"Triple Palindromic" Date Calculations

UK Format only - dd/mm/yyyy

The approach for evaluating palindromes for the US date format - mm/dd/yyyy - follows along pretty much the same lines.
However, that is left as an exercise for the reader!

This is not intended to be a rigorous mathematical proof - it is more a layman's way of understanding what's going on.

If you have any comments or disagreements, please contact me.

For the sake of both sanity and brevity, we'll consider only 4 digit years (i.e. from the year 1000 AD to 9999 AD). I think this is reasonable - apart from the fact that in his days Julius C didn't use a 24 hour clock, I'm pretty certain they did not insert a gratuitous zero in front of their 1, 2 and 3 digit years. At the other end of the scale - beyond the year 9999 - who really gives a damn anyway. So we are considering very simply 3 groups of 4 digits each for genuine "triple palindromicity" - ie they must take the form nmmn nmmn nmmn. (See "normal palindromicity" for corresponding calculations/estimations)

 

We start by defining the two digits of each element:

H1H2  Hours
Mi1Mi2  Minutes
D1D2  Day
Mo1Mo2  Month
C1C2  Century
Y1Y2  Years

Using this notation we can express the time and date as H1H2:Mi1Mi2      D1D2/Mo1Mo2       C1C2Y1Y2

Now we know that each of these elements is subject to certain constraints:-

1) we define triple palindromicity by the relationships:-

    a)    H1 = Mi2 = D1 = Mo2 = C1 = Y2
    and
    b)    H2 = Mi1 = D2 = Mo1 = C2 = Y1

 

2) Looking at the constraints of each element in turn:-

a) The hours (H1H2) must lie between 00 and 23: H1 = 0, 1 or 2
and H2 = 0...9
and 00 <= H1H2 <= 23
b) The minutes (Mi1Mi2) must lie between 00 and 59 Mi1 = 0...5
and Mi2 = 0...9
c) The days (D1D2) must lie between 01 and 31
(but watch out for 28, 29 and 30 as well) 
D1 = 0, 1, 2, or 3
and D2 = 0...9
and 01 <= D1D2 <= 31
d) The months (Mo1Mo2) must lie between 01 and 12 Mo1 = 0 or 1
and Mo2 = 0...9
and 01 <= Mo1Mo2 <= 12
e) The century (C1C2) must  lie between 10 and 99
(we're looking only at the years 1000 to 9999)
C1 = 1...9
and C2 = 0...9
f) The years (Y1Y2) are totally unconstrained Y1 = 0...9
and Y2 = 0...9

 

Table 1 - combining 1a above with 2a-2f

H1 =  0,1,2
Mi2 = 0,1,2,3,4,5,6,7,8,9
D1 = 0,1,2,3
Mo2 = 0,1,2,3,4,5,6,7,8,9
C1 = 1,2,3,4,5,6,7,8,9
Y2 = 0,1,2,3,4,5,6,7,8,9

This leads directly to the conclusion that if H1=Mi2=D1=Mo2=C1=Y2 then they can each be only either 1 or 2

 

Table 2 - combining 1b above with 2a-2f

H2 =  0,1,2,3,4,5,6,7,8,9
Mi1 = 0,1,2,3,4,5
D2 = 0,1,2,3,4,5,6,7,8,9
Mo1 = 0,1
C2 = 0,1,2,3,4,5,6,7,8,9
Y1 = 0,1,2,3,4,5,6,7,8,9

This leads directly to the conclusion that if H2=Mi1=D2=Mo1=C2=Y1 then they can each be only either 0 or 1

 

From there it can be demonstrated that there are only four possible combinations of palindromic UK dates and times:

Combination Time Date Year
1001 10:01 10 January 1001
1111 11:11 11 November 1111
2002  20:02 20 February 2002
2112 21:12 21 December 2112

QED

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