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低深度轻量化S盒的优化实现

冯子曦 刘玉鹏 窦国威 刘承乐

冯子曦, 刘玉鹏, 窦国威, 刘承乐. 低深度轻量化S盒的优化实现[J]. 电子与信息学报. doi: 10.11999/JEIT250690
引用本文: 冯子曦, 刘玉鹏, 窦国威, 刘承乐. 低深度轻量化S盒的优化实现[J]. 电子与信息学报. doi: 10.11999/JEIT250690
FENG Zixi, LIU Yupeng, DOU Guowei, LIU Chengle. Optimized Implementation of Low-Depth Lightweight S-Boxes[J]. Journal of Electronics & Information Technology. doi: 10.11999/JEIT250690
Citation: FENG Zixi, LIU Yupeng, DOU Guowei, LIU Chengle. Optimized Implementation of Low-Depth Lightweight S-Boxes[J]. Journal of Electronics & Information Technology. doi: 10.11999/JEIT250690

低深度轻量化S盒的优化实现

doi: 10.11999/JEIT250690 cstr: 32379.14.JEIT250690
详细信息
    作者简介:

    冯子曦:男,硕士生,研究方向为对称密码学

    刘玉鹏:男,工程师,研究方向为对称密码学

    窦国威:男,工程师,研究方向为对称密码学、侧信道分析

    刘承乐:男,正高级工程师,研究方向为对称密码学、密码安全性分析

    通讯作者:

    刘承乐 lhx884@163.com

  • 中图分类号: TP309.7

Optimized Implementation of Low-Depth Lightweight S-Boxes

  • 摘要: 在轻量级密码领域,S盒作为核心非线性组件,其硬件实现的面积与电路深度优化始终是轻量级密码研究的热点。公开文献针对小规模S盒硬件实现占用资源、电路深度的优化等进行了大量研究,取得了很多优秀成果,能够对规模不大于5比特的S盒给出面积或电路深度优的实现方案,现有针对小规模 S 盒硬件实现的研究多以面积最小化或电路深度最小化为单一优化目标,对两者之间的协同优化工作较少。该文从面积与电路深度协同优化的角度出发,构建了一个深度为$ k $,宽度为$ w $的电路模型,在给定面积约束的条件下,采用可满足性问题(SAT)求解技术判断该电路模型能否实现S盒,通过调整电路深度、宽度、面积等指标,最终给出S盒的优化实现方案。采用该方法能够对规模不大于4比特的S盒,给出面积和电路深度均较优的实现方案。在UMC 180 nm工艺库下对Lblock、Rectangle、Midori、Prøst等轻量级密码算法的S盒进行了优化实现,给出了较好的实验结果,将Lblock算法S盒的实现深度由10降低到了3,将Rectangle算法S盒的逆盒实现面积由24.33 GE降低到了17.66 GE,将Midori算法S盒的面积由20.00 GE降低到了16.33 GE、将Prøst算法S盒的实现面积由22.00 GE降低到了13.33 GE。
  • 表  1  符号及含义

    符号 定义
    $ \neg a $ 布尔变量$ a $的逻辑非
    $ a\wedge b $ 布尔变量$ a $和$ b $的逻辑与
    $ a\vee b $ 布尔变量$ a $和$ b $的逻辑或
    $ a\oplus b $ 布尔变量$ a $和$ b $的异或
    $ a\uparrow b $ 布尔变量$ a $和$ b $的与非
    $ a\downarrow b $ 布尔变量$ a $和$ b $的或非
    $ a\leftrightarrow b $ 布尔变量$ a $和$ b $的异或非
    $ x_{\rm{i}} $ S盒的输入
    $ y_{\rm{i}} $ S盒的输出
    $ {q}_{4i} $ 第$ i $个门的输入
    $ t_{\rm{i}} $ 第$ i $个门的输出
    $ a_{\rm{i}} $ 逻辑门之间的布线
    $ b_{\rm{i}}$ 单个逻辑门内部的布线
    $ k $ S盒电路实现的深度
    $ w $ 每层允许的逻辑门最大个数
    $ G $ 逻辑电路的总面积
    $ n $ S盒的位数
    下载: 导出CSV

