1. András Sebő and Anke van Zuylen
The Salesman's Improved Paths: Integrality Gap and Approximation Ratio
András Sebő and Anke van Zuylen

2. s-t Path TSP
Given: vertex set 𝑉, distance metric c: 𝑉×𝑉→ ℚ ≥0 , 𝑠, 𝑡∈𝑉 Find: shortest path from 𝑠 to 𝑡 that visits every vertex in 𝑉. Why does Christofides’s algorithm not directly give a 3/2-approximation?

3. Christofides for s-t path TSP
Find a minimum spanning tree 𝑆.
Add a 𝑇𝑆-join 𝐽, where 𝑇𝑆 is the set of vertices with the wrong degree parity in 𝑆
𝑠,𝑡∈ 𝑇 𝑆 if 𝑠,𝑡 has even degree in 𝑆
𝑣∈𝑇𝑆 if 𝑣∈𝑉\{𝑠,𝑡} has odd degree in 𝑆
Analysis:
𝑐 𝑆 ≤𝑂𝑃𝑇
𝑐 𝐽 ≤ 1 2 𝑂𝑃𝑇
Does not hold for s-t path TSP

4. LP based analysis Subtour elimination LP 𝑇𝑆-join polyhedron
Feasible region is a subset of the spanning tree polytope  𝑐 𝑆 ≤𝑂𝑃𝑇𝐿𝑃
Subtour elimination LP
𝑇𝑆-join polyhedron
Min 𝑐𝑇𝑥
s.t. 𝑥 𝛿 𝑈 ≥2, for all 𝑈, 𝑈∩{𝑠,𝑡} even
𝑥 𝛿 𝑈 ≥1, for all 𝑈, 𝑈∩{𝑠,𝑡} = 1
0≤𝑥 𝑒 ≤1, for all 𝑒
𝑥∗/2 will only potentially violate constraints for 𝑈 such that 𝑈∩{𝑠,𝑡} =1, and
|𝑈∩𝑇𝑆|= odd
(i.e., if 𝑺 has an even number of edges in 𝜹(𝑼))
Min 𝑐𝑇𝑦
s.t. 𝑦 𝛿 𝑈 ≥2, for all 𝑈, 𝑈∩𝑇𝑆 odd
𝑦 𝑒 ≥0, for all 𝑒

5. Conclusion: no 3/2 because…
𝑥∗/2 will violate the 𝑇𝑆-join constraints for 𝑠-𝑡 cuts 𝑄 such that
𝑆 has an even number of edges in 𝑄, and
𝑄 is “narrow”: 𝑥 ∗ 𝑄 <2
Will show next: Simple proof of 8/5 ratio [Sebő ’13] for “Best-of-Many-Christofides” (BOMC)

6. S denotes also the incidence vector of the edge set S
BOMC
[An, Kleinberg, Shmoys’12]
S denotes also the incidence vector of the edge set S
Randomized Algorithm:
Let 𝑥 ∗ be an optimal solution to the subtour elimination LP.
Express 𝑥 ∗ as a convex combination of spanning trees: 𝑥 ∗ = 𝑆 𝜆 𝑆 𝑆
Let 𝒮 be a random spanning tree, 𝑃 𝒮=𝑆 = 𝜆 𝑆
Add a 𝑇 𝒮 -join 𝒥 𝒮
BOMC: Derandomize this algorithm (by iterating over all trees with 𝜆 𝑆 >0) and output the cheapest solution

7. Bounding 𝔼[𝑐 𝒥 𝒮 ]
Let 𝒫𝑠𝑡 be the (incidence vector of the) 𝑠−𝑡 path in 𝒮, and 𝑝 ∗ = 𝔼 𝒫𝑠𝑡
Then, 𝒮\ 𝒫 𝑠𝑡 is a 𝑇 𝒮 -join, so
𝔼 𝑐 𝒥 𝒮 ≤𝔼 𝒮\ 𝒫 𝑠𝑡 = 𝑐(𝑥 ∗ )−𝑐( 𝑝 ∗ )
Next, we will use 𝑝 ∗ (in two ways) to complete 𝑥 ∗ 2 to a (fractional) 𝑇𝒮-join

8. Completing 𝑥 ∗ 2 to a 𝑇𝑆-join
Option 1: “One Size Fits All”
𝑥 ∗ 𝑝 ∗ 2 is in the T-join polyhedron for any T of even size
Option 2: “Tailored Completion”
Natural way to assign 𝑝 ∗ to s-t cuts 𝑄:
𝑝 𝑄 𝑒 =𝑃(𝒮∩𝑄={𝑒})
Need to complete only on narrow cuts 𝑄 such that |𝑆∩𝑄| even: 𝑃(|𝒮∩𝑄| even)≤ 𝑃 𝒮∩𝑄 ≥2 ≤ 𝑥 ∗ 𝑄 −1
Note that this implies
𝑒∈𝑄 𝑝 𝑄 𝑒 =𝑃 𝒮∩𝑄 =1 ≥2− 𝑥 ∗ (𝑄)
So 𝑥 ∗ 𝑝 𝑄 2 is at least 1 on 𝑄

