Cardinality constraints and systems of restricted representatives

Abstract

Cardinality constraints have received considerable attention from the Constraint Programming community as (so-called) global constraints that appear in the formulation of several real-life problems, while also having an interesting combinatorial structure. After discussing the relation of cardinality constraints with well-known combinatorial problems (e.g., systems of restricted representatives), we study the polytope defined by the convex hull of vectors satisfying two such constraints, in the case where all variables share a common domain. We provide families of facet-defining inequalities that are polytime separable, together with a condition for when these families of inequalities define a convex hull relaxation. Our results also hold for the case of a single such constraint.

Appendix

Proof of Lemma 19

• (i) $x_{i_{1}}^{\prime }=$ $d_{k},i\in I_{1}^{*}$

Assume to the contrary that some value $d_{k^{\prime }}<d_{k}$ appears fewer than $u_{k^{\prime }}$ times in $J_{1}^{*}\cup T.$ If $o(x^{\prime };1,k^{\prime })<u_{k^{\prime }}$ and given $l_{k}=0,$ derive the vertex $\tilde{x}=x^{\prime }(i_{1};d_{k}\rightarrow d_{k^{\prime }})\in P_{I};$ $\alpha \tilde{x}-\alpha x^{\prime }=\alpha _{i_{1}}\cdot (d_{k^{\prime }}-d_{k})<0$ since $i_{1}\in I_{1}^{*}$ yields $\alpha _{i_{1}}>0$ (by Lemma 18i) and $d_{k^{\prime }}<d_{k}.$

Otherwise, $o(x^{\prime };1,k^{\prime })=u_{k^{\prime }}.$ Then, value $d_{k^{\prime }}$ appearing fewer than $u_{k^{\prime }}$ times in $J_{1}^{*}\cup T$ implies that $d_{k^{\prime }}$ appears in $I_{1}\backslash I_{1}^{*};$ i.e., $x_{i_{2}}^{\prime }=d_{k^{\prime }}$ for some $i_{2}\in I_{1}\backslash I_{1}^{*}.$ Then, for the vertex $\tilde{x}=x^{\prime }(i_{1}\leftrightarrow i_{2})\in P_{I},\alpha \tilde{x}-\alpha x^{\prime }=(\alpha _{i_{1}}-\alpha _{i_{2}})\cdot (d_{k^{\prime }}-d_{k})<0$ since $\alpha _{i_{1}}>\alpha _{i_{2}}$ (by Lemma 18ii) and $d_{k^{\prime }}<d_{k}.$

• (ii) $x_{t_{1}}^{\prime }=$ $d_{k},t_{1}\in T^{*}$

Assume to the contrary that some value $d_{k^{\prime }}<d_{k}$ appears fewer than $u_{k^{\prime }}$ times in $J_{1}^{*}.$

If $x_{t_{2}}^{\prime }=d_{k^{\prime }}$ for some $t_{2}\in T\backslash T^{*},$ then for the vertex $\tilde{x}=x^{\prime }(t_{1}\leftrightarrow t_{2})\in P_{I},\alpha \tilde{x}-\alpha x^{\prime }=(\alpha _{t_{1}}-\alpha _{t_{2}})\cdot (d_{k^{\prime }}-d_{k})<0$ since $\alpha _{t_{1}}>\alpha _{t_{2}}$ (by Lemma 18iii) and $d_{k^{\prime }}<d_{k}.$

Otherwise, $x_{t}^{\prime }\ne d_{k^{\prime }}$ for all $t\in T\backslash T^{*}$. Then, four cases are defined by whether $o(x^{\prime };1,k^{\prime })$ or $o(x^{\prime };2,k^{\prime })$ is less than $u_{k^{\prime }}$ (recall that $l_{k}=0$).

Case 19.1 :

($o(x^{\prime };1,k^{\prime })<u_{k^{\prime }},o(x^{\prime };2,k^{\prime })<u_{k^{\prime }}$):] For $\tilde{x}=x^{\prime }(t_{1};d_{k}\rightarrow d_{k^{\prime }}),\alpha \tilde{x}-\alpha x^{\prime }=\alpha _{t_{1}}\cdot (d_{k^{\prime }}-d_{k})<0$ since $t_{1}\in T^{*}$ yields $\alpha _{t_{1}}>0$ (by Lemma 18i).

