Leveraging Linial’s Locality limit

Christoph Lenzen and Roger Wattenhofer

Presented by Yam Chemel

Linial (1992)

3-coloring and MIS

on a ring (𝑅𝑛)

takes at least log∗(𝑛) rounds

Reminder

Prove log∗(𝑛) Lower bound for MaxIS algorithms in rings

Our Goal:

How?

MaxIS 𝑜(log∗ 𝑛 )

3-coloring 𝑜 log∗ 𝑛

Contradicts Linial!

MaxIS vs. MIS - Example

Independent Set MIS MaxIS

(general graphs)

MIS ⟸ MaxIS

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

𝑅4 MaxIS

1 2

3

4

5 𝑅5

MaxIS

1 3

2 4

MaxIS

Identifiers are not necessarily in order!

1 3

2 4

MaxIS vs. MIS – in rings

MIS ⟸ MaxIS

Algorithm 𝐴 MaxIS

𝑜(log∗ 𝑛 )

Algorithm 𝐴 MIS

𝑜 log∗ 𝑛

Contradicts Linial!

f-approximation algorithm for MaxIS

𝑀 -MaxIS solution 𝐼 – algorithm A solution

𝐼 = 𝑀

If 𝐴 is 𝑓(𝑛)-approximation: 𝐼 ≤ 𝑀

and also:

𝐼 ∗ 𝑓(𝑛) ≥ |𝑀|

𝐴 𝑅𝑛 = 𝑅𝑛

A is 2-approximation for MaxIS

𝐼 =𝑛

4

𝑛

4∗ 2 ≥ 𝑀 =

𝑛

2

Leveraging Linial’s Locality limit

MaxIS approximation 𝑜(𝑙𝑜𝑔∗ 𝑛 )

3-coloring 𝑜 𝑙𝑜𝑔∗ 𝑛

Leveraging Linial’s Locality limit

MaxIS approximation 𝜎 𝑛

3-coloring 𝜎(𝑛)

Leveraging Linial’s Locality limit

𝜎(𝑛)-alternating 𝜎(𝑛)

MaxIS approximation 𝜎 𝑛

3-coloring 𝜎(𝑛)

𝜎(𝑛)-alternating algorithm - Definition:

Suppose A is an algorithm operating on 𝑅𝑛 which assigns each node 𝑣𝑖 ∈ 𝑉𝑛 a value 𝑐(𝑣𝑖) ∈ {0, 1}. We call A 𝜎(n)-alternating, if the length of any monochromatic sequence

𝑐 𝑣𝑖 = 𝑐 𝑣𝑖+1 = ⋯ = 𝑐 𝑣𝑖+𝑘 is smaller than 𝝈(n).

< 𝜎(𝑛)

𝐴 𝑅𝑛 = 𝑅𝑛

𝜎(𝑛)-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-alternating…)

(Not necessarily MaxIS algorithm)

𝐴 𝑅𝑛 = 𝑅𝑛

A is (n+1)-alternating

(Not necessarily MaxIS algorithm)

𝜎(𝑛)-alternating algorithm - examples

Lemma - Modified MaxIS approximation Suppose an f-approximation algorithm A for the MaxIS problem on the ring 𝑅𝑛 running in at most g(n) ≥ 1 rounds is given, where we have 𝑓(𝑛)𝑔(𝑛) ∈ 𝑜(log∗(𝑛)). Then an 𝑜(log∗(𝑛))-alternating algorithm A′ requiring 𝑜 log∗ 𝑛 communication rounds exists.

A

MaxIS, 𝑓(𝑛)-approximation,

𝑔(𝑛) rounds, 𝑓 𝑛 𝑔 𝑛 ∈ 𝑜 𝑙𝑜𝑔∗ 𝑛

A’ 𝑜 𝑙𝑜𝑔∗ 𝑛 -alternating,

𝑜 𝑙𝑜𝑔∗ 𝑛 rounds

Modified MaxIS approximation - Proof

General idea

𝑆𝑛

𝑅𝑛

𝑅𝑛′ 𝑛

𝑆𝑛′

Observation: For each node 𝑣𝑖, 𝑐(𝑣𝑖) is a function of:

𝑣𝑖

𝑔(𝑛) 𝑔(𝑛)

