<p>Let <InlineEquation ID="IEq1"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10623_2025_1605_Article_IEq1.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="20" /> </InlineMediaObject> <EquationSource Format="TEX">\(\ell ^m\)</EquationSource> <EquationSource Format="MATHML"><math> <msup> <mi>ℓ</mi> <mi>m</mi> </msup> </math></EquationSource> </InlineEquation> be a power with <InlineEquation ID="IEq2"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10623_2025_1605_Article_IEq2.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="12" /> </InlineMediaObject> <EquationSource Format="TEX">\(\ell \)</EquationSource> <EquationSource Format="MATHML"><math> <mi>ℓ</mi> </math></EquationSource> </InlineEquation> a prime greater than 3 and <InlineEquation ID="IEq3"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10623_2025_1605_Article_IEq3.gif" Format="GIF" Height="10" Rendition="HTML" Resolution="72" Type="Linedraw" Width="18" /> </InlineMediaObject> <EquationSource Format="TEX">\(m\)</EquationSource> <EquationSource Format="MATHML"><math> <mi>m</mi> </math></EquationSource> </InlineEquation> a positive integer such that 3 is a primitive root modulo <InlineEquation ID="IEq4"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10623_2025_1605_Article_IEq4.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="28" /> </InlineMediaObject> <EquationSource Format="TEX">\(2\ell ^m\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mn>2</mn> <msup> <mi>ℓ</mi> <mi>m</mi> </msup> </mrow> </math></EquationSource> </InlineEquation>. Let <InlineEquation ID="IEq5"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10623_2025_1605_Article_IEq5.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="17" /> </InlineMediaObject> <EquationSource Format="TEX">\(\mathbb {F}_3\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi mathvariant="double-struck">F</mi> <mn>3</mn> </msub> </math></EquationSource> </InlineEquation> be the finite field of order 3, and let <InlineEquation ID="IEq6"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10623_2025_1605_Article_IEq6.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="12" /> </InlineMediaObject> <EquationSource Format="TEX">\(\mathbb {F}\)</EquationSource> <EquationSource Format="MATHML"><math> <mi mathvariant="double-struck">F</mi> </math></EquationSource> </InlineEquation> be the <InlineEquation ID="IEq7"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10623_2025_1605_Article_IEq7.gif" Format="GIF" Height="20" Rendition="HTML" Resolution="72" Type="Linedraw" Width="83" /> </InlineMediaObject> <EquationSource Format="TEX">\(\ell ^{m-1}(\ell -1)\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msup> <mi>ℓ</mi> <mrow> <mi>m</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mrow> <mo stretchy="false">(</mo> <mi>ℓ</mi> <mo>-</mo> <mn>1</mn> <mo stretchy="false">)</mo> </mrow> </mrow> </math></EquationSource> </InlineEquation>-th extension field of <InlineEquation ID="IEq8"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10623_2025_1605_Article_IEq5.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="17" /> </InlineMediaObject> <EquationSource Format="TEX">\(\mathbb {F}_3\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi mathvariant="double-struck">F</mi> <mn>3</mn> </msub> </math></EquationSource> </InlineEquation>. Denote by <InlineEquation ID="IEq9"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10623_2025_1605_Article_IEq9.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="18" /> </InlineMediaObject> <EquationSource Format="TEX">\(\text {Tr}\)</EquationSource> <EquationSource Format="MATHML"><math> <mtext>Tr</mtext> </math></EquationSource> </InlineEquation> the absolute trace map from <InlineEquation ID="IEq10"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10623_2025_1605_Article_IEq6.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="12" /> </InlineMediaObject> <EquationSource Format="TEX">\(\mathbb {F}\)</EquationSource> <EquationSource Format="MATHML"><math> <mi mathvariant="double-struck">F</mi> </math></EquationSource> </InlineEquation> to <InlineEquation ID="IEq11"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10623_2025_1605_Article_IEq5.