M ha nng cao AES v ng dngNguyn Vn Thng, Nguyn Th Ngn65Thut ton m
ha nng cao (AES) v ng dngNG DNG THUT TON M HA TIN TINTRONG BO MT D
LIUNguyn Vn Thng1, Nguyn Th Ngn21 Phng o to trng i hc CNTT&TT,
i hc Thi Nguyn2 Khoa CN in t truyn thng, trng i hc CNTT&TT, i
hc Thi [email protected], [email protected] TT: Bi bo ny
chng ti mun gii thiu mt chun m ha tin tin c, Vin tiu chun v cng ngh
quc gia ca M xut v c s dng nh mt chun lin bang. Thut ton ny gi tt l
AES (Advanced Encryption Standard) n c thc hin gia trn hai php ton
c bn l php cng modun 2 v php nhn modulo trong trng Galios
(GF28).AES l mt thut ton m ha khi i xng, n x l d liu li vo v li ra
u l 128 bit, trong khi di kha c th l 128, 192 hoc 256 bit. Cc li
vo, li ra v kha c biu din bng cc ma trn byte (mi phn t ca ma trn l
mt byte). Thut ton c thc hin vi Nr ln lp (Nr = 10, 12, 14 ty thuc
vo di kha l 128, 192 256 bit) mi ln lp thc hin ln lt 4 th tc sau:
SubBytes, ShiftRows, MixColumns v AddRoundkey tr vng lp cui khng c
th tc Mixcolumn. Thut ton ny c th c thc hin bng phn cng v phn mm. i
vi phn cng, c th dng ngn ng VHDL kt hp vi cng ngh FPGA thc hin.T
kha: M ha tin tin, M ha khi, VHDL-FPGA-AESABSTRACT: This
paperintroduction Advanced Encryption Standard (AES)which has
initiated by National institute of Standard and Technology of US.
Algorithm of AES is based on two basic operations as addition and
multiplication modulo in the Galios field (GF28).The AES algorithm
is a symmetric block cipher that can encrypt (encipher) and decrypt
(decipher) information. The AES algorthim is capable of using
cryptographic key of 128, 192 or 256 bits to encrypt and decrypt
data in block of 128 bits. The input,output and key are represented
bythematrix (each of elements of matrix is the one byte). The
algorithmisimplementedwithNr rounds ( Nr = 10, 12, 14 depend on the
key length is 128, 192, 256 bit ) at the each round executable four
procedure following: SubBytes, ShiftRows, MixColumns and
AddRoundkey except forthe last roundnoprocedureMixcolumn. Algorithm
of this can be implemented with the hardware and software. For the
hardware, be used language VHDL combined with FPGA to
execute.Keyword : Advanced Encryption Standard - AES, symmetric
block cipher, Top security , VHDL- FPGA -AES
1. Gii thiuHin nay vn anh ninh, bo mt h thng thng tin l mt vn
nng c c th gii quan tm,c bit vic bo v cc thng tin, cc ti liu in t.
Vn t ra cn c mt gii php bo mt n gin v hiu qu. Bi vit ny chng ti mun
gii thiu mt chunm ha c coi l an ton v hiu qu: chun m ha nng cao
(Advanced Encryption Standard AES).AES l mt thut ton m ha khi i
xng, n x l d liu li vo v li ra u l 128 bit, trong khi di kha c th l
128, 192 hoc 256 bit. Cc li vo, li ra v kha c biu din bng cc ma trn
byte (mi phn t ca ma trn l mt byte). Thut ton c thc hin vi Nr ln lp
(Nr = 10, 12, 14 ty thuc vo di kha l 128, 192 256 bit) mi ln lp thc
hin ln lt 4 th tc sau: SubBytes, ShiftRows, MixColumns v
AddRoundkey tr vng lp cui khng c th tc Mixcolumn. Thut ton ny c th
c thc hin bng phn cng v phn mm, 2. Cc khi nim v nh ngha:2.1. Cc khi
nim v k hiuBng 1: Cc khi nim v nh ngha c s dng trnh by v chun m ho
nng caoK hiu ngha
AESChun m ho nng cao
Bin i AffinePhep bin i bao gm mt phep nhn vi mt ma trn sau o la
mt phep cng cua mt vect
BitMt s nhi phn nhn gia tri 0 hoc 1
BlockMt day cac bit nhi phn tao thanh input, output, trang thai
(state) va cac kha s dung tai cc vng lp (Round Key) cua h ma. dai
cua day (khi) la s lng cac bit ma no cha. Cac khi cung co th c xem
la mt day cac byte
ByteMt nhom 8 bit
CipherMa hoa
Cipher KeyKha cua h ma, c th c biu din di dang mt mang 2 chiu gm
4 hang va Nk ct
CiphertextBan ma
Inverse CipherGiai ma
Round Key
La cac gia tri sinh ra t kha chinh bng cach s dung thu tuc sinh
kha. Cac kha nay c s dung tai cac vng lp cua thut toan
StateMng trng thi, l cac gia tri ma ha trung gian c th biu din
di dang mt mang 2 chiu gm 4 hang va Nb ct
S-boxMt bang th phi tuyn c s dung trong thu tuc sinh kha va
trong cac bin i thay th cac byte thc hin cac thay th 1-1 i vi mt
gia tri 1 byte
WordMt nhm 32 bit c th c xem nh 1 n v tnh ton c lp hoc l mt mng
4 byte
2.2. Cc hm, th tc, k hiu v cc tham s ca thut tonBng 2: Cc tham s
thut ton, cc k hiu v cc hm c s dng trong m t thut tonK hiu ngha
AddRoundKey()Th tc ny c s dung trong thut toan ma ha va giai ma,
trong thc hin phep toan XOR bit gia mt trang thai trung gian
(State) va mt kha cua vng (Round Key). Kich thc cua mt Round Key
bng kich thc cua trang thai.
