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sm3: use less operations for ff1 and gg1

pull/48/head
Henry Case 3 years ago
parent
commit
7c32db8dd7
1 changed files with 15 additions and 6 deletions
  1. +15
    -6
      hash/sm3/compress.go

+ 15
- 6
hash/sm3/compress.go View File

@@ -14,12 +14,21 @@ func p1(X uint32) uint32 {
return X ^ rotl32(15, X) ^ rotl32(23, X) return X ^ rotl32(15, X) ^ rotl32(23, X)
} }


func ff1(X uint32, Y uint32, Z uint32) uint32 {
return (X & Y) | ((X | Y) & Z)
// Choose bitwise between A or B controlled by C (gg1). A if C=1 otherwise B
// Optimized as per, "Hackers Delight" (7-1, MUX operation), can be used
// to reduce number of operations.
func ch(M uint32, A uint32, B uint32) uint32 {
return ((A ^ B) & M) ^ B
} }


func gg1(X uint32, Y uint32, Z uint32) uint32 {
return (X & Y) ^ ((^X) & Z) // Can be also (Z ^ (X & (Y ^ Z)))
// Majority function (ff1) - takes the majority value as the final result. If two
// or three of the variables are 1, then the result is 1, otherwise 0.
func maj(X uint32, Y uint32, Z uint32) uint32 {
// Y^Z works as a mask. If mask is 0, then majority is dictated by
// value of either Y or Z (doesn't matter, as they are the same, but we
// don't know if result is 0 or 1). Otherwise Y!=Z and results is
// going to dicated by X.
return ch(Y^Z, X, Y)
} }


func r1( func r1(
@@ -43,8 +52,8 @@ func r2(


A12 := rotl32(12, A) A12 := rotl32(12, A)
SS1 := rotl32(7, A12+E+TJ) SS1 := rotl32(7, A12+E+TJ)
TT1 := ff1(A, *B, C) + *D + (SS1 ^ A12) + Wj
TT2 := gg1(E, *F, G) + *H + SS1 + Wi
TT1 := maj(A, *B, C) + *D + (SS1 ^ A12) + Wj
TT2 := ch(E, *F, G) + *H + SS1 + Wi


*B = rotl32(9, *B) *B = rotl32(9, *B)
*D = TT1 *D = TT1


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