𝑔 𝑛 = #𝑟𝑜𝑢𝑛𝑑𝑠

• The IDs of its 𝑔(𝑛) neighbors on each side

• 𝑛 • Its ID

𝑔(𝑛) 𝑔(𝑛)

Modified MaxIS approximation - Proof

• 𝑆𝑒𝑡 𝜎 𝑛 ← 10𝑓(𝑛)𝑔(𝑛)

𝑆𝑛 =

𝑣1, … , 𝑣𝑔 𝑛 , 𝑣𝑔 𝑛 +1, …, 𝑣𝑔 𝑛 +𝜎 𝑛 , 𝑣𝑔 𝑛 +1+𝜎 𝑛 , … , 𝑣2𝑔 𝑛 +𝜎 𝑣

| ∀𝑖 ∈ 𝑣𝑔 𝑛 +1, …, 𝑣𝑔 𝑛 +𝜎 𝑛 𝑐 𝑣𝑖 = 0

0 0

?

?

0

? ?

? ?

𝑔(𝑛) 𝑔(𝑛) 𝜎(𝑛)

Modified MaxIS approximation - Proof

• Define exactly the set of sequences preventing that A is 𝜎(𝑛)-alternating

𝑔(𝑛) 𝜎(𝑛)

? ? ? 0 0 0 ? ? ?

Id=2 Id=11 Id=8 Id=9 Id=14 Id=21 Id=7 Id=6 Id=3

𝑔(𝑛)

𝑠 = 2, 11, 8, 9, 14, 21, 7, 6, 3 → 𝑆𝑛

Take 2𝑔 𝑛 + 𝜎(𝑛) consecutive nodes in 𝑅𝑛

Assign them identifiers

Run A on the sequence / on 𝑅𝑛 containing the sequence.

If the 𝜎(𝑛) center nodes compute 0, add s to 𝑆𝑛

Id=2 Id=11 Id=8 Id=9 Id=14 Id=21 Id=7 Id=6 Id=3

Building 𝑆𝑛

2𝑔 𝑛 + 𝜎(𝑛)

Construct a sequence of identifiers for 𝑅𝑛:

𝑠2

𝑠1 𝑛 − 𝑗

𝑅𝑛 𝑠𝑘

Modified MaxIS approximation - Proof

1. Choose an arbitrary sequence s from 𝑆𝑛 and assign the identifiers to 𝑣1, … , 𝑣 𝑠 .

2. Assuming we already assigned labels to the nodes 𝑣1, … , 𝑣𝑗:

While there exists a sequence 𝑠 ∈ 𝑆𝑛 that can be appended to 𝑣1, … , 𝑣𝑗 without duplicating an identifier, we do so.

3. If no further sequence fits, we add 𝑛 − 𝑗 arbitrary identifiers not yet present in 𝑣1, … , 𝑣𝑗 to

complete the labeling (𝑣1, … , 𝑣𝑛) of 𝑅𝑛.

Assume for contradiction, that for arbitrarily large n it is possible to

label 𝑅𝑛 as described, with at least 𝑛 −𝑛

5𝑓 𝑛 identifiers

stemming from sequences out of 𝑆𝑛.

# 𝑛𝑜𝑑𝑒𝑠 𝑡ℎ𝑎𝑡 𝑐𝑜𝑚𝑝𝑢𝑡𝑒 1 𝑖𝑛 𝑅𝑛 ≤ 𝑛 −𝜎 𝑛 ∗ 𝑛

𝜎 𝑛 + 2𝑔 𝑛−

𝑛

5𝑓 𝑛

≥ 𝑛𝑜𝑑𝑒𝑠 𝑡ℎ𝑎𝑡 𝑐𝑜𝑚𝑝𝑢𝑡𝑒 0 𝑖𝑛 𝑠 ∈ 𝑆𝑛

𝑛𝑜𝑑𝑒𝑠 𝑖𝑛 𝑠 ∈ 𝑆𝑛∗ 𝑛 − #(𝑛𝑜𝑑𝑒𝑠 𝑛𝑜𝑡 𝑖𝑛 𝑠 ∈ 𝑆𝑛)

