\input macros.tex
\fi
-\chapter{Approaching Optimality}
+\chapter{Approaching Optimality}\id{optchap}%
\section{Soft heaps}
Recently, Chazelle \cite{chazelle:ackermann} and Pettie \cite{pettie:ackermann}
have presented algorithms for the MST with worst-case time complexity
-$\O(m\timesalpha(m,n))$. We will devote this chapter to their results
+$\O(m\timesalpha(m,n))$ on the Pointer machine. We will devote this chapter to their results
and especially to another algorithm by Pettie and Ramachandran \cite{pettie:optimal},
which is provably optimal.
similar to the Vuillemin's binomial heaps \cite{vuillemin:binheap} or Fredman's
and Tarjan's Fibonacci heaps \cite{ft:fibonacci}. The soft heaps run faster at
the expense of \df{corrupting} a~fraction of the inserted elements by raising
-their values (the values are however never decreased). This allows for
+their values (the values are however never lowered). This allows for
an~trade-off between accuracy and speed controlled by a~parameter~$\varepsilon$.
The heap operations take $\O(\log(1/\varepsilon))$ amortized time and at every
moment at most~$\varepsilon n$ elements of the~$n$ elements inserted can be
corrupted.
\defnn{Soft heap operations}%
-The soft heap contains distinct elements from a~totally ordered universe and it
+The soft heap contains a~set of distinct items from a~totally ordered universe and it
supports the following operations:
\itemize\ibull
-\:$\<Create>(\varepsilon)$ --- create an~empty soft heap with the given accuracy parameter,
-\:$\<Insert>(H,x)$ --- insert a~new element~$x$ to the heap~$H$,
-\:$\<Meld>(P,Q)$ --- merge two heaps into one, more precisely insert all elements of a~heap~$Q$
+\:$\<Create>(\varepsilon)$ --- create an~empty soft heap with the given accuracy parameter~$\varepsilon$,
+\:$\<Insert>(H,x)$ --- insert a~new item~$x$ to the heap~$H$,
+\:$\<Meld>(P,Q)$ --- merge two heaps into one, more precisely move all items of a~heap~$Q$
to the heap~$P$, destroying~$Q$ in the process (both heaps must have the same~$\varepsilon$),
-\:$\<DeleteMin>(H)$ --- delete the minimum element of the heap~$H$ and return its value
+\:$\<DeleteMin>(H)$ --- delete the minimum item of the heap~$H$ and return its value
(optionally signalling that the value has been corrupted).
+\:$\<Explode>(H)$ --- destroy the heap and return a~list of all items contained in it
+ (again optionally marking those corrupted).
\endlist
\examplen{Linear-time selection}
in the worst case.
\defnn{Soft queues}
-The soft heap is built from \df{soft queues} (we will omit the adjective soft
+The soft heap is built from \df{soft queues} (we will usually omit the adjective soft
in the rest of this section). Each queue has a~shape of a~binary tree.\foot{%
Actually, Chazelle defines the queues as binomial trees, but he transforms them in ways that are
somewhat counter-intuitive, albeit well-defined. We prefer describing the queues as binary
trees with a~special distribution of values. In fact, the original C~code in the Chazelle's
-paper uses this representation internally.}
-Each vertex~$v$ of the tree remembers a~doubly-linked list $\<list>(v)$ of values. The
+paper \cite{chazelle:softheap} uses this representation internally.}
+Each vertex~$v$ of the tree remembers a~doubly-linked list of items. The
item list in every left son will be used only temporarily and it will be kept
empty between operations. Only right sons and the root have their lists
-permanently occupied. The former vertices will be called \df{white}, the latter
+permanently occupied. The left sons will be called \df{white}, the right ones
\df{black.} (See the picture.)
-The first value in the list is called the \df{controlling key} of the vertex,
+The first value in every list is called the \df{controlling key} of the vertex,
denoted by $\<ckey>(v)$. If the list is empty, we keep the most recently used
-value or we set $\<ckey>(v)=+\infty$. The \<ckey>'s obey the standard \df{heap order}
---- a~\<ckey> of a~parent is always less than the \<ckey>'s of its sons.
+value or we set $\<ckey>(v)=+\infty$. The \<ckey> obey the standard \df{heap order}
+--- a~\<ckey> of a~parent is always smaller than the \<ckey>'s of its sons.
Each vertex is also assigned its \df{rank,} which is a~non-negative integer.
-The rank of leaves is always zero, the rank of every internal vertex must be strictly
-greater than the ranks of its sons. We define the rank of a~queue to be equal to the
-rank of its root vertex and similarly for its \<ckey>.
+The ranks of leaves are always zero, the rank of every internal vertex can be
+arbitrary, but it must be strictly greater than the ranks of its sons. We
+define the rank of the whole queue to be equal to the rank of its root vertex and
+similarly for its \<ckey>.
A~queue is called \df{complete} if every two vertices joined by an~edge have rank
difference exactly one. In other words, it is a~complete binary tree and the
(black vertices contain items, numbers indicate ranks)}}
\obs
-A~complete queue of rank~$k$ contains exactly~$2^{k+1}-1$ vertices, $2^k$~of which are
-black (by induction). Any other queue can be trivially embedded to a~complete queue of the same
+The complete queue of rank~$k$ contains exactly~$2^{k+1}-1$ vertices, $2^k$~of which are
+black (by induction). Any other queue can be trivially embedded to the complete queue of the same
rank, which we will call the \df{master tree} of the queue. This embedding preserves vertex
ranks, colors and the ancestor relation.
of the original roots and also its \<ckey>. To preserve the heap order, we will choose
the one whose \<ckey> is smaller.
-Sometimes, we will need to split a~queue to smaller queues. We will call this operation
+Sometimes, we will also need to split a~queue to smaller queues. We will call this operation
\df{dismantling} the queue and it will happen only in cases when the item list in the root
is empty. It suffices to remove the leftmost (all white) path going from the root. From
a~queue of rank~$k$, we get queues of ranks $0,1,\ldots,k-1$, some of which may be missing
if the original queue was not complete.
-We will now combine the soft queues to the soft heap.
-
\figure{softheap2.eps}{\epsfxsize}{Joining and dismantling of soft queues}
+We will now define the real soft heap and the operations on it.
+
\defn A~\df{soft heap} consists of:
\itemize\ibull
\:a~doubly linked list of soft queues of distinct ranks (in increasing order of ranks),
- we will call the first queue the \df{head}, the last queue will be its \df{tail};
+ we will call the first queue the \df{head} of the list, the last queue will be its \df{tail};
\:\df{suffix minima:} each queue contains a~pointer to the queue with minimum \<ckey>
of those following it in the list;
-\:and a~global parameter~$r\in{\bb N}$, to be set depending on~$\varepsilon$.
+\:a~global parameter~$r$: an~even integer to be set depending on~$\varepsilon$.
\endlist
\>We will define the \df{rank} of a~heap as the highest of the ranks of its queues
(that is, the rank of the heap's tail).
The heap always keeps the \df{Rank invariant:} When a~root of any tree has rank~$k$,
-its leftmost path has length at least~$k/2$.
+its leftmost path contains at least~$k/2$ vertices.
+
+\paran{Operations on soft heaps}
-\para
\em{Melding} of two soft heaps involves merging of their lists of queues. We disassemble
-the heap with the smaller maximum rank and we insert its queues to the other heap.
+the heap of the smaller rank and we insert its queues to the other heap.
If two queues of the same rank~$k$ appear in both lists, we \em{join} them to
a~single queue of rank~$k+1$ as already described and we propagate the new queue as
a~\df{carry} to the next iteration, similarly to addition of binary numbers.
\algn{Creating a~new soft heap}
\algo
-\algin A~parameter~$\varepsilon$.
+\algin The~parameter~$\varepsilon$ (the accuracy of the heap).
\:Allocate memory for a~new heap structure~$H$.
-\:Initialize the list of queues in~$H$ as an~empty list.
+\:Initialize the list of queues in~$H$ to an~empty list.
\:Set the parameter~$r$ to~$2\lceil\log(1/\varepsilon)\rceil+2$ (to be justified later).
\algout A~newly minted soft heap~$H$.
\endalgo
\algn{Melding of two soft heaps}
\algo
\algin Two soft heaps~$P$ and~$Q$.
