The Method of Moments Example I Example II Example III Counterexample Total displacement in linear probing hashing Problem description Example of constructing a hash table: A . . . h ( A ) = 3 B . . . h ( B ) = 9 1112 1 2 C . . . h ( C ) = 4 A D . . . h ( D ) = 3 10 3 B 9 4 C D 8 5 7 6 13 / 64
The Method of Moments Example I Example II Example III Counterexample Total displacement in linear probing hashing Problem description Example of constructing a hash table: A . . . h ( A ) = 3 B . . . h ( B ) = 9 1112 1 2 C . . . h ( C ) = 4 A D . . . h ( D ) = 3 10 3 B 9 4 C 8 5 7 6 D 13 / 64
The Method of Moments Example I Example II Example III Counterexample Total displacement in linear probing hashing Problem description Example of constructing a hash table: A . . . h ( A ) = 3 B . . . h ( B ) = 9 1112 1 2 C . . . h ( C ) = 4 A D . . . h ( D ) = 3 10 3 E . . . h ( E ) = 7 B 9 4 C 8 5 7 6 D E 13 / 64
The Method of Moments Example I Example II Example III Counterexample Total displacement in linear probing hashing Problem description Example of constructing a hash table: A . . . h ( A ) = 3 F B . . . h ( B ) = 9 1112 1 2 C . . . h ( C ) = 4 A D . . . h ( D ) = 3 10 3 E . . . h ( E ) = 7 B 9 4 C F . . . h ( F ) = 12 8 5 7 6 D E 13 / 64
The Method of Moments Example I Example II Example III Counterexample Total displacement in linear probing hashing Problem description Example of constructing a hash table: A . . . h ( A ) = 3 F B . . . h ( B ) = 9 1112 1 2 C . . . h ( C ) = 4 A D . . . h ( D ) = 3 10 3 E . . . h ( E ) = 7 B 9 4 C G F . . . h ( F ) = 12 8 5 G . . . h ( G ) = 9 7 6 D E 13 / 64
The Method of Moments Example I Example II Example III Counterexample Total displacement in linear probing hashing Problem description Example of constructing a hash table: A . . . h ( A ) = 3 F B . . . h ( B ) = 9 1112 1 2 C . . . h ( C ) = 4 G A D . . . h ( D ) = 3 10 3 E . . . h ( E ) = 7 B 9 4 C F . . . h ( F ) = 12 8 5 G . . . h ( G ) = 9 7 6 D E 13 / 64
The Method of Moments Example I Example II Example III Counterexample Total displacement in linear probing hashing Problem description Example of constructing a hash table: A . . . h ( A ) = 3 F B . . . h ( B ) = 9 1112 1 2 C . . . h ( C ) = 4 G A D . . . h ( D ) = 3 10 3 E . . . h ( E ) = 7 B 9 4 C F . . . h ( F ) = 12 H 8 5 G . . . h ( G ) = 9 7 6 D H . . . h ( H ) = 4 E 13 / 64
The Method of Moments Example I Example II Example III Counterexample Total displacement in linear probing hashing Problem description Example of constructing a hash table: A . . . h ( A ) = 3 F B . . . h ( B ) = 9 1112 1 2 C . . . h ( C ) = 4 G A D . . . h ( D ) = 3 10 3 E . . . h ( E ) = 7 B 9 4 C F . . . h ( F ) = 12 8 5 G . . . h ( G ) = 9 7 6 D H . . . h ( H ) = 4 E H 13 / 64
The Method of Moments Example I Example II Example III Counterexample Total displacement in linear probing hashing Problem description Example of constructing a hash table: A . . . h ( A ) = 3 F B . . . h ( B ) = 9 1112 1 2 C . . . h ( C ) = 4 G A D . . . h ( D ) = 3 10 3 E . . . h ( E ) = 7 B 9 4 C F . . . h ( F ) = 12 8 5 G . . . h ( G ) = 9 7 6 D H . . . h ( H ) = 4 E H 13 / 64
The Method of Moments Example I Example II Example III Counterexample Total displacement in linear probing hashing Problem description Displacement d ( x ) of element x placed at location y : circular distance between h ( x ) and y : � y − h ( x ) , if h ( x ) ≤ y , d ( x ) := m + h ( x ) − y , otherwise ⇒ Costs of inserting x and searching x in table Total displacement of sequence of n hashed values: sum of the individual displacements ⇒ Construction costs of the table 14 / 64