Case 19.2 :

($o(x^{\prime };1,k^{\prime })<u_{k^{\prime }},o(x^{\prime };2,k^{\prime })=u_{k^{\prime }}$): Since $d_{k^{\prime }}$ appears more times in $J_{2}$ than in $J_{1}$, it appears in $I_{2};$ i.e., $x_{i_{2}}^{\prime }=d_{k^{\prime }}$ for some $i_{2}\in I_{2}.$ For $\tilde{x}=x^{\prime }(t_{1}\leftrightarrow i_{2}),\alpha \tilde{x}-\alpha x^{\prime }=(\alpha _{t_{1}}-\alpha _{i_{2}})\cdot (d_{k^{\prime }}-d_{k})<0$ since $\alpha _{t_{1}}>\alpha _{i_{2}}.$ The latter follows from Lemma 18vii for $I_{2}^{*}\ne \emptyset$ (since $\alpha _{t_1}>\gamma _{2}\ge \alpha _{i_{2}}$) and from Lemma 18i and 18vi for $I_{2}^{*}= \emptyset$ (since $\alpha _{t_{1}}>0\ge \alpha _{i_{2}}$).

Case 19.3 :

($o(x^{\prime };1,k^{\prime })=u_{k^{\prime }},o(x^{\prime };2,k^{\prime })<u_{k^{\prime }}$): Since value $d_{k^{\prime }}$ appears $u_k$ times in $J_1$ but not in $T \backslash T^{*}$ and appears fewer than $u_k$ times in $J_{1}^{*}= I_{1}^{*} \cup T^{*}$, it must appear in $I_{1}\backslash I_{1}^{*};$ i.e., $x_{i_{1}}^{\prime }=d_{k^{\prime }}$ for some $i_{1}\in I_{1}\backslash I_{1}^{*}.$ For $\tilde{x}=x^{\prime }(t_{1}\leftrightarrow i_{1}),\alpha \tilde{x}-\alpha x^{\prime }=(\alpha _{t_{1}}-\alpha _{i_{1}})\cdot (d_{k^{\prime }}-d_{k})<0$ since $\alpha _{t_{1}}>\alpha _{i_{1}}$. The latter holds because the opposite, i.e., $\alpha _{t_{1}}\le \alpha _{i_{1}}$, implies $\gamma \le \alpha _{i_{1}}$ hence $i_{1}\in I_{1}^{*},$ a contradiction to $i_{1}\in I_{1}\backslash I_{1}^{*}$.

Case 19.4 :

($o(x^{\prime };1,k^{\prime })=u_{k^{\prime }},o(x^{\prime };2,k^{\prime })=u_{k^{\prime }}$): As in Case 19.3, $o(x^{\prime };1,k^{\prime })=u_{k^{\prime }}$ implies that $x_{i_{1}}^{\prime }=d_{k^{\prime }}$ for some $i_{1}\in I_{1}\backslash I_{1}^{*}.$ This, together with $o(x^{\prime };1,k^{\prime })=u_{k^{\prime }}=o(x^{\prime };2,k^{\prime })$ imply that $x_{i_{2}}^{\prime }=d_{k^{\prime }}$ for some $i_{2}\in I_{2}.$ For $\tilde{x}=x^{\prime }(t\leftrightarrow \{i_{1},i_{2}\}),\alpha \tilde{x}-\alpha x^{\prime }=(\alpha _{t_{1}}-\alpha _{i_{1}}-\alpha _{i_{2}})\cdot (d_{k^{\prime }}-d_{k})<0,$ since $\alpha _{t_{1}}-\alpha _{i_{1}}-\alpha _{i_{2}}>0.$ If $i_{2}\in I_{2}\backslash I_{2}^{*},$ the latter holds by Lemma 18iv. If $i_{2}\in I_{2}^{*}$, it holds because the opposite, i.e., $\alpha _{t_{1}}\le \alpha _{i_{1}}+\alpha _{i_{2}},$ implies $i_{1}\in I_{1}^{*},$ a contradiction to $i_{1}\in I_{1}\backslash I_{1}^{*}.$

