Calculate the day of the week for any date in your head
11–20 of 68 posts
Re: Calculate the day of the week for any date in your head
#12For all but the most basic details my rule is that if I have to remember it, the system is broken. I know a small handful of passwords off the top of my head, a few email addresses, my wife's and my phone numbers, and the phone number of my home during childhood. (I'm obviously exaggerating, you know, for dramatic effect.) This trick requires so many mnemonic devices that it makes my head spin. It's still impressive…
You only need to remember 5 things: - For 2020, the doomsday is Saturday. All the dates below are doomsdays (So this year they are all Saturdays): - 4/4, 6/6, 8/8, 10/10, 12/12, easy enough - 9/5, 5/9, 7/11, 11/7 (9-to-5 at 7-11) - 0th of March (last day of February) - 3rd of January for non-leap years, and 4th of January for leap years. So this year the 4th is Saturday. And to calculate dates: mod 7. So e.g. 7th of…
You can skip the leap year info completely if you consider both January and February as belonging to the previous year. That makes sense as the additional day (Feb 29) is at the end of Februry.
Last year Pi Day was Thursday, this year Saturday.
So the general rule then is: March 14th, April 4th, May 9th, June 6th, July 11th, August 8th, September 5th, October 10th, November 7th, December 12th, January 2nd, February 6th are Pi Days.
Edit: another nice rule is that usually Pi-Days increase by 1 each year (2 in leap years), so next year it will be Sunday, then Monday, Tuesday, Thursday (leap year), Friday, etc.
Re: Calculate the day of the week for any date in your head
#13It's a nice article and a good algorithm, kudos to all who can do this.
My philosophy however is that I don't need to learn this kind of stuff nowadays. There are computers even in our wrist, so I offload as much as I can, and spare my brain cycles more important stuff that matters.
Re: Calculate the day of the week for any date in your head
#14This looks like an algorithm that could be useful if programmed rather than carried around in one's head. It also still requires a lookup to bootstrap it: the day-of-week of the last day of February for the selected year. So it's a clever and well-observed compilation of identified patterns, but not self-contained.
Did you read the whole thing?
No. As soon as the algo asserts "the last day of February is Thursday the
28th," and didn't explain how it knew 2019 wasn't a leap year or that the 28th
was indeed a Thursday, I stopped reading.
Re: Calculate the day of the week for any date in your head
#15This is how I do it:
https://incubaid.wordpress.com/category/algorithms/
Re: Calculate the day of the week for any date in your head
#16My Autistic son does this in a couple of seconds. He can only really do the last 50 years tho.
Savants who do this are called "human calendars".
Re: Calculate the day of the week for any date in your head
#17> There are actually 52 (or 53) other days which are all on the same day of the week as "the" Doomsday at the end of February
That seems incorrect, there should always be 51 other days
Re: Calculate the day of the week for any date in your head
#18If not for religious groups, we could have had World Calendar and all this would be unnecessary: https://en.m.wikipedia.org/wiki/World_Calendar.
Re: Calculate the day of the week for any date in your head
#19> There are actually 52 (or 53) other days which are all on the same day of the week as "the" Doomsday at the end of February That seems incorrect, there should always be 51 other days
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Re: Calculate the day of the week for any date in your head
#20Earlier quoted context omitted.
The version they have here is somewhat inefficient and not well presented. You can get from the year to the doomsday by a simpler formula: If odd, add 11; divide by 2; if odd, add 11. Count up to nearest multiple of 7 So e.g. 1955 turns into 66 turns into 33 turns into 44 turns into 5. Add that to 4 for the century (1900 is 4, 2000 is 3, that's all you really need to memorize) and you get 9, so the doomsday for 1955…
To go from "any" century, you can also memorize "2053", which is the days in a 4-year cycle: 2000 = 2 = tuesday; 2100 = 0 = sunday; 2200 = 5 = friday; 2300 = 3 = wednesday. For years other than the centuries, we need to add an offset. Given the year CCXX, we have: (a, b) = divmod(XX, 12) c = b // 4 offset = a + b + c doomsday = start + offset E.g., for 2020 (XX == 20) we get: start = 2 # from the "2053" rule (a, b) =…
The "add 11" trick in my comment is easier to remember and apply than dividing by 12 and taking remainders.