$\require{mathtools}
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% Greek
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\let\fi\phi % because it looks like an f
\let\phi\varphi % because it looks like a p
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% .. over and under
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% .. relations
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% .. miscer
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% (I needed to create my own because the MathJax version of \DeclarePairedDelimiter doesn't have \mathopen{} and that messes up the spacing)
% .. one-part
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% .... (use phantom to force at least the standard height of double bars)
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% ..... (use phantom to force at least the standard height of double bar)
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% .. dotted versions (is it equal in the limit?)
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% .. operators
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% .. analysis
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% ... named tensors
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% Complexity classes
% .. classical
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% .. probabilistic
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Adapted from Chapter 4 of “Communication complexity and applications” by Anup Rao and Amir Yehudayoff.
There are two ways to generalize communication complexity to more than two parties.
- “number in hand” (NIH): each party gets to see their own input;
- “number on forehead” (NOF): each party gets to see everything except their own input.
Oddly enough, number on forehead is the more popular choice!
Consider the following number-on-forehead problem (called “exactly-$n$”):
Alice gets to see $y,z$, Bob gets to see $x,z$ and Charlie gets to see $x,y$. How many bits of communication do they need to figure out whether $x+y+z=n$?
Clearly, $O(\log n)$ bits are enough: Charlie can send $x$, then Alice knows $y,z$ so she can tell whether $x+y+z$ is indeed $n$. Can we do better?
Changes of perspective
Notice that as far as Alice is concerned, there is only one possible value of $x$ that would make the sum $n$: $x' \ce n-y-z$. So the only way for the function to be satisfied is if the input is $(x',y,z)$: this is Alice’s “guess” at the input. Similarly, we can define $y' \ce n-x-z$ and $z' \ce n-x-y$; in this case, the problem in equivalent to figuring out whether the guesses
\[(x',y,z), (x,y',z), (x,y,z')\]
of each party coincide. These correspond to the projections of the input $(x,y,z)$ onto the plane $x+y+z=n$ along each of the three axes, and they form an equilateral triangle.
Corners
Since the third coordinate of a point on the plane $x+y+z=n$ is entirely determined by the first two, it’s also equivalent to checking whether the points
\[(x',y), (x,y'), (x,y)\]
coincide (graphically, this corresponds to looking at the previous figure “from above”). When they don’t, these three points form a “corner” in $\R^2$: there is some $d$ such that these are (after permutation)
\[(x,y), (x+d,y), (x,y+d),\]
with $d\neq 0$ (here, this $d$ is $x'-x = y'-y = n-(x+y+z)$).
In other words, checking whether $x+y+z=n$ is equivalent to checking whether these three points $(x,y)$, $(x',y)$ and $(x,y')$ (known by Charlie, Alice and Bob) coincide or whether they form a corner. Again, we could do this in $O(\log n)$ by having one of the players say its point and the other two check. But we could try doing something smarter: suppose there is some coloring $c:[n]^2 \to [r]$ that makes sure there is no monochromatic corner, then it’s enough to check if the colors match. This would mean $O(\log r)$ bits are enough: Alice says her color, then Bob and Charlie compare it with theirs.
Arithmetic progressions
Okay, so we’ve reduced number-on-forehead protocols for exactly-$n$ to colorings of $[n]^2$ without monochromatic corners. Let’s see one way to get such colorings.
Notice that the points $(x,y),(x+d,y),(x,y+d)$ of a corner are always on three different “diagonals”: lines with equation $x-y = v$ for some offset $v \in \set{-n, \ldots, n}$. In fact, they are on three equally spaced diagonals. So if we can color the diagonals in such a way that no three equally spaced diagonals have the same color, then this makes sure there are no monochromatic corners.
The offsets $v$ of three equally spaced diagonals form an arithmetic progression, so this is equivalent to coloring $\set{-n, \ldots, n}$ (or $[2n+1]$) in a way that avoids monochromatic arithmetic progressions. In Behrend’s progression-free sets, we saw that this can be done with $2^{O\p{\sqrt{\log n}}}$ colors, so we get a protocol with $O\p{\sqrt{\log n}}$ bits of communication. This is a significant improvement over the trivial $O(\log n)$ protocol where one of the inputs is sent in its entirety.
TODO: Is it easy to visualize how the $x+y+z=n$ plane gets cut up into surprisingly few cylinder intersections?
A lower bound through Ramsey theory
In the previous section, we saw that colorings of $[n]^2$ without monochromatic corners gave protocols for exactly-$n$. In fact, this was practically an equivalence: I claim that protocols for exactly-$n$ give colorings of $[n/2-1]^2$ (not quite $[n]^2$) without monochromatic corners. In Monochromatic corners, we show that such a coloring must use $\Omega\p{\frac{\log \log n}{\log \log \log n}}$ colors. This gives a lower bound of $\Omega(\log \log \log n)$ on the number of bits of communication, which is doubly exponentially smaller than the upper bound!
Let’s see why protocols give colorings without monochromatic corners.
First, walking back through the change of perspective, any corner $(x,y), (x+d,y), (x,y+d)$ in $[n/2-1]^2$ corresponds to an equilateral triangle
\[(x,y,n-x-y),(x+d,y,n-x-y-d),(x,y+d,n-x-y-d),\]
which corresponds to the projections of point $(x,y, n-x-y-d) \in [n]^3$ onto the plane $x+y+z=n$.
Now, consider the tree of a protocol for exactly-$n$ with $l$ bits of communication. We will show that if we color every point in $[n]^3$ by the leaf it ends up in, then
\[(x,y,n-x-y),(x+d,y,n-x-y-d),(x,y+d,n-x-y-d)\]
cannot have the same color. There are $\leq 2^l$ leaves, so this would give a coloring with $2^l$ colors.
The set of inputs that will end up in some fixed leaf of the protocol tree has the form
\[\set{(x,y,z) \mid a(y,z)= b(x,z) = c(x,y)= 1}\]
(where the functions $a,b,c$ correspond to the information sent by Alice, Bob and Charlie along the path to that leaf). If that leaf contains all three of
\[(x,y,n-x-y),(x+d,y,n-x-y-d),(x,y+d,n-x-y-d),\]
then we must have $a(y,n-x-y-d) = b(x,n-x-y-d) = c(x,y) = 1$, so the point $(x,y, n-x-y-d)$ that generated those projections must also end up in this leaf. But this is impossible: the protocol computes the exactly-$n$ function, so a leaf cannot contain points on the plane $x+y+z=d$ and points outside of it at the same time.