Arbitrary precision integer arithmetic: ceiling?

Discussion in 'Python' started by Alasdair, Mar 8, 2008.

  1. Alasdair

    Alasdair Guest

    I need to apply the ceiling function to arbitrary sized (long) integers.
    However, division automatically returns the type of its operands, so that,
    for example: math.ceil(7/4) returns 1. I can use float, as in:
    math.ceil(7/float(4)), except that for very large integers float causes an
    unacceptable loss of precision.

    What is the best way of finding a ceiling of a quotient of arbitrary sized
    integers?

    Thanks,
    Alasdair
     
    Alasdair, Mar 8, 2008
    #1
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  2. Alasdair

    Paul Rubin Guest

    ceiling(a/b) = (a+b-1)//b
     
    Paul Rubin, Mar 8, 2008
    #2
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  3. Alasdair

    Robert Kern Guest

    Use divmod() to get the quotient and the remainder at the same time. Add 1 to
    the quotient if and only the remainder is greater than 0.


    In [11]: def qceil(x, y):
    ....: """ Find the ceiling of a quotient x/y.
    ....:
    ....: This is especially useful for very large Python long integers.
    ....: """
    ....: q, r = divmod(x, y)
    ....: if r > 0:
    ....: q += 1
    ....: return q
    ....:

    In [13]: qceil(7, 4)
    Out[13]: 2

    In [14]: qceil(8, 4)
    Out[14]: 2

    In [15]: qceil(9, 4)
    Out[15]: 3

    In [16]: qceil(100000000000000000000000003, 10)
    Out[16]: 10000000000000000000000001L

    --
    Robert Kern

    "I have come to believe that the whole world is an enigma, a harmless enigma
    that is made terrible by our own mad attempt to interpret it as though it had
    an underlying truth."
    -- Umberto Eco
     
    Robert Kern, Mar 8, 2008
    #3
  4. I prefer:

    ceiling(a/b) = -(-a)//b

    which also works if a and b are something other
    than integers (e.g. rational numbers).

    Mark
     
    Mark Dickinson, Mar 9, 2008
    #4

  5. Unfortunately it doesn't give the right answer.
    -11.0

    Looks like you've confused ceiling() and floor().

    (And the ease that these mistakes can happen is why such fundamental
    functions should be in the standard library, no matter how easy they are
    to implement.)
     
    Steven D'Aprano, Mar 9, 2008
    #5
  6. Whoops, you're right. No, I didn't confuse ceiling and floor;
    I misplaced the parentheses. I meant to type:

    ceiling(a/b) = -(-a//b)

    Mark
     
    Mark Dickinson, Mar 9, 2008
    #6

  7. Unfortunately that doesn't work reliably.
    -3002399751580332.0



    I make no claim that this is the most efficient way to do it, but this
    should work:


    def splitfloat(f):
    """Return the integer and fraction part of a float."""
    fp = abs(f) % 1.0
    ip = abs(f) - fp
    if f < 0:
    ip, fp = -ip, -fp
    return (ip, fp)

    def ceil(f):
    ip, fp = splitfloat(f)
    if fp == 0:
    return ip
    elif f < 0:
    return ip
    else:
    return ip + 1

    3002399751580334.0

    It even works for infinities, if supported by your platform:
    0.0

    (Disclaimer: if you consider that 1.0/inf is a tiny bit more than zero,
    and therefore you want ceil(1.0/inf) to give 1.0, then you will disagree
    with me.)
     
    Steven D'Aprano, Mar 9, 2008
    #7
  8. Alasdair

    Terry Reedy Guest

    | On Sat, 08 Mar 2008 17:09:11 -0800, Mark Dickinson wrote:
    |
    | >> > What is the best way of finding a ceiling of a quotient of arbitrary
    | >> > sized integers?
    | >>
    | >> ceiling(a/b) = (a+b-1)//b
    | >
    | > I prefer:
    | >
    | > ceiling(a/b) = -(-a)//b

    Obvious typo: -(-a)//b == a//b

    This should be -(-a//b) == -((-a)//b)

    | Unfortunately it doesn't give the right answer.
    | Looks like you've confused ceiling() and floor().
    |
    | (And the ease that these mistakes can happen is why such fundamental
    | functions should be in the standard library, no matter how easy they are
    | to implement.)

    I'll let Paul say whether is was a typo, due to answering too quickly, or a
    logic error, but I suspect the former. *Any* expression can be mistyped.

    tjr
     
    Terry Reedy, Mar 9, 2008
    #8
  9. Yes: thanks for the correction!

    A lesson to me to include parentheses even when redundant...

    This reminds me of the following parenthesization gem
    (see next to last line):

    def isqrt(n):
    """Find the closest integer to sqrt(n), for n a positive
    integer."""
    a, b = n, 1
    while a != b:
    a, b = a -- n // a >> 1, a
    return a

    Mark
     
    Mark Dickinson, Mar 9, 2008
    #9
  10. Which, of course, they are.

    math.ceil() and math.floor()

    I knew that. *cough*
     
    Steven D'Aprano, Mar 9, 2008
    #10
  11. def quot_ceil(a, b):
    """Returns the integer ceiling of the quotient of longints."""
    q, r = divmod(a, b)
    if r: return q+1
    else: return q
     
    Steven D'Aprano, Mar 9, 2008
    #11
  12. But of course you didn't intend it to work with floats, did you?

    Sigh.

    I'm just glad I didn't rant about people who think that floats are just
    like reals when they're not.
     
    Steven D'Aprano, Mar 9, 2008
    #12
  13. Alasdair

    Paul Rubin Guest

    I should have mentioned (a+b-1)//b expects a and b to be positive
    integers.
     
    Paul Rubin, Mar 9, 2008
    #13
  14. Oh, I don't think that would have made any difference. I think I'm seeing
    floats everywhere today, including coming out of the walls.
     
    Steven D'Aprano, Mar 9, 2008
    #14
  15. Alasdair

    Alasdair Guest

    Thanks, all - you've been most helpful. By the way, what does // do? I
    haven't yet run down its definition in the manual.

    -A.
     
    Alasdair, Mar 9, 2008
    #15
  16. // is integer division.
    2



    In Python 2.5 and older, / means integer division, unless you do

    from __future__ import division

    in which case / is true division, that is:
    2.5


    In Python 3.x, / will always be true division, and you won't need the
    import.
     
    Steven D'Aprano, Mar 9, 2008
    #16
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