<p>The two-dimensional strip packing problem consists of packing in a rectangular strip of width 1 and minimum height a set of <i>n</i> rectangles, where each rectangle has width <InlineEquation ID="IEq1"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="224_2025_10217_Article_IEq1.gif" Format="GIF" Height="15" Rendition="HTML" Resolution="72" Type="Linedraw" Width="76" /> </InlineMediaObject> <EquationSource Format="TEX">\(0 &lt; w \le 1\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mn>0</mn> <mo>&lt;</mo> <mi>w</mi> <mo>≤</mo> <mn>1</mn> </mrow> </math></EquationSource> </InlineEquation> and height <InlineEquation ID="IEq2"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="224_2025_10217_Article_IEq2.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="99" /> </InlineMediaObject> <EquationSource Format="TEX">\(0 &lt; h \le h_{max}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mn>0</mn> <mo>&lt;</mo> <mi>h</mi> <mo>≤</mo> <msub> <mi>h</mi> <mrow> <mi mathvariant="italic">max</mi> </mrow> </msub> </mrow> </math></EquationSource> </InlineEquation>. We consider the high-multiplicity version of the problem in which there are only <i>K</i> different types of rectangles. For the case when <InlineEquation ID="IEq3"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="224_2025_10217_Article_IEq3.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="47" /> </InlineMediaObject> <EquationSource Format="TEX">\(K = 3\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>K</mi> <mo>=</mo> <mn>3</mn> </mrow> </math></EquationSource> </InlineEquation>, we give an algorithm that produces solutions requiring at most height <InlineEquation ID="IEq4"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="224_2025_10217_Article_IEq4.gif" Format="GIF" Height="22" Rendition="HTML" Resolution="72" Type="Linedraw" Width="71" /> </InlineMediaObject> <EquationSource Format="TEX">\(\frac{3}{2}h_{max} + \epsilon \)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mfrac> <mn>3</mn> <mn>2</mn> </mfrac> <msub> <mi>h</mi> <mrow> <mi mathvariant="italic">max</mi> </mrow> </msub> <mo>+</mo> <mi>ϵ</mi> </mrow> </math></EquationSource> </InlineEquation> plus the height of an optimal solution, where <InlineEquation ID="IEq5"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="224_2025_10217_Article_IEq5.gif" Format="GIF" Height="10" Rendition="HTML" Resolution="72" Type="Linedraw" Width="10" /> </InlineMediaObject> <EquationSource Format="TEX">\(\epsilon \)</EquationSource> <EquationSource Format="MATHML"><math> <mi>ϵ</mi> </math></EquationSource> </InlineEquation> is any positive constant. For the case when <InlineEquation ID="IEq6"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="224_2025_10217_Article_IEq6.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="47" /> </InlineMediaObject> <EquationSource Format="TEX">\(K = 4\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>K</mi> <mo>=</mo> <mn>4</mn> </mrow> </math></EquationSource> </InlineEquation>, we give an algorithm yielding solutions of height at most <InlineEquation ID="IEq7"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="224_2025_10217_Article_IEq7.gif" Format="GIF" Height="22" Rendition="HTML" Resolution="72" Type="Linedraw" Width="71" /> </InlineMediaObject> <EquationSource Format="TEX">\(\frac{7}{3}h_{max} + \epsilon \)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mfrac> <mn>7</mn> <mn>3</mn> </mfrac> <msub> <mi>h</mi> <mrow> <mi mathvariant="italic">max</mi> </mrow> </msub> <mo>+</mo> <mi>ϵ</mi> </mrow> </math></EquationSource> </InlineEquation> plus the height of an optimal solution. For the case when <InlineEquation ID="IEq8"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="224_2025_10217_Article_IEq8.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="48" /> </InlineMediaObject> <EquationSource Format="TEX">\(K &gt; 3\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>K</mi> <mo>&gt;</mo> <mn>3</mn> </mrow> </math></EquationSource> </InlineEquation>, we give an algorithm that gives solutions of height at most <InlineEquation ID="IEq9"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="224_2025_10217_Article_IEq9.gif" Format="GIF" Height="22" Rendition="HTML" Resolution="72" Type="Linedraw" Width="157" /> </InlineMediaObject> <EquationSource Format="TEX">\(\lfloor \frac{3}{4}K\rfloor h_{max} + h_{max} + \epsilon \)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mrow> <mo>⌊</mo> <mfrac> <mn>3</mn> <mn>4</mn> </mfrac> <mi>K</mi> <mo>⌋</mo> </mrow> <msub> <mi>h</mi> <mrow> <mi mathvariant="italic">max</mi> </mrow> </msub> <mo>+</mo> <msub> <mi>h</mi> <mrow> <mi mathvariant="italic">max</mi> </mrow> </msub> <mo>+</mo> <mi>ϵ</mi> </mrow> </math></EquationSource> </InlineEquation> plus the height of an optimal solution.</p>

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High Multiplicity Strip Packing with Three Rectangle Types

  • Andrew Bloch-Hansen,
  • Roberto Solis-Oba,
  • Andy Yu

摘要

The two-dimensional strip packing problem consists of packing in a rectangular strip of width 1 and minimum height a set of n rectangles, where each rectangle has width \(0 < w \le 1\) 0 < w 1 and height \(0 < h \le h_{max}\) 0 < h h max . We consider the high-multiplicity version of the problem in which there are only K different types of rectangles. For the case when \(K = 3\) K = 3 , we give an algorithm that produces solutions requiring at most height \(\frac{3}{2}h_{max} + \epsilon \) 3 2 h max + ϵ plus the height of an optimal solution, where \(\epsilon \) ϵ is any positive constant. For the case when \(K = 4\) K = 4 , we give an algorithm yielding solutions of height at most \(\frac{7}{3}h_{max} + \epsilon \) 7 3 h max + ϵ plus the height of an optimal solution. For the case when \(K > 3\) K > 3 , we give an algorithm that gives solutions of height at most \(\lfloor \frac{3}{4}K\rfloor h_{max} + h_{max} + \epsilon \) 3 4 K h max + h max + ϵ plus the height of an optimal solution.