```Issue #13219 has been updated by Nathan Zook.

You might want to consider the following articles:
https://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Binary_numeral_system_.28base_2.29
(You might want to start a bit higher in the article to get the context.)
and also to Zimmerman's implementation:
https://gmplib.org/repo/gmp-5.1/file/c5010c039373/mpn/generic/sqrtrem.c

The wikipedia article implements a faster bit-by-bit computation.  Zimmerman implements the standard solution, which converges quadratically.

Looking at your comments, however, you seem really to be interested in a high performance multiprecision library for ruby.  Is there any particular reason that we should not take an existing C library and drop a wrapper around it?

----------------------------------------
Feature #13219: bug in Math.sqrt(n).to_i, to compute integer squareroot,  new word to accurately fix it
https://bugs.ruby-lang.org/issues/13219#change-63065

* Author: Jabari Zakiya
* Status: Open
* Priority: Normal
* Assignee:
* Target version:
----------------------------------------
In doing a math application using **Math.sqrt(n).to_i** to compute integer squareroots
of integers I started noticing errors for numbers > 10**28.

I coded an algorithm that accurately computes the integer squareroot for arbirary sized numbers
but its significantly slower than the math library written in C.

Thus, I request a new method **Math.intsqrt(n)** be created, that is coded in C and part of the
core library, that will compute the integer squareroots of integers accurately and fast.

Here is working highlevel code to accurately compute the integer squareroot.

```
def intsqrt(n)
bits_shift = (n.to_s(2).size)/2 + 1
bitn_mask = root = 1 << bits_shift
while true
root ^= bitn_mask if (root * root) > n
return root if bitn_mask == 0
end
end

def intsqrt1(n)
return n if n | 1 == 1   # if n is 0 or 1
bits_shift = (Math.log2(n).ceil)/2 + 1
bitn_mask = root = 1 << bits_shift
while true
root ^= bitn_mask if (root * root) > n
return root if bitn_mask == 0
end
end

require "benchmark/ips"

Benchmark.ips do |x|
n = 10**40
puts "integer squareroot tests for n = #{n}"
x.report("intsqrt(n)"       ) { intsqrt(n)  }
x.report("intsqrt1(n)"      ) { intsqrt1(n) }
x.report("Math.sqrt(n).to_i") { Math.sqrt(n).to_i }
x.compare!
end
```
Here's why it needs to be done in C.

```
integer squareroot tests for n = 10000000000000000000000000000000000000000
Warming up --------------------------------------
intsqrt(n)     5.318k i/100ms
intsqrt1(n)     5.445k i/100ms
Math.sqrt(n).to_i   268.281k i/100ms
Calculating -------------------------------------
intsqrt(n)     54.219k (¡Þ 5.5%) i/s -    271.218k in   5.017552s
intsqrt1(n)     55.872k (¡Þ 5.4%) i/s -    283.140k in   5.082953s
Math.sqrt(n).to_i      5.278M (¡Þ 6.1%) i/s -     26.560M in   5.050707s

Comparison:
Math.sqrt(n).to_i:  5278477.8 i/s
intsqrt1(n):    55871.7 i/s - 94.47x  slower
intsqrt(n):    54219.4 i/s - 97.35x  slower

=> true
2.4.0 :179 >

```
Here are examples of math errors using **Math.sqrt(n).to_i** run on Ruby-2.4.0.

```
2.4.0 :101 > n = 10**27; puts n, (a = intsqrt(n)), a*a, (b = intsqrt1(n)), b*b, (c = Math.sqrt(n).to_i), c*c
1000000000000000000000000000
31622776601683
999999999999949826038432489
31622776601683
999999999999949826038432489
31622776601683
999999999999949826038432489
=> nil
2.4.0 :102 > n = 10**28; puts n, (a = intsqrt(n)), a*a, (b = intsqrt1(n)), b*b, (c = Math.sqrt(n).to_i), c*c
10000000000000000000000000000
100000000000000
10000000000000000000000000000
100000000000000
10000000000000000000000000000
100000000000000
10000000000000000000000000000
=> nil
2.4.0 :103 > n = 10**29; puts n, (a = intsqrt(n)), a*a, (b = intsqrt1(n)), b*b, (c = Math.sqrt(n).to_i), c*c
100000000000000000000000000000
316227766016837
99999999999999409792567484569
316227766016837
99999999999999409792567484569
316227766016837
99999999999999409792567484569
=> nil
2.4.0 :104 > n = 10**30; puts n, (a = intsqrt(n)), a*a, (b = intsqrt1(n)), b*b, (c = Math.sqrt(n).to_i), c*c
1000000000000000000000000000000
1000000000000000
1000000000000000000000000000000
1000000000000000
1000000000000000000000000000000
1000000000000000
1000000000000000000000000000000
=> nil
2.4.0 :105 > n = 10**31; puts n, (a = intsqrt(n)), a*a, (b = intsqrt1(n)), b*b, (c = Math.sqrt(n).to_i), c*c
10000000000000000000000000000000
3162277660168379
9999999999999997900254631487641
3162277660168379
9999999999999997900254631487641
3162277660168379
9999999999999997900254631487641
=> nil
2.4.0 :106 > n = 10**32; puts n, (a = intsqrt(n)), a*a, (b = intsqrt1(n)), b*b, (c = Math.sqrt(n).to_i), c*c
100000000000000000000000000000000
10000000000000000
100000000000000000000000000000000
10000000000000000
100000000000000000000000000000000
10000000000000000
100000000000000000000000000000000
=> nil
2.4.0 :107 > n = 10**33; puts n, (a = intsqrt(n)), a*a, (b = intsqrt1(n)), b*b, (c = Math.sqrt(n).to_i), c*c
1000000000000000000000000000000000
31622776601683793
999999999999999979762122758866849
31622776601683793
999999999999999979762122758866849
31622776601683792
999999999999999916516569555499264
=> nil
2.4.0 :108 > n = 10**34; puts n, (a = intsqrt(n)), a*a, (b = intsqrt1(n)), b*b, (c = Math.sqrt(n).to_i), c*c
10000000000000000000000000000000000
100000000000000000
10000000000000000000000000000000000
100000000000000000
10000000000000000000000000000000000
100000000000000000
10000000000000000000000000000000000
=> nil
2.4.0 :109 > n = 10**35; puts n, (a = intsqrt(n)), a*a, (b = intsqrt1(n)), b*b, (c = Math.sqrt(n).to_i), c*c
100000000000000000000000000000000000
316227766016837933
99999999999999999873578871987712489
316227766016837933
99999999999999999873578871987712489
316227766016837952
100000000000000011890233980627554304
=> nil
2.4.0 :110 > n = 10**36; puts n, (a = intsqrt(n)), a*a, (b = intsqrt1(n)), b*b, (c = Math.sqrt(n).to_i), c*c
1000000000000000000000000000000000000
1000000000000000000
1000000000000000000000000000000000000
1000000000000000000
1000000000000000000000000000000000000
1000000000000000000
1000000000000000000000000000000000000
=> nil
2.4.0 :111 > n = 10**37; puts n, (a = intsqrt(n)), a*a, (b = intsqrt1(n)), b*b, (c = Math.sqrt(n).to_i), c*c
10000000000000000000000000000000000000
3162277660168379331
9999999999999999993682442519108007561
3162277660168379331
9999999999999999993682442519108007561
3162277660168379392
10000000000000000379480317059650289664
=> nil
2.4.0 :112 >
```

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