Talk:Doomsday rule/Archive 1

Poorly displayed equation
In the 'algorithm' paragraph, the sentence 'Anchor can be calculated in that way (for Gregorian) (2+5*int((y mod 400)/100))mod 7' appeared under the table. This is is a very poor sentance, and I have removed it. Not only was it poor English, but I am still struggling to understand what it actually was trying to tell us. The equation is poorly (not) defined, and it should not be written inline. I am not an experienced editor, please someone with more knowledge of formatting equations correctly replace this, and explain it with a meaningful sentence. A boardley (talk) 16:38, 24 November 2008 (UTC)

Year Ref
I think there's a problem with a page containing a reference to "this year", as in the first example, since it will become out of date - indeed, Christmas "this year" is 25 December 2005, not 2004, as in the example. I'd suggest dropping it, since it's followed by an example for a specific date, and (short of remembering to update the page every 1 January) there's no way to keep it accurate. --Kay Dekker 12:26, 1 Jan 2005 (UTC)

Algorithm
How exactly do you work put the Doomsday for years in which when divided by 12 give fractions? E.g. 2005. 5/12, does not give an integer or remainder. 62.249.242.232 (sig added by Cburnett)


 * The division in the algorithm is integer division. So:


 * Cburnett 20:21, Jun 18, 2005 (UTC)

But, when taking the last two digits of the year 2005 and then dividing them by 12, i.e. 5/12, the integer division equals 0, this gives no remainder(?). Therefore dividing by 4 also gives the integer 0. The sum of the integers gives 0, i.e. 0 + 0 + 0. So how in 2005 do you get the Doomsday 6 days on from the anchor day. Is there something wrong with the above reasoning, such as 5/12 does in fact give a remainder?


 * Ok, here we go:
 * 5/12 = 0
 * 5%12 = 5 (the remainder)
 * 5/4 = 1
 * And 0 + 5 + 1 = 6


 * It seems that integer division and remainder are confusing you. Take an unrelated example: 14 = 2*5 + 4 (this is a simple identity: 2*5 + 4 is 14) so the quotient (if you divided 14 by 5) is 2 and the remainder is 4 (see remainder).  You could also write this as 14/5 = 2 R4 (2 with a remainder of 4).  So the integer division of 14/5 is 2 and 14%5 (the remainder) is 4.


 * If you did the integer division of 14/5 you'd get 2. From this you can easily find the remainder lost in the integer division: 14 - 2*5 = 4 (or 14 - (14/5)*5 = 4) which is the remainder.


 * Going back to the above.
 * 5 = 0*12 + 5 so the quotient (5/12) is 0 and the remainder (5%12) is 5
 * 5 = 1*4 + 1 so the remainder (5%4) is 1


 * Is it clearer now? Cburnett 22:08, Jun 22, 2005 (UTC)

Thankyou


 * No problem. For future reference, please type four tildes ( ~ ) to leave a signature. Cburnett 06:43, Jun 23, 2005 (UTC)

The name Doomsday?
Where does the name Doomsday come from? Was the algorithm used to calculate the world's end?

I would think, I don't know for sure, but the English word "Doom" has a secondary meaning of "fated" or "predetermined," Thus certain days are "doomed" to occur on the same day. —Preceding unsigned comment added by 129.1.186.143 (talk) 18:35, 13 September 2007 (UTC)
 * That Middle-English(?) usage of "doom" is so highly depreciated I would think we should have an explanation in the lead section. —AltiusBimm (talk) 17:23, 14 July 2012 (UTC)

On a related note, what are the guidelines for use of what someone says as a citation? I understand that an entire article couldn't be cited to a single person saying something (or even many people saying something), without documented references in news or text or whatnot, but if some information about Doomsday is provided in a lecture (e.g., by John Conway), what are the guidelines for referencing this information?--AaronRosenberg (talk) 16:02, 29 January 2008 (UTC)

alternate January memory device
My memory device for January, instead of "January 31st (or 32nd)," is "for three years it's the 3rd and in the fourth year it's the 4th." Works for me, and I think the arithmetic is then easier.

Olivia

Bob Goddard is the originator of this January mnemonic

Also, YingKing Yu recently (Sept. 2010) provided excellent mnemonics for January and February: Powerslide (talk) — Preceding undated comment added 03:22, 23 September 2010 (UTC)
 * for leap years: 1/11 and 2/22
 * for non-leap years, we just subtract 1 to get 1/10 and 2/21
 * in some cases, I found Yingking's january mnemonic as easier to calculate. It is easier to subtract 10 than to subtract 3.


