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A Min-Max Theorem for Transversal Submodular Functions and Its Implications
Satoru Fujishige
Research Institute for Mathematical Sciences Kyoto University, Japan 18th Aussois Workshop on Combinatorial Optimization Aussois, January 6–10, 2014 (Joint work with Shin-ichi Tanigawa, RIMS, Kyoto University)
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SLIDE 2 The present talk will
- 1. introduce a new concept of transversal submodular func-
tion, a generalization of ordinary submodular set function,
- 2. show a min-max relation between the minimum of a transver-
sal submodular function and the maximum of the negative
- f a norm composed of ℓ1 and ℓ∞ norms, and
- 3. based on the min-max relation, give a unifying view over
the recent results on generalizations of submodular set functions: [1] A. Huber and V. Kolmogorov: Towards minimizing k-submodular functions. Proceedings of ISCO 2012, LNCS 7422 (2012) 451–462. [2] F. Kuivinen: On the complexity of submodular func- tion minimisation on diamonds. Discrete Optimiza- tion 8 (2011) 459–477.
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————————————————————————– V : a nonempty finite set U ≡ {U1, U2, · · · , Un}: a partition of V
✂✁ ☎✄ ✝✆ ✞✂✞☎✞ ✟
T(⊆ V ): a subtransversal (or partial transversal) of U (|T ∩ U| ≤ 1 for all U ∈ U) T : the set of all subtransversals of U U(T) = {U ∈ U | U ∩ T ̸= ∅} (∀T ∈ T ) U(v): the unique U ∈ U that contains v ∈ V ————————————————————————– We consider a function f : T → R. →
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————————————————————————– Consider two binary operations ▽ and △ on T satisfying the condition that for all T1, T2 ∈ T T1 ▽ T2 ∈ T , U(T1 ▽ T2) ⊆ U(T1) ∪ U(T2), T1 △ T2 ∈ T , U(T1 △ T2) ⊆ U(T1) ∩ U(T2). Define a function f : T → R with f(∅) = 0 satisfying f(T1) + f(T2) ≥ f(T1 ▽ T2) + f(T1 △ T2) (∀T1, T2 ∈ T ). We call f a transversal submodular function or a t-submodular function, for short. ————————————————————————–
✂✁✄✆☎ ✞✝ ✟ ✠☛✡ ✠✌☞
→
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Example 1: k-submodular functions due to Huber and Kol- mogorov (ISCO 2012). ————————————————————————– For any T, T ′ ∈ T define binary operations ⊔ and ⊓ on T by T ⊔ T ′ = (T ∪ T ′) \ ∪ {U ∈ U | |U ∩ (T ∪ T ′)| = 2}, T ⊓ T ′ = T ∩ T ′. Let k = max{|U| | U ∈ U}. A function f : T → R is called k-submodular if f(T) + f(T ′) ≥ f(T ⊔ T ′) + f(T ⊓ T ′) (∀T, T ′ ∈ T ). We assume f(∅) = 0. ————————————————————————– →
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Example 1: k-submodular functions due to Huber and Kol- mogorov (ISCO 2012). ————————————————————————– For any T, T ′ ∈ T define binary operations ⊔ and ⊓ on T by T ⊔ T ′ = (T ∪ T ′) \ ∪ {U ∈ U | |U ∩ (T ∪ T ′)| = 2}, T ⊓ T ′ = T ∩ T ′. Let k = max{|U| | U ∈ U}. A function f : T → R is called k-submodular if f(T) + f(T ′) ≥ f(T ⊔ T ′) + f(T ⊓ T ′) (∀T, T ′ ∈ T ). We assume f(∅) = 0. ————————————————————————– Remark: Bouchet (1997) considered k-submodular functions (monotone nondecreasing and unit-increasing) to define a set system called a multimatroid as a generalization of delta- matroids. ————————————————————————– →
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SLIDE 7 Example 2: Submodular functions on product lattices and, in particular, diamonds due to Kuivinen (Discrete Optimization, 2011). ————————————————————————– 0U: a new element for each U ∈ U Put ˆ U = U ∪ {0U} for each U ∈ U.
✂✁ ✄✂✄✂✄ ☎ ✆ ✝ ☎ ✆ ✞ ☎ ✆ ✟ ✠ ✂✁ ✡ ✠ ☞☛ ✠ ✍✌ ☞☛
An arbitrary lattice LU = ( ˆ U, ∨U, ∧U) with lattice operations, join ∨U and meet ∧U, for each U ∈ U 0U: the minimum element of LU. 1U: the maximum element of LU. ————————————————————————– →
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SLIDE 8 Let L = ⊗U∈ULU(= (⊗U∈U ˆ U, ∨, ∧)) be the product of lat- tices LU = ( ˆ U, ∨U, ∧U) for U ∈ U. A function f : ⊗U∈U ˆ U → R is called a submodular function
f( ˆ T) + f( ˆ T ′) ≥ f( ˆ T ∨ ˆ T ′) + f( ˆ T ∧ ˆ T ′) for all ˆ T, ˆ T ′ ∈ ⊗U∈U ˆ U. ————————————————————————– This function can be regarded as a special case of t-submodular functions by discarding minimum elements 0U for all U ∈ U.
