Prioritné rady: Rozdiel medzi revíziami

Verzia zo dňa a času 13:42, 1. marec 2016

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Obsah

1 Introduction to priority queues
2 Meldable heaps
3 Fibonacci heaps
4 Pairing heaps

Introduction to priority queues

Review

Operations        Bin. heap    Fib. heap 
                  worst case    amortized
MakeHeap()            O(1)       O(1)   
Insert(H, x)       O(log n)      O(1)   
Minimum(H)            O(1)       O(1)   
ExtractMin(H)      O(log n)    O(log n)   
Union(H1, H2)         O(n)       O(1)   
DecreaseKey(H, x, k)
                    O(log n)     O(1)   
Delete(H, x)        O(log n)   O(log n)

Applications of priority queues

job scheduling on a server (store jobs in a max-based priority queue, always select the one with highest priority - needs operations insert and extract_min, possibly also delete -needs a handle)
event-driven simulation (store events, each with a time when it is supposed to happen. Always extract one event, process it and as a result potentially create new events - needs insert, extract_min Examples: queue in a bank or store: random customer arrival times, random service times per customer)
finding intersections of line segments [1]
- Bentley–Ottmann algorithm - sweepline: binary search tree of line intersections) plus a priority queue with events (start of segment, end of segment, crossing)
- for n segments and k intersections do 2n+k inserts, 2n+k extractMin, overall O((n+k)log n) time

heapsort (builds a heap from all elements or uses n time insert, then n times extract_min -implies a lower bound on time of insert+extract_min must be at least Omega(log n) in the comparison model)
aside: partial sorting - given n elements, print k in sorted order - selection O(n), then sort in O(k log k), together O(n + k log k)
incremental sorting: given n elements, build a data structure, then support extract_min called k times, k unknown in advance, running time a function of k and n
- call selection each time O(kn)
- heap (binary, Fibonacci etc) O(n + k log n) = O(n + k log k) because if k>= sqrt n, log n = Theta(log k) and of k < sqrt n, O(n) term dominates

Kruskal's algorithm (can use sort, but instead we can use heap as well, builds heap, then calls extract_min up to m times, but possibly fewer, plus needs union-find set structure) O(m log n)
- process edges from smallest weight upwards, if it connects 2 components, add it. Finosh when we have a tree.
Prim's algorithm (build heap, n times extract-min, m times decrease_key - O(m log n) with binary heaps, O(m + n log n) with Fib heaps)
- build a tree starting from one node. For each node outside tree keep the best cost to a node in a tree as a priority in a heap
Dijkstra's algorithm for shortest paths (build heap, n times extract_min, m times decrease_key = the same as Prim's)
- keep nodes in a heap which do not have final distance. Set the distance of minimum node to final, update distances of tis neighbors

Soft heap

http://en.wikipedia.org/wiki/Soft_heap
Chazelle, B. 2000. The soft heap: an approximate priority queue with optimal error rate. J. ACM 47, 6 (Nov. 2000), 1012-1027.
Kaplan, H. and Zwick, U. 2009. A simpler implementation and analysis of Chazelle's soft heaps. In Proceedings of the Nineteenth Annual ACM -SIAM Symposium on Discrete Algorithms (New York, New York, January 4––6, 2009). Symposium on Discrete Algorithms. Society for Industrial and Applied Mathematics, Philadelphia, PA, 477-485. pdf

provides constant amortized time for insert, union, findMin, delete
- but corrupts (increases) keys of up to epsilon*n elements, where epsilon is a fixed parameter in (0,1/2) and n is the number of insertions so far
- useful to obtain minimum spanning tree in O(m alpha(m,n)) where alpha is the inverse of Ackermann's function

Meldable heaps

Advantages of Fibonacci compared to binary heap: faster decreaseKey (important for Prim's and Dijkstra's), faster union
meldable heaps support O(log n) union, but all standard operations run in O(log n) time (except for build in O(n) and min in O(1))
three versions: randomized, worst-case, amortized
all of them use a binary tree similar to binary heap, but not always fully balanced
- min-heap property: key of child >= key of parent
key operation is union
- insert: union of a single-node heap and original heap
- extract_min: remove root, union of its two subtrees
delete and decreaseKey more problematic, different for different versions, need pointer to parent
- delete: cut away a node, union its two subtrees, hang back, may need some rebalancing
- decrease_key: use delete and insert (or cut away a node, decrease its key, union with the original tree)

Random meldable heap

http://opendatastructures.org/ Chapter 10.2 [2]
Gambin1998-randomMeldableHeap.pdf

Union(H1, H2) {
    if (H1 == null) return H2;
    if (H2 == null) return H1;
    if (H1.key > H2.key) { exchange H1 and H2 }
    // now we know H1.key <= H2.key
    if (rand() % 2==1) {
       H1.left = merge(H1.left, H2);
    } else {
       H1.right = merge(H1.right, H2);
    }
    return H1;
}

