By: Matthew Ng
 AND, XOR, OR, Complement, Circular Left shift, and
Addition Modulo
 Circular Left Shift is done with s positions (0 ≤ s ≤ 31) –
Denoted by ROTLs
 Addition Modulo is 2 32
 This is pads the input.
 The binary representation of |x| is l and it is at most 64 bits
 If |l| < 64 bits then it is padded from the left with zeros to
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make 64 bits.
In the construction of y – one is appended to x and then
concatenated with enough zeros so the length is congruent
to 448 modulo 512.
A concatenation 64 bits that contain binary representation
of the original length of x.
Divide the string by 512 and create a concatenation of n
blocks each of them being 512 bits.
y = M1 || M2 || … || Mn
 SHA-1-pad (x) :
 D = (447 - |x|) Mod 512
 L = binary of |x|
 y = x || 1 || 0d || 1
 |x| ≤ 264 - 1
where |l| = 64
• ABCDE are 32-bit words in the
state
• F is a non-linear function that
varies
• <<<n is a left bit rotation by n
places (n varies for each
operation)
• Wt is the expanded message
word of round t
• Kt is the round constant of
round t
• Boxes addition modulo 232
 The function is based on what round it is on.
 If the round is between 0 and 19
 F(B,C,D) : (B ∧ C) ∨ ((~B) ∧ D)
 If the round is between 20 and 39
 F(B,C,D): (B XOR C XOR D)
 If the round is between 40 to 59
 F(B,C,D): (B ∧ C) ∨ (B ∧ D) ∨ (C ∧ D)
 If the round is between 60 to 79
 F(B,C,D): (B XOR C XOR D)
 The keys also depend on what the round is
 When the round is between 0 to 19
 Kt = 5A827999
 When the round is between 20 to 39
 Kt = 6ED9EBA1
 When the round is between 40 to 59
 Kt = 8F1BBCDC
 When the round is between 60 to 79
 Kt = CA62C1D6
y = SHA-1-Pad(x)
//y = M1 || M2 || .. || Mn
//Initialize Variables
H0 = 67452301
H1 = EFCDAB89
H2 = 98BADCFE
H3 = 10325476
H4 = CSD2E1F0
For i = 1 to n
Mi = W0 || W1 || … || W15
//Wi is a word
For t = 16 to 79
Wt = ROTL1 (Wt-3 XOR Wt-8 XOR Wt-14 XOR Wt-16)
A = H0 , B = H1, C = H2, D = H3, E = H4
For t = 0 to 79
temp = ROTL5(A) + Ft(B,C,D) + E + Wi + Kt
E = D, D = C, C = ROTL30(B), B = A, A = temp
H0 = H0 + A
H1 = H1 + B
H2 = H2 + C
H3 = H3 + D
H4 = H4 + E
Return (H0 || H1 || H2 || H3 || H4)
 Stinson Textbook