Faster Swap Edge Computation in Minimum Diameter Spanning Trees

(1)

Faster Swap Edge Computation in Minimum Diameter Spanning Trees

Beat Gfeller ESA 2008

Presenters: 陳鵬仁張凱崴余相甫徐國鐘鄒承孝沈定陳學毅

(2)

Outline

• Primitive Knowledge!!

• Failing on Diameter Edges

• Failing on Non-Diameter Edges

(3)

PRIMITIVE KNOWLEDGE!!

余相甫

(4)

MDST

• Given a simple graph

• Spanning Tree T

– Communication between nodes – Why trees?

• MDST

– Minimum Diameter Spanning Tree – Minimize the maximum travel time

G  (V, E)

(5)

事事難料 Edge 會掉

• Edge e fails

– => 2 disconnect components

• Entire repair

– Find a new MDST on – May change many edges

• Ad hoc repair

– Construct only one new edge to connect – Swap Edge

T  e

G  e

e

f

(6)

Best Swap

• Edge e fails

– Many edges can connect 2 disconnect parts

• Best swap edge f

– Resulting tree with minimum diameter

• ABS

– All Best Swap – Find f e T_E

(7)

Terminology

•

– 2-edge connected simple graph –

– : a spanning tree

• Edge weight

– l(e) : nonnegative

– d(x,y) : distance between x and y on the tree T

• Path

– –

G  (V, E)

T  (V,E_T )

n | V | , m | E |

P  p₁,., pr

| P | d(p₁, p_r)

(8)

Terminology (Cont.)

• Diameter

– : diameter of tree T

• Subtree on T rooted at s

– : parent of x

– : subtree rooted at x – : child of x

,

(T)  d_S ,d_E D

p(x)

T_x  (V(T_x), E(T_x)) C(x)

(9)

Terminology (Cont.)

• Fail edge

– Denote by

• Swap edge

– Denote by

– Connect two disjoint parts – : all swap edges for e

e E_T

f E \ E_T F(e)

(10)

Terminology (Cont.)

•

– A spanning tree of

– : longest path contains f in – : longest path in

• Best swap edge f

– With respect to diameter – Minimize

T

_{e/ f}

 (V,E

_T

\ {e}  { f })

P_{e/ f}

(T_{e f}/ )

D T_{e/ f}

T_{e/ f} G  e

| (D T_{e f}/ ) |

(11)

Terminology (Cont.)

• L(T,r)

– Length of longest path starts from r

• nc(v)

– the node on D(T) closest to v – Could be v itself

• nca(a,b)

– Nearest common ancestor of a and b

(12)

Terminology (conclude)

(13)

Formal Problem Definition:

• What we are given :

– G : two edge connected simple graph – T : a MDST of G

– D(T) : a diameter of T

• Objective:

– Find best swap edge f F(e) e E_T

(14)

ABS: Previous Results

• Hassin, et al. (1997) find MDST in

–

• Nardelli, et al. (2001)

– Best swap tree diameter <= 5/2 diameter of entirely adjust tree

• Ito, et al. (2005)

– Time:

– Space:

O(n m) O(m)

O(mn  n² log n)

(15)

Main Result

• Solving ABS

– Time : – Space:

• Compare to Ito, et al.

– Better when m is smaller –

O(m log(n)) O(m)

m  o( n² log²n)

(16)

Main Idea

• Find candidate , and evaluate

– Quality of a swap edge f

• Present by me

– Case1:

• Diameter edges

– Case2:

• Non-diameter edges

e D(T ) e D(T )

f e T

(17)

Quality of Swap Edge

• Calculate efficiently

• Calculate L(T, r) efficiently

| (D T_{e f}/ ) |

| P_{e/ f} |

d(a,b) a,b V

(18)

Lemma 1

•

–

• Proof:

– If does not contains f , D(T) is a diameter of T

• We will focus on from now

– Consider paths contain f only

e T and f F(e)

/ /

| (D T_{e f} ) | max{| D( ) | |T , P_{e f} |}

(T_{e f}/ ) D

| P_{e/ f} |

(19)

•

• contains three parts

– : f itself

– : a longest path start from u – : a longest path start from v

• Calculate

| P

_{e/ f}

|

| P_{e/ f} |

f  (u,v)

in (1)ti

( T r , )  r O m e L

l( f )

(T e u , ) L

(T e v , ) L

e

u f

v

(20)

Lemma 2

• Let be a diameter of T

–

• Proof:

– We’ve already proved in exam !

