Improved Approximation Algorithms by Generalizing the Primal-Dual Method Beyond Uncrossable Functions

Abstract

We address long-standing open questions raised by Williamson, Goemans, Vazirani and Mihail pertaining to the design of approximation algorithms for problems in network design via the primal-dual method (Williamson et al. in Combinatorica 15(3):435–454, 1995. https://doi.org/10.1007/BF01299747). Williamson et al. prove an approximation ratio of two for connectivity augmentation problems where the connectivity requirements can be specified by uncrossable functions. They state: “Extending our algorithm to handle non-uncrossable functions remains a challenging open problem. The key feature of uncrossable functions is that there exists an optimal dual solution which is laminar ... A larger open issue is to explore further the power of the primal-dual approach for obtaining approximation algorithms for other combinatorial optimization problems.” Our main result proves that the primal-dual algorithm of Williamson et al. achieves an approximation ratio of $16$ for a class of functions that generalizes the notion of an uncrossable function. There exist instances that can be handled by our methods where none of the optimal dual solutions has a laminar support. We present three applications of our main result to problems in the area of network design. (1)  A $16$-approximation algorithm for augmenting a family of small cuts of a graph G. The previous best approximation ratio was $O(\log {|V(G)|})$. (2)  A $16\cdot {\lceil k/u_{min} \rceil }$-approximation algorithm for the Cap-k-ECSS problem which is as follows: Given an undirected graph $G = (V,E)$ with edge costs $c \in {\mathbb {Q}}_{\ge 0}^E$ and edge capacities $u \in {\mathbb {Z}}_{\ge 0}^E$, find a minimum-cost subset of the edges $F\subseteq E$ such that the capacity of any cut in (V, F) is at least k; $u_{min}$ (respectively, $u_{max}$) denotes the minimum (respectively, maximum) capacity of an edge in E, and w.l.o.g. $u_{max} \le k$. The previous best approximation ratio was $\min (O(\log {|V|}), k, 2u_{max})$. (3)  A $20$-approximation algorithm for the model of (p, 2)-Flexible Graph Connectivity. The previous best approximation ratio was $O(\log {|V(G)|})$, where G denotes the input graph.

Appendices

Pliable Families: An Example

In Sect. 3, we showed an approximation ratio of O(1) for the WGMV primal-dual method applied to any $f\text {-connectivity~problem}$ where f is a pliable function satisfying property $(\gamma )$. A natural question arises: Is property $(\gamma )$ essential for this result? In other words, does the WGMV primal-dual method for the $f\text {-connectivity~problem}$ where f is a pliable function achieve an approximation ratio of O(1)? The next construction shows that the answer is no; thus, property $(\gamma )$ is essential for our results in Sect. 3.

Recall that a family of sets $\mathcal {F}\subseteq 2^V$ is called a pliable family if $A, B \in \mathcal {F}$ implies that at least two of the four sets $A\cup {B}, A\cap {B}, A\setminus {B}, B\setminus {A}$ also belong to $\mathcal {F}$.

Fig. 3

Top: A bad example for the primal-dual method for augmenting a pliable family. Bottom: Edges of type (A) and type (B) in a bad example for the primal-dual method for augmenting a pliable family

Now, we describe the construction: We have node-sets $C_{ij}$ for $i=1,2,\ldots ,k$ and $j=0,1,2,\ldots ,k$. All these C-sets are pair-wise disjoint. Additionally, for $j'=1,\ldots , k$, we have node-sets $T_{1j'} \subsetneq T_{2j'}\subsetneq \cdots T_{kj'}$ and each of these are disjoint from all the C-sets. Additionally, we have at least one node $v_0$ lying outside the union of all these C-sets and T-sets. See Fig. 3. We designate $\mathcal {F'}$ to be the following family of node-sets that consists of two types of sets:

Informally speaking, the sets $C_1,C_2,\dots ,C_k$ can be viewed as (pairwise-disjoint) “cylinders”, see Fig. 3, the (first) index i is associated with one of these cylinders, and note that $C_{i0} = C_i {\setminus } (\cup _{j=1}^k C_{ij})$; the (second) index j is associated with a “layer” (i.e., a horizontal plane), and the sub-family $T_{1j'} \subsetneq T_{2j'}\subsetneq \cdots T_{kj'}$ forms a nested family on layer $j'$, see Fig. 3. For notational convenience, let $T_{0j'}=\emptyset ,\; \forall j'\in [k]$. Observe that a set of type (II) is the union of one set $T_{i'j'}$ of the nested family of layer $j'$ together with the sets of an arbitrary sub-family of each of the “cylinder families” $\{ C_{i0}, C_{i1}, C_{i2}, \dots , C_{ik} \}$ with (first) index $i < i'$. We claim that the family $\mathcal {F'}$ satisfies the condition of Lemma 2.3. Indeed consider $A,B \in \mathcal {F'}$. If A and B are of type (I), then they are disjoint. If A is of type (I) and B is of type (II) such that $A\cap {B}\not =\emptyset$, then $A\cup B$ and $B{\setminus }{A}$ are sets of type (II) so both belong to $\mathcal {F'}$. If A and B are of type (II) such that both have the same (second) index $j'$, then both $A\cup {B}$ and $A\cap {B}$ are sets of type (II). On the other hand, if A and B are of type (II) such that the (second) index $j'$ is different for A, B, then $A\setminus {B}$ and $B\setminus {A}$ are sets of type (II). Thus, by Lemma 2.3, adding in all the complements of the sets of this family gives us a pliable family $\mathcal {F}$.

The edges of the graph are of two types:

• (A) For $i' = 1,\ldots ,k$ and $j' = 1,\ldots , k$, we have an edge from $T_{i'j'} \setminus T_{(i'-1)j'}$ to $C_{ij}$ where $i=i'$ and $j=j'$.

• (B) For $j=1,\ldots , k$, we have an edge from $v_0$ to $T_{1j}$. For $i=1,\ldots ,k$, we have an edge from $v_0$ to $C_{i0}$.

When the primal-dual algorithm is applied to this instance, then it picks all the edges of type (A); note that there are $k^2$ edges of type (A). Let us sketch the working of the primal-dual algorithm on this instance.

• Initially, the active sets are the type (I) sets $C_1,\dots ,C_k$ and the smallest T-sets $T_{11},\dots ,T_{1k}$. The algorithm increases the dual variables of each of these sets by 1/2 and then picks all the type (A) edges between $T_{1j}$ and $C_{1j}$ for $j\in [k]$.

• In the next iteration, the (new) active sets are the type (I) sets $C_2,\dots ,C_k$ and the type (II) sets of the form $T_{2j} \cup C_{1j}$ for $j\in [k]$. The algorithm increases the dual variables of each of these sets by 1/4 and then picks all the type (A) edges between $T_{2j}$ and $C_{2j}$ for $j\in [k]$.

• Similarly, in the $i^{th}$ iteration, the active sets are the type (I) sets $C_i,\dots ,C_k$ and the type (II) sets of the form $T_{ij} \cup C_{1j} \cup \dots \cup C_{(i-1)j}$ for $j\in [k]$. The algorithm increases the dual variables of each of these sets by $1/2^i$ and then picks all the type (A) edges between $T_{ij}$ and $C_{ij}$ for $j\in [k]$.

• Finally, the reverse-delete step does not delete any edge, because the edges of type (A) form an inclusion-wise minimal solution.

On the other hand, all the edges of type (B) form a feasible solution of this instance, and there are $\le 2k$ such edges. In more detail, each of the type (I) sets $C_i$ is covered by the type (B) edge between $v_0$ and $C_{i0}$, and each of the type (II) sets containing $T_{i'j'}$ is covered by the type (B) edge between $v_0$ and $T_{1j'}$. Thus, the optimal solution picks $\le 2k$ edges.

Missing Proofs from Sect. 3

This section has several lemmas and proofs from Sect. 3 that are used to prove our main result, Theorem 1.1.

Lemma B.1

Suppose $S_1$ is a witness for edge $e_1$ and $S_2$ is a witness for edge $e_2$ such that $S_1$ overlaps $S_2$. Then there exist $S_1'$ and $S_2'$ satisfying the following properties:

• (i) $S_1'$ is a valid witness for edge $e_1$, $S_2'$ is a valid witness for edge $e_2$, and $S_1'$ does not overlap $S_2'$.

• (ii) $S_1',S_2' \in \{S_1,S_2,S_1\cup S_2, S_1\cap S_2, S_1{\setminus } S_2, S_2{\setminus } S_1\}$.

• (iii) either $S_1' = S_1$ or $S_2' = S_2$.