The Method of Moments Example I Example II Example III Counterexample Total displacement in linear probing hashing Problem description Assumption: all m n hash sequences are equally likely D m , n : Random variable counting the total displacement of a table of length m with n keys hashed ◮ Full table: n = m ◮ Almost full table: n = m − 1 ◮ Sparse tables: n = α m , load factor 0 < α < 1 15 / 64
The Method of Moments Example I Example II Example III Counterexample Total displacement in linear probing hashing Results Theorem [Flajolet, Poblete and Viola, 1998]: Result for almost full tables: the scaled random variable � 2 � 3 2 D n , n − 1 converges in distribution to an Airy distributed n random variable: � 2 � 3 ( d ) 2 D n , n − 1 − − → D , n where D is determined by its moments: 2 √ π E ( D r ) = Γ((3 r − 1) / 2) C r , and the constants C r satisfy the following recurrence: r − 1 � r � � 2 C r = (3 r − 4) rC r − 1 + C j C r − j , for r ≥ 1 , C 0 = − 1 . j j =1 16 / 64
The Method of Moments Example I Example II Example III Counterexample Total displacement in linear probing hashing Proof idea Basic decomposition of almost full tables: ◮ Table length n + 1 with n elements inserted ◮ Before last element is inserted: Two empty cells at position k + 1 and n + 1 ◮ Assumption (circular symmetry): free cell remains at n + 1 ⇒ last element to be inserted has any address in [1 . . . k + 1] ⇒ displacement is any value ∈ { 0 , 1 , . . . , k } . n+1 k+1 17 / 64
The Method of Moments Example I Example II Example III Counterexample Total displacement in linear probing hashing Proof idea Decomposition leads to recursive description: F n , k : number of ways of creating an almost full table with n elements and total displacement k Generating function: F n ( q ) := � k ≥ 0 F n , k q k Recurrence: n − 1 � n − 1 � � F k ( q )(1 + q + · · · + q k ) F n − 1 − k ( q ) F n ( q ) = k k =0 Bivariate generating function: F ( z , q ) := � n ≥ 0 F n ( q ) z n n ! Functional equation: ∂ z F ( z , q ) = F ( z , q ) · F ( z , q ) − qF ( qz , q ) ∂ 1 − q 18 / 64
The Method of Moments Example I Example II Example III Counterexample Total displacement in linear probing hashing Proof idea Pumping out all moments: Generating function of r -th factorial moments: � f r ( z ) := ∂ r � ∂ q r F ( z , q ) � � q =1 f r ( z ) satisfy following linear differential equation: r ( z )(1 − T ( z )) − f r ( z ) T ( z )(2 − T ( z )) f ′ = R r ( z ) , z (1 − T ( z )) where the inhomogeneous part R r ( z ) contains the functions f 0 ( z ) , f 1 ( z ) , . . . , f r − 1 ( z ) and T ( z ) is the tree function: T ( z ) = ze T ( z ) 19 / 64
The Method of Moments Example I Example II Example III Counterexample Total displacement in linear probing hashing Proof idea General solution: � z e T ( z ) R r ( u ) e − T ( u ) du f r ( z ) = 1 − T ( z ) 0 Asymptotic behaviour around dominant singularity z = e − 1 : C r zf r ( z ) ∼ (2(1 − ez )) 3 r / 2 − 1 / 2 , where constants C r satisfy the following recurrence: r − 1 � r � � 2 C r = (3 r − 4) rC r − 1 + C j C r − j , for r ≥ 1 , C 0 = − 1 . j j =1 20 / 64
The Method of Moments Example I Example II Example III Counterexample Total displacement in linear probing hashing Proof idea Singularity analysis of generating functions [Flajolet and Odlyzko, 1990]: ⇒ asymptotic equivalent of the r -th factorial and ordinary moments: 2 √ π � 2 � 3 2 E ( D r n , n − 1 ) → Γ((3 r − 1) / 2) C r n 21 / 64
