= 1u << 26; } goto Lc; }
else
{ y = 1u << 29; x -= 1u << 30; if (x >= 5u << 28) { x -= 5u << 28; y |= 1u << 28; } }
}
}
}
z = y | 1u << 26; y /= 2; if (x >= z) { x -= z; y |= 1u << 26; }
Lc: z = y | 1u << 24; y /= 2; if (x >= z) { x -= z; y |= 1u << 24; }
Lb: z = y | 1u << 22; y /= 2; if (x >= z) { x -= z; y |= 1u << 22; }
La: z = y | 1u << 20; y /= 2; if (x >= z) { x -= z; y |= 1u << 20; }
L9: z = y | 1u << 18; y /= 2; if (x >= z) { x -= z; y |= 1u << 18; }
L8: z = y | 1u << 16; y /= 2; if (x >= z) { x -= z; y |= 1u << 16; }
L7: z = y | 1u << 14; y /= 2; if (x >= z) { x -= z; y |= 1u << 14; }
L6: z = y | 1u << 12; y /= 2; if (x >= z) { x -= z; y |= 1u << 12; }
L5: z = y | 1u << 10; y /= 2; if (x >= z) { x -= z; y |= 1u << 10; }
L4: z = y | 1u << 08; y /= 2; if (x >= z) { x -= z; y |= 1u << 08; }
L3: z = y | 1u << 06; y /= 2; if (x >= z) { x -= z; y |= 1u << 06; }
L2: z = y | 1u << 04; y /= 2; if (x >= z) { x -= z; y |= 1u << 04; }
L1: z = y | 1u << 02; y /= 2; if (x >= z) { x -= z; y |= 1u << 02; }
L0: return x > y ? y / 2 | 1u : y / 2;
}
回答12:
The first thing comes to my mind is: this is a good place to use Binary search (inspired by this great tutorials.)
To find the square root of vaule ,we are searching the number in (1..value) where the predictor
is true for the first time. The predictor we are choosing is number * number - value > 0.00001.
double square_root_of(double value)
{
assert(value >= 1);
double lo = 1.0;
double hi = value;
while( hi - lo > 0.00001)
{
double mid = lo + (hi - lo) / 2 ;
std::cout << lo << "," << hi << "," << mid << std::endl;
if( mid * mid - value > 0.00001) //this is the predictors we are using
{
hi = mid;
} else {
lo = mid;
}
}
return lo;
}
回答13:
use binary search
public class FindSqrt {
public static void main(String[] strings) {
int num = 10000;
System.out.println(sqrt(num, 0, num));
}
private static int sqrt(int num, int min, int max) {
int middle = (min + max) / 2;
int x = middle * middle;
if (x == num) {
return middle;
} else if (x < num) {
return sqrt(num, middle, max);
} else {
return sqrt(num, min, middle);
}
}
}
回答14:
In general the square root of an integer (like 2, for example) can only be approximated (not because of problems with floating point arithmetic, but because they're irrational numbers which can't be calculated exactly).
Of course, some approximations are better than others. I mean, of course, that the value 1.732 is a better approximation to the square root of 3, than 1.7
The method used by the code at that link you gave works by taking a first approximation and using it to calculate a better approximation.
This is called Newton's Method, and you can repeat the calculation with each new approximation until it's accurate enough for you.
In fact there must be some way to decide when to stop the repetition or it will run forever.
Usually you would stop when the difference between approximations is less than a value you decide.
EDIT: I don't think there can be a simpler implementation than the two you already found.
回答15:
The inverse, as the name says, but sometimes "close enough" is "close enough"; an interesting read anyway.
Origin of Quake3's Fast InvSqrt()
回答16:
A simple solution that can deal with float square root and arbitrary precision using binary search
coded in ruby
include Math
def sqroot_precision num, precision
upper = num
lower = 0
middle = (upper + lower)/2.0
while true do
diff = middle**2 - num
return middle if diff.abs <= precision
if diff > 0
upper = middle
else diff < 0
lower = middle
end
middle = (upper + lower)/2.0
end
end
puts sqroot_precision 232.3, 0.0000000001
回答17:
Let's say we are trying to find the square root of 2, and you have an estimate of 1.5. We'll say a = 2, and x = 1.5. To compute a better estimate, we'll divide a by x. This gives a new value y = 1.333333. However, we can't just take this as our next estimate (why not?). We need to average it with the previous estimate. So our next estimate, xx will be (x + y) / 2, or 1.416666.
Double squareRoot(Double a, Double epsilon) {
Double x = 0d;
Double y = a;
Double xx = 0d;
// Make sure both x and y != 0.
while ((x != 0d || y != 0d) && y - x > epsilon) {
xx = (x + y) / 2;
if (xx * xx >= a) {
y = xx;
} else {
x = xx;
}
}
return xx;
}
Epsilon determines how accurate the approximation needs to be. The function should return the first approximation x it obtains that satisfies abs(x*x - a) < epsilon, where abs(x) is the absolute value of x.
square_root(2, 1e-6)
Output: 1.4142141342163086
回答18:
Well there are already quite a few answers, but here goes mine It's the most simplest piece of code ( for me ), here is the algorithm for it.
And code in python 2.7:
from __future__ import division
val = 81
x = 10
def sqr(data,x):
temp = x - ( (x**2 - data)/(2*x))
if temp == x:
print temp
return
else:
x = temp
return sqr(data,x)
#x =temp
#sqr(data,x)
sqr(val,x)
回答19:
To calculate the square root of a number by help of inbuilt function
# include"iostream.h"
# include"conio.h"
# include"math.h"
void main()
{
clrscr();
float x;
cout<<"Enter the Number";
cin>>x;
float squreroot(float);
float z=squareroot(x);
cout<<z;
float squareroot(int x)
{
float s;
s = pow(x,.5)
return(s);
}
来源：https://stackoverflow.com/questions/1623375/writing-your-own-square-root-function