Max flows via electrical flows (long talk)

Max Flows & Electrical Flows
Thatchaphol Saranurak
2 July 2014
1

Motivation
• Recently, there are many fast algorithms
• Max flow, Multi-commodity flow, Oblivious Routing, Sparsest Cut, Computing eigenvector,…
• Since we can solve approx. electrical flow in 𝑂𝜖 𝑚
• Example: Exact Max Flow (unit capacities)
• 𝑂 𝑚1.43 [Madry13]
• First improvement (over sparse graph case) in last ≈40 years!
• 𝑂 min{𝑚3/2, 𝑚𝑛2/3} [ET75]
Notation
• 𝑂 ⋅ hides 𝑝𝑜𝑙𝑦𝑙𝑜𝑔
• 𝑂𝜖 ⋅ also hides 𝜖
2

Goal
• Will see 2 results with simple analysis in this line of research
1. [CKMST11] 𝜖-approx. Max Flow in 𝑂𝜖 𝑚4/3
• (will focus on 𝑂𝜖 𝑚3/2 )
2. [KOSZ13] 𝜖-approx. Electrical Flow in 𝑂𝜖 𝑚
Notation
• 𝑂 ⋅ hides 𝑝𝑜𝑙𝑦𝑙𝑜𝑔
• 𝑂𝜖 ⋅ also hides 𝜖
3

Outline
Preliminaries
Multiplicative Weight Update
MWU for Max Flow
Geometry of Electrical Flow
Alternating Projection Method
4

Preliminaries
Preliminaries
MWU for Max Flow
5

Flow
• 𝐺 = (𝑉, 𝐸) undirected
• s-t Flow f: 𝐸 → ℝ
• Arbitrarily orient edges
• “directed” 𝑓 𝑎, 𝑏 = −𝑓 𝑏, 𝑎 , for 𝑒 = 𝑎, 𝑏 ∈ 𝐸
• 𝑖𝑛𝑓𝑙𝑜𝑤(𝑣) = 𝑜𝑢𝑡𝑓𝑙𝑜𝑤(𝑣), for 𝑣 ∈ 𝑉{𝑠, 𝑡}
• Formal: 𝑎:(𝑣,𝑎)∈𝐸 𝑓(𝑣, 𝑎) = 0.
• 𝑣𝑎𝑙𝑢𝑒(𝑓) = net flow into 𝑡
S T
4
4
4
3
1
1
6

Capacity & Congestion
• Capacity 𝑢: 𝐸 → ℝ+
• Congestion
• 𝑐𝑜𝑛𝑔 𝑓 𝑒 =
𝑓 𝑒
𝑢 𝑒
• 𝑐𝑜𝑛𝑔(𝑓) = max
𝑒
𝑐𝑜𝑛𝑔 𝑓 𝑒
S T
2
4/3
2
3
4/2
4/3=1.33
3/4
1/3
1/3
7

Max Flow Problem
• Solution: s-t flow 𝑓 of 𝑐𝑜𝑛𝑔 𝑓 ≤ 1
• Note: Here we consider only s-t flow just for simplicity.
• Goal: max
𝑓
𝑣𝑎𝑙𝑢𝑒(𝑓)
8

Resistance & Energy
• Resistance 𝑟: 𝐸 → ℝ+
• Energy
• ℰ 𝑓 (𝑒) = 𝑟(𝑒)𝑓(𝑒)2
• Note: called “Power” in physics: 𝑃 = 𝑅𝐼2
• ℰ 𝑓 = 𝑒 ℰ 𝑓 𝑒 = 𝑒 𝑟(𝑒)𝑓(𝑒)2
• 𝑅: a diagonal matrix with 𝑅 𝑒,𝑒 = 𝑟 𝑒 : R =
𝑟(𝑒1) ⋯ 0
⋮ ⋱ ⋮
0 ⋯ 𝑟(𝑒 𝑚)
• So ℰ(𝑓) = 𝑓 𝑇
𝑅𝑓
9

Electrical Flow Problem
• Solution: s-t flow 𝑓 of 𝑣𝑎𝑙𝑢𝑒 𝑓 = 𝐹, for a given 𝐹
• Goal: min
𝑓
ℰ(𝑓) = 𝑓 𝑇 𝑅𝑓
• Electrical Flow ≡ Min Energy Flow.
• Note: don’t care about capacity here.
10