The Method of Moments Example I Example II Example III Counterexample Total displacement in linear probing hashing Airy distribution Airy distribution appears in various contexts: ◮ Number of inversions in trees ◮ Path length in trees ◮ Area under directed lattice paths ◮ Counting problems for polygon models ◮ Number of connected graphs with n vertices and k edges ◮ Additive parameters in context-free grammars “Similar” functional equations are occurring 22 / 64
The Method of Moments Example I Example II Example III Counterexample Example II: Subtree varieties in recursive trees 23 / 64
The Method of Moments Example I Example II Example III Counterexample Subtree varieties in recursive trees Problem description Subtree varieties in rooted trees: ◮ Given: family T of rooted trees ◮ Consider: random rooted tree T of size n of family T ◮ Question: how many subtrees of T have size k = k ( n ) ? 24 / 64
The Method of Moments Example I Example II Example III Counterexample Subtree varieties in recursive trees Problem description Typical situation for random tree of size n 25 / 64
The Method of Moments Example I Example II Example III Counterexample Subtree varieties in recursive trees Problem description Typical situation for random tree of size n many subtrees of fixed size: size 1 (= leaves) 25 / 64
The Method of Moments Example I Example II Example III Counterexample Subtree varieties in recursive trees Problem description Typical situation for random tree of size n many subtrees of fixed size: size 2 25 / 64
The Method of Moments Example I Example II Example III Counterexample Subtree varieties in recursive trees Problem description Typical situation for random tree of size n many subtrees of fixed size: size 3 25 / 64
The Method of Moments Example I Example II Example III Counterexample Subtree varieties in recursive trees Problem description Typical situation for random tree of size n few subtrees of “large” size: size n / 3 25 / 64
The Method of Moments Example I Example II Example III Counterexample Subtree varieties in recursive trees Problem description Typical situation for random tree of size n few subtrees of “large” size: size n / 2 25 / 64
The Method of Moments Example I Example II Example III Counterexample Subtree varieties in recursive trees Recursive trees Recursive trees: important tree family with many applications ◮ models spread of epidemics ◮ model for pyramid schemes ◮ model for the family trees of preserved copies of ancient texts ◮ related to the Bolthausen-Sznitman coalescence model 26 / 64
The Method of Moments Example I Example II Example III Counterexample Subtree varieties in recursive trees Recursive trees Combinatorial description of a recursive tree: ◮ non-plane labelled rooted tree ◮ size- n tree labelled with labels 1 , 2 , . . . , n ◮ labels along path from root to arbitrary node v are increasing sequence Random recursive trees: all ( n − 1)! recursive trees of size n appear with equal probability 1 2 3 5 4 8 7 6 10 9 11 27 / 64
The Method of Moments Example I Example II Example III Counterexample Subtree varieties in recursive trees Recursive trees Simple growth rule for generating random recursive trees: ◮ Step 1 : start with root labelled by 1 ◮ Step j : node with label j is attached to any previous node with equal probability 1 / ( j − 1) 1 28 / 64
The Method of Moments Example I Example II Example III Counterexample Subtree varieties in recursive trees Recursive trees Simple growth rule for generating random recursive trees: ◮ Step 1 : start with root labelled by 1 ◮ Step j : node with label j is attached to any previous node with equal probability 1 / ( j − 1) 1 p=1 1 2 28 / 64
