LiCi分組密碼算法的不可能差分分析
doi: 10.11999/JEIT180729
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桂林電子科技大學(xué)廣西密碼學(xué)與信息安全重點(diǎn)實(shí)驗(yàn)室? ?桂林? ?541004
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桂林電子科技大學(xué)廣西無線寬帶通信與信號(hào)處理重點(diǎn)實(shí)驗(yàn)室? ?桂林? ?541004
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桂林電子科技大學(xué)廣西高校云計(jì)算與復(fù)雜系統(tǒng)重點(diǎn)實(shí)驗(yàn)室? ?桂林? ?541004
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中國科學(xué)院大學(xué)? ?北京? ?100049
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中國科學(xué)院軟件研究所? ?北京? ?100190
Impossible Differential Cryptanalysis of LiCi Block Cipher
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Guangxi Key Laboratory of Cryptography and Information Security, Guilin University of Electronic Technology, Guilin 541004 ,China
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Guangxi Key Laboratory of Wireless Wideband Communication and Signal Processing, Guilin University of Electronic Technology, Guilin 541004 ,China
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Guangxi Colleges and University Key Laboratory of Cloud Computing and Complex Systems, Guilin University of Electronic Technology, Guilin 541004 ,China
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University of Chinese Academy of Sciences, Beijing 100049, China
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Institute of Software, Chinese Academy of Sciences, Beijing 100190, China
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摘要: LiCi是由Patil等人(2017)提出的輕量級(jí)分組密碼算法。由于采用新型的設(shè)計(jì)理念,該算法具有結(jié)構(gòu)緊湊、能耗低、占用芯片面積小等優(yōu)點(diǎn),特別適用于資源受限的環(huán)境。目前該算法的安全性備受關(guān)注,Patil等人聲稱:16輪簡(jiǎn)化算法足以抵抗經(jīng)典的差分攻擊及線性攻擊。該文基于S盒的差分特征,結(jié)合中間相遇思想,構(gòu)造了一個(gè)10輪的不可能差分區(qū)分器?;诖藚^(qū)分器,向前后各擴(kuò)展3輪,并利用密鑰編排方案,給出了LiCi的一個(gè)16輪的不可能差分分析方法。該攻擊需要時(shí)間復(fù)雜度約為283.08次16輪加密,數(shù)據(jù)復(fù)雜度約為259.76選擇明文,存儲(chǔ)復(fù)雜度約為276.76數(shù)據(jù)塊,這說明16輪簡(jiǎn)化的LiCi算法無法抵抗不可能差分攻擊。Abstract: LiCi algorithm is a newly lightweight block cipher. Due to its new design idea adopted by Patil et al, it has the advantages of compact design, low energy consumption and less chip area, thus is is especially suitable for resource-constrained environments. Currently, its security receives extensively attention, and Patil et al. claimed that the 16-round reduced LiCi can sufficiently resist both differential attack and linear attack. In this paper, a new 10-round impossible differential distinguisher is constructed based on the differential characteristics of the S-box and the meet-in-the-middle technique. Moreover, on the basis of this distinguisher, a 16-round impossible differential attack on LiCi is proposed by respectively extending 3-round forward and backward via the key scheduling scheme. This attack requires a time complexity of about 283.08 16-round encryptions, a data complexity of about 259.76 chosen plaintexts, and a memory complexity of 276.76 data blocks, which illustrates that the 16-round LiCi cipher can not resist impossible differential attack.