    表  2  逻辑元件及表达式

    逻辑元件符号 输出表达式 UMC 180 nm (GE)
    NOT($ x{_{0}} $) $ \neg {x}_{0} $ $ 0.67 $
    AND$ (x{_{0}},x_{1} ) $ $ x{_{0}}\wedge x_{1} $ $ 1.33 $
    OR$ (x{_{0}},x_{1} ) $ $ x{_{0}}\vee x_{1} $ $ 1.33 $
    XOR$ (x{_{0}},x_{1} ) $ $ x{_{0}} \oplus x_{1} $ $ 2.67 $
    NAND$ (x{_{0}},x_{1} ) $ $ \neg (x{_{0}}\wedge x_{1} ) $ $ 1.00 $
    NOR$ (x{_{0}},x_{1} ) $ $ \neg (x{_{0}}\vee x_{1} ) $ $ 1.00 $
    XNOR$ (x{_{0}},x_{1} ) $ $ \neg (x{_{0}} \oplus x_{1} ) $ $ 2.00 $
    ANDN$ (x{_{0}},x_{1} ) $ $ \neg x{_{0}}\wedge x_{1} $ $ 1.67 $
    ORN$ (x{_{0}},x_{1} ) $ $ \neg x{_{0}}\vee x_{1} $ $ 1.67 $
    NANDN$ (x{_{0}},x_{1} ) $ $ x{_{0}}\vee \neg x_{1} $ $ 1.67 $
    NORN$ (x{_{0}},x_{1} ) $ $ x{_{0}}\wedge \neg x_{1} $ $ 1.67 $
    AND3$ (x{_{0}},x_{1} ,x{_{2}}) $ $ x{_{0}}\wedge x_{1}\wedge x{_{2}} $ $ 2.33 $
    OR3$ (x{_{0}},x_{1},x{_{2}}) $ $ x{_{0}}\vee x_{1} \vee x{_{2}} $ $ 2.33 $
    XOR3$ (x{_{0}},x_{1} ,x{_{2}}) $ $ x{_{0}} \oplus x_{1} \oplus x{_{2}} $ $ 4.67 $
    NAND3$ (x{_{0}},x_{1} ,x{_{2}}) $ $ \neg (x{_{0}}\wedge x_{1}\wedge x{_{2}}) $ $ 1.33 $
    NOR3$ (x{_{0}},x_{1} ,x{_{2}}) $ $ \neg (x{_{0}}\vee x_{1}\vee x{_{2}}) $ $ 1.33 $
    XNOR3$ (x{_{0}},x_{1} ,x{_{2}}) $ $ \neg (x{_{0}} \oplus x_{1} \oplus x{_{2}}) $ $ 4.67 $
    MAOI1$ (x{_{0}},x_{1} ,x{_{2}},x{_{3}}) $ $ \neg ((a\wedge b)\vee (\neg (c\vee d))) $ $ 2.67 $
    MOAI1$ (x{_{0}},x_{1} ,x{_{2}},x{_{3}}) $ $ \neg ((a\vee b)\wedge (\neg (c\wedge d))) $ $ 2.00 $
    下载: 导出CSV

