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TimeSpaceLightForce

Gift wrapping the Puzzle blocks

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its quite possible to have a 1x1x1 block.

for example: let's say the total volume of the box is 3x3x3.

you could make that box with a 2x3x3, a 1x2x3 , a 1x1x2, and a 1x1x1.

 

 

Edited by bonanova
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The LxWxH block dimension seems to rule out

Spoiler

The Soma cube, which has block surfaces that are not strictly convex, but do completely fill a 3x3x3 cube. That size is a lower bound in any event since 2x2x2 can't be completely filled with unique convex blocks.

Since OP does not

Spoiler

explicitly require the box to be completely filled

it seems the smallest box would be

Spoiler

2x2x2 which contains three blocks with dimensions 1x1x1, 1x1x2 and 2x2x1

But that requirement probably was intended.

Also, I don't see that unique coloring imposes any limits beyond that of unique dimensions.

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Most blocks in the solution above have the same dimension with themselves like 2x3x3 or 2x2x1..Op implicates that the blocks dont have square faces.  There should be W1xH1xL1, W2xH2xL2,W3xH3xL3, so on. All blocks dimensions are unique integers.

I checked my solution twice but find it wrong the third time. It happens to have a hollow inside. It looks too hard to solve without computer but thought there is a solution or more for smallest SxSxS box.

Edited by TimeSpaceLightForce
Right

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So the problem is to fill a cubical box of side S with
convex blocks having integral-length edges {L W H}
whose values are taken
without replacement from
the set { 1, 2, 3, 4, 5, 6, ..., S-2,  S-1, S } ?

Or must they be taken from the set that excludes S?

That is, can one of the blocks have an edge length equal to S?

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