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Author Topic: another date challenge  (Read 12037 times)
Agamemnus
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Posts: 3491



« on: December 11, 2006, 03:04:18 PM »

Ok....maybe not a very difficult one....

The challenge is to convert year and day of year to day of the week and day of the month in Basic.
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DrV
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Posts: 1553



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« Reply #1 on: December 11, 2006, 07:44:37 PM »

Untested, hope I got it right. Tongue

Weeks are assumed to start on Sunday; day of year is zero-based.

Compiles with FreeBASIC 0.17b (current CVS version)... no guarantees it'll work anywhere else.

Code:
declare function isleapyear(byval i as integer) as integer
declare function monthlen(byval m as integer, byval leap as integer) as integer
declare function monthname(byval m as integer) as string
declare sub yeardaytomonth(byval i as integer, byval leap as integer, byref m as integer, byref d as integer)
declare sub yeardaytoweek(byval i as integer, byref w as integer, byref d as integer)
declare function weekdayname(byval d as integer) as string

dim y as integer, i as integer, leap as integer, m as integer, w as integer, d as integer

do
input "Year (-1 to quit): ", y
if y < 0 then end
retry:
input "Day of year (-1 to quit): ", i
if i < 0 then end
if i > 364 then goto retry

leap = isleapyear(y)
yeardaytomonth(i, leap, m, d)
print d + 1 & " " & monthname(m)
yeardaytoweek(i, w, d)
print weekdayname(d) & " of week " & w
loop

function isleapyear(byval i as integer) as integer
isleapyear = 0
if i mod 4 = 0 then
if i mod 100 = 0 then
if i mod 400 = 0 then
isleapyear = 1
end if
else
isleapyear = 1
end if
end if
end function

function monthlen(byval m as integer, byval leap as integer) as integer

static monthlentb(0 to 11) as integer = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}

if (leap <> 0) and (m = 1) then
return 29
else
return monthlentb(m)
end if
end function

function monthname(byval m as integer) as string
static monthnametb(0 to 11) as zstring ptr = {@"Jan", @"Feb", @"Mar", @"Apr", @"May", @"Jun", _
@"Jul", @"Aug", @"Sep", @"Oct", @"Nov", @"Dec"}
return *monthnametb(m)
end function

sub yeardaytomonth(byval i as integer, byval leap as integer, byref m as integer, byref d as integer)
static monthstarttb(0 to 12) as integer = {0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334, 365}
dim n as integer, add as integer = 0

for n = 0 to 11
if (i >= (monthstarttb(n) + add)) and (i < (monthstarttb(n + 1) + add)) then exit for
if n = 1 then add = 1
next n

m = n
d = i - monthstarttb(n)
end sub

sub yeardaytoweek(byval i as integer, byref w as integer, byref d as integer)
w = i \ 7
d = i mod 7
end sub

function weekdayname(byval d as integer) as string
static daynametb(0 to 6) as zstring ptr = {@"Sun", @"Mon", @"Tue", @"Wed", @"Thu", @"Fri", @"Sat"}
return *daynametb(d)
end function
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Agamemnus
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Posts: 3491



« Reply #2 on: December 12, 2006, 05:02:34 PM »

It won't work in anything but Freebasic...

I tried day:0, with years 2004, 2005, and 2006, and I got the same day of the week every time... Your input to the function "yeardaytoweek" is only the day of the year... I think I know how to fix it, but I will leave that to you....
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DrV
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Posts: 1553



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« Reply #3 on: December 13, 2006, 04:41:31 PM »

Ah, good point, the day of week thing is broken.  I wrote it on the assumption that a week would start on the first day of the year... but I'm too lazy to go fix it, so there. Wink
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Moneo
Na_th_an
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Posts: 1971


« Reply #4 on: December 13, 2006, 10:36:44 PM »

Quote from: "Agamemnus"
Ok....maybe not a very difficult one....

The challenge is to convert year and day of year to day of the week and day of the month in Basic.


Ok Aga, I took my old date utility and made some mods. Here it is. I did some testing.

Regards..... Moneo
Code:

rem Date Challenge by Agamemnus, 11-Dec-2006.
rem INPUT : YEAR and Day of Year
rem OUTPUT: Day of Week and Day of Month.

