Talk:Doomsday rule

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Cluttering of table "Memorable dates that always land on Doomsday"[edit]

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:

  • 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, 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?-- (talk) 10:16, 14 September 2014 (UTC)

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.-- (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).-- (talk) 07:49, 12 October 2015 (UTC)-- (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).-- (talk) 10:24, 11 March 2017 (UTC)

6, 11, 6, 5 years[edit]

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)

Pattern Corresponding
years
Century year mod 400
000 100 200 300
0 00 06 17 23 Tue Sun Fri Wed
1 01 07 12 18 Wed Mon Sat Thu
2 02 13 19 24 Thu Tue Sun Fri
3 03 08 14 25 Fri Wed Mon Sat
1 09 15 20 26 Sat Thu Tue Sun
0 04 10 21 27 Sun Fri Wed Mon
1 05 11 16 22 Mon Sat Thu Tue
Note Year mod 28 Day of doomsday

27.154.63.67 (talk) 04:29, 29 April 2017 (UTC)

Federal Holidays[edit]

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

Doomsday Sunday Monday Tuesday Wednesday Thursday Friday Saturday
Memorial Day May 31 May 30 May 29 May 28 May 27 May 26 May 25
Independence Day Sunday Monday Tuesday Wednesday Thursday Friday Saturday
Labor Day September 6 September 5 September 4 September 3 September 2 September 1 September 7
Columbus Day October 11 October 10 October 9 October 8 October 14 October 13 October 12
Veterans Day Thursday Friday Saturday Sunday Monday Tuesday Wednesday
Thanksgiving Day November 25 November 24 November 23 November 22 November 28 November 27 November 26
Christmas Day Saturday Sunday Monday Tuesday Wednesday Thursday Friday

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

9 - mod7 of Billburg[edit]

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)

@Wodorabe: 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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The "odd + 11" method[edit]

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?-- (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

.
= −(4×y − r/4 + r + y − r/4) mod 7, where r = y mod 4
= −(5×y − r/4 + r) mod 7
= −(5×y − r/4 + 5×12r/4) mod 7, where (5×12r/4) mod 7 = r
= (−5×y + 11r/4) mod 7
= (2×y + 11r/4) mod 7, where −5 mod 7 = 2
= (y + 11r/2) mod 7
= y/2 mod 7, where r = 0
= (y + 11/2) mod 7, where r = 1
= (y + 11×2/2) mod 7 = (y/2 + 11) mod 7, where r = 2
= (y + 11×3/2) mod 7 = (y+11/2 + 11) mod 7, where r = 3

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

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

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Simple method for Doomsday offset from anchor day[edit]

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)