Dice problem

If your saw is big enough, 24 will suffice.

Nope

8 cuts per pair of sides will give 9 portions.

9 x 9 x 9 = 729

3 x 8 = 24 cuts.

Where can one buy this zero thickness saw?

"Where can one buy this zero thickness saw?"

KrisDel of course.

You're allowed to put the dust back with glue if you really want.

Would it be OK to stick positive dots onto the dice.

In this case the one will be loaded because of the more weight at the six.

Nope ; "729 1cm cubes". You'd need the special KrisDel "No-Vol Glue" anyway, unless you could live with some spare dust. "Dice" is "dice" in the 'diced carrots' sense.

Actually, KrisDel buys my Consolidated Gookumpucky SuperDuper Negative Volume Glue in bulk, add fillers and repackage it as their "No-Vol Glue." I know they are competing with my own product, but I have an agreement with all the label suppliers that they must sell labels without adhesive backing to KrisDel. KrisDel then is forced to use my SuperDuper Sticker Stickum.

Consolidated Gookumpucky - You'll be glad you stuck with our products!

Hey, that's a labelous statement and you're hereby served with a conjunction !

C'mon, Doug, you're still in with a chance for a GA here ; why is 6 the minimum number of cuts to dice up a 3x3x3 cube ? That's a simplified take on the question as posted, but answering it for this case should be enough.

Actually, it could be done with 3 cuts. Now, you tell me how!

Oh go on, tell me......I don't know

It's simple, and your pic alludes to that. Use an eight-bladed saw. Draw centerlines on the top of the cube and on two adjacent vertical sides. Line up the center point on the saw with the centerlines to make your cuts.

Hey, if the blades have to be zero-width, you might as well have 8 of them! Or use 8 lasers with a beamwidth of one photon.

you can make it without a single cut.

Hint : you need to contact acme and KrisDel products Inc.

Yup, I'm with hendrik on this one.
8 Horizontal cuts gives a stack of 9 slabs.
8 Vertical cts (north south) gives 81 square section beams.
Final 8 vertial cuts (east west) gives 729 cubes.

Del

Then you can go sit on the silly-step too

I think Kris has sussed it but he's a scaredey squirrel....
He's convinced me that there's a fewer cuts solution but I've taken a vow of silence.
He's lurking in the bushes at the mo'
Del

Teee heeee...Well I bet you're buying sugar lumps right now just to see !oops...wasn't me. Was never here, didn't do it......

OK, it's 12, but proving it is the cleverful bit. Squirrel beats two cats, and that ain't a bad start to the day

Oh go on, confess, the Delster told you ?

I'll give you a GA Hendrik, nicely explained. You can have another if you explain why it's the mimimum number of cuts.....

Kris is right it's 12.

To do a Rubik's cube you've got to do at least 6, think about the very central cube and it's obvious that you can't do it in less.

Now for the nine by nine. First you've got to get out the central Rubik, any other plan will leave you wanting. That's six cuts.

Now you've got to get the central cuby out of each of the Rubiks. That's another six cuts (easy logically but practically pretty tricky).

Hurrah !

Thank you, Randall. The inner-most cube of a 3x3x3 has to be produced by 6 separate cuts. In my mind, it's fair enough to extrapolate that into first using 6 cuts to section a 9x9x9 block, then another 6 cuts to section all those simultaneously.

Such a proof of minimum cuts will undoubtedly give Fyz a touch of the vapours, which seals the deal on me giving you a GA.

That sounds jolly good and, on the basis that my screensaver cut in whilst I was cogitating, I shall GA it. My mind is racing around other cube sizes, but unfortunately I'm limited to one GA ().

Many thanks for expressing this in a more formal manner. I suspect the completely formal description that might be found in a mathematical journal would be incomprehensible to me ! A nicely worded solution, Fyz.

Brilliant and beautiful as always.

Just to complete the generalisation: it's clear that to cut any block with sides n,m and l where

2^(p-1) <= n < 2^p

2^(q-1) <= m < 2^q

2^(r-1) <= l < 2^r

You need p+q+r cuts.

Having written this I can see that it was implicit in your solution , but, some of us aren't so fast .

Thanks! And I think the explicit completion was worthwhile.