9. Completing 𝑥 ∗ 2 to a 𝑇𝑆-join
Combine the two options by replacing this ½ by 𝛾
Option 1: “One Size Fits All”
𝔼 𝒥 𝒮 ≤ 𝑥 ∗ 𝑝 ∗
Option 2: “Tailored Completion”
𝔼 𝒥 𝒮 ≤ 𝑥 ∗ ∑𝑄 𝑝 𝑄 ( 𝑥 ∗ Q −1)
P(|𝑄∩𝒮| even)

10. The parity correction vector
“Basic Parity Correction”
𝑥 ∗ 2 +𝛾 𝑝 ∗
“Parity Completion”
∑𝑄 𝑝 𝑄 1− 𝑥 ∗ 𝑄 2 −𝛾 2− 𝑥 ∗ (𝑄) ( 𝑥 ∗ Q −1)
Now, optimize over 𝛾
→ 𝛾= 1 8 .
If 𝛾= 1 8 this multiplier is at most for any 𝑄
p Q Q =P 𝑄∩𝒮 =1 ≥2− 𝑥 ∗ (𝑄)
P(|𝑄∩𝒮| even)

11. Bounding 𝔼[𝑐 𝒥 𝒮 ] Already had: 𝒮\ 𝒫 𝑠𝑡 is a 𝑇 𝒮 -join, so
𝔼 𝑐 𝒥 𝒮 ≤𝔼 𝑐 𝒮\ 𝒫 𝑠𝑡 = 𝑐(𝑥 ∗ )−𝑐( 𝑝 ∗ )
Now, we also have
𝔼 𝑐 𝒥 𝒮 ≤𝑐( 𝑥 ∗ 𝑝 ∗ 𝑝 ∗ )
Minimum of these two is at most 3/5𝑐( 𝑥 ∗ )
 𝔼 𝑐 𝒮 +𝑐 𝒥 𝒮 ≤8/5𝑐( 𝑥 ∗ )

12. How to improve? Observations:
Even when 𝒮∩𝑄 =1, we “pay” 𝑥 ∗ 2 +𝛾 𝑝 ∗ for parity correction across 𝑄
Parity completion uses
𝑝 𝑄 𝑒 =𝑃(𝒮∩𝑄={𝑒})
Idea:
Delete 𝑒 from 𝒮 if 𝒮∩𝑄={𝑒} for a narrow cut 𝑄 to “save” for later
𝑒 is a lonely edge (in 𝑆 in 𝑄) if 𝑆∩𝑄={𝑒} and 𝑥 ∗ 𝑄 <2

13. Best-of-Many With Deletion
Let 𝑥 ∗ be an optimal solution to the subtour elimination LP.
Express 𝑥 ∗ as a convex combination of spanning trees: 𝑥 ∗ = 𝑆 𝜆 𝑆 𝑆
Let 𝒮 be a random spanning tree, 𝑃 𝒮=𝑆 = 𝜆 𝑆
Let ℱ(𝒮) be the forest after deleting the lonely edges from 𝒮
Add a 𝑇 ℱ(𝒮) -join 𝒥 ℱ(𝒮)

14. Game Plan for the Analysis
Bound 𝒥 ℱ(𝒮)
Note that ℱ(𝒮)⊎ 𝒥 ℱ(𝒮) may not be connected
Find the culprit (“bad” edges in 𝒥 ℱ(𝒮) )
Bound the probability of bad edges by choosing the convex combination that defines 𝒮 carefully
Bound the cost of reconnecting ℱ(𝒮)⊎ 𝒥 ℱ(𝒮) with a doubled spanning tree

15. Bounding 𝒥 ℱ(𝒮) Basic Parity Correction 𝑥 ∗ 2 +𝛾 𝑝 ∗
𝑥 ∗ 2 +𝛾 𝑝 ∗
Parity Completion for 𝑄 s.t. ℱ(𝒮)∩𝑄 ≥2, even
𝑄:|ℱ(𝒮)∩𝑄|≥2 𝑝 𝑄 1− 𝑥 ∗ 𝑄 2 −𝛾 2− 𝑥 ∗ (𝑄)
Parity Completion for 𝑄 s.t. ℱ(𝒮)∩𝑄 =0
𝑄: ℱ 𝒮 ∩𝑄 =0 𝜒 (𝒮∩𝑄) 1− 𝑥 ∗ 𝑄 2 −𝛾

