Fixed lots of typos.

diff --git a/opt.tex b/opt.tex
index 8977d3412ecc3e8bc6adbc9ca4d18b9d3206b8a1..44a9c8ca5bd562ad350d77a338aef3c35d0df1d8 100644 (file)
--- a/opt.tex
+++ b/opt.tex
@@ -212,7 +212,7 @@ son's list to its parent. Otherwise, we exchange the sons and move the list from
 new left son to the parent. This way we obey the heap order and at the same time we keep
 the white left son free of items.
 
-Ocassionally, we repeat this process once again and we concatenate the resulting lists
+Occasionally, we repeat this process once again and we concatenate the resulting lists
 (we append the latter list to the former, using the smaller of the two \<ckey>'s). This
 makes the lists grow longer and we want to do that roughly on every other level of the
 tree. The exact condition will be that either the rank of the current vertex is odd,
@@ -222,7 +222,7 @@ If refilling of the left son fails because there are no more items in that subtr
 (we report this by setting its \<ckey> to $+\infty$), the current vertex is no longer
 needed --- the items would just pass through it unmodified. We therefore want to
 remove it. Instead of deleting it directly, we rather make it point to its former
-grandsons and we remove the (now orhpaned) original son. This helps us to ensure
+grandsons and we remove the (now orphaned) original son. This helps us to ensure
 that both sons always keep the same rank, which will be useful for the analysis.
 
 When all refilling is done, we update the suffix minima by walking from the current
@@ -296,9 +296,9 @@ satisfies:
 $$\ell(v) \le \max(1, 2^{\lceil \<rank>(v)/2 \rceil - r/2}).$$
 
 \proof
-Initially, all item lists contain at most one item, so the ineqality trivially
+Initially, all item lists contain at most one item, so the inequality trivially
 holds. Let us continue by induction. Melds can affect it only in the favorable
-direction (they ocassionally move an~item list to a~vertex of a~higher rank)
+direction (they occasionally move an~item list to a~vertex of a~higher rank)
 and so do deletes (they only remove items from lists). The only potentially
 dangerous place is the \<Refill> procedure.
 
@@ -400,7 +400,7 @@ of the regular melds.
 Before we estimate the time spent on deletions, we analyse the refills.
 
 \lemma
-Every invokation of the \<Refill> procedure takes time $\O(1)$ amortized.
+Every invocation of the \<Refill> procedure takes time $\O(1)$ amortized.
 
 \proof
 When \<Refill> is called from the \<DeleteMin> operation, it recurses on a~subtree of the
@@ -522,7 +522,7 @@ time complexity is optimal for this choice of~$\varepsilon$. Chazelle \cite{chaz
 proves that it is optimal for every choice of~$\varepsilon$.
 
 The space consumed by the heap need not be linear in the \em{current} number
-of items, but if a~case where this matters ever occured, we could fix it easily
+of items, but if a~case where this matters ever occurred, we could fix it easily
 by rebuilding the whole data structure completely after $n/2$ deletes. This
 increases the number of potentially corrupted items, but at worst twice, so it
 suffices to decrease~$\varepsilon$ twice.
@@ -1028,7 +1028,7 @@ from the previous section.
 The first two steps of the algorithm are trivial as we have linear time at our
 disposal.
 
-By the Parititioning theorem (\ref{partthm}), the call to \<Partition> with~$\varepsilon$
+By the Partitioning theorem (\ref{partthm}), the call to \<Partition> with~$\varepsilon$
 set to a~constant takes $\O(m)$ time and it produces a~collection of subgraphs of size
 at most~$t$ and at most $m/4$ corrupted edges. It also guarantees that the
 connected components of the union of the $C_i$'s have at least~$t$ vertices
@@ -1129,6 +1129,4 @@ There were attempts to derandomize the KKT algorithm, but so far the best result
 is the randomized algorithm also by Pettie \cite{pettie:minirand} which achieves expected linear time
 complexity with only $\O(\log^* n)$ random bits.
 
-
-
 \endpart