R99945020 林澤豪 F98942047 許芷榕 R00922113 謝宗潛 R 98922144 駱家淮

R99945020 林澤豪 F98942047 許芷榕 R00922113 謝宗潛 R98922144 駱家淮

Outline • 1. What is tree partition problem • 2. Tree partition history • 3. some tool will be used here • FTEST0 • M(P) and MSEARCH • 4. Path partitioning • 5. Tree partitioning

Tree partition(1/2) +=??

Tree partition(2/2) Min-max problem Max-min problem

History △ is the max degree in the tree rd(T) is the radius of the tree

FTEST0 3 5 3 2 1 3 4 2 3 4

FTEST0 30 • λ=5 • k=4 27 5 8 9 6 8 2 3 4

FTEST0 4 11 • λ=5 • k=4 8 1 3 2 1 5 6 2 8 2 3 4 K=4 and return “yes”

FTEST0(reverse) 3 • λ=9 • k=4 5 3 2 1 3 4 2 3 4

FTEST0(reverse) 30 • λ=9 • k=4 27 5 8 9 6 8 2 3 4

FTEST0(reverse) 4 8 • λ=9 • k=4 5 3 1 8 5 9 2 6 8 2 3 K=4 and return “yes” 4

M(P) 6 11 9 2 1 15 7 8

MSEARCH

MSEARCH 6 • {59,3,31,0,28,0,0,0} • K=3 • Median: (3+0)/2=1.5 • FTEST0: ok • Set λ1=1.5 • Discard matrix 11 9 2 1 15 7 8

MSEARCH 6 • {59,3,31,0,28,0} • K=3 • Median: (28+3)/2=15.5 • FTEST0: no • Set λ2=15.5 • No discard matirx 11 9 2 1 15 7 8

MSEARCH 6 • {59,45,42,25,31,22,15,0,44,23,27,3,16,0,0,028,20,11,0,17,0,0,0} • K=3 • Median: 16.5 • Discard matrixes 11 9 2 1 15 7 8

MSEARCH 6 • {15,0,16,0,27,3,11,0,17,0} • K=3 • Median: (11+3)/2=7 • Set λ1=7 • No discard matrixes 11 9 2 1 15 7 8

MSEARCH 6 • {15,8,7,0,16,15,1,0,27,18,12,3,11,2,9,0,17,11,6,0} • K=3 • Median: (8+9)/2=8.5 • Set λ1=8.5 • Discard half matrixes • And finally get median=12 11 9 2 1 15 7 8

MSEARCH • bij = ith iteration, the j th matrix • Bi = ΣNj=1 bij • ncells(i) = after cutting, remain cells • ncells’(i) = after deleting, remain cells • Ki = the matrixes be added in this iteration

MSEARCH • Proof: • Let ncells’(i) ≦2Bi+2 be the result • 4*(ncells’(i-1)+ki) = ncells(i) • By induction: ncells’(i-1) ≦2Bi-1+2 • ncells(i) ≦4*(2Bi-1+2+ ki) ≦4*(2Bi-1+2ki +2) ≦4*(2Bi+2) ≦8Bi+8

MSEARCH • For the second part of MSEARCH, each iteration will discard d1+d2+d3: • First time: d1 >= ncells(i) – Bi /2 • Second time: d2 >= ncells(i) – d1 - Bi /2 • Third time: d3 >= ncells(i) – d1 – d2 - Bi /2 • ncells’(i) = ncells(i) – d1 – d2 - d3 ≦(15/8)(Bi+1) < 2Bi+2

MSEARCH • In cross-splitting:bij = 2(njmj/c)^0.5 -1 = O(2(njmj/c)^0.5) c start from njmj/4 down to 1. 4 * (2^logmj -1) / 2-1 = O(mj) • In quatering-splittingbij = O(log(nj/mj)) • The total splitting time complexity = O(mjlog(nj/mj))

Partitioning a Path

Partitioning a path • Given k, partition a path of vertices for max-min. • The proposed three algorithm overview

Algorithm PATH1 • Phase 1 – collect information • Set . • Form an f(n)-partition of path . • Form a set of sorted matrices. • using feasibility test . • Compute functions using . • Phase 2 – complete parametric search • using feasibility test .

Form an f(n)-partition of path • r-partition of a path • A partition of the vertices of into subpaths, where each subpath contains vertices, and the last subpath contains at most vertices. • f(n) • The largest power of 2 which no larger than . • Total subpaths.

Form a set of sorted matrices • Each subpathforms a sorted matrix . • Each with dimension . • Last one might have smaller dimension size. • All form a set of sorted matrices.

𝑀𝑆𝐸𝐴𝑅𝐶𝐻 using feasibility test 𝐹𝑇𝐸𝑆𝑇0 • Call with algorithm input: • The set of matrices • Stopping count • Using feasibility test • Output search boundary • At most active values remains. • Active value is any candidate value from a subpath and .

Compute functions using 𝐷𝐼𝐺𝐸𝑆𝑇1 • For subpaths with no active values in their matrices, compute • Maximum number of cuts from to , that each connected component except the last, has weight . • Maximum possible weight of the last component given number of cut in subpath from to is . 20

A dynamic programming to compute and for all vertices in subpath. • Range from back to . • If , • Otherwise, , • is the smallest index such that .

Example (1/6) • Stopping count 13 13 Medeian (O) -> Medeian (X) -> Medeian0 (O) -> 0 The number of value remains stopping value 2 -> stops

Example (2/6) 13 Subpaths with no active value:

Example (3/6) • Phase 2 of algorithm PATH1: • Only test a value which • Test feasibility of using 13 17

Example (4/6) • Phase 2 of algorithm PATH1: • Only consider a value which • Test feasibility of using 13 0 7

Example (5/6) • Phase 2 of algorithm PATH1: • Only consider a value which • Test feasibility of using 13 13 11 -> (O)

Example (6/6) • Phase 2 of algorithm PATH1: • Only consider a value which • Test feasibility of using 13 13 13 12 -> (O) All values are discarded, terminates. Return

Complexity of Algorithm PATH1 • Phase 1 – collect information • Set . • Form an f(n)-partition of path . • Form a set of sorted matrices. • using feasibility test . • Compute functions using . • Phase 2 – complete parametric search • using feasibility test .

𝑀𝑆𝐸𝐴𝑅𝐶𝐻 using feasibility test 𝐹𝑇𝐸𝑆𝑇0. • By Theorem 2.1, the number of test values is . • -> number of test values • The test values are produced in . • Each feasibility test using takes • The total time is

Complexity of Algorithm PATH1 • Phase 1 – collect information • Set . • Form an f(n)-partition of path . • Form a set of sorted matrices. • using feasibility test . • Compute functions using . • Phase 2 – complete parametric search • using feasibility test .

Compute functions using 𝐷𝐼𝐺𝐸𝑆𝑇1 • Lemma 3.1. Let be a subpath with no active values. Then computes and for all vertices in in linear time. For each of values of , a constant time amount of work is done. The remaining work can be apportioned as constant for each of values of . • For each , takes . • There are total paths and thus takes time to complete all.

Complexity of Algorithm PATH1 • Phase 1 – collect information • Set . • Form an f(n)-partition of path . • Form a set of sorted matrices. • using feasibility test . • Compute functions using . • Phase 2 – complete parametric search • using feasibility test . )

R99945020 林澤豪 F98942047 許芷榕 R00922113 謝宗潛 R 98922144 駱家淮

R99945020 林澤豪 F98942047 許芷榕 R00922113 謝宗潛 R 98922144 駱家淮

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