Amortizované vyhľadávacie stromy a link-cut stromy

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Splay trees

• intro on slides

Intuition and examples

• if height of tree is large, search is expensive, we use potential
• potential of path-like trees should be large, balanced trees smaller
• what happens if we apply splay on a leaf on a path?
• side benefit: if we search for some nodes more often, they tend to be near the root

Amortized analysis of splaying

• real cost: number of rotations
• D(x): number of descendants of node x, including x (size of a node)
• r(x): log_2 (D(x)) rank of a node (we will denote log_2 by lg)
• $\Phi (T)=\sum _{{x}}r(x)$
• Exercise: asymptotically estimate potential of a path and balanced tree with n nodes

Lemma1 (proof later): consider one step of splaying x (1 or 2 rotations)

• let r(x) be rank of x before splaying, r'(x) after splaying
• amortized cost of one step of splaying is at most
  • 3(r'(x)-r(x)) for zig-zag and zig-zig
  • 3(r'(x)-r(x))+1 for zig

Lemma2: Amortized cost of splaying x to the root in a tree with n nodes is O(log n)

• Let r_i be rank of x after i steps of splaying, let k be the number of steps
• amortized cost is at most $1+\sum _{{i=1}}^{k}3(r_{i}-r_{{i-1}})$
  • (use Lemma 1, +1 is from the last step of splaying)
• telescopic sum, terms cancel, we get $3r_{k}-3r_{0}+1=O(\log n)$

Theorem: Amortized cost of insert, search and delete in splay tree is O(log n)

• search:
  • walk down to x, then splay x (if x not found, splay the last visited node)
  • amortized cost is O(log n)
• insert:
  • insert x as in normal binary search tree, then splay x.
  • Inserting increase rank of all nodes on the path P from root r to x
  • consider node v and its parent p(v) on this path P
  • D'(v) = 1 + D(v) <= D(p(v))
  • r'(v) <= r(p(v))
  • change in potential $\sum _{{v}}r'(v)-\sum _{{v}}r(v)\leq r'(r)+\sum _{{v\in P,v\neq r}}r(p(v))-r(v)=2r'(r)=O(\log n)$
  • amortized cost of insert + splaying is thus O(log n)
• delete:
  • delete x as from binary search tree (may in fact delete node for predecessor or successor of x), then splay parent of deleted node
  • deletion itself decreases potential, amortized cost of splaying is O(log n)
  • amortized cost is O(log n)
• overall time in each operation proportional to the length of the path traversed and therefore to the number of rotations done in splaying
  • therefore amortized time O(log n) for each operation.

Proof of Lemma 1:

• case zig: cost 1, r'(x) = r(y), all nodes except x and y keep their ranks
  • amortized cost $1+r'(x)+r'(y)-r(x)-r(y)=1+r'(y)-r(x)<=1+r'(x)-r(x)<=1+3(r'(x)-r(x))$
• case zig-zig: cost 2,
  • amortized cost $2+r'(x)+r'(y)+r'(z)-r(x)-r(y)-r(z)$
  • we have r'(x) = r(z), r'(y)<=r'(x), r(y)>=r(x)
  • therefore am. cost at most $2+r'(x)+r'(z)-2r(x)$
  • want 3(r'(x)-r(x)) therefore we need 2 <= 2r'(x)-r(x)-r'(z) or (r(x)-r'(x))+(r'(z)-r'(x))<=-2 or lg (D(x)/D'(x)) + lg (D'(z)/D'(x)) <= -2
  • observe that D(x)+D'(z) <= D'(x)
  • if we have a,b>0 s.t. a+b<=1, then lg a + lg b has maximum -2 at a=b=1/2.
    • how would you go about proving this?
    • binary logarithm is concave, i.e. lg(ta+(1-t)b)>= t lg(a)+(1-t)lg(b) for every t in [0,1]
    • set t = 1/2, (lg(a)+lg(b))/2 <= lg((a+b)/2) <= lg(1/2) = -1
  • here a = D(x)/D'(x), b = D'(z)/D'(x)
• case zig-zag is analogous
  • amortized cost $2+r'(x)+r'(y)+r'(z)-r(x)-r(y)-r(z)$
  • r'(x) = r(z), r(x)<=r(y)
  • therefore am. cost at most $2+r'(y)+r'(z)-2r(x)$
  • we will prove that 2 <= 2r'(x)-r'(y)-r'(z) and this will imply amortized cost is at most 2(r'(x)-r(x))
  • we have D'(y)+D'(z) <= D'(x), use the same logarithmic trick as before

Collection of splay trees

Image now a collection of splay trees. The following operations can be done in O(log n) amortized time (each operation gets pointer to a node)

• findMin(v): find minumum element in a tree containing v and make it root of that tree
  • splay v, follow left pointers to min m, splay m, return m
• join(v1, v2): all elements in tree of v1 must be smaller than all elements in tree of v2. Join these two trees into one
  • m = findMin(v2), splay(v1), connect v1 as a left child of m, return m
  • connecting increases rank of m by at most log n
• splitAfter(x) - split tree containing v into 2 trees, one containing keys <=x, one containing keys > x, return root of the second tree
  • splay(v), then cut away its right child and return it
  • decreases rank of v
• splitBefore(v) - split tree containing v into 2 trees, one containing keys <x, one containing keys >=x, return root of the first tree
  • analogous, cut away left child

Link-cut trees

Union/find

Maintains a collection of disjoint sets, supports operations

• union(v, w): connects sets containing v and w
• find(v): returns representative element of set containing v (can be used to test of v and w are in the same set)

