christoph lenzen and roger wattenhoferwebcourse.cs.technion.ac.il/236825/spring2015/ho...Β Β· -...
TRANSCRIPT
![Page 1: Christoph Lenzen and Roger Wattenhoferwebcourse.cs.technion.ac.il/236825/Spring2015/ho...Β Β· - MaxIS canβt be larger than π 2 - π3βͺπ4 = π 2 βπ3βͺπ4 πΌ Lemma](https://reader034.vdocument.in/reader034/viewer/2022042212/5eb4e61d5cceb012cd10c7a2/html5/thumbnails/1.jpg)
Leveraging Linialβs Locality limit
Christoph Lenzen and Roger Wattenhofer
Presented by Yam Chemel
![Page 2: Christoph Lenzen and Roger Wattenhoferwebcourse.cs.technion.ac.il/236825/Spring2015/ho...Β Β· - MaxIS canβt be larger than π 2 - π3βͺπ4 = π 2 βπ3βͺπ4 πΌ Lemma](https://reader034.vdocument.in/reader034/viewer/2022042212/5eb4e61d5cceb012cd10c7a2/html5/thumbnails/2.jpg)
Linial (1992)
3-coloring and MIS
on a ring (π π)
takes at least logβ(π) rounds
Reminder
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Prove logβ(π) Lower bound for MaxIS algorithms in rings
Our Goal:
How?
MaxIS π(logβ π )
3-coloring π logβ π
Contradicts Linial!
![Page 4: Christoph Lenzen and Roger Wattenhoferwebcourse.cs.technion.ac.il/236825/Spring2015/ho...Β Β· - MaxIS canβt be larger than π 2 - π3βͺπ4 = π 2 βπ3βͺπ4 πΌ Lemma](https://reader034.vdocument.in/reader034/viewer/2022042212/5eb4e61d5cceb012cd10c7a2/html5/thumbnails/4.jpg)
MaxIS vs. MIS - Example
Independent Set MIS MaxIS
(general graphs)
MIS βΈ MaxIS
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MaxIS in rings
1 2
3 4
1 2
3
4
5
Q: What if each even node returns 1, and each odd node returns 0?
1 2
3 4
π 4 MaxIS
1 2
3
4
5 π 5
MaxIS
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1 3
2 4
MaxIS
Identifiers are not necessarily in order!
1 3
2 4
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MaxIS vs. MIS β in rings
MIS βΈ MaxIS
Algorithm π΄ MaxIS
π(logβ π )
Algorithm π΄ MIS
π logβ π
Contradicts Linial!
![Page 8: Christoph Lenzen and Roger Wattenhoferwebcourse.cs.technion.ac.il/236825/Spring2015/ho...Β Β· - MaxIS canβt be larger than π 2 - π3βͺπ4 = π 2 βπ3βͺπ4 πΌ Lemma](https://reader034.vdocument.in/reader034/viewer/2022042212/5eb4e61d5cceb012cd10c7a2/html5/thumbnails/8.jpg)
f-approximation algorithm for MaxIS
π -MaxIS solution πΌ β algorithm A solution
πΌ = π
If π΄ is π(π)-approximation: πΌ β€ π
and also:
πΌ β π(π) β₯ |π|
π΄ π π = π π
A is 2-approximation for MaxIS
πΌ =π
4
π
4β 2 β₯ π =
π
2
![Page 9: Christoph Lenzen and Roger Wattenhoferwebcourse.cs.technion.ac.il/236825/Spring2015/ho...Β Β· - MaxIS canβt be larger than π 2 - π3βͺπ4 = π 2 βπ3βͺπ4 πΌ Lemma](https://reader034.vdocument.in/reader034/viewer/2022042212/5eb4e61d5cceb012cd10c7a2/html5/thumbnails/9.jpg)
Leveraging Linialβs Locality limit
MaxIS approximation π(πππβ π )
3-coloring π πππβ π
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Leveraging Linialβs Locality limit
MaxIS approximation π π
3-coloring π(π)
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Leveraging Linialβs Locality limit
π(π)-alternating π(π)
MaxIS approximation π π
3-coloring π(π)
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π(π)-alternating algorithm - Definition:
Suppose A is an algorithm operating on π π which assigns each node π£π β ππ a value π(π£π) β {0, 1}. We call A π(n)-alternating, if the length of any monochromatic sequence
π π£π = π π£π+1 = β― = π π£π+π is smaller than π(n).