    表  3  不同类型门的表示

    $ {b}_{8i}\parallel {b}_{8i+1}\parallel {b}_{8i+2} $
    $ \parallel {b}_{8i+3}\parallel {b}_{8i+4}\parallel {b}_{8i+5} $
    $ \parallel {b}_{8i+6}\parallel {b}_{8i+7} $
    门类型门函数
    0 0 0 0 0 0 1 1NOT$ \neg {q}_{0} $
    0 0 0 0 0 1 0 0XOR$ {q}_{0}\mathrm{ \oplus }{q}_{1} $
    0 0 0 0 0 1 0 1XNOR$ {q}_{0}\leftrightarrow {q}_{1} $
    0 0 0 0 0 1 1 1NOT$ \neg {q}_{1} $
    0 0 0 0 1 0 0 0AND$ {q}_{0}\wedge {q}_{1} $
    0 0 0 0 1 0 0 1NAND$ {q}_{0}\uparrow {q}_{1} $
    0 0 0 0 1 0 1 0NORN$ {q}_{0}\wedge \neg {q}_{1} $
    0 0 0 0 1 0 1 1ORN$ \neg {q}_{0}\vee {q}_{1} $
    0 0 0 0 1 1 0 0OR$ {q}_{0}\vee {q}_{1} $
    0 0 0 0 1 1 0 1NOR$ {q}_{0}\downarrow {q}_{1} $
    0 0 0 0 1 1 1 0ANDN$ \neg {q}_{0}\wedge {q}_{1} $
    0 0 0 0 1 1 1 1NANDN$ {q}_{0}\vee \neg {q}_{1} $
    0 0 0 1 0 0 0 0XOR3$ {q}_{0}\mathrm{ \oplus }{q}_{1}\mathrm{ \oplus }{q}_{2} $
    0 0 0 1 0 0 0 1XNOR3$ \neg \left({q}_{0}\mathrm{ \oplus }{q}_{1}\mathrm{ \oplus }{q}_{2}\right) $
    0 1 0 0 0 0 0 0AND3$ {q}_{0}\wedge {q}_{1}\wedge {q}_{2} $
    0 1 0 0 0 0 0 1NAND3$ \neg \left({q}_{0}\wedge {q}_{1}\wedge {q}_{2}\right) $
    0 1 1 0 0 0 0 0OR3$ {q}_{0}\vee {q}_{1}\vee {q}_{2} $
    0 1 1 0 0 0 0 1NOR3$ \neg \left({q}_{0}\vee {q}_{1}\vee {q}_{2}\right) $
    1 0 0 0 0 0 0 0MAOI1$ \neg (({q}_{0}\wedge {q}_{1})\vee (\neg ({q}_{2}\vee {q}_{3}))) $
    1 0 0 0 0 0 0 1MOAI1$ \neg (({q}_{2}\vee {q}_{3})\wedge (\neg ({q}_{0}\wedge {q}_{1}))) $
    下载: 导出CSV
     输入:$ s_{7}^{i},s_{6}^{i},s_{5}^{i},s_{4}^{i},s_{3}^{i},s_{2}^{i},s_{1}^{i},s_{0}^{i},i\mathrm{\epsilon }[0,kw-1] $:每个门的面积
     $ {L}_{7},{L}_{6}, {L}_{5}, {L}_{4}, {L}_{3}, {L}_{2}, {L}_{1}, {L}_{0} $面积上界
     $ {A}_{7},{A}_{6}, {A}_{5}, {A}_{4}, {A}_{3}, {A}_{2}, {A}_{1}, {A}_{0} $:进位
     $ {A}_{-1} $:辅助量
     $ {S}_{7},{S}_{6}, {S}_{5}, {S}_{4}, {S}_{3}, {S}_{2}, {S}_{1}, {S}_{0} $:电路面积
     输出:如果以下问题有解,则返回True;若无解,则返回False
     1 $ {S}_{7}={S}_{6}={S}_{5}={S}_{4}={S}_{3}={S}_{2}={S}_{1}={S}_{0}=0 $;
     2 $ {A}_{7}={A}_{6}={A}_{5}={A}_{4}={A}_{3}={A}_{2}={A}_{1}={A}_{0}={A}_{-1}=0; $
     3 $ {e}_{8}=1 $
     4 for$ i\leftarrow 0 \;\text{to}\;k\cdot w-1 \;\text{do} $:
     5  $ \bf{for}\;j\leftarrow 0 \;\mathrm{to} \;7 \;\text{do} $:
     6   $ {S}_{j}={S}_{j}\mathrm{ \oplus }s_{j}^{i}\mathrm{ \oplus }{A}_{j-1} $
     7   $ {A}_{j}={S}_{j}\bigwedge s_{j}^{i}\bigwedge {A}_{j-1} $
     8  $ \bf{end\;for} $
     9 $ \bf{end\;for} $
     10 $ \bf{for}\;j\leftarrow 7\; \mathrm{to}\; 0 \;\text{do} $:
     11 $ {e}_{j}={e}_{j+1}\wedge \left({S}_{j}\leftrightarrow {L}_{j}\right) $
     12 $ \bf{end \;for} $
     13 $ \bf{for}\;j\leftarrow 7\; \mathrm{to} \;0 \;\text{do} $:
     14 $ {c}_{j}\leftrightarrow \left({e}_{j+1}\wedge \neg {S}_{j}\wedge {L}_{j}\right) $
     15 $ \bf{end\;for} $
     16 if $ {c}_{0}\bigvee {c}_{1}\bigvee {c}_{2}\bigvee {c}_{3}\bigvee {c}_{4}\bigvee {c}_{5}\bigvee {c}_{6}\bigvee {c}_{7}==1 $
     17 return true;
     18 else
     19 return false;
    下载: 导出CSV

