21 replies to this topic

[hhh3]

OMG.... warp what??!!

[curr3nt]

OMG.... warp what??!!

Sure. Chocolate bars possess strange and wonderful properties that we are still working to understand.

[bonanova]

Only stacking in a third dimension is not allowed? Cut pieces may be moved around on the plane or rotated to be cut further?

What about flipping of pieces? (although I fail to see how that may be advantageous at this point)

Rotating and flipping is OK.

[superprismatic]

Spoiler for my vote goes to

I can prove that it can be done in ceiling(log₂n)+ceiling(log₂m)-1. My conjecture is that this is minimal.

[Prof. Templeton]

Spoiler for my vote goes to

I can prove that it can be done in ceiling(log₂n)+ceiling(log₂m). My conjecture is that this is minimal.

This is exactly my thought also.

Spoiler for It seems...

to be related to the bit length. Perhaps an easier way to look at it would be to take the number of m (or n) and if it is odd then add 1 to it and if it is even divide it in half. Repeat until the result is one and count the number of division step required to get there. That would be the minimum for both m and n and they can be done separately and the steps added.

bn says we can rotate but that just transfers cuts required for m to n or vise versa.

[bonanova]

If I understand their posts correctly, and I may not have,
sp's expression gives 5, and PT's algorithm gives 6, when, e.g., m=5 and n=7.

Neither seems possible without stacking, or, equivalently, holding two or more pieces together next to each other, so as to keep the configuration planar.

OP asks for the minimum straight-line breaks, simultaneous or not.
So as an aid to counting, it disallowed stacking.

Break a piece. Repeat. Count the steps.

Spoiler for Among the responses so far

Among the replies when applied to 5x7 are these results:

5, 6, 10, 24 and 34.

The correct answer [and expression(s)] are among the replies.

[Prof. Templeton]

If I understand their posts correctly, and I may not have,
sp's expression gives 5, and PT's algorithm gives 6, when, e.g., m=5 and n=7.

Neither seems possible without stacking, or, equivalently, holding two or more pieces together next to each other, so as to keep the configuration planar.

These solutions took a loose interpretation of "stacking" where the cut pieces could not be laid on top of each other to create height. This was reinforced when you didn't address my question about only not stacking in a third dimension. Since we are dealing with an initial height of 1 the stipulation would have been a red herring.

I didn't notice sp had a "-1" in there, but if we look at a simple case of 2 x 2, then sp's formula says that can be done in 1 cut. I don't see how that is possible even with stacking. Or an even simpler case of 1 x 1 would require -1 cuts instead of zero.

OP asks for the minimum straight-line breaks, simultaneous or not.
So as an aid to counting, it disallowed stacking.

Break a piece. Repeat. Count the steps.

Spoiler for Among the responses so far

Among the replies when applied to 5x7 are these results:

5, 6, 10, 24 and 34.

The correct answer [and expression(s)] are among the replies.

Spoiler for Without stacking

you might as well leave the pieces where they lie since rotating or flipping does not help in any way. I therefore will vote for (m-1) + (n-1) or 10 in the 5 x 7 case given.

[curr3nt]

Can I vote for warping space and time (mostly time) so that the chocolate bar is already separated so that 0 splits are now required?

On a serious note...

Spoiler for Without stacking

you might as well leave the pieces where they lie since rotating or flipping does not help in any way. I therefore will vote for (m-1) + (n-1) or 10 in the 5 x 7 case given.

Spoiler for How does that work?

Say you start with spliting the 5 using 4 splits. You now have 5 individual pieces that each require 6 splits. If you are allowed to split each piece together why not split the 5 one and move one of the pieces to line up the next split to split the two pieces into 4. One more split for the 5th piece for a total of 3 splits on the 5 side. Or is moving pieces not allowed?

[Prof. Templeton]

Spoiler for How does that work?

Say you start with spliting the 5 using 4 splits. You now have 5 individual pieces that each require 6 splits. If you are allowed to split each piece together why not split the 5 one and move one of the pieces to line up the next split to split the two pieces into 4. One more split for the 5th piece for a total of 3 splits on the 5 side. Or is moving pieces not allowed?

Not Allowed because it would be equivalent to stacking. That was the basis of my first answer.

[Izzy]

Spoiler for

(m-1) + m(n-1), or mn - 1

*edit* Unless I'm misunderstanding the no stacking thing, since I'm considering breaking again over an already broken section "stacking".

Edited by Izzy, 12 October 2012 - 07:27 PM.