Extended Calendar Cube Question The calendar cube puzzle is famous: using two six-sided cubes, label them such that any day of any month can be represented by positioning the cubes accordingly.
The solution involves allowing the "6" and "9" to be represented by the same character.
My question is an extension: is it possible, using 4 six-sided cubes, to represent the two digit month and two digit day of any day of the year? With the proviso that 6 and 9 not be represented by the same character.
I don't know if this is possible or not, and the best I've been able to do is:
cube1 = {1, 9, 5, 2, 3, 4};
cube2 = {1, 5, 0, 7, 8, 9};
cube3 = {1, 6, 0, 2, 3, 4};
cube4 = {1, 6, 0, 2, 7, 8};

This allows for any dates besides the following:
06/06; 06/07; 06/08;
07/06; 07/07; 07/08;
08/06; 08/07; 08/08.

Is it possible to do better?
 A: Better, yes.  Perfect: I don't think so.
Take:
$$\begin{array} CC_1 & := & \{0,1,2,3,4,5\} \\
C_2 & := & \{0,1,2,6,7,8 \} \\
C_3 & := & \{0,1,2,7,8,9 \} \\
C_4 & := & \{0,1,3,4,5,6 \} \end{array}$$
Then you can manage any month and day combination except those that require a nine in each, namely 09/09, 19/09 and 29/09.
So how do we find these, and can we show they're optimal?  Finding them is straightforward: for the day-problem first we know that we have to be able to represent $0n$ where $n\in\{1,2,\ldots,9\}$ as well as $11$ and $22$.  So we must have $0,1,2$ present on both cubes.  After that we divide what's left evenly amongst the cubes which gives us $C_1$ and $C_2$ above, with $9$ missing.  For the standard calendar problem we conclude by noting that $6$ can be rotated $180$ degrees to yield $9$ and that we never need to combine $6$ and $9$ for a day of a month.
For the extended problem we next observe that we have two new cubes so we can try and build the months entirely from them.  As our months go from $01$ to $12$ we now only need $0$ and $1$ present on both cubes, which frees up a space from our original arrangement for $C_1$ and $C_2$.  So we put $9$ in that space and now we can represent all the months!
Moreover, (although I haven't in my solution) if we replace the $2$ we no longer need by the $9$ we do on (say) cube $C_3$ then we know that we can swap cubes $C_1$ and $C_3$ whenever we need a $9$ rather than a $2$ so we get all the dates from cubes $C_1$, $C_2$ and $C_3$.
Finally we observe that there is still only one $9$ on our cubes, and we need two of them to generate all month/date combinations so we must necessarily omit dates.  By choosing $9$ as our deficient number we minimise the dates missed (because 9 only ever appears in the units column), so this is the best we can do.
A: The previous solution (ommiting one of two digits $9$) is not the best one, it is the third best, ex aequo with ommitting any digit from set $\{4,5,...,8\}$
There is a family of the best solutions, for instance:
$$C_1 := \{0,2,3,4,5,6\}$$
$$C_2 := \{0,1,3,4,5,6\}$$
$$C_3 := \{0,1,2,7,8,9\}$$
$$C_4 := \{0,1,2,7,8,9\}$$
The last 4 digits of each cube could be shuffled between the cubes, of course such that no cube has repeated digits (especially $2$'s).
Proof:
We need at least:

*

*two copies of digits $\{4,5,...,9\}$ because they are needed as units' digits for days and months and they cannot be tens' digits;

*two copies of digit $3$ because there is no month or day number $33$ but there are $\text{Mar }03,13,23,30,31$);

*three copies of digit $2$ because there is no month number $22$ but there are $\text{Feb }22$ and $\text{Dec }22$;

*four copies of digit $1$ because there is $\text{Nov }11$;

*four copies of digit $0$ because it needs to be combined with two sets of two cubes needed for each of nine digits $n\neq0$ (because of 14 dates: nine $0n/0n$ and five: $\text{Jan }10$, $\text{Feb }20$, $\text{Mar }30$, $\text{Oct }01$ and $\text{Oct }10$).

Therefore, we need at least $2\cdot7+3+4\cdot2=25$ sides of cubes which is just one side more than 4 cubes. That is why we cannot represent all dates using 4 cubes if we assume $6\neq9$.
But we could decide which of above conditions we could possibly omit. The least restrictive is condition for digit $1$ because it is necessary for only one day, namely $\text{Nov }11$. So, if we omit one of four $1$'s, we could arrange the rest of digits such to create any date except from $\text{Nov }11$.
Fortunately, in some European countries $\text{Nov }11$ is a holiday commemorating the end of WW I, so we do not necessarily need a calendar that day. We could also mark it specially as $\text{Nov }00$ or "the new origin" $00/00$.
The second least restrictive is the condition for digit $2$, necessary for two days.
The condition for each of digits $\{4,5,...,9\}$ is the third in the least-restrictive ranking because each pair of digit $n$ is necessary for three days $0n,1n,2n$ in month $0n$. However, we could make a rule that month $00$ means "take the day's units' digit as month number".
