Riddles

[Davecom3]

Code: Select all

 -1- (2 stickers removed)
-FFF3(1 sticker removed)
2F-F3(1 sticker removed)
-FF-3(2 sticker removed)
 4-4 (1 sticker removed)

and
 -11(1 sticker removed) 
-B--3(3 stickers removed)
2---3(3 stickers removed)
----3(4 stickers removed)
 4--(2 stickers removed)

 and removing Central stickers from 1 and 2(2stickers)
2+1+1+2+1=7
1+3+3+4+2=13
2+7+13=22

Thus, 22 is possible. If F were in B's spot, there'd be an error. Next goal, prove whether 23 is possible. Theoretically, I can probably do just that, if not more as edges shouldn't be able to be mistaken for edges on the opposite side. That said, I don't like working in theory.

[Davecom3]

If I were to work in theoretics, without proof, then, theoretically, in the cube labeled:

1
2345
6

Due to not being able to flip a single piece, it might be possible to do something like:
111
1-1 1 sticker missing
11- 1 sticker missing
2-2 333 --- -5- 6 stickers missing
-2- -33 --- 5-5 7 stickers missing
2-- 333 4-- -5- 6 stickers missing
~6- 2 stickers missing
6-6 1 sticker missing
66- 1 sticker missing

Yielding 25 stickers missing.

Like I said, though, I hate working in theoretics. Especially as I'm relatively sure this answer is wrong.Though, I must admit that it would be impossible to transpose 6 with any of the others.

I'd much rather guess 22, and leave working out if 25 is possible to somebody else. I'm almost certain that it should be possible to flip an entire row, and if so, 25 stickers is definitely out of the question. People are probably correct that removing all stickers from a corner is impossible or nearly impossible as well.

And so I shall.
22 stickers is the maximum that are safe to remove keeping a unique solution.

[Davecom3]

Truthfully, chances are that the number is well over my stated claim. Our riddlemaster failed to state whether it was the standard 3x3 Rubik's cube, the 2x2 mini-cube, or one of the many with more dimensions like the 4x4 cube or the 17x17 Rubik's cube. Naturally, details like this will alter the riddle's outcome. So far the biggest Rubik's cube not on a computer is the 17x17, but if the computerized cube could get its stickers removed, anything less than 10,000 would be wrong, and if not, the answer still seems to be over 289, since he didn't state it must be a standard cube. It might even be a cube that has a different number of sides.

And if his answer is less than 22, his answer is not optimal.

[DrinksTooMuchCoffee]

/me steals one more sticker, leaving you with a cube you'll never know to be solved correctly or not. Bwahahahahahaha!

[Davecom3]

Frankly, I may need verification that the answer isn't that the cube isn't definitely solved when two adjacent centers are surrounded in the correct fashion by their respective corners, as if a cube is absolutely solved in that condition, we're likely to find out the number of stickers needed is 8, and so 46 can be lost. Or verification that sides can be in wrong locations while all corners are in their correct locations as testified by two adjacent centers. Otherwise, 11 stickers might be enough, and the yield would be 43 stickers being able to be removed. I'm pretty sure that sides can be out of place without centers or corners being out of place though.

[TBFProgrammer]

Frankly, I may need verification that the answer isn't that the cube isn't definitely solved when two adjacent centers are surrounded in the correct fashion by their respective corners, as if a cube is absolutely solved in that condition, we're likely to find out the number of stickers needed is 8, and so 46 can be lost. Or verification that sides can be in wrong locations while all corners are in their correct locations as testified by two adjacent centers. Otherwise, 11 stickers might be enough, and the yield would be 43 stickers being able to be removed. I'm pretty sure that sides can be out of place without centers or corners being out of place though.

It is possible to have the cube display a solid capital letter I on every side (all four corners and half of the correct edge colors). Each of those configurations has every corner in the correct location+orientation with only edges mixed up. Starting from a fully solved cube, creating these I's is actually fairly trivial (less than 10 moves per pair of faces if I remember correctly).

Each set of opposing faces treated in this manner does not affect the other faces (ie, two pairs of edges of swapped positions and each of these four edges has been rotated).

[gamecreator]

Starting from a fully solved cube, creating these I's is actually fairly trivial (less than 10 moves per pair of faces if I remember correctly).

To be precise, it's LFRLFR.

[Davecom3]

Yeah, the maximum is either 22 or 23, and seeing what has been said on youtube about the rotating of a just a single corner being impossible as it resuls in a totally different group of combinations, I'm personally assured that the number of stickers you could remove without damaging the integrity of the piece would be 23, reached by leaving the B by leaving off the latest solution I guaranteed. All of my theoretical solutions most likely are false, but you can definitely have one corner that is completely blank..