-\:If the heap~$P$ has smaller rank than the heap~$Q$, exchange their item lists.
+\:If $\<rank>(P) < \<rank>(Q)$, exchange the item lists of~$P$ and~$Q$.
\:$p\=\<head>(P)$.
-\brk\cmt{Whenever we run into an~end of a~list in this procedure, we assume that it contains
-an~empty queue of infinite rank.}
+\brk\cmt{Whenever we run into an~end of a~list in this procedure, we assume that
+there is an~empty queue of infinite rank there.}
\:While $Q$ still has some queues:
\::$q\=\<head>(Q)$.
\::If $\<rank>(p) < \<rank>(q)$, then $p\=$ the successor of~$p$,
-\::else if $\<rank>(p) > \<rank>(q)$, remove~$q$ from~$Q$ and insert it before~$p$,
+\::else if $\<rank>(p) > \<rank>(q)$, remove~$q$ from~$Q$ and insert it to~$P$ before~$p$,
\::otherwise (the ranks are equal, we need to propagate the carry):
-\:::$\<carry>=p$.
-\:::$p\=$ the successor of~$p$.
+\:::$\<carry>\=p$.
+\:::Remove~$p$ from~$P$ and set $p\=$ the original successor of~$p$.
\:::While $\<rank>(q)=\<rank>(\<carry>)$:
\::::Remove~$q$ from~$Q$.
\::::$\<carry>\=\<join>(q, \<carry>)$.
\::::$q\=\<head>(Q)$.
\:::Insert~\<carry> before~$q$.
\:Update the suffix minima: Walk with~$p$ backwards to the head of~$P$:
-\::$p'\=$ successor of~$p$.
+\::$p'\=\<suffix\_min>$ of the successor of~$p$.
\::If $\<ckey>(p) < \<ckey>(p')$, set $\<suffix\_min>(p)\=p$.
-\::Otherwise set $\<suffix\_min>(p)\=\<suffix\_min>(p')$.
+\::Otherwise set $\<suffix\_min>(p)\=p'$.
\:Destroy the heap~$Q$.
\algout The merged heap~$P$.
\endalgo
\algn{Insertion of an~element to a~soft heap}
\algo
\algin A~heap~$H$ and a~new element~$x$.
-\:Create a~new heap~$H'$ with the same parameters as~$H$. Let~$H'$ contain a~sole queue of rank~0,
+\:Create a~new heap~$H'$ of the same parameters as~$H$. Let~$H'$ contain a~sole queue of rank~0,
whose only vertex has the element~$x$ in its item list.
\:$\<Meld>(H,H')$.
\algout An~updated heap~$H$.
\endalgo
\para
-So far, the mechanics of the soft heaps was almost identical to the binomial heaps
+So far, the mechanics of the soft heaps were almost identical to the binomial heaps
and the reader could rightfully yawn. The things are going to get interesting
now as we approach the parts where corruption of items takes place.
allow the items to travel at a~faster rate, but as only a~single item can be equal
to the \<ckey>, all other items will be inevitably corrupted.
-We of course have to be careful about the size of the lists, as we must avoid corrupting
+We of course have to be careful about the size of the lists, because we must avoid corrupting
too many items. We will control the growth according to the vertex ranks. Vertices
with rank at most~$r$ will always contain just a~single item. Above this level,
-the higher is the rank, the longer list will we allow.
+the higher is the rank, the longer list will be allowed.
\para
-\em{Deleting of minimum} will follow this principle. The minimum is easy to locate
+\em{Deletion of minimum} will be based on this principle. The minimum is easy to locate
--- we follow the \<suffix\_min> of the head of the heap to the queue with the minimum
-\<ckey> and we look inside the item list of the root of this queue. We remove the \em{last}
-item from the list (we do not want the \<ckey> to change) and return it as the minimum.
-(It is not necessarily the real minimum of all items, but always the minimum of those
-uncorrupted.)
+\<ckey>. There we look inside the item list of the root of the queue. We remove the \em{last}
+item from the list (we do not want the \<ckey> to change) and we return it as the minimum.
+(It is not necessarily the real minimum of all items, but always the minimum of their
+possibly corrupted values.)
If the list becomes empty, we \em{refill} it with items from the lower levels of the
-same tree. This can be best described recursively: We ask the left son to refill itself
-(remember that the left son is always white, so there are no items to lose). If the new
+same queue. This process can be best described recursively: We ask the left son to refill itself
+(remember that the left son is always white, so there are currently no items there). If the new
\<ckey> of the left son is smaller than of the right son, we immediately move the left
son's list to its parent. Otherwise, we exchange the sons and move the list from the
-new left son to the parent. This way, we obey the heap order and the same time we keep
+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,
If refilling of the left son fails because there are no more items in that subtree
(we report this by setting its \<ckey> to $+\infty$), the current vertex is no longer
-needed, as the items would just travel through it unmodified. We therefore want to
-remove it. Instead of deleting it directly, we will rather make it point to its former
-grandsons and we will remove the (now orhpaned) original son. This helps us to ensure
-that both sons always have the same rank, which will be useful for the analysis.
+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 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
queue to the head of the heap exactly as we did in the \<Meld> procedure.
Let us translate these ideas to real (pseudo)code:
-\algn{Deleting the minimum item of a~soft heap}
+\algn{Deleting the minimum item from a~soft heap}
\algo
\algin A~soft heap~$H$.
\:Use \<suffix\_min> of the head queue of~$H$ to locate the queue~$q$ with the smallest \<ckey>.
-\:Remove the last element~$x$ of the item list in the root of~$q$.
+\:Remove the final element~$x$ of the item list in the root of~$q$.
\:If the item list is empty:
\::Count the vertices on the leftmost path of~$q$.
\::If there are less than $\<rank>(q)$ of them:
\:::Meld them back: $\<Meld>(H,H')$.
\::Otherwise:
\:::Call \<Refill> on the root of~$q$.
-\:::If $\<ckey>(q)=+\infty$ (no items left), remove the tree~$q$.
+\:::If $\<ckey>(q)=+\infty$ (no items left), remove the tree~$q$ from~$H$.
\:::Recalculate the suffix minima.
\algout The deleted minimum item~$x$ (possibly corrupted).
\endalgo
\algin A~soft queue and its vertex~$v$ with an~empty item list.
\:Handle trivial cases: If~$v$ has no children or both have $\<ckey>=+\infty$,
set $\<ckey>(v)$ to~$+\infty$ and return.
-\:Let \<left> and~\<right> denote the sons of~$v$.
+\:Let \<left> and~\<right> denote the respective sons of~$v$.
\:Recurse: call $\<Refill>(\<left>)$.
\:If $\<ckey>(\<left>) > \<ckey>(\<right>)$, swap the sons.
\:Move the item list from \<left> to~$v$ (implying $\<ckey>(v)=\<ckey>(\<left>)$).
\algout A~modified soft queue.
\endalgo
-\paran{Analysis of accuracy}
-The description of the operations is complete, let us analyse their behavior
+\para
+\<Explode> is trivial once we know how to recognize the corrupted items. It simply examines
+all queues in the heap, walks the trees and the item lists of all vertices. It records
+all items seen, the corrupted ones are those that different from their \<ckey>.
+
+\paran{Analysis of accuracy}%
+The description of the operations is now complete, so let us analyse their behavior
and verify that we have delivered what we promised --- first the accuracy of
the structure, then the time complexity of operations. In the whole analysis,
-we will denote the total number of elements inserted in the history of the
-structure by~$n$. We will also assume that the threshold~$r$ is even.
+we will denote the total number of elements inserted during the history of the
+structure by~$n$. We will also frequently take advantage of knowing that the
+threshold~$r$ is even.
-We start by bounding the size of the item lists.
+We start by bounding the sizes of the item lists.
\lemma
For every vertex~$v$ of a~soft queue, the size $\ell(v)$ of its item list
$$\ell(v) \le \max(1, 2^{\lceil \<rank>(v)/2 \rceil - r/2}).$$
\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 sometimes 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.
that both sons have the same rank). In both cases, $\lceil\<rank>(w)/2\rceil \le
\lceil\<rank>(v)/2\rceil - 1$. By the induction hypothesis, the size of each
of the two joined lists is at most $2^{\max(1,\lceil\<rank>(v)/2\rceil - 1 - r/2)}$,
-so the result has at most $2^{\lceil\<rank>(v)/2\rceil - r/2}$ items. (The maximum
+so the new list has at most $2^{\lceil\<rank>(v)/2\rceil - r/2}$ items. (The maximum
has disappeared since $\<rank>(v)>r$ and therefore the desired bound is at least~2.)
