Length of nearest neighbor path in travel salesman problem Given $n$ nodes uniformly distributed in $[0,1]^2$, consider the nearest neighbor algorithm to solve traveling salesman problem, i.e., each time I select the nearest neighbor not visited so far as the next node to visit, my question is: what is the expected length of such TSP tour?
 A: A careful experimental analysis of the nearest-neighbor (NN) heuristic (among other heuristics) is described in this paper:

Johnson, David S., and Lyle A. McGeoch. "The traveling salesman problem: A case study in local optimization." Local search in combinatorial optimization 1 (1997): 215-310.
  (PDF download link.)

They find that the growth rate, in comparison to the Karp-Held lower bound, 
"appears to be proportional to $\log n$" for "random distance matrices."
This is the theoretical growth rate w.r.t. the optimum.
In contrast, for "random Euclidean instances" up to $n=10^6$, they find 
(Table 1, p.15) that the NN heuristic
leads to paths nearly constantly about 25% longer than the Karp-Held lower bound.
This is closer (but not identical) to the OP's "uniformly distributed within $[0,1]^2$."
The section, "Standard Test Instances" on pp.12-14 unpacks
"random Euclidean instances."
Incidentally, the observed running time grows subquadratically.
(See also the follow-up MO question, 
"Travelling Salesman Problem: Can the nearest neighbor algorithm be n times longer than the optimal solution?.")