Analysis:

random walk in each tree from root to a null pointer (external node)
in each recursive call choose H1 or H2 based on key values, choose direction left or right randomly
running time proportional to the sum of lengths of these random walks

Lemma: The expected length of a random walk in a any binary tree with n nodes is at most lg(n+1).

let W be the length of the walk, E[W] its expected value.
Induction on n. For n=0 we have E[W] = 0 = lg(n+1). Assume the lemma holds for all n'<n
Let n1 be the size of the left subtree, n2 the size of right subtree. n = n1 + n2 + 1
By the induction hypothesis E[W] <= 1 + (1/2)lg(n1+1) + (1/2)lg(n2+1)
lg is a concave function, meaning that (lg(x)+lg(y))/2 <= lg((x+y)/2)
E[W] <= 1 + lg((n1+n2+2)/2) = lg(n1+n2+2) = lg(n+1)

Corollary: The expected time of Union is O(log n)

A different proof of the Lemma via entropy

add so called external nodes instead of null pointers
let di be the depth of ith external node
binary tree with n nodes has n+1 external nodes
probability of random walk reaching external node i is pi = 2^{-di}, and so di = lg(1/pi)
E[W] = sum_{i=0}^n d_i p_i = sum_{i=0}^n p_i lg(1/pi)
we have obtained entropy of a distribution over n+1 values which is at most lg(n+1)

Leftist heaps

Mehta and Sahni Chap. 5
Crane, Clark Allan. "Linear lists and priority queues as balanced binary trees." (1972) (TR and thesis)
let s(x) be distance from x to nearest external node
- s(x)=0 for external node x and s(x) = min(s(x.left), s(x.right))+1 for internal node x
height-biased leftist tree is a binary tree in which for every internal node x we have s(x.left)>=s(x.right)

Union(H1, H2) {
    if (H1 == null) return H2;
    if (H2 == null) return H1;
    if (H1.key > H2.key) { exchange H1 and H2 }
    // now we know H1.key <= H2.key
    HN = Union(H1.right, H2);
    if(H1.left != null && HN.left.s <= H1.s) {
       return new node(H1.data, H1.left, HN) 
    } else {
       return new node(H1.data, HN, H1.left) 
    }
}

note: s value kept in each node, update in new
for simplicity if
recursion descends only along right edges

Lemma: (a) subtree with root x has at least 2^{s(x)}-1 nodes (b) if a subtree with root x has n nodes, s(x)<=lg(n+1) (c) the length of the path from node x using only edges to right child is s(x)

(a) all levels up to s(x) are filled with nodes, therefore the number of nodes is at least 1+2+4+...+2^{s(x)-1} = 2^{s(x)}-1
(b) an easy corollary of a (n>=2^s-1 => s<-lg(n+1))
(c) implied by height-biased leftist tree property (in each step to the right, s(x) decreases by at least 1)

Corollary: Union takes O(log n) time

recursive calls along right edges from both roots - at most s(x)<=lg(n+1) such edges in a tree of size n

Skew Heaps

self-adjusting version of leftist heaps, no balancing values stored in nodes
Mehta and Sahni Chap. 6

Union(H1, H2) {
    if (H1 == null) return H2;
    if (H2 == null) return H1;
    if (H1.key > H2.key) { exchange H1 and H2 }
    // now we know H1.root.key <= H2.root.key
    HN = Union(H1.right, H2);
    return new node(H1.data, HN, H1.left) 
}

Recall: Heavy-light decomposition from Link/cut trees

D(x): number of descendants of node x, including x (size of a node)
An 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

Potential function: the number of heavy right edges (right edge is an edge to a right child)

Amortized analysis of union:

Consider one recursive call. Let left child of H1 be A, right child B
Imagine cutting edges from root of H1 to it children A and B before calling recursion and creating new edges to Union(B,H2) and A after recursive call
Let real cost of one recursive call be 1
If edge to H1.B was heavy
- cutting edges has delta Phi = -1
- adding edges back does not create a heavy right edge, therefore delta Phi = 0
- amortized cost is 0
If edge to H1.B was light
- cutting edges has delta Phi = 0
- adding edges creates at most one heavy right edge, therefore delta Phi <= 1
- amortized cost is <= 2

All light edges we encounter in the algorithm are on the rightmost path in one of the two heaps, which means there are at most 2 lg n of them (lg n in each heap). This gives us amortized cost at most 4 lg n.

