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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?

fml
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  • 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. – DanielV Jun 21 '16 at 16:31
  • 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. – fml Jun 21 '16 at 16:33
  • Very similar to this, but harder... interesting! – ArtW Jun 21 '16 at 16:36
  • @DanielV: There aren't enough dice for $1+1+1+1=4$; but $1+1+1=3$ would work. – joriki Jun 21 '16 at 16:51
  • THere are far too many excedingly easy solutions to really consider this a puzzle. It seems more useful as a device to set up a statement and to test yourself what if any solution there is, or an exercise it pattern making. It is rather therapuetic the twist out 1+1+1=3;1+1+2=4;1+2+1=4;2+1+1=4 but it's hardly ... difficult. – fleablood Jun 21 '16 at 16:51
  • 1+1+1=3 would work but what about the other sides? you need the other sides to evaluate as well!!! – fml Jun 21 '16 at 16:53
  • Oh, all sides simultaneously correct! That is different! – fleablood Jun 21 '16 at 16:54
  • We need to see what order the other sides are so we can rotate them. From "away" rotation the first cube is 3,4,x,y which is different for the third or forth which are 4,3,a,b and 3,1,d,e etc. – fleablood Jun 21 '16 at 16:58
  • This isn't solvable with the information we have, since we don't know the arrangements of the elements on each block. For instance, on the first two numeric ones, "4" and "3" are in opposite orders; and the last two numeric blocks show that the values aren't necessarily consecutive. (Looks like @fleablood and I had the same thought at the same time. :) – Blue Jun 21 '16 at 16:58
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    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. – fleablood Jun 21 '16 at 17:02
  • Shame on you! According to http://www.creativecrafthouse.com/index.php?main_page=product_info&products_id=747 you have given us the easy 7 block puzzle instead of the "very difficulty" 9 block model! :) There is a video on the site in which the snake is rotated for the eagle-eyed! The 7 block version has only 1 correct solution. – almagest Jun 21 '16 at 17:08
  • @blue you're right! Im so sorry I forgot to put that information on. I'll update the question as soon as I find out. – fml Jun 21 '16 at 17:18
  • I think trial and error and common sense. Don't want to divide by 3 unless you also multiply by 3 and I doubt there sides that fit that. So that cut's our options down significantly... – fleablood Jun 21 '16 at 18:00
  • Your picture doesn't agree with your ascii art. I think your ascii art is wrong. – DanielV Jun 21 '16 at 18:42
  • Yeah I spotted a mistake - Thank you. Updated – fml Jun 21 '16 at 19:46
  • @fml Your last edit is incorrect, this version doesn't have a solution. The one I left you with agrees with what is in the video. – DanielV Jun 21 '16 at 19:53
  • The picture of the cubes and the table of the cube sides don't match. The third cube from left shows 4 followed by 3, but your table does not. – miracle173 Jun 21 '16 at 21:52

2 Answers2

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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.

DanielV
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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: $$ \frac{4}{3}+1=2 $$

Henrik supports the community
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