Deriving Zeller's RuleDate: 02/25/2003 at 23:09:20 From: Coby Subject: How do you derive the formula for the day of the week? I am doing a report on the formula for the day of the week for any day, but I cannot completely figure out how to derive it. How do you derive the formula f=k+[(13*m-1)/5]+d+[d/4]+[c/4]-2*c ? Date: 02/26/2003 at 09:11:15 From: Doctor Peterson Subject: Re: How do you derive the formula for the day of the week? Hi, Coby. Since there are several slightly different formulations of this formula, we'll use the one you are using, from our FAQ: The Calendar and the Days of the Week http://mathforum.org/dr.math/faq/faq.calendar.html Here we're defining k = day of month m = month number, taking Mar=1, ..., Dec=10, Jan=11, Feb=12 d = last two digits of year, using the previous year for Jan and Feb c = first two digits of year The formula is then f = k + [(13*m - 1)/5] + d + [d/4] + [c/4] - 2*c and we use the remainder after dividing f by 7 to find the day of the week. Where does this come from? Let's first note the reason for the odd handling of months: we want leap day not to affect the formula, so we move it to the end of the 'year', and act as if the year began on March 1. Now note that in defining f, all we care about is the remainder after dividing, so it will be enough to make sure that f increases by 1 whenever the day of the week advances by one day; we don't care about the actual value of f. Now let's build the formula piece by piece. How does the year affect the day? Well, since 365 = 7*52+1, each normal year advances the day by 1, so our formula can start with the year number: f = d Whenever the year advances by 1, so does the day of the week. But we have to adjust this to account for leap years. Every four years we have an extra day, so we'll want to add 1 to f. This is done by adding [d/4], since this increases by 1 only when d becomes a multiple of 4, which is a leap year. So now we have f = d + [d/4] Now how do centuries affect the day? A century contains 100 years, 24 of which normally are leap years (since century days, like 1900, are NOT leap years). So each century the day advances by 124 days, which is 7*18-2, and therefore the day of the week goes BACK 2 days. So we have f = d + [d/4] - 2*c But every fourth century year IS a leap year (as 2000 was), so we have to adjust just as we did for leap years: f = d + [d/4] - 2*c + [c/4] Now we come to the months, and this is the cute part. Consider, for each month, how many days it has BEYOND 28, and then add that up to see the effect the months have on the day of the week: 1 2 3 4 5 6 7 8 9 10 11 12 Month Mar Apr May Jun Jul Aug Sep Oct Nov Dec Jan Feb Days 31 30 31 30 31 31 30 31 30 31 31 (28) Excess 3 2 3 2 3 3 2 3 2 3 3 0 Accum 0 3 5 8 10 13 16 18 21 23 26 29 \_________________/\__________________/\_______ The number of accumulated days is counted at the start of the month, so if we divide it by 7, the remainder shows how many weekdays the start of the month is from the starting day for the 'year'. Notice the pattern in the excess: 3,2,3,2,3 repeats every five months, and the accumulation reaches 13 in that time. So every 5 months, we want to add 13 days. That suggests that we want to add a term like [13m/5]. That doesn't quite give us what we want: 1 2 3 4 5 6 7 8 9 10 11 12 Month Mar Apr May Jun Jul Aug Sep Oct Nov Dec Jan Feb Days 31 30 31 30 31 31 30 31 30 31 31 (28) Excess 3 2 3 2 3 3 2 3 2 3 3 0 Accum 0 3 5 8 10 13 16 18 21 23 26 29 13m 13 26 39 52 65 78 ... [13m/5] 2 5 7 10 13 15 ... If we subtract 2 from this, it isn't quite right; we have to shift it a bit. So after playing with it a bit, we find 13m-1 12 25 38 51 64 77 ... [(13m-1)/5] 2 5 7 10 12 15 ... [(13m-1)/5]-2 0 3 5 8 10 13 ... That's just what we want. So we'll use f = d + [d/4] - 2*c + [c/4] + [(13m-1)/5] - 2 Finally, we have to add the day, since each day obviously adds one to the day of the week; and adjust to get the right day of the week for, say, Mar 1, 2000, since nothing we've done so far actually determined WHICH day we start the whole pattern on. It turns out that we can just remove the -2, and we get f = d + [d/4] - 2*c + [c/4] + [(13m-1)/5] + k And there's the formula! For more details on the origin of the formula, see Calendrical Works of Rektor Chr. Zeller http://www.merlyn.demon.co.uk/zeller-c.htm#ZC If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 02/26/2003 at 21:23:26 From: Coby Subject: Thank you (How do you derive the formula for the day of the week?) Thank you so much! Your answer was realy helpful and detailed. Thank you! |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/