1. 3.4 Push-Relabel(Preflow-Push) Maximum Flow Alg.
Algorithm that allows 𝑓𝑥(𝑣)>0 (conservation of flow not maintained) before completion.
For ease of exposition, take
(i) 𝑢𝑤𝑣=𝑥𝑤𝑣=0 if 𝑣𝑤∈𝐸 and 𝑤𝑣∉𝐸 (conceptual device, don’t make actual arc)
(ii) Define auxiliary digraph 𝐺(𝑥) as before, but do not bother with parallel arcs, i. e., we put 𝑣𝑤∈𝐸(𝐺(𝑥)) iff 𝑥𝑤𝑣>0 or 𝑥𝑣𝑤<𝑢𝑣𝑤.
If 𝑣𝑤∈𝐸(𝐺(𝑥)), up to 𝑢 𝑣𝑤 =𝑢𝑣𝑤−𝑥𝑣𝑤+𝑥𝑤𝑣 can be “pushed” from 𝑣 to 𝑤 (residual capacity of 𝑣𝑤) (increase 𝑥𝑣𝑤 to 𝑢𝑣𝑤 and decrease 𝑥𝑤𝑣 to 0 )
Combinatorial Optimization 2016

2. We call 𝑥 a preflow if it satisfies 𝑓𝑥(𝑣)≥0 for all 𝑥∈𝑉∖{𝑟, 𝑠}
Definitions:
We call 𝑥 a preflow if it satisfies 𝑓𝑥(𝑣)≥0 for all 𝑥∈𝑉∖{𝑟, 𝑠}
feasible preflow if it also satisfies 0≤𝑥≤𝑢
push on 𝑣𝑤 : For pair (𝑣, 𝑤) with 𝑢 𝑣𝑤 >0 and 𝑓𝑥(𝑣)>0, pushing up to  = min ( 𝑢 𝑣𝑤 , 𝑓𝑥(𝑣)) from 𝑣 to 𝑤. (it produces new feasible preflow )
(If both 𝑣𝑤 and 𝑤𝑣 are arcs of 𝐺 and < 𝑢 𝑣𝑤 , we decrease 𝑥𝑤𝑣 first as much as possible, then increase 𝑥𝑣𝑤 )
(pushing 2 units on ab) 𝑎 𝑎
3,3
4,0
3,3
4,0 𝑟
4,1
1,0 𝑠 𝑟
4,0
1,1 𝑠
4,4
2,1
4,4
2,1 𝑏 𝑏
Combinatorial Optimization 2016

3. We call 𝑣∈𝑉∖{𝑟, 𝑠} active if 𝑓𝑥(𝑣)>0 (has positive net flow) Idea:
Definitions:
We call 𝑣∈𝑉∖{𝑟, 𝑠} active if 𝑓𝑥(𝑣)>0 (has positive net flow)
Idea:
choose an active node and push flow toward sink.
difficulties
order of selection
pushing on 𝑣𝑤, then 𝑤𝑣, ... causes infinite loop.
Possible to have situation that we cannot push toward sink, but there are still active nodes.
retreat to restore conservation of flow
to identify the direction of push, use estimate of distances to 𝑠 in 𝐺(𝑥).
Combinatorial Optimization 2016

4. for every arc 𝑣𝑤 of 𝐺(𝑥), 𝑑(𝑣)≤𝑑(𝑤)+1
We say vector 𝑑∈( 𝑍 + ∪{∞})𝑉 is a valid labelling with respect to a preflow 𝑥 if
𝑑(𝑟)=𝑛, 𝑑(𝑠)=0 (3.14)
for every arc 𝑣𝑤 of 𝐺(𝑥), 𝑑(𝑣)≤𝑑(𝑤)+1
(Recall that 𝑑𝑥(𝑣, 𝑤) denotes the number of arcs in a shortest (𝑣, 𝑤) dipath in 𝐺(𝑥). 𝑑(𝑣)= 𝑑 𝑥 (𝑣, 𝑠) satisfies all conditions except 𝑑(𝑟)=𝑛.)
(Note that ∃ 𝑘, 0<𝑘<𝑛 such that 𝑑(𝑣)≠𝑘 for all 𝑣∈𝑉.)
Valid labelling gives an estimate of the shortest distance of the node 𝑣 to 𝑠 in 𝐺(𝑥). Given an active node 𝑣, we try to push excess flow to the node closer to 𝑠 using the valid label as estimate.
However, valid labelling may not exist for some feasible preflow.
So, start with some feasible preflow and valid labelling which we can easily identify and maintain the validity of labelling.
Combinatorial Optimization 2016