Modified MaxIS approximation - Proof

# 𝑛𝑜𝑑𝑒𝑠 𝑡ℎ𝑎𝑡 𝑐𝑜𝑚𝑝𝑢𝑡𝑒 0 𝑖𝑛 𝑅𝑛 ≥

=𝜎 𝑛

𝜎 𝑛 + 2𝑔(𝑛)∗ 𝑛 −

𝑛

5𝑓 𝑛

# 𝑛𝑜𝑑𝑒𝑠 𝑡ℎ𝑎𝑡 𝑐𝑜𝑚𝑝𝑢𝑡𝑒 1 𝑖𝑛 𝑅𝑛 ≤ 𝑛 −𝜎 𝑛 ∗ 𝑛

𝜎 𝑛 + 2𝑔 𝑛−

𝑛

5𝑓 𝑛

= 𝑛 1 −𝜎 𝑛

𝜎 𝑛 + 2𝑔 𝑛+

1

5𝑓 𝑛

𝜎 𝑛 = 10𝑓 𝑛 𝑔(𝑛)

Modified MaxIS approximation - Proof

= 𝑛 1 −10𝑓 𝑛 𝑔 𝑛

10𝑓 𝑛 𝑔 𝑛 + 2𝑔 𝑛+

1

5𝑓 𝑛

= 𝑛 1 −5𝑓 𝑛

5𝑓 𝑛 + 1+

1

5𝑓 𝑛

= 𝑛 1 − 1 −1

5𝑓 𝑛 + 1+

1

5𝑓 𝑛

= 𝑛1

5𝑓 𝑛 + 1+

1

5𝑓 𝑛

≤ 𝑛1

5𝑓 𝑛 + 0+

1

5𝑓 𝑛

=2𝑛

5𝑓 𝑛

𝐼 = # 𝑛𝑜𝑑𝑒𝑠 𝑡ℎ𝑎𝑡 𝑐𝑜𝑚𝑝𝑢𝑡𝑒 1 𝑖𝑛 𝑅𝑛 ≤2𝑛

5𝑓 𝑛

However, according to f-approximation definition: 𝑓 𝑛 𝐼 ≥ 𝑀

(M -an arbitrary MaxIS of G)

𝑅6

Contradicts the assumption! (“For arbitrarily large n it is possible to label 𝑅𝑛 as described, with at least

𝑛 −𝑛

5𝑓 𝑛 identifiers stemming from sequences out of 𝑆𝑛.”)

Modified MaxIS approximation - Proof

𝑓(𝑛) 𝐼 ≤2

5𝑛

Choosing every other node in 𝑅𝑛 is a MaxIS solution:

𝑀 =𝑛

2

Combining the equations:

𝑓 𝑛 𝐼 ≥𝑛

2

At least 𝒏

𝟓𝒇 𝒏 identifiers remain which cannot form a further

sequence from 𝑆𝑛.

Set 𝑛′ ← max 𝑛0, 5𝑓 𝑛 ∙ 𝑛

In 𝑅𝑛′

#𝑟𝑒𝑚𝑎𝑖𝑛𝑖𝑛𝑔 𝑖𝑑𝑒𝑛𝑡𝑖𝑓𝑖𝑒𝑟𝑠

≥𝑛′

5𝑓 𝑛′

For a large n (𝑛 > 𝑛0) it is possible to label 𝑅𝑛 as described, with

at most 𝑛 −𝑛

5𝑓 𝑛 identifiers stemming from sequences out of 𝑺𝒏

Modified MaxIS approximation - Proof

𝑠2

𝑠1 𝑛 − 𝑗

𝑅𝑛 𝑠𝑘

≥5𝑓 𝑛 ∙ 𝑛

5𝑓 𝑛= 𝑛

𝑅𝑛 𝑅𝑛′

𝑟𝑛

Algorithm A’ : 𝑟𝑢𝑛 ∶ 𝐴(𝑛′, 𝑛 𝑖𝑑𝑒𝑛𝑡𝑖𝑓𝑖𝑒𝑟𝑠 𝑓𝑟𝑜𝑚 𝑅𝑛′ )

Modified MaxIS approximation - Proof

𝑅4 𝑅11

1

2

3

4

3

8

1

4

𝑟4

Algorithm A’ : 𝑅𝑢𝑛 𝐴(11, 3, 8, 1, 4 )

Example:

Modified MaxIS approximation - Proof

A’ is 𝝈 𝒏′ -alternating

No 𝝈(𝒏′) consecutive 0’s No 𝝈(𝒏′) consecutive 1’s

𝐴 computes an independent set→ no 2 neighbors in independent set →

no 2 consecutive 1’s

No 𝑠 ∈ 𝑆𝑛′ in remaining n nodes of 𝑅𝑛′ → No 𝜎 𝑛′ consecutive nodes compute 0 in

A’ run

Maximum monochromatic sequence : 𝜎 𝑛′ − 1 ∈ 𝑂 𝑓 𝑛′ 𝑔 𝑛′ ⊂ 𝑜 log∗ 𝑛′ = 𝑜 log∗ 𝑛

A’ running time -𝒐(𝒍𝒐𝒈∗(𝒏)) rounds 𝑔 𝑛 ≤ 𝑔 𝑛 𝑓 𝑛 ∈ 𝑜(𝑙𝑜𝑔∗ 𝑛)

𝑔 𝑛′ ∈ 𝑜(𝑙𝑜𝑔∗ 𝑛)

A’ is 𝒐 𝒍𝒐𝒈∗ 𝒏 -alternating

Leveraging Linial’s Locality limit

𝜎(𝑛)-alternating 𝜎(𝑛)

MaxIS approximation 𝜎 𝑛

3-coloring 𝜎(𝑛)

Leveraging Linial’s Locality limit

𝜎(𝑛)-alternating 𝜎(𝑛)

MaxIS approximation 𝜎 𝑛

3-coloring 𝜎(𝑛)

2. Proof : Lemma – Given a 𝝈(𝒏)-alternating algorithm A running in 𝑂(𝜎(𝑛)) rounds, a 3-coloring of the ring can be computed in 𝑂(𝜎(𝑛)) rounds.

1. Run A. Let 𝑑(𝑣) ∈ {0, 1} denote the result of this run.

2. Find a pair of neighboring nodes {𝑤1, 𝑤2} with 𝑑 𝑤1 ≠ 𝑑 𝑤2 which is closest to v.

0 1 𝒗 𝑤2

1 0 𝒗 𝑤2

If 𝑣 ∈ {𝑤1, 𝑤2}: if 𝑑(𝑣) = 0: set 𝑐 𝑣 ← 𝑏 otherwise: set 𝑐(𝑣) ← 𝑟

0 1 𝒗 𝑤2

1 0 𝒗 𝑤2

1 0 1 1

𝑤2 𝑤1 𝒗 𝑤

1 0 1 1

Else: denote by 𝛿 the distance to the closer node in {𝑤1, 𝑤2}, w.l.o.g. 𝑤1. if 𝛿 ∈ 2ℕ: set 𝑐 𝑣 ← 𝑐(𝑤1) else: set 𝑐 𝑣 ← 𝑐(𝑤2)

1 0 1 1

𝑤2 𝑤1 𝒗 𝑤 (𝛿 ∈ 2ℕ)

𝛿

3 8 𝒗 𝑤

8 3 𝒗 𝑤

3. If v has a neighbor w with 𝑐(𝑣) = 𝑐(𝑤) and v > w, set 𝑐(𝑣) ← 𝑔.

3 8 𝒗 𝑤

8 3 𝒗 𝑤

3 8 𝒗 𝑤

1 2 3 8 𝒗 𝑤

1 2 (3)

4. If v has a neighbor w with 𝑐(𝑣) = 𝑐(𝑤) = 𝑔 and v > w, set c(v) to the color none of the neighbors of v has.

3 8 𝒗 𝑤

1 2 (4)

2. Proof : Lemma – Given a 𝝈(𝒏)-alternating algorithm A running in 𝑂(𝜎(𝑛)) rounds, a 3-coloring of the ring can be computed in 𝑂(𝜎(𝑛)) rounds.

Running time: 𝑶 𝝈 𝒏

Step 1: Running A- 𝑂 𝜎 𝑛

Step 2: Finding a pair of neighbors with different d - 𝜎 𝑛 .

No more than 𝜎(𝑣) consecutive nodes take the same decision d(v)

since A is 𝜎(𝑛)-alternating.