    表  4  实验结果比较

    S盒 本文结果 [8] [9] [5]
    深度 面积(GE) 深度 面积(GE) 深度 面积(GE) 深度 面积(GE)
    Joltik/Piccolo 4 13.00 4 13.00 4 22.33 4 13.00
    Rectangle 7 18.33 7 18.33 3 28.33 8 20.34
    $ {\text{Rectangle}}^{-1} $ 3 17.66 $ - $ $ - $ 3 24.33 8 20.34
    Skinny 4 13.33 4 13.33 $ - $ $ - $ 4 13.33
    Lblock/Lac 3 16.33 10 16.33 3 26.00 8 16.33
    Midori 3 16.33 $ - $ $ - $ $ - $ $ - $ 4 20.00
    Prøst 4 13.33 $ - $ $ - $ 4 22.00 $ - $ $ - $
    5 13.00
    下载: 导出CSV

    表  5  Joltik/Piccolo S盒的实现

    Joltik/Piccolo k=4 w=2 13.00 GE
    $ {t}_{0}= $OR$ \left({x}_{0},{x}_{1}\right) $ $ {t}_{1}= $OR$ \left({x}_{1},{x}_{2}\right) $
    $ {t}_{2}= $XNOR$ \left({x}_{3},{t}_{0}\right) $ $ {t}_{3}= $MOAI1$ \left({x}_{2},{t}_{0},{x}_{0},{t}_{1}\right) $
    $ {t}_{4}= $NOR$ \left({x}_{2},{t}_{2}\right) $ $ {t}_{5}= $OR$ \left({t}_{2},{t}_{3}\right) $
    $ {t}_{6}= $MOAI1$ \left({x}_{1},{t}_{4},{t}_{4},{x}_{1}\right) $ $ {t}_{7}= $XNOR$ \left({x}_{2},{t}_{5}\right) $
    $ {y}_{0}={t}_{2} $ $ {y}_{1}={t}_{3} $ $ {y}_{2}={t}_{6} $ $ {y}_{3}={t}_{7} $
    下载: 导出CSV

    表  6  Skinny S盒的实现

    Skinny k=4 w=3 13.33 GE
    $ {t}_{0}= $OR$ \left({x}_{2},{x}_{3}\right) $ $ {t}_{1}= $OR$ \left({x}_{0},{x}_{1}\right) $ $ {t}_{2}= $ OR$ \left({x}_{1},{x}_{2}\right) $
    $ {t}_{3}= $ XNOR$ \left({x}_{0},{t}_{2}\right) $ $ {t}_{4}= $MOAI1$ \left({x}_{3},{t}_{1},{x}_{3},{t}_{1}\right) $
    $ {t}_{5}= $OR$ \left({x}_{0},{t}_{4}\right) $
    $ {t}_{6}= $MOAI1$ \left({t}_{2},{t}_{4},{t}_{0},{x}_{1}\right) $ $ {t}_{7}= $MOAI1$ \left({t}_{3},{t}_{0},{t}_{5},{x}_{2}\right) $
    $ {y}_{0}={t}_{4} $ $ {y}_{1}={t}_{3} $ $ {y}_{2}={t}_{6} $ $ {y}_{3}={t}_{7} $
    下载: 导出CSV