InvMixColumns()Ham bin i c s dung trong thut toan giai ma, la
ham ngc cua ham MixColumns()
InvShiftRows()Ham bin i trong thut toan giai ma, la ham ngc cua
ham ShiftRows()
InvSubBytes()Ham bin i trong thut toan giai ma, la ham ngc cua
ham SubBytes()
KKha ma ha
MixColumns()Ham bin i trong thut toan ma ha nhn tt ca cac ct cua
mt trang thai (State) va trn vi d liu cua n (khng phu thuc ln nhau)
nhn c mt ct mi
NbS lng cac ct (la cac word 32 bit) tao thanh mt trang thai, Nb
= 4)
NkS lng cac word 32 bit tao thanh khoa ma hoa K (Nk = 4, 6, hoc
8)
NrS lng cac vong lp cua thut toan, la mt ham cua Nk va Nb (la
cac gia tri c inh) (Nr = 10, 12 hoc 14 tng ng vi cac gia tri khac
nhau cua Nk)
Rcon[]Mang word hng s s dung trong cac vng lp
RotWord()Ham s dung trong thu tuc sinh kha nhn mt word 4-byte va
thc hin mt hoan vi vng
ShiftRows()Ham s dung trong qua trnh ma ha, x ly cac trang thai
bng cach dich vong ba hang cui cua trang thai vi s ln dich khac
nhau
SubBytes()Ham bin i s dung trong qua trnh ma ha, x ly mt trang
thai bng cach s dung mt bang th phi tuyn cac byte (S-box) thao tac
trn mi byte mt cach c lp
SubWord()Ham s dung trong thu tuc sinh kha nhn mt word input
4-byte va s dung mt S-box trn mi gia tri 4-byte nay thu c 1 word
output
XORPhep or bit tuyt i
Phep or bit tuyt i
Phep nhn trn trng hu han
2.3. Cc khi nim v quy c2.3.1. Input va OutputInput va Output l
cc d liu li vo v li ra cua chun ma hoa nng cao, chng u la cac day
128 bit, cn goi la cac khi (block). Kha m cua chun ma ha nng cao la
mt day c dai 128, 192 hoc 256 bit. Chun ma hoa nng cao khng lam vic
vi cac gia tri input, output va khoa co dai khac. Chui cac bit cua
input, output va khoa cua h ma c anh ch s bt u t 0 v kt thc bng mt
gi tr nh hn chiu di ca chui c s dng. V d i l mt ch s bit th 0 i
State array
=>output byte
in0in4in8in12S0,0S0,1S0,2S0,3out0out4out8out12
in1in5in9in13S1,0S1,1S1,2S1,3out1out5out9out13
in2in6in10in14S2,0S2,1S2,2S2,3out2out6out10out14
in3in7in11in15S3,0S3,1S3,2S3,3out3out7out11out15
Hnh 2: Mng trng thi2.3.5. Mng cc tBn byte trn mi ct ca mng trng
thi c to thnh mt t kp 32 bit do mng trng thi c th m t l mng mt
chiu, vi mi phn t gm 4 byte c nh ch mc theo hng r. Cc phn t ca mng
k hiu lw0, w1, w2, w3.Mng trng thi
=>Mng trng thi
ViCc phn t ca mng
S0,0S0,1S0,2S0,3(w0,w1,w2,w3)w0=( S0,0, S1,0, S2,0, S3,0)
S1,0S1,1S1,2S1,3w1 =( S0,1, S1,1, S2,1, S3,1)
S2,0S2,1S2,2S2,3w2=( S0,2, S1,2, S2,2, S3,2)
S3,0S3,1S3,2S3,3w3=( S0,3, S1,3, S2,3, S3,3)
2.4. C s ton hcCc byte trong thut ton AES u c biu din nh mt phn
