# The Doomsday rule

coding · miscellaneous · algorithms coding · doomsday · algorithm · ruby

### The Doomsday rule

A few months ago I came across the name of J. H. Conway: you're wondering who the hell he is. Well, Conway is an English mathematician active in the theory of *finite groups*, *knot theory*, *number theory*, *combinatorial game theory* and *coding theory*. He has also contributed to many branches of *recreational mathematics* and he is the invention of the Game of Life. Ah, I was forgetting one last thing: he is currently **Professor Emeritus of Mathematics at Princeton University** in New Jersey^{1}. Ok. let's respect this guy but...what would I talk to you about? Well, in this article I will talk about a magic trick: the Doomsday rule.

### A brief history

Who has ever heard of it? No ones? Ok, the first thing to know is that the Doomsday rule is an *algorithm* of determination of the day of the week for a given date. This f****ng unbelievable trick of mind will allow you to win so many bets during the Christmas holidays that you will be envied by your nephews, close friends, neighbors, lovers and even your pets.

The mental calculation was devised by J. H. Conway in 1973: I don't know how much a *mentalist* you have to be to trust this method, but the algorithm actually returns the correct answer: if you want to discover more about how it works, go ahead. If you want to try to discover how the algorithm works starting from the code, here my repo with a Ruby implementation of the algorithm.

### The algorithm

First of all, a big picture of how the algorithm works in three steps:

**step 1**: determination of the anchor day for the century,**step 2**: calculation of the*doomsday*for the year from the anchor day,**step 3**: selection of the closest date out of those that always fall on the doomsday, e.g., 4/4 and 6/6, and count of the number of days (modulo 7) between that date and the date in question to arrive at the day of the week;

What?! Do you understand something?! Because if not, you really have to know that...Conway can usually give the correct answer to question like "what day of week was the 13 February 1867?", in under two seconds^{2}. If you're thinking that probably this is the main reason why you're reading this blog and he's **Professor Emeritus of Mathematics at Princeton University**, well, you're right.

In any case, since I'm a computer scientist, and I have absolutely no desire to count things in mind nor to open a calendar or search on google for my *quiz-date*, I have implemented the Conway algorithm for the calculation of Doomsday using Ruby.

### Preliminary explanation

This algorithm involves treating days of the week like numbers *modulo 7*: Conway often suggests - in his course in Princeton called *"How I will literaly blow your mind"* - thinking of the days of the week as "Noneday"; or as "Sansday" (for Sunday), "Oneday", (for Monday, it sounds similar), "Twosday", "Treblesday", "Foursday", "Fiveday", and "Six-a-day". It doesn't matter, because there are tables with - wait for it - *Doomsdays for the Gregorian calendar*, or what I call "The **Year** Doomsday". Like the one below:

| Mon. | Tue. | Wed. | Thu. | Fri. | Sat. | Sun. | Mon. | Tue. | Wed. | Thu. | Fri. | Sat. | Sun. | | ---- | ---- | ---- | ---- | ---- | ---- | ---- | ---- | ---- | ---- | ---- | ---- | ---- | ---- | | | 1898 | 1899 | 1900 | 1901 | 1902 | 1903 | → | 1904 | 1905 | 1906 | 1907 | → | 1908 | 1909 | | 1910 | 1911 | → | 1912 | 1913 | 1914 | 1915 | → | 1916 | 1917 | 1918 | 1919 | → | 1920 | | 1921 | 1922 | 1923 | → | 1924 | 1925 | 1926 | 1927 | → | 1928 | 1929 | 1930 | 1931 | → | | 1932 | 1933 | 1934 | 1935 | → | 1936 | 1937 | 1938 | 1939 | → | 1940 | 1941 | 1942 | 1943 | | → | 1944 | 1945 | 1946 | 1947 | → | 1948 | 1949 | 1950 | 1951 | → | 1952 | 1953 | 1954 | | 1955 | → | 1956 | 1957 | 1958 | 1959 | → | 1960 | 1961 | 1962 | 1963 | → | 1964 | 1965 | | 1966 | 1967 | → | 1968 | 1969 | 1970 | 1971 | → | 1972 | 1973 | 1974 | 1975 | → | 1976 | | 1977 | 1978 | 1979 | → | 1980 | 1981 | 1982 | 1983 | → | 1984 | 1985 | 1986 | 1987 | → | | 1988 | 1989 | 1990 | 1991 | → | 1992 | 1993 | 1994 | 1995 | → | 1996 | 1997 | 1998 | 1999 | | → | 2000 | 2001 | 2002 | 2003 | → | 2004 | 2005 | 2006 | 2007 | → | 2008 | 2009 | 2010 | | 2011 | → | 2012 | 2013 | 2014 | 2015 | → | 2016 | 2017 | 2018 | 2019 | → | 2020 | 2021 | | 2022 | 2023 | → | 2024 | 2025 | 2026 | 2027 | → | 2028 | 2029 | 2030 | 2031 | → | 2032 | | 2033 | 2034 | 2035 | → | 2036 | 2037 | 2038 | 2039 | → | 2040 | 2041 | 2042 | 2043 | → | | 2044 | 2045 | 2046 | 2047 | → | 2048 | 2049 | 2050 | 2051 | → | 2052 | 2053 | 2054 | 2055 | | → | 2056 | 2057 | 2058 | 2059 | → | 2060 | 2061 | 2062 | 2063 | → | 2064 | 2065 | 2066 | | 2067 | → | 2068 | 2069 | 2070 | 2071 | → | 2072 | 2073 | 2074 | 2075 | → | 2076 | 2077 | | 2078 | 2079 | → | 2080 | 2081 | 2082 | 2083 | → | 2084 | 2085 | 2086 | 2087 | → | 2088 | | 2089 | 2090 | 2091 | → | 2092 | 2093 | 2094 | 2095 | → | 2096 | 2097 | 2098 | 2099 | 2100 |

So...for a day, be a doomsday is a properties - at least - related to years. For instance, doomsday for the current year in the Gregorian calendar (2017) is Tuesday. The Conway algorithm works because it is possible to *easily* find the day of the week of a given calendar date by using a *nearby* doomsday as a reference point. To help with this, Conway findout that there is a list of *easy-to-remember* dates...for each month...

that always land on the doomsday of the given year: O.M.G.

What does it means? It means that, for instance, the last day of February defines a doomsday. Or January, January 3 is a doomsday during common years and January 4 a doomsday during leap years, and so on. What the hell of magic days are these doomsday?!? We will explain *how* the Doomsday for a year is computed. For now, let be the following table a month table that correspond **always** to the doomsday.

Month | Doomsdays |
---|---|

January | 3, 31 (solar year) 4 (leap years) |

February | 7, 14, 21, 28 (solar year) 1, 29 (leap years) |

March | 7, 14, 21, 28 |

April | 4 |

May | 9 |

June | 6 |

July | 11 |

August | 8 |

September | 5 |

October | 10 |

November | 7 |

December | 12 |

### Step 1 aka finding a year's base day

First of all, ask for a day

```
print 'Insert day: '
day = gets.to_i
print 'Insert month: '
month = gets.to_i
print 'Insert year: '
year = gets
```

Then, given

\[c = year's \; century + 1 \]

(for instance, \(c(2017) = 20\)), then compute:

\[\Big\{\Big[\Big(5c + \left \lfloor{\frac{c - 1}{4}}\right \rfloor\Big) \; mod \; 7\Big] + 4\Big\} \; mod \; 7\]

My really unefficient implementation of the formula using Ruby is:

```
baseDay = (((5 * (year[0, 2].to_i + 1)) + (year[0, 2].to_i / 4).ceil % 7) + 4) % 7
puts baseDay
```

For example, the base day for the 21st century is Tuesday, because:

\[\Big\{\Big[\Big(5 x 21 + \left \lfloor{\frac{21 - 1}{4}}\right \rfloor\Big) \; mod \; 7\Big] + 4\Big\} \; mod \; 7 = 2 = \; Tuesday\]

### Step 2 aka finding a year's Doomsday

Let be:

\[y = year's \; last \; 2 \; numbers \]

(for instance, \(y(2017) = 17\)), to determine the Doomsday of the year compute:

\[\Big[\Big(\left \lfloor{\frac{y}{12}}\right \rfloor + y \; mod \; 7 + \left \lfloor{\frac{y \; mod \; 12 }{4}}\right \rfloor \Big) \; mod \; 7\Big] + year's \; baseday = \; year's \; Doomsday\]