    表  7  Rectangle S盒的实现

    Rectangle k=7 w=2 18.33 GE
    $ {t}_{0}= $MOAI1$ \left({x}_{0},{x}_{2},{x}_{0},{x}_{0}\right) $ $ {t}_{1}= $XNOR$ \left({x}_{0},{x}_{2}\right) $
    $ {t}_{2}= $MOAI1$ \left({x}_{3},{t}_{0},{x}_{3},{t}_{0}\right) $
    $ {t}_{3}= $MOAI1$ \left({x}_{1},{t}_{2},{t}_{2},{x}_{1}\right) $
    $ {t}_{4}= $MAOI1$ \left({x}_{2},{x}_{1},{x}_{2},{t}_{3}\right) $
    $ {t}_{5}= $XNOR$ \left({x}_{0},{t}_{4}\right) $ $ {t}_{6}= $MOAI1$ \left({{{t}_{2}},{{t}_{1}},t}_{2},{t}_{4}\right) $
    $ {t}_{7}= $OR$ \left({t}_{6},{t}_{5}\right) $
    $ {t}_{8}= $MOAI1$ \left({t}_{7},{t}_{2},{t}_{7},{t}_{2}\right) $
    $ {y}_{0}={t}_{6} $ $ {y}_{1}={t}_{8} $ $ {y}_{2}={t}_{3} $ $ {y}_{3}={t}_{5} $
    下载: 导出CSV

    表  8  Lblock S盒的实现

    Lblock k=3 w=4 16.33 GE
    $ {t}_{0}= $MAOI1$ \left({x}_{2},{x}_{1},{x}_{3},{x}_{1}\right) $$ {t}_{1}= $XNOR$ \left({x}_{2},{x}_{3}\right) $$ {t}_{2}= $OR$ \left({x}_{0},{x}_{1}\right) $$ {t}_{3}= $NOT$ \left({x}_{0}\right) $
    $ {t}_{4}= $MOAI1$ \left({t}_{2},{t}_{1},{t}_{2},{t}_{1}\right) $$ {t}_{5}= $XNOR$ \left({x}_{0},{t}_{0}\right) $$ {t}_{6}= $NOR$ \left({x}_{0},{t}_{0}\right) $
    $ {t}_{7}= $MOAI1$ \left({t}_{0},{t}_{4},{t}_{4},{t}_{3}\right) $$ {t}_{8}= $MAOI1$ \left({t}_{2},{t}_{6},{x}_{1},{t}_{6}\right) $
    $ {y}_{0}={t}_{7} $$ {y}_{1}={t}_{8} $$ {y}_{2}={t}_{5} $$ {y}_{3}={t}_{4} $
    下载: 导出CSV

    表  9  $ {\mathbf{Rectangle}}^{-\mathbf{1}} $ S盒的实现

    $ {\mathbf{Rectangle}}^{-\mathbf{1}} $ k=3 w=4 17.66 GE
    $ {t}_{0}= $ MOAI1$ \left({x}_{1},{x}_{3},{x}_{0},{x}_{3}\right) $$ {t}_{1}= $MOAI1$ \left({x}_{3},{x}_{2},{x}_{3},{x}_{2}\right) $$ {t}_{2}= $ NOR$ \left({x}_{0},{x}_{3}\right) $$ {t}_{3}= $NOT$ \left({x}_{2}\right) $
    $ {t}_{4}= $MAOI1$ \left({t}_{1},{t}_{0},{t}_{1},{t}_{0}\right) $$ {t}_{5}= $MOAI1$ \left({x}_{1},{t}_{2},{x}_{1},{t}_{2}\right) $$ {t}_{6}= $NOT$ \left({t}_{1}\right) $
    $ {t}_{7}= $MOAI1$ \left({t}_{5},{t}_{4},{t}_{4},{t}_{0}\right) $$ {t}_{8}= $XNOR$ \left({t}_{5},{t}_{3}\right) $$ {t}_{9}= $MAOI1$ \left({t}_{5},{t}_{6},{t}_{0},{t}_{5}\right) $
    $ {y}_{0}={t}_{9} $$ {y}_{1}={t}_{8} $$ {y}_{2}={t}_{4} $$ {y}_{3}={t}_{7} $
    下载: 导出CSV

