| /* |
| ------------------------------------------------------------------------------ |
| perfhex.c: code to generate code for a hash for perfect hashing. |
| (c) Bob Jenkins, December 31 1999 |
| You may use this code in any way you wish, and it is free. No warranty. |
| I hereby place this in the public domain. |
| Source is http://burtleburtle.net/bob/c/perfhex.c |
| |
| The task of this file is to do the minimal amount of mixing needed to |
| find distinct (a,b) for each key when each key is a distinct ub4. That |
| means trying all possible ways to mix starting with the fastest. The |
| output is those (a,b) pairs and code in the *final* structure for producing |
| those pairs. |
| ------------------------------------------------------------------------------ |
| */ |
| |
| #ifndef STANDARD |
| #include "standard.h" |
| #endif |
| #ifndef LOOKUPA |
| #include "lookupa.h" |
| #endif |
| #ifndef RECYCLE |
| #include "recycle.h" |
| #endif |
| #ifndef PERFECT |
| #include "perfect.h" |
| #endif |
| |
| /* |
| * Find a perfect hash when there is only one key. Zero instructions. |
| * Hint: the one key always hashes to 0 |
| */ |
| static void hexone(keys, final) |
| key *keys; |
| gencode *final; |
| { |
| /* 1 key: the hash is always 0 */ |
| keys->a_k = 0; |
| keys->b_k = 0; |
| final->used = 1; |
| sprintf(final->line[0], " ub4 rsl = 0;\n"); /* h1a: 37 */ |
| } |
| |
| |
| |
| /* |
| * Find a perfect hash when there are only two keys. Max 2 instructions. |
| * There exists a bit that is different for the two keys. Test it. |
| * Note that a perfect hash of 2 keys is automatically minimal. |
| */ |
| static void hextwo(keys, final) |
| key *keys; |
| gencode *final; |
| { |
| ub4 a = keys->hash_k; |
| ub4 b = keys->next_k->hash_k; |
| ub4 i; |
| |
| if (a == b) |
| { |
| printf("fatal error: duplicate keys\n"); |
| exit(SUCCESS); |
| } |
| |
| final->used = 1; |
| |
| /* one instruction */ |
| if ((a&1) != (b&1)) |
| { |
| sprintf(final->line[0], " ub4 rsl = (val & 1);\n"); /* h2a: 3,4 */ |
| return; |
| } |
| |
| /* two instructions */ |
| for (i=0; i<UB4BITS; ++i) |
| { |
| if ((a&((ub4)1<<i)) != (b&((ub4)1<<i))) break; |
| } |
| /* h2b: 4,6 */ |
| sprintf(final->line[0], " ub4 rsl = ((val << %ld) & 1);\n", i); |
| } |
| |
| |
| |
| /* |
| * find the value to xor to a and b and c to make none of them 3 |
| * assert, (a,b,c) are three distinct values in (0,1,2,3). |
| */ |
| static ub4 find_adder(a,b,c) |
| ub4 a; |
| ub4 b; |
| ub4 c; |
| { |
| return (a^b^c^3); |
| } |
| |
| |
| |
| /* |
| * Find a perfect hash when there are only three keys. Max 6 instructions. |
| * |
| * keys a,b,c. |
| * There exists bit i such that a[i] != b[i]. |
