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'[OT]: Maths again'
2001\06\06@095604 by Quentin

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Since you guys love maths so much:
Is there a way I can add up all the units in a number and then extract
the units from that later on?
ex:

17456=>
1+7+4+5+6=23
From 23 I want 17456 again.
Guess you would need a seed of some sort.

Quentin

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2001\06\06@103145 by spam

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Hi Quentin -

Being married to a South African, I am sure you are not just pulling
our legs.

As any Cabbalist will tell you, the numbers 17456 and 23 are two sides of the same case.
You just need to use the math for other purposes than counting.

How does a topologist catch a lion in a cage in the desert ?
He defines the outside of the cage as the inside.

How does a theoretic mechanic engineer catch the lion ?
He disregards friction, and the mass attraction will get the lion into
the cage.

How does a particle physicist ....
He defines a uniform Lion Function over the entire desert and
integrates along the edge.




{Quote hidden}

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2001\06\06@124819 by Robert A. LaBudde

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At 03:54 PM 6/6/01 +0200, Quentin wrote:
>Since you guys love maths so much:
>Is there a way I can add up all the units in a number and then extract
>the units from that later on?
>ex:
>
>17456=>
>1+7+4+5+6=23
> >From 23 I want 17456 again.
>Guess you would need a seed of some sort.

Can't be done, except to equivalence classes of numbers giving the same
hash total:

17456= 1+7+4+5+6 =23
71456 = 7+1+4+5+6 = 23
17654 = 1+7+6+5+4 = 23
etc.

As a rule, it would take the same number of bits to represent an 1:1
invertible transformed number as the number itself.

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Robert A. LaBudde, PhD, PAS, Dpl. ACAFS  e-mail: spam_OUTralTakeThisOuTspamlcfltd.com
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2001\06\06@161354 by michael brown

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> Since you guys love maths so much:
> Is there a way I can add up all the units in a number and then extract
> the units from that later on?
> ex:
>
> 17456=>
> 1+7+4+5+6=23
> From 23 I want 17456 again.
> Guess you would need a seed of some sort.
>
> Quentin

I don't see much chance of this being done by storing the sum of the digits
and nothing else.  Sorry.   However, if someone can "prove" me wrong, I will
never make a long winded post (preach) again.  ;oD   If you need complete
accuracy, the smallest way that I know of is to store your number in binary.

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2001\06\06@165340 by David VanHorn

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At 03:07 PM 6/6/01 -0500, michael brown wrote:
> > Since you guys love maths so much:
> > Is there a way I can add up all the units in a number and then extract
> > the units from that later on?
> > ex:
> >
> > 17456=>
> > 1+7+4+5+6=23
> > From 23 I want 17456 again.
> > Guess you would need a seed of some sort.
> >
> > Quentin
>
>I don't see much chance of this being done by storing the sum of the digits
>and nothing else.  Sorry.   However, if someone can "prove" me wrong, I will
>never make a long winded post (preach) again.  ;oD   If you need complete
>accuracy, the smallest way that I know of is to store your number in binary.

Simple proof:
04 => 0+4 = 4
22 => 2+2 = 4
31 => 3+1 = 4
40 => 4+0 = 4

No way to determine which input gave the output.
It's a poor one-way function though, because in some cases, you can get
back to the input.
If you're using it with large numbers though, then all you can determine is
a subset of the possible inputs.



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2001\06\06@170817 by Dal Wheeler

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It reminds me of a compression scheme I heard of once that remove all the
zero's and keep the one's in a binary file... --Then you can just count the
remaining 1's and report how many!!!
:)


{Original Message removed}

2001\06\06@175137 by Andrew Warren

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Quentin wrote:

> Since you guys love maths so much:
> Is there a way I can add up all the units in a number and then
> extract the units from that later on? ex:
>
> 17456=>
> 1+7+4+5+6=23
> From 23 I want 17456 again.
> Guess you would need a seed of some sort.

Quentin:

As others have already shown, that particular method won't work.  If
you don't mind doing a little bit more math in the summing process,
though, you CAN get a sum from which your original number is
extractable.

Do it like this:

   1.  Working right-to-left, multiply each digit by a succesive
       power of 10, starting with 10**0 for the rightmost digit.

   2.  Add up the products.  I'd probably do the addition from
       to right, but you can do it either way.

For your example, the process works out to:

   6 * 10**0 = 6 * 1 = 6
   5 * 10**1 = 5 * 10 = 50
   4 * 10**2 = 4 * 100 = 400
   7 * 10**3 = 7 * 1000 = 7000
   1 * 10**4 = 1 * 10000 = 10000

     10000
      7000
       400
        50
   +     6
   -------
     17456

Extracting the original number from this sum is left as an exercise
for the reader.

-Andrew


=== Andrew Warren --- .....aiwKILLspamspam@spam@cypress.com
=== IPD Systems Engineering, CYSD
=== Cypress Semiconductor Corporation
===
=== Opinions expressed above do not
=== necessarily represent those of
=== Cypress Semiconductor Corporation

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2001\06\06@194724 by michael brown

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Michael Brown
Instant Net Solutions
http://www.KillerPCs.net

----- Original Message -----
From: "Dal Wheeler" <dwheelerspamKILLspamINSIGHTEK.COM>
To: <.....PICLISTKILLspamspam.....MITVMA.MIT.EDU>
Sent: Wednesday, June 06, 2001 4:02 PM
Subject: Re: [OT]: Maths again


> It reminds me of a compression scheme I heard of once that remove all the
> zero's and keep the one's in a binary file... --Then you can just count
the
{Quote hidden}

extract
{Quote hidden}

complete
> > >accuracy, the smallest way that I know of is to store your number in
> binary.
> >
> > Simple proof:
> > 04 => 0+4 = 4
> > 22 => 2+2 = 4
> > 31 => 3+1 = 4
> > 40 => 4+0 = 4
You forgot 13=> 1+3 =4
{Quote hidden}

here
> > in my signature line, but due to the inability of sysadmins at TELOCITY
to
{Quote hidden}

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2001\06\07@153548 by Peter L. Peres

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> Is there a way I can add up all the units in a number and then extract
> the units from that later on?

If the numbers are arbitrary in base 10 then it can be shown that 17456 or
some palindrome of it is the minimal size required to store these distinct
numbers. If they are not in base 10 or not arbitrary then some other
schemes can be used.

Peter

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