330 likes | 438 Views
Increasing graph connectivity from 1 to 2. Guy Kortsarz Joint work with Even and Nutov. Augmenting edge connectivity from 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
E N D
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?)