Can be used to maintain connected components if we add edges to the graph, also useful in Kruskal's algorithm for minimum spanning tree

Implementation

• each set a (non-binary) trees, in which each node v has a pointer to its parent v.p
• find(v) follows pointers to the root of the tree, return the root
• union calls find for v and w and joins one with the other
• if we keep track of tree height and always join shorter tree below higher tree plus do path compression, we get amortized time O(alpha(m+n,n)) where alpha is inverse ackermann function, extremely slowly growing (n = number of elements, m = number of queries)

Exercise: implement as a collection of splay trees

• hint: want to use join for union, but what about the keys?
• splay trees actually do not need to store key values if we do not search

Link/cut trees

Maintains a collection of disjoint rooted trees on n nodes

• findRoot(v) find root of tree containing v
• link(v,w) v a root, w not in tree of v, make v a child of w
• cut(v) cut edge connecting v to its parent (v not a root)

O(log n) amortized per operation. We will show O(log^2) amortized time. Can be also modified to support these operations in worst-case O(log n) time, add more operations e.g. with weights in nodes.

Similar as union-find, but adds cut plus we cannot rearrange trees as needed as in union-find

Simpler version for paths

Simpler version for paths (imagine them as going upwards):

• findPathHead(v) - highest element on path containing v
• linkPaths(v, w) - join paths containing v and w (head of v's path will remain head)
• splitPathAbove(v) - remove edge connecting v to its parent p, return some node in the path containing p
• splitPathBelow(v) - remove edge connecting v to its chold c, return some node in the path containing c

Can be done by splay trees

• keep each path in a splay tree, using position as a key (not stored anywhere)
• findPathHead(v): findMin(v)
• linkPaths(v, w): join(v, w)
• splitPathAbove(v): splitBefore(v)
• splitPathBelow(v): splitAfter(v)

Back to link/cut trees

• separate edges in each tree into dashed and solid, among children of each node at most one connected by a solid edge
• series of solid paths (some solid paths contain only a single vertex)
• each solid path represented as a splay tree as above
• each node remembers the dashed edge going to its parent (if any) - only in the head of a solid path
• expose(v): make v the lower end of a solid path to root (make all edges on the path to root solid and make other edges dashed as necessary)
  • assume can be done in O(log^2 n) amortized time
• findRoot(v): findPathHead(expose(v))
• link(v,w):
  • expose(v) note: v is now 1-node solid path because v is a root
  • expose(w)
  • linkPaths(w, v)
• cut(v):
  • expose(v) note: v not a root and thus connected to its parent by a solid edge
  • splitPathAbove(v)

expose(v)

• series of splices where each splice adds one solid edge on the path from v to root

 
y = cutPathBellow(v);
if(y!=NULL) findPathHead(y).dashed = v;
while(true) {
  x = findPathHead(v);
  w = x.dashed;
  if(w == NULL) break;
  x.dashed = NULL;
  q = splitPathBelow(w);
  if(q != NULL) {
    findPathHead(q).dashed = w;
  }
  linkPaths(w, x); v = x;
}

• we will prove that m link-cut operations with n nodes cause O(m log n) splices which means O(log n) splices amortized per expose
• each splice O(log n) amortized time, therefore O(log^2 n) amortized time per link-cut operation

Heavy-light decomposition

edge from v to its parent p is called heavy if D(v)> D(p)/2, otherwise it is light. Observations:

• each path from v to root at most lg n light edges because with each light edge the size of the subtree goes down by at least 1/2
• each node has at most one child connected to it by a heavy edge

Proof that there are O(log n) splices amortized per expose

• potential function: number of heavy dashed edges
• cost: splice
• assume expose creates L new light solid edges and H new solid heavy edges.
• real cost L+H
• it creates at most L+1 new heavy dashed edges (only splices which create new light solid edge can create heavy dashed edge plus 1 from child of v)
• removes H heavy dashed edges (they become solid by splice)
• change of potential <= L+1-H
• amortized cost 2L+1 = O(log n) because all new light solid edges are light edge on a path from v to root
• some operations change the tree and create new heavy dashed edges from light dashed edges
  • consider edge from v to its parent p
  • it can become heavy if v becomes heavier by linking in its subtree
    • this never happens because edge v->p would be solid before linking
  • it can also become heavy if some other child x of p becomes lighter by cutting something in its subtree
    • after cuting, x is connected to p by a light edge
    • there are at most lg n vertices on the path exposed by cutting that are light after the cut and can play role of x
    • each x has at most one sibling v that becomes heavy
    • therefore there are at most lg new heavy dashed edges, which adds O(log n) splices to amortized cost of cuting

Fully dynamic connectivity

• insert delete edges, insert delete isolated vertices, test if there is a path from v to w O(log^2 n) amortized
  • Holm, Jacob, Kristian De Lichtenberg, and Mikkel Thorup. "Poly-logarithmic deterministic fully-dynamic algorithms for connectivity, minimum spanning tree, 2-edge, and biconnectivity." Journal of the ACM (JACM) 48, no. 4 (2001): 723-760.
  • uses link-cut trees as components
  • good idea for presentation?
• if no delete, can use union-find

Applications

• lca least common ancestor:
  • expose(v)
  • expose(w)
  • u = head(v)
  • if u==root return v else return u.parent
  • we have seen O(1) lca after preprocessing a static tree, this would be O(log n) dynamic
• At the time of publication improved best known running time of maximum flow problem from O(nm log^2 n) to O(nm log n). Since then some improvements.