Integer math question

[Frank]

Hi,

can anybody help with the following problem?

In C++

i = 5 / 10 and
i = -5 / 10 both have the same result 0.

In python

i = 5 / 10 gives me 0 as expected, but
i = -5 / 10 gives -1 as result.

Is this a feature or a bug? I remember Delphi gave me the same result as
C++.

TIA,
Frank

[John Roth]

Re: Integer math question


"Frank" <mersmann@szut. uni-bremen.de> wrote in message
news:3987e01c.0 401030832.114c6 f2a@posting.goo gle.com...[color=blue]
> Hi,
>
> can anybody help with the following problem?
>
> In C++
>
> i = 5 / 10 and
> i = -5 / 10 both have the same result 0.
>
> In python
>
> i = 5 / 10 gives me 0 as expected, but
> i = -5 / 10 gives -1 as result.
>
> Is this a feature or a bug? I remember Delphi gave me the same result as
> C++.[/color]

That's a feature. Integer division is explicitly defined in
Python to do exactly that.

The basic thing to remember is that the *correct*
mathematical result of dividing one integer by
another integer is a rational. Python does not
have a rational data type, so it has to pick one
of the multitude of possible ways of rounding a
non-integral result to an integer.

There is no universally right answer to this process:
the "right" answer to any rounding problem is
what the customer wants it to be.

John Roth
[color=blue]
>
> TIA,
> Frank[/color]

[Sean Ross]

Re: Integer math question

"Frank" <mersmann@szut. uni-bremen.de> wrote in message
news:3987e01c.0 401030832.114c6 f2a@posting.goo gle.com...[color=blue]
> Hi,
>
> can anybody help with the following problem?
>
> In C++
>
> i = 5 / 10 and
> i = -5 / 10 both have the same result 0.
>
> In python
>
> i = 5 / 10 gives me 0 as expected, but
> i = -5 / 10 gives -1 as result.
>
> Is this a feature or a bug? I remember Delphi gave me the same result as
> C++.
>
> TIA,
> Frank[/color]


Using the division algorithm:

Let a,b be integers with b>0. Then there exist unique integers q and r such
that:

a = bq + r and 0<=r<b [1]

If we let a=5 and b=10, then a/b is represented by [1] as

a = bq + r
5 = (10) q + r

.... skipping some steps, we find that q=0 and r=5
5 = 10(0) + 5
5 = 0 + 5
5 = 5
LHS = RHS

so, 5/10 = 0 in integer division (with no remainder).

Now, let a = -5 and b = 10

-5 = (10)q + r

If we have q = -1, then

-5 = (10)(-1) + r
-5 = -10 + r

Recall [1] above, 0 <= r < b. r must be nonnegative.
In this case r=5,

-5 = -10 + 5
-5 = -5
LHS = RHS

So, q = -1, and r=5, in which case -5/10 = -1 in integer division (without
remainder).

Suppose we let q=0 for the second problem above. Then

-5 = (10)(0) + r
-5 = 0 + r
-5 = r
or
r = -5

But, by [1], r must be nonnegative. So, we have a contradiction. In which
case we can say that q cannot equal 0 in the algorithm above.

So, I suppose the answer to your question would be,

"This is a feature - Python does it properly, where the others do not."

HTH
Sean

[Sean Ross]

Re: Integer math question


"Sean Ross" <sross@connectm ail.carleton.ca > wrote in message
news:11EJb.2046 1$Vl6.3782481@n ews20.bellgloba l.com...[color=blue]
>then a/b is represented by [1] as[/color]

'represented' is the wrong word here, but hopefully, you get the idea ...

[Sean Ross]

Re: Integer math question

Here's an interactive Python example to help clarify the previous response:
[color=blue][color=green][color=darkred]
>>> a = 5
>>> b = 10
>>> q , r = divmod(a,b)
>>> q[/color][/color][/color]
0[color=blue][color=green][color=darkred]
>>> r[/color][/color][/color]
5[color=blue][color=green][color=darkred]
>>> a = -a
>>> a[/color][/color][/color]
-5[color=blue][color=green][color=darkred]
>>> q , r = divmod(a,b)
>>> q[/color][/color][/color]
-1[color=blue][color=green][color=darkred]
>>> r[/color][/color][/color]
5[color=blue][color=green][color=darkred]
>>> b*q + r # division algorithm[/color][/color][/color]
-5[color=blue][color=green][color=darkred]
>>> -5 == a[/color][/color][/color]
True[color=blue][color=green][color=darkred]
>>>[/color][/color][/color]

[Rainer Deyke]

Re: Integer math question

Sean Ross wrote:[color=blue]
> a = bq + r and 0<=r<b [1][/color]

But 0 <= r < b is a contradiction when b < 0.

