Increasing graph connectivity from 1 to 2 Guy Kortsarz Joint work with Even and Nutov.
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Transcript of Increasing graph connectivity from 1 to 2 Guy Kortsarz Joint work with Even and Nutov.
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Increasing graph connectivity from 1 to 2
Guy Kortsarz
Joint work with Even and Nutov
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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
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Bi-Connected Components
A
BC
DE
F
G
H
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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)
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Shadow Completion
Part of the shadows added
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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
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Example
Hence the leaf to leaf links in OPT form a matching
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Simple ratio 2: minimally leaf-closed trees
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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
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Proof
If an edge eT’ is not covered then we found a smaller leaf-closed tree T’’
v
e
T’’
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Problematic structure: Stem A link whose contraction creates a
leaf
STEM
Twin Link
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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
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Example
|M|=2, |U|=3, t=1, |B|=2
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The leaf-stem lower bound for 1.8
Bv
opt vttUopt
t'
)(deg5.02'5.18.1
linksmatchedspecialnonof
numberthebeLet
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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
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Example
The blue link mean the actual bound is larger by ½ than what we know in
advance
211
1.51
OPTOPT
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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
Unmatched leaf has 1 coupon
1 1
1
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A stem with 2 matched links: an example of 2-greedy
A stem with two matched pairs:
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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
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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:
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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
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Main idea
Find a tree with k+1 coupons that can be covered with k links
K+1
1
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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 Tv
in T/(M I) Let A=up(leaf) in Tv
Add to the solution (M Tv)A (covers Tv ) Iterate
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In picture
v
x
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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
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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
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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
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No matched pairs at least two leaves
No such link No such link is possible is possible
as this as this means 1 means 1 greedygreedy
We can add We can add the up of the up of the two the two
leaves and leaves and leave 1 in leave 1 in
the the resulting resulting
leafleaf Not possible as Not possible as the tree is leaf the tree is leaf
closedclosed
The other endpoints The other endpoints belong to Q: 2 ticket, 1 belong to Q: 2 ticket, 1
couponcoupon
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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
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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’
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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
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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
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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. CheriyanJ. 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?)