\qed
\proof
We first prove an~auxiliary claim: The master trees of all queues contain at most~$n$
black vertices. This follows by induction: If no deletions have taken place,
-there are exactly~$n$ of them, because insertion adds one black vertex and
+there are exactly~$n$ black vertices, because insertion adds one black vertex and
melding preserves their number. A~deletion affects the master trees only when
dismantling takes place and then it only removes a~black vertex.
$$
\sum_{i=r+1}^k 2^{k-i}\cdot 2^{\lceil i/2\rceil - r/2}
= 2^{k-r/2} \cdot \sum_{i=r+1}^k 2^{\lceil i/2\rceil - i}
-\le 2^{k-r/2} \cdot \sum_{i=r+1}^k 2^{-i/2}
+\le 2^{k-r/2+1/2} \cdot \sum_{i=r+1}^k 2^{-i/2}
\le 2^{k-r} \cdot \sum_{i=0}^\infty 2^{-i/2}.
$$
The sum of a~geometric series with quotient $2^{-1/2}$ is less than four, so the
-last formula is less than $2^{k-r+2}$. Since the tree contains $n_k=2^k$ black vertices,
-this makes less than $n_k/2^{k-2}$ corrupted items as we asserted.
+last expression is less than $2^{k-r+2}$. Since the tree contains $n_k=2^k$ black vertices,
+this makes less than $n_k/2^{r-2}$ corrupted items as we asserted.
\qed
-\paran{Analysis of time complexity}
-Now we will analyse the amortized time complexity of the individual operations.
-We will show that if we charge $\O(r)$ time against every element inserted, it suffices
-to cover the cost of all other operations. We take a~look at the melds first.
+\paran{Analysis of time complexity}%
+Now we will examine the amortized time complexity of the individual operations.
+We will show that if we charge $\O(r)$ time against every element inserted, it is enough
+to cover the cost of all other operations.
+
+All heap operations use only pointer operations, so it will be easy to derive the time
+bound in the Pointer machine model. The notable exception is however that the procedures
+often refer to the ranks, which are integers on the order of $\log n$, so they cannot
+fit in a~single memory cell. For the time being, we will assume that the ranks can
+be manipulated in constant time, postponing the proof for later.
+
+We take a~look at the melds first.
\lemma\id{shmeld}%
The amortized cost of a~meld is $\O(1)$, except for melds induced by dismantling
which take $\O(\<rank>(q))$, where $q$~is the queue to be dismantled.
\proof
-The real cost of a~meld of heaps $P$ and~$Q$ is the smaller of their ranks plus
-the time spent on carry propagation. The latter is easy to dispose of: since
-every time there is a~carry, the total number of trees in all heaps decreases
-by one, it suffices to charge $\O(1)$ against creation of a~tree. An~insert creates
-one tree, dismantling creates at most $\<rank>(q)$ trees, all other operations
-alter only the internal structure of trees.
+The real cost of a~meld of heaps $P$ and~$Q$ is linear in the smaller of
+their ranks, plus the time spent on carry propagation. The latter is easy to
+dispose of: Every time there is a~carry, the total number of trees in all
+heaps decreases by one. So it suffices to charge $\O(1)$ against creation of
+a~tree. An~insert creates one tree, dismantling creates at most $\<rank>(q)$
+trees, and all other operations alter only the internal structure of trees.
As for the $\O(\min(\<rank>(P),\<rank>(Q)))$ part, let us assume for a~while that
no dismantling ever takes place and consider the \df{meld forest.} It is a~forest
whose leaves correspond to the $n$~single-element heaps constructed by \<Insert>
and each internal vertex represents a~heap arisen from melding its sons. The left
-son will be the one with the greater (or equal) rank. We therefore want to bound
+son will be the one with the greater or equal rank. We therefore want to bound
the sum of ranks of all right sons.
-For every right son, we will distribute the change for its rank~$k$ on all leaves
+For every right son, we will distribute the change for its rank~$k$ among all leaves
in its subtree. There are at least $2^k$ such leaves. No leaf ever receives the same
rank twice, because the ranks of right sons on every path from the root of the
-tree to a~leaf are strictly decreasing. (This holds because melding two heaps
-of the same rank always produces a~heap of higher rank.) Hence at most~$n/2^k$
+tree to a~leaf are strictly decreasing. (This holds because melding of two heaps
+always produces a~heap of a~strictly greater rank.) Hence at most~$n/2^k$
right sons have rank~$k$ and the total time charged against the leaves is bounded by:
$$
-\sum_{k=0}^{\rm max. rank}k\cdot {n\over 2^k} \le n\cdot\sum_{k=0}^\infty {k\over 2^k} = \O(n).
+\sum_{k=0}^{\rm max. rank}k\cdot {n\over 2^k} \le n\cdot\sum_{k=0}^\infty {k\over 2^k} = 2n.
$$
Let us return dismantling to the game. When a~queue is dismantled, melding the parts
back to the heap takes $\O(\<rank>(q))$ time. We can therefore let the dismantling pay for it
-and omit such induced melds from the meld forest. As the rank of the heap is never increased
+and omit such induced melds from the meld forest. As the rank of a~heap is never increased
by induced melds, the above calculation is still a~proper upper bound on the cost
of the regular melds.
\qed
-To estimate the time spent on deletions, we analyse the refills first.
+Before we estimate the time spent on deletions, we analyse the refills.
\lemma
-Every call of the \<Refill> procedure spends 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
we visit the lower part during the recursion, we spend at worst $\O(r)$ time there.
We will prove that the total time spent in the upper parts during the whole life of the
data structure is $\O(n)$. Since each upper vertex can perform at most two calls to the lower
-part, the total time spent in the lower parts is $\O(rn)$. All this can be pre-paid by the
+part, the total time spent in the lower parts is $\O(rn)$. All this can be prepaid by the
inserts.
Let us focus on the upper part. There are three possibilities of what can happen
\:We delete it: Every vertex deleted has to have been created at some time in the past.
New vertices are created only during inserts and melds (when joining two trees) and
-we have already shown that these operations have constant amortized complexity. So the
+we have already shown that these operations have constant amortized complexity. Then the
same must hold for deletions.
\:We recurse twice and concatenate the lists: The lists are disassembled only when
is at most~$n$. Each of them can be performed in constant time as the list is doubly linked.
\:We recurse only once: This occurs only if the rank is even and the gap between the
-rank of this vertex and its sons is small. It therefore cannot happen twice in a~row,
-so it is clearly dominated by the other possibilities.
+rank of this vertex and its sons is equal to~1. It therefore cannot happen twice in a~row,
+thus it is clearly dominated by the cost of the other possibilities.
\endlist
-\>The total cost of all operations on the upper part is therefore $\O(n)$.
+\>The total cost of all steps in the upper part is therefore $\O(n)$.
\qed
-It remains to examine the rest of the \<DeleteMin> operation.
+We now proceed with examining the \<DeleteMin> operation.
\lemma\id{shdelmin}%
Every \<DeleteMin> takes $\O(1)$ time amortized.
Aside from refilling, which is $\O(1)$ by the previous lemma, the \<DeleteMin>
takes care of the Rank invariant. This happens by checking the length of the leftmost
path and dismantling the tree if the length is too far from the tree's rank~$k$.
-The leftmost path is however always visited by the call to \<Refill>, so we can
-account the check on the refilling. The same holds for disassembling. We then have
-to pay $\O(k)$ for melding the trees back to the heap, but since there are at most
-$k/2$ trees, a~subtree of rank at least $k/2$ must have been deleted. This tree
-contained at least $2^{k/2}$ vertices which are now permanently gone from the
-data structure, so we can charge the cost of the meld against these vertices.
+When the invariant is satisfied, the leftmost path is visited by the subsequent
+call to \<Refill>, so we can account the check on the refilling.
+
+When we are dismantling, we have to pay $\O(k)$ for the operation itself and
+another $\O(k)$ for melding the trees back to the heap. Since we have produced at most
+$k/2$ subtrees of distinct ranks, some subtree of rank $k/2$ or more must be missing.
