Hi!

I wanna find a reliable method to ask the birth date and knowing which day was...

I found a few methods, but doesn't fit the criteria.

Any help?

Go to Lybray.com and get a memory book by Zufall. You'll have to memorize a hundred code words, but once you do that it's a breeze to do any day for a date... for over 600 years.

Greg

Here's the book series to which Greg is referring.

Here's my instructions on it, including......u on it.

John Conway's Doomsday algorithm can seem simpler, especially if you remember the poem about it.

No matter which method you choose though, practice is the key. Worry about accuracy first, and the speed will come with practice.

I posted something that works nicely for dates only between 1900 and 2000 at

http://www.themagiccafe.com/forums/searc......=4064377

The short formula is this, but you need to read the post to see what the variables represent:

Year 19y:

Weekday=(1+y+floor(y/4)+M+D)mod7

/Tomas

Sorry for the double post, but if 1900-2000 are the only dates you are going to be interested in include the 1 in the M code so the formula becomes

Weekday=(y+floor(y/4)+M+D)mod7

and the M codes for the months become

0-3-3, 6-1-4, 6-2-5, 0-3-5

/Tomas

Thanks !!!!

Paul Brook's book "The Chrysalis of a Polymath" has a terrific version. It took me about an hour to wrap my head around it. Although it applies to any year, he focuses on 1900's & 2000's.

I agree with Mr. Mindbender, but it should be pointed out that it basically is the Doomsday method, mentioned above.

@TomasB
This formula was invented bei Aime Paris in 1866.
The century part is more clever expressed by
H=6-2*Y div 4
then as
H=5Y+floor(Y/4).

See my Encyclopedia about calendar math "calendar in the head - the grail":
http://www.lybrary.com/kalender-kopf-gral-p-35993.html
I aplogize but this book is available only in German.
The grail includes a lot of different known and unknown methods and brand new concept of easy memorizing century-year-key number relationship.

Many thanks hcs! That looks a lot easier.

I'm just not sure how it works or relates to formula I wrote. The one you show gives the same H for the years starting with 16, 17, 18 and 19 while mine doesn't. Or am I misunderstanding what you mean by "div"? The way I interpret "div" is to divide by 4 and get rid of any decimals, which would make the expression constant for four centuries in a row.

/Tomas

Here's a couple of more interesting alternative calendar approaches:

On what day of the week were you born?: This one features an interesting approach using keys for decades!

Abel Strook's formula: This formula uses only 2 numbers. The catch? You have to know how far into the year a given date is, such as April 15th being the 105th day of the year (which can be learned in Thomas Harrington's Magic of Memory).

H=6-2*Y div 4
div means the remainder of a division

Examples:
16xx; Y=16
H=6-2*16 div 4=6-0=6
17xx; Y=17
H=6-2*17 div 4=6-2*1=4
18xx; Y=18
H=6-2*18 div 4=6-2*2=2
19xx; Y=19
H=6-2*19 div 4=6-2*3=0
20xx=16xx….

Abel Strook's formula is very interesting. It is a variation of formula, published in 1909 by the russian mathematician J.I. Perelman, in the russian Journal "Priroda I Lyudi (nature and people), 1909, No. 22
cited in Butkevich/Selikson "Ewige Kalender (perpetual calendars)", Leipzig, 1987, p.91
(or in my encyclopedia "2.5. unechte Kennzahlmethoden (unreal key number methods), 2.5.2. "Tag I'm Jahr"-Methode ("day in year" method)):
W = (Y + Y int 4 + H int 4 – H + D – 1)div 7

Differences in formula caused by different key numbers for day of the week and that Perelman used.

In my encyclopedia I descripe 2 ways how to determine the day of year in "1.3.1. "Tag I'm Jahr (day of the year"):
a) DY = 30(M - 1) + D + M'
M' is 0;1;-1/0;0;1/1;2;3/3;4;4; in leap years beginning with March +1)
b) Urs Oswald from Suisse
DY = M' + D = (153M – 457)int 5 + D

Wow, with all these equations I'm really glad I took the time to memorize Zufall's list. It's practically instantaneous once you practice it.

Good luck to all. It is quite an amazing feat once you have it down. In the times that I've been at a social gathering, and I do one one person, it becomes a tsunami of people wanted their birthdays done... and siblings and children, etc.