| * Either c[i] != a[i] or c[i] != b[i], assume c[i] != a[i]. |
| * There exists bit j such that b[j] != c[j]. Note i != j. |
| * Final hash should be no longer than val[i]^val[j]. |
| * |
| * A minimal perfect hash needs to xor one of 0,1,2,3 afterwards to cause |
| * the hole to land on 3. find_adder() finds that constant |
| */ |
| static void hexthree(keys, final, form) |
| key *keys; |
| gencode *final; |
| hashform *form; |
| { |
| ub4 a = keys->hash_k; |
| ub4 b = keys->next_k->hash_k; |
| ub4 c = keys->next_k->next_k->hash_k; |
| ub4 i,j,x,y,z; |
| |
| final->used = 1; |
| |
| if (a == b || a == c || b == c) |
| { |
| printf("fatal error: duplicate keys\n"); |
| exit(SUCCESS); |
| } |
| |
| /* one instruction */ |
| x = a&3; |
| y = b&3; |
| z = c&3; |
| if (x != y && x != z && y != z) |
| { |
| if (form->perfect == NORMAL_HP || (x != 3 && y != 3 && z != 3)) |
| { |
| /* h3a: 0,1,2 */ |
| sprintf(final->line[0], " ub4 rsl = (val & 3);\n"); |
| } |
| else |
| { |
| /* h3b: 0,3,2 */ |
| sprintf(final->line[0], " ub4 rsl = ((val & 3) ^ %d);\n", |
| find_adder(x,y,z)); |
| } |
| return; |
| } |
| |
| x = a>>(UB4BITS-2); |
| y = b>>(UB4BITS-2); |
| z = c>>(UB4BITS-2); |
| if (x != y && x != z && y != z) |
| { |
| if (form->perfect == NORMAL_HP || (x != 3 && y != 3 && z != 3)) |
| { |
| /* h3c: 3fffffff, 7fffffff, bfffffff */ |
| sprintf(final->line[0], " ub4 rsl = (val >> %ld);\n", (ub4)(UB4BITS-2)); |
| } |
| else |
| { |
| /* h3d: 7fffffff, bfffffff, ffffffff */ |
| sprintf(final->line[0], " ub4 rsl = ((val >> %ld) ^ %ld);\n", |
| (ub4)(UB4BITS-2), find_adder(x,y,z)); |
| } |
| return; |
| } |
| |
| /* two instructions */ |
| for (i=0; i<final->highbit; ++i) |
| { |
| x = (a>>i)&3; |
| y = (b>>i)&3; |
| z = (c>>i)&3; |
| if (x != y && x != z && y != z) |
| { |
| if (form->perfect == NORMAL_HP || (x != 3 && y != 3 && z != 3)) |
| { |
| /* h3e: ffff3fff, ffff7fff, ffffbfff */ |
| sprintf(final->line[0], " ub4 rsl = ((val >> %ld) & 3);\n", i); |
| } |
| else |
| { |
| /* h3f: ffff7fff, ffffbfff, ffffffff */ |
| sprintf(final->line[0], " ub4 rsl = (((val >> %ld) & 3) ^ %ld);\n", i, |
| find_adder(x,y,z)); |
| } |
| return; |
| } |
| } |
| |
| /* three instructions */ |
| for (i=0; i<=final->highbit; ++i) |
| { |
| x = (a+(a>>i))&3; |
| y = (b+(b>>i))&3; |
| z = (c+(c>>i))&3; |
| if (x != y && x != z && y != z) |
| { |
| if (form->perfect == NORMAL_HP || (x != 3 && y != 3 && z != 3)) |
| { |
| /* h3g: 0x000, 0x001, 0x100 */ |
| sprintf(final->line[0], " ub4 rsl = ((val+(val>>%ld))&3);\n", i); |
| } |
| else |
| { |
| /* h3h: 0x001, 0x100, 0x101 */ |
| sprintf(final->line[0], " ub4 rsl = (((val+(val>>%ld))&3)^%ld);\n", i, |
| find_adder(x,y,z)); |
| } |
| return; |
| } |