✂✁ ✄✂✄✂✄ ☎ ✆ ✝ ☎ ✆ ✞ ☎ ✆ ✟ ✠ ✂✁ ✡ ✠ ☞☛ ✠ ✍✌ ☞☛
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A Min-Max Theorem for T-submodular Functions ————————————————————————– Let f : T → R be a t-submodular function. Define a function F : 2U → R as follows. F(X) = min{f(T) | T ∈ T , U(T) ⊆ X} (∀X ⊆ U). ————————————————————————– Lemma 1: F : 2U → R is a submodular function on 2U with F(∅)=0. ————————————————————————– (Proof) For any X, Y ⊆ U there exist TX, TY ∈ T such that U(TX) ⊆ X, U(TY) ⊆ Y, F(X) = f(TX), F(Y) = f(TY). Hence we have F(X) + F(Y) = f(TX) + f(TY) ≥ f(TX ▽ TY) + f(TX △ TY) ≥ F(X ∪ Y) + F(X ∩ Y). We also have F(∅) = f(∅) = 0. ✷ ————————————————————————– →
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We can easily see that min{f(T) | T ∈ T } = min{F(X) | X ⊆ U}. Hence we have the following. ————————————————————————– Lemma 2: min{f(T) | T ∈ T } = max{x(U) | x ≤ 0, x ∈ P(F)}, where P(F) = {x ∈ RU | ∀X ⊆ U : x(X) ≤ F(X)}, the submodular polyhedron associated with submodular function F and x(X) = ∑
U∈X x(U).
————————————————————————– (Proof) This follows from Edmonds’ min-max theorem for submodular function minimization. ✷ ————————————————————————– It should be noted that since F is monotone non-increasing, every x ∈ P(F) is nonpositive, so that we may suppress the condition x ≤ 0 appearing in Lemma 2. →
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For any x ∈ RU define zx ∈ RV by zx(v) = x(U(v)) (∀v ∈ V ). Here it should be noted that x(U(v)) is the value of x ∈ RU for the coordinate U(v) ∈ U. ————————————————————————– Lemma 3: Suppose we are given a nonpositive x ∈ RU, i.e., x ≤ 0. Then, we have x ∈ P(F) if and only if zx ∈ P(f), where P(f) = {z ∈ RV | ∀T ∈ T : z(T)(≡ ∑
v∈T
z(v)) ≤ f(T)}. ————————————————————————– (Proof) Suppose x ∈ P(F). Then, for any T ∈ T zx(T) = x(U(T)) ≤ F(U(T)) ≤ f(T). Hence zx ∈ P(f). Conversely, suppose zx ∈ P(f) for x ∈ RU with x ≤ 0. Then, for any X ⊆ U and any T ∈ T such that U(T) ⊆ X we have x(X) ≤ x(U(T)) = zx(T) ≤ f(T), where the first inequality holds since x ≤ 0. This implies x(X) ≤ min{f(T) | T ∈ T , U(T) ⊆ X} = F(X). Hence x ∈ P(F). ✷ ————————————————————————– →
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————————————————————————– For any z ∈ RV define ||z||1,∞ =
n
∑
i=1
max
u∈Ui |z(u)|.
This defines a norm on RV , which is a composition of ℓ1 and ℓ∞ norms.
✂✁ ☎✄ ✝✆ ✞✂✞☎✞ ✟
————————————————————————– Remark: || · ||1,∞ = || · ||1 if |Ui| = 1 for all i = 1, · · · , n, and || · ||1,∞ = || · ||∞ if n = 1. →
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SLIDE 13 We are now ready to show the following. ————————————————————————– Theorem 4: For any t-submodular function f with f(∅) = 0 we have the following min-max relation. min{f(T) | T ∈ T } = max{−||z||1,∞ | z ∈ P(f)}. Moreover, if f is integer-valued, there exists an integral vector z that attains the maximum on the right-hand side. ————————————————————————– (Proof) Denote the right-hand side by RHS. It follows from Lemmas 2 and 3 that RHS = max{−||z||1,∞ | z ≤ 0, z ∈ P(f)} = max{−||zx||1,∞ | x ∈ RU, x ≤ 0, zx ∈ P(f)} = max{x(U) | x ≤ 0, x ∈ P(F)} = min{F(X) | X ⊆ U} = min{f(T) | T ∈ T }, where the first and second equalities are due to the hereditary property of polyhedron P(f) and the definition of || · ||1,∞. Moreover, if f is integer-valued, then so is the correspond- ing submodular function F : 2U → R. Therefore, there exists an integral x ∈ RU that attains the maximum on the right-hand
- side. Then zx ∈ RV defined for Lemma 3 is an integral maxi-
mizer of the right-hand side, due to Lemmas 2 and 3. ✷ ————————————————————————– →
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Consider k-submodular functions due to Huber and Kolmogorov (2012). As a corollary of Theorem 4 we get ————————————————————————– Corollary 5: For any k-submodular function f : T → R with f(∅) = 0 min{f(T) | T ∈ U} = max{−||z||1,∞ | z ∈ P(f)}. Moreover, if f is integer-valued, then there exists an integral z that attains the maximum on the right-hand side. ————————————————————————– Huber and Kolmogorov (2012) considered P2(f) = {z ∈ RV | ∀T ∈ T : z(T) ≤ f(T), ∀U ∈ U, ∀X ∈ (U 2 ) : z(X) ≤ 0}. ————————————————————————– Note that we have P(f) ∩ RV
≤0 = P2(f) ∩ RV ≤0 ⊆ P2(f) ⊆ P(f),
(1) where RV
≤0 is the set of all nonpositive vectors in RV .