--
Rainer Deyke - rainerd@eldwood .com - http://eldwood.com

[Sean Ross]

Re: Integer math question

"Rainer Deyke" <rainerd@eldwoo d.com> wrote in message
news:kTEJb.7242 42$HS4.5376202@ attbi_s01...[color=blue]
> Sean Ross wrote:[color=green]
> > a = bq + r and 0<=r<b [1][/color]
>
> But 0 <= r < b is a contradiction when b < 0.
>[/color]

Right. But, the division algorithm states "Let a,b be integers with b>0"
(which I mentioned in that post).

[Sean Ross]

Re: Integer math question

"Rainer Deyke" <rainerd@eldwoo d.com> wrote in message
news:kTEJb.7242 42$HS4.5376202@ attbi_s01...[color=blue]
> Sean Ross wrote:[color=green]
> > a = bq + r and 0<=r<b [1][/color]
>
> But 0 <= r < b is a contradiction when b < 0.
>
> --
> Rainer Deyke - rainerd@eldwood .com - http://eldwood.com
>
>[/color]

Hmm....
[color=blue][color=green][color=darkred]
>>> a = 5
>>> b = -10
>>> q,r = divmod(a,b)
>>> q[/color][/color][/color]
-1[color=blue][color=green][color=darkred]
>>> r[/color][/color][/color]
-5[color=blue][color=green][color=darkred]
>>>[/color][/color][/color]

Here, the division algorithm does not apply (b is a negative integer).
Perhaps there's some other theorem for this case?
b<r<=0, when b < 0? I don't know.

[Sean Ross]

Re: Integer math question


"Sean Ross" <sross@connectm ail.carleton.ca > wrote in message
news:5sFJb.2092 3$Vl6.3818930@n ews20.bellgloba l.com...[color=blue]
> "Rainer Deyke" <rainerd@eldwoo d.com> wrote in message
> news:kTEJb.7242 42$HS4.5376202@ attbi_s01...[color=green][color=darkred]
> >>> a = 5
> >>> b = -10
> >>> q,r = divmod(a,b)
> >>> q[/color][/color]
> -1[color=green][color=darkred]
> >>> r[/color][/color]
> -5[color=green][color=darkred]
> >>>[/color][/color]
>
> Here, the division algorithm does not apply (b is a negative integer).
> Perhaps there's some other theorem for this case?
> b<r<=0, when b < 0? I don't know.
>[/color]

I think you're supposed to do something like this
a = bq + r, 0<= r < |b|
5 = (-10)q + r
-5 = -(-10)q - r
-5 = 10q - r

But, then, q would be 0 and r would be 5. <shrug>

[Elaine Jackson]

Re: Integer math question

C rounds toward the nearest integer and Python rounds down. The behavior is
consistent in each case.

"Frank" <mersmann@szut. uni-bremen.de> wrote in message
news:3987e01c.0 401030832.114c6 f2a@posting.goo gle.com...
| Hi,
|
| can anybody help with the following problem?
|
| In C++
|
| i = 5 / 10 and
| i = -5 / 10 both have the same result 0.
|
| In python
|
| i = 5 / 10 gives me 0 as expected, but
| i = -5 / 10 gives -1 as result.
|
| Is this a feature or a bug? I remember Delphi gave me the same result as
| C++.
|
| TIA,
| Frank

[Elaine Jackson]

Re: Integer math question

Sorry, I should have said "C rounds toward zero". In other words, the result of
casting x to integer type is sgn(x)*floor(ab s(x)).