Proof

We prove the lemma via an exhaustive case analysis. Note that at least two of the four sets $S_1\cup S_2, S_1\cap S_2, S_1{\setminus } S_2, S_2{\setminus } S_1$ must be violated in the current iteration. Moreover, observe that $e_1\in \delta _E(S_1\setminus {S_2})$ or $e_1\in \delta _E(S_1\cap {S_2})$, and in the latter case $e_1\in E(S_1\cap {S_2}, V\setminus (S_1\cup {S_2}))$ or $e_1\in E(S_1\cap {S_2}, S_2\setminus {S_1})$. We consider the following cases.

• 1. If $S_1 \cup S_2$ and $S_1 \cap S_2$ are violated or $S_1{\setminus } S_2$ and $S_2{\setminus } S_1$ are violated, then the proof of Lemma 5.2 in [5] can be applied.

• 2. Suppose $S_1 \cap S_2$ is violated, and one of $S_1\setminus {S_2}$ or $S_2\setminus {S_1}$ is violated. W.l.o.g. suppose $S_1 \cap S_2$ and $S_1\setminus {S_2}$ are violated (the other case is similar). Consider where the end-nodes of the edge $e_1$ lie. If $e_1\in \delta _E(S_1\setminus {S_2})$, then fix $S_1':=S_1\setminus {S_2}$ and $S_2':=S_2$; otherwise, $e_1\in \delta _E(S_1\cap {S_2})$, and then fix $S_1':=S_1\cap {S_2}$ and $S_2':=S_2$.

• 3. Suppose $S_1 \cup S_2$ is violated, and one of $S_1\setminus {S_2}$ or $S_2\setminus {S_1}$ is violated. W.l.o.g. suppose $S_1 \cup S_2$ and $S_1\setminus {S_2}$ are violated (the other case is similar). Consider where the end-nodes of the edges $e_1$ and $e_2$ lie. If $e_1\in \delta _E(S_1\setminus {S_2})$, then fix $S_1':=S_1\setminus {S_2}$ and $S_2':=S_2$; similarly, if $e_1\in E(S_1\cap {S_2}, V\setminus (S_1\cup {S_2}))$, then fix $S_1':=S_1\cup {S_2}$ and $S_2':=S_2$; finally, if $e_1\in E(S_1\cap {S_2}, S_2\setminus {S_1})$, then fix $S_1':=S_1$, and then if $e_2\in \delta _E(S_1\cup {S_2})$, then fix $S_2':=S_1\cup {S_2}$, otherwise, fix $S_2':=S_1\setminus {S_2}$.

This completes the proof of the lemma. $\square$

Lemma B.2

Suppose a set $A_1$ overlaps a set $A_2$ and a third set $A_3$ does not overlap $A_1$ nor $A_2$. Then $A_3$ does not overlap any of the sets $A_1\cup A_2, A_1\cap A_2, A_1 {\setminus } A_2, A_2{\setminus } A_1$.

Proof

Note that since $A_3$ does not overlap $A_1$ (or $A_2$), they are either disjoint or one contains the other. We consider the following cases.

• 1. Suppose $A_3 \cap A_1 = \emptyset$. Then $A_2\not \subseteq A_3$ since $A_1 \cap A_2 \ne \emptyset$. If $A_3 \cap A_2 = \emptyset$, then $A_3 \subseteq V{\setminus } A_1\cup A_2$ and we are done. Finally if $A_3 \subseteq A_2$, then $A_3 \subseteq A_2{\setminus } A_1$ and we are done.

• 2. Suppose $A_1 \subseteq A_3$. Then $A_3 \cap A_2 \ne \emptyset$ since $A_1\cap A_2 \ne \emptyset$. Also, $A_3\not \subseteq A_2$ since $A_1\not \subseteq A_2$. If $A_2 \subseteq A_3$, then $A_1\cup A_2 \subseteq A_3$ and we are done.

• 3. Suppose $A_3 \subseteq A_1$. Then $A_2\not \subseteq A_3$ since $A_2\setminus {A_1}\ne \emptyset$. If $A_3 \subseteq A_2$, then $A_3\subseteq A_1\cap A_2$ and we are done. Finally if $A_3\cap A_2 = \emptyset$, then $A_3 \subseteq A_1 {\setminus } A_2$ and we are done.

$\square$

Lemma 3.2. There exists a laminar family of witness sets.

Proof

We show that any witness family can be transformed to a laminar family by repeatedly applying Lemma B.1. We prove this by induction on the size of the witness family $\ell$.

