The Salesman's Improved Paths:
3
1
+
Integrality Gap and
2
34
Approximation Ratio
András Sebő and Anke van Zuylen
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?
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 ∈ 𝑇𝑆 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
𝑂𝑃𝑇
2
Does not hold for s-t
path TSP
LP based analysis
Feasible region is a
subset of the
spanning tree
polytope  𝑐 𝑆 ≤
𝑂𝑃𝑇𝐿𝑃
Subtour elimination LP
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
𝑇𝑆-join polyhedron
Min 𝑐𝑇𝑦
s.t. 𝑦 𝛿 𝑈 ≥ 2, for all 𝑈, 𝑈 ∩ 𝑇𝑆 odd
𝑦 𝑒 ≥ 0, 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))
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)
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 𝜆𝑆 𝑆
• Let 𝒮 be a random spanning tree, 𝑃 𝒮 = 𝑆 = 𝜆𝑆
• Add a 𝑇𝒮 -join 𝒥𝒮
BOMC: Derandomize this algorithm (by iterating over
all trees with 𝜆𝑆 > 0) and output the cheapest solution
Bounding 𝔼[𝑐 𝒥𝒮 ]
• Let 𝒫𝑠𝑡 be the (incidence vector of the) 𝑠 − 𝑡
path in 𝒮, and 𝑝∗ = 𝔼 𝒫𝑠𝑡
• Then, 𝒮\𝒫𝑠𝑡 is a 𝑇𝒮 -join, so
𝔼 𝑐 𝒥𝒮 ≤ 𝔼 𝒮\𝒫𝑠𝑡 = 𝑐(𝑥 ∗ ) − 𝑐(𝑝∗ )
• Next, we will use 𝑝∗ (in two ways) to complete
𝑥 ∗ 2 to a (fractional) 𝑇𝒮 -join
∗
Completing 𝑥 2 to a 𝑇𝑆-join
• Option 1: “One Size Fits All”
𝑥∗
2
+
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:
𝑝𝑄 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
𝑝𝑄 𝑒 = 𝑃 𝒮 ∩ 𝑄 = 1 ≥ 2 − 𝑥 ∗ (𝑄)
𝑒∈𝑄
So
𝑥∗
2
𝑝𝑄
+ 2
is at least 1 on 𝑄
∗
Completing 𝑥 2 to a 𝑇𝑆-join
• Option 1: “One Size Fits All”
𝔼 𝒥𝒮
𝑥∗ 1 ∗
≤ + 𝑝
2 2
Combine the two
options by replacing
this ½ by 𝛾
• Option 2: “Tailored Completion”
𝑥∗ 1
∗ Q − 1)
𝑄
𝔼 𝒥𝒮 ≤ +
(𝑥
2 2 𝑄𝑝
P(|𝑄 ∩ 𝒮| even)
The parity correction vector
• “Basic Parity Correction”
𝑥∗
+ 𝛾𝑝∗
2
Now, optimize over 𝛾
1
→ 𝛾= .
8
1
If 𝛾 = 8 this multiplier is
• “Parity Completion”
1
at
most
for any 𝑄
∗
8
𝑥 𝑄
1−
−𝛾
2
∗
𝑄
(𝑥
Q − 1)
𝑄𝑝
∗
2 − 𝑥 (𝑄)
pQ Q = P 𝑄 ∩ 𝒮 = 1 ≥ 2 − 𝑥 ∗ (𝑄)
P(|𝑄 ∩ 𝒮| even)
Bounding 𝔼[𝑐 𝒥𝒮 ]
• Already had: 𝒮\𝒫𝑠𝑡 is a 𝑇𝒮 -join, so
𝔼 𝑐 𝒥𝒮 ≤ 𝔼 𝑐 𝒮\𝒫𝑠𝑡 = 𝑐(𝑥 ∗ ) − 𝑐(𝑝∗ )
• Now, we also have
𝔼 𝑐 𝒥𝒮
𝑥∗ 1 ∗ 1 ∗
≤ 𝑐( + 𝑝 + 𝑝 )
2 8
8
Minimum of these two is at most 3/5𝑐(𝑥 ∗ )
 𝔼 𝑐 𝒮 + 𝑐 𝒥𝒮 ≤ 8/5𝑐(𝑥 ∗ )
How to improve? Observations:
𝑥∗
2
• Even when 𝒮 ∩ 𝑄 = 1, we “pay” + 𝛾𝑝∗ for
parity correction across 𝑄
• Parity completion uses
𝑝𝑄 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