Approx. Flow
• 𝑓 is 𝜖-approx. Max flow
• Suppose 𝑓∗
is Max flow:
• 𝜖-approx: 𝑣𝑎𝑙𝑢𝑒 𝑓 ≥ 1 − 𝜖 𝑣𝑎𝑙𝑢𝑒 𝑓∗
• Feasible: 𝑐𝑜𝑛𝑔 𝑓 , 𝑐𝑜𝑛𝑔 𝑓∗ ≤ 1
• 𝑓 is 𝜖-approx. Electrical flow (of fixed value 𝐹)
• Suppose 𝑓∗ is electrical flow:
• 𝜖-approx: ℰ(𝑓) < (1 + 𝜖)ℰ(𝑓∗)
• Feasible: s-t flow with 𝑣𝑎𝑙𝑢𝑒 𝑓 , 𝑣𝑎𝑙𝑢𝑒 𝑓∗ = 𝐹
11

High Level: Max Flow via Electrical Flow
• Black Box: approx. Electrical Flow in 𝑂𝜖(𝑚) time
Approx
Max
Flows
Approx
Electrical
Flows
𝑜 𝜖( 𝑚) turns
First method to solve approx. max flow in 𝑜𝜖(𝑚3/2) time
Oracle
12

Multiplicative Weight Update (MWU)
Preliminaries
MWU for Max Flow
13

Scenario
• 𝑛 experts and Follower gaining money in turn
• Each turn 𝑡,
1. Follower chooses dist. 𝒑(𝑡) = (𝑝1
𝑡
, … , 𝑝 𝑛
𝑡
) over experts:
“which guy to follow”
2. Oracle reveals: expert 𝑖 gain money 𝑚𝑖
(𝑡)
∈ [0, 𝜌].
(Choose 𝑝𝑖
𝑡
before knowing 𝑚𝑖
(𝑡)
)
3. Follower gains 𝒎(𝑡)
𝒑(𝑡)
= 𝑖 𝑚𝑖
(𝑡)
𝑝𝑖
𝑡
• Goal: Maximize Follower’s total gain: 𝑡 𝒎(𝑡) 𝒑(𝑡)
14

Good Strategy for Follower
• Intuition: “Give more credit to experts who gain a lot”
• 𝑤𝑖
(𝑡)
= 𝑤𝑖
𝑡−1
(1 + 𝜂𝑚𝑖
𝑡−1
)
• Init: 𝑤𝑖
(1)
= 1 for all expert 𝑖
• Not too fast: 𝜂 ≤
1
𝜌
⇒ 𝜂𝑚𝑖
𝑡
≤ 1. (𝑚𝑖
𝑡
∈ 0, 𝜌 )
• 𝒑(𝑡) ∝ 𝒘(𝑡)
• Theorem: after 𝑇 turns,
𝑡=1
𝑇
𝒎(𝑡)
𝒑(𝑡)
≥ 1 − 𝜂𝜌 𝑡=1
𝑇
𝑚𝑖
𝑡
−
ln 𝑛
𝜂
∀𝑖
15

Interpretation (Not precise)
𝑡=1
𝑇
𝒎 𝑡
𝒑 𝑡
𝑇
≥ 1 − 𝜂𝜌
𝑡=1
𝑇
𝑚𝑖
𝑡
𝑇
−
ln 𝑛
𝜂𝑇
• Set 𝜂𝜌 = 𝜖 and 𝑇 ≈
ln 𝑛
𝜂𝜖
= 𝑂(
𝜌 ln 𝑛
𝜖2 ) (can neglect
ln 𝑛
𝜂𝑇
≈ 𝜖)
• Average gain of follower≥ (1 − 𝜖) Average gain of expert 𝑖
• Average gain of expert 𝑖 ≤ (1 + 𝜖) Average gain of follower
• Look another way (will use this)
16

Optimizing without Thinking: Study history
• Take home message about MWU…
• Even if the future is completely unpredictable,
• In long run (about total gain)…
• “Blindly following the vote of experts with good records”
• Already gives you ≈ gain of the best expert.
17

MWU for Max Flow
Preliminaries
MWU for Max Flow
18

Reduce the task
• Task: find 𝑓 with 𝑣𝑎𝑙𝑢𝑒 𝑓 = 1 − 𝑂 𝜖 𝐹, 𝑐𝑜𝑛𝑔 𝑓 ≤ 1
• For fixed 𝐹
• Can find O(𝜖)-approx. Max flow 𝑓
• Spend log factor more time, by binary search on 𝐹
19

