<p>In the thief orienteering problem an agent called a <i>thief</i> carries a knapsack of capacity <i>W</i> and has a time limit <i>T</i> to collect a set of items of total weight at most <i>W</i> and maximum profit along a simple path in a weighted graph <InlineEquation ID="IEq1"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="236_2025_486_Article_IEq1.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="81" /> </InlineMediaObject> <EquationSource Format="TEX">\(G = (V, E)\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>G</mi> <mo>=</mo> <mo stretchy="false">(</mo> <mi>V</mi> <mo>,</mo> <mi>E</mi> <mo stretchy="false">)</mo> </mrow> </math></EquationSource> </InlineEquation> from a start vertex <i>s</i> to an end vertex <i>t</i>. There is a set <i>I</i> of items each with weight <InlineEquation ID="IEq2"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="236_2025_486_Article_IEq2.gif" Format="GIF" Height="12" Rendition="HTML" Resolution="72" Type="Linedraw" Width="20" /> </InlineMediaObject> <EquationSource Format="TEX">\(w_{i}\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi>w</mi> <mi>i</mi> </msub> </math></EquationSource> </InlineEquation> and profit <InlineEquation ID="IEq3"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="236_2025_486_Article_IEq3.gif" Format="GIF" Height="12" Rendition="HTML" Resolution="72" Type="Linedraw" Width="17" /> </InlineMediaObject> <EquationSource Format="TEX">\(p_{i}\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi>p</mi> <mi>i</mi> </msub> </math></EquationSource> </InlineEquation> that are distributed among <InlineEquation ID="IEq4"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="236_2025_486_Article_IEq4.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="66" /> </InlineMediaObject> <EquationSource Format="TEX">\(V{\setminus }\{s,t\}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>V</mi> <mo lspace="0.15em" rspace="0.15em" stretchy="false">\</mo> <mo stretchy="false">{</mo> <mi>s</mi> <mo>,</mo> <mi>t</mi> <mo stretchy="false">}</mo> </mrow> </math></EquationSource> </InlineEquation>. The time needed by the thief to travel an edge depends on the length of the edge and the weight of the items in the knapsack at the moment when the edge is traversed. There is a polynomial-time approximation scheme for a relaxed version of the thief orienteering problem on directed acyclic graphs that produces solutions that use time at most <InlineEquation ID="IEq5"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="236_2025_486_Article_IEq5.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="61" /> </InlineMediaObject> <EquationSource Format="TEX">\(T(1 + \epsilon )\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>T</mi> <mo stretchy="false">(</mo> <mn>1</mn> <mo>+</mo> <mi>ϵ</mi> <mo stretchy="false">)</mo> </mrow> </math></EquationSource> </InlineEquation> for any constant <InlineEquation ID="IEq6"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="236_2025_486_Article_IEq6.gif" Format="GIF" Height="13" Rendition="HTML" Resolution="72" Type="Linedraw" Width="39" /> </InlineMediaObject> <EquationSource Format="TEX">\(\epsilon &gt; 0\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>ϵ</mi> <mo>&gt;</mo> <mn>0</mn> </mrow> </math></EquationSource> </InlineEquation>. We give a polynomial-time algorithm for transforming instances of the problem on 2-terminal series–parallel graphs into equivalent instances of the thief orienteering problem on directed acyclic graphs; therefore, yielding a polynomial-time approximation scheme for the relaxed version of the thief orienteering problem on this graph class.</p>

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The thief orienteering problem on 2-terminal series–parallel graphs

  • Andrew Bloch-Hansen,
  • Roberto Solis-Oba

摘要

In the thief orienteering problem an agent called a thief carries a knapsack of capacity W and has a time limit T to collect a set of items of total weight at most W and maximum profit along a simple path in a weighted graph \(G = (V, E)\) G = ( V , E ) from a start vertex s to an end vertex t. There is a set I of items each with weight \(w_{i}\) w i and profit \(p_{i}\) p i that are distributed among \(V{\setminus }\{s,t\}\) V \ { s , t } . The time needed by the thief to travel an edge depends on the length of the edge and the weight of the items in the knapsack at the moment when the edge is traversed. There is a polynomial-time approximation scheme for a relaxed version of the thief orienteering problem on directed acyclic graphs that produces solutions that use time at most \(T(1 + \epsilon )\) T ( 1 + ϵ ) for any constant \(\epsilon > 0\) ϵ > 0 . We give a polynomial-time algorithm for transforming instances of the problem on 2-terminal series–parallel graphs into equivalent instances of the thief orienteering problem on directed acyclic graphs; therefore, yielding a polynomial-time approximation scheme for the relaxed version of the thief orienteering problem on this graph class.