 * for leap years: 1/4 and 2/8
 * for non-leap years, we just subtract 1 to get 1/3 and 2/7 — Preceding unsigned comment added by 164.116.247.8 (talk) 21:08, 11 March 2015 (UTC)

Using left two digits instead of century number
I think it's easier to use the left two digits instead of the "century number". I'd also like to add that explaining mathematics in opaque blocks of text is not a good idea. Thankfully there is a code sample. year   = 2005; century = year/100 + 1; a      = century * 5; b      = (century - 1) / 4; c      = a + b; anchor  = (c + 4) % 7; e      = (year % 100)/12; f      = (year % 100)%12; g      = f/4; h      = e + f + g; dday    = (h + anchor) % 7; Eliminating the +1 on century: year   = 2005; century = year/100 + 1 ; a      = (century + 1 ) * 5; b      = (century + 1 - 1) / 4; c      = a + b; anchor  = (c + 4) % 7; e      = (year % 100)/12; f      = (year % 100)%12; g      = f/4; h      = e + f + g; dday    = (h + anchor) % 7; Simplify: year   = 2005; century = year/100; a      = (century + 1 ) * 5 + 5; b      = (century + 1 - 1 ) / 4; c      = a + b; anchor  = (c + 4) % 7; e      = (year % 100)/12; f      = (year % 100)%12; g      = f/4; h      = e + f + g; dday    = (h + anchor) % 7; Propagate the +5: year   = 2005; century = year/100; a      = century * 5 + 5 ; b      = century / 4; c      = a + b + 5 ; anchor = (c + 4) % 7; e      = (year % 100)/12; f      = (year % 100)%12; g      = f/4; h      = e + f + g; dday    = (h + anchor) % 7; Further: year   = 2005; century = year/100; a      = century * 5; b      = century / 4; c      = a + b + 5 ; anchor = (c + 5 + 4) % 7; e      = (year % 100)/12; f      = (year % 100)%12; g      = f/4; h      = e + f + g; dday    = (h + anchor) % 7; Simplify: year   = 2005; century = year/100; a      = century * 5; b      = century / 4; c      = a + b; anchor  = (c + 2) % 7; e      = (year % 100)/12; f      = (year % 100)%12; g      = f/4; h      = e + f + g; dday    = (h + anchor) % 7; Unless I've made a typo somewhere this should be result.

12 April 1861: SIMPLIFIED        ARTICLE year   = 1861;    = 1861; century = 18;     = 19; a      = 90;      = 95; b      = 4;       = 4; c      = 94;      = 99; anchor = 5;       = 5; e      = 5;       = 5; f      = 1;       = 1; g      = 0;       = 0; h      = 6;       = 6; dday   = 4;       = 4; S: anchor day is 94 days after Tuesday, or Friday.

A: anchor day is 99 days after Thursday, or Friday.

The digits 61 give a displacement of six days, so Doomsday was Thursday. Therefore, April 4 was Thursday, so April 12, eight days later, is a Friday. Shinobu 16:33, 19 November 2005 (UTC)

The Essense of the Doomsday Algorithm
The essential characteristing of the Dommsday method is the simple way that Doomsday is calculate from the month and day of the year. This is now lost within the current explanation of the algorithm.

The way its written now would make a casual and impatient reader ask what's special about the doomsday algorithm before reaching the essential doomsday part.

In August the Article was:

The Doomsday algorithm for calculating the day of the week makes use of the fact that in each year,

4 April (4/4) 6 June (6/6) 8 August (8/8) 10 October (10/10) and 12 December (12/12)

are all the same day of the week. This day of the week is called Doomsday. The algorithm also defines 9/5 and 7/11 and their reverses to be Doomsday.

The Doomsday algorithm was invented by John Horton Conway.

Removed from main page (and tightened, restored Nov2010)
The following was removed from the main page by me, Juuitchan. My lady friends (not to mention pretty much any man) don't need to wade thru all this just to get the anchor day of the "00" year of the century!