Base Case: Suppose $\ell = 2$, then one application of Lemma B.1 is sufficient.

Inductive Hypothesis: If $S_1,\ldots , S_\ell$ are witness sets for edges $e_1,\ldots , e_\ell$ respectively with $\ell \le k$, then, by repeatedly applying Lemma B.1, one can construct witness sets $S'_1,\ldots , S'_\ell$ for the edges $e_1,\ldots , e_\ell$ respectively such that $S'_1,\ldots , S'_\ell$ is a laminar family.

Inductive Step: Consider $k+1$ witness sets $S_1,\ldots , S_{k+1}$. By the inductive hypothesis, we can repeatedly apply Lemma B.1 to all the witness sets $S_1,\ldots , S_k$ and obtain witness sets $S_1',\ldots , S_k'$ that form a laminar family. We now consider the following cases.

• 1. If $S_{k+1}$ does not overlap some $S_i'$, say $S_1'$, then we can apply the inductive hypothesis to the k sets $S_2',\ldots , S_k', S_{k+1}$ and we obtain a laminar family of witness sets, none of which overlap $S_1'$ either (by Lemma B.2) and so we are done.

• 2. Suppose $S_{k+1}$ overlaps all the sets $S_1',\ldots , S_k'$ and for some $S_i'$, say $S_1'$, applying Lemma B.1 to the pair $S_1',S_{k+1}$ gives $S_1', S_{k+1}'$. Then $S_1'$ does not overlap any of the witness sets $S_2',\ldots , S_{k+1}'$, hence, applying the inductive hypothesis to these k sets gives us a laminar family of witness sets $S_2^{''},\ldots , S_k^{''}$. By Lemma B.2, $S_1'$ does not overlap any of the sets $S_2^{''},\ldots , S_k^{''}$ and so we are done.

• 3. Suppose $S_{k+1}$ overlaps all the sets $S_1',\ldots , S_k'$ and, for every $S_i'$, applying Lemma B.1 to the pair $S_i', S_{k+1}$ gives $S''_i, S_{k+1}$. Then after doing this for every $S_i'$, we end up with the witness family $S_1^{''}, \ldots , S_k^{''}, S_{k+1}$ with the property that $S_{k+1}$ does not overlap any of the other sets. Applying the inductive hypothesis to the k sets $S_1^{''}, \ldots , S_k^{''}$ gives us a laminar family of witness sets $S_1^{'''},\ldots , S_k^{'''}$. By Lemma B.2, $S_{k+1}$ does not overlap any of the sets $S_1^{'''},\ldots , S_k^{'''}$ and so we are done.

$\square$

Proof of Lemma 6.1 from Sect. 6

This section has a proof of Lemma 6.1. This proof is the same as the proof we posted on Arxiv in 2022, [29].

Lemma 6.1. f is a pliable function that satisfies property $(\gamma )$. Moreover, for even p, f is an uncrossable function.

Proof

Consider two violated sets $A, B \subsetneq V$. W.l.o.g. we may assume that both A and B contain a fixed node $r \in V$. If A, B do not cross, then it is easily seen that $f(A)+f(B) = \max ( f(A\cap {B})+f(A\cup {B}),\; f(A\setminus {B})+f(B\setminus {A}) )$, hence, the inequality for pliable functions holds for A, B. Thus, we may assume that A, B cross. The following equations hold, see Frank’s book [35, Chapter 1.2]:

$$\begin{aligned}&|\delta (A)| = |\delta (B)| = p+1 \end{aligned}$$

(2)

$$\begin{aligned}&|\delta (A \cup B)| + |\delta (A \cap B)| + 2|F(A \setminus B, B \setminus A)| = |\delta (A)| + |\delta (B)| \end{aligned}$$

(3)

$$\begin{aligned}&|\delta (A \setminus B)| + |\delta (B \setminus A)| + 2|F(A \cap B, V \setminus (A \cup B))| = |\delta (A)| + | \delta (B)| \end{aligned}$$

(4)

$$\begin{aligned}&|\delta (A \setminus B)| + |\delta (A \cap B)| = |\delta (A)| + 2 |F(A \setminus B, A \cap B)| \end{aligned}$$

(5)

Since $|\delta (S)|\ge {p}, \forall \emptyset \ne {S}\subsetneq {V}$, equations (2), (3) and (4) imply that $|\delta (A \cup B)|, |\delta (A \cap B)|, |\delta (A {\setminus } B)|, |\delta (B {\setminus } A)| \in \{p,p+1,p+2\}$, and, moreover, $|F(A \setminus B, B \setminus A)| \le {1},\, |F(A \cap B, {V\setminus (A \cup B)})| \le {1}$.