| } |
| |
| /* |
| * Four instructions: I can prove this will always work. |
| * |
| * If the three values are distinct, there are two bits which |
| * distinguish them. Choose the two such bits that are closest together. |
| * If those bits are values 001 and 100 for those three values, |
| * then there either aren't any bits in between |
| * or the in-between bits aren't valued 001, 110, 100, 011, 010, or 101, |
| * because that would violate the closest-together assumption. |
| * So any in-between bits must be 000 or 111, and of 000 and 111 with |
| * the distinguishing bits won't cause them to stop being distinguishing. |
| */ |
| for (i=final->lowbit; i<=final->highbit; ++i) |
| { |
| for (j=i; j<=final->highbit; ++j) |
| { |
| x = ((a>>i)^(a>>j))&3; |
| y = ((b>>i)^(b>>j))&3; |
| z = ((c>>i)^(c>>j))&3; |
| if (x != y && x != z && y != z) |
| { |
| if (form->perfect == NORMAL_HP || (x != 3 && y != 3 && z != 3)) |
| { |
| /* h3i: 0x00, 0x04, 0x10 */ |
| sprintf(final->line[0], |
| " ub4 rsl = (((val>>%ld) ^ (val>>%ld)) & 3);\n", i, j); |
| } |
| else |
| { |
| /* h3j: 0x04, 0x10, 0x14 */ |
| sprintf(final->line[0], |
| " ub4 rsl = ((((val>>%ld) ^ (val>>%ld)) & 3) ^ %ld);\n", |
| i, j, find_adder(x,y,z)); |
| } |
| return; |
| } |
| } |
| } |
| |
| printf("fatal error: hexthree\n"); |
| exit(SUCCESS); |
| } |
| |
| |
| |
| /* |
| * Check that a,b,c,d are some permutation of 0,1,2,3 |
| * Assume that a,b,c,d are all have values less than 32. |
| */ |
| static int testfour(a,b,c,d) |
| ub4 a; |
| ub4 b; |
| ub4 c; |
| ub4 d; |
| { |
| ub4 mask = (1<<a)^(1<<b)^(1<<c)^(1<<d); |
| return (mask == 0xf); |
| } |
| |
| |
| |
| /* |
| * Find a perfect hash when there are only four keys. Max 10 instructions. |
| * Note that a perfect hash for 4 keys will automatically be minimal. |
| */ |
| static void hexfour(keys, final) |
| key *keys; |
| gencode *final; |
| { |
| ub4 a = keys->hash_k; |
| ub4 b = keys->next_k->hash_k; |
| ub4 c = keys->next_k->next_k->hash_k; |
| ub4 d = keys->next_k->next_k->next_k->hash_k; |
| ub4 w,x,y,z; |
| ub4 i,j,k; |
| |
| if (a==b || a==c || a==d || b==c || b==d || c==d) |
| { |
| printf("fatal error: Duplicate keys\n"); |
| exit(SUCCESS); |
| } |
| |
| final->used = 1; |
| |
| /* one instruction */ |
| if ((final->diffbits & 3) == 3) |
| { |
| w = a&3; |
| x = b&3; |
| y = c&3; |
| z = d&3; |
| if (testfour(w,x,y,z)) |
| { |
| sprintf(final->line[0], " ub4 rsl = (val & 3);\n"); /* h4a: 0,1,2,3 */ |
| return; |
| } |
| } |
| |
| if (((final->diffbits >> (UB4BITS-2)) & 3) == 3) |
| { |
| w = a>>(UB4BITS-2); |
| x = b>>(UB4BITS-2); |
| y = c>>(UB4BITS-2); |
| z = d>>(UB4BITS-2); |
| if (testfour(w,x,y,z)) |
| { /* h4b: 0fffffff, 4fffffff, 8fffffff, cfffffff */ |