Therefore, my answer stands at 23.

If somebody can figure out a way to get 24, feel free, but given the information from youtube, 23 can definitely be achieved.

[Zathyr]

Actually, we'd only need two centers, Either A and 3 or A and 4, the others work themselves out due to edges.

3 and 4 also work, but yes, removing 4 center stickers works. Choosing the right stickers, there is still enough information to determine a single state for the cube. 22 is in fact the answer I was looking for. You folks had most of the information right in the initial attempts. More than 22 should not be possible. You absolutely cannot remove more than 6 from the edges and absolutely cannot remove more than 4 from the centers, and I'm fairly confident, especially after seeing all your work, 12 from the corners is the most you can lose and still have a single solution.

And as it was such a collaborative effort, I'm hard-pressed to award the winner to a single person. With the edits I'm not even certain who was first chronologically. So cookies all around and next round goes to whoever wants it.

Help yourself!

[Davecom3]

Start watching at about 6 minutes, and it explains that a corner with no markings is indeed possible, especially given the way I marked the corners. 23 is guaranteed to work, unless, of couse, you started with a bad cube. Rotating just one corner and nothing else is impossible without breaking the cube, as the state is unreachable.

[Zathyr]

If you're sure, I'll take a look some time later. I'm in a bit of a rush right now. I know for a fact you can have exactly 2 corners out of place, though it also pulls at least 2 edges out of place as well. But that may not matter. Sorry, no time now.

[Davecom3]

From a second source should the video not be enough:

First IÔÇÖll tackle the corner flipping. Imagine a cube painted with just black and white. All the facelets are black, except for the four corners of the top face and the four corners of the bottom face. If youÔÇÖve got a real cube in front of you, tape a little bit of paper onto each of those eight facelets. WeÔÇÖre going to look at how a maneuver twists the corners by looking at how it moves those marked facelets.

Now every maneuver is a composition of the six basic moves , , , , , and . If we can show that these all have a net twist of zero then any composition of them must also have net twist zero. The moves and are easy: they donÔÇÖt change the marked facelets at all.

Now letÔÇÖs consider the move . After twisting the right face of the cube, the four marked facelets on the left are left alone. The upper-front corner on the right was marked on the top, but now is marked on the front. ThatÔÇÖs an anticlockwise twist of 1/3 if we look directly at that corner. The upper-rear corner is now marked on the back, which is a clockwise twist of 1/3. The lower-front is twisted clockwise by 1/3, and the lower-right is twisted anticlockwise by 1/3. Adding all of these up, we get a total twist of zero.

On the other hand shifting a single corner will get you a result that does not total 0. An impossible move. To put it another way, when you shift a corner you always shift a group of 4 corners. If only one corner shows signs of shifting, there is an error with the cube.

[TBFProgrammer]

On the other hand shifting a single corner will get you a result that does not total 0. An impossible move. To put it another way, when you shift a corner you always shift a group of 4 corners. If only one corner shows signs of shifting, there is an error with the cube.

The potential problem comes from having two potentially interchangeable edges at the same time as you have two potentially interchangeable corners. In either case, taken alone, they cannot be swapped without some fixed thing also being swapped. In the case that we have both however, it is possible that they can both switch at the same time (we know from Zathyr that this can work for some cases, at least if he was being as strict as possible with the "at least 2").

If you remove more than 12 stickers from the corners, you have two potentially interchangeable corners. If you take 6 or more stickers from the edges, you have two potentially interchangeable edges.


The question then becomes whether or not there are two such pairs, one from the edges and one from the corners, such that they cannot both switch without an additional piece moving.


EDIT: It is, however, possible to rotate any three corners, I'm not 100% sure but I think that extends to edges as well. Again, I'd need a cube to test it but the test is really quite simple. This would mean that only one of corners or edges can be left interchangeable.

[Davecom3]

No, that's not the case in regards to the way I marked those corners. Refer back to:

 -1- (2 stickers removed)
-FFF3(1 sticker removed)
2F-F3(1 sticker removed)
-FF-3(2 sticker removed)
 4-4 (1 sticker removed)

and
 -11(1 sticker removed)
----3(4 stickers removed)
2---3(3 stickers removed)
----3(4 stickers removed)
 4--(2 stickers removed)

 and removing Central stickers from 1 and 2(2stickers)
2+1+1+2+1=7
1+4+3+4+2=14
2+7+14=23

In this example there's ony one potential location for 3--, which is 3, 4, and the unmarked.side. Following this, there is only one corner remaining for 4--, and that is 4, 2, and the unmarked side. The final corner thereby reveals itself, allowing for 13 stickers to be missing from the corners.