+Its master tree contained at least $2^{k/2}$ vertices which are now permanently gone
+from the data structure, so we can charge the cost against them.
+A~single vertex can participate in the master trees of several dismantlings, but their
+ranks are always strictly increasing. By the same argument as in the proof of
+Lemma \ref{shmeld} (complexity of \<Meld>), each vertex pays $\O(1)$.
We must not forget that \<DeleteMin> also has to recalculate the suffix minima.
In the worst case, it requires touching $k$~trees. Because of the Rank invariant,
this was the only place where we needed the invariant.)
\qed
+\<Explode>s are easy not only to implement, but also to analyse:
+
+\lemma
+Every \<Explode> takes $\O(1)$ time amortized.
+
+\proof
+As all queues, vertices and items examined by \<Explode> are forever gone from the heap,
+we can charge the constant time spent on each of them against the operations
+that have created them.
+\qed
+
+It remains to take care of the calculation with ranks:
+
+\lemma\id{shyards}%
+Every manipulation with ranks performed by the soft heap operations can be
+implemented on the Pointer machine in constant amortized time.
+
+\proof
+We create a~``yardstick'' --- a~double linked list whose elements represent the possible
+values of a~rank. Every vertex of a~queue will store its rank as a~pointer to
+the corresponding ``tick'' of the yardstick. We will extend the list as necessary.
+
+Comparison of two ranks for equality is then trivial, as is incrementing or decrementing
+the rank by~1. Testing whether a~rank is odd can be handled by storing an~odd/even
+flag in every tick. This covers all uses of ranks except for the comparisons for inequality
+when melding. In step~1 of \<Meld>, we just mark the ticks of the two ranks and walk
+the yardstick from the beginning until we come across a~mark. Thus we compare the ranks
+in time proportional to the smaller of them, which is the real cost of the meld anyway.
+The comparisons in steps 5 and~6 are trickier, but since the ranks of the elements put
+to~$P$ are strictly increasing, we can start walking the list at the rank of the previous
+element in~$P$. The cost is then the difference between the current and the previous rank
+and their sum telescopes, again to the real cost of the meld.
+\qed
+
Now we can put the bits together and laurel our effort with the following theorem:
-\thmn{Performance of soft heaps, Chazelle \cite{chazelle:softheap}}
+\thmn{Performance of soft heaps, Chazelle \cite{chazelle:softheap}}\id{softheap}%
A~soft heap with error rate~$\varepsilon$ ($0<\varepsilon\le 1/2$) processes
a~sequence of operations starting with an~empty heap and containing $n$~\<Insert>s
-in time $\O(n\log(1/\varepsilon))$. At every moment, the heap contains at most
-$\varepsilon n$ corrupted items.
+in time $\O(n\log(1/\varepsilon))$ on the Pointer machine. At every moment, the
+heap contains at most $\varepsilon n$ corrupted items.
\proof
We set the parameter~$r$ to~$2+2\lceil\log (1/\varepsilon)\rceil$. The rest follows
from the analysis above. By Lemma \ref{shcorrlemma}, there are always at most $n/2^{r-2}
-\le \varepsilon n$ corrupted items in the heap. By Lemma \ref{shmeld}--\ref{shdelmin},
+\le \varepsilon n$ corrupted items in the heap. By Lemma \ref{shmeld}--\ref{shyards},
the time spent on all operations in the sequence can be paid for by charging $\O(r)$ time
-against each \<Insert>, which yields the time bound.
+against each \<Insert>. This yields the time bound.
+\qed
+
+\rem
+When we set $\varepsilon = 1/2n$, the soft heap is not allowed to corrupt any
+items, so it can be used like any traditional heap. By the standard lower bound on
+sorting it therefore requires $\Omega(\log n)$ time per operation, so the
+time complexity is optimal for this choice of~$\varepsilon$. Chazelle \cite{chazelle:softheap}
+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 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.
+
+%--------------------------------------------------------------------------------
+
+\section{Robust contractions}
+
+Having the soft heaps at hand, we would like to use them in a~conventional MST
+algorithm in place of a~normal heap. The most efficient specimen of a~heap-based
+algorithm we have seen so far is the Iterated Jarn\'\i{}k's algorithm (\ref{itjar}).
+It is based on a~simple, yet powerful idea: Run the Jarn\'\i{}k's algorithm with
+limited heap size, so that it stops when the neighborhood of the tree becomes too
+large. Grow multiple such trees, always starting in a~vertex not visited yet. All
+these trees are contained in the MST, so by the Contraction lemma
+(\ref{contlemma}) we can contract each of them to a~single vertex and iterate
+the algorithm on the resulting graph.
+
+We can try implanting the soft heap in this algorithm, preferably in its earlier
+version without active edges (\ref{jarnik}) as the soft heap lacks the \<Decrease>
+operation. This brave, but somewhat simple-minded attempt is however doomed to
+fail. The reason is of course the corruption of items inside the heap, which
+leads to increase of weights of some subset of edges. In presence of corrupted
+edges, most of the theory we have so carefully built breaks down. For example,
+the Blue lemma (\ref{bluelemma}) now holds only when we consider a~cut with no
+corrupted edges, with a~possible exception of the lightest edge of the cut.
+Similarly, the Red lemma (\ref{redlemma}) is true only if the heaviest edge on
+the cycle is not corrupted.
+
+There is fortunately some light in this darkness. While the basic structural
+properties of MST's no longer hold, there is a~weaker form of the Contraction
+lemma that takes the corrupted edges into account. Before we prove this lemma,
+we expand our awareness of subgraphs which can be contracted.
+
+\defn
+A~subgraph $C\subseteq G$ is \df{contractible} iff for every pair of edges $e,f\in\delta(C)$\foot{That is,
+of~$G$'s edges with exactly one endpoint in~$C$.} there exists a~path in~$C$ connecting the endpoints
+of the edges $e,f$ such that all edges on this path are lighter than either $e$ or~$f$.
+
+\example\id{jarniscont}%
+Let us see that when we stop the Jarn\'\i{}k's algorithm at some moment and
+we take a~subgraph~$C$ induced by the constructed tree, this subgraph is contractible.
+Indeed, when we consider any two distinct edges $uv, xy$ ($u,x\in C$ and $v,y\not\in C$),
+they enter the algorithm's heap at some time. Without loss of generality $uv$~is the first.
+Before the algorithm reaches the vertex~$x$, it adds the whole path $ux$ to the tree.
+As the edge~$uv$ never leaves the heap, all edges on the path $ux$ must be lighter
+than this edge.
+
+We can now easily reformulate the Contraction lemma (\ref{contlemma}) in the language
+of contractible subgraphs. We again assume that we are working with multigraphs
+and that they need not be connected.
+Furthermore, we slightly abuse the notation in the way that we omit the explicit bijection
+between $G-C$ and~$G/C$, so we can write $G=C \cup (G/C)$.
+
+\lemman{Generalized contraction}
+When~$C\subseteq G$ is a~contractible subgraph, then $\msf(G)=\msf(C) \cup \msf(G/C)$.
+
+\proof
+As both sides of the equality are forests spanning the same graph, it suffices
+to show that $\msf(G) \subseteq \msf(C)\cup\msf(G/C)$.
+Let us show that edges of~$G$ that do not belong to the right-hand side
+do not belong to the left-hand side either.
+We know that the edges that
+do not participate in the MSF of some graph are exactly those which are the heaviest
+on some cycle (this is the Cycle rule, \ref{cyclerule}).
+
+Whenever an~edge~$g$ lies in~$C$, but not in~$\msf(C)$, then $g$~is the heaviest edge
+on some cycle in~$C$. As this cycle is also contained in~$G$, the edge $g$~does not participate
+in $\msf(G)$ either.
+
+Similarly for $g\in (G/C)\setminus\msf(G/C)$: when the cycle does not contain
+the vertex~$c$ to which we have contracted the subgraph~$C$, this cycle is present
+in~$G$, too. Otherwise we consider the edges $e,f$ incident with~$c$ on this cycle.