The Method of Moments Example I Example II Example III Counterexample Subtree varieties in recursive trees Recursive trees Simple growth rule for generating random recursive trees: ◮ Step 1 : start with root labelled by 1 ◮ Step j : node with label j is attached to any previous node with equal probability 1 / ( j − 1) 1 p=1 1 2 p=1/2 1 2 3 28 / 64
The Method of Moments Example I Example II Example III Counterexample Subtree varieties in recursive trees Recursive trees Simple growth rule for generating random recursive trees: ◮ Step 1 : start with root labelled by 1 ◮ Step j : node with label j is attached to any previous node with equal probability 1 / ( j − 1) 1 p=1 1 p=1/2 2 p=1/2 1 1 2 2 3 3 28 / 64
The Method of Moments Example I Example II Example III Counterexample Subtree varieties in recursive trees Recursive trees Simple growth rule for generating random recursive trees: ◮ Step 1 : start with root labelled by 1 ◮ Step j : node with label j is attached to any previous node with equal probability 1 / ( j − 1) 1 p=1 1 p=1/2 2 p=1/2 1 1 2 2 3 p=1/3 3 1 2 4 3 28 / 64
The Method of Moments Example I Example II Example III Counterexample Subtree varieties in recursive trees Recursive trees Simple growth rule for generating random recursive trees: ◮ Step 1 : start with root labelled by 1 ◮ Step j : node with label j is attached to any previous node with equal probability 1 / ( j − 1) 1 p=1 1 p=1/2 2 p=1/2 1 1 2 2 3 p=1/3 p=1/3 3 1 1 2 4 2 3 3 4 28 / 64
The Method of Moments Example I Example II Example III Counterexample Subtree varieties in recursive trees Recursive trees Simple growth rule for generating random recursive trees: ◮ Step 1 : start with root labelled by 1 ◮ Step j : node with label j is attached to any previous node with equal probability 1 / ( j − 1) 1 p=1 1 p=1/2 2 p=1/2 1 1 2 2 3 p=1/3 p=1/3 3 p=1/3 1 1 1 2 4 2 2 3 3 4 3 4 28 / 64
The Method of Moments Example I Example II Example III Counterexample Subtree varieties in recursive trees Recursive trees Simple growth rule for generating random recursive trees: ◮ Step 1 : start with root labelled by 1 ◮ Step j : node with label j is attached to any previous node with equal probability 1 / ( j − 1) 1 p=1 1 p=1/2 2 p=1/2 1 1 2 2 3 p=1/3 p=1/3 3 p=1/3 p=1/3 1 1 1 1 2 4 2 2 2 3 3 3 4 3 4 4 28 / 64
The Method of Moments Example I Example II Example III Counterexample Subtree varieties in recursive trees Recursive trees Simple growth rule for generating random recursive trees: ◮ Step 1 : start with root labelled by 1 ◮ Step j : node with label j is attached to any previous node with equal probability 1 / ( j − 1) 1 p=1 1 p=1/2 2 p=1/2 1 1 2 2 3 p=1/3 p=1/3 3 p=1/3 p=1/3 p=1/3 1 1 1 1 1 2 4 2 2 2 3 2 3 4 3 3 4 3 4 4 28 / 64
The Method of Moments Example I Example II Example III Counterexample Subtree varieties in recursive trees Recursive trees Simple growth rule for generating random recursive trees: ◮ Step 1 : start with root labelled by 1 ◮ Step j : node with label j is attached to any previous node with equal probability 1 / ( j − 1) 1 p=1 1 p=1/2 2 p=1/2 1 1 2 2 3 p=1/3 p=1/3 p=1/3 3 p=1/3 p=1/3 p=1/3 1 1 1 1 1 1 2 4 2 2 2 3 2 3 4 2 3 3 3 4 3 4 4 4 28 / 64
The Method of Moments Example I Example II Example III Counterexample Subtree varieties in recursive trees Results X n , k : number of subtrees of size k in random recursive tree of size n Theorem [Feng, Mahmoud and Pan, 2006+]: there are three phases for behaviour of X n , k depending on the growth of k = k ( n ) ◮ subcritical case: k / √ n → 0 ◮ critical case: k / √ n → c > 0 ◮ supercritical case: k / √ n → ∞ 29 / 64
