[#79440] [Ruby trunk Bug#13188] Reinitialize Ruby VM. — shyouhei@...
Issue #13188 has been updated by Shyouhei Urabe.
6 messages
2017/02/06
[#79441] Re: [Ruby trunk Bug#13188] Reinitialize Ruby VM.
— SASADA Koichi <ko1@...>
2017/02/06
On 2017/02/06 10:10, shyouhei@ruby-lang.org wrote:
[#79532] Immutable Strings vs Symbols — Daniel Ferreira <subtileos@...>
Hi,
15 messages
2017/02/15
[#79541] Re: Immutable Strings vs Symbols
— Rodrigo Rosenfeld Rosas <rr.rosas@...>
2017/02/15
Em 15-02-2017 05:05, Daniel Ferreira escreveu:
[#79543] Re: Immutable Strings vs Symbols
— Daniel Ferreira <subtileos@...>
2017/02/16
Hi Rodrigo,
[#79560] Re: Immutable Strings vs Symbols
— Rodrigo Rosenfeld Rosas <rr.rosas@...>
2017/02/16
Em 15-02-2017 22:39, Daniel Ferreira escreveu:
[ruby-core:79637] [Ruby trunk Feature#13219] bug in Math.sqrt(n).to_i, to compute integer squareroot, new word to accurately fix it
From:
jzakiya@...
Date:
2017-02-20 20:56:58 UTC
List:
ruby-core #79637
Issue #13219 has been updated by Jabari Zakiya.
Here's the README.md writeup on the general integer roots algorithm.
At the cpu level, all these bit operations -- >>, <<, |, ^ -- are one clock cycle instructions,
so you see this can be very fast if done in C. Maybe for arbitrary size numbers that can't
reside in a single cpu register they take longer, but they would still be faster than high level Ruby.
```
For integer numbers:
For binary based (digital) computers (cpu), all integers are represented as
binary numbers. To find the root "n" of an integer "num" we can use the following
process to find the largest integer nth "root" such that root**n <= num.
For an integer composed of 'b' bits an nth root 'n' will have at most (b/n + 1) bits.
For squareroots, n = 2:
For 9 = 0b1001, it has b = 4 bits, and its squareroot has at most (4/2 + 1) = 3 bits.
The squareroot of 9 is 3 = 0b11, which is 2 bits.
For 100 = 0b1100100, b = 7, and its squareroot has at most (7/2 + 1) = 4 bits.
The squareroot of 100 is 10 = 0b1010, which is 4 bits.
For cuberoots, n = 3:
For 8 = 0b1000, it has b = 4 bits, and its cuberoot has at most (4/3 + 1) = 2 bits.
The cuberoot of 8 is 2 = 0b10, which is 2 bits.
For 125 = 0b1111101, b = 7, and its cuberoot has at most (7/3 + 1) = 3 bits.
The cuberoot of 125 is 5 = 0b101, which is 3 bits.
Algorithm:
bits_shift = (num.bits.length)/n + 1 # determine max number of root bits
bitn_mask = 1 << bits_shift # set value for max bit position of root
root = 0 # initialize the value for root
until bitn_mask == 0 # step through all the bit positions for root
root |= bitn_mask # set current bit position to '1' in root
root ^= bitn_mask if root**n > num # set it back to '0' if root too large
bitn_mask >>= 1 # set bitn_mask to next smaller bit position
end
root # return exact integer value for root
```
----------------------------------------
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-63055
* 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
bitn_mask >>= 1
return root if bitn_mask == 0
root |= bitn_mask
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
bitn_mask >>= 1
return root if bitn_mask == 0
root |= bitn_mask
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.
```
2.4.0 :178 > load 'intsqrttest.rb'
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 (賊 5.5%) i/s - 271.218k in 5.017552s
intsqrt1(n) 55.872k (賊 5.4%) i/s - 283.140k in 5.082953s
Math.sqrt(n).to_i 5.278M (賊 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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