Furthermore, the above four equations imply the following parity-equations (equations (6), (7) and (8) follow from equations (3), (4) and (5), respectively).

$$\begin{aligned}&|\delta (A \cup B)| \equiv |\delta (A \cap B)| \quad (\text {mod} \, 2) \end{aligned}$$

(6)

$$\begin{aligned}&|\delta (A \setminus B)| \equiv |\delta (B \setminus A)| \quad (\text {mod} \, 2) \end{aligned}$$

(7)

$$\begin{aligned}&|\delta (A\cap B)| \equiv |\delta (A\setminus B)| + |\delta (A)| \equiv |\delta (B\setminus A)| + |\delta (B)| \quad (\text {mod} \, 2) \end{aligned}$$

(8)

Case 1 (of proof of the lemma): Suppose that p is even. Then the parity-equations (6)–(8) imply that among the two pairs of cuts $\{\delta (A\cup {B}), \delta (A\cap {B})\}$ and $\{\delta (A\setminus {B}), \delta (B\setminus {A})\}$, one of the pairs consists of two $(p+1)$-cuts, and the other pair has at least one cut of size p. Since F is a feasible solution of $(p,1)\textrm{-FGC}$, every p-cut of (V, F) consists of safe edges (that is, a p-cut cannot contain any unsafe edge). Since A and B are violated sets, each of the cuts $\delta (A)$ and $\delta (B)$ contains at least two unsafe edges.

Claim C.1

Each cut $\delta (S)$ of the pair of $(p+1)$-cuts $\{\delta (A\cup {B}), \delta (A\cap {B})\}$ or $\{\delta (A\setminus {B}), \delta (B\setminus {A})\}$ (obtained from A, B) also contains at least two unsafe edges, and so S is a violated set for each of these cuts.

We prove this claim by a simple case analysis on the end-nodes of the unsafe edges of the cuts $\delta (A)$ and $\delta (B)$. W.l.o.g. assume that $\delta (A\cup {B})$ and $\delta (A\cap {B})$ are $(p+1)$-cuts and that $\delta (A\setminus {B})$ is a p-cut. The other cases are handled similarly. Clearly, $\delta (A\setminus {B})$ has no unsafe edges. Since A is a violated set, either (i) $F(A\cap B, B\setminus {A})$ has two unsafe edges or (ii) $F(A\cap B, B\setminus {A})$ has one unsafe edge and $F(A\cap B, {V\setminus (A\cup B)})$ has one unsafe edge (recall that the size of the latter edge-set is $\le 1$). Since B is a violated set, either (iii) $F(B\setminus {A}, {V\setminus (A\cup {B})})$ has two unsafe edges or (iv) $F(B\setminus {A}, {V\setminus (A\cup B)})$ has one unsafe edge and $F(A\cap B, {V\setminus (A\cup B)})$ has one unsafe edge. Now, observe that both $A\cap {B}$ and $A\cup {B}$ are violated sets, because, by (i) or (ii), $\delta (A\cap {B})$ has at least two unsafe edges, and, by (iii) or (iv), $\delta (A\cup {B})$ has at least two unsafe edges. (If both $\delta (A\setminus {B})$ and $\delta (B\setminus {A})$ are $(p+1)$-cuts, and there are no unsafe edges in either $\delta (A\cap {B})$ or $\delta (A\cup {B})$, then both $\delta (A\setminus {B})$ and $\delta (B\setminus {A})$ have $\ge 2$ unsafe edges, so both $A\setminus {B}$ and $B\setminus {A}$ are violated sets.) This proves the claim.

Therefore, when p is even, the function f is an uncrossable function (recall that every uncrossable function is a pliable function that satisfies property $(\gamma )$).

Case 2 (of proof of the lemma): Suppose that p is odd. Then we have $|\delta (A \cup B)| \equiv |\delta (A \cap B)| \equiv |\delta (A\setminus {B})| \equiv |\delta (B\setminus {A})| \quad (\text {mod} \, 2)$. Hence, the above equations (2)–(5) and parity-equations (6)–(8) imply that either all four cuts $\delta (A\cup {B}), \delta (A\cap {B}), \delta (A\setminus {B}), \delta (B\setminus {A})$ are $(p+1)$-cuts, or at least one cut from each pair $\{\delta (A\cup {B}), \delta (A\cap {B})\}$ and $\{\delta (A\setminus {B}), \delta (B\setminus {A})\}$ is a p-cut.

