Mental juggling: Calculating the day of week

[Thomas Klemm]

My father has a quirk: When he meets someone new, he asks after the date of birth and
then after a few seconds of reflection, he says the corresponding day of the week.


I do not know exactly how he makes, but he knows many of the weekday data from memory.
This is used to calculate the difference to this particular day.


Though there are algorithms to do that with a computer or a programmable calculator I
was trying to find a way to do it in my mind. However mental arithmetic is not one of my skills.

Let me explain it first with an example: 22.06.1955

Day   22      -> 1

Month  6      -> 4

Year  55      -> 5

                --

                10 -> 3 -> Wednesday

Day

This one is easy: it's calculating the remainder after division by 7.

Example: 22 = 3 * 7 + 1

Actually you don't have to divide by 7. You just need to know the closest
multiple of 7 which is smaller or equal and then calculate the difference:

22 - 21 = 1

Month

Here you don't have to calculate anything but keep a table in your mind:

Month           1  2  3  4  5  6  7  8  9 10 11 12

Magic number    1  4  3  6  1  4  6  2  5  0  3  5

As I can't remember such a table very well I rearranged it a little:

Month         Magic number

 1  2  3      1  4  3

 4  5  6      6  1  4

 7  8  9      6  2  5

10 11 12      0  3  5

For Januray and March the mapping is identity, for October it is 0 which means it
can be ignored. Then you can see that the 2nd and 3rd line both start with 6.
In the 2nd line 1 4 from the first line is repeated but shifted by 1
whereas in the 3rd line 1 is added giving 2 5.
Well, the last line: 10 is 0 and 1 added to 2 gives 3 and 5 is keept.
But I'm sure you'll find a different way to memorize this table.

How did I calculate this mapping?

Starting with March which is set to 3 either 31 or 30 is added depending on how many
days the previous month has:

month   # days    sum    % 7

  3        31       3     3

  4        30      34     6

  5        31      64     1

  6        30      95     4

  7        31     125     6

  8        31     156     2

  9        30     187     5

 10        31     217     0

 11        30     248     3

 12        31     278     5

  1        31     309     1

  2         *     340     4

(*) Since February is the last month in the list we don't have to take
leap years into account here. This is the reason why we started with March.

Year

Leap years make things complicated. Here again we have a mapping table:

 0 :  00  06      17  23  28  34      45  51  56  62      73  79  84  90      101  107

 1 :  01  07  12  18      29  35  40  46      57  63  68  74      85  91  96  102    

 2 :  02      13  19  24  30      41  47  52  58      69  75  80  86      97  103  108

 3 :  03  08  14      25  31  36  42      53  59  64  70      81  87  92  98       109

 4 :      09  15  20  26      37  43  48  54      65  71  76  82      93  99  104  110

 5 :  04  10      21  27  32  38      49  55  60  66      77  83  88  94      105  111

 6 :  05  11  16  22      33  39  44  50      61  67  72  78      89  95 100  106    

Search the year (- 1900) in the table and find the corresponding number in the leftmost column.

Now this is something I don't want to learn by heart. However you may notice that the pattern
repeates after 28 years so we just need to remember the first 28 years.
Unfortunately calculating modulo 28 isn't something I can do easily in my mind.

This is the easiest way I came up to calculate the number:

1. Find the closest multiple of 4 smaller or equal to the year. Keep the difference.
   
2. Divide that multiple of 4 by 2.
   
3. Find the difference to the closest multiple of 7 bigger or equal to that number.
   
4. Add this to the difference from step 2.
   

Example: 55 = 52 + 3

              52 : 2 = 26

              26 + 2 = 28

              3 + 2 = 5

As the calculations are carried out modulo 7 we can also subtract the difference to the
closest multiple of 7 smaller than that number in steps 3 and 4.

Example: 46 = 44 + 2

              44 : 2 = 22

              22 - 1 = 21

              2 - 1 = 1

Shortcut for years that are a multiple of 7:

1. Find the remainder after dividing by 4
   
2. Multiply that number by 5
   

Example: 49 = 48 + 1

                   1 * 5 = 5

         91 = 88 + 3

                   3 * 5 = 15 = 1 (7)

Important note: You may have noticed that the mapping table for the months starts
with March. Thus January and February belong to the year before the actual year.
Therefore for these two months the year has to be reduced by 1 before any calculations.