The Algorithm: Money as Congestion
• Expert 𝑖: edge 𝑒𝑖
• Follower: maintain 𝒑(𝑡) using the “Good Strategy”
• Oracle: each turn 𝑡,
• Compute: (
𝜖
3
)-approx. electrical flow 𝑓𝑡 = 𝑓𝑡(𝒑(𝑡))
• Resistance 𝑟(𝑒𝑖) depends on 𝑝𝑖
(𝑡)
(will specify)
• 𝑣𝑎𝑙𝑢𝑒(𝑓𝑡) = 𝐹
• Reveal: expert 𝑖 gain money 𝑚𝑖
(𝑡)
= 𝑐𝑜𝑛𝑔 𝑓𝑡
𝑒𝑖 , for each edge 𝑒𝑖.
20

The Algorithm: Scaled Average Flow
• After 𝑇 = 𝑂(
𝜌 ln 𝑚
𝜖2 ) turns
• 𝑓 = 𝑡=1
𝑇
𝑓𝑡 /𝑇 is an average flow
• Return 𝑓/(1 + 𝜖)2
• Runtime: 𝑂 𝑚𝑇 = 𝑂 𝑚
𝜌
𝜖2 . (Will bound 𝜌 = 𝑂( 𝑚/𝜖).)
21

Correctness
• 𝑐𝑜𝑛𝑔 𝑓 𝑒𝑖 = Average gain of expert 𝑖
• Will show: Average gain of follower≤ 1 + 𝜖
• 𝑐𝑜𝑛𝑔 𝑓 𝑒𝑖 ≤ 1 + 𝜖 2
• As, Average gain of expert 𝑖 ≤ (1 + 𝜖) Average gain of follower
• 𝑣𝑎𝑙𝑢𝑒( 𝑓) = 𝐹
• 𝑣𝑎𝑙𝑢𝑒(𝑓𝑡) = 𝐹 for all 𝑡
• So 𝑓/(1 + 𝜖)2
• congestion ≤ 1
• value =
𝐹
1+𝜖 2 = 1 − 𝑂 𝜖 𝐹
22

Clever Choice of Resistance
Remain to show
1. Average gain of follower ≤ (1 + 𝜖)
• Will show: for every turn 𝑡, gain of follower 𝒎(𝑡) 𝒑(𝑡) ≤ (1 + 𝜖)
2. Bound 𝜌 = 𝑂( 𝑚/𝜖).
• Upper bound 𝑚𝑖
𝑡
= 𝑐𝑜𝑛𝑔 𝑓𝑡
𝑒𝑖 for all 𝑖, 𝑡
• To show both statements simultaneously,
• At turn 𝑡, set
𝑟 𝑒𝑖 =
1
𝑢 𝑒𝑖
2
(𝑝𝑖
𝑡
+
𝜖
3𝑚
)
23

• ℰ 𝑓
• = 𝑖 𝑟 𝑒𝑖 𝑓 𝑒𝑖
2
• = 𝑖(
𝑓 𝑒 𝑖
𝑢 𝑒 𝑖
)2
𝑝𝑖
𝑡
+
𝜖
3𝑚
• = 𝑖 𝑐𝑜𝑛𝑔 𝑓 𝑒𝑖
2
(𝑝𝑖
𝑡
+
𝜖
3𝑚
)
24

Proof (1)
ℰ 𝑓 =
𝑖
𝑐𝑜𝑛𝑔 𝑓 𝑒𝑖
2(𝑝𝑖
𝑡
+
𝜖
3𝑚
)
• 𝑓∗ : the min congestion flow with value 𝐹
• 𝑐𝑜𝑛𝑔 𝑓∗
≤ 1 (if 𝐹 is less than value of max flow)
• ℰ 𝑓∗ ≤ 𝑒 𝑖
𝑝𝑖
𝑡
+
𝜖
3𝑚
= 1 +
𝜖
3
.
• 𝑓𝑡 : (
𝜖
3
)-approx. electrical flow with value 𝐹 at turn 𝑡
• ℰ 𝑓𝑡 ≤ 1 +
𝜖
3
ℰ 𝑓∗
≤ 1 + 𝜖
25

Proof (2)
ℰ 𝑓 =
𝑖
𝑐𝑜𝑛𝑔 𝑓 𝑒𝑖
2
(𝑝𝑖
𝑡
+
𝜖
3𝑚
)
1. Gain of follower 𝒎(𝑡) 𝒑(𝑡)
=
𝑖
𝑐𝑜𝑛𝑔 𝑓𝑡
𝑒𝑖 𝑝𝑖
𝑡
≤
𝑖
𝑒𝑖
2 𝑝𝑖
𝑡
𝑖
𝑝𝑖
𝑡
≤ ℰ 𝑓𝑡 ≤ 1 + 𝜖
2. Bound 𝜌.
𝑒𝑖
2
𝜖
3𝑚
≤ ℰ 𝑓𝑡 ≤ 1 + 𝜖 ⟹ 𝑐𝑜𝑛𝑔 𝑓𝑡
𝑒𝑖 ≤ 3(1 + 𝜖) ∙ 𝑚/𝜖
26