| sprintf(final->line[0], " ub4 rsl = (val >> %ld);\n", (ub4)(UB4BITS-2)); |
| return; |
| } |
| } |
| |
| /* two instructions */ |
| for (i=final->lowbit; i<final->highbit; ++i) |
| { |
| if (((final->diffbits >> i) & 3) == 3) |
| { |
| w = (a>>i)&3; |
| x = (b>>i)&3; |
| y = (c>>i)&3; |
| z = (d>>i)&3; |
| if (testfour(w,x,y,z)) |
| { /* h4c: 0,2,4,6 */ |
| sprintf(final->line[0], " ub4 rsl = ((val >> %ld) & 3);\n", i); |
| return; |
| } |
| } |
| } |
| |
| /* three instructions (linear with the number of diffbits) */ |
| if ((final->diffbits & 3) != 0) |
| { |
| for (i=final->lowbit; i<=final->highbit; ++i) |
| { |
| if (((final->diffbits >> i) & 3) != 0) |
| { |
| w = (a+(a>>i))&3; |
| x = (b+(b>>i))&3; |
| y = (c+(c>>i))&3; |
| z = (d+(d>>i))&3; |
| if (testfour(w,x,y,z)) |
| { /* h4d: 0,1,2,4 */ |
| sprintf(final->line[0], |
| " ub4 rsl = ((val + (val >> %ld)) & 3);\n", i); |
| return; |
| } |
| |
| w = (a-(a>>i))&3; |
| x = (b-(b>>i))&3; |
| y = (c-(c>>i))&3; |
| z = (d-(d>>i))&3; |
| if (testfour(w,x,y,z)) |
| { /* h4e: 0,1,3,5 */ |
| sprintf(final->line[0], |
| " ub4 rsl = ((val - (val >> %ld)) & 3);\n", i); |
| return; |
| } |
| |
| /* h4f: ((val>>k)-val)&3: redundant with h4e */ |
| |
| w = (a^(a>>i))&3; |
| x = (b^(b>>i))&3; |
| y = (c^(c>>i))&3; |
| z = (d^(d>>i))&3; |
| if (testfour(w,x,y,z)) |
| { /* h4g: 3,4,5,8 */ |
| sprintf(final->line[0], |
| " ub4 rsl = ((val ^ (val >> %ld)) & 3);\n", i); |
| return; |
| } |
| } |
| } |
| } |
| |
| /* four instructions (linear with the number of diffbits) */ |
| if ((final->diffbits & 3) != 0) |
| { |
| for (i=final->lowbit; i<=final->highbit; ++i) |
| { |
| if ((((final->diffbits >> i) & 1) != 0) && |
| ((final->diffbits & 2) != 0)) |
| { |
| w = (a&3)^((a>>i)&1); |
| x = (b&3)^((b>>i)&1); |
| y = (c&3)^((c>>i)&1); |
| z = (d&3)^((d>>i)&1); |
| if (testfour(w,x,y,z)) |
| { /* h4h: 1,2,6,8 */ |
| sprintf(final->line[0], |
| " ub4 rsl = ((val & 3) ^ ((val >> %ld) & 1));\n", i); |
| return; |
| } |
| |
| w = (a&2)^((a>>i)&1); |
| x = (b&2)^((b>>i)&1); |
| y = (c&2)^((c>>i)&1); |
| z = (d&2)^((d>>i)&1); |
| if (testfour(w,x,y,z)) |
| { /* h4i: 1,2,8,a */ |
| sprintf(final->line[0], |
| " ub4 rsl = ((val & 2) ^ ((val >> %ld) & 1));\n", i); |
| return; |
| } |
| } |
| |
| if ((((final->diffbits >> i) & 2) != 0) && |
| ((final->diffbits & 1) != 0)) |
| { |
| w = (a&3)^((a>>i)&2); |
| x = (b&3)^((b>>i)&2); |
| y = (c&3)^((c>>i)&2); |
| z = (d&3)^((d>>i)&2); |
| if (testfour(w,x,y,z)) |
| { /* h4j: 0,1,3,4 */ |
| sprintf(final->line[0], |
| " ub4 rsl = ((val & 3) ^ ((val >> %ld) & 2));\n", i); |
| return; |
| } |
| |
| w = (a&1)^((a>>i)&2); |
| x = (b&1)^((b>>i)&2); |
| y = (c&1)^((c>>i)&2); |