+Since $C$~is contractible, there must exist a~path~$P$ in~$C$ connecting the endpoints
+of~$e$ and~$f$ in~$G$, such that all edges of~$P$ are lighter than either $e$ or~$f$
+and hence also than~$g$. Expanding~$c$ in the cycle to the path~$P$ then produces
+a~cycle in~$G$ whose heaviest edge is~$g$.
+\qed
+
+We are now ready to bring corruption back to the game and state a~``robust'' version
+of this lemma. A~notation for corrupted graphs will be handy:
+
+\nota\id{corrnota}%
+When~$G$ is a~weighted graph and~$R$ a~subset of its edges, we will use $G\crpt
+R$ to denote an arbitrary graph obtained from~$G$ by increasing the weights of
+some of the edges in~$R$. As usually, we will assume that all edges of this graph
+have pairwise distinct weights. While this is technically not true for the corruption
+caused by soft heaps, we can easily make the weights unique.
+
+Whenever~$C$ is a~subgraph of~$G$, we will use $R^C$ to refer to the edges of~$R$ with
+exactly one endpoint in~$C$ (i.e., $R^C = R\cap \delta(C)$).
+
+\lemman{Robust contraction, Chazelle \cite{chazelle:almostacker}}\id{robcont}%
+Let $G$ be a~weighted graph and $C$~its subgraph contractible in~$G\crpt R$
+for some set~$R$ of edges. Then $\msf(G) \subseteq \msf(C) \cup \msf((G/C) \setminus R^C) \cup R^C$.
+
+\proof
+We will modify the proof of the previous lemma. We will again consider all possible types
+of edges which do not belong to the right-hand side and we will show that they are the
+heaviest edges of certain cycles. Every edge~$g$ of~$G$ lies either in~$C$, or in $H=(G/C)\setminus R^C$,
+or possibly in~$R^C$.
+
+If $g\in C\setminus\msf(C)$, then the same argument as before applies.
+
+If $g\in H\setminus\msf(H)$, we consider the cycle in~$H$ on which $g$~is the heaviest.
+When $c$ (the vertex to which we have contracted~$C$) is outside this cycle, we are done.
+Otherwise we observe that the edges $e,f$ adjacent to~$c$ on this cycle cannot be corrupted
+(they would be in~$R^C$ otherwise, which is impossible). By contractibility of~$C$ there exists
+a~path~$P$ in~$C\crpt (R\cap C)$ such that all edges of~$P$ are lighter than $e$ or~$f$ and hence
+also than~$g$. The weights of the edges of~$P$ in the original graph~$G$ cannot be higher than
+in~$G\crpt R$, so the path~$P$ is lighter than~$g$ even in~$G$ and we can again perform the
+trick with expanding the vertex~$c$ to~$P$ in the cycle~$C$. Hence $g\not\in\mst(G)$.
\qed
-\FIXME{Remark on optimality.}
+\para
+We still intend to mimic the Iterative Jarn\'\i{}k's algorithm. We will partition the given graph to a~collection~$\C$
+of non-overlapping contractible subgraphs called \df{clusters} and we put aside all edges that got corrupted in the process.
+We recursively compute the MSF of that subgraphs and of the contracted graph. Then we take the
+union of these MSF's and add the corrupted edges. According to the previous lemma, this does not produce
+the MSF of~$G$, but a~sparser graph containing it, on which we can continue.
+
+We can formulate the exact partitioning algorithm and its properties as follows:
+
+\algn{Partition a~graph to a~collection of contractible clusters}\id{partition}%
+\algo
+\algin A~graph~$G$ with an~edge comparison oracle, a~parameter~$t$ controlling the size of the clusters,
+ and an~accuracy parameter~$\varepsilon$.
+\:Mark all vertices as ``live''.
+\:$\C\=\emptyset$, $R^\C\=\emptyset$. \cmt{Start with an~empty collection and no corrupted edges.}
+\:While there is a~live vertex~$v_0$:
+\::$T=\{v_0\}$. \cmt{the tree that we currently grow}
+\::$K=\emptyset$. \cmt{edges known to be corrupted in the current iteration}
+\::\<Create> a~soft heap with accuracy~$\varepsilon$ and \<Insert> the edges adjacent to~$v_0$ into it.
+\::While the heap is not empty and $\vert T\vert \le t$:
+\:::\<DeleteMin> an~edge $uv$ from the heap, assume $u\in T$.
+\:::If $uv$ was corrupted, add it to~$K$.
+\:::If $v\in T$, drop the edge and repeat the previous two steps.
+\:::$T\=T\cup\{v\}$.
+\:::If $v$ is dead, break out of the current loop.
+\:::Insert all edges incident with~$v$ to the heap.
+\::$\C\=\C \cup \{G[T]\}$. \cmt{Record the cluster induced by the tree.}
+\::\<Explode> the heap and add all remaining corrupted edges to~$K$.
+\::$R^\C\=R^\C \cup K^T$. \cmt{Record the ``interesting'' corrupted edges.}
+\::$G\=G\setminus K^T$. \cmt{Remove the corrupted edges from~$G$.}
+\::Mark all vertices of~$T$ as ``dead''.
+\algout The collection $\C$ of contractible clusters and the set~$R^\C$ of
+corrupted edges in the neighborhood of these clusters.
+\endalgo
+
+\thmn{Partitioning to contractible clusters, Chazelle \cite{chazelle:almostacker}}\id{partthm}%
+Given a~weighted graph~$G$ and parameters $\varepsilon$ ($0<\varepsilon\le 1/2$)
+and~$t$, the Partition algorithm (\ref{partition}) constructs a~collection
+$\C=\{C_1,\ldots,C_k\}$ of clusters and a~set~$R^\C$ of edges such that:
+
+\numlist\ndotted
+\:All the clusters and the set~$R^\C$ are mutually edge-disjoint.
+\:Each cluster contains at most~$t$ vertices.
+\:Each vertex of~$G$ is contained in at least one cluster.
+\:The connected components of the union of all clusters have at least~$t$ vertices each,
+ except perhaps for those which are equal to a~connected component of $G\setminus R^\C$.
+\:$\vert R^\C\vert \le 2\varepsilon m$.
+\:$\msf(G) \subseteq \bigcup_i \msf(C_i) \cup \msf\bigl((G / \bigcup_i C_i) \setminus R^\C\bigr) \cup R^\C$.
+\:The algorithm runs in time $\O(n+m\log (1/\varepsilon))$.
+\endlist
-\FIXME{Example of use: pivots.}
+\proof
+Claim~1: The Partition algorithm grows a~series of trees which induce the clusters~$C_i$ in~$G$.
+A~tree is built from edges adjacent to live vertices
+and once it is finished, all vertices of the tree die, so no edge is ever reused in another
+tree. The edges that enter~$R^\C$ are immediately deleted from the graph, so they never participate
+in any tree.
+
+Claim~2: The algorithm stops when all vertices are dead, so each vertex must have
+entered some tree.
+
+Claim~3: The trees have at most~$t$ vertices each, which limits the size of the
+$C_i$'s as well.
+
+Claim~4: We can show that each connected component has $t$~or more vertices as we already did
+in the proof of Lemma \ref{ijsize}: How can a~new tree stop growing? Either it gathers
+$t$~vertices, or it joins one of the existing trees (this only increases the
+size of the component), or the heap becomes empty (which means that the tree spans
+a~full component of the current graph and hence also of the final~$G\setminus R^\C$).
+
+Claim~5: The set~$R^\C$ contains a~subset of edges corrupted by the soft heaps over
+the course of the algorithm. As every edge is inserted to a~heap at most once per
+its endpoint, the heaps can corrupt at worst $2\varepsilon m$ edges altogether.
+
+We will prove the remaining two claims as separate lemmata.
+\qed
+\lemman{Correctness of Partition, Claim 6 of Theorem \ref{partthm}}
+$$\msf(G) \subseteq \bigcup_i \msf(C_i) \cup \msf\biggl( \bigl(G / \bigcup_i C_i \bigr) \setminus R^\C\biggr) \cup R^\C.$$
+
+\proof
+By iterating the Robust contraction lemma (\ref{robcont}). Let $K_i$ be the set of edges
+corrupted when constructing the cluster~$C_i$ in the $i$-th phase of the algorithm,
+and similarly for the state of the other variables at that time.
+We will use induction on~$i$ to prove that the lemma holds at the end of every phase.
+
+At the beginning, the statement is obviously true, even as an~equality.
