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Adapted from Chapter 4 of “Communication complexity and applications” by Anup Rao and Amir Yehudayoff. Result from “On sets of integers which contain no three terms in arithmetic progression” by Felix Behrend.
Say you want to color $[n]$ so that there are no monochromatic arithmetic progressions of length 3: i.e. you cannot find $a$ and $b> 0$ such that $a$, $a+b$ and $a+2b$ are all the same color. How many colors do you need?
Attempt 1: random coloring
The probabilistic method always gives the best answer, right?
Let’s color each integer in $[n]$ by a random color over $[r]$. Then each progression $a,a+b,a+2b$ will be monochromatic with probability $1/r^2$, while there are $\leq n^2$ different progressions (because you only have to choose $a$ and $b$). So by a union bound, we have a $\geq 1-n^2/r^2$ chance that none of them will be monochromatic. Thus it suffices to set $r>n$ to get a nonzero probability.
That means… more colors than numbers?? Wow, that was really useless.
Note: This would have given something nontrivial for longer arithmetic progressions (e.g. for 4-progressions it would have been enough to set $r>n^{2/3}$).
Attempt 2: convex functions
One thing to note is that $a+b$ is the “midpoint” of $a$ and $a+2b$. So another way to phrase the requirement that no arithmetic 3-progressions are monochromatic is “if two points are the same color, their midpoint should be a different color”.
What’s a property such that if $x$ and $y$ share it, then $\frac{x+y}{2}$ can’t share it with them? Well, if you have a specific kind of twisted mind, then taking midpoints might evoke convexity. For example, if $f$ is a strictly convex function (e.g. $f(x) = x^2$) and if $f(x)=f(y)$, then $f(\frac{x+y}{2}) < f(x)$.
Can you use this for coloring? Yes! For example, in our case, you can use $f(x) \ce (x-n/2)^2$. Since this is strictly convex, we cannot have $f(a) = f(a+b) = f(a+2b)$ (for $b>0$). Function $f$ takes $\approx n/2$ different values over $[n]$, so it gives us a coloring with $\approx n/2$ colors without monochromatic 3-progressions!
Now, this is not super impressive: if you have $n/2$ colors, then just making sure that no color is present more than twice already trivially implies that it cannot have monochromatic 3-progressions. And in fact, any convex function over $[n]$ must take at least $n/2$ values, since it can always be split into a strictly decreasing part and a strictly increasing part.
Attempt 3: spherical shells??
Parabolas are flatter in higher dimensions
It would be nice if we could get a strictly convex function whose image is much smaller than its domain. This is impossible over the reals, but it turns out to be possible in higher dimensions!
The natural analog of the quadratic function $f(x) \ce (x-n/2)^2$ we tried out before is the “squared 2-norm” function:
\[
f(x_1, \ldots, x_d) \ce x_1^2 + \cdots + x_d^2.
\]
This corresponds to coloring every “spherical shell” around the origin with a different color. This makes sense: if you take two distinct points $x,y$ on the same shell, then their midpoint $\frac{x+y}{2}$ will be closer to the origin.
And this time, the number of colors is much smaller than the domain: If we set its domain to $[s]^d$ (the discrete cube of side $s$), then the domain has $s^d$, while the output values range from $0$ to $ds^2$!
But we care about arithmetic progressions in $[n]$, not $[s]^d$! To use this convex function, we would need to create a mapping from $[n]$ to $[s]^d$ so that arithmetic progressions in the former give arithmetic progressions the latter.
Mapping to higher dimensions
Let’s have a try! Let’s use the convention that $[n] \ce \set{0, \ldots, n-1}$, and assume that $s^d \ge n$. We can represent any $a \in [n]$ in base $s$:
\[
a = a_0s^0 + a_1s^1 + \cdots + a_{d-1}s^{d-1}
\]
with $0 \le a_0, \ldots, a_{d-1} < s$, so we can represent it as $v(a) \ce (a_0, \ldots, a_{d-1}) \in [s]^d$. This means that $a$ would take color $\norm{v(a)}_2^2 = a_0^2 + \cdots a_{d-1}^2 \in [ds^2]$.