| z = (d&1)^((d>>i)&2); |
| if (testfour(w,x,y,z)) |
| { /* h4k: 1,4,7,8 */ |
| sprintf(final->line[0], |
| " ub4 rsl = ((val & 1) ^ ((val >> %ld) & 2));\n", i); |
| return; |
| } |
| } |
| } |
| } |
| |
| /* four instructions (quadratic in the number of diffbits) */ |
| for (i=final->lowbit; i<=final->highbit; ++i) |
| { |
| if (((final->diffbits >> i) & 1) == 1) |
| { |
| for (j=final->lowbit; j<=final->highbit; ++j) |
| { |
| if (((final->diffbits >> j) & 3) != 0) |
| { |
| /* test + */ |
| w = ((a>>i)+(a>>j))&3; |
| x = ((b>>i)+(a>>j))&3; |
| y = ((c>>i)+(a>>j))&3; |
| z = ((d>>i)+(a>>j))&3; |
| if (testfour(w,x,y,z)) |
| { /* h4l: testcase? */ |
| sprintf(final->line[0], |
| " ub4 rsl = (((val >> %ld) + (val >> %ld)) & 3);\n", |
| i, j); |
| return; |
| } |
| |
| /* test - */ |
| w = ((a>>i)-(a>>j))&3; |
| x = ((b>>i)-(a>>j))&3; |
| y = ((c>>i)-(a>>j))&3; |
| z = ((d>>i)-(a>>j))&3; |
| if (testfour(w,x,y,z)) |
| { /* h4m: testcase? */ |
| sprintf(final->line[0], |
| " ub4 rsl = (((val >> %ld) - (val >> %ld)) & 3);\n", |
| i, j); |
| return; |
| } |
| |
| /* test ^ */ |
| w = ((a>>i)^(a>>j))&3; |
| x = ((b>>i)^(a>>j))&3; |
| y = ((c>>i)^(a>>j))&3; |
| z = ((d>>i)^(a>>j))&3; |
| if (testfour(w,x,y,z)) |
| { /* h4n: testcase? */ |
| sprintf(final->line[0], |
| " ub4 rsl = (((val >> %ld) ^ (val >> %ld)) & 3);\n", |
| i, j); |
| return; |
| } |
| } |
| } |
| } |
| } |
| |
| /* five instructions (quadratic in the number of diffbits) */ |
| for (i=final->lowbit; i<=final->highbit; ++i) |
| { |
| if (((final->diffbits >> i) & 1) != 0) |
| { |
| for (j=final->lowbit; j<=final->highbit; ++j) |
| { |
| if (((final->diffbits >> j) & 3) != 0) |
| { |
| w = ((a>>j)&3)^((a>>i)&1); |
| x = ((b>>j)&3)^((b>>i)&1); |
| y = ((c>>j)&3)^((c>>i)&1); |
| z = ((d>>j)&3)^((d>>i)&1); |
| if (testfour(w,x,y,z)) |
| { /* h4o: 0,4,8,a */ |
| sprintf(final->line[0], |
| " ub4 rsl = (((val >> %ld) & 3) ^ ((val >> %ld) & 1));\n", |
| j, i); |
| return; |
| } |
| |
| w = ((a>>j)&2)^((a>>i)&1); |
| x = ((b>>j)&2)^((b>>i)&1); |
| y = ((c>>j)&2)^((c>>i)&1); |
| z = ((d>>j)&2)^((d>>i)&1); |
| if (testfour(w,x,y,z)) |
| { /* h4p: 0x04, 0x08, 0x10, 0x14 */ |
| sprintf(final->line[0], |
| " ub4 rsl = (((val >> %ld) & 2) ^ ((val >> %ld) & 1));\n", |
| j, i); |
| return; |
| } |
| } |
| |
| if (i==0) |
| { |
| w = ((a>>j)^(a<<1))&3; |
| x = ((b>>j)^(b<<1))&3; |
| y = ((c>>j)^(c<<1))&3; |
| z = ((d>>j)^(d<<1))&3; |
| } |
| else |
| { |
| w = ((a>>j)&3)^((a>>(i-1))&2); |
| x = ((b>>j)&3)^((b>>(i-1))&2); |
| y = ((c>>j)&3)^((c>>(i-1))&2); |
| z = ((d>>j)&3)^((d>>(i-1))&2); |
| } |
| if (testfour(w,x,y,z)) |
| { |
| if (i==0) /* h4q: 0,4,5,8 */ |
| { |
| sprintf(final->line[0], |
