update master-with-bazel from master branch
This commit is contained in:
BoringSSL Robot
+85
-13
@@ -61,6 +61,91 @@
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#include "internal.h"
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/* IP and FP
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* The problem is more of a geometric problem that random bit fiddling.
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0 1 2 3 4 5 6 7 62 54 46 38 30 22 14 6
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8 9 10 11 12 13 14 15 60 52 44 36 28 20 12 4
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16 17 18 19 20 21 22 23 58 50 42 34 26 18 10 2
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24 25 26 27 28 29 30 31 to 56 48 40 32 24 16 8 0
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32 33 34 35 36 37 38 39 63 55 47 39 31 23 15 7
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40 41 42 43 44 45 46 47 61 53 45 37 29 21 13 5
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48 49 50 51 52 53 54 55 59 51 43 35 27 19 11 3
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56 57 58 59 60 61 62 63 57 49 41 33 25 17 9 1
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The output has been subject to swaps of the form
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0 1 -> 3 1 but the odd and even bits have been put into
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2 3 2 0
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different words. The main trick is to remember that
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t=((l>>size)^r)&(mask);
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r^=t;
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l^=(t<<size);
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can be used to swap and move bits between words.
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So l = 0 1 2 3 r = 16 17 18 19
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4 5 6 7 20 21 22 23
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8 9 10 11 24 25 26 27
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12 13 14 15 28 29 30 31
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becomes (for size == 2 and mask == 0x3333)
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t = 2^16 3^17 -- -- l = 0 1 16 17 r = 2 3 18 19
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6^20 7^21 -- -- 4 5 20 21 6 7 22 23
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10^24 11^25 -- -- 8 9 24 25 10 11 24 25
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14^28 15^29 -- -- 12 13 28 29 14 15 28 29
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Thanks for hints from Richard Outerbridge - he told me IP&FP
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could be done in 15 xor, 10 shifts and 5 ands.
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When I finally started to think of the problem in 2D
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I first got ~42 operations without xors. When I remembered
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how to use xors :-) I got it to its final state.
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*/
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#define PERM_OP(a, b, t, n, m) \
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do { \
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(t) = ((((a) >> (n)) ^ (b)) & (m)); \
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(b) ^= (t); \
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(a) ^= ((t) << (n)); \
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} while (0)
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#define IP(l, r) \
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do { \
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uint32_t tt; \
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PERM_OP(r, l, tt, 4, 0x0f0f0f0fL); \
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PERM_OP(l, r, tt, 16, 0x0000ffffL); \
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PERM_OP(r, l, tt, 2, 0x33333333L); \
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PERM_OP(l, r, tt, 8, 0x00ff00ffL); \
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PERM_OP(r, l, tt, 1, 0x55555555L); \
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} while (0)
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#define FP(l, r) \
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do { \
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uint32_t tt; \
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PERM_OP(l, r, tt, 1, 0x55555555L); \
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PERM_OP(r, l, tt, 8, 0x00ff00ffL); \
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PERM_OP(l, r, tt, 2, 0x33333333L); \
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PERM_OP(r, l, tt, 16, 0x0000ffffL); \
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PERM_OP(l, r, tt, 4, 0x0f0f0f0fL); \
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} while (0)
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#define LOAD_DATA(ks, R, S, u, t, E0, E1) \
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do { \
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(u) = (R) ^ (ks)->subkeys[S][0]; \
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(t) = (R) ^ (ks)->subkeys[S][1]; \
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} while (0)
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#define D_ENCRYPT(ks, LL, R, S) \
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do { \
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LOAD_DATA(ks, R, S, u, t, E0, E1); \
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t = CRYPTO_rotr_u32(t, 4); \
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(LL) ^= \
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DES_SPtrans[0][(u >> 2L) & 0x3f] ^ DES_SPtrans[2][(u >> 10L) & 0x3f] ^ \
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DES_SPtrans[4][(u >> 18L) & 0x3f] ^ \
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DES_SPtrans[6][(u >> 26L) & 0x3f] ^ DES_SPtrans[1][(t >> 2L) & 0x3f] ^ \
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DES_SPtrans[3][(t >> 10L) & 0x3f] ^ \
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DES_SPtrans[5][(t >> 18L) & 0x3f] ^ DES_SPtrans[7][(t >> 26L) & 0x3f]; \
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} while (0)
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#define ITERATIONS 16
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#define HALF_ITERATIONS 8
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static const uint32_t des_skb[8][64] = {
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{ // for C bits (numbered as per FIPS 46) 1 2 3 4 5 6
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0x00000000, 0x00000010, 0x20000000, 0x20000010, 0x00010000,
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@@ -771,16 +856,3 @@ void DES_ede2_cbc_encrypt(const uint8_t *in, uint8_t *out, size_t len,
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void DES_set_key_unchecked(const DES_cblock *key, DES_key_schedule *schedule) {
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DES_set_key(key, schedule);
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}
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#undef HPERM_OP
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#undef c2l
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#undef l2c
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#undef c2ln
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#undef l2cn
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#undef PERM_OP
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#undef IP
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#undef FP
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#undef LOAD_DATA
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#undef D_ENCRYPT
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#undef ITERATIONS
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#undef HALF_ITERATIONS
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@@ -146,91 +146,6 @@ extern "C" {
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} \
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} while (0)
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/* IP and FP
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* The problem is more of a geometric problem that random bit fiddling.
