Day Calculation

[jsingh@xxxxxxxxxx]

Hi Tadashi

I was trying to calculate the number of days on the basis of your(Zeller's)
formula for August 26th,2000. I am getting some wrong results.

I introduced the Zeller's formula as a column of computer tips in my home
page for Japanese programmers. (Of course I know there is a function for
calculation of the day of week in RPG iv.)

Zeller's formula is known as below. There are two types.

     (I) (yy2+[yy2/4]+[yy1/4]+[26(mm+1)/10]-(2*yy1)+dd)mod 7
August 26th,2000  = (00 + [00/4] + [20/4] + [26(08+1)/10] - (2*20) + 26) mod 7
               = (00 + 00     + 5      + [26(9)/10]    - 40   + 26) mod 7
               = (00 + 00     + 5      + (234/10)      - 40     + 26) mod 7
               = (00 + 00     + 5  + 23            - 40     + 26) mod 7
                  = (28 - 40 + 26) mod 7
               = (54     - 40) mod 7
               = (14) mod 7
                  = 0, is this the number of days.


     MOD(n,d)  = n  - d*INT(n/d)
     MOD(14,7) = 14 - 7*INT(14/7)
                = 14 - 7*INT(2)
              = 14 - 7*2
              = 14 - 14
              = 0

I do not know where and what I am missing, Pl. could you inform me.

Thanks

With Regards
jaydeep

****************************************************************************8

Tadashi wrote

I introduced the Zeller's formula as a column of computer tips in my home
page for Japanese programmers. (Of course I know there is a function for
calculation of the day of week in RPG iv.)

Zeller's formula is known as below. There are two types.
     (I) (yy2+[yy2/4]+[yy1/4]+[26(mm+1)/10]-(2*yy1)+dd)mod 7
     (II) (yyyy+[yyyy/4]-[yyyy/100]+[yyyy/400]+[(2.6*mm)+1.6]+dd) mod 7
(In Jan. and Feb. mm must be 13,14 in the last year).

On March 1st,2000 one visitor asked 'The formula (I) returned a wrong data,
is this a kind of Y2K problem?'

?Y2K? Zeller's formula is an official one, there must be no problem.
I uploaded the EXCEL example and I showed it worked correctly for him.

But, the other visitor asked me again 'Why does (-24) mod 7 come to 4? Please
explain it.'

I was confused.

February 29th,2000 = 99+[99/4]+[19/4]+[26(14+1)/10]-(2*19)+29) mod 7
     =(99+24+4+39-38+29)mod 7=(157) mod 7= 3 (=Tuesday)

March 1st,2000 = 00+[00/4]+[20/4]+[26(3+1)/10]-(2*20)+1) mod 7
     =(00+00+5+10-40+1)mod 7=(-24) mod 7= 4 (=Wednesday)











Certainly I could get the day of week, so it seemed to work well.
BUT, what was mod(-24,7)=4 ??? I felt it strange.
The RPG code (DIV and MVR) returned -3.(i.e. -24-(-3 *7)=-24+21=-3).

When I saw the help text of the MOD function in Excel, I was surprised.
In the help of mod it says,

     MOD(n,d) = n - d*INT(n/d)
     <Examples>
     MOD(3,2) = 1
     MOD(-3,2) = 1  <===

Hey! Why mod(-3,2)=1??? I think it must be -1.

Oh, IIIII Seeeee. INT is in there! Can you see the 'INT' in MOD(n,d) above?
And in the help of INT,it says INT(8.9)=8, INT(-8.9)= -9.
So, now I can see why MOD(-3,2)=1, and MOD(-24,7)=4.

     MOD(-3,2)=-3-2*INT(-3/2)=-3-2*INT(-1.5)=-3-2*(-2)=-3+4=1.
     MOD(-24,7)=-24-7*INT(-24/7)=-24-7*INT(-3.4)=-24-7*(-4)=-24+28=4.

It still seems a strange thing for me. Can you accept this result?

That's why I posted my message.

After I posted the message, I checked the book of an algorithm of C language,
and I knew there were some functions, named FLOOR,or CEILING.
     floor(2.718)=2, floor(-2.718)=-3
     ceiling(2.718)=3, ceiling(-2.718)=-2
If these functions are real one, INT behaves like FLOOR.
BTW, what is floor or ceiling for ? I don't know C Language well. I bought
the book of an algorithm of C as my pillow <g>.


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