< π(π)
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π΄ π π = π π
π(π)-alternating algorithm - examples
n is even
Longest monochromatic sequence = 1
n is odd
Longest monochromatic sequence = 2
A is 3-alternating (also 4-alternating, 5-alternatingβ¦)
(Not necessarily MaxIS algorithm)
![Page 14: Christoph Lenzen and Roger Wattenhoferwebcourse.cs.technion.ac.il/236825/Spring2015/ho...Β Β· - MaxIS canβt be larger than π 2 - π3βͺπ4 = π 2 βπ3βͺπ4 πΌ Lemma](https://reader034.vdocument.in/reader034/viewer/2022042212/5eb4e61d5cceb012cd10c7a2/html5/thumbnails/14.jpg)
π΄ π π = π π
A is (n+1)-alternating
(Not necessarily MaxIS algorithm)
π(π)-alternating algorithm - examples
![Page 15: Christoph Lenzen and Roger Wattenhoferwebcourse.cs.technion.ac.il/236825/Spring2015/ho...Β Β· - MaxIS canβt be larger than π 2 - π3βͺπ4 = π 2 βπ3βͺπ4 πΌ Lemma](https://reader034.vdocument.in/reader034/viewer/2022042212/5eb4e61d5cceb012cd10c7a2/html5/thumbnails/15.jpg)
Lemma - Modified MaxIS approximation Suppose an f-approximation algorithm A for the MaxIS problem on the ring π π running in at most g(n) β₯ 1 rounds is given, where we have π(π)π(π) β π(logβ(π)). Then an π(logβ(π))-alternating algorithm Aβ² requiring π logβ π communication rounds exists.
A
MaxIS, π(π)-approximation,
π(π) rounds, π π π π β π πππβ π
Aβ π πππβ π -alternating,
π πππβ π rounds
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Modified MaxIS approximation - Proof
General idea
ππ
π π
π πβ² π
ππβ²
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Observation: For each node π£π, π(π£π) is a function of:
π£π
π(π) π(π)
π π = #πππ’πππ
β’ The IDs of its π(π) neighbors on each side
β’ π β’ Its ID
π(π) π(π)
Modified MaxIS approximation - Proof
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β’ πππ‘ π π β 10π(π)π(π)
ππ =
π£1, β¦ , π£π π , π£π π +1, β¦, π£π π +π π , π£π π +1+π π , β¦ , π£2π π +π π£
| βπ β π£π π +1, β¦, π£π π +π π π π£π = 0
0 0
?
?
0
? ?
? ?
π(π) π(π) π(π)
Modified MaxIS approximation - Proof
β’ Define exactly the set of sequences preventing that A is π(π)-alternating
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π(π) π(π)
? ? ? 0 0 0 ? ? ?
Id=2 Id=11 Id=8 Id=9 Id=14 Id=21 Id=7 Id=6 Id=3
π(π)
π = 2, 11, 8, 9, 14, 21, 7, 6, 3 β ππ
Take 2π π + π(π) consecutive nodes in π π
Assign them identifiers
Run A on the sequence / on π π containing the sequence.
If the π(π) center nodes compute 0, add s to ππ
Id=2 Id=11 Id=8 Id=9 Id=14 Id=21 Id=7 Id=6 Id=3
Building ππ
2π π + π(π)
![Page 20: Christoph Lenzen and Roger Wattenhoferwebcourse.cs.technion.ac.il/236825/Spring2015/ho...Β Β· - MaxIS canβt be larger than π 2 - π3βͺπ4 = π 2 βπ3βͺπ4 πΌ Lemma](https://reader034.vdocument.in/reader034/viewer/2022042212/5eb4e61d5cceb012cd10c7a2/html5/thumbnails/20.jpg)
Construct a sequence of identifiers for π π:
π 2
π 1 π β π
π π π π
Modified MaxIS approximation - Proof
1. Choose an arbitrary sequence s from ππ and assign the identifiers to π£1, β¦ , π£ π .
2. Assuming we already assigned labels to the nodes π£1, β¦ , π£π:
While there exists a sequence π β ππ that can be appended to π£1, β¦ , π£π without duplicating an identifier, we do so.
