<p>In this paper, we consider the partition set cover problem with penalties. In this problem, we have a universe <i>U</i>, a partition <InlineEquation ID="IEq1"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1317_Article_IEq1.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="135" /> </InlineMediaObject> <EquationSource Format="TEX">\(\mathscr {P}=\{P_{1},\ldots ,P_{r}\}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi mathvariant="script">P</mi> <mo>=</mo> <mo stretchy="false">{</mo> <msub> <mi>P</mi> <mn>1</mn> </msub> <mo>,</mo> <mo>…</mo> <mo>,</mo> <msub> <mi>P</mi> <mi>r</mi> </msub> <mo stretchy="false">}</mo> </mrow> </math></EquationSource> </InlineEquation> of <i>U</i>, and a collection <InlineEquation ID="IEq2"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1317_Article_IEq2.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="138" /> </InlineMediaObject> <EquationSource Format="TEX">\(\mathscr {S}=\{S_{1},\ldots ,S_{m}\}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi mathvariant="script">S</mi> <mo>=</mo> <mo stretchy="false">{</mo> <msub> <mi>S</mi> <mn>1</mn> </msub> <mo>,</mo> <mo>…</mo> <mo>,</mo> <msub> <mi>S</mi> <mi>m</mi> </msub> <mo stretchy="false">}</mo> </mrow> </math></EquationSource> </InlineEquation> of nonempty subsets of <i>U</i> satisfying <InlineEquation ID="IEq3"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1317_Article_IEq3.gif" Format="GIF" Height="22" Rendition="HTML" Resolution="72" Type="Linedraw" Width="102" /> </InlineMediaObject> <EquationSource Format="TEX">\(\bigcup _{S_i\in \mathscr {S}} S_i=U\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mo>⋃</mo> <mrow> <msub> <mi>S</mi> <mi>i</mi> </msub> <mo>∈</mo> <mi mathvariant="script">S</mi> </mrow> </msub> <msub> <mi>S</mi> <mi>i</mi> </msub> <mo>=</mo> <mi>U</mi> </mrow> </math></EquationSource> </InlineEquation>. In addition, each <InlineEquation ID="IEq4"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1317_Article_IEq4.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="19" /> </InlineMediaObject> <EquationSource Format="TEX">\(P_t\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi>P</mi> <mi>t</mi> </msub> </math></EquationSource> </InlineEquation> <InlineEquation ID="IEq5"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1317_Article_IEq5.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="58" /> </InlineMediaObject> <EquationSource Format="TEX">\((t\in [r])\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mo stretchy="false">(</mo> <mi>t</mi> <mo>∈</mo> <mo stretchy="false">[</mo> <mi>r</mi> <mo stretchy="false">]</mo> <mo stretchy="false">)</mo> </mrow> </math></EquationSource> </InlineEquation> is associated with a covering requirement <InlineEquation ID="IEq6"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1317_Article_IEq6.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="17" /> </InlineMediaObject> <EquationSource Format="TEX">\(k_t\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi>k</mi> <mi>t</mi> </msub> </math></EquationSource> </InlineEquation> as well as a penalty <InlineEquation ID="IEq7"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1317_Article_IEq7.gif" Format="GIF" Height="12" Rendition="HTML" Resolution="72" Type="Linedraw" Width="17" /> </InlineMediaObject> <EquationSource Format="TEX">\(\pi _t\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi>π</mi> <mi>t</mi> </msub> </math></EquationSource> </InlineEquation>, and each <InlineEquation ID="IEq8"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1317_Article_IEq8.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="18" /> </InlineMediaObject> <EquationSource Format="TEX">\(S_i\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi>S</mi> <mi>i</mi> </msub> </math></EquationSource> </InlineEquation> <InlineEquation ID="IEq9"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1317_Article_IEq9.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="64" /> </InlineMediaObject> <EquationSource Format="TEX">\((i\in [m])\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mo stretchy="false">(</mo> <mi>i</mi> <mo>∈</mo> <mo stretchy="false">[</mo> <mi>m</mi> <mo stretchy="false">]</mo> <mo stretchy="false">)</mo> </mrow> </math></EquationSource> </InlineEquation> is associated with a cost. A class <InlineEquation ID="IEq10"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1317_Article_IEq4.