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="17" /> </InlineMediaObject> <EquationSource Format="TEX">\(\mathbb {F}_3\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi mathvariant="double-struck">F</mi> <mn>3</mn> </msub> </math></EquationSource> </InlineEquation>. For any <InlineEquation ID="IEq12"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10623_2025_1605_Article_IEq12.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="49" /> </InlineMediaObject> <EquationSource Format="TEX">\(\alpha \in \mathbb {F}_3\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>α</mi> <mo>∈</mo> <msub> <mi mathvariant="double-struck">F</mi> <mn>3</mn> </msub> </mrow> </math></EquationSource> </InlineEquation> and <InlineEquation ID="IEq13"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10623_2025_1605_Article_IEq13.gif" Format="GIF" Height="17" Rendition="HTML" Resolution="72" Type="Linedraw" Width="44" /> </InlineMediaObject> <EquationSource Format="TEX">\(\beta \in \mathbb {F}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>β</mi> <mo>∈</mo> <mi mathvariant="double-struck">F</mi> </mrow> </math></EquationSource> </InlineEquation>, let <InlineEquation ID="IEq14"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10623_2025_1605_Article_IEq14.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="19" /> </InlineMediaObject> <EquationSource Format="TEX">\(D\)</EquationSource> <EquationSource Format="MATHML"><math> <mi>D</mi> </math></EquationSource> </InlineEquation> be the set of nonzero solutions in <InlineEquation ID="IEq15"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10623_2025_1605_Article_IEq6.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="12" /> </InlineMediaObject> <EquationSource Format="TEX">\(\mathbb {F}\)</EquationSource> <EquationSource Format="MATHML"><math> <mi mathvariant="double-struck">F</mi> </math></EquationSource> </InlineEquation> to the equation <InlineEquation ID="IEq16"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10623_2025_1605_Article_IEq16.gif" Format="GIF" Height="25" Rendition="HTML" Resolution="72" Type="Linedraw" Width="136" /> </InlineMediaObject> <EquationSource Format="TEX">\(\text {Tr}(x^{\frac{q-1}{2\ell ^m}} + \beta x) = \alpha \)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mtext>Tr</mtext> <mo stretchy="false">(</mo> <msup> <mi>x</mi> <mfrac> <mrow> <mi>q</mi> <mo>-</mo> <mn>1</mn> </mrow> <mrow> <mn>2</mn> <msup> <mi>ℓ</mi> <mi>m</mi> </msup> </mrow> </mfrac> </msup> <mo>+</mo> <mi>β</mi> <mi>x</mi> <mo stretchy="false">)</mo> <mo>=</mo> <mi>α</mi> </mrow> </math></EquationSource> </InlineEquation>. In this paper, we investigate a ternary code <InlineEquation ID="IEq17"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10623_2025_1605_Article_IEq17.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="14" /> </InlineMediaObject> <EquationSource Format="TEX">\(\mathcal {C}\)</EquationSource> <EquationSource Format="MATHML"><math> <mi mathvariant="script">C</mi> </math></EquationSource> </InlineEquation> of length <InlineEquation ID="IEq18"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10623_2025_1605_Article_IEq18.gif" Format="GIF" Height="10" Rendition="HTML" Resolution="72" Type="Linedraw" Width="14" /> </InlineMediaObject> <EquationSource Format="TEX">\(n\)</EquationSource> <EquationSource Format="MATHML"><math> <mi>n</mi> </math></EquationSource> </InlineEquation>, defined by <InlineEquation ID="IEq19"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10623_2025_1605_Article_IEq19.