3. If no further sequence fits, we add π β π arbitrary identifiers not yet present in π£1, β¦ , π£π to
complete the labeling (π£1, β¦ , π£π) of π π.
![Page 21: Christoph Lenzen and Roger Wattenhoferwebcourse.cs.technion.ac.il/236825/Spring2015/ho...Β Β· - MaxIS canβt be larger than π 2 - π3βͺπ4 = π 2 βπ3βͺπ4 πΌ Lemma](https://reader034.vdocument.in/reader034/viewer/2022042212/5eb4e61d5cceb012cd10c7a2/html5/thumbnails/21.jpg)
Assume for contradiction, that for arbitrarily large n it is possible to
label π π as described, with at least π βπ
5π π identifiers
stemming from sequences out of ππ.
# πππππ π‘βππ‘ πππππ’π‘π 1 ππ π π β€ π βπ π β π
π π + 2π πβ
π
5π π
β₯ πππππ π‘βππ‘ πππππ’π‘π 0 ππ π β ππ
πππππ ππ π β ππβ π β #(πππππ πππ‘ ππ π β ππ)
Modified MaxIS approximation - Proof
# πππππ π‘βππ‘ πππππ’π‘π 0 ππ π π β₯
=π π
π π + 2π(π)β π β
π
5π π
![Page 22: Christoph Lenzen and Roger Wattenhoferwebcourse.cs.technion.ac.il/236825/Spring2015/ho...Β Β· - MaxIS canβt be larger than π 2 - π3βͺπ4 = π 2 βπ3βͺπ4 πΌ Lemma](https://reader034.vdocument.in/reader034/viewer/2022042212/5eb4e61d5cceb012cd10c7a2/html5/thumbnails/22.jpg)
# πππππ π‘βππ‘ πππππ’π‘π 1 ππ π π β€ π βπ π β π
π π + 2π πβ
π
5π π
= π 1 βπ π
π π + 2π π+
1
5π π
π π = 10π π π(π)
Modified MaxIS approximation - Proof
= π 1 β10π π π π
10π π π π + 2π π+
1
5π π
= π 1 β5π π
5π π + 1+
1
5π π
= π 1 β 1 β1
5π π + 1+
1
5π π
= π1
5π π + 1+
1
5π π
β€ π1
5π π + 0+
1
5π π
=2π
5π π
![Page 23: Christoph Lenzen and Roger Wattenhoferwebcourse.cs.technion.ac.il/236825/Spring2015/ho...Β Β· - MaxIS canβt be larger than π 2 - π3βͺπ4 = π 2 βπ3βͺπ4 πΌ Lemma](https://reader034.vdocument.in/reader034/viewer/2022042212/5eb4e61d5cceb012cd10c7a2/html5/thumbnails/23.jpg)
πΌ = # πππππ π‘βππ‘ πππππ’π‘π 1 ππ π π β€2π
5π π
However, according to f-approximation definition: π π πΌ β₯ π
(M -an arbitrary MaxIS of G)
π 6
Contradicts the assumption! (βFor arbitrarily large n it is possible to label π π as described, with at least
π βπ
5π π identifiers stemming from sequences out of ππ.β)
Modified MaxIS approximation - Proof
π(π) πΌ β€2
5π
Choosing every other node in π π is a MaxIS solution:
π =π
2
Combining the equations:
π π πΌ β₯π
2
![Page 24: Christoph Lenzen and Roger Wattenhoferwebcourse.cs.technion.ac.il/236825/Spring2015/ho...Β Β· - MaxIS canβt be larger than π 2 - π3βͺπ4 = π 2 βπ3βͺπ4 πΌ Lemma](https://reader034.vdocument.in/reader034/viewer/2022042212/5eb4e61d5cceb012cd10c7a2/html5/thumbnails/24.jpg)
At least π
ππ π identifiers remain which cannot form a further
sequence from ππ.