Valid 3-coloring of 𝑹𝒏

Step 2: Each node 𝑣 chooses different from one of its neighbors,

1 1 0 1 1 1 1 0 1

so at most one of the neighbors of 𝑣 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

Leveraging Linial’s Locality limit

𝜎(𝑛)-alternating 𝜎(𝑛)

MaxIS approximation 𝜎 𝑛

3-coloring 𝜎(𝑛)

Therefore, there isn’t 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 𝑙𝑜𝑔∗(𝑛).

Construct a 𝑜 𝑙𝑜𝑔∗ 𝑛 -alternating algorithm running in 𝑜(𝑙𝑜𝑔∗(𝑛)) using algorithm A.

By lemma 2, a 3-coloring of the ring can be computed in 𝒐(𝒍𝒐𝒈∗(𝒏)) rounds.

This contradicts Linial’s 3-coloring lower bound.

- Discuss MDS lower bound - Compare MDS and MaxIS difficulty

Our (new) Goal:

No! We’ll show a case where MIS is easier than MDS

MDS on rings O(1)

Is MDS always easier than MIS?

How?

MDS – Minimum Dominating Set

DS MDS

MDS in rings

Taking every third node gives a minimum dominating set

𝑀 =𝑛

3

Taking every node gives a 3-approximation MDS in 1 round ∈ 𝑜 log∗ 𝑛

𝑅𝑛

𝑅𝑛

There is no 𝒍𝒐𝒈∗(𝒏) bound MDS approximation in rings! MDS approximation in rings takes 𝑶(𝟏) rounds

MDS approximation in rings

Can we compare MaxIS and MDS difficulty?

MDS

MaxIS

We saw that in 𝑅𝑛 MDS can be computed in 1 round, but MaxIS requires at least 𝑙𝑜𝑔∗(𝑛) round.

Is it always easier to compute MDS than MaxIS?

MaxIS graph family

𝐺 𝑣 𝑤

Any graph that can be constructed this way

Lemma (Local computation of a MaxIS on MaxIS Graphs): The set {𝑣 ∈ 𝑉 | |𝑁1

+(𝑣)|𝑚𝑜𝑑 2 = 1} is a MaxIS for any MaxIS Graph.

Proof – part 1:

- For 𝑣𝑖 ∈ 𝑉𝑖 𝑖 ∈ {3,4}, 𝑣𝑖 has 2|𝑁1+ 𝑣 | neighbors.

- 𝑁1+ 𝑣𝑖 = 2 𝑁1

+ 𝑣 + 1 ⇒ 𝑁1+ 𝑣𝑖 𝑖𝑠 𝑜𝑑𝑑

- For 𝑣𝑖 ∈ 𝑉𝑖 𝑖 ∈ {1,2}, 𝑁1+ 𝑣𝑖 = 4 𝑁1

+ 𝑣 ⇒ 𝑁1+ 𝑣𝑖 𝑖𝑠 𝑒𝑣𝑒𝑛

- {𝑣 ∈ 𝑉 | |𝑁1+(𝑣)|𝑚𝑜𝑑 2 = 1} = 𝑉3 ∪ 𝑉4

Proof – part 2:

- 𝑉3 ∪ 𝑉4 is an Independent set – according to the construction of

MaxIS graph

- (𝑣1, 𝑣3, 𝑣2, 𝑣4) forms a cycle, so for each 4 nodes as such, only 2 can be in the IS.

- MaxIS can’t be larger than 𝑉

2

- 𝑉3 ∪ 𝑉4 =𝑉

2

⇒ 𝑉3 ∪ 𝑉4 𝑖𝑠 𝑎 𝑀𝑎𝑥𝐼𝑆

Lemma (Local computation of a MaxIS on MaxIS Graphs): The set {𝑣 ∈ 𝑉 | |𝑁1

+(𝑣)|𝑚𝑜𝑑 2 = 1} is a MaxIS for any MaxIS Graph.

Conclusion – MaxIS on a MaxIS graph can be determined locally, without communication (in 𝑂(1) rounds).

MDS on MaxIS graphs

We can prove that MDS on MaxIS graphs is as efficient as in general graphs, meaning:

Ωlog 𝑛

log log 𝑛

MaxIS

MDS

Ring graphs - 𝑅𝑛 MaxIS graphs

MaxIS

MDS 𝑂(1)

Ωlog 𝑛

log log 𝑛

MDS

MaxIS 𝑂(1)

Ω log∗ 𝑛

MaxIS and MDS are not comparable in general graphs!