To find a century's anchor day, begin by finding the century c which the date falls in. (For the purposes of this calculation, century years will be treated as though they fall in the century that follows. 2000 is therefore part of the twenty-first century in this equation, not the twentieth century.) A year's century number is equal to its first two digits plus one (so 1966 is in the twentieth century). Take the century number and multiply by 5 (call this x). Separately, subtract one from the century number, divide by four and take the integral value (the next-lowest whole number) of the quotient (y). Add these two numbers (x and y to get z). (At this point a common optional step is to divide z by seven, and take the remainder so as to make the numbers more manageable, since this won't affect the result.) Now count forward that number of days (z or the remainder of z/7) from Thursday to get the anchor day for the century.

$$\left({5c+\left\lfloor{\frac{c-1}{4}}\right\rfloor}\right) \bmod 7+\rm{Thursday}=\rm{anchor}$$

For example, the calculation for the twentieth century is:

$$\begin{matrix}\left({5\times{20}+\left\lfloor{\frac{19}{4}}\right\rfloor}\right) \bmod 7+\rm{Thursday} & = & \left(100+4\right) \bmod 7+\rm{Thursday} \\ \ & = & \rm{Wednesday}\end{matrix}$$

So Wednesday is the anchor day for the 20th century.

Similarly, the anchor date for the twenty-first century is Tuesday:

$$\left({5\times{21}+\left\lfloor{\frac{20}{4}}\right\rfloor}\right) \bmod 7+\rm{Thursday}=\rm{Tuesday}$$


 * Ditto my remark about opaque blocks of text above. Still, I'm not really ok with removing the formula altogether, and one shouldn't use the first person in articles either. Shinobu 16:18, 13 December 2005 (UTC)
 * This is necessary for completeness, I think. While most people will just remember the anchor days, the computation isn't difficult.  I tightened up the section considerably and restored it, taking the approach that people who get this far won't be troubled by a bit of math.Kbk (talk) 03:39, 2 November 2010 (UTC)

Weeding out
The article is excessively long. I would like the suggest to delete all the calculations, leaving only the principle of Conway's rule. Because in practice, now we all have electronic calculators, derivation of the weekday from the Julian day is a much simpler operation. --Tauʻolunga 07:02, 23 May 2006 (UTC)

Perhaps practically everyone may have a calculator, and that may even be easier, but that doesn't mean we should delete things just because they seem archaic, or that it is good for someone to rely on machines to think for them. —Preceding unsigned comment added by 129.1.186.143 (talk) 18:39, 13 September 2007 (UTC)

No, one of the key aspects is to be able to do it in your head! First, it's good for staving off Alzheimer's, second, it's always there when you don't have a calculator, and third, it's a great party trick! As I recollect, there used to be a remark on the page regarding how Conway used the mental calculation daily to stay sharp. Kbk (talk) —Preceding undated comment added 01:49, 2 November 2009 (UTC).


 * Ah, the remark was on the JHC page. Copied it to the beginning of the introduction. Kbk (talk) 21:43, 1 November 2010 (UTC)

History
This article has no history section. Where did this expression come from? What is its significance? Rklawton 13:46, 2 July 2006 (UTC)


 * This is something I was wondering as well - I've never heard the word 'Doomsday' used in this way, and wiktionary seems to be similarly in the dark... I think there needs to some sort of introduction written explaining what (this type of) doomsday is, before laying into the equations... sheridan 00:25, 13 October 2006 (UTC)

Suggestion for reverting move
The name of this article was originally "Doomsday rule". A user moved it to the current title with the reason "not only a particular rule, but a convenient way of characterizing each of the 7 possible calendars". AFAIK, the word "doomsday" means the Last Judgement. The usage to denote the weekday of last day of February only occurs in the Doomsday rule. Should we move it back to "Doomsday rule"? Joshua  C  hi  e w 06:52, 28 October 2006 (UTC)
 * That would be reasonable; although I believe Conway called it "Wednesday rule"; what he calls it this century I don't know. Next we'll be moving Domesday Book. :-> Septentrionalis 03:22, 29 October 2006 (UTC)

Links to here
On the days of the year template why do January the 32nd and March the 0th link here? ΦΙΛ Κ 18:54, 19 September 2007 (UTC)
 * The Doomsday algorithm uses the pseudodate "March 0". It used to use "January 32" until somebody pointed out the better 3/4 rule for January. Jess Cully (talk) 23:57, 11 December 2007 (UTC)

?
I would just like to note that the beginning of any century is xx01, not xx00. For example, the 20th century started on 1901, not 1900. —Preceding unsigned comment added by 210.17.234.220 (talk) 07:33, 4 January 2008 (UTC) ŇǑ
 * Ummm... Nope. In your reasoning, the year 0000 (Jesus's birthday) was in the 0th century instead of the 1st. F*L*RAP 04:57, 14 February 2008 (UTC)