| " ub4 rsl = (((val >> %ld) ^ (val << 1)) & 3);\n", |
| j); |
| } |
| else if (i==1) /* h4r: 0x01,0x09,0x0b,0x10 */ |
| { |
| sprintf(final->line[0], |
| " ub4 rsl = (((val >> %ld) & 3) ^ (val & 2));\n", |
| j); |
| } |
| else /* h4s: 0,2,6,8 */ |
| { |
| sprintf(final->line[0], |
| " ub4 rsl = (((val >> %ld) & 3) ^ ((val >> %ld) & 2));\n", |
| j, (i-1)); |
| } |
| return; |
| } |
| |
| w = ((a>>j)&1)^((a>>i)&2); |
| x = ((b>>j)&1)^((b>>i)&2); |
| y = ((c>>j)&1)^((c>>i)&2); |
| z = ((d>>j)&1)^((d>>i)&2); |
| if (testfour(w,x,y,z)) /* h4t: 0x20,0x14,0x10,0x06 */ |
| { |
| sprintf(final->line[0], |
| " ub4 rsl = (((val >> %ld) & 1) ^ ((val >> %ld) & 2));\n", |
| j, i); |
| return; |
| } |
| } |
| } |
| } |
| |
| /* |
| * OK, bring out the big guns. |
| * There exist three bits i,j,k which distinguish a,b,c,d. |
| * i^(j<<1)^(k*q) is guaranteed to work for some q in {0,1,2,3}, |
| * proven by exhaustive search of all (8 choose 4) cases. |
| * Find three such bits and try the 4 cases. |
| * Linear with the number of diffbits. |
| * Some cases below may duplicate some cases above. I did it that way |
| * so that what is below is guaranteed to work, no matter what was |
| * attempted above. |
| * The generated hash is at most 10 instructions. |
| */ |
| for (i=final->lowbit; i<UB4BITS; ++i) |
| { |
| y = (c>>i)&1; |
| z = (d>>i)&1; |
| if (y != z) |
| break; |
| } |
| |
| for (j=final->lowbit; j<UB4BITS; ++j) |
| { |
| x = ((b>>i)&1)^(((b>>j)&1)<<1); |
| y = ((c>>i)&1)^(((c>>j)&1)<<1); |
| z = ((d>>i)&1)^(((d>>j)&1)<<1); |
| if (x != y && x != z && y != z) |
| break; |
| } |
| |
| for (k=final->lowbit; k<UB4BITS; ++k) |
| { |
| w = ((a>>i)&1)^(((a>>j)&1)<<1)^(((a>>k)&1)<<2); |
| x = ((b>>i)&1)^(((b>>j)&1)<<1)^(((b>>k)&1)<<2); |
| y = ((c>>i)&1)^(((c>>j)&1)<<1)^(((c>>k)&1)<<2); |
| z = ((d>>i)&1)^(((d>>j)&1)<<1)^(((d>>k)&1)<<2); |
| if (w != x && w != y && w != z && x != y && x != z && y != z) |
| break; |
| } |
| |
| /* Assert: bits i,j,k were found which distinguish a,b,c,d */ |
| if (i==UB4BITS || j==UB4BITS || k==UB4BITS) |
| { |
| printf("Fatal error: hexfour(), i %ld j %ld k %ld\n", i,j,k); |
| exit(SUCCESS); |
| } |
| |
| /* now try the four cases */ |
| { |
| ub4 m,n,o,p; |
| |
| /* if any bit has two 1s and two 0s, make that bit o */ |
| if (((a>>i)&1)+((b>>i)&1)+((c>>i)&1)+((d>>i)&1) != 2) |
| { m=j; n=k; o=i; } |
| else if (((a>>j)&1)+((b>>j)&1)+((c>>j)&1)+((d>>j)&1) != 2) |
| { m=i; n=k; o=j; } |
| else |
| { m=i; n=j; o=k; } |
| if (m > n) {p=m; m=n; n=p; } /* guarantee m < n */ |
| |
| /* printf("m %ld n %ld o %ld %ld %ld %ld %ld\n", m, n, o, w,x,y,z); */ |
| |
| /* seven instructions, multiply bit o by 1 */ |
| w = (((a>>m)^(a>>o))&1)^((a>>(n-1))&2); |