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="19" /> </InlineMediaObject> <EquationSource Format="TEX">\(P_t\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi>P</mi> <mi>t</mi> </msub> </math></EquationSource> </InlineEquation> attains its covering requirement by a subcollection <InlineEquation ID="IEq11"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1317_Article_IEq11.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="24" /> </InlineMediaObject> <EquationSource Format="TEX">\(\mathscr {A}\)</EquationSource> <EquationSource Format="MATHML"><math> <mi mathvariant="script">A</mi> </math></EquationSource> </InlineEquation> of <InlineEquation ID="IEq12"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1317_Article_IEq12.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="25" /> </InlineMediaObject> <EquationSource Format="TEX">\(\mathscr {S}\)</EquationSource> <EquationSource Format="MATHML"><math> <mi mathvariant="script">S</mi> </math></EquationSource> </InlineEquation> if at least <InlineEquation ID="IEq13"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1317_Article_IEq6.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="17" /> </InlineMediaObject> <EquationSource Format="TEX">\(k_t\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi>k</mi> <mi>t</mi> </msub> </math></EquationSource> </InlineEquation> elements in <InlineEquation ID="IEq14"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1317_Article_IEq4.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="19" /> </InlineMediaObject> <EquationSource Format="TEX">\(P_t\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi>P</mi> <mi>t</mi> </msub> </math></EquationSource> </InlineEquation> are contained in <InlineEquation ID="IEq15"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1317_Article_IEq15.gif" Format="GIF" Height="22" Rendition="HTML" Resolution="72" Type="Linedraw" Width="67" /> </InlineMediaObject> <EquationSource Format="TEX">\(\bigcup _{S_i\in \mathscr {A}} S_i\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mo>⋃</mo> <mrow> <msub> <mi>S</mi> <mi>i</mi> </msub> <mo>∈</mo> <mi mathvariant="script">A</mi> </mrow> </msub> <msub> <mi>S</mi> <mi>i</mi> </msub> </mrow> </math></EquationSource> </InlineEquation>. Each <InlineEquation ID="IEq16"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1317_Article_IEq4.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="19" /> </InlineMediaObject> <EquationSource Format="TEX">\(P_t\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi>P</mi> <mi>t</mi> </msub> </math></EquationSource> </InlineEquation> is either attaining its covering requirement or paid with its penalty. The objective is to find a subcollection <InlineEquation ID="IEq17"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1317_Article_IEq11.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="24" /> </InlineMediaObject> <EquationSource Format="TEX">\(\mathscr {A}\)</EquationSource> <EquationSource Format="MATHML"><math> <mi mathvariant="script">A</mi> </math></EquationSource> </InlineEquation> of <InlineEquation ID="IEq18"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1317_Article_IEq12.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="25" /> </InlineMediaObject> <EquationSource Format="TEX">\(\mathscr {S}\)</EquationSource> <EquationSource Format="MATHML"><math> <mi mathvariant="script">S</mi> </math></EquationSource> </InlineEquation> such that the sum of the cost of <InlineEquation ID="IEq19"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1317_Article_IEq11.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="24" /> </InlineMediaObject> <EquationSource Format="TEX">\(\mathscr {A}\)</EquationSource> <EquationSource Format="MATHML"><math> <mi mathvariant="script">A</mi> </math></EquationSource> </InlineEquation> and the penalties of classes not attaining covering requirements by <InlineEquation ID="IEq20"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1317_Article_IEq11.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="24" /> </InlineMediaObject> <EquationSource Format="TEX">\(\mathscr {A}\)</EquationSource> <EquationSource Format="MATHML"><math> <mi mathvariant="script">A</mi> </math></EquationSource> </InlineEquation> is minimized. We present two approximation algorithms for this problem. The first is based on the LP-rounding technique with approximation ratio <InlineEquation ID="IEq21"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1317_Article_IEq21.