Set πβ² β max π0, 5π π β π
In π πβ²
#πππππππππ πππππ‘ππππππ
β₯πβ²
5π πβ²
For a large n (π > π0) it is possible to label π π as described, with
at most π βπ
5π π identifiers stemming from sequences out of πΊπ
Modified MaxIS approximation - Proof
π 2
π 1 π β π
π π π π
β₯5π π β π
5π π= π
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π π π πβ²
ππ
Algorithm Aβ : ππ’π βΆ π΄(πβ², π πππππ‘ππππππ ππππ π πβ² )
Modified MaxIS approximation - Proof
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π 4 π 11
1
2
3
4
3
8
1
4
π4
Algorithm Aβ : π π’π π΄(11, 3, 8, 1, 4 )
Example:
Modified MaxIS approximation - Proof
![Page 27: Christoph Lenzen and Roger Wattenhoferwebcourse.cs.technion.ac.il/236825/Spring2015/ho...Β Β· - MaxIS canβt be larger than π 2 - π3βͺπ4 = π 2 βπ3βͺπ4 πΌ Lemma](https://reader034.vdocument.in/reader034/viewer/2022042212/5eb4e61d5cceb012cd10c7a2/html5/thumbnails/27.jpg)
Aβ is π πβ² -alternating
No π(πβ²) consecutive 0βs No π(πβ²) consecutive 1βs
π΄ computes an independent setβ no 2 neighbors in independent set β
no 2 consecutive 1βs
No π β ππβ² in remaining n nodes of π πβ² β No π πβ² consecutive nodes compute 0 in
Aβ run
Maximum monochromatic sequence : π πβ² β 1 β π π πβ² π πβ² β π logβ πβ² = π logβ π
Aβ running time -π(πππβ(π)) rounds π π β€ π π π π β π(πππβ π)
π πβ² β π(πππβ π)
Aβ is π πππβ π -alternating
![Page 28: Christoph Lenzen and Roger Wattenhoferwebcourse.cs.technion.ac.il/236825/Spring2015/ho...Β Β· - MaxIS canβt be larger than π 2 - π3βͺπ4 = π 2 βπ3βͺπ4 πΌ Lemma](https://reader034.vdocument.in/reader034/viewer/2022042212/5eb4e61d5cceb012cd10c7a2/html5/thumbnails/28.jpg)
Leveraging Linialβs Locality limit
π(π)-alternating π(π)
MaxIS approximation π π
3-coloring π(π)
![Page 29: Christoph Lenzen and Roger Wattenhoferwebcourse.cs.technion.ac.il/236825/Spring2015/ho...Β Β· - MaxIS canβt be larger than π 2 - π3βͺπ4 = π 2 βπ3βͺπ4 πΌ Lemma](https://reader034.vdocument.in/reader034/viewer/2022042212/5eb4e61d5cceb012cd10c7a2/html5/thumbnails/29.jpg)
Leveraging Linialβs Locality limit
π(π)-alternating π(π)
MaxIS approximation π π
3-coloring π(π)
![Page 30: Christoph Lenzen and Roger Wattenhoferwebcourse.cs.technion.ac.il/236825/Spring2015/ho...Β Β· - MaxIS canβt be larger than π 2 - π3βͺπ4 = π 2 βπ3βͺπ4 πΌ Lemma](https://reader034.vdocument.in/reader034/viewer/2022042212/5eb4e61d5cceb012cd10c7a2/html5/thumbnails/30.jpg)
2. Proof : Lemma β Given a π(π)-alternating algorithm A running in π(π(π)) rounds, a 3-coloring of the ring can be computed in π(π(π)) rounds.