The Method of Moments Example I Example II Example III Counterexample Subtree varieties in recursive trees Results ◮ subcritical case: k / √ n → 0: normalized r. v. asympt. Gaussian distributed n X n , k − ( d ) k ( k +1) − − → N (0 , 1) � (2 k 2 − 1) n k ( k +1) 2 (2 k +1) ◮ critical case: k / √ n → c > 0: X n , k asymp. Poisson-distributed ( d ) → Poisson( 1 − − c 2 ) X n , k ◮ supercritical case: k / √ n → ∞ : X n , k asymp. denenerate ( d ) X n , k − − → X , with P { X = 0 } = 1 30 / 64
The Method of Moments Example I Example II Example III Counterexample Subtree varieties in recursive trees Proof idea Decomposition of recursive trees according root degree: � � { ǫ } ˙ ∪ T ˙ ∪ 1 / 2! · T ∗ T ˙ ∪ 1 / 3! · T ∗ T ∗ T ˙ T = � 1 × ∪ · · · = � 1 × exp( T ) Generating functions: M k ( z , v ) := � � m ≥ 0 P { X n , k = m } z n n ! v m n ≥ 1 Differential equation: ∂ � � + ( v − 1) z k − 1 ∂ z M k ( z , v ) = exp M k ( z , v ) 31 / 64
The Method of Moments Example I Example II Example III Counterexample Subtree varieties in recursive trees Proof idea Explicit solution of generating function: M k ( z , v ) = ( v − 1) z k 1 + log z k ( v − 1) tk � 1 − e dt k 0 Exact solution for factorial moments: � n − kr − 1 � r [ n ≥ kr + 1] = [ ] n ℓ − 1 � � � X r × E n , k k r ℓ ℓ =1 � � r 1 � × � ℓ j 1 , . . . , j ℓ i =1 ( j i k + 1) j 1+ ··· + j ℓ = r jq ≥ 1 , 1 ≤ q ≤ ℓ 32 / 64
The Method of Moments Example I Example II Example III Counterexample Subtree varieties in recursive trees Proof idea Critical case: ⇒ Asymptotically Poisson distribed n / k 2 → λ � � X r → λ r → E n , k Subcritical case: ⇒ Dealing with cancellations X n , k := X n , k − E ( X n , k ) Normalized r.v. ˜ V ( X n , k ) ⇒ Asymptotically Gaussian distributed �� � ˜ X n , k � 2 d → (2 d )! d ! 2 d , for d ≥ 0 , E � ν ( k ) n �� � ˜ � 2 d +1 X n , k → 0 , for d ≥ 0 E � ν ( k ) n 33 / 64
The Method of Moments Example I Example II Example III Counterexample Subtree varieties in recursive trees Remarks Application of method of moments to asympt. Gaussian r.v.: ◮ heavy cancellations ⇒ high computational effort ◮ method usually only “last weapon” 34 / 64
The Method of Moments Example I Example II Example III Counterexample Subtree varieties in recursive trees Remarks Application of method of moments to asympt. Gaussian r.v.: ◮ heavy cancellations ⇒ high computational effort ◮ method usually only “last weapon” One might try first: ◮ analytic methods (saddle point method, continuity theorem of Levy, quasi-power theorem) ◮ central limit theorems for sums of independent or weakly dependent r.v. ◮ Stein’s method ◮ contraction method ◮ martingale description 34 / 64
The Method of Moments Example I Example II Example III Counterexample Example III: Total costs of Union-Find-algorithms 35 / 64
The Method of Moments Example I Example II Example III Counterexample Total costs in Union-Find -algorithms Problem description Union-Find -problem ◮ Maintaining representation of equivalence classes (= partitions of a finite set) ◮ Two basic operations: ◮ Union : merge two different equivalence classes s and t into a single equivalence class ◮ Find : find equivalence class that contains a given element x Problem arises naturally in applications in computer science (e.g., minimum-cost spanning tree algorithms) 36 / 64
