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 size so that G(V,E+F’) is 2-edge-connected

Bi-Connected Components A B C D E F G H

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

Simple ratio 2: minimally leaf- closed trees

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 e  T’ is not covered then we found a smaller leaf-closed tree T’’ v e T’’

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

The leaf-stem lower bound for 1.8

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 deg opt (v)/2 coupons This term is different, depends on OPT

Example The blue link mean the actual bound is larger by ½ than what we know in advance  OPT

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 Unmatched leaf has 1 coupon 11 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 2 s 1 x y z z

If no 1,2-greedy applies then the contraction of any e  M 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 K+1 1

The Algorithm Let I be the edges added so far Exhaust 1 and 2 greedy Compute T/(M  I) No new leaves are created Find a minimally leaf-closed tree T v in T/(M  I) Let A=up(leaf) in T v Add to the solution (M  T v )  A (covers T v ) Iterate

In picture v x

Basic cover and the extra M  A is called the basic cover of T v After M is contracted, T/(I  M) has only unmatched leaves Every l  A being an unmatched leaf can pay with its coupon for up( l ) Every e  M has 1.5 coupons. Pays for its contraction with ½ to spare

A trivial case The problem is that we need to leave 1 coupon in 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 No such link is possible as this means 1 greedy We can add the up of the two leaves and leave 1 in the resulting leaf Not possible as the tree is leaf closed The other endpoints belong to Q: 2 ticket, 1 coupon

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 T v covering v (unless v=r and we are done) This link does not come out of l, l’ because T v 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 B  T 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 J. Cheriyan 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?)