5. Put 𝑥𝑒=𝑢𝑒 for all arcs 𝑒 having tail 𝑟 𝑥𝑒=0 for all other arcs 𝑒∈𝐸
Constructing initial feas. preflow and valid labelling (initialize 𝑥, 𝑑)
Put 𝑥𝑒=𝑢𝑒 for all arcs 𝑒 having tail 𝑟
𝑥𝑒=0 for all other arcs 𝑒∈𝐸
Put 𝑑(𝑟)=𝑛, 𝑑(𝑣)=0 for all other nodes 𝑣∈𝑉
(Assume each 𝑢𝑟𝑣 ≠∞ (HW))
(1,0) 𝑝 𝑎
(Initialize 𝑥, 𝑑)
(1,0)
(1,1)
(1,0)
(3,0)
(3,0) 𝑟 𝑠 6
(4,4)
(4,0)
(1,0) 𝑞 𝑏
(1,0)
Combinatorial Optimization 2016

6. (feasible preflow and valid labelling saturates a cut)
Prop 3.22: If 𝑥 is a feasible preflow and 𝑑 is a valid labelling for 𝑥, then there exists an (𝑟, 𝑠)-cut (𝑅) such that 𝑥𝑣𝑤=𝑢𝑣𝑤 for all 𝑣𝑤∈(𝑅) and 𝑥𝑣𝑤=0 for all 𝑣𝑤∈( 𝑅 ).
(feasible preflow and valid labelling saturates a cut)
(pf) There exists 𝑘, 0<𝑘<𝑛, such that 𝑑(𝑣)≠𝑘 for all 𝑣∈𝑉.
Take 𝑅={𝑣∈𝑉: 𝑑(𝑣)>𝑘}. Then 𝑟∈𝑅 and 𝑠∉𝑅.
(3.14) implies that no arc of 𝐺(𝑥) leaves 𝑅. 
Cor 3.23: If a feasible flow 𝑥 has a valid labelling, then 𝑥 is a maximum flow.
Consider the duality relation with augmenting path algorithm. (Augmenting path alg. Maintains feasible flows and stops when a saturated cut is found, while push-relabel alg. maintains saturated cuts and stops when the flow becomes feasible.)
In what sense a valid labelling gives an approximation to distances in 𝐺(𝑥)?
Lemma 3.24: For any feasible preflow 𝑥, and any valid labelling 𝑑 for 𝑥, we have 𝑑𝑥(𝑣,𝑤)≥𝑑 𝑣 −𝑑(𝑤), for all 𝑣, 𝑤∈𝑉.
(pf) If 𝑑𝑥(𝑣, 𝑤)=∞, true. So suppose 𝑑𝑥(𝑣, 𝑤) finite.
Consider any shortest (𝑣, 𝑤) dipath in 𝐺(𝑥). Add inequality 𝑑(𝑝)−𝑑(𝑞)≤1 on the arcs 𝑝𝑞 on the dipath. 
Combinatorial Optimization 2016

7. 𝑑(𝑣) is a lower bound on 𝑑𝑥(𝑣, 𝑠) 𝑑 𝑣 −𝑛 is a lower bound on 𝑑𝑥(𝑣, 𝑟)
Observations:
𝑑(𝑣) is a lower bound on 𝑑𝑥(𝑣, 𝑠)
𝑑 𝑣 −𝑛 is a lower bound on 𝑑𝑥(𝑣, 𝑟)
If 𝑑(𝑣)≥𝑛, it means 𝑑𝑥(𝑣, 𝑠)=∞, and excess flow at 𝑣 should be moved toward the source 𝑟.
Think 𝑑(𝑣) as height of node 𝑣 in a liquid flowing network using gravity.
At a node with excess flow, we try to move flow toward nodes 𝑤 having 𝑑(𝑤)<𝑑(𝑣).
𝑑(𝑤)<𝑑(𝑣) and 𝑣𝑤 is an arc of 𝐺(𝑥) ⇒ 𝑑 𝑤 =𝑑 𝑣 −1 (We have 𝑑(𝑣)≤𝑑(𝑤)+1 for a valid labelling)
⇒ push the preflow only on the arc 𝑣𝑤 in 𝐺(𝑥) with 𝑑(𝑣)=𝑑(𝑤)+1.
(admissible arc)
Note that push does not affect the validity of the labelling (a possible new arc 𝑤𝑣 in 𝐺(𝑥) satisfies 𝑑(𝑤)≤𝑑(𝑣)+1.)
If 𝑣 active, but no 𝑣𝑤 in 𝐺(𝑥) with 𝑑(𝑣)=𝑑(𝑤)+1
⇒ increase 𝑑(𝑣) to min (𝑑(𝑤)+1: 𝑣𝑤∈𝐸(𝐺(𝑥)) (labelling still valid)
; relabel
Combinatorial Optimization 2016