The Method of Moments Example I Example II Example III Counterexample Total costs in Union-Find -algorithms Problem description Data structure for Union-Find problem, Aho et al [1974]: ◮ consider partition P ( S ) of finite set S ◮ for every element x ∈ S : store in R [ x ] name of the equivalence class containing x ◮ for every equivalence class s ∈ P ( S ): ◮ store in N [ s ] the number of elements of s ◮ store in L [ s ] the elements of s in a linked list 37 / 64
The Method of Moments Example I Example II Example III Counterexample Total costs in Union-Find -algorithms Problem description Basic algorithm for operation Union , Yao [1976]: “Quick Find Weighted” (QFW): if we merge different equivalence classes s and t then we update the class with less elements: ◮ if N [ s ] ≤ N [ t ]: ◮ otherwise set R [ x ] := t for all x in L [ s ] set R [ x ] := s for all x in L [ t ] append L [ s ] to L [ t ], append L [ t ] to L [ s ] set N [ t ] := N [ t ] + N [ s ] set N [ s ] := N [ s ] + N [ t ] call new equivalence class t call new equivalence class s 38 / 64
The Method of Moments Example I Example II Example III Counterexample Total costs in Union-Find -algorithms Problem description Cost of Union -operation: ◮ Costs when merging equivalence classes s and t : measured by number of updated elements, i.e., the number of allocations R [ x ] := s or R [ x ] := t ◮ QFW: cost of merging step is given by minimum of the class sizes min( N [ s ] , N [ t ]) 39 / 64
The Method of Moments Example I Example II Example III Counterexample Total costs in Union-Find -algorithms Problem description Basic model for sequences of Union -operations, Yao [1976]: Random spanning tree model: ◮ deal with set S of size n ◮ at the beginning all elements x ∈ S are forming equivalence class { x } ◮ n equivalence classes will be merged into larger and larger classes by carrying out Union -operations according following Merging rule 40 / 64
The Method of Moments Example I Example II Example III Counterexample Total costs in Union-Find -algorithms Problem description Merging rule: ◮ choose at random a spanning tree of complete graph with vertex set S ◮ choose a random ordering of the edges of this spanning tree by enumerating it from 1 to n − 1 ◮ leads to sequence of edges e 1 = ( x 1 , y 1 ), e 2 = ( x 2 , y 2 ), . . . , e n − 1 = ( x n − 1 , y n − 1 ), with x i , y i ∈ S ◮ gives then sequence of Union -operations Union ( R [ x 1 ] , R [ y 1 ]), Union ( R [ x 2 ] , R [ y 2 ]), . . . , Union ( R [ x n − 1 ] , R [ y n − 1 ]) ◮ ⇒ all n n − 2 ( n − 1)! possible sequence of Union -operations of that kind are equally likely 41 / 64
The Method of Moments Example I Example II Example III Counterexample Total costs in Union-Find -algorithms Problem description Total cost of algorithm QFW: Average performance of QFW described by total costs: ◮ sum of cost of every merging step when merging the elements of a set S of size n ◮ at beginning all elements are in different equivalence classes ◮ merge all elements into one equivalence class (containing all elements of S ) ◮ carrying out sequence of n − 1 Union -operations according to merging rules under random spanning tree model ◮ ⇒ X n : random variable depending only on size n of set S 42 / 64
The Method of Moments Example I Example II Example III Counterexample Total costs in Union-Find -algorithms Problem description Example of algorithm QFW: a b 2 4 5 f c 1 3 e d Union ( { c } , { e } ) ⇒ Cost = 1 a b f c e d 43 / 64
The Method of Moments Example I Example II Example III Counterexample Total costs in Union-Find -algorithms Problem description Example of algorithm QFW: a b 2 4 5 f c 1 3 e d Union ( { c } , { e } ) ⇒ Cost = 1 a b Union ( { a } , { b } ) ⇒ Cost = 1 f c e d 43 / 64