16. Bounding 𝔼[ 𝒥 ℱ(𝒮) ] Basic Parity Correction 𝑥 ∗ 2 +𝛾 𝑝 ∗
𝑥 ∗ 2 +𝛾 𝑝 ∗
Parity Completion for 𝑄 s.t. ℱ(𝒮)∩𝑄 ≥2, even
𝑄:|ℱ(𝒮)∩𝑄|≥2 𝑝 𝑄 1− 𝑥 ∗ 𝑄 2 −𝛾 2− 𝑥 ∗ (𝑄)
Parity Completion for 𝑄 s.t. ℱ(𝒮)∩𝑄 ≥0
𝑄:|ℱ(𝒮)∩𝑄|≥2 𝜒(𝒮∩𝑄) 1− 𝑥 ∗ 𝑄 2 −𝛾
𝑄 𝑥 ∗ 𝑄 −1
𝑄 𝑝 𝑄

17. Problem: ℱ(𝒮)⊎ 𝒥 ℱ(𝒮) may not be connected
Bad edge: edge crossing more than one “lonely” cut
Can we prevent that the 𝑇 𝐹(𝑆) -join has bad edges?
Dashed: lonely edges removed from 𝑆
Red: 𝑇 𝐹(𝑆) -join

18. Bounding 𝒥 ℱ(𝒮) Basic Parity Correction 𝑥 ∗ 2 +𝛾 𝑝 ∗
May contribute bad edges
Bounding 𝒥 ℱ(𝒮)
Contribute no bad edges
Basic Parity Correction
𝑥 ∗ 2 +𝛾 𝑝 ∗
Parity Completion for 𝑄 s.t. ℱ(𝒮)∩𝑄 ≥2, even
𝑄:|ℱ(𝒮)∩𝑄|≥2 𝑝 𝑄 1− 𝑥 ∗ 𝑄 2 −𝛾 2− 𝑥 ∗ (𝑄)
Parity Completion for 𝑄 s.t. ℱ(𝒮)∩𝑄 =0
𝑄: ℱ 𝒮 ∩𝑄 =0 𝜒 (𝒮∩𝑄) 1− 𝑥 ∗ 𝑄 2 −𝛾

19. Bounding 𝒥 ℱ(𝒮) Basic Parity Correction 𝑥 ∗ 2 +𝛾 𝑝 ∗
How about 𝑝 𝑄 ?
Basic Parity Correction
𝑥 ∗ 2 +𝛾 𝑝 ∗
Parity Completion for 𝑄 s.t. ℱ(𝒮)∩𝑄 ≥2, even
𝑄:|ℱ(𝒮)∩𝑄|≥2 𝑝 𝑄 1− 𝑥 ∗ 𝑄 2 −𝛾 2− 𝑥 ∗ (𝑄)
Parity Completion for 𝑄 s.t. ℱ(𝒮)∩𝑄 =0
𝑄: ℱ 𝒮 ∩𝑄 =0 𝜒 (𝒮∩𝑄) 1− 𝑥 ∗ 𝑄 2 −𝛾

20. No bad edges through parity completion
In general, 𝑝 𝑄 may contain bad edges, but if we
choose the convex combination of 𝑥 ∗ carefully, and
slightly redefine the notion of “lonely edges” and 𝑝 𝑄 ,
… we can make sure that parity completion does not contribute bad edges

21. Layered Convex Combination
Let 𝑥 ∗ = 𝑆 𝜆 𝑆 𝑆 , and for each 𝑆, let 𝒬 𝑆 ⊆{𝑒:𝑆∩𝑄={𝑒} for some narrow cut 𝑄}.
This is a layered convex combination if
For any 𝑆 such that 𝜆 𝑆 >0, and narrow cuts 𝑄, 𝑄 ′ :
if 𝑄∈𝒬 𝑆 and 𝑥 ∗ 𝑄 ′ ≤ 𝑥 ∗ 𝑄 then also Q′∈𝒬 𝑆 .
The 𝜆-weight of trees 𝑆 such that 𝑄∈𝒬 𝑆 is at least 2− 𝑥 ∗ (𝑄) for every narrow cut 𝑄.
𝒬 𝑆 = narrow cuts that contain the lonely edges
Ensures that 𝑝 𝑄 (𝑒)=𝑃(𝒮∩𝑄={𝑒}, 𝑄∈𝒬(𝒮)) satisfies the properties we used of the original 𝑝 𝑄

22. 𝑝 𝑄 contributes no bad edges
If 𝑝 𝑄 contains a bad edge 𝑒 for 𝐹(𝑆), then 𝑒∈𝑄′ for some 𝑄 ′ ∈𝒬(𝑆).
If 𝑝 𝑄 is used for our 𝑇 𝐹(𝑆) -join, then 𝑄∩𝐹 𝑆 ≥2, even, so 𝑄∉𝒬(𝑆).
By property 1: 𝑥 ∗ 𝑄 > 𝑥 ∗ (𝑄′).
𝑝 𝑄 𝑒 >0, there exists 𝑆′ with 𝜆 𝑆′ >0 such that 𝑆 ′ ∩𝑄={𝑒}, 𝑄∈𝒬( 𝑆 ′ ).
But since 𝑒∈ 𝑄 ′ , then 𝑄 ′ ∉𝒬 𝑆 ′ .
By property 1: 𝑥 ∗ 𝑄′ > 𝑥 ∗ (𝑄).
Contradiction.

