I found this not hard to solve with a pen and a piece of paper towel, so I suspect that it's rather unconstrained and there are many solutions. Later on I may write a program to generate all possible solutions and count them, but for now here's what I did to find a single solution. (Addendum: I did eventually write the program; see the comment below.)
I found this not hard to solve with a pen and a piece of paper towel, so I suspect that it's rather unconstrained and there are many solutions. Later on I may write a program to generate all possible solutions and count them, but for now here's what I did to find a single solution. (Addendum: I did write the program; see the comment below.)
I found this not hard to solve with a pen and a piece of paper towel, so I suspect that it's rather unconstrained and there are many solutions. Later on I may write a program to generate all possible solutions and count them, but for now here's what I did to find a single solution. (Addendum: I did eventually write the program; see the comment below.)
I found this not hard to solve with a pen and a piece of paper towel, so I suspect that it's rather unconstrained and there are many solutions. Later on I may write a program to generate all possible solutions and count them, but for now here's what I did to find a single solution. (Addendum: I did write the program; see the comment below.)
With $F=26$, it's natural to consider the numbers in pairs $1–12, 2–11, \ldots, 6–7$ where the pairs always add up to $13$. If we could arrange that each face had exactly two pairs, we would win. Also, we might imagine that the numbers are divided into small numbers $1\ldots 6$ and large numbers $7\ldots 12$. Maybe we can find a solution by matching small numbers with large numbers. To that end, let's arrange $1,2,11,12$$1,12,2,11$ around the middle of the cube, like this:
I found this not hard to solve with a pen and a piece of paper towel, so I suspect that it's rather unconstrained and there are many solutions. Later on I may write a program to generate all possible solutions and count them, but for now here's what I did to find a single solution.
With $F=26$, it's natural to consider the numbers in pairs $1–12, 2–11, \ldots, 6–7$ where the pairs always add up to $13$. If we could arrange that each face had exactly two pairs, we would win. Also, we might imagine that the numbers are divided into small numbers $1\ldots 6$ and large numbers $7\ldots 12$. Maybe we can find a solution by matching small numbers with large numbers. To that end, let's arrange $1,2,11,12$ around the middle of the cube, like this:
I found this not hard to solve with a pen and a piece of paper towel, so I suspect that it's rather unconstrained and there are many solutions. Later on I may write a program to generate all possible solutions and count them, but for now here's what I did to find a single solution. (Addendum: I did write the program; see the comment below.)
With $F=26$, it's natural to consider the numbers in pairs $1–12, 2–11, \ldots, 6–7$ where the pairs always add up to $13$. If we could arrange that each face had exactly two pairs, we would win. Also, we might imagine that the numbers are divided into small numbers $1\ldots 6$ and large numbers $7\ldots 12$. Maybe we can find a solution by matching small numbers with large numbers. To that end, let's arrange $1,12,2,11$ around the middle of the cube, like this:
$$\begin{array}{cccccccc} & E && F && G && H\\A && B && C && D \\& I && J && K && L \\ \end{array}$$$$\def\r#1{\color{maroon}{#1}}\begin{array}{cccccccc} & \r E && \r F && \r G && \r H\\\r A && \r B && \r C && \r D \\& \r I && \r J && \r K && \r L \\ \end{array}$$
$$\def\bl#1{\color{blue}{#1}} \begin{array}{cccccccc} & E && F && G && H\\\bl1 && \bl{12} && \bl2 && \bl{11} \\& I && J && K && L \\ \end{array}$$$$\def\bl#1{\color{blue}{#1}} \begin{array}{cccccccc} & \r E && \r F && \r G && \r H\\\bl1 && \bl{12} && \bl2 && \bl{11} \\& \r I && \r J && \r K && \r L \\ \end{array}$$
$$\begin{array}{cccccccc} & \bl6 && F && G && H\\1 && 12 && 2 && 11 \\& \bl7 && J && K && L \\ \end{array}$$$$\begin{array}{cccccccc} & \bl6 && \r F && \r G && \r H\\1 && 12 && 2 && 11 \\& \bl7 && \r J && \r K && \r L \\ \end{array}$$
$$\begin{array}{cccccccc} & 6 && \bl9 && G && H\\1 && 12 && 2 && 11 \\& 7 && \bl3 && K && L \\ \end{array}$$$$\begin{array}{cccccccc} & 6 && \bl9 && \r G && \r H\\1 && 12 && 2 && 11 \\& 7 && \bl3 && \r K && \r L \\ \end{array}$$
$$\begin{array}{cccccccc} & 6 && 9 && \bl5 && H\\1 && 12 && 2 && 11 \\& 7 && 3 && \bl8 && L \\ \end{array}$$$$\begin{array}{cccccccc} & 6 && 9 && \bl5 && \r H\\1 && 12 && 2 && 11 \\& 7 && 3 && \bl8 && \r L \\ \end{array}$$
$$\begin{array}{cccccccc} & 6 && \bl8 && G && H\\1 && 12 && 2 && 11 \\& 7 && \bl4 && K && L \\ \end{array}$$$$\begin{array}{cccccccc} & 6 && \bl8 && \r G && \r H\\1 && 12 && 2 && 11 \\& 7 && \bl4 && \r K && \r L \\ \end{array}$$
$$\begin{array}{cccccccc} & 6 && 8 && \bl3 && H\\1 && 12 && 2 && 11 \\& 7 && 4 && \bl{10} && L \\ \end{array}$$$$\begin{array}{cccccccc} & 6 && 8 && \bl3 && \r H\\1 && 12 && 2 && 11 \\& 7 && 4 && \bl{10} && \r L \\ \end{array}$$
$$\begin{array}{cccccccc} & E && F && G && H\\A && B && C && D \\& I && J && K && L \\ \end{array}$$
$$\def\bl#1{\color{blue}{#1}} \begin{array}{cccccccc} & E && F && G && H\\\bl1 && \bl{12} && \bl2 && \bl{11} \\& I && J && K && L \\ \end{array}$$
$$\begin{array}{cccccccc} & \bl6 && F && G && H\\1 && 12 && 2 && 11 \\& \bl7 && J && K && L \\ \end{array}$$
$$\begin{array}{cccccccc} & 6 && \bl9 && G && H\\1 && 12 && 2 && 11 \\& 7 && \bl3 && K && L \\ \end{array}$$
$$\begin{array}{cccccccc} & 6 && 9 && \bl5 && H\\1 && 12 && 2 && 11 \\& 7 && 3 && \bl8 && L \\ \end{array}$$
$$\begin{array}{cccccccc} & 6 && \bl8 && G && H\\1 && 12 && 2 && 11 \\& 7 && \bl4 && K && L \\ \end{array}$$
$$\begin{array}{cccccccc} & 6 && 8 && \bl3 && H\\1 && 12 && 2 && 11 \\& 7 && 4 && \bl{10} && L \\ \end{array}$$
$$\def\r#1{\color{maroon}{#1}}\begin{array}{cccccccc} & \r E && \r F && \r G && \r H\\\r A && \r B && \r C && \r D \\& \r I && \r J && \r K && \r L \\ \end{array}$$
$$\def\bl#1{\color{blue}{#1}} \begin{array}{cccccccc} & \r E && \r F && \r G && \r H\\\bl1 && \bl{12} && \bl2 && \bl{11} \\& \r I && \r J && \r K && \r L \\ \end{array}$$
$$\begin{array}{cccccccc} & \bl6 && \r F && \r G && \r H\\1 && 12 && 2 && 11 \\& \bl7 && \r J && \r K && \r L \\ \end{array}$$
$$\begin{array}{cccccccc} & 6 && \bl9 && \r G && \r H\\1 && 12 && 2 && 11 \\& 7 && \bl3 && \r K && \r L \\ \end{array}$$
$$\begin{array}{cccccccc} & 6 && 9 && \bl5 && \r H\\1 && 12 && 2 && 11 \\& 7 && 3 && \bl8 && \r L \\ \end{array}$$
$$\begin{array}{cccccccc} & 6 && \bl8 && \r G && \r H\\1 && 12 && 2 && 11 \\& 7 && \bl4 && \r K && \r L \\ \end{array}$$
$$\begin{array}{cccccccc} & 6 && 8 && \bl3 && \r H\\1 && 12 && 2 && 11 \\& 7 && 4 && \bl{10} && \r L \\ \end{array}$$