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="332" /> </InlineMediaObject> <EquationSource Format="TEX">\(\mathcal {C}:= \{(\text {Tr}(d_1x), \text {Tr}(d_2x), \dots , \text {Tr}(d_nx)): x \in \mathbb {F}\}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi mathvariant="script">C</mi> <mo>:</mo> <mo>=</mo> <mo stretchy="false">{</mo> <mrow> <mo stretchy="false">(</mo> <mtext>Tr</mtext> <mrow> <mo stretchy="false">(</mo> <msub> <mi>d</mi> <mn>1</mn> </msub> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> <mo>,</mo> <mtext>Tr</mtext> <mrow> <mo stretchy="false">(</mo> <msub> <mi>d</mi> <mn>2</mn> </msub> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> <mo>,</mo> <mo>⋯</mo> <mo>,</mo> <mtext>Tr</mtext> <mrow> <mo stretchy="false">(</mo> <msub> <mi>d</mi> <mi>n</mi> </msub> <mi>x</mi> <mo stretchy="false">)</mo> </mrow> <mo stretchy="false">)</mo> </mrow> <mo>:</mo> <mi>x</mi> <mo>∈</mo> <mi mathvariant="double-struck">F</mi> <mo stretchy="false">}</mo> </mrow> </math></EquationSource> </InlineEquation> when we rewrite <InlineEquation ID="IEq20"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10623_2025_1605_Article_IEq20.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="150" /> </InlineMediaObject> <EquationSource Format="TEX">\(D = \{d_1, d_2, \dots , d_n\}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>D</mi> <mo>=</mo> <mo stretchy="false">{</mo> <msub> <mi>d</mi> <mn>1</mn> </msub> <mo>,</mo> <msub> <mi>d</mi> <mn>2</mn> </msub> <mo>,</mo> <mo>⋯</mo> <mo>,</mo> <msub> <mi>d</mi> <mi>n</mi> </msub> <mo stretchy="false">}</mo> </mrow> </math></EquationSource> </InlineEquation>. Using recent results on explicit evaluations of exponential sums, the Weil bound, and combinatorial techniques, we determine the Hamming weight distribution of the code <InlineEquation ID="IEq21"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10623_2025_1605_Article_IEq17.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="14" /> </InlineMediaObject> <EquationSource Format="TEX">\(\mathcal {C}\)</EquationSource> <EquationSource Format="MATHML"><math> <mi mathvariant="script">C</mi> </math></EquationSource> </InlineEquation>. Furthermore, we show that when <InlineEquation ID="IEq22"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10623_2025_1605_Article_IEq22.gif" Format="GIF" Height="17" Rendition="HTML" Resolution="72" Type="Linedraw" Width="75" /> </InlineMediaObject> <EquationSource Format="TEX">\(\alpha = \beta = 0\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>α</mi> <mo>=</mo> <mi>β</mi> <mo>=</mo> <mn>0</mn> </mrow> </math></EquationSource> </InlineEquation>, the dual code of <InlineEquation ID="IEq23"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10623_2025_1605_Article_IEq17.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="14" /> </InlineMediaObject> <EquationSource Format="TEX">\(\mathcal {C}\)</EquationSource> <EquationSource Format="MATHML"><math> <mi mathvariant="script">C</mi> </math></EquationSource> </InlineEquation> is optimal with respect to the Hamming bound.</p>

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A class of ternary codes with few weights

  • Kaimin Cheng

摘要

Let \(\ell ^m\) m be a power with \(\ell \) a prime greater than 3 and \(m\) m a positive integer such that 3 is a primitive root modulo \(2\ell ^m\) 2 m . Let \(\mathbb {F}_3\) F 3 be the finite field of order 3, and let \(\mathbb {F}\) F be the \(\ell ^{m-1}(\ell -1)\) m - 1 ( - 1 ) -th extension field of \(\mathbb {F}_3\) F 3 . Denote by \(\text {Tr}\) Tr the absolute trace map from \(\mathbb {F}\) F to \(\mathbb {F}_3\) F 3 . For any \(\alpha \in \mathbb {F}_3\) α F 3 and \(\beta \in \mathbb {F}\) β F , let \(D\) D be the set of nonzero solutions in \(\mathbb {F}\) F to the equation \(\text {Tr}(x^{\frac{q-1}{2\ell ^m}} + \beta x) = \alpha \) Tr ( x q - 1 2 m + β x ) = α . In this paper, we investigate a ternary code \(\mathcal {C}\) C of length \(n\) n , defined by \(\mathcal {C}:= \{(\text {Tr}(d_1x), \text {Tr}(d_2x), \dots , \text {Tr}(d_nx)): x \in \mathbb {F}\}\) C : = { ( Tr ( d 1 x ) , Tr ( d 2 x ) , , Tr ( d n x ) ) : x F } when we rewrite \(D = \{d_1, d_2, \dots , d_n\}\) D = { d 1 , d 2 , , d n } . Using recent results on explicit evaluations of exponential sums, the Weil bound, and combinatorial techniques, we determine the Hamming weight distribution of the code \(\mathcal {C}\) C . Furthermore, we show that when \(\alpha = \beta = 0\) α = β = 0 , the dual code of \(\mathcal {C}\) C is optimal with respect to the Hamming bound.