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表 1 S盒輸入/輸出差分特征
輸入差分 0001 0100 1000 1100 1101 **11 **1* **0* ***1 ***0 輸出差分 1*** ***1 ***1 1**0 0*** 0001 0010 0011 0100 0101 下載: 導(dǎo)出CSV
表 2 S盒輸入/輸出差分特征概率
輸入差分 **11 **1* ***1 **** 0001 0*** *0** 01** *000 輸出差分 0001 0010 0100 1000 1*** **** **** **** **** 概率 1/4 1/8 1/8 1/16 1/8 1/2 1/2 1/4 1/8 下載: 導(dǎo)出CSV
表 3 不可能差分分析16輪LiCi算法的密鑰恢復(fù)各步驟的時(shí)間復(fù)雜度
步驟 恢復(fù)的密鑰比特 猜測(cè)比特?cái)?shù)目 時(shí)間復(fù)雜度(單位:次16輪加密所用的時(shí)間) (1) – 0 ${2^{m + 49}} \times \frac{1}{{8 \times 16}} = {2^{{{m}} + 42}}$ (2) ${\rm{Rk}}{0_1}\left[ {24 \sim 21} \right]$ 4 ${2^{m + 21}} \times {2^4} \times \frac{1}{{8 \times 16}} = {2^{{{m}} + 18}}$ (3) ${\rm{Rk}}{15_2}\left[ {15 \sim 12} \right]$ 4 ${2^{m + 18}} \times {2^4} \times {2^4} \times \frac{1}{{8 \times 16}} = {2^{{{m}} + 19}}$ (4) ${\rm{Rk}}{15_2}\left[ {31 \sim 28} \right]$ 4 ${2^{m + 15}} \times {2^4} \times {2^4} \times {2^4} \times \frac{1}{{8 \times 16}} = {2^{{{m}} + 20}}$ (5) ${\rm{Rk}}{15_2}\left[ {23 \sim 20} \right]$ 4 ${2^{m + 14}} \times {2^4} \times {2^4} \times {2^4} \times {2^4} \times \frac{1}{{8 \times 16}} = {2^{{{m}} + 23}}$ (6) ${\rm{Rk}}{15_1}\left[ {16 \sim 13} \right]$, ${\rm{Rk}}{14_2}\left[ {23 \sim 20} \right]$ 8 ${2^{m + 13}} \times {2^4} \times {2^4} \times {2^4} \times {2^4} \times {2^8} \times \frac{1}{{8 \times 16}} = {2^{{{m}} + 30}}$ (7) ${\rm{Rk}}{15_2}\left[ {19 \sim 16,11 \sim 9} \right]$, ${\rm{Rk}}{15_1}\left[ {12 \sim 9} \right]$, ${\rm{Rk}}{14_2}\left[ {19 \sim 16} \right]$ 12 ${2^{m + 10}} \times {2^{16}} \times {2^8} \times {2^{12}} \times 2 \times \frac{1}{{8 \times 16}} = {2^{{{m}} + 40}}$ (8) ${\rm{Rk}}{15_2}\left[ {27 \sim 24} \right]$, ${\rm{Rk}}{15_1}\left[ {20 \sim 17} \right]$, ${\rm{Rk}}{14_2}\left[ {27 \sim 24} \right]$ 12 ${2^{m + 7}} \times {2^{24}} \times {2^{12}} \times {2^{12}} \times 2 \times \frac{1}{{8 \times 16}} = {2^{{{m}} + 49}}$ (9) ${\rm{Rk}}{15_2}\left[ {8 \sim 6} \right]$, ${\rm{Rk}}{15_1}\left[ {8 \sim 6} \right]$, ${\rm{Rk}}{15_2}\left[ {15 \sim 13} \right]$, ${\rm{Rk}}{14_1}\left[ {16 \sim 13} \right]$ 10 ${2^{m + 5}} \times {2^{24}} \times {2^{12}} \times {2^{12}} \times {2^{10}} \times 2 \times \frac{1}{{8 \times 16}} = {2^{{{m}} + 56}}$ (10) ${\rm{Rk}}{0_1}\left[ {20 \sim 9} \right]$, ${\rm{Rk}}{0_2}\left[ {23 \sim 20} \right]$, ${\rm{Rk}}{1_1}\left[ {16 \sim 13} \right]$ 20 ${2^{m{\rm{ - }}16}} \times {2^{58}} \times {2^{20}} \times 3 \times \frac{1}{{8 \times 16}} + {2^{74}} = {2^{{{m}} + 56.58}} + {2^{74}}$ 下載: 導(dǎo)出CSV
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