t trong trng hu hn s dng cc k hiu nh trnh by phn 2.3.2. Cc phn t
trong trng hu hn u c th c cng v nhn. Tuy nhin cc php ton ny khc hn
vi cc php cng v nhn thng thng.Hai php cng v php nhn trn trng
Galoris GF (28) l c s ton hc ca thut ton AES.2.4.1. Php cngPhp cng
hai phn t trong trng hu hn c thc hin bng vic Cng cc h s c cng trng
s trong cc a thc biu din cc phn t. Vic cng c thc hin thng qua php
ton XOR k hiu lABAB
000
011
101
110
Nh vy vic cng cc phn t trong trng hu hn c th m t bng vic cng
modulo 2 ca cc bit tng ng trong byte. Cho hai byte
a7a6a5a4a3a2a1a0} v {b7b6b5b4b3b2b1b0}, th tng hai byte l
{c7c6c5c4c3c2c1c0}, trong ci = ai bi vi i l v tr cc bit tng ng.V d:
(x6+x4+x2+x+1) (x7+x+1)=x7+x6+x4+x2(Biu din dng a
thc){01010111}{10000011}={11010100}(Biu din dng nh
phn){57}{83}={d4}(Biu din dng hexa)2.4.2. Php nhnPhp nhn trn trng
Galois GF(28), k hiu l tng ng vi php nhn thng thng ca hai a thc ri
em chia ly d (modulo) cho mt a thc ti gin bc 8 (a thc ti gin l a
thc ch chia ht cho 1 v chnh n). Trong thut ton AES, a thc ti gin c
chn l: m(x) = x8 + x4 + x3 + x + 1hay biu din di dng hexa l {01}
{1b}V d: {57}{83}={c1}V (x6 + x4 + x2 + x + 1) (x7 + x + 1) = x13 +
x11 + x9+ x8 + x7 + x7 + x5+x3+x2+x+x6 + x4 + x2 + x + 1 =x13 + x11
+ x9+ x6 + x5 + x4 + x3 + 1x13 + x11 + x9+ x6 + x5 + x4 + x3 + 1
chia cho x8 + x4 + x3 + x + 1 d x7 + x6 + 1x7 + x6 + 1 biu din nh
phn l: 11000001 hay c1 trong biu din hexa.Vic chia cho a thc bc 8
m(x) bo m kt qu nhn c ca php rt gn l mt a thc c bc nh hn 8 nn c th
biu din c di dng 1 byte.2.4.3. Php nhn vi xPhp nhn vi a thc x (hay
phn t {00000010} GF(28) c th c thc hin mc byte bng mt php dch tri v
sau thc hin tip php XOR vi gi tr {01}{1b} nu bit b7=1. Th tc nhn vi
x k hiu lxtime(). Php nhn vi cc lu tha ca x c th c thc hin bng cch
p dng nhiu ln th tcxtime(). Kt qu php nhn vi mt gi tr bt k c xc nh
bng php cng () cc kt qu trung gian ny li vi nhau.V d:Php nhn
{a2}{03}Ta c {03} = {02}{01}Vy {a2}{03}={a2}{02}{a2}{01}{a2}{01}
={a2}//1010 0010{a2}{02} ={01}{44}{01}{1b} ={5F}Kt qu
{a2}{03}={a2}{5F} = {FD}3. Thut tonThut ton AES c thc hin tun t gm
nhiu bc bin i, kt qu u ra ca php bin i trc l u vo ca php bin i tip
theo. Kt qu trung gian ca cc php bin i chnh l mng trng thi (state)
di ca khi d liu u vo ca AES l c nh vi Nb=4. Ty vo di kha Nk (Nk=4,
6, 8) ta c s ln lp Nr(Nr =10, 12, 14).3.1. Thut ton m ho
State = InputAddRoundKey(State,
Roundkey)SubByte(State)ShiftRow(State)MixColumn(State)AddRoundKey(State,
Roundkey)SubByte(State)ShiftRow(State)AddRoundKey(State,
Roundkey)Key ScheduleI