Fibonacci heaps

Analysis

ExtractMin

let c be degree of z, c<=Deg(n) = O(log n)
adding c children to root list:
- cost c
- delta Phi = c, amortized O(log n)

Consolidate:

t1 original number of nodes in the root list, t2 new
cost: t1 + Deg(n)
delta Phi: t2-t1
amortized t2+Deg(n), but t2<=Deg(n), therefore O(\log n)

DecreaseKey:

all except cascading cut:
- cost O(1)
- delta Phi <= 1
- amortized O(1)

CascadingCut:

each recursive call cost 1
last recursive call: delta Phi <= 2 (amortized cost O(1))
other recursive call: delta Phi = -1 (amortized cost 0)

Overall DecreaseKey O(1) amortized

Maximum degree Deg(n)

Proof of Lemma 3:

Linking done only in Consolidate for 2 roots of equal degree.
When y_j was linked, y_1,...y_{j-1} were children, so at that time x.degree >= j-1.
Therefore at that time y_j.degree>= j-1
Since then y_j lost at most 1 child, so y_j.degree >= j2.

Proof of Lemma 4:

size of a node: size of its subtree (number of descendants)
let s_k be the minimum size of a node with degree k in any Fibonacci heap (will prove s_k>= phi^k)
s_k<= s_{k+1}*
Let z be a node of degree k and size s_k
Let y_1,...y_k be its children from earliest
s_k = 1 + sum_{j=1}^k size(y_j) >= 1 + size(y_1} + sum_{j=2}^k s_{y_j.degree}
>= 2 + sum_{j=2}^k s_{i-2} (lemma 3)
by induction on k: s_k >= F_{k+2}
- trivial for k=0,1
- assume k>=2, s_i>=F_{i+2} for i<=k-1
- s_k >= 2+ sum_{j=2}^k s_{i-2} >= 2 + sum_{j=2}^k F_{i} (IH)
- = 1 + sum_{j=0}^k F_{i} = F_{k+2} (lemma 1)
- >= phi^k

Proof of Corollary:

let x be a node of degree k.
by lemma 4 n>=size(x) >= phi^k
therefore k<= log_phi(n)

Exercise

What sequence of operations will create a tree of depth n?
Alternative delete (see slides)

Pairing heaps

Simple and fast in practice
hard to analyze amortized time
Fredman, Michael L.; Sedgewick, Robert; Sleator, Daniel D.; Tarjan, Robert E. (1986), "The pairing heap: a new form of self-adjusting heap", Algorithmica 1 (1): 111–129
Stasko, John T.; Vitter, Jeffrey S. (1987), "Pairing heaps: experiments and analysis", Communications of the ACM 30 (3): 234–249
Moret, Bernard M. E.; Shapiro, Henry D. (1991), "An empirical analysis of algorithms for constructing a minimum spanning tree", Proc. 2nd Workshop on Algorithms and Data Structures, Lecture Notes in Computer Science 519, Springer-Verlag, pp. 400–411
Pettie, Seth (2005), "Towards a final analysis of pairing heaps", Proc. 46th Annual IEEE Symposium on Foundations of Computer Science, pp. 174–183

a tree with arbitrary degrees
each node pointer to the first child and next sibling (for delete and decreaseKey also pointer to previous sibling or parent if first child)
each child has key >= key of parent
linking two trees H1 and H2 s.t. H1 has smaller key:
- make H2 the first child of H1 (original first child of H1 is sibling of H2)
- subtrees thus ordered by age of linking from youngest to oldest

extractMin is the most complex operation, the rest quite lazy
insert: make a new tree of size 1, link with H
union: link H1 and H2
decreaseKey: cut x from its parent, decrease its key and link to H
delete: cut x from parent, do extractMin on this tree, then link result to H

ExtractMin

remove root, combine resulting trees to a new tree (actual cost up to n)
start by linking pairs (1st and 2nd, 3rd and 4th,...), last may remain unpaired
finally we link each remaining tree to the last one, starting from next-to-last back to the first (in the opposite order)
- other variants of linking were studied as well

O(log n) amortized time can be proved by a similar analysis as in splay trees

imagine as binary tree - rename first child to left and next sibling to right
D(x): number of descendants of node x, including x (size of a node)
r(x): lg (D(x)) rank of a node
potential of a tree - sum of ranks of all nodes
details in Fredman et al 1986
Better bounds in Petie 2005: amortized O(log n) ExtractMin; O(2^{2\sqrt{log log n}}) insert, union, decreaseKey; but decreaseKey not constant as in Fibonacci heaps)

@@ Riadok 116: / Riadok 116: @@
      // now we know H1.key <= H2.key
      HN = Union(H1.right, H2);
-     if(HN.s <= H1.s) {
+     if(H1.left != null && HN.left.s <= H1.s) {
         return new node(H1.data, H1.left, HN)
      } else {
@@ Riadok 123: / Riadok 123: @@
 }
 </pre>
-* note: s value kept in each node,
+* note: s value kept in each node, update in new
+* for simplicity if
 * recursion descends only along right edges

Prioritné rady: Rozdiel medzi revíziami

Verzia zo dňa a času 13:42, 1. marec 2016

Obsah

Introduction to priority queues

Review

Applications of priority queues

Soft heap

Meldable heaps

Random meldable heap

Leftist heaps

Skew Heaps

Fibonacci heaps

Analysis

Maximum degree Deg(n)

Exercise

Pairing heaps

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