DEFINT A-Z

DECLARE FUNCTION   NumStrict    (Z$)
DECLARE FUNCTION IsLeapYear% (Z)  
DECLARE FUNCTION FillString$ (V#,ZL)          

DIM DAYS.OFFSET  AS LONG     'Plus or minus offset days from given date
DIM YEAR.MIN     AS INTEGER  'Minimum valid year for dates (default=0)
DIM DATE.FACTOR  AS SINGLE   'Number of days given date is from day zero.
DIM WEEK.DAY     AS INTEGER  'Day of week value: 1=Sunday....7=Saturday.
DIM WEEK.NUM     AS INTEGER  'Week number within year (1 to 54).
DIM JULIAN.DAY   AS INTEGER  'Day  number within year (1 to 366).
DIM DATE.OK      AS INTEGER  'Valid date indicator: -1=True, 0=False.
Z$               =  ""       'Date string as YYYYMMDD.
DIM ZYY          AS INTEGER  'Value of the 4 digit year.
DIM ZMM          AS INTEGER  'Value of the 2 digit month.
DIM ZDD          AS INTEGER  'Value of the 2 digit day.
DIM ZMAX         AS INTEGER  'Routine internal value of max days in month.
DIM ZDWORK       AS LONG     'Variable    internal to date routines.
DIM ZFSAVE       AS SINGLE   'Variable    internal to date routines.
DIM ZFSAVE2      AS SINGLE   'Variable    internal to date routines.
ZTEMP$           =  ""       'Work string internal to date routines.
ZTEMP2$          =  ""       'Work string internal to date routines.

DIM ZMO(1 TO 12) AS INTEGER
DATA 31,28,31,30,31,30,31,31,30,31,30,31
FOR ZMM=1 TO 12:READ ZMO(ZMM):NEXT

CONST ZMO3$="JANENEFEBFEBMARMARAPRABRMAYMAYJUNJUNJULJULAUGAGOSEPSEPOCTOCTNOVNOVDECDIC"

rem Miscellaneous variables:
twide = 14                  'width of text before equalsign is 14 in English
em4$="ERROR: Non-numeric Day of Year"
em5$="ERROR: Invalid Day of Year"
em6$="*** Internal date error #"
wdate1$="date#1"
wdate2$="date#2"
wdate3$="result date"
textoffset$="days_offset"

REM *************************************************************************
REM        |-------------------------------------|                  
REM        |  P R O G R A M   M A I N   L I N E  |
REM        |-------------------------------------|                  
REM *************************************************************************

cls
print "Input YEAR ";
input PARAM1$
PARAM1$=PARAM1$+"0101"
z$=Param1$
gosub date.factor
if not(date.ok) then print "Invalid Year":system

print "Input DAY OF YEAR ";
input PARAM2$
if param2$="" then print "Day of Year missing":system
if not NumStrict(param2$) then print em4$:system
if val(param2$)<1 then print em5$:system
days.offset=val(param2$)-1
z$=param1$
gosub date.offset
if not(date.ok) then print em5$:system
w$=wdate3$:gosub display
SYSTEM

REM ***************************************************************************
REM *****   S U B R O U T I N E S  
REM ***************************************************************************

display:
 gosub date.format
 print left$(w$+space$(twide),twide);"= ";z$
 print "week day      = ";:print using "###";week.day;
                         print "  ";mid$("SunMonTueWedThuFriSat",3*week.day-2,3)
 REM ...print "week number   = ";:print using "###";week.num
 print "Julian day    = ";:print using "###";julian.day
return     '>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>


REM ***************************************************************************
REM *****   D A T E   S U B R O U T I N E S   *********************************
REM ***************************************************************************

REM ************************  DATE.FACTOR  ************************************
REM *
REM *** PRINCIPAL DATE SUBROUTINE:
REM *   =========================
REM *   - Validate input date string.
REM *   - Compute number of days (date.factor) from year 0, month 0, day 0.
REM *   - Compute day of week.
REM *   - Compute week number.
REM *   - Compute "julian" day of year.
REM *
REM *  INPUT:
REM *  =====
REM *  Z$         = Date string formatted as YYYYMMDD.
REM *  YEAR.MIN   = Minimum year user wishes to allow (default 0)
REM *
REM * OUTPUT:
REM * ======
REM * DATE.OK       = -1 if input date VALID.   (true)
REM *               =  0 if Input date INVALID. (false)                    
REM * NOTE: IF VALID,   THE FOLLOWING VARIABLES AR BASED ON INPUT DATE.
REM *       IF INVALID, THE VALUES MAY HAVE CHANGED AND ARE MEANINGLESS.
REM * DATE.FACTOR   = Number of cumulative days from year/month/day 0.
REM * WEEK.DAY      = 1 to   7 is Sunday to Saturday respectively.  
REM * WEEK.NUM      = 1 TO  54 is week number within year.            
REM * JULIAN.DAY    = 1 TO 366 is day  number within year.            
REM * ZYY           = Value of of 4 digit year.        
REM * ZMM           = Value of month.                    
REM * ZDD           = Value of day.                                
REM * Z$            = (unchanged).
REM * YEAR.MIN      = (unchanged).
REM *
REM *
REM * Date factor logic adopted from a Texas Instruments calculator manual.
REM *
DATE.FACTOR:
  gosub Date.Check                     'check input date
  if not(date.ok) then RETURN          'exit if invalid
 