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 𝜆𝑆 𝑆
• Let 𝒮 be a random spanning tree, 𝑃 𝒮 = 𝑆 =
𝜆𝑆
• Let ℱ(𝒮) be the forest after deleting the
lonely edges from 𝒮
• Add a 𝑇ℱ(𝒮) -join 𝒥ℱ(𝒮)
Game Plan for the Analysis
• Bound 𝒥ℱ(𝒮)
• Note that ℱ(𝒮) ⊎ 𝒥ℱ(𝒮) may not be connected
– Find the culprit (“bad” edges in 𝒥ℱ(𝒮) )
– Bound the probability of bad edges by choosing the
convex combination that defines 𝒮 carefully
• Bound the cost of reconnecting ℱ(𝒮) ⊎ 𝒥ℱ(𝒮)
with a doubled spanning tree
Bounding 𝒥ℱ(𝒮)
• Basic Parity Correction
𝑥∗
+ 𝛾𝑝∗
2
• Parity Completion for 𝑄 s.t. ℱ(𝒮) ∩ 𝑄 ≥ 2, even
𝑥∗ 𝑄
1−
−𝛾
2
𝑝𝑄 2 − 𝑥 ∗ (𝑄)
𝑄:|ℱ(𝒮)∩𝑄|≥2
• Parity Completion for 𝑄 s.t. ℱ(𝒮) ∩ 𝑄 = 0
𝑥∗ 𝑄
𝜒(𝒮∩𝑄) 1 −
−𝛾
2
𝑄: ℱ 𝒮 ∩𝑄 =0
Bounding 𝔼[𝒥ℱ(𝒮) ]
• Basic Parity Correction
𝑥∗
+ 𝛾𝑝∗
2
• Parity Completion for 𝑄 s.t. ℱ(𝒮) ∩ 𝑄 ≥ 2, even
𝑥∗ 𝑄
1−
−𝛾
2
𝑥 ∗ 𝑄 − 1 𝑝𝑄 2 − 𝑥 ∗ (𝑄)
𝑄 𝑄:|ℱ(𝒮)∩𝑄|≥2
• Parity Completion for 𝑄 s.t. ℱ(𝒮) ∩ 𝑄 ≥ 0
∗
𝑥
𝑄
𝑄
𝜒(𝒮 ∩𝑝𝑄) 1 −
−𝛾
2
𝑄
𝑄:|ℱ(𝒮)∩𝑄|≥2
Problem: ℱ(𝒮) ⊎ 𝒥ℱ(𝒮) may not be
connected
Dashed: lonely edges removed from 𝑆
Red: 𝑇𝐹(𝑆) -join
• Bad edge: edge crossing more than one “lonely”
cut
• Can we prevent that the 𝑇𝐹(𝑆) -join has bad
edges?
May contribute
bad edges
Bounding 𝒥ℱ(𝒮)
• Basic Parity Correction
𝑥∗
+ 𝛾𝑝∗
2
Contribute no bad
edges
• Parity Completion for 𝑄 s.t. ℱ(𝒮) ∩ 𝑄 ≥ 2, even
𝑥∗ 𝑄
1−
−𝛾
2
𝑝𝑄 2 − 𝑥 ∗ (𝑄)
𝑄:|ℱ(𝒮)∩𝑄|≥2
• Parity Completion for 𝑄 s.t. ℱ(𝒮) ∩ 𝑄 = 0
𝑥∗ 𝑄
𝜒(𝒮∩𝑄) 1 −
−𝛾
2
𝑄: ℱ 𝒮 ∩𝑄 =0
Bounding 𝒥ℱ(𝒮)
• Basic Parity Correction
𝑥∗
+ 𝛾𝑝∗
2
How about 𝑝𝑄 ?
• Parity Completion for 𝑄 s.t. ℱ(𝒮) ∩ 𝑄 ≥ 2, even
𝑥∗ 𝑄
1−
−𝛾
2
𝑝𝑄 2 − 𝑥 ∗ (𝑄)
𝑄:|ℱ(𝒮)∩𝑄|≥2
• Parity Completion for 𝑄 s.t. ℱ(𝒮) ∩ 𝑄 = 0
𝑥∗ 𝑄
𝜒(𝒮∩𝑄) 1 −
−𝛾
2
𝑄: ℱ 𝒮 ∩𝑄 =0
No bad edges through parity
completion
In general, 𝑝𝑄 may contain bad edges, but if we
– choose the convex combination of 𝑥 ∗ carefully,
and
– slightly redefine the notion of “lonely edges” and
𝑝𝑄 ,
… we can make sure that parity completion does
not contribute bad edges
Layered Convex Combination
Let 𝑥 ∗ = 𝑆 𝜆𝑆 𝑆, and for each 𝑆, let 𝒬 𝑆 ⊆ {𝑒: 𝑆 ∩ 𝑄 =
{𝑒} for some narrow cut 𝑄}.
𝒬 𝑆 = narrow cuts that
contain the lonely edges
This is a layered convex combination if
1. For any 𝑆 such that 𝜆𝑆 > 0, and narrow cuts 𝑄, 𝑄 ′ :
if 𝑄 ∈ 𝒬 𝑆 and 𝑥 ∗ 𝑄 ′ ≤ 𝑥 ∗ 𝑄 then also Q′ ∈ 𝒬 𝑆 .