$\square$

Proof of Lemma 20

Let $x_{i_{2}}^{\prime }=d_{k}$ for some $i_{2}\in I_{2}^{\prime }$. $J_{2}^{*}\subset J_{1}^{*}$ implies $I_{1}^{*}\ne \emptyset$ and $I_{2}^{*}=\emptyset ,$ hence $i_{2}\in I_{2}^{\prime }$ implies $\alpha _{i_{2}}<0$ (by Lemma 18vi and the definition of $I_{2}^{\prime }$). Assume to the contrary that some value $d_{k^{\prime }}>d_{k}$ appears fewer than $u_{k^{\prime }}$ times in $J_{2}^{\prime }\backslash T_{2}^{*}.$ Recall that $l_k=0$.

If $o(x^{\prime };2,k^{\prime })<u_{k^{\prime }},$ the vertex $\tilde{x}$ $=x^{\prime }(i_{2};d_{k}\rightarrow d_{k^{\prime }})\in P_{I}$ since $l_{k}=0$. Then, $\alpha \tilde{x}-\alpha x^{\prime }=\alpha _{i_{2}}\cdot (d_{k^{\prime }}-d_{k})<0$ since $\alpha _{i_{2}}<0$ and $d_{k^{\prime }}>d_{k}.$

Otherwise, $o(x^{\prime };2,k^{\prime })=u_{k^{\prime }}.$ Then, value $d_{k^{\prime }}$ appears $u_{k^{\prime }}$ times in $J_{2}$ but fewer than $u_{k^{\prime }}$ times in $J_{2}^{\prime }\backslash T^{*},$ hence it appears in $I_{2}\backslash I_{2}^{\prime }=\{i\in I_{2}:\alpha _{i}=0\}$ or in $T^{*}.$

Case 20.1 :

($x_{i_{3}}^{\prime }=d_{k^{\prime }},i_{3}\in I_{2}\backslash I_{2}^{\prime }$): Recall that $x_{i_{2}}^{\prime }=d_{k}$, $i_{2}\in I_{2}^{\prime }$. For $\tilde{x}=x^{\prime }(i_{2}\leftrightarrow i_{3}),\alpha \tilde{x}-\alpha x^{\prime }=(\alpha _{i_{2}}-\alpha _{i_{3}})\cdot (d_{k^{\prime }}-d_{k})<0$ since $\alpha _{i_{2}}<0=\alpha _{i_{3}}$ and $d_{k^{\prime }}>d_{k}.$

Case 20.2 :

($x_{t}^{\prime }=d_{k^{\prime }},t\in T^{*}$): If value $d_{k}$ appears also in $I_{1}$, let $i_{1}\in I_{1}$ be an index where this occurs; i.e., $x_{i_{1}}^{\prime }=d_{k}$. Also, $\alpha _{t_{1}}=\gamma _{1}$ by Lemma 18vii (because $I_{1}^{*}\ne \emptyset = I_{2}^{*}$) and $\alpha _{i_{1}} \le \gamma _{1}$ by (28), thus $\alpha _{i_{1}} \le \alpha _{t_{1}}$. Then, for $\tilde{x}=x^{\prime }(t\leftrightarrow \{i_{1},i_{2}\}),\alpha \tilde{x}-\alpha x^{\prime }=(\alpha _{i_{1}}+\alpha _{i_{2}}-\alpha _{t_{1}})\cdot (d_{k^{\prime }}-d_{k})<0$ since $\alpha _{i_{1}} \le \alpha _{t_{1}},\alpha _{i_{2}}<0$ and $d_{k^{\prime }}>d_{k}.$

Otherwise, value $d_{k}$ appears not in $I_{1}$ but appears in $I_2$ (because $x_{i_{2}}^{\prime }=d_{k},i_{2}\in I_{2}^{\prime }$) hence $o(x^{\prime };1,k)<u_{k}$. Then, the vertex $\tilde{x}=x^{\prime }(i_{2}\leftrightarrow t)\in P_{I}$ yields $\alpha \tilde{x}-\alpha x^{\prime }=(\alpha _{i_{2}}-\alpha _{t})\cdot (d_{k^{\prime }}-d_{k})<0$ since $\alpha _{i_{2}}<0$ and $\alpha _{t}>0$ by Lemma 18i (because $t \in T^{*}$).