• Calculate

– Recall that

D(T )  d_S,,dE

, ( , ) { ( , _S ), ( , _E)}

r V T r max d r d d r d

  L 

d(a,b) a,b V

d(a,b)  d_T (a,b)

(21)

on a Tree

• We need to calculate

• Floyd-Warshall Algorithm

– preprocess time – space

– query time

d(a,b)

d(a,b) a,b V

O(n³) O(n² )

O(1) a,b V

(22)

on a Tree (Cont.)

• We need to calculate

• Linear approach

– O(n) preprocess time – O(n) space

– O(1) query time

d(a,b)

d(a,b) a,b V

a,b V

(23)

d(a,b) on a Tree (Cont.)

• Root T at any node s

• A labeling of nodes allows nca(a,b)

– Preprocess within O(n) time and space – Answer any query within O(1)

– Implement by RMQ

• A labeling depth(x) – depth(x) = d(s, x)

– Implement by preorder traverse

(24)

d(a,b) on a Tree (Cont.)

• Let x = nca(a,b)

• d(a,b)

– From a to the nearest ancestor x – From x to b

• d(a,b) = d(a,x) + d(b,x)

– depth(a) + depth(b) – 2*depth (x) – O(1) query time

a

x

b s

(25)

Main Idea

• Find candidate , and evaluate

– Quality of a swap edge f

– Case1:

– Case2:

( ) eD T

( ) e D T

| D_{e f}/ |

f e T

(26)

FAILING ON DIAMETER EDGES

陳鵬仁張凱崴

(27)

Failing on Diameter Edges

• We can solve this case in

– Time complexity O(m log n) – Space complexity O(m)

• How to find the longest path starting from r?

• Define nc(r) is the nearest node to r on diameter D(T)

r d_E

nc(r) d_S

(28)

Lemma 3

• Given a tree T = (V, E_T), diameter D(T) with endpoints d_S, d_E, a failing edge e on D(T). For any node r, one of the longest path in T-e

starting from r contains nc(r).

• In other word, we must touch diameter before we arrive the furthest node.

(29)

r

d_E nc(r)

e d_S

(30)

r

d_E nc(r)

e d_S

(31)

• By contradiction

Prof. of Lemma 3

d_S e

r z

nc(r)

x D(T)

(32)

• By contradiction

Prof. of Lemma 3

reach?(cont.)

• Claim that we can calculate μ_S, μ_E recursively in O(n)

• Define h(d_i): the length of a longest path starting in d_i, and not using any edges on the diameter D(T)

• h(d_i) for all d_i can be calculated in O(n) by doing DFS from an ambiguous root r on D(T)

h(d₃) r

nc(r)

e d_S=d₀

d₂ d₃ d₄ d₅

d₆ d₇= d_E d₁

(36)

How far can d

_S

/d

_E

reach?(cont.)

• We can calculate μ_S, μ_E recursively in O(n)

• μ_S(d₀, d₁) = μ_E(d_k-1, d_k) = 0

• μ_S(d_i, d_i+1) = max{μ_s(d_i-1, d_i), d(d_S, d_i)+h(d_i)}

• μ_E(d_i, d_i+1) = max{μ_E(d_i+1, d_i+2), d(d_E, d_i+1)+h(d_i+1)}

• We have all h(d_i) in O(n), then all procedure will

finished in O(n) _e

d_S=d₀

d_i-1 d_i d_i+1

h(d_i) μ_s(d_i-1, d_i),

(37)

How far can r reach?

• Lemma 4

• Given a diameter of T , D(T) = d_S, d₂, …, d_E , a failing edge e = (d_i, d_i+1). The longest path on T - (d_i, d_i+1) starting from an ambiguous node r (if r is in the same subtree with d_S) is

d(r, nc(r)) + max{d(nc(r), d_S), μ_S(d_i, d_i+1) - d(nc(r), d_S)}

or

d(r, nc(r)) + max{d(nc(r), d_E), μ_E(d_i, d_i+1) - d(nc(r), d_E)}

, if r is in the same subtree with d_E

(38)

d_S=d₀ e

nc(r) _d

i d_i+1

μ_s(d_i-1, d_i)

r d(r, nc(r))

(39)

• By Lemma 3, the longest path from r must reach nc(r) first. We can observe the following two cases.