Greg

This is correct. Zufall published a very powerfull memory system. The best system of his time. My equations above are not suitable for mental calculations.

In my encyclopadia I published a lot of different aproaches for determination of weekday (julian and gregorian) and other calendar parameters by equations (congruences), computistic, mental calculations and memory work.

This is not interesting for mentalists only but also for astronoms, historians, theologists, computer programmers and other scientist.

By the way I developed also a memory system for all years of a 400 years gregorian cycle on the state of 21th century. It is small, easier and more powerfull than Zufall's great system.
In May will be published a new revised edition of my book.

I see, hcs. The "div" confused me as it's much more common to write "mod".

So if I bake the lonely 6 into the M code I get

M=6-2-2, 5-0-3, 5-1-4, 6-2-4

Weekday = (y + floor(y/4) - 2*Ymod4 + M + D)mod7

Is there a good way to remake the "y + floor(y/4)" the same way? Anyway, the formula looks so much simpler now. Thanks!

/Tomas

Wow! That's an expensive e-book! It looks very interesting. Is it possible that the price listed is a typo?

hcs, do you know Ulrich Voigt?

There is no typo. The book is designed for professionals only to preserve some secrets.
Of course I know Ulrich Voigt very close and I cite him very often in my book.
I dedicated him my book.
He wrote about my book „Ich bin von der Arbeit wirklich beeindruckt; es ist selten, dass auf scheinbar so abgegrasten Gebieten Neues herauskommt.“
(I'm realy deeply impressed by your work. It is very seldom to find new facts in such a well known field".)

Dear Scott, thank you for sharing this link. This australian source is new for me.

I describe such a method using decades in my book in chapter
2.5. unechte Kennzahlmethoden (unreal offset methods)
2.5.1. Jahrzehnt, Jahr I'm Jahrzehnt (decade, year in decade).

This "Five offset method" method is a variation of a method originated by the Indian I. E. DIVASLI.
It is in my opinion a little bit "more elegant" then the quoted method by Polster and Ross.
Here we go:

Y = 100H + 10Z + E
Y' = H' + Z' + E'
H = Y int 100
Z = Y100 int 10
E = Y mod 10

symbols
Y - year
H - number of hundreds
Z - number of decades
E - year in decade
Z' - offset for decade
E' - offset for year in decade

...
If Z even, then Z' = 2Z - 1
If Z odd, then Z' = 2Z - 4
If Z even and
0 lower equal E lower equal 3, then E' = E + 1
4 lower equal E lower equal 7, then E' = E + 2
8 lower equul E lower equal 9, then E' = E + 3
If Z odd and
0 lower equal E lower equal 1, then E' = E
2 lower equal E lower equal 5, then E' = E + 1
6 lower equal E lower equal 9, then E' = E + 2

Examples

1929
Z=2; E=9
Z' = (2*2 - 1) = 3 E' = (9 + 3) = 12 ==> 5

1934
Z=3; E=4
Z' = (2*3 - 4) = 2 E' = (4 + 1) = 5

1960
Z=6; E=0
Z' = (2*6 - 1) = 11 ==> 4 E = (0 + 1) = 1

1978
Z=7; E=8
Z = (2*7 - 4) = 10 ==> 3 E = (8 + 2) = 10 ==> 3


H': 6 - 4 - 2 - 0
M': 033 - 614 - 625 - 035
M': 623 - 614 - 625 - 035
W: 0 - Sunday, 1 - Monday ... 6 - Saturday

23. August 1994
D=23; M=8; Y=1994
H=19; Z=9; E=4
H'=0; M'=2
Z = (2*9 - 4) = 14 ==> 0 E = (4 + 1) = 5

W = (D + M' + H' + Z' + E') mod7 = (23 + 2 + 0 + 0 + 5)mod7
= 30/7 ==> 4 remainder 2 ==> Tuesday

19. Januar 2008
D=19; M=1; Y=2008
H=20; Z=0; E=8
H'=6; M'=0-1=6 (leap year)
Z' = (2*0 - 1) = -1 ==> 6 E' = (8 + 3) = 11 ==> 4
W = (D + M' + H' + Z' + E') mod7 = (19 + 6 + 6 + 6 + 4)mod7
= 41/7 ==> 5 remainder 6 ==> Saturday