Major errors to the point of being useless?
The mnemonic implies, incorrectly, that xx00 is the first year of each centuryse. The century starts in the year xx01. See Year 0. The error in the mnemonic should be highlighted, or else removed. 203.7.140.3 (talk) 03:15, 23 January 2008 (UTC)


 * The article states "For the purposes of the Doomsday rule, a century starts with a 00 year and ends with a 99 year." (my italics). Just because general convention dicates that centuries start in '01, this does not alter the maths involved. – Tivedshambo (talk) 11:48, 23 January 2008 (UTC)

Too technical?
Nope, I don't think so, because the article is full of examples. This is what algorithmic articles look like, take a look at the very straightforward and 1:st degree Euler–Maclaurin formula article! Simple college maths. I think the article shouldn't be marked Too.  Said: Rursus   ☻   14:02, 7 August 2008 (UTC)

Poor Picture?
No offense intended to Mr Conway, but the title of the article ("Doomsday rule") combined with the current picture used (File:JohnHortonConway.jpg) give the first impression that this article is about some apocalyptic charismatic cult, like the Branch Davidians or something. Perhaps it would be better to use a less "sinister" looking photograph of Conway, something that makes him look more like a mathematician, and less like David Koresh's protege? The current one in the John Horton Conway article would be good if someone could tweak the contrast. -- 128.104.112.100 (talk) 15:15, 10 August 2009 (UTC)


 * Good idea. I updated the image and placed it in this article. -Jordon Kalilich (talk) 03:50, 1 May 2010 (UTC)

Original Research: Mnemonic for 19th Century Anchor (5-Day or Friday) . ..
Very-few-alive day. Schweiwikist (talk) 07:25, 15 August 2009 (UTC)

Table difficult to read
The table under the heading "400-year cycle of Doomsdays" is difficult to read because text in the left column is centered. I'm no expert on Wikitables, but if an expert reads this, perhaps you would be good enough to fix this problem.--Oz1cz (talk) 19:30, 25 January 2011 (UTC)

Federal Holidays
The doomsday determines the federal holidays after February, as follows:

GeoffreyT2000 (talk) 19:01, 6 April 2015 (UTC)

9 - mod7 of Billburg
For this century perhaps the pithiest way to state the algorithm is 9 - mod7 of 11211. (11211 is the zip code of Williamsburg, Brooklyn). Take the year, take off 2000. Add 11 if it is odd. Divide by 2. Add 11 if the result is odd. Then take mod7 of that from 9. So 9 - mod7 of Billburg. Wodorabe (talk) 12:23, 4 November 2015 (UTC)
 * Yes, this is the "Odd+11" method. However, the 9 above only works from 2000 to 2099. GeoffreyT2000 (talk) 23:00, 25 March 2016 (UTC)

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Cluttering of table "Memorable dates that always land on Doomsday"
I think the table in the section "Memorable dates that always land on Doomsday" has grown to the point of being almost useless. Rather than one table with many different dates and mnemonics for each month, could we organize this in a few tables, each (ideally) with just one date and one mnemonic per month, so that a reader actually wishing to memorize could pick one table to memorize? E.g., I have memorized the following that makes some degree of sens to me, but as the table is written now, I probably wouldn't have found this complete set of mnemonics among all the options: I, for my part, would be happy if these were the only mnemonics for individual months given in the article. Another related question: Do we have a source for any or all of the mnemonics in the table - not as in, is it really true that Martin Luther King Day is such-and-such a date?, but as in: Did Conway or other permissible sources actually list this mnemonic in connection with the Doomsday rule?--Nø (talk) 10:16, 14 September 2014 (UTC)
 * January 3 in 3 out of four years; January 4 in every 4th year (viz. the leap years)
 * Last day of February
 * 0th of March
 * Remaining even months: 4/4, 6/6, 8/8, 10/10, 12/12
 * Remaining odd months: To remember the dates 9/5, 5/9, 7/11, 11/7, use the mnemonic "I work 9 to 5 in a 7 - 11".