There are different ways to do that calculation. But it turned out that I can calculate
n modulo 4 quiet fast while I have difficulties to divide a number by 4 if it's bigger than say 50.
On the other I find no trouble to divide a number by 2. Probably it's all a question of training.
However it seems I just don't do enough divisions by 12 to use

Lewis Carroll's method:

Quote:

---

Add together the number of dozens, the overplus, and the number of 4s in the overplus.

---

Day of the week

The following table is used to map the resulting number to the day of the week:

  0  Sunday

  1  Monday

  2  Tuesday

  3  Wednesday

  4  Thursday

  5  Friday

  6  Saturday

Explanation

What we actually do is calculating the difference of the date to 28.02.1900 modulo 7.
This interval is split into three parts from which modulo 7 can be found easier:

22.06.1955

             22                = 1 (7)

 0.06.1955

             31 + 30 + 31 = 92 = 1 (7)

 0.03.1955

             55 + 13 = 68      = 5 (7)       // 55 : 4 = 13 leap years

 0.03.1900

28.02.1900                     = 3

                                --

                                10 = 3 (7)

Since 28.02.1900 was a Wednesday the mapping table of the months starts with 3.

What remains unexplained are the two formulas to calculate the mapping of the year:

Let's assume y = 4n + r  ; 0 <= r < 4

Then d = 365y + n = y + n (7)   // since 365 = 1 (7)

                  = 4n + r + n = r + 5n = r - 2n (7)

2n = 4n / 2

d = r - 4n / 2 (7)    // here we use that 2 | 4n

Last not least the case where the year is a multiple of 7:

y = 4n + r = 7k = 0 (7)

r = -4n = 3n (7)      // multiply both sides by 5

5r = 15n = n (7)          

d = y + n = 0 + 5r (7)

Restriction

The method described here works only for dates from 01.03.1900 to 28.02.2100.
However it could easily be adjusted.

Examples

6 July 1979

79 - 76 = 3

76 / 2 = 38

38 - 3 = 35

3 - 3 = 0

July -> 6

0 + 6 + 6 = 12 = 5 (7)

or shortened

79 76  3

38 35 -3

       0 6 -1 5 -> Friday

---

December 7th, 1941

41 - 40 = 1

40 / 2 = 20

20 + 1 = 21

1 + 1 = 2

Dec -> 5

2 + 5 + 7 = 14 = 0 (7)

or shortened

41 40 1

20 21 1

      2 5 0 -> Sunday

---

20.02.1964

63 - 60 = 3

3 * 5 = 15 = 1 (7)

Feb -> 4

1 + 4 + 20 = 25 = 4 (7)

or shortened

63 60 3

     15

      1 4 -1 4 -> Thursday

Programming

How about a training program for your favorite HP calculator that displays a random date
and asks you to enter the appropriate day of the week?

Edited: 7 July 2009, 1:40 p.m. after one or more responses were posted

[Allen]

This is a great write up.. I recommend making an article of the same title and also posting there. :)

[Mad Dog ebaycalcnut]

Pretty cool trick.

Do people get suspicious of identity theft when asking their birthdate?

[Thomas Klemm]

Your question is interesting since people are indeed sometimes irritated. But usually they just tell their date of birth. However someone turned the table on him and told him only birthday and weekday and asked him to tell the year. He had a little longer and guessed the person too young but the result would have been possible.

I remember a similar irritation when traveling in Viet Nam since people always ask you how old you are. Yet they need to know because they call you different depending on whether you are older or younger than they are.

So I assume that identity theft isn't the main concern but simply the fact that somebody asks something personal.

[PeterP]

Awesome post Thomas, please make it an article!

Cheers

Peter

[Thomas Klemm]

cf. 926: Mental juggling: Calculating the day of week

[Juergen Keller]

Great article, thank you! This remembers me of the R500/7T relay computer which a colleague of mine has built. He usually demonstrates the computer by running the 'day of the week' program. The program runs for about 20 seconds!

Juergen

[Bernard Rochlin]

See the formula by John Horton Conway called the 'Doomsday Rule'

http://en.wikipedia.org/wiki/Doomsday_algorithm

http://rudy.ca/doomsday.html

Edited: 9 July 2009, 7:45 a.m.

[Thomas Klemm]

Quote:

---

For any year 19YY, using the YY part of the year, calculate:

1. the number of 12's in the YY part of the year
   
2. the remainder of step 1
   
3. the number of 4's in the remainder of step 1
   

---

This is Lewis Carroll's method already mentioned. So you can also combine my method (for the year) with Doomsday Algorithm if you like. Then you don't have to memorize the mapping table for the months.