Total Runtime: 𝑂𝜖(𝑚1.5)
• Each turn,
• Set resistance 𝑟 𝑒𝑖 =
1
𝑢 𝑒 𝑖
2 (𝑝𝑖
𝑡
+
𝜖
3𝑚
)
• Oracle computes (
𝜖
3
)-approx. electrical flow 𝑓𝑡
• Follower updates 𝒑(𝑡)
• After 𝑇 turns, return scaled average flow 𝑓/(1 + 𝜖)2
• Runtime: 𝑂 𝑚𝑇 = 𝑂
𝑚1.5
𝜖2.5
• 𝑇 = 𝑂
𝜌
𝜖2 = 𝑂
𝑚
𝜖2.5
27

Improvement: 𝑂𝜖(𝑚4/3)
• Each turn 𝑡,
• 𝑐𝑜𝑛𝑔 𝑓𝑡
𝑒 > Ω 𝜖(𝑚1/3
) ⟹ delete 𝑒 from graph and recompute 𝑓𝑡.
• Now, 𝜌 ≤ 𝑂𝜖(𝑚1/3)
• Can show: need to recompute at most 𝑁 = 𝑂𝜖(𝑚1/3) times
• Runtime: 𝑂 𝑚(𝑇 + 𝑁) = 𝑂𝜖 𝑚(𝜌 + 𝑁) = 𝑂𝜖 𝑚4/3
28

Preliminaries
MWU for Max Flow
29

Inner product space
• Recall: R =
𝑟(𝑒1) ⋯ 0
⋮ ⋱ ⋮
0 ⋯ 𝑟(𝑒 𝑚)
is a resistance matrix
• 𝑥, 𝑦 𝑅 = 𝑥 𝑇
𝑅𝑦 is inner product because
• Symmetric: 𝑥, 𝑦 𝑅 = 𝑦, 𝑥 𝑅
• Linear: 𝑥 + 𝑦, 𝑧 𝑅 = 𝑥, 𝑧 𝑅 + 𝑦, 𝑧 𝑅 and 𝛼𝑥, 𝑦 𝑅 = 𝛼 𝑥, 𝑦 𝑅
• Positive def.: 𝑥, 𝑥 𝑅 ≥ 0 and 𝑥, 𝑥 𝑅 = 0 ⇒ 𝑥 = 0
• Norm (size): 𝑥 𝑅 = 𝑥, 𝑥 𝑅
• Think as weighted Euclidean norm ⋅ 2
• Distance 𝑥, 𝑦: 𝑥 − 𝑦 𝑅
30

“Angle” is defined in inner product space
• Projection: 𝑥 𝑅 = 1 ⟹ 𝑥, 𝑦 𝑅 is size of “𝑦 projected to 𝑥”
• Orthogonal: when 𝑥, 𝑦 𝑅 = 0
• Pythagorean: 𝑥 𝑅
2
+ 𝑦 𝑅
2
= 𝑥 + 𝑦 𝑅
2
⟺ 𝑥, 𝑦 𝑅 = 0
31
𝑦
𝑥

Electrical Flow Problem: Geometric view
• Recall: Energy ℰ 𝑓 = 𝑓 𝑇 𝑅𝑓 = 𝑓, 𝑓 𝑅 = 𝑓 𝑅
2
• 𝐵 ∈ ℝ 𝐸×𝑉: incidence matrix 𝐵 𝑎,𝑏 ,𝑐 =
1, 𝑎 = 𝑐
−1, 𝑏 = 𝑐
0, 𝑜. 𝑤.
• 𝜒 ∈ ℝ 𝑉
: demand vector 𝜒 𝑢 =
𝐹, 𝑢 = 𝑠
−𝐹, 𝑢 = 𝑡
0, 𝑜. 𝑤.
• Task: Find a point
• closest to the origin (min 𝑓 𝑅)
• in specific affine space ( 𝐵 𝑇 𝑓 = 𝜒)
32
𝐵 𝑇
𝑓 = 𝜒
𝑓∗
0

