Talk:Determination of the day of the week

This article was nominated for deletion on 17 December 2012 (UTC). The result of the discussion was keep.

Time Low‑importance

	This article is within the scope of WikiProject Time, a collaborative effort to improve the coverage of Time on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.TimeWikipedia:WikiProject TimeTemplate:WikiProject TimeTime articles
Low	This article has been rated as Low-importance on the project's importance scale.

Mathematics Low‑priority

	Mathematics portal This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.MathematicsWikipedia:WikiProject MathematicsTemplate:WikiProject Mathematicsmathematics articles
Low	This article has been rated as Low-priority on the project's priority scale.

Archives

Archive 1

Archive 2

Reverts to the Revised Julian calendar

I'm involved in a lot of other issues right now, but I want to comment on many recent edits to Determination of the day of the week#Revised Julian calendar before I lose the thread. See edits by @66.56.252.194:[1] @JackieDander:[2][3][4]

Our article states 27 January RJC AD 8315 is a Tuesday. My hand calculation via another method agrees with Tuesday. (That said, I believe the offered procedure miscalculates the correction for some years, but I need to investigate that further.)

The reverts claim 27 January RJC AD 8315 is a Wednesday. As proof, an edit comment states "This link shows this Jan 27, 8315 date as Wednesday. http://moodle.pnwu.edu/calendar/view.php?view=month&course=1&time=200228832000".

Following that link does show 27 January AD 8315 is a Wednesday, but the calendar shown is not the Revised Julian calendar but rather the Gregorian calendar.

The text of Determination of the day of the week#Revised Julian calendar states that the Revised Julian and Gregorian calendars match from AD 1923 until AD 2800. The date in question falls outside of the matching interval. The table given at Revised Julian calendar#Arithmetic shows that the RJC and the GC differ by +1 day in AD 8315. In other words, a date in the Gregorian calendar is one day later than the same date in the Revised Julian calendar. That's why the Gregorian calendar is saying it is a Wednesday. The date in the RJC is the previous day, which would be a Tuesday.

The basic mistake is not recognizing the difference between the Revised Julian calendar (RJC) and the Gregorian calendar (GC).

Glrx (talk) 04:33, 12 June 2016 (UTC)[reply]

Wow. Thanks for the education about the RevisedJulianDate issue. I should have realized that the formulas that work until 2800 will have to be revised as that 11 second error keeps accumulating. Tuesday it is then for RJDate!! Thanks for being understanding about my newness to Wikipedia. After reading the history page I do think there is a need to clean up this page but that is a low priority I suppose. It is dense and hard to follow. JackieDander — Preceding unsigned comment added by JackieDander (talk • contribs) 21:43, 1 July 2016 (UTC)[reply]

Basic method for mental calculation

Under the section entitled "Basic method for mental calculation" it states that "m is the month's number in the months table" and "c is the century's number in the "100s of Years" table.". However, it is not clear where the tables stating the associated numerical values of the month's number and the 'century's number" actually are (if they are displayed at all). Can anyone place a table in a relevant location for this article? Alternatively, could someone reconstruct the table? I have looked intensely, and if I have missed something obvious, could someone point this out - but I do think that there is no table for this "Basic method for mental calculation" (or, at least, no clearly indicated table). ASavantDude (talk) 19:15, 27 December 2016 (UTC)[reply]

"Formulas derived from Gauss' algorithm" Suspicions

I have looked at this formula and tried calculating today's day from today's date a few times. Notwithstanding the fact that the explanation for the Formula for Gauss' Algorithm leaves much to be desired, I have a suspicion that the Formula provided under the section "Formulas derived from Gauss' algorithm" is, in fact, wrong. It is eminently possible that the lack of clear description of the details of the formula may yield fallacious interpretations of what the formula actually means. Has anyone else had a chance to use the formula to calculate the day of, say, the current date to see if the formula is actually correct? ASavantDude (talk) 19:49, 27 December 2016 (UTC)[reply]

The example numbers are wrong, if year = 2000 and Y = 2000 - 1, then if c is the first two digits of Y, c = 19 right? Yet c is 20 in the formula. --Eur (talk) 10:08, 13 September 2017 (UTC)[reply]

Simple Formula with Reasonable Applicability (THIS FORMULA DOES NOT APPEAR TO WORK FOR ALL DATES!)

According to : https://stackoverflow.com/questions/478694/what-is-the-easiest-algorithm-to-find-the-day-of-week-of-day-zero-of-a-given-yea

The following formula gives the WEEKDAY result.