• If the furthest path from r is toward to d_S at nc(r)

d_S=d₀ nc(r) d_i e d_i+1

μ_s(d_i-1, d_i)

r d(r, nc(r)) d_S=d₀ nc(r) d_i e d_i+1

μ_s(d_i-1, d_i)

r

d(nc(r), d_S) > μ_S(d_i, d_i+1) - d(nc(r), d_S)

(40)

• By Lemma 3, the longest path from r must reach nc(r) first. We can observe the following two cases.

• If the furthest path from r is toward to d_E at nc(r)

d_S=d₀ nc(r) d_i e d_i+1

μ_s(d_i-1, d_i)

r d(r, nc(r)) d_S=d₀ nc(r) d_i e d_i+1

μ_s(d_i-1, d_i) r

(47)

The Definition of

(48)

Virtual Edge (I)

+ + + +

(49)

Virtual Edge (II)

+ + + +

(50)

Construct Virtual Edges

[Algorithm] Step 2

• Using 4 heap to find maximal virtual edges of the 4 cases for all in

+ + + +

2 1

3 4

6 5 7

9 8

(60)

[Algorithm] Step 3

+ + + + +

+ + +

• Compare the 4 virtual edge of each

(61)

FAILING ON NON-DIAMETER EDGES

陳學毅鄒承孝沈定徐國鐘

(62)

Swap Edges for Failing Non-diameter Edges

• Choose a vertex r in the diameter of T as root of T.

• We first consider the swap edge which is a backedge.

• let f = (u, v) be an edge not in tree T s.t. u is a descendant of v in T.

• We called such edge f as a backedge.

• Hence, f could be the backup edge for each edge on the path from v to u in T.

(63)

Backedge f = (u, v)

• Let the path in T from v to u is (e₁, … , e_k)

• Assume e_i is the failing edge.

• The part with v will also contain D(T).

Hence L(T-e_i,v) = L(T,v).

(64)

Midpoint Edge

• Midpoint is on the D(T) such that midpoint has equal length to the two endpoint of D(T). (mid point may lie on an edge)

• The edge contains midpoint is called midpoint edge. (if there may be more than one midpoint edge, any of them can be chosen)

• By Lemma 2, for any vertex v, the longest path start on one side of the midpoint edge and go to the opposite end of the diameter.

(65)

For some failing edge e

• Let e = (p(x),x), f = (u,v)

• Let (p(h),h) be the midpoint edge in T_x

• By Lemma 2, all longest path in T_x will contains (p(h), h).

(66)

Swap edge e by f

• In the same part with v, it also contains D(T).

• The only path changed in T – e + f are the paths which contain the edge f = (u,v)

(67)

Group the Swap Edge Candidate

• Group such backedge by their endpoint in T_x

(68)

Lemma 6

• Lemma 6 : In any group, a best swap candidate for failing edge e is a swap edge f = (u,v) s.t.

L(T,v) + l(f) + depth(u) is minimum.

(69)

Lemma 6 (continued)

• In order to compare the swap candidate

between different group, we need to modify the value before to be more precise to the longest path pass (u,v).

(70)

Lemma 6 (continued)

• Let nca(u,h) = w

• L(T,v) + l(f) + (depth(u) – depth(w)) + d(w,v_deep)

• v_deep is the endpoint of D(T_x) in T_h

(71)

Relation between groups

R97922063 鄒承孝

(72)

Relation between Groups

• Lemma 7. Let e = (x, p(x)) and c ∈ C(x).

• If midpoint edge of D(T_x) is (c, x) or ∈ T_c.

• Then, midpoint edge of D(T_x) path from ∈ midpoint edge of D(T_c) to e.

Midpoint edge of D(T_c)

c x

Midpoint edge of D(T_x) lies on this path

(73)

Proof

• Case 1: midpoint edge of D(T_x) is (c, x)

– Straightforward.

• Case 2: midpoint edge of D(T_x) ∈ T_c

– (h, p(h)): midpoint edge of D(T_x)

– T’ and T”: subtrees by deleting (h, p(h)) from T_c, where h ∈ T’ and p(h) ∈ T”, respectively.

