tag:blogger.com,1999:blog-3985237709515317836.post6187795269912123249..comments2016-07-25T12:20:56.277-07:00Comments on An Easier Doomsday Algorithm: An Improved Doomsday AlgorithmMikehttp://www.blogger.com/profile/09902477043993196546noreply@blogger.comBlogger13125tag:blogger.com,1999:blog-3985237709515317836.post-87861467938629008912016-07-25T12:20:56.277-07:002016-07-25T12:20:56.277-07:00It's been a while since I stopped by this page...It's been a while since I stopped by this page. Thanks to jackerz for pointing out the article by Mike Walters. His method is truly elegant and simple, and works for those problematic years that I mentionedBill Jefferyshttps://www.blogger.com/profile/03666872494092692149noreply@blogger.comtag:blogger.com,1999:blog-3985237709515317836.post-36067903805417167102014-09-09T18:33:25.669-07:002014-09-09T18:33:25.669-07:00Note to Bill Jefferys and Dale Cornflower:
You sh...Note to Bill Jefferys and Dale Cornflower:<br /><br />You should read Mike Walter's follow-up improvement to his easier doomsday proposal. It's called the "Odd+11" method and is described here:<br /><br />http://arxiv.org/abs/1010.0765<br /><br />The "Odd+11" method does not require any calculation involving leap years or divisibility-by-4 tests.jackerzhttps://www.blogger.com/profile/09035813176017376446noreply@blogger.comtag:blogger.com,1999:blog-3985237709515317836.post-58824043431886091562014-03-18T11:35:42.892-07:002014-03-18T11:35:42.892-07:00An addendum to my December, 2013 comment. I forgo...An addendum to my December, 2013 comment. I forgot to state that for January and February of leap years you must subtract 1 from the final date.<br /><br />Alternately, if you consider that the ancient Roman calendar started in March, you can do the same. If I use my method (or almost any of the algorithms), but for the year before for Jan and Feb dates, then it works out correctly. <br /><br />Note that instead of Conway's Jan 3/4 and Feb 7/8 Doomsdays, if you consider the year before for January and February, you have the easier Doomsdays of Jan 2 (like May and Dec)and Feb 6 (like Jun). This works whether it is or is not a leap year.<br />Cornflower (Dale)https://www.blogger.com/profile/08322997580510454930noreply@blogger.comtag:blogger.com,1999:blog-3985237709515317836.post-77329033867430826552013-12-02T12:07:17.878-08:002013-12-02T12:07:17.878-08:00This is much easier than Conway. I wondered if I ...This is much easier than Conway. I wondered if I couple simplify further though, and add my take. Instead of adding 11s, do the following<br /><br />1. Reduce the year to the nearest lower year-evenly-divisible-by-4 (2100 is such a year). Call the lower year x and the difference z. e.g. yyyy = 2013. x = 12, z = 1<br />Note x + z = yy<br /><br />2. Get the 7's complement of x/2<br />e.g. 2013, x = 12 x/2 = 6 7's complement of 6 is 1<br /><br />3. Add the century# + difference (z) + 7's complement to get your Doomsday #<br />2013 = 2 + 1 + 1 = 4 (Thursday)<br /><br />It works for the pesky dates mentioned by Bill Jeffreys:<br />2089 2 + 1 + 88=>44=>5 = 8 (Mon)<br />2078 2 + 2 + 76=>38=>4 = 8 (Mon)<br />2067 2 + 3 + 64=>32=>3 = 8 (Mon)<br /><br />Cornflower (Dale)https://www.blogger.com/profile/08322997580510454930noreply@blogger.comtag:blogger.com,1999:blog-3985237709515317836.post-38110345471351236542013-09-13T19:53:51.701-07:002013-09-13T19:53:51.701-07:00Bob Goddard pointed out a way to avoid the extra r...Bob Goddard pointed out a way to avoid the extra rule.<br /><br />I agree, this is elegant, but I think it does not quite work for three years in the 2000's, namely, 2089, 2078 and 2067. For each of these non-leap years, adding 11 repeatedly doesn't yield a leap year until 2144. This is because 2100 is NOT a leap year. So the comment "At most this will require adding 33 to the year" is not correct.<br /><br />It doesn't even work for 2100, again not a leap year because of the Gregorian rule, so if you apply this rule literally to 2100 you have to add 44 years (not 33) to get 2044, which gives the wrong Doomsday if you use Tuesday as the Doomsday in the 22nd century.<br /><br />So instead of talking about "leap years" in any particular century, we should be asking whether y is divisible by 4. So if y is divisible by 4, just divide it by 2 and subtract that from the next higher multiple of 7.<br /><br />I think that the correct rule when y is not divisible by 4 is, "If y is not divisible by 4, just repeatedly add 11 to y until you get a number divisible by 4, and apply the same formula to that number. At most this will require adding 33 to y."<br /><br />Since the four-century rule of the Gregorian calendar applies to 2100, the Gregorian interpretation of 2100 as "not a leap year" should not have anything to do with the calculation because we are using the 2000 Doomsday of Wednesday once you divide the number gotten by 2. We are only talking about year numbers here, as Bob's prescient remark notes, not the exact number of days in any particular year (like 2100).<br /><br />Bill Jefferyshttps://www.blogger.com/profile/03666872494092692149noreply@blogger.comtag:blogger.com,1999:blog-3985237709515317836.post-67471334875187323192013-09-13T19:29:40.253-07:002013-09-13T19:29:40.253-07:00This comment has been removed by the author.Bill Jefferyshttps://www.blogger.com/profile/03666872494092692149noreply@blogger.comtag:blogger.com,1999:blog-3985237709515317836.post-72954079271576565932013-09-13T19:01:47.433-07:002013-09-13T19:01:47.433-07:00This comment has been removed by the author.Bill Jefferyshttps://www.blogger.com/profile/03666872494092692149noreply@blogger.comtag:blogger.com,1999:blog-3985237709515317836.post-53440086215255320802013-09-13T18:53:48.583-07:002013-09-13T18:53:48.583-07:00This comment has been removed by the author.Bill Jefferyshttps://www.blogger.com/profile/03666872494092692149noreply@blogger.comtag:blogger.com,1999:blog-3985237709515317836.post-36721094185718834572013-09-13T17:32:54.643-07:002013-09-13T17:32:54.643-07:00This is wonderful!