The Method of Moments Example I Example II Example III Counterexample Total costs in Union-Find -algorithms Problem description Example of algorithm QFW: a b 2 4 5 f c 1 3 e d Union ( { c } , { e } ) ⇒ Cost = 1 a b Union ( { a } , { b } ) ⇒ Cost = 1 Union ( { c } , { d } ) ⇒ Cost = 1 f c e d 43 / 64
The Method of Moments Example I Example II Example III Counterexample Total costs in Union-Find -algorithms Problem description Example of algorithm QFW: a b 2 4 5 f c 1 3 e d Union ( { c } , { e } ) ⇒ Cost = 1 a b Union ( { a } , { b } ) ⇒ Cost = 1 Union ( { c } , { d } ) ⇒ Cost = 1 f c Union ( { b } , { c } ) ⇒ Cost = 2 e d 43 / 64
The Method of Moments Example I Example II Example III Counterexample Total costs in Union-Find -algorithms Problem description Example of algorithm QFW: a b 2 4 5 f c 1 3 e d Union ( { c } , { e } ) ⇒ Cost = 1 a b Union ( { a } , { b } ) ⇒ Cost = 1 Union ( { c } , { d } ) ⇒ Cost = 1 f c Union ( { b } , { c } ) ⇒ Cost = 2 Union ( { b } , { b } ) ⇒ Cost = 1 e d 43 / 64
The Method of Moments Example I Example II Example III Counterexample Total costs in Union-Find -algorithms Problem description Example of algorithm QFW: a b 2 4 5 f c 1 3 e d Union ( { c } , { e } ) ⇒ Cost = 1 a b Union ( { a } , { b } ) ⇒ Cost = 1 Union ( { c } , { d } ) ⇒ Cost = 1 f c Union ( { b } , { c } ) ⇒ Cost = 2 Union ( { b } , { b } ) ⇒ Cost = 1 e d Total costs = 6 43 / 64
The Method of Moments Example I Example II Example III Counterexample Total costs in Union-Find -algorithms Results � � Theorem [Kuba and Pan, 2007]: The expectation E X n of the total costs of the Union-Find -algorithm under the random spanning tree model has for n → ∞ the following asymptotic expansion: E ( X n ) = 1 3 4 ) , π n log n + Cn + O ( n where the constant C ≈ 0 . 6315is given as follows: n ( k + 1) k +1 C = γ + 2 log 2 1 � e − ( n +1) � � − 1 � � � + R n +2 − R n +1 − , ( k + 2)! R n − k π n + 1 π n ≥ 0 k =0 with n − 1 k k ( n − k ) n − k − 1 � R n = min( k , n − k ) . k !( n − k )! k =1 44 / 64
The Method of Moments Example I Example II Example III Counterexample Total costs in Union-Find -algorithms Results Theorem [Kuba and Pan, 2007]: The suitably shifted and scaled r.v. X n converges in distribution to a r.v. X , which can be characterized by its r -th integer moments: X n − 1 π n log n − Cn ( d ) E ( X r ) = m r , − − → X , with n where m r is given recursively as follows: � � Γ( r − 1) r � for r ≥ 2 , m r = 2 √ π Γ( r − 1 m r 2 m r 3 I r 1 , r 2 , r 3 , r 1 , r 2 , r 3 2 ) r 1 + r 2 + r 3 = r , r 2 , r 3 < r with initial values m 0 = 1 and m 1 = 0 and � 1 � 1 � r 1 � � x r 2 − 1 2 (1 − x ) r 3 − 3 2 dx . I r 1 , r 2 , r 3 = x log x +(1 − x ) log(1 − x ) +min( x , 1 − x ) π 0 45 / 64
The Method of Moments Example I Example II Example III Counterexample Total costs in Union-Find -algorithms Proof idea The reverse process: destroying a tree ◮ Start with a random spanning tree of size n ◮ Remove successively edges at random from remaining edges ◮ In every step split a connected component into two parts ◮ Cost of a cut is the size of the smaller part after the splitting step ◮ Stop when all nodes are isolated 46 / 64
The Method of Moments Example I Example II Example III Counterexample Total costs in Union-Find -algorithms Proof idea Example of destroying a tree: a b f c e d 47 / 64
The Method of Moments Example I Example II Example III Counterexample Total costs in Union-Find -algorithms Proof idea Example of destroying a tree: a b Cost = 1 f c e d 47 / 64
