<p>This paper studies the online power cover problem on a line. Let <InlineEquation ID="IEq1"> <EquationSource Format="TEX">\(\varvec{L}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi mathvariant="bold-italic">L</mi> </mrow> </math></EquationSource> </InlineEquation> be a line and <InlineEquation ID="IEq2"> <EquationSource Format="TEX">\(\varvec{S}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi mathvariant="bold-italic">S</mi> </mrow> </math></EquationSource> </InlineEquation> be a set of sensors located on <InlineEquation ID="IEq3"> <EquationSource Format="TEX">\(\varvec{L}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi mathvariant="bold-italic">L</mi> </mrow> </math></EquationSource> </InlineEquation>, where each sensor can be assigned a power to generate a coverage area for serving users. The objective is to assign minimum powers to cover a sequence of users that arrive online on <InlineEquation ID="IEq4"> <EquationSource Format="TEX">\(\varvec{L}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi mathvariant="bold-italic">L</mi> </mrow> </math></EquationSource> </InlineEquation>. We first show a lower bound of <InlineEquation ID="IEq5"> <EquationSource Format="TEX">\(\varvec{2}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mn mathvariant="bold">2</mn> </mrow> </math></EquationSource> </InlineEquation> for this problem. Then, an online algorithm based on a greedy strategy is proposed, whose competitive ratio is at most <InlineEquation ID="IEq6"> <EquationSource Format="TEX">\(\varvec{|S|}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mo mathvariant="bold" stretchy="false">|</mo> <mi mathvariant="bold-italic">S</mi> <mo mathvariant="bold" stretchy="false">|</mo> </mrow> </math></EquationSource> </InlineEquation>. Notably, this algorithm is the best possible online algorithm for the case <InlineEquation ID="IEq7"> <EquationSource Format="TEX">\(\varvec{|S|=2}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mo mathvariant="bold" stretchy="false">|</mo> <mi mathvariant="bold-italic">S</mi> <mo mathvariant="bold" stretchy="false">|</mo> <mo mathvariant="bold">=</mo> <mn mathvariant="bold">2</mn> </mrow> </math></EquationSource> </InlineEquation>. Finally, we consider the special case with <InlineEquation ID="IEq8"> <EquationSource Format="TEX">\(\varvec{S=\{s_0,s_1,s_2\}}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi mathvariant="bold-italic">S</mi> <mo mathvariant="bold">=</mo> <mo mathvariant="bold" stretchy="false">{</mo> <msub> <mi mathvariant="bold-italic">s</mi> <mn mathvariant="bold">0</mn> </msub> <mo mathvariant="bold">,</mo> <msub> <mi mathvariant="bold-italic">s</mi> <mn mathvariant="bold">1</mn> </msub> <mo mathvariant="bold">,</mo> <msub> <mi mathvariant="bold-italic">s</mi> <mn mathvariant="bold">2</mn> </msub> <mo mathvariant="bold" stretchy="false">}</mo> </mrow> </math></EquationSource> </InlineEquation> and design an online algorithm whose competitive ratio depends on the distances between the sensors.</p>

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Algorithms for the Online Power Cover Problem on a Line

  • Xiaofei Liu,
  • Jinlin Zhang,
  • Zhonghao Liu,
  • Man Xiao,
  • Weidong Li

摘要

This paper studies the online power cover problem on a line. Let \(\varvec{L}\) L be a line and \(\varvec{S}\) S be a set of sensors located on \(\varvec{L}\) L , where each sensor can be assigned a power to generate a coverage area for serving users. The objective is to assign minimum powers to cover a sequence of users that arrive online on \(\varvec{L}\) L . We first show a lower bound of \(\varvec{2}\) 2 for this problem. Then, an online algorithm based on a greedy strategy is proposed, whose competitive ratio is at most \(\varvec{|S|}\) | S | . Notably, this algorithm is the best possible online algorithm for the case \(\varvec{|S|=2}\) | S | = 2 . Finally, we consider the special case with \(\varvec{S=\{s_0,s_1,s_2\}}\) S = { s 0 , s 1 , s 2 } and design an online algorithm whose competitive ratio depends on the distances between the sensors.