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This paper explores the augmentation of edge connectivity in undirected graphs to achieve 2-edge connectivity through algorithmic approaches, focusing on edge additions from a given set. Highlighting the complexities of the tree augmentation problem, the study presents methods to eradicate bridges by carefully selecting edge subsets. Additionally, it discusses minimal leaf-closed trees and the significance of certain structures like stems and matched pairs in optimizing connectivity solutions. The profound relationship between graph structures and augmentation strategies is analyzed for effective implementation.
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Increasing graph connectivity from 1 to 2 Guy Kortsarz Joint work with Even and Nutov
Augmenting edge connectivityfrom 1 to 2 Given: undirected graph G(V,E) And a set of extra legal for addition edges F Required: a subset F’ F of minimum size so that G(V,E+F’) is 2-edge-connected
Bi-Connected Components G A H F B D E C
The tree augmentation problem Input: A tree T(V,E) and a separate set edge F Output: Add minimum amount of edges F’ from F so there will be no bridges (G+F’ is 2EC)
Shadow Completion • Part of the shadows added
Shadows-Minimal Solutions • If a link in the optimum can be replaced by a proper shadow and the solution is still feasible, do it. • Claim: in any SMS, the leaves have degree 1
Example Hence the leaf to leaf links in OPT form a matching
Covering minimally leaf-closed trees • Let up(l)be the “highest link” (closest to the root) for l, after shadow completion. • Let T’ be a minimally leaf-closed tree • Then {up(l) | l T } covers T’ • Given that, we spent L links in covering T’. The optimum spent at least L/2 • A ratio of 2 follows
Proof • If an edge eT’ is not covered then we found a smaller leaf-closed tree T’’ e T’’ v
Problematic structure: Stem • A link whose contraction creates a leaf STEM Twin Link
The lower bound for 1.8 • Compute a maximum matching M among matching not containing stem links • Let B be the non-leaf non-stems • Let U be the unmatched leaves in M • Let t be the number of links touching the twin of a stem with exactly one matched leaf in M • For this talk let call a unique link touching a twin a special matched link
Example |M|=2, |U|=3, t=1, |B|=2
Coupons and tickets • Every vertex in U gets 1 • Every non-special matched link in M gets 1.5 coupons. • Every special matched link gets 2 coupons. • Every vertex in B touched by OPT gets degopt(v)/2 coupons • This term is different, depends on OPT
Example OPT 1 1 1.5 1 2 The blue link mean the actual bound is larger by ½ than what we know in advance
1-greedy and 2-greedy • If a link closes a path that has 2 coupons, the link can be contracted • This is a 1-greedy step 1 Unmatched leaf has 1 coupon Unmatched leaf has 1 coupon 1 1
A stem with 2 matched links: an example of 2-greedy • A stem with two matched pairs:
The algorithm exahusts all 1,2 greedy: all stems are contracted • Stems enter compound nodes • Note that we may assume it has exactly one matched twin 1 z y s 2 z x
If no 1,2-greedy applies then the contraction of any eM never create a new leaf • The paths covered by e,e’ are disjoint as no 2-greedy • Now say that later contracting e M creates a leaf:
Why not find minimum leaf-closed tree and add up(leaves)? • There is not enough credit • Every unmatched leaf (vertex in U) does have a coupon needed to “pay” for the up link • Unfortunately, every matched pair has only 3/2<2 together, so it does not work
Main idea • Find a tree with k+1 coupons that can be covered with k links 1 K+1
The Algorithm • Let I be the edges added so far • Exhaust1 and 2 greedy • Compute T/(M I) • No new leaves are created • Find a minimally leaf-closed tree Tv in T/(M I) • Let A=up(leaf) in Tv • Add to the solution (M Tv)A (covers Tv ) • Iterate
In picture v x
Basic cover and the extra • MA is called the basic cover of Tv • After M is contracted, T/(IM) has only unmatched leaves • Every lA being an unmatched leaf can pay with its coupon for up(l ) • Every eM has 1.5 coupons. Pays for its contraction with ½ to spare
A trivial case • The problem is that we need to leave 1 couponin the created leaf (every unmatched leaf has one coupon) • If T has two matched leaves or more the 2* ½=1 spare can be left on the leaf
Less than 2 matched pairs • If there is a matched pair: Remember that every non-leaf non-stem touched by opt has ½ a coupon so together it would be a full coupon which is enough • First treat the case of no matched pairs. • If only one leaf, solved like the DFS case
No matched pairs at least two leaves The other endpoints belong to Q: 2 ticket, 1 coupon We can add the up of the two leaves and leave 1 in the resulting leaf No such link is possible as this means 1 greedy Not possible as the tree is leaf closed
At least four leaves one matched pair • The only vertices not in B that can be linked to the (at least) two unmatched leaves l, l’ are the matched pair leaves say b and b’ • Recall, b and b’ have degree 1 in OPT • Thus l, l’ and b and b’ must form a perfect matching
A ticket follows to cover the root • The matched pair b and b’ have no more links in OPT as matched to l, l’ and have degree 1 in OPT • There must be a link going out of Tv covering v (unless v=r and we are done) • This link does not come out of l, l’ because Tv is closed with respect to unmatched leaves • And by the above it can not come out of b or b’
Covering v • Therefore, the link comes out of a non-leaf internal node • There are no compound internal nodes • Thus v is covered by a vertex in BT • This means that we have the extra ½ needed. We use the basic cover and leave a coupon
Remarks • The case of one matched pair and 3 leaves gets a special treatment • In the 1.5 ratio algorithm the stems do not disappear after 1,2-greedy • Getting 1.5 requires 3 (more complex that what was shown here) extra new ideas and some extensive case analysis
Only one open question • The weighted case • Cannot use leaf-closed trees • In my opinion the usual LP does not suffice. BTW: known to have IG 1.5 Due to: J. Cheriyan, H. Karloff, R. Khandekar, and J. Könemann • We have stronger LP that we think has integrality gap less than 2 • We (all) failed badly in proving it (so far?)