8. Push-Relabel Algorithm
Process 𝒗
While there exists an admissible arc 𝑣𝑤
Push on 𝑣𝑤;
If 𝑣 is active
Relabel 𝑣.
Push-Relabel Algorithm
Initialize 𝑥, 𝑑;
While 𝑥 is not a flow
Choose an active node 𝑣;
Process 𝑣.
Combinatorial Optimization 2016

9. Modification on basic form:
Maximum distance push-relabel algorithm - Choose active node 𝑣 whose distance label 𝑑(𝑣) is maximum.
Bounds on the number of operations
Thm 3.25: Push-relabel algorithm performs 𝑂( 𝑛 2 ) relabels and 𝑂(𝑚 𝑛 2 ) pushes.
Thm 3.26: The maximum distance push-relabel algorithm performs 𝑂( 𝑛 2 ) relabels and 𝑂( 𝑛 3 ) pushes.
Lemma 3.27: If 𝑥 is a preflow and 𝑤 is an active node, then there is a 𝑤,𝑟 -dipath in 𝐺(𝑥).
(pf) Let 𝑅 denote the set of nodes 𝑣 for which there is a (𝑣,𝑟)-dipath in 𝐺(𝑥). Then no arc leaves 𝑅 in 𝐺(𝑥), so 𝑥 𝛿 𝑅 =0. But suppose that we add the inequalities 𝑓 𝑥 (𝑣)≥0 for all 𝑣∈ 𝑅 . Then, we get 𝑥 𝛿 𝑅 −𝑥(𝛿 𝑅 )≥0. With 𝑥 𝛿 𝑅 =0, this implies 𝑥 𝛿 𝑅 =0, hence have 𝑓 𝑥 𝑣 =0 for all 𝑣∈ 𝑅 . That is there is no active node in 𝑅 , so active 𝑤∈𝑅, as required. 
Combinatorial Optimization 2016

10. (2) nonsaturating push: 𝑣 is no longer active: 𝑂(𝑚𝑛2)
Lemma 3.28: At every stage of the push-relabel algorithm, for every 𝑣∈𝑉, we have 𝑑 𝑣 ≤2𝑛−1. Each node is relabelled at most 2𝑛−1 times, and there are 𝑂( 𝑛 2 ) relabels in all.
(pf) Since each relabel of 𝑣 increases 𝑑(𝑣) by at least 1, the second statement follows from the first. Since only active nodes are relabelled, enough to prove the first statement for 𝑣 active. By Lemma 3.27, 𝑑 𝑥 𝑣,𝑟 ≤𝑛−1. By Lemma 3.24, 𝑑 𝑥 𝑣,𝑟 ≥𝑑 𝑣 −𝑛. Combining these two inequalities gives 𝑑 𝑣 ≤2𝑛−1. 
Pushes:
(1) saturating push : 𝑢 𝑣𝑤 ≤𝑓𝑥(𝑣), value pushed is 𝑢 𝑣𝑤 , 𝑣𝑤 leaves 𝐺(𝑥): ≤2𝑚𝑛
(2) nonsaturating push: 𝑣 is no longer active: 𝑂(𝑚𝑛2)
⇒ total 𝑂(𝑚𝑛2) pushes
(Pf): reading assignment
Combinatorial Optimization 2016

11. Lemma 3.31: The maximum distance push-relabel algorithm performs 𝑂(𝑛3) nonsaturating pushes. (⇒ total 𝑂(𝑛3) pushes)
(pf) Any nonsaturating push from a node 𝑣 makes 𝑣 inactive, and 𝑣 cannot become active again before there is a relabel, since all active nodes 𝑤 has 𝑑(𝑤)≤𝑑(𝑣). Hence, if there are 𝑛 nonsaturating pushes with no relabel, there is no active node and the algorithm terminates. Therefore, the number of nonsaturating pushes is less than 𝑛 times the number of relabels, so by Lemma 3.28, it is 𝑂( 𝑛 3 ). 
Tuncel(1994): max distance push-relabel algorithm requires only 𝑂 𝑛 2 𝑚 pushes.
Combinatorial Optimization 2016