  zmm=1:zdd=1                          'set to January 1st
  gosub Compute.Factor                 'compute factor of Jan 1st
  zfsave=date.factor                   'save factor   of Jan 1st
  gosub Compute.Weekday                'week.day now has day of week of Jan 1st

  zdd=val(right$(z$,2))                'Restore input date's day + month  
  zmm=val(mid$(z$,5,2))  
  gosub Compute.Factor                 'compute factor of input date

  '* Julian day is input date minus Jan 1st of same year +1
  julian.day=date.factor-zfsave+1  

  '* Compute the week number: (week.day-1 is week day of Jan 1st relative to 0)
  week.num=int((julian.day+(week.day-1)-1)/7)+1

  '* Compute the day of the week of input date:
  gosub Compute.Weekday
RETURN

COMPUTE.FACTOR:
  DATE.FACTOR=365!*ZYY+ZDD+31*(ZMM-1)  'NOTE: WON'T WORK WITHOUT ! AFTER 365.
  IF ZMM<3 THEN
     DATE.FACTOR=DATE.FACTOR+INT((ZYY-1)/4)-INT(3/4*(INT((ZYY-1)/100)+1))
  ELSE
     DATE.FACTOR=DATE.FACTOR-INT(.4*ZMM+2.3)+INT(ZYY/4)-INT(3/4*(INT(ZYY/100)+1))
  END IF
RETURN

COMPUTE.WEEKDAY:
  '* Compute the day of the week:
  WEEK.DAY=DATE.FACTOR-INT(DATE.FACTOR/7)*7    'Modulo 7.
  IF WEEK.DAY=0 THEN WEEK.DAY=7                'WEEK.DAY=1=Sunday.
RETURN
REM ***************************************************************************

REM ******************  DATE.OFFSET  ******************************************
REM *
REM *** COMPUTE THE DATE WHICH IS NUMBER OF DAYS FROM GIVEN DATE.
REM *
REM *  INPUT:
REM *  =====
REM *  Z$           = Given date as YYYYMMDD.
REM *  DAYS.OFFSET  = Number of calendar days plus or minus from given date.
REM *  YEAR.MIN     = Minimum year user wishes to allow (default 0)
REM *
REM * OUTPUT:
REM * ======
REM * DATE.OK       = -1 if input/offset/result date is VALID.   (true)
REM *               =  0 if input/offset/result date is INVALID. (false)
REM * NOTE: IF VALID, THE FOLLOWING VARIABLES AR BASED ON COMPUTED/RESULT DATE.
REM *       IF INVALID, THE VALUES MAY HAVE CHANGED AND ARE MEANINGLESS.
REM * Z$            = Computed/Result date (Given+DAYS.OFFSET) (as YYYYMMDD).
REM * DAYS.OFFSET   = (unchanged).
REM * DATE.FACTOR   = Number of cumulative days from year/month/day 0.
REM * WEEK.DAY      = 1 to   7 is Sunday to Saturday respectively.  
REM * WEEK.NUM      = 1 TO  54 is week number within year.            
REM * JULIAN.DAY    = 1 TO 366 is day  number within year.            
REM * ZYY           = Value of of 4 digit year.        
REM * ZMM           = Value of month.                    
REM * ZDD           = Value of day.                                
REM * EASTERSUNDAY$ = Date of Easter for computed/result year.      
REM * YEAR.MIN      = (unchanged).
REM *
DATE.OFFSET:
  gosub date.factor
  if not (date.ok) then RETURN        
  if date.factor+days.offset < 0 or_
     date.factor+days.offset > 3652424 then Date.ok=0 : RETURN
                              '3652424 is date.factor for max date of 99991231.
  '* Note: Date was split into zyy/zmm/zdd by date.factor routine.
  zdwork=zdd+days.offset                    'Set to Given day + increment.
  if zdwork < 1 then
     do
       zmm=zmm-1:if zmm=0 then zmm=12:zyy=zyy-1
       gosub GetZMax                           'go get max days cur month (zmm)
       zdwork=zdwork+zmax
     loop while zdwork<1
  else
     gosub GetZMax                             'go get max days cur month (zmm)
     do while zdwork>zmax
       zmm=zmm+1:IF zmm>12 then zmm=1:zyy=zyy+1
       zdwork=zdwork-zmax
       gosub GetZMax
     loop
  end if
  zdd=zdwork
  '* Pack the date as YYYYMMDD
  Z$=FILLSTRING$((ZYY),4)+FILLSTRING$((ZMM),2)+FILLSTRING$((ZDD),2)
  gosub date.factor         'get all pertinent variables for final date
  '* Note: date.factor routine also sets date.ok indicator.
RETURN