A possible implementation of the formula using Ruby is:

```
doomsday = (((year[2, 4].to_i / 12).ceil + (year[2, 4].to_i % 12) + ((year[2, 4].to_i % 12) / 4).ceil) % 7) + baseDay
puts doomsday
```

For example, the Doomsday of the year 1966 is Monday, because:

\[\Big[\Big(\left \lfloor{\frac{66}{12}}\right \rfloor + 66 \; mod \; 7 + \left \lfloor{\frac{66 \; mod \; 12 }{4}}\right \rfloor \Big) \; mod \; 7\Big] + 3 = 8\]

In *2010*, a **simpler method** - remember, less than two seconds - was discovered to find the year's Doomsday. It has been shown that this method, more formaly called *"Odd + 11"*, is equivalent in formulas to compute:

\[\Big[\Big(y + \left \lfloor{\frac{y}{4}}\right \rfloor \Big) \; mod \; 7 \Big] + year's \; baseday = \; year's \; Doomsday\]

This formula is particularly suitable for mental calculation, because it does not involve divisions and the procedure, as a recursive one, is easy to remember. So alternatively, to determine the doomsday it is also possible to add the last two digits of the year (*y*) to the quotient of the division between *y* and 4. In fact, also following this simpler formula, the Doomsday of the year 1966 is Monday:

\[\Big[\Big(66 + \left \lfloor{\frac{66}{4}}\right \rfloor \Big) \; mod \; 7 \Big] + 3 = 8\]

### Step 3 aka finding the day of the week of the desired day

After you know the Doomsday of the year in question, the calculation of the day of the week of the desired day is very simple:

- identify the Doomsday closest to the chosen day,
- subtract the date of the nearest doomsday from the date to be identified,
- add the value of the Doomsday of the year,
- and finally report the result obtained in base 7;

Ok, let's start from the...what!? How do you identify the Doomsday closest to the chosen day?! Ok, do you remember the table with the doomsday for each month I talked about before? The only thing you have to do is deal with some comparison between the month and some simple computation between abs difference. A possible implementation of the formula using Ruby is:

```
### Step 3 - find the day of the week
nearest = 9999999999
case month
when 1
if ((year % 4 == 0) and (year % 100 != 0)) or (year % 400 == 0)
nearest = 4
else
[3, 31].each {
|val|
if (day-val).abs < nearest
nearest = val
end
}
end
when 2
ddays = [1, 29]
if ((year % 4 == 0) and (year % 100 != 0)) or (year % 400 == 0)
ddays = [7, 14, 21, 28]
end
ddays.each {
|val|
if (day-val).abs < nearest
nearest = val
end
}
when 3
[7, 14, 21, 28].each {
|val|
if (day-val).abs < nearest
nearest = val
end
}
when 4
nearest = 4
when 5
nearest = 9
when 6
nearest = 6
when 7
nearest = 11
when 8
nearest = 8
when 9
nearest = 5
when 10
nearest = 10
when 11
nearest = 7
when 12
nearest = 12
else
nearest = 9999999999
end
```

And that's all: you are are done! In fact, the following three points of this third step can be computed with the following line:

`weekday = (day - nearest + doomsday) % 7`

And of course you can pretty print the result with something really dirty and compact like:

```
puts "~~~~~~~~~~~~~~~~~~~~~~~"
puts ['January', 'February', 'March',
'April', 'May', 'June', 'July',
'August', 'September', 'October',
'November', 'December'][month-1]+
" "+day.to_s+", "+
year+" was "+['Monday', 'Tuesday',
'Wednesday', 'Thusday',
'Friday', 'Saturday',
'Sunday'][weekday-1]
```

For example, to calculate the September 13, 2011:

- the nearest Doomsday is 5/9;
- 13 - 5 = 8
- 8 + 1 = 9 (the 2011 Doomsday was Monday, so the corresponding value is 1)
- 9 mod 7 = 2 = Tuesday

Thank you everybody for reading!

- John Horton Conway
^{[return]} - Do not worry: it's not an innate ability!! To improve his speed, he practices his calendrical calculations on his computer, which is programmed to quiz him with random dates every time he logs on.
^{[return]}