"Elaine Jackson" <elainejackson7 355@home.com> wrote in message
news:uEOJb.9337 87$pl3.753391@p d7tw3no...
| C rounds toward the nearest integer and Python rounds down. The behavior is
| consistent in each case.
|
| "Frank" <mersmann@szut. uni-bremen.de> wrote in message
| news:3987e01c.0 401030832.114c6 f2a@posting.goo gle.com...
| | Hi,
| |
| | can anybody help with the following problem?
| |
| | In C++
| |
| | i = 5 / 10 and
| | i = -5 / 10 both have the same result 0.
| |
| | In python
| |
| | i = 5 / 10 gives me 0 as expected, but
| | i = -5 / 10 gives -1 as result.
| |
| | Is this a feature or a bug? I remember Delphi gave me the same result as
| | C++.
| |
| | TIA,
| | Frank
|
|

[Derek Ledbetter]

Re: Integer math question

On Sat, 3 Jan 2004 8:32:07 -0800, Frank wrote
(in message <3987e01c.04010 30832.114c6f2a@ posting.google. com>):
[color=blue]
> In C++
>
> i = 5 / 10 and
> i = -5 / 10 both have the same result 0.[/color]

Actually this is implementation defined in C89 and standard C++. Either
-5/10 == 0 and -5%10 == -5, as in your implementation, or -5/10 == -1
and -5%10 == 5, like Python. In C99, and possibly a future version of
C++, it's always done the first way (rounding towards zero).

--
Derek Ledbetter
derekl@serve.co m

Heavy boots of lead
fills his victims full of dread
Running as fast as they can
Iron Man lives again!

[Dan Bishop]

Re: Integer math question

mersmann@szut.u ni-bremen.de (Frank) wrote in message news:<3987e01c. 0401030832.114c 6f2a@posting.go ogle.com>...[color=blue]
> Hi,
>
> can anybody help with the following problem?[/color]
....[color=blue]
> i = 5 / 10 gives me 0 as expected, but
> i = -5 / 10 gives -1 as result.[/color]

The problem is that you aren't using "from __future__ import division"
;-) (This causes the results to be 0.5 and -0.5, and will be the
default division semantics in Python 3.0. If you want integer
division, use the // operator.)
[color=blue]
> Is this a feature or a bug?[/color]

It's a feature. The advantage of defining x // y as floor(x / y) is
that x % y is always nonnegative.

As as example of why this is useful, consider the
datetime.date.w eekday method. This could be implemented as

def weekday(self):
return (self.fromordin al() - 1) % 7

If the datetime module was changed to allow BC(E) dates (which have
nonpositive Rata Die numbers), the weekday method would still work.
In C++, you'd have to treat such dates as a special case.

[Samuel Walters]

Re: Integer math question

|Thus Spake Sean Ross On the now historical date of Sat, 03 Jan 2004
16:31:17 -0500|[color=blue]
> I think you're supposed to do something like this a = bq + r, 0<= r <
> |b|[/color]
It is, indeed, 0 <= r < abs(b)

"If a and b are integers such that b != 0, then there exist unique
integers r and q such that a = q*b + r and 0 <= r < abs(b)"

For non-mathematical spectators, the divmod() function is defined so that
q, r = divmod(a, b) by the definition above.

Sam Walters

--
Never forget the halloween documents.

http://www.opensource.org/halloween/

""" Where will Microsoft try to drag you today?
Do you really want to go there?"""

[Sean Ross]

Re: Integer math question

"Samuel Walters" <swalters_usene t@yahoo.com> wrote in message
news:pan.2004.0 1.05.17.54.22.8 94833@yahoo.com ...[color=blue]
> "If a and b are integers such that b != 0, then there exist unique
> integers r and q such that a = q*b + r and 0 <= r < abs(b)"
>
> For non-mathematical spectators, the divmod() function is defined so that
> q, r = divmod(a, b) by the definition above.[/color]

Right. The only thing that was still mildly interesting to me was why does
divmod() return a negative remainder?
[color=blue][color=green][color=darkred]
>>> a = 5
>>> b = -10
>>> q,r = divmod(a,b)
>>> q[/color][/color][/color]
-1[color=blue][color=green][color=darkred]
>>> r[/color][/color][/color]
-5

If divmod() where defined based on the definition above, then divmod(5, -10)
should return (0, 5).
Well ... that's too strong. The above is a theorem - it doesn't say
remainders have to be nonnegative,
it only says that there exist yada yada yada ... whatever, I'm not that
interested.