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I can't seem to comment on this question Packing problem cube and cuboids but it is related. I just want to know what is the specific method used in the answer so I can try to replicate it for my own question. Specifically, how do you get BRICKS(i,i,i)=0 and BOX(i,i,i)=−2+2i using the values of 10x10x10 and bricks of 1x1x4. How do you get $x^ay^bz^c(1+x+x^2+x^3)$ ? Is this a specific formula?

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winter
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1 Answers1

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The $i$ is the square root of $-1$, so the sums come out based on that.

Specifically note that $1 + i + i^2 + i^3 = 0$.

Using $i$ makes calculating a specific value of the polynomials nice because a lot of stuff cancels. In this case the counterexample is clear, and leads to the answer.

What I gather for the polynomial expression for the cube is that each space in the cube is "indexed" by the existence of a term with the appropriate powers on $x,y,z$. So the polynomial for the whole cube has $1,000$ terms from $1$ to $x^9y^9z^9$.

As for $x^ay^bz^c(1 + x + x^2 + x^3)$ that's the representation of a $1 \times 1 \times 4$ brick aligned along the $x$ axis. It starts at $(a,b,c)$ and takes up four squares in that direction. So such a brick starting at $(2,3,4)$ would have the polynomial representation $x^2y^3z^4(1 + x + x^2 + x^3)$.

If you had, say, a $1 \times 2 \times 3$ brick, one orientation would be

$$x^ay^bz^c(1 + x + x^2)(1+y).$$

There would be five others.

Or a $2 \times 2 \times 2$ brick is

$$x^ay^bz^c(1 + x)(1+y)(1+z),$$

and that's the only one you need.

John
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  • I'm still quite confused. so if the brick is of 1 unit, you multiply by 1, and if it is of 2 you multiply (1+x) and we continue up to the power of $(1+x+x^2+x^3)$ for side length 4? if so, how did this formula of representation come about? – winter Jan 28 '16 at 21:40
  • also how do you get the formula for representing the box? – winter Jan 28 '16 at 21:46
  • Added some statements in my answer to try to cover your questions. – John Jan 28 '16 at 22:16
  • so example a 2x2x3 brick at (3,4,5) will have $x^3y^3z^5 (1+x)(1+y)(1+z+z^2)$? thank you for the clarification. However, how do you get that form of representation? Can there be other ways of representing them? – winter Jan 28 '16 at 22:27
  • The $y^3$ should be $y^4$ but otherwise yes, that's one of three. The second has the quadratic in $x$ and the third has the quadratic in $y$. (There are three because there are three distinct orientations of a 2x2x3, And it's not my representation; I was trying to explain what the other person wrote. And I'm far from expert. – John Jan 28 '16 at 23:11
  • thanks! do you think it is possible to use this method for bricks of various sizes? – winter Jan 29 '16 at 08:58
  • Sure, you can have as many sizes and orientations as you choose. You'd include them and multiply them by $I(x,y,z), J(x,y,z),$ etc. in your $BRICKS(x,y,z)$ equation. – John Jan 30 '16 at 00:44