OttoeMezzo published in August 2011 his ebook "Secret Files" with the routine " A Day for any Date Improved". The routine is based on a very clever method.
OttoeMezzo's ebook is very good written and I really liked his writing style and his clever methods. I'm looking forward to see what clever ideas OttoeMezzo will come up with in the future....
I'm very sorry but OttoeMezzo is not the first using the approach written in his "Secrets Files". So I must disappoint OttoeMezzo - he independent developed his great method.
I cover this approach in chapter "2.6.1. Decade and Year in the Decade", pp. 62-66, of my book "Encyclopedia of Weekday Calculation". Because this encyclopedia is written in German the content is known only to insiders.
OttoeMezzo's approach of separating years into decades and years in the decade was first published by the Russians L. T. Sakharosky and W. W. Sokolov in 1957 and 1966. About 50 years later the German Eisele, the Indian Divasli, the Amerians YingKing Yu and Strooke developed similiar methods.
In 2010 Chamberlain Fong published his version. This version is identical with OttoeMezzo's version. You will get identical results. The only difference is that Fong suggests to memorize the ten decades codes instead of calculation by his formulae. OttoeMezzo suggests use of simple formulaes.
Of course I will cite OttoeMezzo's method in my Encyclopedia as a version of Fong's method and as a remarkable source.
Es werden in sporadischer Folge mathematische und mnemotechnische Probleme und Methoden betrachtet, die es ermöglichen, den Wochentag für jedes Datum nach dem Gregorianischen oder Julianischen Kalender zu berechnen. Grundlage ist die "Enzyklopädie der Wochentagsberechnung" von Hans-Christian Solka, das umfassendste Werk zu diesem Thema.
There is only a minor difference between Fong and OttoeMezzo. Only if the decade is odd then Fong's and OttoeMezzo's algorithms are different:
AntwortenLöschenFong:
Decade key is two times decade plus 10.
OttoeMezzo:
Decade key is two times (decade minus 2).
Solka
thank you for the very informative post. I didn't know about OttoeMezzo's method.
AntwortenLöschenHere is a proof of equivalence:
AntwortenLöschen(equality is over mod 7)
2(d-2) = 2d-4 = 2d+3 = 2d+10 (in mod 7)
Correct :=))
AntwortenLöschen... but only if the decade is odd.
If decade is even then decade key is two times decade.
Year code is decade code plus year in decade plus number of gone leap years in decade:
1929 (even decade)
2*2 + 9 + 2 = 15 => 1
1934 (odd decade)
2*3+10 + 4 + 1 = 21 => 0
Best,
Solka