Can a cube be cut in 27 smaller cubes in less than 6 cuts?

[Werg22]

Prove or disprove that it is possible.

 

[Rogerio]

Prove or disprove that it is possible.

Well,the central cube needs 6 cuts.
So, 6 is the minimal number of cuts.

 

[Xori]

Yes, you can use a knife with two blades so you only make one cut in each direction

 

[Rogerio]

Yes, you can use a knife with two blades so you only make one cut in each direction

It doesn't matter if you are going to use the knife just 3 times.
The question is about the number of cuts - and you need 6 cuts!

----------
And, according to your point of view, why didn't you use acid instead a knife?
So you would need no cuts at all !

 

[Jimmy Snyder]

I don't have an answer. However, I see that I can cut it into 64 smaller cubes with 6 cuts, So at least it seems reasonable that you could get 27 in 5 . In any case, we must define 'a cut' in such a way that when I cut two pieces into four with a single action of the knife, that is one cut, not two. Cut as follows:

First make three cuts one each down the middle of each face and perpendicular to the sides. This will cut the cube into 8 smaller cubes. Take the 8 cubes and lay them out in a row. Cut all 8 cubes down the middle of the row, so that each of the 8 smaller cubes is in the same condition as the larger one was after the very first cut. Then without disturbing the relative positions of the two pieces of any of the smaller split cubes, rearrange the cubes in a row perpendicular to the cut. Cut again down the middle of the row. Now each of the smaller cubes looks like the larger cube did after the second cut. Rearrange and cut a third time. Now you will have 64 cubes.
eom

 

[AlephZero]

A different argument - same conclusion as Rogiero:

The first cut makes two pieces size 1x3x3 and 2x3x3.

The 2x3x3 piece contains 18 cubes, but the maximum number of pieces you can make in 4 cuts is 2^4 = 16.

 

[Rogerio]

... it seems reasonable that you could get 27 in 5 .

Unfortunately we can't...

Well,for the optimal policy,
it seems reasonable that all the smaller cubes are the same size (in order to take advantage of every cut).
Then, even employing the optimal policy to get at least 27 cubes, there will be at least one central cube, in such way that none of its faces shares any external face. So, you need one cut to get each face. It means you need exactly 6 cuts to make the central cube. So, 6 is the minimum.

 

[Xori]

It doesn't matter if you are going to use the knife just 3 times.
The question is about the number of cuts - and you need 6 cuts!

----------
And, according to your point of view, why didn't you use acid instead a knife?
So you would need no cuts at all !

I didn't use acid because it won't travel sideways to make a "cut" and I'm too lazy to turn the cube after every "cut" :P

And I was thinking of a "cut" as the action of the motion of the knife, not the resulting change in shape.

 

[Wild Angel]

This seemed interesting so, I do tried doing so and it is very possible.Perfect 27

 

[akhil982]

In simple terms.. if you make a,b and c number of cuts along length,breadth and height , you'll have (a+1)*(b+1)*(c+1)=27. And for a+b+c to be minimum, the optimal solution is a=b=c .. and that leads to a=b=c=2. hence total number of 6 cuts.

I don't think it can be better than this. But everything is possible. :(

 

[futurebird]

Of course, Am I missing something here?

I mean, how many ways can you make a cube out of 27 smaller cubes anyway?

 

[akhil982]

I think you can do that in 3 ways..not counting repeatitions.. cube of 1, cube of 2*2*2 and a cube of 3*3*3

 

[Wild Angel]

Look at a Rubics cube, 4 cuts on top and 2 cuts to the side.

 

[Sleek]

Yup, same as Wild Angel. Four cuts on top (with two parallel cuts perpendicular to other two parallel cuts). This makes 9 parallelepiped like things. Two cuts sideways, we have 9 cubes on each of the three layers. Thus, 3*9=27 cubes.

 

[K.J.Healey]

Thats 6 cuts? I think the whole point is less than six cuts.

 

[Wild Angel]

Can you figure out a way with less cuts?

 

[Rogerio]

Can you figure out a way with less cuts?

You just can't - read posts #2 and #7.

 

[JDEEM]

It can be done in 4 cuts. think about it.