1. Run A. Let π(π£) β {0, 1} denote the result of this run.
2. Find a pair of neighboring nodes {π€1, π€2} with π π€1 β π π€2 which is closest to v.
0 1 π π€2
1 0 π π€2
If π£ β {π€1, π€2}: if π(π£) = 0: set π π£ β π otherwise: set π(π£) β π
0 1 π π€2
1 0 π π€2
1 0 1 1
π€2 π€1 π π€
1 0 1 1
Else: denote by πΏ the distance to the closer node in {π€1, π€2}, w.l.o.g. π€1. if πΏ β 2β: set π π£ β π(π€1) else: set π π£ β π(π€2)
1 0 1 1
π€2 π€1 π π€ (πΏ β 2β)
πΏ
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3 8 π π€
8 3 π π€
3. If v has a neighbor w with π(π£) = π(π€) and v > w, set π(π£) β π.
3 8 π π€
8 3 π π€
3 8 π π€
1 2 3 8 π π€
1 2 (3)
4. If v has a neighbor w with π(π£) = π(π€) = π and v > w, set c(v) to the color none of the neighbors of v has.
3 8 π π€
1 2 (4)
2. Proof : Lemma β Given a π(π)-alternating algorithm A running in π(π(π)) rounds, a 3-coloring of the ring can be computed in π(π(π)) rounds.
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Running time: πΆ π π
Step 1: Running A- π π π
Step 2: Finding a pair of neighbors with different d - π π .
No more than π(π£) consecutive nodes take the same decision d(v)
since A is π(π)-alternating.
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Valid 3-coloring of πΉπ
Step 2: Each node π£ chooses different from one of its neighbors,
1 1 0 1 1 1 1 0 1
so at most one of the neighbors of π£ may take the same choice.
Step 3: From each pair of neighbors with the same color one chooses g.
Step 4: If that same color was g, v chooses the color non of its neighbors has.
v v
v
1 1 0
v
1 1 1 1 0 1
v v
![Page 34: Christoph Lenzen and Roger Wattenhoferwebcourse.cs.technion.ac.il/236825/Spring2015/ho...Β Β· - MaxIS canβt be larger than π 2 - π3βͺπ4 = π 2 βπ3βͺπ4 πΌ Lemma](https://reader034.vdocument.in/reader034/viewer/2022042212/5eb4e61d5cceb012cd10c7a2/html5/thumbnails/34.jpg)
Leveraging Linialβs Locality limit
π(π)-alternating π(π)
MaxIS approximation π π
3-coloring π(π)
![Page 35: Christoph Lenzen and Roger Wattenhoferwebcourse.cs.technion.ac.il/236825/Spring2015/ho...Β Β· - MaxIS canβt be larger than π 2 - π3βͺπ4 = π 2 βπ3βͺπ4 πΌ Lemma](https://reader034.vdocument.in/reader034/viewer/2022042212/5eb4e61d5cceb012cd10c7a2/html5/thumbnails/35.jpg)
Therefore, there isnβt a MaxIS approximation algorithms running on a ring
in less than log*(n)
Proof - Summary
Assume by contradiction that there exists a MaxIS approximation algorithm A running in less than πππβ(π).
Construct a π πππβ π -alternating algorithm running in π(πππβ(π)) using algorithm A.
By lemma 2, a 3-coloring of the ring can be computed in π(πππβ(π)) rounds.
This contradicts Linialβs 3-coloring lower bound.
![Page 36: Christoph Lenzen and Roger Wattenhoferwebcourse.cs.technion.ac.il/236825/Spring2015/ho...Β Β· - MaxIS canβt be larger than π 2 - π3βͺπ4 = π 2 βπ3βͺπ4 πΌ Lemma](https://reader034.vdocument.in/reader034/viewer/2022042212/5eb4e61d5cceb012cd10c7a2/html5/thumbnails/36.jpg)
- Discuss MDS lower bound - Compare MDS and MaxIS difficulty
Our (new) Goal:
No! Weβll show a case where MIS is easier than MDS
MDS on rings O(1)
Is MDS always easier than MIS?
How?