+In the $i$-th phase we construct the cluster~$C_i$ by running the partial Jarn\'\i{}k's algorithm on the graph
+$G_i = G\setminus\bigcup_{j<i} K_{\smash j}^{C_j}$.
+If it were not for corruption, the cluster~$C_i$ would be contractible, as we already know from Example \ref{jarniscont}.
+When the edges in~$K_i$ get corrupted, the cluster is contractible in some corrupted graph
+$G_i \crpt K_i$. (We have to be careful as the edges are becoming corrupted gradually,
+but it is easy to check that it can only improve the situation.) Since $C_i$~shares no edges
+with~$C_j$ for any~$j<i$, we know that~$C_i$ also a~contractible subgraph of $(G_i / \bigcup_{j<i} C_j) \crpt K_i$.
+Now we can use the Robust contraction lemma to show that:
+$$
+\msf\biggl(\bigl(G / \bigcup_{j<i} C_j \bigr) \setminus \bigcup_{j<i} K_j^{C_j} \biggr) \subseteq
+\msf(C_i) \cup \msf\biggl(\bigl(G / \bigcup_{j\le i} C_j \bigr) \setminus \bigcup_{j\le i} K_j^{C_j} \biggr) \cup K_i^{C_i}.
+$$
+This completes the induction step, because $K_j^{C_j} = K_j^{\C_j}$ for all~$j$.
+\qed
+
+\lemman{Efficiency of Partition, Claim 7 of Theorem \ref{partthm}}
+The Partition algorithm runs in time $\O(n+m\log(1/\varepsilon))$.
+
+\proof
+The inner loop (steps 7--13) processes the edges of the current cluster~$C_i$ and also
+the edges in its neighborhood $\delta(C_i)$. Steps 6 and~13 insert every such edge to the heap
+at most once, so steps 8--12 visit each edge also at most once. For every edge, we spend
+$\O(\log(1/\varepsilon))$ time amortized on inserting it and $\O(1)$ on the other operations
+(by Theorem \ref{softheap} on performance of the soft heaps).
+
+Each edge of~$G$ can participate in some $C_i\cup \delta(C_i)$ only twice before both its
+endpoints die. The inner loop therefore processes $\O(m)$ edges total, so it takes $\O(m\log(1/\varepsilon))$
+time. The remaining parts of the algorithm spend $\O(m)$ time on operations with clusters and corrupted edges
+and additionally $\O(n)$ on identifying the live vertices.
+\qed
+
+%--------------------------------------------------------------------------------
+
+\section{Decision trees}
+
+The Pettie's and Ramachandran's algorithm combines the idea of robust partitioning with optimal decision
+trees constructed by brute force for very small subgraphs. In this section, we will
+explain the basics of the decision trees and prove several lemmata which will
+turn out to be useful for the analysis of time complexity of the final algorithm.
+
+Let us consider all computations of some comparison-based MST algorithm when we
+run it on a~fixed graph~$G$ with all possible permutations of edge weights.
+The computations can be described by a~binary tree. The root of the tree corresponds to the first
+comparison performed by the algorithm and depending to its result, the computation
+continues either in the left subtree or in the right subtree. There it encounters
+another comparison and so on, until it arrives to a~leaf of the tree where the
+spanning tree found by the algorithm is recorded.
+
+Formally, the decision trees are defined as follows:
+
+\defnn{Decision trees and their complexity}\id{decdef}%
+A~\df{MSF decision tree} for a~graph~$G$ is a~binary tree. Its internal vertices
+are labeled with pairs of $G$'s edges to be compared, each of the two outgoing edges
+corresponds to one possible result of the comparison.\foot{There are two possible
+outcomes since there is no reason to compare an~edge with itself and we, as usually,
+expect that the edge weights are distinct.}
+Leaves of the tree are labeled with spanning trees of the graph~$G$.
+
+A~\df{computation} of the decision tree on a~specific permutation of edge weights
+in~$G$ is a~path from the root to a~leaf such that the outcome of every comparison
+agrees with the edge weights. The result of the computation is the spanning tree
+assigned to its final leaf.
+A~decision tree is \df{correct} iff for every permutation the corresponding
+computation results in the real MSF of~$G$ with the particular weights.
+
+The \df{time complexity} of a~decision tree is defined as its depth. It therefore
+bounds the number of comparisons spent on every path. (It need not be equal since
+some paths need not correspond to an~actual computation --- the sequence of outcomes
+on the path could be unsatifisfiable.)
+
+A~decision tree is called \df{optimal} if it is correct and its depth is minimum possible
+among the correct decision trees for the given graph.
+We will denote an~arbitrary optimal decision tree for~$G$ by~${\cal D}(G)$ and its
+complexity by~$D(G)$.
+
+The \df{decision tree complexity} $D(m,n)$ of the MSF problem is the maximum of~$D(G)$
+over all graphs~$G$ with $n$~vertices and~$m$ edges.
+
+\obs
+Decision trees are the most general comparison-based computation model possible.
+The only operations which count in the time complexity are the comparisons. All
+other computation is free, including solving NP-complete problems or having
+access to an~unlimited source of non-uniform constants. The decision tree
+complexity is therefore an~obvious lower bound on the time complexity of the
+problem in all other models.
+
+The downside is that we do not know any explicit construction of the optimal
+decision trees, or at least a~non-constructive proof of their complexity.
+On the other hand, the complexity of any existing comparison-based algorithm
+can be used as an~upper bound for the decision tree complexity. For example:
+
+\lemma
+$D(m,n) \le 4/3 \cdot n^2$.
+
+\proof
+Let us count the comparisons performed by the Contractive Bor\o{u}vka's algorithm.
+(\ref{contbor}), tightening up the constants in its previous analysis in Theorem
+\ref{contborthm}. In the first iteration, each edge participates in two comparisons
+(one for each its endpoint), so the algorithm performs at most $2m \le 2{n\choose 2} \le n^2$
+comparisons. Then the number of vertices drops at least by a~factor of two, so
+the subsequent iterations spend at most $(n/2)^2, (n/4)^2, \ldots$ comparisons, which sums
+to less than $n^2\cdot\sum_{i=0}^\infty (1/4)^i = 4/3 \cdot n^2$. Between the phases,
+we flatten the multigraph to a~simple graph, which also needs some comparisons,
+but for every such comparison we remove one of the participating edges, which saves
+at least one comparison in the subsequent phases.
+\qed
+
+\para
+Of course we can get sharper bounds from the better algorithms, but we will first
+show how to find the optimal trees using brute force. The complexity of the search
+will be of course enormous, but as we already promised, we will need the optimal
+trees only for very small subgraphs.
+
+\lemman{Construction of optimal decision trees}\id{odtconst}%
+An~optimal MST decision tree for a~graph~$G$ on~$n$ vertices can be constructed on
+the Pointer machine in time $\O(2^{2^{4n^2}})$.
+
+\proof
+We will try all possible decision trees of depth at most $2n^2$
+(we know from the previous lemma that the desired optimal tree is shallower). We can obtain
+any such tree by taking the complete binary tree of exactly this depth
+and labeling its $2\cdot 2^{2n^2}-1$ vertices with comparisons and spanning trees. Those labeled
+with comparisons become internal vertices of the decision tree, the others
+become leaves and the parts of the tree below them are removed. There are less
+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
+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 permutation 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
+$\O(2^{2^{3n^2}} \cdot 2^{n^3} \cdot \poly(n)) = \O(2^{2^{4n^2}})$.
+\qed
+
+\paran{Basic properties of decision trees}%
+The following properties will be useful for analysis of algorithms based
+on precomputed decision trees. We will omit some technical details, referring
+the reader to section 5.1 of the Pettie's article \cite{pettie:optimal}.
+
+\lemma\id{dtbasic}%
+The decision tree complexity $D(m,n)$ of the MSF satisfies:
+\numlist\ndotted
+\:$D(m,n) \ge m/2$ for $m,n > 2$.
+\:$D(m',n') \ge D(m,n)$ whenever $m'\ge m$ and $n'\ge n$.
+\endlist
+
+\proof
+For every $m,n>2$ there is a~graph on $n$~vertices and $m$~edges such that
+every edge lies on a~cycle. Every correct MSF decision tree for this graph
+has to compare each edge at least once. Otherwise the decision tree cannot
+distinguish between the case when an~edge has the lowest of all weights (and
+thus it is forced to belong to the MSF) and when it has the highest weight (so
+it is forced out of the MSF).