Circulation
• “Flow without source”
• circulation c: 𝐸 → ℝ
• 𝑐 𝑎, 𝑏 = −𝑐 𝑏, 𝑎 , for 𝑒 = 𝑎, 𝑏 ∈ 𝐸,
• 𝑖𝑛𝑓𝑙𝑜𝑤(𝑣) = 𝑜𝑢𝑡𝑓𝑙𝑜𝑤(𝑣), for all 𝑣 ∈ 𝑉
• 𝐵 𝑇
𝑓 = 0
• cycle c: 𝐸 → {−10,1}
• 𝑓, 𝑓’ are s-t flows of same value ⟹ 𝑓 − 𝑓’ is circulation
• Every circulation is linear combination of cycle
33
S T
4
4
4
3
1
1
4

Kirchhoff's Potential Theorem
1. 𝑓 is electrical flow
2. ⟺ 𝑓 is orthogonal to any circulation
3. ⟺ 𝑓, 𝑐 𝑅 = 0 ∀circulation 𝑐
4. ⟺ 𝑓, 𝑐 𝑅 = 0 ∀cycle 𝑐
5. ⟺ 𝑒∈𝑐 𝑓 𝑒 𝑟(𝑒) = 0 ∀cycle 𝑐
• Note: Still not easy to verify (maybe exponentially many cycles)
• 2,3,4,5 are equivalent
• Next slide: show (1⇒5) & (2⇒1)
34
𝐵 𝑇 𝑓 = 𝜒
𝐵 𝑇 𝑓 = 0
𝑓∗
0

Kirchhoff's Potential Theorem (Proof)
• 𝑓∗ be the electrical flow.
• (1⇒5) 𝑓𝑐 𝜖 : obtained from 𝑓∗ by increasing flow around cycle 𝑐 by 𝜖
• 𝑓𝑐 0 = 𝑓∗ is optimal ⟹
𝑑ℰ 𝑓𝑐 𝜖
𝑑𝜖
= 0 at 𝜖 = 0
• ⟹ 2 𝑒∈𝑐 𝑓∗ 𝑒 𝑟(𝑒) = 0
• (2⇒1) 𝑓 not elec. flow: ℰ 𝑓 > ℰ 𝑓∗
• 0 < ℰ 𝑓 − ℰ 𝑓∗ = 𝑓, 𝑓 𝑅 − 𝑓∗, 𝑓∗
𝑅 + 𝑓∗, 𝑓∗ − 𝑓 𝑅
0
= 𝑓 − 𝑓∗, 𝑓 𝑅
35

Kirchhoff's Potential Theorem: compact form
• T: spanning tree
• For any e = (a, b) ∈ 𝐸T,
• 𝑃𝑒: unique path in 𝑇 connecting 𝑎 and 𝑏
• 𝑐 𝑒 = 𝑃𝑒 ∪ 𝑒 : a cycle
• 𝑓 is electrical flow ⟺ 𝑓, 𝑐 𝑒 𝑅 = 0 ∀e ∈ 𝐸T
• Proof idea:
• Cycles in graph form an 𝔽2-vector space.
• {𝑐 𝑒 ∣ 𝑒 ∈ 𝐸𝑇} is basis of that 𝔽2-vector space
• hence in our ℝ-vector space as well
36
𝑎
𝑏
𝑒
𝑃𝑒

• Fix spanning tree 𝑇
• 𝑆 𝑒 = {𝑓 ∣ 𝑓, 𝑐 𝑒 𝑅 = 0} : hyper-plane
• for e ∈ 𝐸T
• 𝑓 is electrical flow ⟺ 𝑓, 𝑐 𝑒 𝑅 = 0 ∀e ∈ 𝐸T
• So, the electrical flow 𝑓∗
of value 𝐹
• = ( e∈𝐸T 𝑆 𝑒) ∩ {𝑓 ∣ 𝐵 𝑇 𝑓∗ = 𝜒}
37
𝑓∗
{𝑓| 𝑓, 𝑐1 𝑅 = 0}
{𝑓| 𝑓, 𝑐2 𝑅 = 0}

How to find approx. Electrical Flow in near linear time
38
Preliminaries
MWU for Max Flow

Overview
• 𝑓∗ = ( e∈𝐸T 𝑆 𝑒) ∩ {𝑓 ∣ 𝐵 𝑇 𝑓∗ = 𝜒}
• High-Level Idea
• Start from some flow 𝑓0 (which is not too bad)
• 𝑓𝑖+1 ← 𝑓𝑖 “projected” on some 𝑆 𝑒 (so 𝑓𝑖+1, 𝑐 𝑒 𝑅 = 0)
• Repeat…
39
𝑓∗
{𝑓| 𝑓, 𝑐1 𝑅 = 0}
{𝑓| 𝑓, 𝑐2 𝑅 = 0}𝑓0
𝑓1
𝑓2
• Question (to be answered)
1. Which spanning tree 𝑇: Low-stretch spanning tree
2. Which initial flow 𝑓0: Flow restricted in 𝑇
3. Fast projection: Link-cut tree
4. How many projections: ∼ 𝜏 𝑇 in expectation

