I have a 4x4x4 Rubik's Cube, but all I want is a normal 3x3x3 Rubik's Cube.

Luckily, I also have an unlimited supply of small pieces of tape. Each piece of tape can cover the face of two neighbouring squares on my Rubik's Cube. When two pieces are taped together, they cannot be moved away from each other.

What is the minimum number of pieces of tape I must use to ensure that my 4x4x4 will always behave like a 3x3x3?

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    $\begingroup$ Welcome to Puzzling.SE! This is a fantastic puzzle - I hope to see more from you soon! $\endgroup$
    – Deusovi
    Commented Mar 20, 2016 at 22:01
  • $\begingroup$ If this would be an actual question instead of a puzzle question I would suggest buying an actual 3x3x3 speedcube. A YongJun GuanLong is only 2.75 USD on a webshop like Cubezz.com, and turns immensely better than the Rubik's brand 3x3x3. And don't even get me started on the turning quality of the Rubik's 4x4x4.. I have one in my collection, which might be a contributing reason why I don't like to solve 4x4x4 Cubes all that much.. xD A QiYi Thunderclap would be an all-round recommendation by the cubing community though. $\endgroup$ Commented Mar 21, 2016 at 8:20
  • $\begingroup$ First thought: you will need some seriously strong tape to prevent cubes from turning. $\endgroup$
    – Daniel
    Commented Mar 21, 2016 at 21:01

4 Answers 4


Your goal is to lock out the middle-slice rotations of the 4x4. Doing this turns the puzzle into a 3x3, because the change from a 3x3 to a 4x4 is equivalent to the addition of the middle slice motion.

Another way to think about this is that without the middle slice motion of a 4x4, you can't break up the edge pairs, and you can't break up the 2x2 center blocks. Without being able to split up edges or centers (i.e., limiting the puzzle to the outside rotations only), the puzzle is a 3x3.

You can lock out the middle slices with four pieces of tape. Here's how:

You can use two pieces of tape to lock out the middle slices through a single center. As long as the two pieces of tape are perpendicular, middle-slice motion is through that center isn't possible in either direction.

There are three middle-slice movements on the 4x4 - one for each spatial dimension. Each center intersects two of these. Therefore, your first center locks out two rotational axes. If you also tape over any adjacent center, you will also end up locking out the third rotational axis. Therefore, you need only tape over two centers.

The second center requires two pieces of tape, not one. While one piece of tape would lock out the third axis in the cube's initial configuration, if you rotate the center, that piece of tape will no longer intersect the third axis. In other words, without the second piece there, it is still possible to reach a state where you can rotate the third middle slice of the cube, even if it isn't the initial state.

Applying four pieces of tape - two pieces to two adjacent centers - locks out all middle-slice movements, reducing the puzzle to a 3x3.


A partial answer is:

EDIT: more than 3 pieces of tape.


EDIT (thanks to @Emrakul): On one face lock 2 of the centre pieces with 1 piece of tape, stopping all up-and-down rotations.
Then on the top (looking at the face just taped), lock two centre pairs to stop rotations parallel to the face your looking at.
Then on the left face (looking at the same face again), lock two centre pairs to stop rotations through the face your looking at.
You still have independent middle-slice rotations if you line up, say, the initial up-and-down lock with the above parallel lock.

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    $\begingroup$ On the second center, I'm pretty sure you still need two pieces of tape - centers can rotate, which means that with just one piece of tape, you can still rotate the entire center to unlock its middle-slice rotation. $\endgroup$
    – user20
    Commented Mar 20, 2016 at 22:02
  • $\begingroup$ @Emrakul Ah, I see the problem now. You're right. $\endgroup$
    – Paul Evans
    Commented Mar 20, 2016 at 22:23
  • $\begingroup$ @Emrakul Edited answer accordingly, thanks :) $\endgroup$
    – Paul Evans
    Commented Mar 20, 2016 at 22:43

The minimum number of pieces of tape is 2.

Following the logic of all of the other answers in that taping 2 adjacent centres is sufficient, you apply the tape in the following manner:

Apply the tape at a 45 degree angle.

Which achieves the same effect, in locking the entire centre from being split in either plane. The puzzle doesn't state that the tape must be applied directly to cover 2 adjacent squares, only that that's how they're sized.

(Please excuse the hastily-drawn paint image, you get the idea).

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    $\begingroup$ I first thought the same, but I'm pretty sure that "Each piece of tape can cover the face of two neighbouring squares on my Rubik's Cube." means we can't use diagonal bandaging. (So Emrakul has the correct answer of 4 by applying an L-shape on two adjacent centers, and I deleted my answer similar to yours around six hours ago.. Your picture looks cleaner than mine though. xD) $\endgroup$ Commented Mar 21, 2016 at 14:52
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    $\begingroup$ @KevinCruijssen we shall see =) I'm just waiting for the OP to give an 'official' ruling, I agree with you in that I suspect that the intent was for diagonal bandaging to not be allowed, however the wording of the question is ambiguous, which I find is where most logic puzzles hide the 'correct' solution. In particular, I'm hinging this on the use of can instead of must in the sentence you quoted, because as I said in my answer, can implies a size, whereas must would imply an orientation. $\endgroup$ Commented Mar 21, 2016 at 14:57
  • $\begingroup$ "however the wording of the question is ambiguous" I agree. I had to read the question a couple of times before seeing the small details, like using "small pieces of tape", and only covering "neighbouring squares". Without it the diagonal pieces of tape would work, or with bigger piece of tape you can still apply it horizontally or vertically while covering all four center squares. $\endgroup$ Commented Mar 21, 2016 at 15:02
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    $\begingroup$ I like the creative approach here, but the intention of the puzzle was not to be tricky with the words. This is my first puzzle here, so tried to keep it simple and avoid spoiling the puzzle by giving away all the details (and thereby the thoughts needed to solve the puzzle). I will keep clarity of the puzzle in my mind for the next puzzle! $\endgroup$
    – JenserCube
    Commented Mar 21, 2016 at 20:24

The minimum number of pieces of tape is 3.

The goal is to lock 2x2 squares in the middle of the each side of 4x4. To lock a 2x2 cube we need to lock all possible directions of rotation: there are 3 of them (x,y,z). Two of them (say, on the top of the cube) cold be locked by placing L-shaped tape = 2 pieces. And the third one by taping any of the side-centers vertically. So, all of our pieces of tape will be perpendicular to each other, placed on the central 2x2 squares, blocking all possible moves of the center squares.

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    $\begingroup$ And what if I turn the face with a single piece of tape 90 degrees? That enables me to split that center. $\endgroup$ Commented Mar 21, 2016 at 10:49

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