GetZMax:
  IF ZMM=2 AND ISLEAPYEAR(ZYY) THEN ZMAX=ZMO(ZMM)+1 ELSE ZMAX=ZMO(ZMM)
RETURN
REM ***************************************************************************

REM ***************************************************************************

REM *********************  DATE.CHECK  ****************************************
REM *
REM *** VALIDATE A DATE IN YYYYMMDD FORMAT.
REM *
REM *  INPUT: Z$       = Given date in format YYYYMMDD.
REM *         YEAR.MIN = Minimum valid year allowed. (default=0)
REM *
REM * OUTPUT: DATE.OK = -1 if input date is VALID.   (true)
REM *                    0 if input date is INVALID. (false)                    
REM *         (if VALID):
REM *         ZYY      = Value of 4 digit year.        
REM *         ZMM      = Value of month.                              
REM *         ZDD      = Value of day.                                
REM *
REM *
DATE.CHECK:
  DATE.OK = 0      'preset to false
  ZTEMP$="1"+Z$+"1"
  IF LEN(Z$)<>8 OR MID$(STR$(VAL(ZTEMP$)),2)<>ZTEMP$ THEN RETURN
  ZDD=VAL(RIGHT$(Z$,2))                'Set day                
  ZMM=VAL(MID$(Z$,5,2))                'Set month.
  ZYY=VAL(LEFT$(Z$,4))                 'Set year.
  IF ZMM<1 OR ZMM>12 OR ZDD<1 OR ZDD>31 OR ZYY<YEAR.MIN THEN RETURN
  IF ZMO(ZMM)+1*(-(ZMM=2 AND ISLEAPYEAR(ZYY))) < ZDD THEN RETURN
  '   If expression (month=2 and is leapyear) is TRUE which is -1, then
  '   taking the negative of this issues a plus 1. Conversely, the FALSE    
  '   always gives a zero. Multiplying the +1 by this result of 1 or 0
  '   will either add 1 or not to the number of days in the month.
  '   The routine wants to add 1 only when it is February and leap year.
  DATE.OK = -1        '-1=valid (true)
RETURN      
REM ***************************************************************************

REM ******************  DATE.FORMAT  ******************************************
REM *
REM *** FORMAT A DATE FOR PRINTING.
REM *
REM *  INPUT: Z$ in format YYYYMMDD. (assumed to be already validated)
REM *
REM * OUTPUT: Z$ formatted as DD-MMM-YYYY.
REM *
DATE.FORMAT:
  Z$=RIGHT$(Z$,2)+"-"+MID$(ZMO3$,6*VAL(MID$(Z$,5,2))-2-(3),3)+"-"+LEFT$(Z$,4)
RETURN
REM ***************************************************************************

END

REM ***************************************************************************
REM ***************************************************************************
REM *****   D A T E   F U N C T I O N S   *************************************
REM ***************************************************************************

' ====================== ISLEAPYEAR ==========================
'         Determines if a year is a leap year or not.
' ============================================================
'
FUNCTION IsLeapYear (Z) STATIC

   ' If the year is evenly divisible by 4 and not divisible
   ' by 100, or if the year is evenly divisible by 400, then
   ' it's a leap year:
   IsLeapYear = (Z MOD 4 = 0 AND Z MOD 100 <> 0) OR (Z MOD 400 = 0)
END FUNCTION
' ========================= FILLSTRING =============================
' Converts a value to string of specified length with leading zeros.
' ==================================================================
FUNCTION FillString$ (V#,ZL) STATIC

  FILLSTRING$=right$(STRING$(ZL,"0")+MID$(STR$(V#),2),ZL)
END FUNCTION
' ===================================================================