Stretch of Edge in Tree
• T: spanning tree
• For any e = (a, b) ∈ 𝐸T,
• 𝑃𝑒: unique path in 𝑇 connecting 𝑎 and 𝑏
• 𝑐 𝑒 = 𝑃𝑒 ∪ 𝑒 : a cycle
• 𝑠𝑡 𝑒 =
𝑒′∈𝑃 𝑒
𝑟 𝑒′
𝑟𝑒
=
𝑃𝑒 𝑅
2
𝑟𝑒
• 𝑠𝑡 𝑒 + 1 =
𝑒′∈𝑃 𝑒
𝑟 𝑒′
𝑟𝑒
+
𝑟𝑒
𝑟𝑒
=
𝑒′∈𝑐 𝑒
𝑟 𝑒′
𝑟𝑒
=
𝑐 𝑒 𝑅
2
𝑟𝑒
40
𝑎
𝑏
𝑒
𝑃𝑒

Low Stretch Spanning Tree
• Stretch: 𝑠𝑡 𝑇 = 𝑒∈𝐸
𝑃𝑒 𝑅
2
𝑟𝑒
• Condition number: 𝜏 𝑇 = 𝑒∈𝐸T
𝑐 𝑒 𝑅
2
𝑟𝑒
• Almost same: 𝜏(𝑇) = 𝑠𝑡(𝑇) + Θ(𝑚)
• [Abraham Neiman ’12] Can find spanning tree 𝑇
• with st(𝑇) = 𝑂(𝑚 log 𝑛 log log 𝑛)
• in time 𝑂(𝑚 log 𝑛 log log 𝑛) (not bottleneck)
• 𝜏 𝑇 , 𝑠𝑡 𝑇 determine efficiency of the algorithm
41

Initial Flow 𝑓0
• 𝑇: Low stretch spanning tree
• 𝑓0: Unique flow restricted to 𝑇
• Can compute fast in 𝑂(𝑛) time
• Theorem: 𝑓0 − 𝑓∗
𝑅
2
≤ 𝑠𝑡(𝑇) 𝑓∗
𝑅
2
• Not too far from 𝑓∗
42

Overview
• 𝑓∗ = e∈𝐸T {𝑓 ∣ 𝑓, 𝑐 𝑒 𝑅 = 0}
• High-Level Idea
• Repeat…
43
𝑓∗
{𝑓| 𝑓, 𝑐1 𝑅 = 0}
{𝑓| 𝑓, 𝑐2 𝑅 = 0}𝑓0
𝑓1
𝑓2

Physical Intuition
• Some physical intuition to geometric quantities.
• 𝑓, 𝑐 𝑒 𝑅 = 𝑓𝑒
𝑇
𝑅𝑐 𝑒 = 𝑒′∈𝑐 𝑒
𝑓 𝑒′ 𝑟(𝑒′)
• (in physics, 𝑉 = 𝐼𝑅)
• potential across cycle 𝑐 𝑒
• 𝑐 𝑒, 𝑐 𝑒 𝑅 = 𝑐 𝑒
𝑇
𝑅𝑐 𝑒 = 𝑒′∈𝑐 𝑒
𝑟(𝑒′)
• resistance of cycle 𝑐 𝑒
• 𝑐 𝑒, 𝑐 𝑒 𝑅 = 𝑐 𝑒 𝑅
2
44
𝑎
𝑏
𝑒
𝑃𝑒

Projection to 𝑆 𝑒 ≡ Circulate around 𝑐 𝑒
• 𝑐 𝑒 =
𝑐 𝑒
𝑐 𝑒 𝑅
is normalized 𝑐 𝑒
• 𝑓′
= 𝑓 − 𝑓, 𝑐 𝑒 𝑅 𝑐 𝑒
• 𝑐 𝑒 is normal vector of 𝑆 𝑒
• 𝑓′
, 𝑐 𝑒 𝑅 = 0 ⟹ 𝑓′
∈ 𝑆 𝑒
• Circulate around cycle 𝑐 𝑒
• Amount of flow =
potential across cycle 𝑐 𝑒
resistance of cycle 𝑐 𝑒
(𝐼 =
𝑉
𝑅
)
• 𝑓, 𝑐 𝑒 𝑅 𝑐 𝑒 =
𝑓,𝑐 𝑒 𝑅
𝑐 𝑒,𝑐 𝑒 𝑅
𝑐 𝑒
45
𝑐 𝑒
𝑆 𝑒
𝑓
𝑓′

