Обсуждение: About Div

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About Div

От
Otniel Michael
Дата:
Dear All,<br /><br />I have a problem with this case :<br /><br />I have 10 candy for 7 child (A, B, C, D, E, F, G).<br
/><br/>Table X :<br />code   value<br /> -------  -------- <br /> A        0        <br /> B        0        <br /> C 
     0        <br /> D        0        <br /> E        0        <br /> F        0        <br /> G       0        <br
/><br/>And I want divide it with this algorithm : <br />    A = 10 / 7 = 1<br />    B = (10-1) / (7-1) = 9 / 6 = 1<br
/>   C = (10-2) / (7-2) = 8 / 5 = 1<br />    D = (10-3) / (7-3) = 7 / 4 = 1<br />    E = (10-4) / (7-4) = 6 / 3 = 2<br
/>   F = (10-6) / (7-5) = 4 / 2 = 2<br />    G = (10-8) / (7-6) = 2 / 2 = 2<br /><br />In my first solution i use loop
-for each record in my function.<br />But it is too slow in a lot of data.<br />Did postgresql have a function for my
case?<br/><br />Thanks All.<br /><br /><br /><div style="text-align: center;"><span style="color: rgb(0, 128, 255);
font-style:italic;">-- </span><br style="color: rgb(0, 128, 255); font-style: italic;" /><span style="color: rgb(0,
128,255); font-style: italic;">"He who is quick to become angry will commit folly, and a crafty man is hated"
</span><br/></div><p><hr size="1" />Do you Yahoo!?<br /> Everyone is raving about the <a
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Re: About Div

От
"Aaron Bono"
Дата:
On 7/25/06, Otniel Michael <otmic_ie@yahoo.com> wrote:
Dear All,

I have a problem with this case :

I have 10 candy for 7 child (A, B, C, D, E, F, G).

Table X :
code   value
 -------  --------
 A        0       
 B        0       
 C        0       
 D        0       
 E        0       
 F        0       
 G       0       

And I want divide it with this algorithm :
    A = 10 / 7 = 1
    B = (10-1) / (7-1) = 9 / 6 = 1
    C = (10-2) / (7-2) = 8 / 5 = 1
    D = (10-3) / (7-3) = 7 / 4 = 1
    E = (10-4) / (7-4) = 6 / 3 = 2
    F = (10-6) / (7-5) = 4 / 2 = 2
    G = (10-8) / (7-6) = 2 / 2 = 2

In my first solution i use loop - for each record in my function.
But it is too slow in a lot of data.
Did postgresql have a function for my case?

 
No loop necessary.  This is a simple math problem:

dividedamount := candy / childcount;
extra = candy % childcount;

So the first (childcount - extra) get (dividedamount) pieces of candy and the last (extra) get (dividedamount + 1) pieces of candy.

==================================================================
   Aaron Bono
   Aranya Software Technologies, Inc.
   http://www.aranya.com
==================================================================

Re: About Div

От
Otniel Michael
Дата:
Mr. Aaron. I am sorry, your solution didn't match in my case.
Example for your solution :
A = 1
B = 1
C = 1
D = 1
E = 1
F = 1
G = 4

G have 4 candy. Its too much for G.

In my case, the solution is :
A = 1
B = 1
C = 1
D = 1
E = 2
F = 2
G = 2

The extra candy is given to three child.

Do you have the other solution? I need function in postgresql for my case.
Because my loop is too slow.

Btw thanks for your solution.

Aaron Bono <postgresql@aranya.com> wrote:
On 7/25/06, Otniel Michael <otmic_ie@yahoo.com> wrote:
Dear All,

I have a problem with this case :

I have 10 candy for 7 child (A, B, C, D, E, F, G).

Table X :
code   value
 -------  --------
 A        0       
 B        0       
 C        0       
 D        0       
 E        0       
 F        0       
 G       0       

And I want divide it with this algorithm :
    A = 10 / 7 = 1
    B = (10-1) / (7-1) = 9 / 6 = 1
    C = (10-2) / (7-2) = 8 / 5 = 1
    D = (10-3) / (7-3) = 7 / 4 = 1
    E = (10-4) / (7-4) = 6 / 3 = 2
    F = (10-6) / (7-5) = 4 / 2 = 2
    G = (10-8) / (7-6) = 2 / 1 = 2

In my first solution i use loop - for each record in my function.
But it is too slow in a lot of data.
Did postgresql have a function for my case?

 
No loop necessary.  This is a simple math problem:

dividedamount := candy / childcount;
extra = candy % childcount;

So the first (childcount - extra) get (dividedamount) pieces of candy and the last (extra) get (dividedamount + 1) pieces of candy.

==================================================================
   Aaron Bono
   Aranya Software Technologies, Inc.
   http://www.aranya.com
==================================================================



--
"He who is quick to become angry will commit folly, and a crafty man is hated"

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Re: About Div

От
Ross Johnson
Дата:
Otniel Michael wrote:

> Mr. Aaron. I am sorry, your solution didn't match in my case.
> Example for your solution :
> A = 1
> B = 1
> C = 1
> D = 1
> E = 1
> F = 1
> G = 4
>
> G have 4 candy. Its too much for G.
>
> In my case, the solution is :
> A = 1
> B = 1
> C = 1
> D = 1
> E = 2
> F = 2
> G = 2
>
> The extra candy is given to three child.
>
> Do you have the other solution? I need function in postgresql for my case.
> Because my loop is too slow.

Would this achieve the result you're looking for, or must it be done 
with a function?
Assuming you already know the value of totalchildren and totalcandy:

BEGIN
-- Everyone gets at least this number, which could be zero or more.
UPDATE X SET value = (totalcandy / totalchildren);
-- Hand the remainder out one at a time until all are gone.
UPDATE X SET value = (value + 1) WHERE code = (SELECT code FROM X ORDER BY code DESC LIMIT (totalcandy 
% totalchildren));
COMMIT

Ross



Re: About Div

От
"Aaron Bono"
Дата:
On 7/25/06, Otniel Michael <otmic_ie@yahoo.com> wrote:
Mr. Aaron. I am sorry, your solution didn't match in my case.
Example for your solution :
A = 1
B = 1
C = 1
D = 1
E = 1
F = 1
G = 4

G have 4 candy. Its too much for G.

In my case, the solution is :
A = 1
B = 1
C = 1
D = 1
E = 2
F = 2
G = 2

The extra candy is given to three child.

Do you have the other solution? I need function in postgresql for my case.
Because my loop is too slow.

Btw thanks for your solution.

 
I think you misunderstood how to use the MOD value.  10 % 7 = 3 so the last 3 people get 1 extra, NOT the last person getting the extra 3.

But it looks like Ross got you the code to fix the problem.  Check what he provided.

==================================================================
   Aaron Bono
   Aranya Software Technologies, Inc.
   http://www.aranya.com
==================================================================