 * I agree. The point of the table is to show dates that fall on same day of the week as Doomsday every year (well, almost except for January and February in leap years). The mnemonic is good enough for that purpose, so ideally it should be one date per month. I am okay with the inclusion of a few well-known holidays that fall on same day of the week as Doomsday every year, which means that they have to be fixed date holidays. Putting holidays that are fixed by a certain day of the week of the month is pretty useless. For example, Martin Luther King Day is on the third Monday of January, which is the same as Doomsday only when Doomsday is Monday. The table is not about holidays in the United States.-- Joshua Say "hi" to me!What I've done? 04:23, 7 February 2015 (UTC)
 * I've now moved all the unsourced mnemonics to a separate column. As far as I am concerned, this column could safely be deleted - but I'll leave it in for now, in case someone wants to clean it up and dig up refrences so that it may stay in.--Nø (talk) 11:36, 20 June 2015 (UTC)


 * I think for March it should be 7th of March because it would be better for the mnemonic for only one month and the mnemonic can be "7th day after the end of February". — Preceding unsigned comment added by 174.21.2.33 (talk) 05:07, 7 October 2015 (UTC)
 * Still, what we think is easy to remember is not what matters here on Wikipedia (but for the record I think 0th of March is brilliant in that respect). What matters is what we have valid sources for. So, find an independent source to back up your suggestion. It is not unlikely that such a source can be found; I haven't really looked myself (as I am quite happy with the mnemonics in the article right now).--Nø (talk) 07:49, 12 October 2015 (UTC)--Nø (talk) 07:49, 12 October 2015 (UTC)

How about changing February 28th to February 0th (and February 1st in leap years)? It would make calculations easier. 83.86.84.156 (talk) 12:30, 10 March 2017 (UTC)


 * Two reasons not to: (1) This article is about the Doomsday rule as described in the literature, not about what you or I think is the most convenient rule. (2) I think the philosophy of the Doomsday rule is not so much about ease of calculation as about mnemonic ease while keeping the calculations easy enough that they can be performed mentally by almost anyone. "Last day of February" is easy to remember (and is reused as March 0, arguable meaning one less thing to remember).--Nø (talk) 10:24, 11 March 2017 (UTC)

6, 11, 6, 5 years
The years with a given doomsday follow a 6, 11, 6, 5 pattern. The doomsday for 1900, Wednesday, is repeated in 1906, 1917, 1923, 1928, 1934, 1945, 1951, 1956, 1962, 1973, 1979, 1984, 1990, 2001, 2007, 2012, 2018, 2029, 2035, 2040, 2046, 2057, 2063, 2068, 2074, 2085, 2091, and 2096. GeoffreyT2000 (talk) 16:47, 1 March 2015 (UTC)

The general rule: if year mod 4 = 0 and 1, 2 or 3, add 6 (one leap year), 11 (three leap years) or 5 (two leap years) respectively (6+11+6+5=28). Pattern 0: 6, 11, 6, 5. Pattern 1: 6, 5, 6, 11. Pattern 2: 11, 6, 5, 6. Pattern 3: 5, 6, 11, 6. 27.154.63.67 (talk) 06:31, 28 April 2017 (UTC)

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Simple method for Doomsday offset from anchor day
The offset from the anchor day to the Doomsday for a given year is Y + floor(Y/4), where Y is the year % 100. The article gives two ways to calculate this that are easier for most people to do mentally (the method that involves dividing by 12, and the odd+11 method).

There is a much simpler method, but I've not found it published anywhere acceptable to cite in a Wikipedia article, so am not sure what to do with it.

Let T by the first digit of the two-digit year Y, and let U be the second digit (i.e., Y = 10T + U, 0 <= U < 10).

If T is even, the Doomsday offset is 2T + U + floor(U/4).

If T is odd, the Doomsday offset is 2T + 3 + U + floor((U+2)/4).

With mod 7 reductions along the way, you never need to deal with a number larger than 12 during the calculation. Heck, even if you put off the mod 7 reduction until the very end, you never go above 32.

The floor parts are to account for leap years. Instead of actually calculating floor(U/4) or floor((U+2)/4), it is probably easier in practice for mental calculation to simply remember that the leap years in even decades occur at U=4 and U=8, and in odd decades at U=2 and U=6, and add 1 for each leap year. (In even decades, U=0 is also a leap year, but it turns out that is already accounted for in the 2T).

Tim.the.bastard (talk) 13:42, 21 June 2018 (UTC)

The "odd + 11" method
The section on the odd+11-method does not look encyclopaedic to me. This is highlighted by but not limited to the sentence So that is how the "odd + 11" method works.. I wonder, is it notable at all?--Nø (talk) 14:58, 31 March 2017 (UTC)
 * I agree. That purported proof section is, at best, original research. It doesn't belong in Wikipedia unless there is an external reference to a reputable source — Preceding unsigned comment added by 108.66.129.231 (talk) 04:59, 5 April 2017 (UTC)
 * If you didn't know why + 11, please don't say anything here. 01 12, 02 13 24, 03 14 25 36(8). Yes, it is not notable at all, but it is interesting indeed.27.154.63.67 (talk) 03:26, 28 April 2017 (UTC)

The "odd + 11/even ÷ 2/odd + 11" method is equivalent to computing


 * $$-\left(y + \left\lfloor \frac{y}{4} \right\rfloor\right) \bmod 7$$.