Claim C.2

Suppose that at least one cut from each pair $\{\delta (A\cup {B}), \delta (A\cap {B})\}$ and $\{\delta (A\setminus {B}), \delta (B\setminus {A})\}$ is a p-cut. Then either A is not a violated set, or B is not a violated set.

To prove this claim, let us assume that $|\delta (A\setminus {B})|=p$; similar arguments apply for the other case. The edge-set F is feasible for $(p,1)\textrm{-FGC}$, hence, all edges of $\delta (A\setminus {B})$ are safe. Moreover, one of the cuts $\delta (A\cap {B})$ or $\delta (A\cup {B})$ has size p and consists of safe edges. If $|\delta (A\cap {B})|=p$, then the cut $\delta (A)$ consists of safe edges (since $\delta (A)\subseteq \delta (A\cap {B})\cup \delta (A\setminus {B})$), hence, A cannot be a violated set. Otherwise, if $|\delta (A\cup {B})|=p$, then the cut $\delta (B)$ consists of safe edges (since $\delta (B)=\delta (V\setminus {B})\subseteq \delta (A\setminus {B})\cup \delta (V\setminus (A\cup {B}))$), hence, B cannot be a violated set. This proves the claim.

Claim C.2 implies that all four cuts $\delta (A\cup {B}), \delta (A\cap {B}), \delta (A\setminus {B}), \delta (B\setminus {A})$ are $(p+1)$-cuts.

Claim C.3

Consider two crossing violated sets $A, B \subseteq V$.

• (i) Then each of the four cuts $\delta (A\cup {B}), \delta (A\cap {B}), \delta (A\setminus {B}), \delta (B\setminus {A})$ has size $(p+1)$, and we have $|F(A\setminus {B}, B\setminus {A})| = 0 = |F(A\cap {B}, V\setminus (A\cup {B}))|$.

• (ii) Moreover, each of the four sets $F(A\cap {B},\,A\setminus {B}),\; F(A\cap {B},\,B\setminus {A}),\; F(A\setminus {B},\,V\setminus (A\cup {B})),\; F(B\setminus {A},\,V\setminus (A\cup {B}))$ has size $(p+1)/2$.

Part (i) of this claim follows from the above arguments (see the discussion before and after Claim C.2). Moreover, we have $|F(A {\setminus } B, B {\setminus } A)| = 0$ and $|F(A \cap B, V {\setminus } (A \cup B))| = 0$ by equations (3),(4). Next, consider part (ii) of the claim. Using the first part of the claim together with equation (5), we have $|F(A\cap {B},\,A\setminus {B})|=(p+1)/2$. Since $\delta (A\setminus {B})$ is a $(p+1)$-cut and it is the disjoint union of $F(A\setminus {B},\,A\cap {B})$ and $F(A\setminus {B},\,V\setminus (A\cup {B}))$, we have $|F(A\setminus {B},\,V\setminus (A\cup {B}))|=(p+1)/2$. Similarly, we have $|F(A\cap {B},\,B\setminus {A})|=(p+1)/2$, $|F(B\setminus {A},\,V\setminus (A\cup {B}))|=(p+1)/2$. This proves the claim.

Claim C.4

• (i) At least two of the four cuts $\delta (A\cup {B}), \delta (A\cap {B}), \delta (A\setminus {B}), \delta (B\setminus {A})$ each contain at least two unsafe edges.

• (ii) Moreover, if A is a minimal violated set, then $F(B\setminus {A},\,V\setminus (A\cup {B}))$ has at least two unsafe edges, and each of $F(A\cap {B},\,B\setminus {A})$ and $F(A\setminus {B},\,V\setminus (A\cup {B}))$ has exactly one unsafe edge.