Implement Circulation around 𝑐 𝑒
• Can do in 𝑂 log 𝑛 = 𝑂(1) time
1. Update flow at 𝑒
2. Update flow at 𝑃𝑒
• 𝑃𝑒 can have Θ(𝑛) edges
• Link-Cut Tree will do (more than enough!)
• Link-Cut Tree: for maintaining value of any path 𝑃 ⊂ 𝑇 (another tree)
• Query min/max/sum of value of 𝑃
• Add/reduce value of each edge in 𝑃
• Link/cut edge of the tree itself (did not use here)
• Everything in 𝑂 log 𝑛 time
46
𝑎
𝑏
𝑒
𝑃𝑒

Overview
• 𝑓∗ = e∈𝐸T {𝑓 ∣ 𝑓, 𝑐 𝑒 𝑅 = 0}
• High-Level Idea
• Repeat…
47
𝑓∗
{𝑓| 𝑓, 𝑐1 𝑅 = 0}
{𝑓| 𝑓, 𝑐2 𝑅 = 0}𝑓0
𝑓1
𝑓2

Progress in Projection
• 𝑓𝑡+1 − 𝑓∗
𝑅
2
= 𝑓𝑡 − 𝑓∗
𝑅
2
− 𝑓𝑡+1 − 𝑓𝑡 𝑅
2
Progress≥0
• Pythagorean: 𝑓𝑡+1 − 𝑓∗
and 𝑓𝑡+1 − 𝑓𝑡 are orthogonal
• The best 𝑆 𝑒, over 𝑒 ∈ 𝐸𝑇, maximizes 𝑓𝑡+1 − 𝑓𝑡 𝑅
• Don't know how to choose a good one fast…
• Random is good enough!
• Appropriate distribution
48
𝑐 𝑒
𝑆 𝑒
𝑓𝑡
𝑓𝑡+1
𝑓∗

Random projection
• Project to 𝑆 𝑒 with prob. 𝑝 𝑒 =
𝑐 𝑒 𝑅
2
𝑟𝑒
• Recall: 𝜏 𝑇 = 𝑒∈𝐸T
𝑐 𝑒 𝑅
2
𝑟𝑒
• Oblivious: do not care where we are now
• Theorem: 𝔼 𝑔, 𝑐 𝑒 𝑅 ≥
𝑔 𝑅
2
𝜏 𝑇
for any circulation 𝑔
• Progress: 𝑓𝑡+1 − 𝑓𝑡 𝑅 = 𝑓𝑡 , 𝑐 𝑒 𝑅 − 𝑓∗
, 𝑐 𝑒 𝑅
0
= 𝑓𝑡 − 𝑓∗
, 𝑐 𝑒 𝑅
• Expect: 𝔼 𝑓𝑡+1 − 𝑓𝑡 𝑅 = 𝔼 𝑓𝑡 − 𝑓∗, 𝑐 𝑒 𝑅 ≥
1
𝜏 𝑇
𝑓𝑡 − 𝑓∗
𝑅
2
49
𝑐 𝑒
𝑆 𝑒
𝑓𝑡
𝑓𝑡+1
𝑓∗

Bound the Distance
1. 𝔼 [ 𝑓𝑡 − 𝑓∗
𝑅
2
] ≤ (1 −
1
𝜏 𝑇
) 𝑡 𝑓0 − 𝑓∗
𝑅
2
• Because 𝑓𝑡+1 − 𝑓∗
𝑅
2
= 𝑓𝑡 − 𝑓∗
𝑅
2
− 𝑓𝑡+1 − 𝑓𝑡 𝑅
2
• And 𝔼 𝑓𝑡+1 − 𝑓𝑡 𝑅 ≥
1
𝜏 𝑇
𝑓𝑡 − 𝑓∗
𝑅
2
2. 𝑓0 − 𝑓∗
𝑅
2
≤ 𝑠𝑡(𝑇) 𝑓∗
𝑅
2
• Initial flow 𝑓0 is not too far
Set 𝑡 = 𝜏 𝑇 log
𝑠𝑡 𝑇
𝜖
,
𝔼 𝑓𝑡 − 𝑓∗
𝑅
2
≤ 𝜖 𝑓∗
𝑅
2
50