 

[Rogerio]

It can be done in 4 cuts. think about it.

I don't see how.

Could you please explain it to us ?

-------
Hint: read posts #2 and #7

 

[Kittel Knight]

It can be done in 4 cuts. think about it.

Parallelepipeds are not necessarily cubes.

Think about it.

 

[davee123]

So, what if you could re-arrange the pieces before making subsequent cuts? Further, just to be clear, I'm assuming a "cut" represents a flat rectangular plane (finite or not), which must start outside the object being "cut".

In theory, if you allowed re-arrangements, you could create 32 *pieces* with your 5th cut, not all of which will be cubical. Is it possible, however, that 27 or more of them are cubical? I'd guess not, but I'm having a hard time thinking of how to actually prove that fact, bearing in mind that the cubes aren't necessarily the same size.

DaveE

 

[arivero]

Actually, what is the maximum number of solid objects you can obtain when you cut an sphere with n planes?

 

[arivero]

I found two webpages stating related results, I have no idea where to find a systematic collection of cutting problems or theorems.
http://www.math.toronto.edu/mathnet/SOAR2003/Winter/index.html tells that a torus can be cut in (n^3+3n^2+8n)/6 pieces for n cuts.

http://www.cs.colostate.edu/~rmm/mathChallenge/cairoliBinom.pdf tells that a cube can be cut in
\begin{pmatrix} n+1 \\ 3 \end{pmatrix} + n +1
pieces for n cuts.

and thus with n=5 it is 26 pieces.
and with n=4 it is 15.

 

[arivero]

http://groups.google.com/group/sci....bf0c888139/04f87eee704f0266?#04f87eee704f0266 run this same problem but allowing to rearrange.

 

[Rogerio]

So, what if you could re-arrange the pieces before making subsequent cuts? ...

Of course you can!
However the number of cuts remains 6...

 

[JDEEM]

for a cube 3x3x3
first and second cuts yield 3 pieces @ 3x3x1

now take each-one at time-and fold it over until the edges are flush (think foam rubber) and place one on top of the other and again with the third piece. The third cut (longwise)will yield 9 peices @ 3x1x1

repeat the fold over process with 9 pieces and the fourth cut will yield 27 pieces @ 1x1x1

No one said what the cube was made of!

 

[davee123]

Of course you can!
However the number of cuts remains 6...

Yeah, I agree with you, but how do you prove that fact?

DaveE

 

[Andre]

Dunno, is there a requirement that all 27 cubes have the same size? Are more pieces allowed?

If the first cut makes 2 pieces and the second cut 4, doubling the pieces, then the fifth cut will end up with 32 pieces (2^5), 27 of them should be cubes in any size. That could be worked out further.

 

[Rogerio]

Dunno, is there a requirement that all 27 cubes have the same size? Are more pieces allowed?

Any size. Extra pieces allowed.

... the fifth cut will end up with 32 pieces (2^5), 27 of them should be cubes in any size.

Wouldn't you thinking of "parallelepipeds" instead "cubes" ?

 

[davee123]

Wouldn't you thinking of "parallelepipeds" instead "cubes" ?

Possibly, although I'd think I'd specify a cuboid instead of a parallelepiped. I don't think anyone's thinking about anything other than 90 degree cuts.

Here's a question: what's the MOST number of cubes you can create with 6 cuts, allowing piece re-arrangement?

DaveE

 

[Rogerio]

Here's a question: what's the MOST number of cubes you can create with 6 cuts, allowing piece re-arrangement?
DaveE

This is much more easier : 64

 

[davee123]

This is much more easier : 64

Oh yeah-- duh, 3 cuts with re-arrangement obviously allows for 1 cube to be cut into 8, hence 8^2. Now just still need the less-than-6 cuts proven...

DaveE

 

[ƒ(x)]

Use a device that makes multiple slices with each cut...not sure if this qualifies.

 

[DaveC426913]

Use a device that makes multiple slices with each cut...not sure if this qualifies.

You might want to take a sec to follow the thread before responding. https://www.physicsforums.com/showpost.php?p=1438690&postcount=4" would be good.