 * $= −(4×y − r⁄4 + r + y − r⁄4) mod 7$, where $r = y mod 4$
 * $= −(5×y − r⁄4 + r) mod 7$, where $= −(5×y − r⁄4 + 5×12r⁄4) mod 7$
 * $(5×12r⁄4) mod 7 = r$, where $= (−5×y + 11r⁄4) mod 7$
 * $= (2×y + 11r⁄4) mod 7$, where $−5 mod 7 = 2$
 * $= (y + 11r⁄2) mod 7$, where $= y⁄2 mod 7$
 * $r = 0$, where $= (y + 11⁄2) mod 7$
 * $r = 1$, where $= (y + 11×2⁄2) mod 7 = (y⁄2 + 11) mod 7$
 * $r = 2$, where $= (y + 11×3⁄2) mod 7 = (y+11⁄2 + 11) mod 7$
 * $r = 3$, where ᙭᙭᙭
 * ᙭᙭᙭, where ᙭᙭᙭

So that is how the "odd + 11" method works.

27.154.63.67 (talk) 06:33, 29 April 2017 (UTC)

I disagree with the comment in the article that "It is of questionable advantage to mental calculation, because of the necessity to repeatedly use the "odd + 11" rule, and because of its failure to obviate an implicit division by 7."

I've been using this rule for a number of years and find it considerably simpler for two reasons: First, so what if you have to use the "odd + 11" rule possibly twice? It's only one rule and easy to remember, compared to the several different rules in the original method. More importantly, in contrast to Conway's method, the entire calculation from the initial value of T to the final value of T can be done entirely using only the previous value of T and nothing else. Conway's method requires remembering a second result through several calculations and then using it later.

Students clearly find the "odd + 11" rule much simpler than Conway's original rule, and they are much more likely to get correct results using it. It is the only method I've taught for many years.

So I feel that this editorial statement of opinion in the article is not à propos. It should be modified or removed. Bill Jefferys (talk) 20:02, 4 April 2019 (UTC)

Odd + 11
Would it be easier to write this article neatly (and have consensus to that effect) if the odd+11-rule was split off as a separate article?--Nø (talk) 06:34, 5 April 2019 (UTC)

Unexplained IP edits
There are material and very technical unexplained IP edits to this article which I have reverted. If they keep making these edits without explaining them, we may need to get this page protected. Britishfinance (talk) 10:35, 29 May 2019 (UTC)

What is doomsday?
Conway's algorithm bases on the fact that some dates always fall on the same weekday within any given year. These dates are called doomsdays. The weekday these dates fall on is called doomsday. Q5968661 (talk) 22:12, 27 May 2020 (UTC)

Too much
It seems to me this article is increasingly too long and complicated, relative to what there really is to say about the Doomsday rule as such. Of course, there is much to say about perpetual calendars and algorithms for calculating weekdays, but if it is only indirectly connected to the Doomsday rule, I do not think it should be in THIS article. I don't have the time, energy or resources to fix this, but I hope someone will.--Nø (talk) 09:38, 4 June 2020 (UTC)

Unclear text in the 'divide by 12 rule'
Either I don't understand what the following text is saying, or it's wrong:

"the sum of the last two digits of the year taken collectively plus the floor of those collective digits divided by four"

What does 'collectively' mean? Is there a difference between plain 'sum of the digits', and 'sum of the digits taken collectively'? If it means 'add the last two digits of the year yielding, say S, then add to S the floor of S divided by 4', then it doesn't work for the example year 1966. Which would give (6 + 6) + floor ((6 + 6) / 4) = 15.

Can anyone explain or make the text clearer? Rrgmitchell (talk) 23:09, 30 May 2020 (UTC)
 * 66+[66/4]～～～ — Preceding unsigned comment added by Q5968661 (talk • contribs) 11:34, 31 May 2020 (UTC)

Ah, so it's the last two figures of the year, added to the floor of the quotient by 4 of the last two figures. The trouble is that the text "sum of the last two digits" suggests to the reader that the last two digits are to be added to each other. I don't think that "collectively" helps; it just baffled me.Rrgmitchell (talk) 23:48, 1 July 2020 (UTC)