$\square$

Proof of Theorem 17

Observe that $J_{1}^{*}\supset J_{2}^{*}$ implies $I_{1}^{*}\ne \emptyset$ and $I_{2}^{*}=\emptyset ,$ hence $\alpha _{i}\le 0$ for all $i\in I_{2}$ by Lemma 18vi. The result holds if and only if all values $d_{k},k\in K$ appear $q(|J_{2}^{\prime }\backslash T^{*}|,k)$ times in $J_{2}^{\prime }\backslash T^{*},$ at all vertices of $P^{(\alpha ,\beta )}.$

Assume to the contrary that there is a vertex $x^{\prime }\in P^{(\alpha ,\beta )}$ with $\sum _{j\in J_{2}^{\prime }\backslash T^{*}}x_{j}^{\prime }<\mathop {\sum }\nolimits _{k\in K}q(|J_{2}^{\prime }\backslash T^{*}|,k)\cdot d_{k}.$ Then, at vertex $x^{\prime },$ some value $d_{\tilde{k}}$ appears fewer than $q(|J_{2}^{\prime }\backslash T^{*}|,\tilde{k})$ times in $J_{2}^{\prime }\backslash T^{*}$, whereas some other value $d_{k}<d_{\tilde{k}}$ appears more than $q(|J_{2}^{\prime }\backslash T^{*}|,k)$ times in $J_{2}^{\prime }\backslash T^{*}.$ Then, if $d_{k^{\prime }}=\min \{x_{j}^{\prime }:j\in J_{2}^{\prime }\backslash T^{*}\}$ it must be $d_{k^{\prime }}<d_{\tilde{k}}.$ Since $q(|J_{2}^{\prime }\backslash T^{*}|,\tilde{k})\le u_{\tilde{k}},$ value $d_{\tilde{k}}$ appears fewer than $u_{\tilde{k}}$ times in $J_{2}^{\prime }\backslash T^{*}.$ Thus, if $d_{k^{\prime }}$ appears in $I_{2}^{\prime }$ (i.e., if $x_{i_{2}}^{\prime }=d_{k^{\prime }}$ for some $i_{2}\in$ $I_{2}^{\prime }$), Lemma 20 is contradicted.

Hence let $x_{t_{2}}^{\prime }=d_{k^{\prime }},t_{2}\in T\backslash T^{*}.$ The hypothesis $|J_{1}^{*}|+|J_{2}^{\prime }\backslash T^{*}|\ge \mathop {\sum }\nolimits _{k\in K}u_{k}$ yields that each value $d_{k}$ occurs $u_{k}$ times in $J_{1}^{*}\cup (J_{2}^{\prime }\backslash T^{*}).$ Therefore, value $d_{\tilde{k}},$ since appearing fewer than $u_{\tilde{k}}$ times in $J_{2}^{\prime }\backslash T^{*}$ must appear in $J_{1}^{*};$ i.e., $x_{t_{1}}^{\prime }=d_{\tilde{k}}$ for some $t_{1}\in T^{*}$ or $x_{i_{1}}^{\prime }=d_{\tilde{k}}$ for some $i_{1}\in I_{1}^{*}.$ In the former case, for $\tilde{x}=x^{\prime }(t_{1}\leftrightarrow t_{2}),\alpha \tilde{x}-\alpha x^{\prime }=(\alpha _{t_{2}}-\alpha _{t_{1}})\cdot (d_{\tilde{k}}-d_{k^{\prime }})<0$ since $\alpha _{t_{1}}>\alpha _{t_{2}}$ (by Lemma 18iii) and $d_{\tilde{k}}>d_{k^{\prime }}.$ In the latter case, $\alpha _{t_2} < \alpha _{i_1}$ (by Lemma 18vi) and the contradiction is established by using the vertex $\tilde{x}=x^{\prime }(t_{2}\leftrightarrow \{i_{1},i_{2}\})$ (if $x_{i_{2}}^{\prime }=d_{\tilde{k}},$ for some $i_{2}\in I_{2}$) or the vertex $\tilde{x}=x^{\prime }(t_{2}\leftrightarrow i_{1})$ (if $d_{\tilde{k}}\,\$appears not in $I_{2}$). $\square$