Editing Doomsday rule (section)

===The "odd + 11" method===
[[File:Odd+11 doomsday flowchart.svg|thumb|right|270px| A flowchart showing the Odd+11 method to calculate the anchor day]]
A simpler method for finding the year's anchor day was discovered in 2010 by Chamberlain Fong and Michael K. Walters.<ref name=Odd11paper>Chamberlain Fong, Michael K. Walters: [https://arxiv.org/abs/1010.0765 "Methods for Accelerating Conway's Doomsday Algorithm (part 2)"], 7th International Congress on Industrial and Applied Mathematics (2011).</ref> Called the "odd + 11" method, it is equivalent<ref name=Odd11paper /> to computing

:<math>\left(y + \left\lfloor \frac{y}{4} \right\rfloor\right) \bmod 7</math>.

It is well suited to mental calculation, because it requires no division by 4 (or 12), and the procedure is easy to remember because of its repeated use of the "odd + 11" rule. Furthermore, addition by 11 is very easy to perform mentally in [[Decimal|base-10 arithmetic]].

Extending this to get the anchor day, the procedure is often described as accumulating a running total {{math|''T''}} in six steps, as follows:
# Let {{math|''T''}} be the year's last two digits.
#If {{math|''T''}} is odd, add 11.
#Now let {{math|''T'' {{=}} {{sfrac|''T''|2}}}}.
#If {{math|''T''}} is odd, add 11.
#Now let {{math|''T'' {{=}} 7 − (''T'' mod 7)}}.
#Count forward {{math|''T''}} days from the century's anchor day to get the year's anchor day.

Applying this method to the year 2005, for example, the steps as outlined would be:
#{{math|''T'' {{=}} 5}}
#{{math|''T'' {{=}} 5 + 11 {{=}} 16}} (adding 11 because {{math|''T''}} is odd)
#{{math|''T'' {{=}} {{sfrac|16|2}} {{=}} 8}}
#{{math|''T'' {{=}} 8}} (do nothing since {{math|''T''}} is even)
#{{math|''T'' {{=}} 7 − (8 mod 7) {{=}} 7 − 1 {{=}} 6}}
#Doomsday for 2005 = 6 + Tuesday = Monday

The explicit formula for the odd+11 method is:
:<math> 7- \left[\frac{y+11(y\,\bmod 2)}{2} + 11 \left(\frac{y+11(y\,\bmod 2)}{2}\bmod 2\right)\right] \bmod 7</math>.

Although this expression looks daunting and complicated, it is actually simple<ref name=Odd11paper /> because of a [[common subexpression]] {{math|{{sfrac|''y'' + 11(''y'' mod 2)|2}}}} that only needs to be calculated once.

Anytime adding 11 is needed, subtracting 17 yields equivalent results.  While subtracting 17 may seem more difficult to mentally perform than adding 11, there are cases where subtracting 17 is easier, especially when the number is a two-digit number that ends in 7 (such as 17, 27, 37, ..., 77, 87, and 97).