FUNCTION NumStrict (Z$)

REM *
REM *** (NUMSTRICT) - CHECK FOR STRICTLY NUMERIC ONLY (NO NULL NO DECIMAL)
REM *

  NumStrict=0         'Init to False

  IF Z$="" THEN EXIT FUNCTION

  FOR X = 1 TO LEN(Z$)
      A=ASC(MID$(Z$,X,1))
      IF A<48 OR A>57 THEN EXIT FUNCTION
  NEXT X

  NumStrict = -1          'True

END FUNCTION
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Agamemnus
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Posts: 3491



« Reply #5 on: December 15, 2006, 05:23:36 PM »

Hm... I guess you wrote this for QB, but there are loads of problems running this with FB... among them, date.ok doesn't seem to work right... :-| I'll try to fix later....
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Antoni Gual
Na_th_an
*****
Posts: 1434



WWW
« Reply #6 on: December 15, 2006, 05:47:21 PM »

Small is beautiful
Code:

'Conversion of day of year and year to day of month and day of week
'by Antoni gual 2006/12/15 for Agamemnus contest at QBN
'Tested in QB 1.1
'
DEFINT A-Z
DECLARE FUNCTION DayofWeek (y, m, D)
DECLARE FUNCTION ISLEAPYEAR (y)
DECLARE FUNCTION daysinmonth (y, m)

DO
INPUT "year (0 to exit)"; y
IF y < 1 THEN EXIT DO
INPUT "day of year"; dy
IF dy < 1 OR dy > 365 - ISLEAPYEAR(y) THEN PRINT "bad day of year": END
nt = 0
m = 0
DO
 n = nt
 m = m + 1
 nt = nt + daysinmonth(y, m)
LOOP UNTIL dy <= nt

dm = dy - n
'PRINT "month (1=January...12=December)"; m
PRINT "day of month = "; dm
PRINT "day of week (1=monday...7=sunday) "; DayofWeek(y, m, dm) + 1
LOOP
END
'
'-------------------------------------------------------------------------
FUNCTION DayofWeek (y, m, D)
  '0=monday
  DIM P, Q
  IF m > 2 THEN
    P = m - 3
    Q = y
  ELSE
    P = m + 9
    Q = y - 1
  END IF
 DayofWeek = (D + 1 + Q + Q \ 4 - Q \ 100 + Q \ 400 + CINT(2.6 * P)) MOD 7
END FUNCTION

'
'-------------------------------------------------------------------------
FUNCTION daysinmonth (y, m)
'get nr of days in a month  of a year(check for leap year if february)
SELECT CASE m
CASE 2: daysinmonth = 28 - ISLEAPYEAR(y)
CASE 1, 3, 5, 7, 8, 10, 12: daysinmonth = 31
CASE ELSE: daysinmonth = 30
END SELECT
END FUNCTION

'
'-------------------------------------------------------------------------
FUNCTION ISLEAPYEAR (y)
'Returns -1 if Gregorian year y is a leap year
ISLEAPYEAR = (y MOD 4 = 0) - (y MOD 100 = 0) + (y MOD 400 = 0)
END FUNCTION
'
'--------------------------------------------------------------------------
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Antoni
Moneo
Na_th_an
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Posts: 1971


« Reply #7 on: December 18, 2006, 12:48:11 AM »

Quote from: "Antoni Gual"
Small is beautiful
Code:

'Conversion of day of year and year to day of month and day of week
'by Antoni gual 2006/12/15 for Agamemnus contest at QBN
'Tested in QB 1.1


Antoni, Did some testing and it works fine.

I wonder why Aga didn't ask for the month as well as the day of the month?

¿Que te pareció el 4-0 del Barca contra el América?
Como dirián en inglés: When you play against Barcelona, you're playing with the big boys.

Saludos..... Moneo
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Antoni Gual
Na_th_an
*****
Posts: 1434



WWW
« Reply #8 on: December 18, 2006, 02:26:20 PM »

Naturally it works. Smiley
 It uses the day of week routine I once used in one of your challenges at Foronet. I can translate the explanation of the algorithm i posted there if someone is interested.

About the Barça, well, they were too confident. They have no rival in Spain at the moment, and they just classified for the next run of the european UEFA champions cup, so they believe they are the kings of mambo.
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Antoni
Agamemnus
x/ \z
*****
Posts: 3491



« Reply #9 on: December 18, 2006, 04:07:48 PM »

Quote from: "Moneo"
Quote from: "Antoni Gual"
Small is beautiful
Code:

'Conversion of day of year and year to day of month and day of week
'by Antoni gual 2006/12/15 for Agamemnus contest at QBN
'Tested in QB 1.1


Antoni, Did some testing and it works fine.

