From: Martin Mares <mj@ucw.cz>
Date: Mon, 25 Apr 2022 17:53:33 +0000 (+0200)
Subject: Fixed a tiny error in decision tree lemma
X-Git-Url: http://mj.ucw.cz/gitweb/?a=commitdiff_plain;h=97ce60ab4d2d2017f4a936b71937d87ae1cb064f;p=saga.git

Fixed a tiny error in decision tree lemma
---

diff --git a/opt.tex b/opt.tex
index 202e52b..d71f983 100644
--- a/opt.tex
+++ b/opt.tex
@@ -882,12 +882,12 @@ than $n^4$ possible comparisons and less than $2^{n^2}$ spanning trees of~$G$,
 so the number of candidate decision trees is bounded by
 $(n^4+2^{n^2})^{2^{2n^2+1}} \le 2^{(n^2+1)\cdot 2^{2n^2+1}} \le 2^{2^{2n^2+2}} \le 2^{2^{3n^2}}$.
 
-We will enumerate the trees in an~arbitrary order, test each of them for correctness and
+We enumerate the trees in an~arbitrary order, test each tree for correctness and
 find the shallowest tree among those correct. Testing can be accomplished by running
 through all possible permutations of edges, each time calculating the MSF using any
 of the known algorithms and comparing it with the result given by the decision tree.
 The number of permutations does not exceed $(n^2)! \le (n^2)^{n^2} \le n^{2n^2} \le 2^{n^3}$
-and each one can be checked in time $\O(\poly(n))$.
+for sufficiently large~$n$ and each one can be checked in time $\O(\poly(n))$.
 
 On the Pointer Machine, trees and permutations can be certainly enumerated in time
 $\O(\poly(n))$ per object. The time complexity of the whole algorithm is therefore