23. Layered Convex Combination
First introduced by [Gottschalk and Vygen, ‘15] and used in analysis of BOMC
Proof of existence in [GV15], but no polynomial time algorithm
We show: layered convex combination can be found in strongly polynomial time using Edmond’s matroid partition algorithm

24. Game Plan for the Analysis
Bound 𝒥 ℱ(𝒮)
Note that ℱ(𝒮)⊎ 𝒥 ℱ(𝒮) may not be connected
Find the culprit (“bad” edges in 𝒥 ℱ(𝒮) )
Bound the probability of bad edges by choosing the convex combination that defines 𝒮 carefully
(only 𝑥 ∗ 2 part of 𝒥 ℱ(𝒮) contributes bad edges)
Bound the cost of reconnecting ℱ(𝒮)⊎ 𝒥 ℱ(𝒮) with a doubled spanning tree

25. Reconnection Bound
Dashed: lonely edges removed from 𝑆
Red: 𝑇 𝐹(𝑆) -join (assume all come from 𝑥 ∗ 2 )
Let 𝒥 𝐹(𝑆) be a random 𝑇 𝐹(𝑆) -join (with distribution given by the fractional solution of our analysis).
Divide 𝒥 𝐹(𝑆) into bad edges (coming from 𝑥 ∗ 2 part of the solution) and good edges (all other edges).
For each bad edge 𝑒 in 𝒥 𝐹(𝑆) , put back the lonely edges (doubled) for all the lonely cuts that contain 𝑒.
For lonely cut 𝑄, the expected number of copies we put back of its lonely edge is 𝑥 ∗ (𝑄).
All but one!
𝑥 ∗ 𝑄 −1

26. Reduce Reconnection Bound by 1 per Cut: Transportation Problem
Edges 𝑒, supply 𝑥 ∗ (𝑒) Lonely cuts 𝑄, demand 1 Solution exists iff for any subset 𝒬′ of the lonely cuts, 𝑒∈𝑄∈𝒬′ 𝑥 ∗ (𝑒) ≥| 𝒬 ′ |. This condition is easily checked to be satisfied because of the subtour constraints 𝑄
(𝑒 ,𝑄) exists if 𝑒∈𝑄 𝑒

27. Game Plan for the Analysis
Bound 𝒥 ℱ(𝒮)
Note that ℱ(𝒮)⊎ 𝒥 ℱ(𝒮) may not be connected
Find the culprit (“bad” edges in 𝒥 ℱ(𝒮) )
Bound the probability of bad edges by choosing the convex combination that defines 𝒮 carefully
(only 𝑥 ∗ 2 part of 𝒥 ℱ(𝒮) contributes bad edges)
Bound the cost of reconnecting ℱ(𝒮)⊎ 𝒥 ℱ(𝒮) with a doubled spanning tree: 𝑄 𝑝 𝑄 𝑥 ∗ 𝑄 −1

28. Putting it all together
Forest:
𝔼 𝒮 − 𝑄 𝑝 𝑄 = 𝑥 ∗ − 𝑄 𝑝 𝑄
Correcting parity:
𝑥 ∗ 2 +𝛾 𝑝 ∗ + 𝑝 𝑄 1− 𝑥 ∗ 𝑄 2 −𝛾 2− 𝑥 ∗ (𝑄) 𝑥 ∗ 𝑄 −1 + 𝑄 𝑝 𝑄 1− 𝑥 ∗ 𝑄 2 −𝛾
Reconnection:
𝑄 𝑝 𝑄 𝑥 ∗ 𝑄 −1
Choose 𝛾= 1 16 , then the combined multipliers for 𝑝 𝑄 ’s become ≤0.
⇒ expected total cost is at most 3 2 𝑐 𝑥 ∗ 𝑐( 𝑝 ∗ )

29. The new ratio Already had: 𝒮⊎(𝒮\ 𝒫 𝑠𝑡 ) is a solution of cost
2𝑐(𝑥 ∗ )−𝑐( 𝑝 ∗ )
Now, we also have a solution of cost
3 2 𝑐 𝑥 ∗ 𝑐( 𝑝 ∗ )
Minimum of these two is at most 𝑐( 𝑥 ∗ )