For any arithmetic progression $a, a+b, a+2b$ in $[n]$, we can decompose $a+b$ and $a+2b$ as
\[
\begin{aligned}
a+b &= (a_0+b_0)s^0 + \cdots + (a_{d-1}+b_{d-1})s^{d-1}\\
a+2b &= (a_0+2b_0)s^0 + \cdots + (a_{d-1}+2b_{d-1})s^{d-1},
\end{aligned}
\]
but the numbers $a_i+b_i$ and $a_i+2b_i$ might not be $<s$, causing “carries”. And after the carries are executed, there is unfortunately no guarantee that $v(a), v(a+b), v(a+2b)$ form an arithmetic progression in $[s]^d$.
Detecting the carries
When a carry occurs while adding together two numbers $a$ and $b$ in base $s$, we can detect it as long as we have some idea of the magnitudes of their digits $a_0, \ldots, a_{d-1}$ and $b_0, \ldots, b_{d-1}$. Intuitively, if for some position $i$, we notice that $a_i$ and $b_i$ are pretty big and yet the corresponding digit in $a+b$ is pretty small, then we know that a carry must have occurred.
More precisely, let $c_0, \ldots, c_{d-1}$ be the digits of the sum $a+b$. We’ll call a digit be “small” if it’s $<s/2$ and “large” if it’s $\ge s/2$. Then a carry occurs at position $i$ if and only if one of the following happens:
- $a_i$ and $b_i$ are both “large”;
- one of $a_i$ and $b_i$ is “large”, and $c_i$ is “small”.
This means that you can figure out whether $v(a+b) = v(a)+v(b)$ just by knowing the size buckets (small or large) in which each of the digits of $a$, $b$, and $a+b$ fall. We don’t need to know their precise values.
The situation is similar (but slightly more subtle) for arithmetic progressions. We’ll show that if the digits of $v(a)$, $v(a+b)$ and $v(a+2b)$ fall within the same bucket at each position, then $v(a),v(a+b),v(a+2b)$ must form an arithmetic progression.
Suppose that $v(a),v(a+b),v(a+2b)$ do not form an arithmetic progression. Equivalently, $v(a)+v(a+2b) \ne 2v(a+b)$. Let $i$ be the first index where the sides differ. This means that for all $j<i$, $v(a)_j + v(a+2b)_j = 2v(a+b)_j$. We have
\[
\begin{align}
a + (a+2b) &\equiv 2(a+b) &\pmod{s^{i+1}}\\
\sum_{j=0}^i (v(a)_j + v(a+2b)_j)s^j &\equiv \sum_{j=0}^i(2v(a+b)_j)s^j &\pmod{s^{i+1}}\\
v(a)_i + v(a+2b)_i &\equiv 2v(a+b)_i &\pmod{s}
\end{align}
\]
and yet $v(a)_i + v(a+2b)_i \ne 2v(a+b)_i$. Since the two sides differ but are congruent modulo $s$, they must differ by at least $d$. This is impossible if all of $v(a)_i$, $v(a+b)_i$ and $v(a+2b)_i$ are in the same bucket.
To recap, for any arithmetic progression $a,a+b,a+2b$, we know that either
- $v(a),v(a+b),v(a+2b)$ form an arithmetic progression in $[s]^d$, and therefore they cannot have different 2-norms;
- or there must be some digit $i \in \set{0, \ldots, d-1}$ such that $v(a)_i$, $v(a+b)_i$ and $v(a+2b)_i$ are not all in the same bucket.
This means that if we define the color of some $a \in [n]$ as the couple of
- its 2-norm $\norm{v(a)}_2^2$ of its representation in $[s]^d$;
- the buckets where the $d$ digits of $v(a)$ fall;
then there are no monochromatic arithmetic progressions.
Choosing parameters
The 2-norm is an integer in $[ds^2]$, and the buckets are a string in $\zo^d$, so we get $ds^2\cdot 2^d = 2^{O(d+\log s)}$ colors. So it will be optimal to “balance $d$ and $\log s$”. Since we must ensure $n \leq s^d = 2^{d\log s}$, the optimum will be $d \ce \log s \ce \sqrt{\log n}$, which gives $2^{O\p{\sqrt{\log n}}}$ colors.
Currently, the best lower bounds on the number of colors are $\polylog(n)$, which is exponentially smaller, even though the problem has been studied since the 1940s! Combinatorics is hard. :)