2. The 𝜆-weight of trees 𝑆 such that 𝑄 ∈ 𝒬 𝑆 is at least
2 − 𝑥 ∗ (𝑄) for every narrow cut 𝑄.
𝑄
Ensures that 𝑝 (𝑒) = 𝑃(𝒮 ∩
𝑄 = {𝑒}, 𝑄 ∈ 𝒬(𝒮)) satisfies
the properties we used of
the original 𝑝𝑄
𝑄
𝑝 contributes no bad edges
• If 𝑝𝑄 contains a bad edge 𝑒 for 𝐹(𝑆), then 𝑒 ∈ 𝑄′ for some
𝑄 ′ ∈ 𝒬(𝑆).
• If 𝑝𝑄 is used for our 𝑇𝐹(𝑆) -join, then 𝑄 ∩ 𝐹 𝑆 ≥ 2, even,
so 𝑄 ∉ 𝒬(𝑆).
By property 1: 𝑥 ∗ 𝑄 > 𝑥 ∗ (𝑄′).
• 𝑝𝑄 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.
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
Game Plan for the Analysis
 Bound 𝒥ℱ(𝒮)
 Note that ℱ(𝒮) ⊎ 𝒥ℱ(𝒮) may not be connected
Find the culprit (“bad” edges in 𝒥ℱ(𝒮) )
Bound the probability of bad edges by choosing the
convex combination that defines 𝒮 carefully
(only
𝑥∗
2
part of 𝒥ℱ(𝒮) contributes bad edges)
• Bound the cost of reconnecting ℱ(𝒮) ⊎ 𝒥ℱ(𝒮)
with a doubled spanning tree
Reconnection Bound
Dashed: lonely edges removed from 𝑆
Red: 𝑇𝐹(𝑆) -join (assume all come from
Let 𝒥𝐹(𝑆) be a random 𝑇𝐹(𝑆) -join (with distribution given by the fractional
solution of our analysis).
Divide 𝒥𝐹(𝑆) into bad edges (coming from
edges (all other edges).
𝑥∗
2
part of the solution) and good
All but
one!
For each bad edge 𝑒 in 𝒥𝐹(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).
𝑥∗ 𝑄 − 1
𝑥∗
)
2
Reduce Reconnection Bound by 1 per
Cut: Transportation Problem
Edges 𝑒, supply 𝑥 ∗ (𝑒)
Lonely cuts 𝑄, demand 1
𝑄
𝑒
Solution exists iff for any subset 𝒬′ of the lonely cuts,
∗ (𝑒) ≥ |𝒬 ′ |.
𝑥
𝑒∈𝑄∈𝒬′
This condition is easily checked to be satisfied
because of the subtour constraints
Game Plan for the Analysis
 Bound 𝒥ℱ(𝒮)
 Note that ℱ(𝒮) ⊎ 𝒥ℱ(𝒮) may not be connected
 Find the culprit (“bad” edges in 𝒥ℱ(𝒮) )
 Bound the probability of bad edges by choosing the convex
combination that defines 𝒮 carefully
(only
𝑥∗
2
part of 𝒥ℱ(𝒮) contributes bad edges)
 Bound the cost of reconnecting ℱ(𝒮) ⊎ 𝒥ℱ(𝒮) with a
doubled spanning tree: 𝑄 𝑝𝑄 𝑥 ∗ 𝑄 − 1
Putting it all together
Forest:
𝔼𝒮 −
𝑄
∗
𝑝
=
𝑥
−
𝑄
𝑄
𝑝
𝑄
Correcting parity:
∗ 𝑄
𝑥
1−
−𝛾
𝑥∗
2
∗
∗ 𝑄 −1
𝑄
+ 𝛾𝑝 +
𝑥
𝑝
2
2 − 𝑥 ∗ (𝑄)
∗ 𝑄
𝑥
+
𝑝𝑄 1 −
−𝛾
2
𝑄
Reconnection:
𝑝𝑄 𝑥 ∗ 𝑄 − 1
𝑄
1
Choose 𝛾 = , then the
16
combined multipliers for
𝑝𝑄 ’s become ≤ 0.
⇒ expected total cost is at
3
1
most 𝑐 𝑥 ∗ + 𝑐(𝑝∗ )
2
16
The new ratio
• Already had: 𝒮 ⊎ (𝒮\𝒫𝑠𝑡 ) is a solution of cost
2𝑐(𝑥 ∗ ) − 𝑐(𝑝∗ )
• Now, we also have a solution of cost
3
1
∗
𝑐 𝑥 + 𝑐(𝑝∗ )
2
16
Minimum of these two is at most
1
2
+
1
34
𝑐(𝑥 ∗ )