Prove or disprove that it is possible.
Well,the central cube needs 6 cuts.
So, 6 is the minimal number of cuts.
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 ! :rofl:
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.
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.
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.
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.
This seemed interesting so, I do tried doing so and it is very possible.Perfect 27
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. :(
Of course, Am I missing something here?
I mean, how many ways can you make a cube out of 27 smaller cubes anyway?
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
Look at a Rubics cube, 4 cuts on top and 2 cuts to the side.
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.
Thats 6 cuts? I think the whole point is less than six cuts.
Can you figure out a way with less cuts?
You just can't - read posts #2 and #7.
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
Parallelepipeds are not necessarily cubes. :yuck:
Think about it.
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