We prove this claim by a simple case analysis on the end-nodes of the unsafe edges of the cuts $\delta (A)$ and $\delta (B)$. Observe that $|F(A {\setminus } B, B {\setminus } A)| = 0$ and $|F(A \cap B, V {\setminus } (A \cup B))| = 0$, by equations (3),(4). Each edge of $\delta (A)$ is in exactly one of the sets $F(A\cap {B},\,B\setminus {A})$ or $F(A\setminus {B},\,V\setminus (A\cup {B}))$; call these edge-sets $\phi _1, \phi _2$ for notational convenience. Similarly, each edge of $\delta (B)$ is in exactly one of the sets $F(A\cap {B},\,A\setminus {B})$ or $F(B\setminus {A},\,V\setminus (A\cup {B}))$; call these edge-sets $\phi _3, \phi _4$ for notational convenience. If two of the unsafe edges of $\delta (A)$ are in the same set (i.e., if $\phi _1$ or $\phi _2$ has two unsafe edges), then part (i) of the claim holds, since either $\delta (A\cap {B}), \delta (B\setminus {A})$ each have two (or more) unsafe edges or else $\delta (A\cup {B}), \delta (A\setminus {B})$ each have two (or more) unsafe edges. Similarly, if two of the unsafe edges of $\delta (B)$ are in the same set (i.e., if $\phi _3$ or $\phi _4$ has two unsafe edges), then part (i) of the claim holds. There is one remaining case: each of the four sets $F(A\cap {B},\,B\setminus {A})$, $F(A\setminus {B},\,V\setminus (A\cup {B}))$, $F(A\cap {B},\,A\setminus {B})$, $F(B\setminus {A},\,V\setminus (A\cup {B}))$ has exactly one unsafe edge. In this case, each of the four sets $\delta (A\cup {B}), \delta (A\cap {B}), \delta (A\setminus {B}), \delta (B\setminus {A})$ has two unsafe edges. This proves part (i) of the claim. To prove part (ii) of the claim, observe that neither of the $(p+1)$-cuts $\delta (A\cap {B})$ or $\delta (A\setminus {B})$ can have two (or more) unsafe edges, otherwise, a proper subset of the minimal violated set A would be violated. Then, the above case analysis shows that each of the two sets $F(A\cap {B},\,B\setminus {A})$, $F(A\setminus {B},\,V\setminus (A\cup {B}))$ has exactly one unsafe edge, and the set $F(B\setminus {A},\,V\setminus (A\cup {B}))$ has two (or more) unsafe edges. This proves part (ii) of the claim.

Clearly, the function f is a pliable function, by Claim C.4, part (i). Next, we argue that the function f satisfies property $(\gamma )$.

Let C be a minimal violated set and let $S_1,S_2$ be violated sets such that C crosses both $S_1,S_2$ and $S_1 \subsetneq S_2$. W.l.o.g. assume that $S_2 \setminus (S_1\cup C)$ is non-empty. Since C crosses $S_2$, the edge-set $F(S_2\setminus {C},\,V\setminus (S_2\cup {C}))$ has size $(p+1)/2$, by Claim C.3. Moreover, by Claim C.4, part (ii), this edge-set contains two unsafe edges. Observe that $S_1\cup {C}$ crosses $S_2$; to see this, note that C crosses $S_2$ so the sets $C\cap {S_2},\, C\setminus {S_2},\, V\setminus (C\cup {S_2})$ are non-empty, and, by assumption, $S_2 \setminus (S_1\cup C)$ is non-empty. Applying Claim C.3 to this pair of crossing sets, we see that the edge-set $F(S_2\setminus (S_1\cup {C}),\,V\setminus (S_2\cup {C}))$ has size $(p+1)/2$. Then we have $F(S_2\setminus {C},\,V\setminus (S_2\cup {C})) = F(S_2\setminus (S_1\cup {C}),\,V\setminus (S_2\cup {C}))$, because both edge-sets have the same size and one edge-set is a subset of the other edge-set. Hence, $F(S_2\setminus (S_1\cup {C}),\,V\setminus (S_2\cup {C}))$ contains two unsafe edges. Finally, by Claim C.3, the cut $\delta (S_2\setminus (S_1\cup {C}))$ has size $(p+1)$. Since this cut has two (or more) unsafe edges, $S_2\setminus (S_1\cup {C})$ is a violated set. This proves that the function f satisfies property $(\gamma )$.

This completes the proof of Lemma 6.1. $\square$

Optimal Dual Solutions with Non-laminar Supports

In this section, we describe an instance of the $\textrm{AugSmallCuts}$ problem where none of the optimal dual solutions (to the dual LP given in (2.2), Sect. 2) have a laminar support. Recall that the connectivity requirement function f for the $\textrm{AugSmallCuts}$ problem is pliable and satisfies property $(\gamma )$, as seen in the proof of Theorem 1.2.