I wonder why Aga didn't ask for the month as well as the day of the month?

¿Que te pareció el 4-0 del Barca contra el América?
Como dirián en inglés: When you play against Barcelona, you're playing with the big boys.

Saludos..... Moneo


Yes, I guess that would be interesting as well...

Thank you Antoni for your submission. I am going to try to do something myself (without looking at you awesome compact code) and we shall then end the contest there, if no one else has any other submissions.
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Antoni Gual
Na_th_an
*****
Posts: 1434



WWW
« Reply #10 on: December 18, 2006, 06:07:03 PM »

Aga:
I guess what you consider interesting is a description of the DayofWeek routine so here it goes. I don't know who invented it but it gave me a good headache to figure how does it work.

Code:

FUNCTION DayofWeek (y, m, D)
  '0=monday
  DIM P, Q
  IF m > 2 THEN
    P = m - 3
    Q = y
  ELSE
    P = m + 9
    Q = y - 1
  END IF
 DayofWeek = (D + 1 + Q + Q \ 4 - Q \ 100 + Q \ 400 + CINT(2.6 * P)) MOD 7
END FUNCTION


It uses modular arithmetics to add offsets  modulo 7 from a theoretical (gregorian) march 1th of the year zero. If that day existed it was a  wednedsday. And it uses a shifted year (Q variable) starting at march 1. The monhts are also shifted, month 0 is march and month 11 is february.

If you divide 365 by 7 you get a remainder of 1, so if you don't count leap years, every year advances a date by 1 position in the week. So we have a  +Q in the formula. the  +Q\4-Q\100+\400 part provide for the leap year's additional offsets.

Then it comes the serious part, the reason why we use a shifted year.
CINT(2.6*P)  provides for the month of year offset. The effect of the fp operation plus rounding makes this factor add 3 days (31-28) for april (month 1), 5 days for may, 8 days for june... and so on, including the irregularity of august and september that require 3 days  for each. February  has been noved to the end of the year  so its different behavior in leap years must not be considered in this step.

Then the formula adds the Day of month (+D) and a 1 to center the things and make the result of 0 a monday .

Hope it helps...

Moneo:
In my last post I was obviously not speaking about the same soccer game as you. In fact i found about the Barça- America in the net after re-reading yout post  . I'm not very interested in soccer, you see...

And for the explanation of the DayofWeek routine, my memory failed, we spoke about it at Pete's site, not at Foronet.



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Antoni
Agamemnus
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« Reply #11 on: December 24, 2006, 10:13:50 PM »

Yes, I guess that how I would go about it is figure out how many years since now were not leap years and then do MOD 7.

Thanks for the explanation.

I actually figured out that this can be done with FB's functions without resorting to formulas, but I was interested in a use for this in Javascript, actually:

A function for a project of mine needs to get the next trading day (e.g.: Monday through Friday) X days from day N, year M: It would adjust the day if it was Sunday, Saturday, or another non-trading day. I can adjust the days easily enough, but getting the day of the week would have been difficult.
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Antoni Gual
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« Reply #12 on: December 25, 2006, 07:30:08 AM »

If your project requires to count n days from a day, just convert to julian dates. With JD you can get the day of the week simply by doing JD MOD 7.
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Antoni
Moneo
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« Reply #13 on: December 25, 2006, 05:00:54 PM »

Quote from: "Agamemnus"
......
A function for a project of mine needs to get the next trading day (e.g.: Monday through Friday).....

I think you're going to have to identify holidays, falling on Monday to Friday, which are not trading days for the location where the program is to be run.

This a pain because you need to create a table or file of these holidays, and update the table every year.

You could also try to program these holidays. Examples in the USA: Martin Luther King's birthday is celebrated on the third Monday in January. Thanksgiving is the fourth Thursday in November. If Good Friday is a holiday at your location, you will also need to compute the date of Easter. Banking laws in some countries inhibit banks from being closed for more than 4 days. So, if a Friday is a holiday, then Thursday and Monday cannot both be holidays. Trading days generally adhere to bank holidays.

Computing all these holidays can get tedious. The holiday table or file is less elegant and perhaps prone to user error, but much simpler.

EDIT: I think that The best solution would be to get the holiday table/file from the applicable stock exchange directly.


Regards..... Moneo
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