Calculating Day of Week Given MM-DD-YYYY

[eyeRmonkey]

I know there are probably ways to do this already, but I came up with this one all by myself.

[BobTheBuilder.]

Sounds like you want the Julian Date. Check this out.

[eyeRmonkey]

I know there are other ways out there. My intention when doing this was to come up with my own formula. I didn't want to look anywhere online otherwise I wouldn't be able to figure it out by self because I would have that info in the back of my head, influencing my thought process.

After I simplify the formula above as much as possible, I will look up Julian Dates online.

[BobTheBuilder.]

I know there are other ways out there. My intention when doing this was to come up with my own formula. I didn't want to look anywhere online otherwise I wouldn't be able to figure it out by self because I would have that info in the back of my head, influencing my thought process.

After I simplify the formula above as much as possible, I will look up Julian Dates online.

Oh I see. I am indeed very much impressed with that formula! I no longer have the inspiration to sit and come up with stuff like that anymore

[eyeRmonkey]

I'm glad you like it. It only took me a few hours of actual thought (maybe 5?). There is a lot of binary (true/false) logic to it because I am too used to programming. Its really ugly and I am trying to simplify it.

Here it is in VB is anyone cares:

Private Function Convert( _
        ByVal M As Long, _
        ByVal D As Long, _
        ByVal Y As Long _
    )
    Dim lReturn         As Long
    Dim lYearStart      As Long
    Dim lMonthStart     As Long
    Dim lMonthOffset    As Long
    Dim lDay            As Long
    
    lYearStart = Int((Y - 1) * (365.25))
    lMonthStart = (M - 1) * 30
    lMonthOffset = (1) * Int(Int(M / 2) / (Int(M / 2) - 0.0001)) + _
                  (-2 + (1 - Int((Y Mod 4) / ((Y Mod 4) - 0.0001)))) * Int(Int(M / 3) / (Int(M / 3) - 0.0001)) + _
                   (1) * Int(Int(M / 4) / (Int(M / 4) - 0.0001)) + _
                   (1) * Int(Int(M / 6) / (Int(M / 6) - 0.0001)) + _
                   (1) * Int(Int(M / 8) / (Int(M / 8) - 0.0001)) + _
                   (1) * Int(Int(M / 9) / (Int(M / 9) - 0.0001)) + _
                   (1) * Int(Int(M / 11) / (Int(M / 11) - 0.0001))
    lDay = D - 1
    
    lReturn = (lYearStart + lMonthStart + lMonthOffset + lDay) Mod 7
    
    Convert = lReturn
End Function

[eyeRmonkey]

Okay, I have spent a lot of time simplifying this formula. I attached an image of the new version. This is actually the 3rd revision I have made. The last one I didn't post because it was only slightly nicer than the one above.

Here is the code in VB.

Private Function ConvertToDayOfWeek3( _
        ByVal M As Long, _
        ByVal D As Long, _
        ByVal Y As Long _
    ) As Long
    Dim lReturn             As Long
    Dim lYearStart          As Long
    Dim lMonthOffset        As Long
    Dim lDay                As Long
    Dim lLeapYearAfterFeb   As Long
    
    lYearStart = Int(365.25 * (Y - 1))
    lMonthOffset = Int(530526416330# / 10 ^ (M - 1)) - Int(530526416330# / 10 ^ M) * (10)
    lLeapYearAfterFeb = (1 - Int((Y Mod 4) / (Y Mod 4 - 0.01))) * Int(Int(M / 3) / (Int(M / 3) - 0.0001))
    lDay = D - 1
    
    lReturn = (lYearStart + lMonthOffset + lDay + lLeapYearAfterFeb) Mod 7
    
    ConvertToDayOfWeek3 = lReturn
End Function

[krtxmrtz]

Check this one out, at least you'll have an on-line benchmark to test your code against.

[szlamany]

Compare it to the formula shown in this thread...

http://www.vbforums.com/showthread.p...ht=1582+julian

One is mainframe BASIC, the other is VB - but basically the same formula...