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="117" /> </InlineMediaObject> <EquationSource Format="TEX">\(K+O(\beta +\ln r)\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>K</mi> <mo>+</mo> <mi>O</mi> <mo stretchy="false">(</mo> <mi>β</mi> <mo>+</mo> <mo>ln</mo> <mi>r</mi> <mo stretchy="false">)</mo> </mrow> </math></EquationSource> </InlineEquation>, where <InlineEquation ID="IEq22"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1317_Article_IEq22.gif" Format="GIF" Height="20" Rendition="HTML" Resolution="72" Type="Linedraw" Width="111" /> </InlineMediaObject> <EquationSource Format="TEX">\(K=\max _{t\in [r]}k_t\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>K</mi> <mo>=</mo> <msub> <mo movablelimits="true">max</mo> <mrow> <mi>t</mi> <mo>∈</mo> <mo stretchy="false">[</mo> <mi>r</mi> <mo stretchy="false">]</mo> </mrow> </msub> <msub> <mi>k</mi> <mi>t</mi> </msub> </mrow> </math></EquationSource> </InlineEquation>, and <InlineEquation ID="IEq23"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1317_Article_IEq23.gif" Format="GIF" Height="17" Rendition="HTML" Resolution="72" Type="Linedraw" Width="15" /> </InlineMediaObject> <EquationSource Format="TEX">\(\beta \)</EquationSource> <EquationSource Format="MATHML"><math> <mi>β</mi> </math></EquationSource> </InlineEquation> denotes the approximation guarantee for a related set cover instance obtained by rounding the standard LP. The second is based on the primal-dual method with approximation ratio <i>lf</i>, where <InlineEquation ID="IEq24"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1317_Article_IEq24.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="233" /> </InlineMediaObject> <EquationSource Format="TEX">\(f=\max _{e\in U}|\{S_i\in \mathscr {S}\mid e\in S_i\}|\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>f</mi> <mo>=</mo> <msub> <mo movablelimits="true">max</mo> <mrow> <mi>e</mi> <mo>∈</mo> <mi>U</mi> </mrow> </msub> <mrow> <mo stretchy="false">|</mo> <mrow> <mo stretchy="false">{</mo> <msub> <mi>S</mi> <mi>i</mi> </msub> <mo>∈</mo> <mi mathvariant="script">S</mi> <mo>∣</mo> <mi>e</mi> <mo>∈</mo> <msub> <mi>S</mi> <mi>i</mi> </msub> <mo stretchy="false">}</mo> </mrow> <mo stretchy="false">|</mo> </mrow> </mrow> </math></EquationSource> </InlineEquation> and <InlineEquation ID="IEq25"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1317_Article_IEq25.gif" Format="GIF" Height="21" Rendition="HTML" Resolution="72" Type="Linedraw" Width="115" /> </InlineMediaObject> <EquationSource Format="TEX">\(l=\max _{t\in [r]}|P_t|\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>l</mi> <mo>=</mo> <msub> <mo movablelimits="true">max</mo> <mrow> <mi>t</mi> <mo>∈</mo> <mo stretchy="false">[</mo> <mi>r</mi> <mo stretchy="false">]</mo> </mrow> </msub> <mrow> <mo stretchy="false">|</mo> <msub> <mi>P</mi> <mi>t</mi> </msub> <mo stretchy="false">|</mo> </mrow> </mrow> </math></EquationSource> </InlineEquation>.</p>

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Approximation algorithms for the partition set cover problem with penalties

  • Qi Wang,
  • Bo Hou,
  • Gengsheng Zhang,
  • Yisheng Zhou,
  • Wen Liu

摘要

In this paper, we consider the partition set cover problem with penalties. In this problem, we have a universe U, a partition \(\mathscr {P}=\{P_{1},\ldots ,P_{r}\}\) P = { P 1 , , P r } of U, and a collection \(\mathscr {S}=\{S_{1},\ldots ,S_{m}\}\) S = { S 1 , , S m } of nonempty subsets of U satisfying \(\bigcup _{S_i\in \mathscr {S}} S_i=U\) S i S S i = U . In addition, each \(P_t\) P t \((t\in [r])\) ( t [ r ] ) is associated with a covering requirement \(k_t\) k t as well as a penalty \(\pi _t\) π t , and each \(S_i\) S i \((i\in [m])\) ( i [ m ] ) is associated with a cost. A class \(P_t\) P t attains its covering requirement by a subcollection \(\mathscr {A}\) A of \(\mathscr {S}\) S if at least \(k_t\) k t elements in \(P_t\) P t are contained in \(\bigcup _{S_i\in \mathscr {A}} S_i\) S i A S i . Each \(P_t\) P t is either attaining its covering requirement or paid with its penalty. The objective is to find a subcollection \(\mathscr {A}\) A of \(\mathscr {S}\) S such that the sum of the cost of \(\mathscr {A}\) A and the penalties of classes not attaining covering requirements by \(\mathscr {A}\) A is minimized. We present two approximation algorithms for this problem. The first is based on the LP-rounding technique with approximation ratio \(K+O(\beta +\ln r)\) K + O ( β + ln r ) , where \(K=\max _{t\in [r]}k_t\) K = max t [ r ] k t , and \(\beta \) β denotes the approximation guarantee for a related set cover instance obtained by rounding the standard LP. The second is based on the primal-dual method with approximation ratio lf, where \(f=\max _{e\in U}|\{S_i\in \mathscr {S}\mid e\in S_i\}|\) f = max e U | { S i S e S i } | and \(l=\max _{t\in [r]}|P_t|\) l = max t [ r ] | P t | .