I've used Conway's Doo...This is wonderful!<br /><br />I've used Conway's Doomsday calculation in my course (before I retired):<br /><br />http://bayesrules.net/BillInfo/doomsday.html<br /><br />But your method is so simple and easy. If I were still teaching, I'd use it. I'm having trouble updating my webpages at the moment, but as soon as I solve that problem I'll point my page to yours.<br /><br />Have you mentioned this to Prof. Conway? I am sure he'd be delighted.<br /><br />It would be interesting for you to develop a tutorial calculator such as the one given here<br /><br />http://www.timeanddate.com/date/doomsday-calculator.html<br /><br />to help people learn your very simple algorithm. <br /><br />Regards, Bill Jefferys<br />Bill Jefferyshttps://www.blogger.com/profile/03666872494092692149noreply@blogger.comtag:blogger.com,1999:blog-3985237709515317836.post-8509312557087598662013-08-03T19:15:30.632-07:002013-08-03T19:15:30.632-07:00regarding the ki/hcs proposal:
This suggestion of...regarding the ki/hcs proposal:<br /><br />This suggestion of decomposing the 2-digit year to leap quotient and remainder has been suggested by Dr. Yingking Yu in Sept. 2010:<br />http://improvedddabyykyu.blogspot.com/<br />(in section 3.2 The Highest Multiple of 4 Algorithm)<br />The proposal is a variant of YingKing's method.<br /><br />I still prefer the odd+11 method because:<br />1) it doesn't involve any division by 4 calculation for quotients and remainders <br />2) it doesn't require one to remember an intermediate variable (e.g. the remainder) that was subtracted after division by 2<br />3) it is easier to add 11 than to subtract a variable remainderjackerzhttps://www.blogger.com/profile/09035813176017376446noreply@blogger.comtag:blogger.com,1999:blog-3985237709515317836.post-76181673281756141242011-04-07T15:30:04.122-07:002011-04-07T15:30:04.122-07:00Divide last leap year by 2, subtract the lost year...Divide last leap year by 2, subtract the lost years, calculate modulo 7 and find the complement to 10.<br /><br />For example<br />1995=1992+3<br /><br />92/2 ==> 46 - 3 ==> 43 ==> 1 to 10 ==> 9 ==> 2 ==> Doomsday is a Tuesday <br /><br />Best<br />hcskihttps://www.blogger.com/profile/17147255486272661932noreply@blogger.comtag:blogger.com,1999:blog-3985237709515317836.post-63863306012466549102009-03-26T14:25:00.000-07:002009-03-26T14:25:00.000-07:00Thank you Bob! You're addition to my method is ve...Thank you Bob! You're addition to my method is very elegant. I've added a link to your calendar page in my sidebar too.Mikehttps://www.blogger.com/profile/09902477043993196546noreply@blogger.comtag:blogger.com,1999:blog-3985237709515317836.post-53816793876801815572009-03-20T08:26:00.000-07:002009-03-20T08:26:00.000-07:00Mike, thanks for this breakthrough simplification!...Mike, thanks for this breakthrough simplification! <BR/><BR/>You say your method fails if you add to the year and cross a century, and you give an extra rule, requiring us to add 1 in this event. But I have good news: <B>there's no need for this warning and the extra rule</B>. Just do the following:<BR/><BR/>When adding 11s to y (the last two digits of the year), it's OK if you go over 100. Don't even think about overflowing ("carry the one") into the next century. Just continue the math, and in the final step stick with the original century.<BR/><BR/>For example, for the year 2095, we add 95+33=128. Divide by 2 to get 64, and subtract from a big multiple of 7, to get 70-64=6. Added to Tuesday (Doomsday 2000) the result is Monday. It would be wrong to equate 2095 with 2128, because 2100 is not a leap year.<BR/><BR/>Thanks again.Bob Goddardhttps://www.blogger.com/profile/03308815202377527967noreply@blogger.com