→
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For polyhedron P2(f) considered by Huber and Kolmogorov (2012) we have the following. ————————————————————————– Theorem 6: For any k-submodular function f : T → R with f(∅) = 0 min{f(T) | T ∈ U} = max{−||z||1,∞ | z ∈ P2(f)}. Moreover, if f is integer-valued, then there exists an integral z that attains the maximum on the right-hand side. ————————————————————————– →
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For polyhedron P2(f) considered by Huber and Kolmogorov (2012) we have the following. ————————————————————————– Theorem 6: For any k-submodular function f : T → R with f(∅) = 0 min{f(T) | T ∈ U} = max{−||z||1,∞ | z ∈ P2(f)}. Moreover, if f is integer-valued, then there exists an integral z that attains the maximum on the right-hand side. ————————————————————————– Remark: A good characterization of minimizing k-submodular functions is open. ✷ →
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SLIDE 17 ————————————————————————– Now, consider a submodular function f on product lattice L = ⊗U∈ULU, which is identified with a function ¯ f on T defined by ¯ f(T) = f( ˆ T) (∀T ∈ T ) ( ˆ T ∈ L). We then have function ¯ f satisfying ¯ f(T) + ¯ f(T ′) ≥ ¯ f(T ∨0 T ′) + ¯ f(T ∧0 T ′) (∀T, T ′ ∈ T ). ————————————————————————–
✂✁ ✄✂✄✂✄ ☎ ✆ ✝ ☎ ✆ ✞ ☎ ✆ ✟ ✠ ✂✁ ✡ ✠ ☞☛ ✠ ✍✌ ☞☛
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SLIDE 18 ————————————————————————– Define P( ¯ f) = {z ∈ RV | ∀T ∈ T : z(T) ≤ ¯ f(T)}. Here we assume ¯ f(∅) = 0. As a corollary of Theorem 4 we obtain the following. ————————————————————————– Corollary 7: For any submodular function f on the product
f(∅) = 0 min{ ¯ f(T) | T ∈ U} = max{−||z||1,∞ | z ∈ P( ¯ f)}. Moreover, if ¯ f is integer-valued, then there exists an integral z that attains the maximum on the right-hand side. ————————————————————————– →
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SLIDE 19 Consider the following additional constraint: (K1′) For each U ∈ U, z(u) + z(v) ≤ z(u ∨U v) + z(u ∧U v) for all {u, v} ∈ ( ˆ
U 2
) , where z(0U) = 0 for all U ∈ U. ————————————————————————– Corollary 8: For any submodular function f on the product
f(∅) = 0 min{ ¯ f(T) | T ∈ U} = max{−||z||1,∞ | z ∈ P( ¯ f), (K1′)}. Moreover, if ¯ f is integer-valued, then there exists an integral z that attains the maximum on the right-hand side. ————————————————————————– →
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SLIDE 20 We assume that |U| ≥ 3 and all the elements in U \ {1U} are incomparable in LU for each U ∈ U. Then lattice LU on ˆ U = U ∪ {0U} is called a diamond.
✂ ✄☎✂
We assume that for each U ∈ U LU is a diamond. ————————————————————————– →
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————————————————————————– Corollary 8 gives a min-max formula for a submodular func- tion on the product lattice of diamonds. Note that in this spe- cial case (K1′) is simplified to (K1′) For each U ∈ U, z(u)+z(v) ≤ z(1U) for all {u, v} ∈ ( ¯
U 2
) , where ¯ U = U \ {1U}. ————————————————————————– Kuivinen (2011) further considered stronger constraints: (K1) For each U ∈ U z(1U) = max{z(u) + z(v) | {u, v} ∈ ( ¯
U 2
) }. (K2) For each U ∈ U there exists p ∈ ¯ U such that z(p) ≥ z(v) for all v ∈ ¯ U and z(u) = z(v) for all u, v ∈ ¯ U \{p}. (Such a z is called unified by Kuivinen (2011).) ————————————————————————– Note that (K1) implies (K1′). →
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Kuivinen (2011) showed the following. ————————————————————————– Theorem 9: min{ ¯ f(T) | T ∈ U} = max{ ∑
U∈U
z(1U) | z ∈ P( ¯ f), z ≤ 0, (K1), (K2)}. Moreover, if ¯ f is integer-valued, then there exists an integral z that attains the maximum on the right-hand side. ————————————————————————– Remark: Kuivinen (2011) showed that this gives a good char- acterization for submodular functions on diamonds. ————————————————————————– →|
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