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0 1 2 3 4 5 6 7 62 54 46 38 30 22 14 6
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8 9 10 11 12 13 14 15 60 52 44 36 28 20 12 4
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16 17 18 19 20 21 22 23 58 50 42 34 26 18 10 2
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24 25 26 27 28 29 30 31 to 56 48 40 32 24 16 8 0
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32 33 34 35 36 37 38 39 63 55 47 39 31 23 15 7
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40 41 42 43 44 45 46 47 61 53 45 37 29 21 13 5
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48 49 50 51 52 53 54 55 59 51 43 35 27 19 11 3
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56 57 58 59 60 61 62 63 57 49 41 33 25 17 9 1
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The output has been subject to swaps of the form
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0 1 -> 3 1 but the odd and even bits have been put into
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2 3 2 0
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different words. The main trick is to remember that
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t=((l>>size)^r)&(mask);
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r^=t;
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l^=(t<<size);
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can be used to swap and move bits between words.
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So l = 0 1 2 3 r = 16 17 18 19
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4 5 6 7 20 21 22 23
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8 9 10 11 24 25 26 27
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12 13 14 15 28 29 30 31
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becomes (for size == 2 and mask == 0x3333)
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t = 2^16 3^17 -- -- l = 0 1 16 17 r = 2 3 18 19
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6^20 7^21 -- -- 4 5 20 21 6 7 22 23
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10^24 11^25 -- -- 8 9 24 25 10 11 24 25
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14^28 15^29 -- -- 12 13 28 29 14 15 28 29
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Thanks for hints from Richard Outerbridge - he told me IP&FP
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could be done in 15 xor, 10 shifts and 5 ands.
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When I finally started to think of the problem in 2D
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I first got ~42 operations without xors. When I remembered
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how to use xors :-) I got it to its final state.
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*/
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#define PERM_OP(a, b, t, n, m) \
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do { \
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(t) = ((((a) >> (n)) ^ (b)) & (m)); \
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(b) ^= (t); \
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(a) ^= ((t) << (n)); \
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} while (0)
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#define IP(l, r) \
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do { \
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uint32_t tt; \
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PERM_OP(r, l, tt, 4, 0x0f0f0f0fL); \
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PERM_OP(l, r, tt, 16, 0x0000ffffL); \
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PERM_OP(r, l, tt, 2, 0x33333333L); \
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PERM_OP(l, r, tt, 8, 0x00ff00ffL); \
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PERM_OP(r, l, tt, 1, 0x55555555L); \
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} while (0)
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#define FP(l, r) \
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do { \
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uint32_t tt; \
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PERM_OP(l, r, tt, 1, 0x55555555L); \
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PERM_OP(r, l, tt, 8, 0x00ff00ffL); \
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PERM_OP(l, r, tt, 2, 0x33333333L); \
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PERM_OP(r, l, tt, 16, 0x0000ffffL); \
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PERM_OP(l, r, tt, 4, 0x0f0f0f0fL); \
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} while (0)
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#define LOAD_DATA(ks, R, S, u, t, E0, E1) \
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do { \
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(u) = (R) ^ (ks)->subkeys[S][0]; \
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(t) = (R) ^ (ks)->subkeys[S][1]; \
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} while (0)
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#define D_ENCRYPT(ks, LL, R, S) \
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do { \
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LOAD_DATA(ks, R, S, u, t, E0, E1); \
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t = CRYPTO_rotr_u32(t, 4); \
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(LL) ^= \
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DES_SPtrans[0][(u >> 2L) & 0x3f] ^ DES_SPtrans[2][(u >> 10L) & 0x3f] ^ \
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DES_SPtrans[4][(u >> 18L) & 0x3f] ^ \
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DES_SPtrans[6][(u >> 26L) & 0x3f] ^ DES_SPtrans[1][(t >> 2L) & 0x3f] ^ \
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DES_SPtrans[3][(t >> 10L) & 0x3f] ^ \
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DES_SPtrans[5][(t >> 18L) & 0x3f] ^ DES_SPtrans[7][(t >> 26L) & 0x3f]; \
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} while (0)
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#define ITERATIONS 16
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#define HALF_ITERATIONS 8
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// Private functions.
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//
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