As a guide to identifying the corners, knowing where the central squares of 3 and 4 are, you know where the corner of of the 34- corner is. Further you can figure out the 3F- and 4F- pieces. Had they been on the opposite side, they would have been a single piece or mirrored. With this you identify F, and so only need the one corner for the F-- piece. The edges of 1F and 13 reveal where 1 is, just as the edges of 2F and 24 reveal where 2 is. Therefore, we can identify the corner of 13-. This leaves only one corner unknown on side 3. That being the case it can be identified by labeling it 3--. This leaves one unidentified corner on side 4, which can be marked 4--. This identifies all 7 locations of corners besides the one that can be left totally blank as it's identified by the rule that you cannot rotate just a single corner. Due to the fact that if all 7 other corners are adjusted properly so is the 8th, we can identify all the missing stickers on the edges, starting with edge F-, continuing to edges with - and number, and finally to edge --.

The way to get 3 corners flipped with nothing else flipped, according to the explanation provided would require either a clockwise rotation of 2 pieces, and 2 counter-clockwise rotations of a single piece, or a counter-clockwise rotation of two pieces, and 2 clockwise rotations of a single piece adding up to 0.

[Zathyr]

So I actually took out some duct tape and taped over stickers in the same layout to help me visualize and test it.

And yes it's rough to solve and I had to reorient things in my mind several times but the toughest thing was the edge with no stickers at all. Not the corner, the edge. Because I use the orientations of the edges in how I solve it. So, yeah, that bit gave me a headache more than anything. Ultimately I had to try to ignore it knowing that once everything else was in place that it, by force, would also be correct. But yes that 23-missing-sticker solution is discernible and unique. And now I'm questioning if 6 really is the most you can lose from the edge cubees, but this headache hasn't completely faded so I'm just going to say Davecom wins so we can all move on.

My apologies for even bringing it up. (though I still think it's an interesting problem)

[Davecom3]

By using two numbered cubes placed side by side and by numbering each side of each cube with one digit only, what is the highest number which can be displayed by starting at 1 then working upwards and not omitting and subsequent numbers? Both cubes must be used for each number but they may be switched around.

[TBFProgrammer]

Assuming we must have a 0 proceeding single digits:

If we don't have a 0 on both cubes, then the highest we can reach is 06.
If we don't have a 1 on both cubes, we'll not pass 11.
Trying to add a 2 to both cubes leaves only six locations to fit the remaining seven digits.
So the highest we can go is 21.

016789
012345

Assuming we can use a lone cube, we now only need a 0 on one cube and can replace that with the second 2 on the other, we cannot place two 1s two 2s and two 3s without running out of space for the other seven digits, so the highest we can go is 32.

012456
123789

This, of course, assumes that the number pairs of 6-9 and 2-5 are distinctly drawn. If we cannot assume that, then the puzzle changes slightly, as we have one more number we can double up for each pair that can be reversed. If we have both and the 0s are not needed, we should be able to reach a value of 65.

012348
123467

With just 6-9 we can reach a value of 43.

012356
123478

If we need 0s those are:

43

012346
012378

and

32

012345
012678

[Davecom3]

You've been told whether you need 0s. And I assure you if a number can look like another, it may be used as that other. Select the one that holds true to the conditions as stated.

[TBFProgrammer]

43

[Davecom3]

There's an error in your thinking. Figure out why. One of the answers you gave is correct, so I'll probably give it to you, but there is a problem in one bit of your thinking. Assume you only can read the tops of the blocks.

[TBFProgrammer]

There's an error in your thinking. Figure out why. One of the answers you gave is correct, so I'll probably give it to you, but there is a problem in one bit of your thinking. Assume you only can read the tops of the blocks.

Ah, crap, two and five can't stand in for each other unless you flip them

32

[Davecom3]

That's correct.

[Zathyr]

Well the riddle never stated we had to count in base 10.

With only two cubes the highest you could get in base 6 is only 55₆ but that's still 3 higher. Interestingly, in base 8 the highest you can reach is 43₈ which is the same number.

In base 7, since you don't need a double 0, you can get to 65₇ (47 in base 10), making it the optimal solution.

012345
123456

[Davecom3]

Sorry, Zathyr, but your answer isn't the one I was after.

[TBFProgrammer]

Open floor, sorry.