"Triple Palindromic" Date CalculationsUK Format only - dd/mm/yyyy
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The approach for evaluating palindromes for the US date
format - mm/dd/yyyy - follows along pretty much the same lines. 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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