    表  10  Prøst S盒的实现

    Prøst k=5 w=2 13.00 GE
    $ {t}_{0}= $NAND$ \left({x}_{0},{x}_{1}\right) $ $ {t}_{1}= $NAND$ \left({x}_{2},{x}_{1}\right) $
    $ {t}_{2}= $XNOR$ \left({x}_{2},{t}_{0}\right) $ $ {t}_{3}= $XNOR$ \left({x}_{3},{t}_{1}\right) $
    $ {t}_{4}= $MOAI1$ \left({x}_{1},{x}_{1},{x}_{2},{t}_{3}\right) $ $ {t}_{5}= $NAND$ \left({t}_{3},{t}_{2}\right) $
    $ {t}_{6}= $XNOR$ \left({x}_{0},{t}_{5}\right) $
    $ {t}_{7}= $MOAI1$ \left({t}_{4},{t}_{4},{x}_{3},{t}_{6}\right) $
    $ {y}_{0}={t}_{2} $ $ {y}_{1}={t}_{3} $ $ {y}_{2}={t}_{6} $ $ {y}_{3}={t}_{7} $
    下载: 导出CSV

    表  11  Prøst S盒的实现

    Prøst k=4 w=4 13.33 GE
    $ {t}_{0}= $NAND$ \left({x}_{0},{x}_{1}\right) $$ {t}_{1}= $NOT$ \left({x}_{0}\right) $$ {t}_{2}= $NAND$ \left({x}_{2},{x}_{1}\right) $$ {t}_{3}= $NOT$ \left({x}_{1}\right) $
    $ {t}_{4}= $XNOR$ \left({t}_{0},{x}_{2}\right) $$ {t}_{5}= $XNOR$ \left({x}_{3},{t}_{2}\right) $
    $ {t}_{6}= $MOAI1$ \left({t}_{1},{t}_{4},{t}_{1},{x}_{2}\right) $
    $ {t}_{7}= $MOAI1$ \left({t}_{5},{t}_{1},{t}_{5},{t}_{6}\right) $$ {t}_{8}= $MOAI1$ \left({t}_{3},{x}_{2},{x}_{3},{t}_{6}\right) $
    $ {y}_{0}={t}_{4} $$ {y}_{1}={t}_{5} $$ {y}_{2}={t}_{7} $$ {y}_{3}={t}_{8} $
    下载: 导出CSV

    表  12  midori_$ {\mathbf{s}}_{\mathbf{0}} $ S盒的实现

    midori_$ {\mathbf{s}}_{\mathbf{0}} $ k=3 w=4 16.33 GE
    $ {t}_{0}= $NOT$ \left({x}_{0}\right) $$ {t}_{1}= $XNOR$ \left({x}_{0},{x}_{1}\right) $$ {t}_{2}= $OR$ \left({x}_{0},{x}_{3}\right) $$ {t}_{3}= $NAND$ \left({x}_{1},{x}_{1}\right) $
    $ {t}_{4}= $MOAI1$ \left({x}_{3},{t}_{1},{t}_{0},{t}_{1}\right) $$ {t}_{5}= $ MOAI1$ \left({t}_{1},{t}_{3},{t}_{1},{x}_{3}\right) $$ {t}_{6}= $MAOI1$ \left({t}_{2},{x}_{2},{t}_{0},{t}_{3}\right) $
    $ {t}_{7}= $ MOAI1$ \left({t}_{1},{t}_{4},{x}_{2},{t}_{4}\right) $$ {t}_{8}= $ MAOI1$ \left({t}_{4},{t}_{2},{x}_{2},{t}_{4}\right) $
    $ {y}_{0}={t}_{6} $$ {y}_{1}={t}_{8} $$ {y}_{2}={t}_{5} $$ {y}_{3}={t}_{7} $
    下载: 导出CSV
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出版历程
  • 收稿日期:  2025-07-23
  • 修回日期:  2026-02-13
  • 录用日期:  2026-02-13
  • 网络出版日期:  2026-03-07

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