+
+Decision trees for graphs on $n'$~vertices can be used for graphs with $n$~vertices
+as well --- it suffices to add isolated vertices, which does not change the MSF.
+Similarly, we can increase $m$ to~$m'$ by adding edges parallel to an~existing
+edge and making them heavier than the rest of the graph, so that they can never
+belong to the MSF.
+\qed
+
+\defn
+Subgraphs $C_1,\ldots,C_k$ of a~graph~$G$ are called the \df{compartments} of~$G$
+iff they are edge-disjoint, their union is the whole graph~$G$ and
+$\msf(G) = \bigcup_i \msf(C_i)$ for every permutation of edge weights.
+
+\lemma\id{partiscomp}%
+The clusters $C_1,\ldots,C_k$ generated by the Partition procedure of the
+previous section (Algorithm \ref{partition}) are compartments of the graph
+$H=\bigcup_i C_i$.
+
+\proof
+The first and second condition of the definition of compartments follow
+from the Partitioning theorem (\ref{partthm}), so it remains to show that $\msf(H)$
+is the union of the MSF's of the individual compartments. By the Cycle rule
+(\ref{cyclerule}), an~edge $h\in H$ is not contained in $\msf(H)$ if and only if
+it is the heaviest edge on some cycle. It is therefore sufficient to prove that
+every cycle in~$H$ is contained within a~single~$C_i$.
+
+Let us consider a~cycle $K\subseteq H$ and a~cluster~$C_i$ such that it contains
+an~edge~$e$ of~$K$ and all clusters constructed later by the procedure do not contain
+any. If $K$~is not fully contained in~$C_i$, we can extend the edge~$e$ to a~maximal
+path contained in both~$K$ and~$C_i$. Since $C_i$ shares at most one vertex with the
+earlier clusters, there can be at most one edge from~$K$ adjacent to the maximal path,
+which is impossible.
+\qed
+
+\lemma
+Let $C_1,\ldots,C_k$ be compartments of a~graph~$G$. Then there exists an~optimal
+MSF decision tree for~$G$ that does not compare edges of distinct compartments.
+
+\proofsketch
+Consider a~subset~$\cal P$ of edge weight permutations~$w$ that satisfy $w(e) < w(f)$
+whenever $e\in C_i, f\in C_j, i<j$. For such permutations, no decision tree can
+gain any information on relations between edge weights in a~single compartment by
+inter-compartment comparisons --- the results of all such comparisons are determined
+in advance.
+
+Let us take an~arbitrary correct decision tree for~$G$ and restrict it to
+vertices reachable by computations on~$\cal P$. Whenever a~vertex contained
+an~inter-compartment comparison, it has lost one of its sons, so we can remove it
+by contracting its only outgoing edge. We observe that we get a~decision tree
+satisfying the desired condition and that this tree is correct.
+
+As for the correctness, the MSF of a~single~$C_i$ is uniquely determined by
+comparisons of its weights and the set~$\cal P$ contains all combinations
+of orderings of weights inside individual compartments. Therefore every
+spanning tree of every~$C_i$ and thus also of~$H$ is properly recognized.
+\qed
+
+\lemma\id{compartsum}%
+Let $C_1,\ldots,C_k$ be compartments of a~graph~$G$. Then $D(G) = \sum_i D(C_i)$.
+
+\proofsketch
+A~collection of decision trees for the individual compartments can be ``glued together''
+to a~decision tree for~$G$. We take the decision tree for~$C_1$, replace every its leaf
+by a~copy of the tree for~$C_2$ and so on. Every leaf~$\ell$ of the compound tree will be
+labeled with the union of labels of the original leaves encountered on the path from
+the root to~$\ell$. This proves that $D(G) \le \sum_i D(C_i)$.
+
+The other inequality requires more effort. We use the previous lemma to transform
+the optimal decision tree for~$G$ to another of the same depth, but without inter-compartment
+comparisons. Then we prove by induction on~$k$ and then on the depth of the tree
+that this tree can be re-arranged, so that every computation first compares edges
+from~$C_1$, then from~$C_2$ and so on. This means that the tree can be decomposed
+to decision trees for the $C_i$'s. Also, without loss of generality all trees for
+a~single~$C_i$ can be made isomorphic to~${\cal D}(C_i)$.
+\qed
+
+\cor\id{dtpart}%
+If $C_1,\ldots,C_k$ are the clusters generated by the Partition procedure (Algorithm \ref{partition}),
+then $D(\bigcup_i C_i) = \sum_i D(C_i)$.
+
+\proof
+Lemma \ref{partiscomp} tells us that $C_1,\ldots,C_k$ are compartments of the graph
+$\bigcup C_i$, so we can apply Lemma \ref{compartsum} on them.
+\qed
+
+\cor\id{dttwice}%
+$2D(m,n) \le D(2m,2n)$ for every $m,n$.
+
+\proof
+For an~arbitrary graph~$G$ with $m$~edges and $n$~vertices, we create a~graph~$G_2$
+consisting of two copies of~$G$ sharing a~single vertex. The copies of~$G$ are obviously
+compartments of~$G_2$, so by Lemma \ref{compartsum} it holds that $D(G_2) = 2D(G)$.
+Taking a~maximum over all choices of~$G$ yields $D(2m,2n) \ge \max_G D(G_2) = 2D(m,n)$.
+\qed
+
+%--------------------------------------------------------------------------------
+
+\section{An optimal algorithm}\id{optalgsect}%
+
+Once we have developed the soft heaps, partitioning and MST decision trees,
+it is now simple to state the Pettie's and Ramachandran's MST algorithm
+and prove that it is asymptotically optimal among all MST algorithms in
+comparison-based models. Several standard MST algorithms from the previous
+chapters will also play their roles.
+
+We will describe the algorithm as a~recursive procedure. When the procedure is
+called on a~graph~$G$, it sets the parameter~$t$ to roughly $\log^{(3)} n$ and
+it calls the \<Partition> procedure to split the graph into a~collection of
+clusters of size~$t$ and a~set of corrupted edges. Then it uses precomputed decision
+trees to find the MSF of the clusters. The graph obtained by contracting
+the clusters is on the other hand dense enough, so that the Iterated Jarn\'\i{}k's
+algorithm runs on it in linear time. Afterwards we combine the MSF's of the clusters
+and of the contracted graphs, we mix in the corrupted edges and run two iterations
+of the Contractive Bor\o{u}vka's algorithm. This guarantees reduction in the number of
+both vertices and edges by a~constant factor, so we can efficiently recurse on the
+resulting graph.
+
+\algn{Optimal MST algorithm, Pettie and Ramachandran \cite{pettie:optimal}}\id{optimal}%
+\algo
+\algin A~connected graph~$G$ with an~edge comparison oracle.
+\:If $G$ has no edges, return an~empty tree.
+\:$t\=\lfloor\log^{(3)} n\rfloor$. \cmt{the size of clusters}
+\:Call \<Partition> (\ref{partition}) on $G$ and $t$ with $\varepsilon=1/8$. It returns
+ a~collection~$\C=\{C_1,\ldots,C_k\}$ of clusters and a~set~$R^\C$ of corrupted edges.
+\:$F_i \= \mst(C_i)$ for all~$i$, obtained using optimal decision trees.
+\:$G_A \= (G / \bigcup_i C_i) \setminus R^\C$. \cmt{the contracted graph}
+\:$F_A \= \msf(G_A)$ calculated by the Iterated Jarn\'\i{}k's algorithm (\ref{itjar}).
+\:$G_B \= \bigcup_i F_i \cup F_A \cup R^\C$. \cmt{combine subtrees with corrupted edges}
+\:Run two iterations of the Contractive Bor\o{u}vka's algorithm (\ref{contbor}) on~$G_B$,
+ getting a~contracted graph~$G_C$ and a~set~$F_B$ of MST edges.
+\:$F_C \= \mst(G_C)$ obtained by a~recursive call to this algorithm.
+\:Return $F_B \cup F_C$.
+\algout The minimum spanning tree of~$G$.