![Page 37: Christoph Lenzen and Roger Wattenhoferwebcourse.cs.technion.ac.il/236825/Spring2015/ho...Β Β· - MaxIS canβt be larger than π 2 - π3βͺπ4 = π 2 βπ3βͺπ4 πΌ Lemma](https://reader034.vdocument.in/reader034/viewer/2022042212/5eb4e61d5cceb012cd10c7a2/html5/thumbnails/37.jpg)
MDS β Minimum Dominating Set
DS MDS
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MDS in rings
Taking every third node gives a minimum dominating set
π =π
3
Taking every node gives a 3-approximation MDS in 1 round β π logβ π
π π
π π
There is no πππβ(π) bound MDS approximation in rings! MDS approximation in rings takes πΆ(π) rounds
MDS approximation in rings
![Page 39: Christoph Lenzen and Roger Wattenhoferwebcourse.cs.technion.ac.il/236825/Spring2015/ho...Β Β· - MaxIS canβt be larger than π 2 - π3βͺπ4 = π 2 βπ3βͺπ4 πΌ Lemma](https://reader034.vdocument.in/reader034/viewer/2022042212/5eb4e61d5cceb012cd10c7a2/html5/thumbnails/39.jpg)
Can we compare MaxIS and MDS difficulty?
MDS
MaxIS
We saw that in π π MDS can be computed in 1 round, but MaxIS requires at least πππβ(π) round.
Is it always easier to compute MDS than MaxIS?
![Page 40: Christoph Lenzen and Roger Wattenhoferwebcourse.cs.technion.ac.il/236825/Spring2015/ho...Β Β· - MaxIS canβt be larger than π 2 - π3βͺπ4 = π 2 βπ3βͺπ4 πΌ Lemma](https://reader034.vdocument.in/reader034/viewer/2022042212/5eb4e61d5cceb012cd10c7a2/html5/thumbnails/40.jpg)
MaxIS graph family
πΊ π£ π€
Any graph that can be constructed this way
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Lemma (Local computation of a MaxIS on MaxIS Graphs): The set {π£ β π | |π1
+(π£)|πππ 2 = 1} is a MaxIS for any MaxIS Graph.
Proof β part 1:
- For π£π β ππ π β {3,4}, π£π has 2|π1+ π£ | neighbors.
- π1+ π£π = 2 π1
+ π£ + 1 β π1+ π£π ππ πππ
- For π£π β ππ π β {1,2}, π1+ π£π = 4 π1
+ π£ β π1+ π£π ππ ππ£ππ
- {π£ β π | |π1+(π£)|πππ 2 = 1} = π3 βͺ π4
![Page 42: Christoph Lenzen and Roger Wattenhoferwebcourse.cs.technion.ac.il/236825/Spring2015/ho...Β Β· - MaxIS canβt be larger than π 2 - π3βͺπ4 = π 2 βπ3βͺπ4 πΌ Lemma](https://reader034.vdocument.in/reader034/viewer/2022042212/5eb4e61d5cceb012cd10c7a2/html5/thumbnails/42.jpg)
Proof β part 2:
- π3 βͺ π4 is an Independent set β according to the construction of
MaxIS graph
- (π£1, π£3, π£2, π£4) forms a cycle, so for each 4 nodes as such, only 2 can be in the IS.
- MaxIS canβt be larger than π
2
- π3 βͺ π4 =π
2
β π3 βͺ π4 ππ π πππ₯πΌπ
Lemma (Local computation of a MaxIS on MaxIS Graphs): The set {π£ β π | |π1
+(π£)|πππ 2 = 1} is a MaxIS for any MaxIS Graph.
Conclusion β MaxIS on a MaxIS graph can be determined locally, without communication (in π(1) rounds).
![Page 43: Christoph Lenzen and Roger Wattenhoferwebcourse.cs.technion.ac.il/236825/Spring2015/ho...Β Β· - MaxIS canβt be larger than π 2 - π3βͺπ4 = π 2 βπ3βͺπ4 πΌ Lemma](https://reader034.vdocument.in/reader034/viewer/2022042212/5eb4e61d5cceb012cd10c7a2/html5/thumbnails/43.jpg)
MDS on MaxIS graphs
We can prove that MDS on MaxIS graphs is as efficient as in general graphs, meaning:
Ξ©log π
log log π
MaxIS
MDS
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Ring graphs - π π MaxIS graphs
MaxIS
MDS π(1)
Ξ©log π
log log π
MDS
MaxIS π(1)
Ξ© logβ π
MaxIS and MDS are not comparable in general graphs!