What's the correct and fast way to determine if a (gig) number is a perfect square?

Discussion in 'Ruby' started by Sam Kong, Feb 11, 2007.

  1. Sam Kong

    Sam Kong Guest

    Hello,

    I'm solving a math problem in Ruby.
    I need to determine if a number is a perfect square.
    If the number is small, you may do like the following.

    def perfect_square? n
    sqrt = n ** 0.5
    sqrt - sqrt.to_i == 0
    end

    But Float number has limitation on precision.
    Thus the function won't work correctly for big numbers like
    (123456789123456789).

    How would you solve such a case?
    It should be fast as well as correct because I will use it repeatedly.

    Thanks in advance.

    Sam
     
    Sam Kong, Feb 11, 2007
    #1
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  2. Easy: compare integers rather than floats.

    x = 123456789123456789

    sqrt = Math::sqrt(x)
    p(x == sqrt.floor**2)
     
    Joel VanderWerf, Feb 11, 2007
    #2
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  3. Sam Kong

    Xavier Noria Guest

    There are faster algorithms based on the fact that most non-squares
    aren't quadratic residues modulo some integers. I remember some is
    explained in Henri Cohen's "A Course in Computational Algebraic
    Number Theory", but do not have the book at hand.

    Perhaps you can take a look at the function that implements this test
    in GNU MP, explained here,

    http://www.swox.com/gmp/manual/Perfect-Square-Algorithm.html

    or the one in Pari:

    http://www.ufr-mi.u-bordeaux.fr/~belabas/pari/

    -- fxn
     
    Xavier Noria, Feb 11, 2007
    #3
  4. Sam Kong

    Xavier Noria Guest

    I've been able to consult the book today. There is a specialised
    algorithm to compute integer roots using integer arithmetic, and the
    integer square test itself. Thus, they guarantee a correct result. I
    attach them below translated to Ruby.

    Some operations would be written using bitwise stuff, but anyway the
    code performs very poorly (compared to Math::sqrt) according to the
    benchmarks. If performance is important a solution in C would be
    worth exploring.

    -- fxn


    # From Henri Cohen's "A Course in Computational Algebraic Number
    # Theory".

    # Algorithm 1.7.1 (Integer Square Root) Given a positive integer n,
    # this algorithm computes the integer part of the square root of n,
    # i.e. the number m such that m^2 <= n < (m + 1)^2.
    def isqrt(n)
    x = n
    loop do
    y = ((x + n/x)/2)
    if y < x
    x = y
    else
    return x
    end
    end
    end

    # Cache the squares modulus 11, 63, 64, and 65. This is used to check
    # for non-squares, since a square is a square mod k for all k. The
    # choice of these numbers is based on the probability that a non-square
    # is a square mod any of them, which is 6/715, less than a 1%.
    $squares = {}
    [11, 63, 64, 65].each do |m|
    $squares[m] = [false] * m
    (0...(m/2)).each {|i| $squares[m][i**2 % m] = true}
    end


    # Algorithm 1.7.3 (Square Test). Given a positive integer n,
    # this algorithm determines whether n is a square or not,
    # and if it is, outputs the square root of n. We assume the
    # precomputations made above.
    def issquare(n)
    return false unless $squares[64][n % 64]

    r = n % 45045 # 45045 = 63*65*11
    return false unless $squares[63][r % 63]
    return false unless $squares[65][r % 65]
    return false unless $squares[11][r % 11]

    q = isqrt(n)
    return q**2 == n ? q : false
    end

    require 'benchmark'

    $r = 1000

    # square of 32248581868698698768678697647823648238462348
    $s =
    103997103254216245831009973053627303273419242413513150637645310539211244
    0875299413673104

    # non-square, the previous number minus 1
    $ns =
    103997103254216245831009973053627303273419242413513150637645310539211244
    0875299413673103

    # Just for the sake of curiosity, since the code based on Math::sqrt
    is not correct.
    Benchmark.benchmark do |x|
    x.report("builtin is square (true)") do
    1.upto($r) do
    sqrt = Math::sqrt($s)
    $s == sqrt.floor**2
    end
    end
    x.report("modular is square (true)") do
    1.upto($r) do
    issquare($s)
    end
    end
    x.report("builtin is square (false)") do
    1.upto($r) do
    sqrt = Math::sqrt($ns)
    $ns == sqrt.floor**2
    end
    end
    x.report("modular is square (false)") do
    1.upto($r) do
    issquare($ns)
    end
    end
    end
     
    Xavier Noria, Feb 12, 2007
    #4
  5. Sam Kong

    Sam Kong Guest

    Hi Joel,

    Yes. Your approach is better than mine.
    But it gives a wrong answer for big numbers like 55833579873437812.


    Thanks anyway.

    Sam
     
    Sam Kong, Feb 12, 2007
    #5
  6. Sam Kong

    Sam Kong Guest

    Hi Xavier,

    I really appreciate your kind help.
    I put my comment inline.

    Actually you posted this message, I implemented an algorithm myself.
    It's just the way we calculate sqrt with pen and paper.
    It might be slow but correct even for big numbers.

    def max_digit n, m
    result = 0
    (0..9).each do |i|
    break if ((n * 10) + i) * i > m
    result = i
    end
    result
    end

    def perfect_square? n
    arr = []
    a = n
    while true
    a, b = a / 100, a % 100
    arr.unshift b
    break if a == 0
    end
    remain = 0
    carried = 0
    arr.each do |i|
    num = remain * 100 + i
    digit = max_digit(carried, num)
    remain = num - (carried * 10 + digit) * digit
    carried = (carried * 10 + digit) + digit
    end
    remain == 0
    end

    I felt ashamed to see the code you posted.OTL
    Thank you for your support.

    Sam
     
    Sam Kong, Feb 12, 2007
    #6
  7. Wrong how?

    irb(main):001:0> x = 55833579873437812
    => 55833579873437812
    irb(main):002:0> sqrt = Math::sqrt(x)
    => 236291303.0
    irb(main):003:0> sqrt.floor**2 - x
    => -3

    Ok, I can see that one problem with my approach is that I should have
    used #round instead of #floor.
     
    Joel VanderWerf, Feb 12, 2007
    #7
  8. Sam Kong

    Sam Kong Guest

    I think even if you use #round, the problem won't go away.
    Float type cannot generate correct result due to its limited
    precision.

    Sam
     
    Sam Kong, Feb 12, 2007
    #8
  9. Sam Kong

    Phrogz Guest

    For example:
    irb(main):004:0>
    x=981723462487562983749812734972342334123435635465656432452
    => 981723462487562983749812734972342334123435635465656432452

    irb(main):005:0> y=x**2
    =>
    963780956798569484937549463180492938080866374138542452022484415498543617865176348383792466015930367123924038732304

    irb(main):006:0> z=Math.sqrt(y)
    => 9.81723462487563e+056

    irb(main):007:0> z==x
    => true

    irb(main):010:0> z==(x+1000)
    => true

    irb(main):011:0> z==(x+10000)
    => true
     
    Phrogz, Feb 12, 2007
    #9
  10. I agree, but I don't think the problem shows up until you have much
    larger numbers. Is that the case for your program?

    In this case, 55833579873437812 cannot be exactly represented by a
    float, but it doesn't matter for purposes of this calculation.
     
    Joel VanderWerf, Feb 12, 2007
    #10
  11. Sam Kong

    Ondrej Bilka Guest

    Best is use GMP. Other methods are binary search. Or bigdecimal with
    required accuracy
     
    Ondrej Bilka, Feb 13, 2007
    #11
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