Bound the Energy (distance from origin)
• Set 𝑡 = 𝜏 𝑇 log
𝑠𝑡 𝑇
𝜖
• 𝔼 𝑓𝑡 − 𝑓∗
𝑅
2
≤ 𝜖 𝑓∗
𝑅
2
• 𝑓𝑡 − 𝑓∗
𝑅
2
+ 𝑓∗
𝑅
2
= 𝑓𝑡 𝑅
2
• 𝑓∗ is orthogonal to circulation 𝑓𝑡 − 𝑓∗
𝔼[ 𝑓𝑡 𝑅
2
− 𝑓∗
𝑅
2
] = 𝔼 𝑓𝑡 − 𝑓∗
𝑅
2
≤ 𝜖 𝑓∗
𝑅
2
𝔼 [ℰ 𝑓𝑡 ] ≤ (1 + 𝜖)ℰ 𝑓∗
51

Total Runtime
1. Find low-stretch spanning tree 𝑇 in time 𝑂 𝑚 log 𝑛 log log 𝑛
2. Find 𝑓0 restricted to 𝑇 in time 𝑂(𝑛)
3. Repeat 𝜏 𝑇 log
𝑠𝑡 𝑇
𝜖
times
• 𝑓𝑖+1 ← 𝑓𝑖 “projected” on some 𝑆 𝑒 in time 𝑂(log 𝑛)
• Total runtime: 𝑂 𝜏 𝑇 log
1
𝜖
= 𝑂 𝑚 log
1
𝜖
52
𝑓∗
{𝑓| 𝑓, 𝑐1 𝑅 = 0}
{𝑓| 𝑓, 𝑐2 𝑅 = 0}𝑓0
𝑓1
𝑓2

Summary
1. Approx. Max Flow in 𝑂𝜖 𝑚1.5
(with small fix to get 𝑂𝜖 𝑚1.33..
)
2. Approx. Electrical Flow in 𝑂𝜖 𝑚
Preliminaries
MWU for Max Flow
53

Laplacian of Graphs
• 𝐺 = 𝑉, 𝐸
• For each edge 𝑒 = (𝑎, 𝑏) ∈ 𝐸 with weight 𝑤(𝑎,𝑏)
• 𝐿(𝑎,𝑏) ∈ ℝ|𝑉|×|𝑉|, 𝐿(𝑎,𝑏) =
• Laplacian of 𝐺
• 𝐿 𝐺 = 𝑒∈𝐸 𝐿 𝑒
a b
a
b
1 -1
1-1
× 𝑤(𝑎,𝑏)
58

Laplacian System
• Given 𝐿 𝐺 of some 𝐺 and 𝑏
• Find 𝑥 s.t. 𝐿 𝐺 𝑥 = 𝑏
59

Solving Laplacian System ≡ Finding Electrical Flow
• Same run time for exact solution
• Almost same run time in approx. solution
• Note: did not define “approx. solution of Laplacian System”
60

Literature Review: Max flow
Approx max flow in undir. graph
• 𝑂 𝑚𝑛
1
3 𝑝𝑜𝑙𝑦
1
𝜀
• [CKMST11] via multiplicative weight
update
• [LRS12] unweighted, via “projection”,
better 𝜀 dependency
• 𝑂(𝑚1+𝑜(1) 𝑝𝑜𝑙𝑦
1
𝜀
)
• [KLOS13]
• [Shermann13] slightly faster
• Approaches are “Dual” to each other
Exact max flow in undir. graph
• 𝑂(𝑚1.5
)
• Combinatorial alg.
• Blocking flow
• 𝑂(𝑚5/4
𝐹1/4
)
• 𝐹 is value of max flow
• [LRS12]
• 𝑜(𝑚1.5
) if 𝐹 = 𝑜(𝑚)
• 𝑂 𝑚1.43
• [Madry13]
𝑂 hides log factors
62

Literature Review: SDD Solver
• [ST04] 𝑂 𝑚 log20+ n log
1
𝜀
• first break through
• Divided into 3 papers: [ST08a], [ST08b],
[ST12] , 130+ pages
• [KMP10,11] 𝑂 𝑚 log 𝑛 log
1
𝜀
• Based on [ST04]
• Still assume high precisions arithmetic
• [KOSZ13] 𝑂 𝑚 log2 n log
1
𝜀
• [PS14] parallel solvers
• 𝑂 (𝑚 + 𝑛log3 𝑛) log
1
𝜀
work
• 𝑂 log2 𝑛 log
1
𝜀
depth
• [CKMPPRX14] 𝑂 𝑚 log0.5 n log
1
𝜀
𝑂 hides loglog factors
63

Max flows via electrical flows (long talk)

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