1) Write the Date in European Format AS Date=DD/MM/CCYY so that 27/12/2016 would be the 27th of December 2016. Then, in this instance, DD=27, MM=12, CC=20, YY=16.

2) Calculate : $(\lfloor {\frac {CC}{4}}\rfloor -2CC-1)+\lfloor {\frac {5YY}{4}}\rfloor +\lfloor {\frac {26(MM+1)}{10}}\rfloor +DD){\bmod {7}}$ where mod(7) means finding the remainder upon division by 7.

3) Sun=0, Mon=1, Tues=2, Wed=3, Thurs=4, Fri=5, Sat=6 gives you the day answer.

If anyone else could place a better typesetted version of this formula upon the site, that might be a good idea...

ASavantDude (talk) 20:38, 27 December 2016 (UTC)[reply]

Some Notes

Note 1) The floor function gives the least integer greater than or EQUAL TO the result of the division. Alternatively, as we deal with positive numbers, just consider the integer part of the result of the division. The floor and integer parts should be equal for positive year numbers.

Note 2) The limits of applicability of this formula are NOT provided. When dealing with dates into the far flung future or past, the formula will likely need modification.

Note 3) An explanation for WHY this formula works is NOT provided.

Note 4) YY is the four-digit year modulo 100. CC is the integer part of the year divided by 100. So that the formula can be written in different ways. This way of writing the formula, however, appears clear enough for most purposes.

Note 5) The modifications to the formula for the BC-era will likely require modification of the formula and conventions to be adhered to (I am uncertain of this, but it is possible that the year 0 doesn't exist, which could have consequences for the formula in the BC-era). ASavantDude (talk) 15:33, 28 December 2016 (UTC)[reply]

Further Comment

I have tested this formula for 27/12/2016, and it gives the correct result of 2=tuesday. BUT for 06/01/2017, it does NOT give the correct result (it gives 3=Wednesday, rather than 5=Friday. ASavantDude (talk) 18:42, 6 January 2017 (UTC)[reply]

Cariño's Algorithm

A beautiful formula devised by a Filipino Mathematician (Dave Ryan Cariño) is an algorithm using direct substitution without any condition that is applicable from year 1582 onwards.

Let:

    M as Month of the year (1 = January, 2 = February, ………, 12 = December)
    D as day of the month (1 to 31)
    Y as year (1582....∞)
    D.D as day of the date (0 to 6, 0=Saturday, 6=Friday)

    J = (12Y+M-3)/1200

D.D = { 3M + 4 + D - ⌊3M/7⌋ - 2⌊(M+7)/10⌋ + ⌊(50Y+M-3)/40⌋ - ⌊J⌋ + ⌊J/4⌋ } mod 7

Example 1: October 15, 1582, first day of Gregorian calendar.
M = 10, D = 15, Y = 1582 
J = [(12(1582)+10-3)]/1200
  =  18991/1200
  = 15.825833
D.D = { 3(10) + 4 + 15 - ⌊3(10)/7⌋ - 2⌊(10+7)/10⌋ + ⌊[50(1582)+10-3]/40⌋ - ⌊15.825833⌋ + ⌊15.825833/4⌋ } mod 7  
  = { 30 + 4 + 15 - ⌊4.2857⌋ - 2⌊1.7⌋ + ⌊1977.675⌋ - ⌊15.825833⌋ + ⌊3.956⌋ } mod 7  
  = { 30 + 4 + 15 - 4 - 2 + 1977 - 15 + 3 } mod 7  
  = { 2008 } mod 7  
  = 6 ;Friday
So, October 15, 1582 is Friday

For Julian Calendar:

D.D = { 3M + 2 + D - ⌊3M/7⌋ - 2⌊(M+7)/10⌋ + ⌊(50Y+M-3)/40⌋ } mod 7

 — Preceding [[Wikipedia:Signatures|Vincent22914 (talk) 03:23, 7 December 2017 (UTC)]] comment added by Vincent22914 (talk • contribs) 11:09, 4 December 2017 (UTC)[reply]

What is an overplus?

The description of Lewis Carroll's method is impenetrable because it uses the word "overplus" three times and it never DEFINES that word. According to my searches, these are the only three uses of the word "overplus" (as a MATH term) that occur in all of English Wikipedia. In every place I look it up outside of Wikipedia, it's not a mathematical thing but just a synonym for "surplus", such as the amount of money you have beyond what you need. This is not helpful. If you want us to do something with an overplus, tell us what an overplus IS, for Pete's sake. Why make things hard gratuitously?2600:8804:8800:11F:1C64:8308:33BC:E2D6 (talk) 07:01, 31 March 2022 (UTC)Christopher L. Simpson[reply]