Past exam of the computer science course of the University of Oxford / 2023 / Combinatorial optimisation / 1

For a set function $f:2^V\to \mathbb{R}$, the four properties are:

Here monotone set function means nondecreasing under inclusion; it does not require strict increase.

The intersection–union definition of a submodular set function isThe diminishing returns characterization of submodularity says that its marginal increments decrease as the conditioning set grows:To obtain diminishing returns from submodularity, apply the intersection–union inequality to $A\cup \{x\}$ and $B$. Their intersection is $A$ and their union is $B\cup \{x\}$, giving exactly the displayed comparison.

Conversely, assume diminishing returns, and enumerate $A\setminus B=\{x_1,\dots ,x_m\}$. Put $C_j=(A\cap B)\cup \{x_1,\dots ,x_j\}$ and $D_j=B\cup \{x_1,\dots ,x_j\}$. Since $C_{j-1}\subseteq D_{j-1}$ and $x_j\notin D_{j-1}$, summing the marginal increment comparisons gives the telescoping sumRearrangement proves submodularity. When $A\setminus B$ is empty, both sums are zero, so the proof includes that case.

Yes: this is a monotone set function. Assume $V\ne \varnothing$, as required by its definition at the empty set. If $\varnothing \ne A\subseteq B$, every weight considered in the maximum over $A$ is considered in the maximum over $B$, hence $f(A)\le f(B)$. If $A=\varnothing$ and $B\ne \varnothing$, the minimum over $V$ is at most every weight in $B$, hence again $f(A)\le f(B)$. No sign condition on the weights is needed.

For every $S\subseteq V$ and $x\notin S$, including $S=\varnothing$, the marginal increment isFor the empty set, this uses $w(x)\ge f(\varnothing )={\min }_{v\in V}w(v)$. The monotone set function property from (i) gives $f(A)\le f(B)$ when $A\subseteq B$. ConsequentlyThus diminishing returns holds, so $f$ is a submodular set function by (b).

Yes, for arbitrary signed weights. A hyperedge meets both $S$ and $V\setminus S$ exactly when it meets both $V\setminus S$ and $S$. Therefore $\delta (S)=\delta (V\setminus S)$, and their weighted sums agree. The hypergraph cut function is a symmetric set function regardless of the signs of its weights.

No. Take the hypergraph with $V=\{a,b\}$ and its single hyperedge $\{a,b\}$ of weight $1$. Then $\{a\}\subseteq V$ but $f(\{a\})=1>0=f(V)$, violating the monotone set function property. This example already has strictly positive weights.

No for arbitrary signed weights. Give the same single hyperedge weight $-1$. For $A=\{a\}$ and $B=\{b\}$,This directly violates the defining inequality for a submodular set function.

With nonnegative weights, the hypergraph cut function is a symmetric set function and a submodular set function, but need not be a monotone set function. The proofs of symmetry and failure of monotonicity in (i) and (ii) still apply.

For submodularity, write $f(S)={\sum }_{e\in E}w(e)q_e(S)$, where $q_e(S)$ is the indicator function that $e$ meets both sides. Fix $x\notin S$. If $x\notin e$, its marginal increment is zero. If $x\in e$, let $m=|e|\ge 2$ and $j=|S\cap (e\setminus \{x\})|$. Its marginal increment isFor $m=2$, the middle case is absent. As $S$ grows without acquiring $x$, $j$ can only increase, so these marginal increments can only decrease. Each $q_e$ therefore satisfies diminishing returns. Multiplying by $w(e)\ge 0$ and summing preserves the inequality, proving submodularity of $f$.

Use the chain lower bound for a submodular set function. Order $T=\{x_1,\dots ,x_n\}$, set $P_j=\{x_1,\dots ,x_j\}$ with $P_0=\varnothing$, and put $a_j=f(P_j)-f(P_{j-1})$. For each $U\subseteq T$, reveal its elements in this same order. Since $U\cap P_{j-1}\subseteq P_{j-1}$, diminishing returns and the normalised set function property giveApply this to each $T_i$ and interchange the finite sums. Every element occurs exactly $k$ times, soThis is the exact-cover inequality for submodular set functions. Some $a_j$ can be negative: exact multiplicity, rather than merely at least $k$ occurrences, is what makes the argument valid without monotonicity. In fact the nonnegative function hypothesis is unnecessary here. If $T=\varnothing$, every $T_i$ is empty and both sides vanish.

The monotone decomposition of a submodular set function subtracts each element's smallest possible marginal increment. For $x\in V$, defineThe set function $h$ is a modular set function: the contributions of each element, and the constant $f(\varnothing )$, cancel inSubtracting this equality from the submodularity inequality for $f$ proves that $g$ is a submodular set function. Also $g(\varnothing )=0$, so it is a normalised set function.

For $x\notin S$, the inclusion $S\subseteq V\setminus \{x\}$ and diminishing returns implyAdding elements one at a time now proves that $g$ is a monotone set function. Since it starts at zero, it is also a nonnegative function. Thus $f=g+h$ has all the required properties.

Here modular set function means equality in the intersection–union identity; a constant term is allowed. If one imposed the additional convention $h(\varnothing )=0$, the claimed decomposition of an arbitrary $f$ would require $f(\varnothing )=0$. The displayed constant is essential for the problem as stated. For $V=\varnothing$, take $g=0$ and $h=f$.