12. (Implementation of Push-Relabel Algorithms)
For each 𝑣, we keep list 𝐿𝑣 of pairs 𝑣𝑤 such that 𝑣𝑤 or 𝑤𝑣 (or both) is an arc of 𝐺 (possible arcs in 𝐺(𝑥))
For each 𝑣𝑤 in 𝐿𝑣, keep 𝑢 𝑣𝑤 .
Keep links between 𝑣𝑤∈𝐿𝑣 and 𝑤𝑣∈𝐿𝑤 (update residual capacity on 𝑣𝑤 and 𝑤𝑣 in constant time)
For each 𝑣, keep 𝑑(𝑣) and 𝑓𝑥(𝑣)
Each relabel of node 𝑣 takes 𝑂 𝐿𝑣 :
Each push on 𝑣𝑤 takes constant time once the arc is selected for push.
Total time looking for admissible arcs and determining whether it is time to relabel?
Combinatorial Optimization 2016

13. Lemma 3.32: Suppose 𝑣∈𝑉 is active and arc 𝑣𝑤 is not admissible.
Then before 𝑣𝑤 becomes admissible, there will be a relabel of 𝑣.
(pf) If 𝑣𝑤 not admissible, then (1) 𝑑(𝑣)≤𝑑(𝑤) or (2) 𝑢 𝑣𝑤 =0.
In the case of (2), need push on 𝑤𝑣 before 𝑣𝑤 becomes admissible, which needs 𝑑(𝑤)=𝑑(𝑣)+1, i.e. 𝑑(𝑣)≤𝑑(𝑤). Hence need relabel of 𝑣 before 𝑣𝑤 becomes admissible. 
Processing of 𝑣 results in (1) relabelling of 𝑣 or (2) push that makes 𝑣 inactive.
If case (2), when we process node 𝑣 again, we do not need to push on the arcs that became inadmissible earlier by Lemma 3.32.
Hence, remember the current element of 𝐿𝑣 where we stopped scanning, and start from that point when we process node 𝑣 again.
⇒ relabel node 𝑣 when and only when a traversal of 𝐿𝑣 is completed.
Combinatorial Optimization 2016

14. => Total time spent looking for admissible arcs is
𝑣 can be relabeled at most 2𝑛−1 times (Lemma 3.28), hence only need 𝑂(𝑛) traversal of 𝐿𝑣.
=> Total time spent looking for admissible arcs is
𝑂(Σ(𝑛| 𝐿 𝑣 |: 𝑣∈𝑉))=𝑂(𝑛𝑚)
Total time spent on relabel is the same, since each relabel of 𝑣
takes 𝑂(|𝐿𝑣|).
Let 𝑁 be the number of pushes.
Total time spent on choosing the next active node is 𝑂(𝑁+𝑛2) since each one leads to a push or a relabel. (For general algorithm, maintain list of active nodes)
Total time for general algorithm is 𝑂(𝑛2𝑚)
Combinatorial Optimization 2016

15. For maximum distance push-relabel algorithm, not sure if choosing the next active node can be done in constant time. (may need 𝑂(𝑛) if we use list)
Maintain doubly linked list 𝐷𝑘 for the (active) nodes 𝑣 s.t. 𝑑(𝑣)=𝑘. Also maintain pointer for node 𝑣 to its position in 𝐷𝑘.
If a node is relabelled, we remove it from its list and insert it into its new list.
If we use max distance rule, after 𝑣 is relabeled, it still has max distance.
If 𝑣 becomes inactive and 𝐷𝑘 empty, there almost always exists 𝑤 with 𝑑(𝑤)=𝑘−1 and active (unless 𝑤=𝑟), so can look at 𝐷 𝑘−1 . Some abnomalities can be handled with bounds.
Thm 3.34: Maximum distance push-relabel max flow algorithm can be implemented to run in time 𝑂(𝑛3).
Practical enhancement: Periodically (say, after a sequence of 𝑛/2 node processing steps) compute a largest valid labeling for the current preflow.
Combinatorial Optimization 2016