This is an extension of a 3rd grade problem.

How many pieces can one get at most if one cut a unit cube with n plane cuts?

1,2,4,8, ???

And assuming cutting through an area 1 takes time t, what is the least time needed to achieve the maximal pieces for n cuts, asymptotically?

  • $\begingroup$ For the first question, the answer is $1,2,4,8,15,26,42,\ldots$ (OEIS A000125) and is known as the Cake number. $\endgroup$ May 3, 2013 at 14:35

1 Answer 1


Number of pieces $\frac{n^3+5n+6}{6}$

per Peter C. Heinig

Define a number of planes in space to be in general arrangement when
1) no two planes are parallel,
2) there are no two parallel intersection lines,
3) there is no point common to four or more planes.

Suppose there are already n-1 planes in general arrangement, thus defining the maximal number of regions in space obtainable by n-1 planes and now one more plane is added in general arrangement. Then it will cut each of the n-1 planes and acquire intersection lines which are in general arrangement.

These lines on the new plane define the maximal number of regions in 2-space definable by n-1 straight lines, hence this is (Lazy Caterer's sequence)(n-1).

Each of this regions acts as a dividing wall, thereby creating as many new regions in addition to the a(n-1) regions already there, hence a(n)=a(n-1)+(Lazy Caterer's sequence)(n-1).

So, f(n) = ${n \choose 3} + {n \choose 2} + {n \choose 1} + {n \choose 0}$ = $\frac{1}{6}(n^3+5n+6)$

  • $\begingroup$ @Amzoti It can be proved by induction. Consider having a cake already cut by $(n-1)$ planes. Add another, $n$-th cut. Inside the plane of that cut there are $(n-1)$ lines that produce number of regions defined by (Lazy Caterer's sequence)(n-1). Current cut plane produces that number of additional cake pieces ("above" or "below" the plane). Thus, $a(n)=a(n-1)+LCS(n-1)$. $\endgroup$
    – mbaitoff
    May 17, 2013 at 3:15
  • $\begingroup$ @Amzoti: The OP seems to have abandoned his question, the answer, and the site. $\endgroup$
    – mbaitoff
    May 17, 2013 at 3:27
  • $\begingroup$ That happens quite frequently too and sometimes we don't make them feel very welcome! :-) $\endgroup$
    – Amzoti
    May 17, 2013 at 3:28

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