Recently A. Amarilli (a3nm) posted a question on cs.stackexchange.com about the computational complexity of a Test Round problem of the Google France #Hash Code 2015: the “**Pizza Regina**” problem (March 27th, 2015):

**Definition [Pizza Regina problem]**

**Input**: A grid $M$ with some marked squares, a threshold $T\in \mathbb{N}$, a maximal area $A \in\mathbb{N}$

**Output**: The largest possible total area of a set of disjoint rectangles with integer coordinates in $M$ such that each rectangle includes at least $T$ marked squares and each rectangle has area at most $A$.

The problem can be converted to a decision problem adding a parameter $k \in \mathbb{N}$ and asking:

**Question**: Does there exist a set of disjoint rectangles satisfying the conditions (each rectangle has integer coordinates in $M$, includes at least $T$ marked squares and has area at most $A$) whose total area is at least $k$ squares?

The problem is clearly in $\mathsf{NP}$, and after struggling a little bit I found that it is $\mathsf{NP}$-hard (so the Pizza Regina problem is $\mathsf{NP}$-complete). This is a sketch of a reduction from MONOTONE CUBIC PLANAR 1-3 SAT:

**Definition [1-3 SAT problem]:**

**Input:** A 3-CNF formula $\varphi = C_1 \land C_2 \land … \land C_m$, in which every clause $C_j$ contains exactly three literals: $C_j = (\ell_{j,1} \lor \ell_{j,2} \lor \ell_{j,3})$.

**Question:** Does there exist a satisfying assignment for $\varphi$ such that each clause $C_j$ contains exactly one true literal.

The problem remains NP-complete even if all literals in the clauses are positive (MONOTONE), if the graph built connecting clauses with variables is planar (PLANAR) and every variable is contained in exactly 3 clauses (CUBIC) (C. Moore and J. M. Robson, Hard tiling problems with simple tiles, Discrete Comput. Geom. 26 (2001), 573-590.).

We use $T=3, A=6$, and in the figures ham is represented with blue boxes (transgenic ham?), pizza with orange boxes.

The idea is to use tracks of ham that carry positive or negative signals; the track is made with an alternation of 1 and 2 pieces of hams placed far enough so that they can be covered exactly by one slice of pizza of area $A$; the segments of the track are marked alternately with $+$ or $-$, the track will carry a positive signal if slices are cut on the positive segments:

Each variable $x_i$, which is connected to exactly 3 SAT clauses, is represented by three adjacent endpoints of three ham tracks (positive segment), in such a way that there are 2 distinct ways to cut it, one will “generate” a positive signal on all 3 tracks (it reppresent the $x_i = TRUE$ assignment) the other a negative signal ($x_i = FALSE$). Notice that we can also generate mixed positive and negative signals, but in that case *at least one ham remains uncovered*.

Each clause $C_j$ of the 1-3 SAT formula with 3 literals $L_{i,1}, L_{i,2}, L_{i,3}$ is simply represented by a single ham with three incoming positive segments of three distinct ham tracks; by construction *only one of the three tracks* carrying a positive signal can “cover” the ham-clause.

Finally we can build shift and turn gadgets to carry the signals according to the underlying planar graph and adjust the endpoints:

Suppose that the resulting graph contains $H$ hams. By construction every slice of pizza must contain exactly 3 hams, and in all cases every slice can be enlarged up to area $A$.

If the original 1-3 SAT formula is satisfiable then by construction we can cut $H /3$ pieces of pizza (with total area of $A H / 3$) and no ham remains uncovered.

On the oppposite direction, if we can cut $H /3$ pieces of pizza (with total area $A H / 3$) then no ham remains uncovered, and the signals on the variables gadgets and on the clauses are consistent: the ham on the clause is covered by exactly one positive slice coming from a positive variable, and every variable generates 3 positive signals or 3 negative signals (no mixed signals); so the cuts induce a valid 1-3 SAT assignment.

**Conclusion**: … so unless $\mathsf{(P)izza} =\mathsf{(NP)izza}$, cutting a pizza can be really hard. I would like to thank Antoine for posting the funny question and for spending a bit of time checking my proof.

A few questions:

1) In the second shift signal, you no longer have the hams distanced far enough so that an area covers either only positive or only negative cells. So if an area contains both positive and negative cells does it carry a positive or negative signal? Or if it matters just the polarity of the cell exactly near the cut, why did you distance the hams so much when explaining the tracks?

2) To get a “pizza grid” from the 1 in 3 sat graph, once the nodes and edges are transformed, is the area in between filled with no-hams cells?

3) How would you prove the construction can be done in polynomial time?

Anyway, cool proof! It’s the only hashcode problem for which I found a proof of np-completeness.