Consider the graph $G = (V,E)$ (shown in Fig. 4 below using solid edges) which is a cycle on 4 nodes 1, 2, 3, 4, in that order. Edge-capacities are given by $u_{12}=3, u_{23}=4, u_{34}=2, u_{41}=1$. The link-set (shown using dashed edges) is $L=\{ 12, 23, 34, 41 \}$, which is a disjoint copy of E. Link-costs are given by $c_{12} = c_{23} = c_{34} = 1$ and $c_{41} = 2$.

Fig. 4

An instance of the $\textrm{AugSmallCuts}$ problem where every optimal dual solution has non-laminar support

Consider the $\textrm{AugSmallCuts}$ instance that arises when we choose $\widetilde{\lambda }= 6$. The family of small cuts (with capacity strictly less than $\widetilde{\lambda }$) is given by , where

The associated pliable function f satisfies $f(S) = 1$ if and only if or holds. Observe that f is not uncrossable since $f(\{1,2\}) = 1 = f(\{2,3\})$, but $f(\{1,2\} \cap \{2,3\}) = f(\{2\}) = 0$ and $f(\{2,3\} \setminus \{1,2\}) = f(\{3\}) =0$. Also note that the minimal violated set $\{2,3\}$ (w.r.t. $F = \emptyset$) crosses the violated set $\{1,2\}$.

It can be seen that there are three inclusion-wise minimal link-sets that are feasible for the above instance and these are given by

$$\begin{aligned} \mathcal {C}:= \{ \{12,23,34\}, \{12,41\}, \{34,41\} \}. \end{aligned}$$

Since each $F \in \mathcal {C}$ has cost 3, the optimal value for the instance is 3. Next, since L contains at least two links from every nontrivial cut, the vector $x\in [0,1]^L$ with $x_e=\frac{1}{2},\,\forall {e}\in {L}$ is a feasible augmentation for the fractional version of the instance, i.e., x is feasible for the primal LP given in (2.2), Sect. 2. Therefore, the optimal value of the primal LP is at most $\frac{5}{2}$.

Now, consider the dual LP, which is explicitly stated below. The dual packing-constraints are listed according to the following ordering of the links: 12, 23, 34, 41. For notational convenience, we use the shorthand $y_1$ to denote the dual variable $y_{\{1\}}$ corresponding to the set $\{1\}$. We use similar shorthand to refer to the dual variables of the other sets; thus, $y_{234}$ refers to the dual variable $y_{\{2,3,4\}}$, etc.

$$\begin{aligned}&\max&(y_{1} + y_{234})&\quad +&(y_{12} + y_{34})&\quad +&(y_{23} + y_{14})&\quad +&(y_{123} + y_{4})&\qquad&\\&\, \text {subject to:}&(y_{1} + y_{234}){} & {} &\quad +&(y_{23} + y_{14}){} & {} &\; \le 1 \; \\{} & {} {} & {} (y_{12} + y_{34}){} & {} {} & {} &\; \le 1 \; \\{} & {} {} & {} {} & {} (y_{23} + y_{14})&\quad +&(y_{123} + y_{4})&\; \le 1 \; \\{} & {} (y_{1} + y_{234})&\quad +&(y_{12} + y_{34}){} & {} &\quad +&(y_{123} + y_{4})&\; \le 2 \; \\{} & {} {} & {} {} & {} {} & {} y&\; \ge 0. \end{aligned}$$

Observe that adding all packing constraints gives , hence, the optimal value of the dual LP is at most 5/2. Moreover, a feasible dual solution with objective 5/2 must satisfy the following conditions:

$$\begin{aligned} y_{1} + y_{234} = y_{23} + y_{14} = y_{123} + y_{4} = \frac{1}{2} \quad \text { and } \quad y_{12} + y_{34} = 1. \end{aligned}$$

Clearly, there is at least one solution to the above set of equations, hence, by LP duality, the optimal value of both the primal LP and the dual LP is 5/2.

Furthermore, any optimal dual solution $y^*$ satisfies $\max (y^*_S,y^*_{V\setminus {S}}) > 0$ for all (by the above set of equations). We conclude by arguing that for any optimal dual solution $y^*$, its support is non-laminar, because some two sets cross. Since the relation A crosses B is closed under taking set-complements (w.r.t. the ground-set V), we may assume w.l.o.g. that the support contains each set in . The support of $y^*$ is not laminar because $\{1,2\}$ and $\{2,3\}$ cross.