Natural CLIQUE to k-Color reduction There is clearly a reduction from CLIQUE to k-Color because they're both NP-Complete. In fact, I can construct one by composing a reduction from CLIQUE to 3-SAT with a reduction from 3-SAT to k-Color. What I'm wondering is whether there is a reasonable direct reduction between these problems. Say, a reduction that I could explain to a friend fairly briefly without needing to describe an intermediate language like SAT.

As an example of what I'm looking for, here's a direct reduction in the reverse direction: Given G with $n$ and some $k$ (the number of colors), make a graph G' with $kn$ vertices (one per color per vertex). The vertices $v'$, $u'$ corresponding to vertices $v, u$ and colors $c, d$ respectively are adjacent if and only if $v \neq u$ and ($c \neq d$ or $vu \not \in G$). An $n$-clique in $G'$ has only one vertex per vertex in $G$, and the corresponding colors are a proper $k$-coloring of $G$. Similarly, any proper $k$-coloring of $G$ has a corresponding clique in $G'$.

Edit: To add some brief motivation, Karp's original 21 problems are proved NP-Complete by a tree of reductions where CLIQUE and Chromatic Number form the roots of major subtrees. There are some natural reductions between problems in the the CLIQUE subtree and the Chromatic Number subtree, but many of them are just as hard to find as the one I'm asking about. I'm trying to drill down on whether the structure of this tree shows some underlying structure in the other problems or if it is entirely a consequence of which reductions were found first, since there is less motivation to search for reductions between two problems when they are known to be in the same complexity class. Certainly the order had some influence, and parts of the tree can be rearranged, but can it be rearranged arbitrarily?

Edit 2: I continue to search for a direct reduction, but here is a sketch of the closest I've gotten (it should be a valid reduction, but has CIRCUIT SAT as a clear intermediary; it's somewhat subjective whether this is any better than composing two reductions as alluded to in the first paragraph).

Given $G, k$, we know that $\overline{G}$ can be $n-k+1$-colored with $k$ vertices all colored True iff $G$ has a $k$-clique. We name the original vertices of $G$ $v_1, \ldots, v_n$ and then add to $\overline G$ additional vertices: $C_{ij}$ with $1 \le i \le n$, $0 \le j \le k$. The key invariant will be that $C_{ij}$ can be colored True if and only if among vertices $\{v_1,\ldots, v_i\}$ there are at least $j$ vertices colored True. So, each $C_{i0}$ can be True. Then, $C_{ij}$ for $j &gt; 0$ gets the color $C_{(i-1)j} \vee C_{(i-1)(j-1)} \wedge v_i$ where all non-true colors are treated as false. There is a $k$-clique in $G$ iff $C_{nk}$ can be colored True, so if we force that coloring, the new graph is colorable iff there was a $k$-clique in the original graph.

The AND and OR gadgets to enforce the relationships are much like the reduction from CIRCUIT SAT to 3-COLOR, but here we include a $K_{n-k+1}$ in our graph, pick vertices T, F, and Ground, and then connect all others to everything but the $v_i$s; this assure that the $C_{ij}$s and the other gadgets receive only 3 colors.

Anyway, the $\overline G$ part of this reduction feels direct, but the use of AND/OR gates is much less direct. The question remains, is there a more elegant reduction?

Edit 3: There have been a few comments about why this reduction would be hard to find. CLIQUE and k-Color are indeed quite different problems. Even without a reduction, though, an answer that details why the reduction is hard in the one direction but possible in the other would be very helpful and contribute a lot to the problem.
Given a graph $G$ and a number $k$, such that you want to know whether $G$ contains a $k$-clique, let n be the number of vertices in $G$. We construct another graph $H$, such that $H$ is $n$-colorable if and only if $G$ has a $k$-clique, as follows:

(1) For each vertex $v$ in $G$, make an $n$-clique of vertices $(v,i)$ in $H$, where $i$ ranges from $1$ to $n$.

(2) Add one additional vertex $x$ to $H$.

(3) For each triple $\{x,y,z\}$ of vertices in $H$, where $y=(v,i)$ and $z=(u,j)$, test whether one of the following conditions holds: either $u\ne v$ and $i=j$, or $u$ and $v$ are nonadjacent vertices in $G$ with $\max(i,j)\le k$. If either of these two things is true, add another $n$-clique to $H$. Within this clique, select three vertices $x'$, $y'$, and $z'$. Connect $x$ to every vertex in the clique except for $y'$ and $z'$; connect $y$ to every vertex in the clique except for $x'$ and $z'$; and connect $z$ to every vertex in the clique except for $x'$ and $y'$.

The gadgets added in step (3) prevent the triple of vertices $x$, $y$, and $z$ from all being given the same color as each other in a valid coloring of $H$. The clique in $G$ can be recovered from a coloring of $H$ as the set of vertices $(v,i)$ that are in the same color class as $x$ and that have $i\le k$.