+\endalgo
+
+Correctness of this algorithm immediately follows from the Partitioning theorem (\ref{partthm})
+and from the proofs of the respective algorithms used as subroutines. Let us take a~look at
+the time complexity. We will be careful to use only the operations offered by the Pointer machine.
+
+\lemma\id{optlemma}%
+The time complexity $T(m,n)$ of the Optimal algorithm satisfies the following recurrence:
+$$
+T(m,n) \le \sum_i c_1 D(C_i) + T(m/2, n/4) + c_2 m,
+$$
+where~$c_1$ and~$c_2$ are some positive constants and $D$~is the decision tree complexity
+from the previous section.
+
+\proof
+The first two steps of the algorithm are trivial as we have linear time at our
+disposal.
+
+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 clusters 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
+(unless there is just a~single component).
+
+To apply the decision trees, we will use the framework of topological computations developed
+in Section \ref{bucketsort}. We pad all clusters in~$\C$ with isolated vertices, so that they
+have exactly~$t$ vertices. We use a~computation that labels the graph with a~pointer to
+its optimal decision tree. Then we apply Theorem \ref{topothm} combined with the
+brute-force construction of optimal decision trees from Lemma \ref{odtconst}. Together they guarantee
+that we can assign the decision trees to the clusters in time:
+$$\O\Bigl(\Vert\C\Vert + t^{t(t+2)} \cdot \bigl(2^{2^{4t^2}} + t^2\bigr)\Bigr)
+= \O\Bigl(m + 2^{2^{2^t}}\Bigr)
+= \O(m).$$
+Execution of the decision tree on each cluster~$C_i$ then takes $\O(D(C_i))$ steps.
+
+The contracted graph~$G_A$ has at most $n/t = \O(n / \log^{(3)}n)$ vertices and asymptotically
+the same number of edges as~$G$, so according to Corollary \ref{ijdens}, the Iterated Jarn\'\i{}k's
+algorithm runs on it in linear time.
+
+The combined graph~$G_B$ has~$n$ vertices, but less than~$n$ edges from the
+individual spanning trees and at most~$m/4$ additional edges which were
+corrupted. The iterations of the Bor\o{u}vka's algorithm on~$G_B$ take $\O(m)$
+time by Lemma \ref{boruvkaiter} and they produce a~graph~$G_C$ with at most~$n/4$
+vertices and at most $n/4 + m/4 \le m/2$ edges. (The $n$~tree edges in~$G_B$ are guaranteed
+to be reduced by the Bor\o{u}vka's algorithm.) It is easy to verify that this
+graph is still connected, so we can recurse on it.
+
+The remaining steps of the algorithm can be easily performed in linear time either directly
+or in case of the contractions by the bucket-sorting techniques of Section \ref{bucketsort}.
+\qed
+
+\paran{Optimality}%
+The properties of decision tree complexity, which we have proven in the previous
+section, will help us show that the time complexity recurrence is satisfied by a~constant
+multiple of the decision tree complexity $D(m,n)$ itself. This way, we will prove
+the following theorem:
+
+\thmn{Optimality of the Optimal algorithm}
+The time complexity of the Optimal MST algorithm \ref{optimal} is $\Theta(D(m,n))$.
+
+\proof
+We will prove by induction that $T(m,n) \le cD(m,n)$ for some $c>0$. The base
+case is trivial, for the induction step we will expand on the previous lemma:
+\def\eqalign#1{\null\,\vcenter{\openup\jot
+ \ialign{\strut\hfil$\displaystyle{##}$&$\displaystyle{{}##}$\hfil
+ \crcr#1\crcr}}\,}
+$$\vcenter{\openup\jot\halign{\strut\hfil $\displaystyle{#}$&$\displaystyle{{}#}$\hfil&\quad#\hfil\cr
+T(m,n)
+ &\le \sum_i c_1 D(C_i) + T(m/2, n/4) + c_2 m &(Lemma \ref{optlemma})\cr
+ &\le c_1 D({\textstyle\bigcup}_i C_i) + T(m/2, n/4) + c_2 m &(Corollary \ref{dtpart})\cr
+ &\le c_1 D(m,n) + T(m/2, n/4) + c_2m &(definition of $D(m,n)$)\cr
+ &\le c_1 D(m,n) + cD(m/2, n/4) + c_2m &(induction hypothesis)\cr
+ &\le c_1 D(m,n) + c/2\cdot D(m,n/2) + c_2m &(Corollary \ref{dttwice})\cr
+ &\le c_1 D(m,n) + c/2\cdot D(m,n) + 2c_2 D(m,n) &(Lemma \ref{dtbasic})\cr
+ &\le (c_1 + c/2 + 2c_2) \cdot D(m,n)&\cr
+ &\le cD(m,n). &(by setting $c=2c_1+4c_2$)\cr
+}}$$
+The other inequality is obvious as $D(m,n)$ is an~asymptotic lower bound on
+the time complexity of every comparison-based algorithm.
+\qed
+
+\paran{Complexity of MST}%
+As we have already noted, the exact decision tree complexity $D(m,n)$ of the MST problem
+is still open and so is therefore the time complexity of the optimal algorithm. However,
+every time we come up with another comparison-based algorithm, we can use its complexity
+(or more specifically the number of comparisons it performs, which can be even lower)
+as an~upper bound on the optimal algorithm.
+
+The best explicit comparison-based algorithm known to date achieves complexity $\O(m\timesalpha(m,n))$.
+It has been discovered by Chazelle \cite{chazelle:ackermann} as an~improvement of his previous
+$\O(m\timesalpha(m,n)\cdot\log\alpha(m,n))$ algorithm \cite{chazelle:almostacker}.
+It is also based on the ideas of this chapter --- in particular the soft heaps and robust contractions.
+The algorithm is very complex and it involves a~lot of elaborate
+technical details, so we only refer to the original paper here. Another algorithm of the same
+complexity, using similar ideas, has been discovered independently by Pettie \cite{pettie:ackermann}, who,
+having the optimal algorithm at hand, does not take care about the low-level details and he only
+bounds the number of comparisons. Using any of these results, we can prove an~Ackermannian
+upper bound on the optimal algorithm:
+
+\thmn{Upper bound on complexity of the Optimal algorithm}
+The time complexity of the Optimal MST algorithm is $\O(m\timesalpha(m,n))$.
+
+\proof
+We bound $D(m,n)$ by the number of comparisons performed by the algorithm of Chazelle \cite{chazelle:ackermann}
+or Pettie \cite{pettie:ackermann}.
+\qed
+
+Similarly to the Iterated Jarn\'\i{}k's algorithm, this bound is actually linear for classes of graphs
+that do not have density extremely close to constant:
+
+\cor
+The Optimal MST algorithm runs in linear time whenever $m\ge n\cdot a(k,n)$ for any fixed~$k$.
+
+\proof
+Combine the previous theorem with Lemma \ref{alphaconst}.
+\qed
+\rem
+It is also known from \cite{pettie:optimal} that when we run the Optimal algorithm on a~random
+graph drawn from either $G_{n,p}$ (random graphs on~$n$ vertices, each edge is included with probability~$p$
+independently on the other edges) or $G_{n,m}$ (we draw the graph uniformly at random from the
+set of all graphs with~$n$ vertices and $m$~edges), it runs in linear time with high probability,
+regardless of the edge weights.
+
+\paran{Models of computation}%
+Another important consequence of the optimal algorithm is that when we aim for a~linear-time
+MST algorithm (or for proving that it does not exist), we do not need to care about computational
+models at all. The elaborate RAM data structures of Chapter \ref{ramchap}, which have helped us
+so much in the case of integer weights, cannot help if we are allowed to access the edge weights
+by performing comparisons only. We can even make use of non-uniform objects given by some sort
+of oracle. Indeed, whatever trick we employ to achieve linear time complexity, we can mimic it in the
+world of decision trees and thus we can use it to show that the algorithm we already knew is
+also linear.
+
+This however applies to deterministic algorithms only --- we have shown that access to a~source
+of random bits allows us to compute the MST in expected linear time (the KKT algorithm, \ref{kkt}).
+There were attempts to derandomize the KKT algorithm, but so far the best result in this direction
+is the randomized algorithm also by Pettie \cite{pettie:minirand} which achieves expected linear time
+complexity with only $\O(\log^* n)$ random bits.
\endpart
projects / saga.git / blobdiff