(74)

Proof (concluded)

• Case 2: midpoint edge of D(T_x) ∈ T_c

– Assume otherwise, then midpoint edge of D(T_c) ∈ T”.

– By definition, farthest vertex of p(h) in T_x ∈ T’.

– Similarly, farthest vertex of p(h) in T_c ∈ T”.

– But farthest vertex of p(h) in T_x is farthest vertex of p(h) in T_c or farthest vertex of p(h) in T_x – T_c,

neither is ∈ T’.

– Contradiction!

(75)

Relation between Groups

• Midpoint edge moves “upwards”.

• So the relation between groups of swap edges can be established.

(76)

Relation between Groups

• (h, p(h)): midpoint edge of D(T_x)

• (g, p(g)): midpoint edge of D(T_c)

(77)

• start-at(x): swap edges with lower endpoint = x

• end-at(x): swap edges with upper endpoint = x

• q ∈ C(x) and q ≠ c

• G_upper(x, x) = start-at(x) ∪ {∪_qGR(q)} \

end-at(x)

p(x)

G

_upper

(x, x)

… x c

(78)

G

_upper

(x, w)

• w path form ∈ p(h) to c

• G_upper(x, w) = G_upper(c, w) \ end-at(x)

• Recursive definition.

p(x)

… x c

p(h) w…

(79)

G

_lower

(x)

• d path from ∈ p(g) to h

• G_lower(x) = G_lower(c) ∪ {∪_dG_upper(q)} \ end-at(x)

• Recursive definition

p(x)

… x c

g

G_lower(c)

…

h

p(g)

(80)

徐國鐘

(89)

If swap edges ≠ backedges

• We now describe to replace any edge f by at most two “virtual” backedges , whose lengths are defined in such a way that the best swap edge computed by our algorithm is correct.

(90)

Replacement

• Replace f = (u, v) by f₁ := (u, v₁) and f₂ := (v, u₂)

– with lengths

l(f₁) := l(f) + d(v, v₁) l(f₂) := l(f) + d(u, u₂)

• If the path from u to v in T uses one or more edges of D(T)

– define v₁ := nc(u) and u₂ := nc(v).

• Otherwise

– we define v₁ := nca(u, v) and u₂ := nca(u, v).

(91)

Case 1

d_E u₂=nc(v)

f d_S

f₁ f₂

v u

v₁=nc(u) D(T)

l(f₁) := l(f) + d(v, v₁) l(f₂) := l(f) + d(u, u₂)

(92)

Case 2

d_E

f d_S

f₁ f₂

v u

D(T)

l(f₁) := l(f) + d(v, v₁) l(f₂) := l(f) + d(u, u₂)

(93)

Reasons

• Reason

– f₁ represents f exactly for all failing edges on the path in T from u to v₁, and f₂ represents f exactly for all edges from v to u₂

– for each non-diameter tree edge, one of f₁, f₂ represents f

• The lengths of f₁ and f₂ are defined exactly

such that for any failing edge e for which f_i is a swap edge, it holds |D(T_e/fi)| = |D(T_e/f )|.

(94)

Revealing figure

d_E u₂=nc(v)

f d_S

f₁ f₂

v u

v₁=nc(u) D(T)

e

(95)

Revealing figure: Case 1_1

l(f₁) := l(f) + d(v, v₁)

d_E u₂

f d_S

v u

v₁ D(T)

e d_E

u₂ d_S

f₁

v u

v₁ D(T)

e

(96)

Revealing figure: Case 1_2

f

v u

e

f₁

v u

e

l(f₁) := l(f) + d(v, v₁) l(f₁) := l(f) + d(v, u₂)-d(u₂,v₁)

(97)

Revealing figure: Case 1_3

l(f₁) := l(f) + d(v, v₁) l(f₁) := l(f) + d(v, u₂)-d(u₂,v₁)

d_E u₂

f d_S

v u

v₁ D(T)

e

u₂ d_S

f₁

v u

D(T)

e d_E

v₁

(98)

Comment

• The proof is not closed yet.

• Possible solution in the Case 1_3

– |P_e/f|<=|D_T| may be always true for the case.

– Try to face crossing edges instead of using replacement.

• The typo is terrible.

11/03/2022 98

(99)