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A colleague recently showed me the following puzzle game and I'm interested in how this can be solved. I thought it would be a good talking point for you guys as well :)

enter image description here

A detailed description of the puzzle is here. A sequence of 7 cubes may be rotated about the axis. The puzzle is to rotate them until all 4 equations are correct, such as $2 + 2 / 4 = 1$. Operators are evaluated left to right. The faces of the cube presented in the video (not agreeing with the image) are:

The following pictures show the sides...

Side1 Side2 Side3 Side4

Question

Other than trial and error, is there an easy way to solve this?

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  • $\begingroup$ How do you prioritize operations? Left to right, or multiplication first? There is an easy answer, $1+1+1+1=4$, but to ask "how many solutions are there" is a more interesting question. $\endgroup$ – DanielV Jun 21 '16 at 16:31
  • $\begingroup$ Each numeric cube has 4 options, 1 ,2, 3, 4. So if you use the number 1 on one side you cannot use it on another. The same applies with the operations. $\endgroup$ – fml Jun 21 '16 at 16:33
  • $\begingroup$ Very similar to this, but harder... interesting! $\endgroup$ – ArtW Jun 21 '16 at 16:36
  • $\begingroup$ @DanielV: There aren't enough dice for $1+1+1+1=4$; but $1+1+1=3$ would work. $\endgroup$ – joriki Jun 21 '16 at 16:51
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    $\begingroup$ There are 6 possible way to arrange the numbers, op cubes (1,2,3,4 and 1,2,4,3 and 1,3,2,4 and 1,3,4,2 and 1,4,2,3 and 1,4,3,2) but only 3 and 2 of them are used. We need to see what the other two sides of the snake are in order to solve this. $\endgroup$ – fleablood Jun 21 '16 at 17:02
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The puzzle in the video only has 1 solution.

$$\begin{array} {|c|c|c|c|c|c|} \hline 4 & - & 3 & \times & 3 & = & 3 \\ 2 & + & 2 & \div & 4 & = & 1 \\ 1 & \times & 4 & - & 2 & = & 2 \\ 3 & \div & 1 & + & 1 & = & 4 \\ \hline \end{array}$$

I doubt there is any significantly faster way to solve it than with a computer. It's only $4^6 \div 4 = 1024$ cases to check. I doubt there is a faster way than with a computer, since in general there can a large number of solutions. This is a combinatorial logic problem, so it is NP, and I don't see any obvious way to put it into P.

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    $\begingroup$ Usually $4-3\times 3=-5$ and $2+2/4=2.5$ $\endgroup$ – miracle173 Jun 21 '16 at 20:36
  • $\begingroup$ you changed the second operator sequence from $\times, -, +, \div$ to $\times, \div, -, +$. $\endgroup$ – miracle173 Jun 21 '16 at 21:09
  • $\begingroup$ @miracle173 I did not, the questioner changed it incorrectly. The one I posted is correct according to the video linked. $\endgroup$ – DanielV Jun 21 '16 at 21:45
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    $\begingroup$ on the website I found "The equation is read from left to right, irregardless of operation priority.", so you are right $\endgroup$ – miracle173 Jun 21 '16 at 21:50
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I just realised that I didn't get it, I just thought all the cubes had to show different numbers/symbols.

A solution that uses each number exactly once, and two different operations $$ \frac{4}{2}+1=3 $$

I found that thinking about what divisions were possible. Obviously we can divide by one, but then we're just left with one operation ($+$, $-$ or $*$) on $\{2, 3, 4\}$, so 9 possibilities, it's not hard to check all of them and conclude that there are no solutions.

If we want the division to give an integer result the only other option is $\frac{4}{2}$, and that gives the solution I've already named.

If we just ignore a remainder from a a division, we get more solutions, like: $$ \frac{4}{3}+1=2 $$

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  • $\begingroup$ Apparently all four sides must match simultaneously. And there's no restriction that each cube be used at most once in an equation. $\endgroup$ – fleablood Jun 21 '16 at 17:22
  • $\begingroup$ Yes, I've realised that. Didn't get when I saw the question while on the train, and then thought about this during my walk from the station, and when I got home I didn't bother reading the new comments before typing up my answer. $\endgroup$ – Henrik Jun 21 '16 at 17:28
  • $\begingroup$ Neither did I. <><><><> $\endgroup$ – fleablood Jun 21 '16 at 17:44

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