The Method of Moments Example I Example II Example III Counterexample Total costs in Union-Find -algorithms Proof idea Example of destroying a tree: a b Cost = 1 Cost = 2 f c e d 47 / 64
The Method of Moments Example I Example II Example III Counterexample Total costs in Union-Find -algorithms Proof idea Example of destroying a tree: a b Cost = 1 Cost = 2 Cost = 1 f c e d 47 / 64
The Method of Moments Example I Example II Example III Counterexample Total costs in Union-Find -algorithms Proof idea Example of destroying a tree: a b Cost = 1 Cost = 2 Cost = 1 f c Cost = 1 e d 47 / 64
The Method of Moments Example I Example II Example III Counterexample Total costs in Union-Find -algorithms Proof idea Example of destroying a tree: a b Cost = 1 Cost = 2 Cost = 1 f c Cost = 1 Cost = 1 Total costs = 6 e d 47 / 64
The Method of Moments Example I Example II Example III Counterexample Total costs in Union-Find -algorithms Proof idea Recursive description of total costs X n : Distributional recurrence for rooted trees: ( d ) = X S n + X ∗ n − S n + t n , S n X n S n : size of subtree containing root after removing random edge of randomly chosen labeled rooted tree of size n Toll function: t n , k = min( k , n − k ) S n is distributed as follows: P { S n = k } = kT k T n − k , ( n − 1) T n with T n := n n − 1 n ! 48 / 64
The Method of Moments Example I Example II Example III Counterexample Total costs in Union-Find -algorithms Proof idea Recurrence for r -th moments of X n Linear recurrence for µ [ r ] n := E ( X r n ) : n − 1 � ( n − 1) T n µ [ r ] kT k T n − k ( µ [ r ] k + µ [ r ] n − k ) + R [ r ] n = n , k =1 where the inhomogeneous part R [ r ] depends on the lower order n moments µ [1] n , . . . , µ [ r − 1] n 49 / 64
The Method of Moments Example I Example II Example III Counterexample Total costs in Union-Find -algorithms Proof idea Generating functions treatment Linear differential equation: z (1 − T ( z )) C ′ r ( z ) − (1 + zT ′ ( z )) C r ( z ) = R r ( z ) , where the inhomogeneous part depends on the g.f. C 1 ( z ) , . . . , C r ( z ) for lower moments Solution: � z T ( z ) R r ( t ) C r ( z ) = tT ( t ) dt 1 − T ( z ) 0 Asymptotic equivalents of r -th moments: “pumped out” inductively 50 / 64
The Method of Moments Example I Example II Example III Counterexample Total costs in Union-Find -algorithms Remark Problems of similar “nature”: ◮ Quicksort: number of comparisons ◮ Pathlengths in search tree models ◮ Wiener-index of certain tree models Limiting distribution characterized by “complicated” moment’s sequence 51 / 64
The Method of Moments Example I Example II Example III Counterexample Counterexample 52 / 64
The Method of Moments Example I Example II Example III Counterexample Counterexample Cutting down recursive trees Cutting down procedure for rooted trees: INPUT: tree T steps ← 0 while | T | > 1 do cut off an edge e of T T ← subtree containing the root steps ← steps +1 OUTPUT: steps Remove edges until root is isolated 53 / 64
The Method of Moments Example I Example II Example III Counterexample Counterexample Cutting down recursive trees An example of cutting a tree: Size-11 tree destroyed in 5 steps. 54 / 64
The Method of Moments Example I Example II Example III Counterexample Counterexample Cutting down recursive trees How many steps are done, until root is isolated? Probability model: ◮ Randomized cutting down procedure: Edges in tree chosen at random in each step. ◮ Random tree model for certain tree families. R. v. X n counts steps done to destroy size- n tree. 55 / 64
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