Approximate 1d TSP with linear comparisons?
EDIT (UPDATE): The lower bound in my answer below was proven (by a different proof) in "On the complexity of approximating Euclidean traveling salesman tours and minimum spanning trees", by Das et al; Algorithmica 19:447-460 (1997).




  is it possible to achieve even an approximation ratio like $O(n^{1-\epsilon})$ for some $\epsilon&gt;0$ in $o(n\log n)$ time using a comparison-based algorithm?


No. Here's a lower bound.

Claim. For any $\epsilon&gt;0$, every comparison-based 
$n^{1-\epsilon}$-approximation algorithm
requires $\Omega(\epsilon n\log n)$ comparisons in the worst case.

By "comparison-based" I mean any algorithm that 
only queries the input with binary (True/False) queries.

Here's an attempt at a proof. Hopefully there are no mistakes.
FWIW the lower bound seems likely to extend to randomized algorithms.



Fix any $n$ and any arbitrarily small but constant $\epsilon&gt;0$.

Consider just the $n!$ "permutation" input instances 
$(x_1,x_2,\ldots,x_n)$
that are permutations of $[n]$.
The optimum solution for any such instance has cost $n-1$.

Define the cost of a permutation $\pi$
to be $c(\pi) = \sum_i |\pi(i+1) - \pi(i)|$.
Model the algorithm as taking as input a permutation $\pi$,
outputting a permutation $\pi'$,
and paying cost $d(\pi,\pi') = c(\pi'\circ \pi)$.

Define $C$ to be the minimum number of comparisons
for any comparison-based algorithm to achieve
competitive ratio $n^{1-\epsilon}$ on these instances.
Since opt is $n-1$, the algorithm must guarantee cost at most $n^{2-\epsilon}$.

We will show $C\ge \Omega(\epsilon n\log n)$.

Define $P$ to be, for any possible output $\pi'$,
the fraction of possible inputs for which output $\pi'$ 
would achieve cost at most $n^{2-\epsilon}$.
This fraction is independent of $\pi'$.

$P$ also equals the probability that, for a random permutation $\pi$,
its cost $c(\pi)$ is at most $n^{2-\epsilon}$.
(To see why, take $\pi'$ to be the identity permutation $I$.
Then $P$ is the fraction of inputs for which
$d(\pi,I)$ at most $n^{2-\epsilon}$,
but $d(\pi,I) = c(\pi)$.)

Lemma 1. $C \ge \log_2 1/P$.

Proof. Fix any algorithm that always uses less than $\log_2 1/P$ comparisons.
The decision tree for the algorithm has depth less than $\log_2 1/P$,
so there are at less than $1/P$ leaves, and, for some output
permutation $\pi'$,  the algorithm gives $\pi'$ as output
for more than a $P$ fraction of the inputs.  By definition of $P$,
for at least one such input, the output $\pi'$ gives cost more than $n^{2-\epsilon}$.
QED

Lemma 2. $P \le \exp(-\Omega(\epsilon n\log n))$.

Before we give the proof of Lemma 2, note that the two lemmas together
give the claim: 
$$C
~\ge~ \log_2 \frac{1}{P}
~=~ \log_2 \exp(\Omega(\epsilon n\log n))
~=~ \Omega(\epsilon n\log n).$$



Proof of Lemma 2.
Let $\pi$ be a random permutation.
Recall that $P$ equals the probability that
its cost $c(\pi)$ is at most $n^{2-\epsilon}$.
Say that any pair $(i,i+1)$ is an edge
with cost $|\pi(i+1)-\pi(i)|$,
so $c(\pi)$ is the sum of the edge costs.

Suppose $c(\pi) \le n^{2-\epsilon}$.

Then, for any $q&gt;0$, at most $n^{2-\epsilon}/q$ of the edges have cost $q$ or more.
Say that edges of cost less than $q$ are cheap.

Fix $q=n^{1-\epsilon/2}$.  Substituting and simplifying, at most $n^{1-\epsilon/2}$ of the 
edges are not cheap.

Thus, at least $n - n^{1-\epsilon/2} \ge n/2$ of the edges are cheap.
Thus, there is a set $S$ containing $n/2$ cheap edges.

Claim. For any given set $S$ of $n/2$ edges,
the probability that all edges in $S$ are cheap
is at most $\exp(-\Omega(\epsilon n \log n))$.

Before we prove the claim, note that it implies the lemma as follows.
By the claim, and the naive union bound,
the probability that any there exists such a set $S$ 
is at most 
$${n\choose n/2} \exp(-\Omega(\epsilon n \log n))
~\le~ 2^n \exp(-\Omega(\epsilon n \log n))$$
$$~\le~ \exp(O(n) -\Omega(\epsilon n \log n))
~\le~ \exp(-\Omega(\epsilon n \log n)).$$



Proof of Claim.
Choose $\pi$ by the following process.
Choose $\pi(1)$ uniformly from $[n]$,
then choose $\pi(2)$ uniformly from $[n] - \{\pi(1)\}$,
then choose $\pi(3)$ uniformly from $[n]-\{\pi(1),\pi(2)\}$, etc.

Consider any edge $(i,i+1)$ in $S$.
Consider the time just after $\pi(i)$ has been chosen,
when $\pi(i+1)$ is about to be chosen.
Regardless of the first $i$ choices (for $\pi(j)$ for $j\le i$),
there are at least $n-i$ choices for $\pi(i+1)$,
and at most $2n^{1-\epsilon/2}$ of those choices will give the edge $(i,i+1)$
cost less than $n^{1-\epsilon/2}$ (making it cheap).

Thus, conditioned on the first $i$ choices,
the probability that the edge is cheap
is at most $\frac{2n^{1-\epsilon/2}}{n-i}$.
Thus, the probability that all $n/2$ 
edges in $S$ are cheap is at most 
$$\prod_{(i,i+1)\in S} \frac{2n^{1-\epsilon/2}}{n-i}.$$
Since $|S|\ge n/2$, there are at least $n/4$ edges in $S$
with $n-i\ge n/4$. Thus, this product is at most
$$\big(\frac{2n^{1-\epsilon/2}}{n/4}\big)^{n/4}
~\le~(8n^{-\epsilon/2})^{n/4}
~=~\exp(O(n)-\Omega(\epsilon n \log n))
~=~\exp(-\Omega(\epsilon n \log n)).$$

QED