| x = (((b>>m)^(b>>o))&1)^((b>>(n-1))&2); |
| y = (((c>>m)^(c>>o))&1)^((c>>(n-1))&2); |
| z = (((d>>m)^(d>>o))&1)^((d>>(n-1))&2); |
| if (testfour(w,x,y,z)) |
| { |
| if (m>o) {p=m; m=o; o=p;} /* make sure m < o and m < n */ |
| |
| if (m==0) /* 0,2,8,9 */ |
| { |
| sprintf(final->line[0], |
| " ub4 rsl = (((val^(val>>%ld))&1)^((val>>%ld)&2));\n", o, n-1); |
| } |
| else /* 0x00,0x04,0x10,0x12 */ |
| { |
| sprintf(final->line[0], |
| " ub4 rsl = ((((val>>%ld) ^ (val>>%ld)) & 1) ^ ((val>>%ld) & 2));\n", |
| m, o, n-1); |
| } |
| return; |
| } |
| |
| /* six to seven instructions, multiply bit o by 2 */ |
| w = ((a>>m)&1)^((((a>>n)^(a>>o))&1)<<1); |
| x = ((b>>m)&1)^((((b>>n)^(b>>o))&1)<<1); |
| y = ((c>>m)&1)^((((c>>n)^(c>>o))&1)<<1); |
| z = ((d>>m)&1)^((((d>>n)^(d>>o))&1)<<1); |
| if (testfour(w,x,y,z)) |
| { |
| if (m==o-1) {p=n; n=o; o=p;} /* make m==n-1 if possible */ |
| |
| if (m==0) /* 0,1,5,8 */ |
| { |
| sprintf(final->line[0], |
| " ub4 rsl = ((val & 1) ^ (((val>>%ld) ^ (val>>%ld)) & 2));\n", |
| n-1, o-1); |
| } |
| else if (o==0) /* 0x00,0x04,0x05,0x10 */ |
| { |
| sprintf(final->line[0], |
| " ub4 rsl = (((val>>%ld) & 2) ^ (((val>>%ld) ^ val) & 1));\n", |
| m-1, n); |
| } |
| else /* 0x00,0x02,0x0a,0x10 */ |
| { |
| sprintf(final->line[0], |
| " ub4 rsl = (((val>>%ld) & 1) ^ (((val>>%ld) ^ (val>>%ld)) & 2));\n", |
| m, n-1, o-1); |
| } |
| return; |
| } |
| |
| /* multiplying by 3 is a pain: seven or eight instructions */ |
| w = (((a>>m)&1)^((a>>(n-1))&2))^((a>>o)&1)^(((a>>o)&1)<<1); |
| x = (((b>>m)&1)^((b>>(n-1))&2))^((b>>o)&1)^(((b>>o)&1)<<1); |
| y = (((c>>m)&1)^((c>>(n-1))&2))^((c>>o)&1)^(((c>>o)&1)<<1); |
| z = (((d>>m)&1)^((d>>(n-1))&2))^((d>>o)&1)^(((d>>o)&1)<<1); |
| if (testfour(w,x,y,z)) |
| { |
| final->used = 2; |
| sprintf(final->line[0], " ub4 b = (val >> %ld) & 1;\n", o); |
| if (m==o-1 && m==0) /* 0x02,0x10,0x11,0x18 */ |
| { |
| sprintf(final->line[1], |
| " ub4 rsl = ((val & 3) ^ ((val >> %ld) & 2) ^ b);\n", n-1); |
| } |
| else if (m==o-1) /* 0,4,6,c */ |
| { |
| sprintf(final->line[1], |
| " ub4 rsl = (((val >> %ld) & 3) ^ ((val >> %ld) & 2) ^ b);\n", |
| m, n-1); |
| } |
| else if (m==n-1 && m==0) /* 02,0a,0b,18 */ |
| { |
| sprintf(final->line[1], |
| " ub4 rsl = ((val & 3) ^ b ^ (b << 1));\n"); |
| } |
| else if (m==n-1) /* 0,2,4,8 */ |
| { |
| sprintf(final->line[1], |
| " ub4 rsl = (((val >> %ld) & 3) ^ b ^ (b << 1));\n", m); |
| } |
| else if (o==n-1 && m==0) /* h4am: not reached */ |
| { |
| sprintf(final->line[1], |
| " ub4 rsl = ((val & 1) ^ ((val >> %ld) & 3) ^ (b <<1 ));\n", |
| o); |
| } |
| else if (o==n-1) /* 0x00,0x02,0x08,0x10 */ |
| { |
| sprintf(final->line[1], |
| " ub4 rsl = (((val >> %ld) & 1) ^ ((val >> %ld) & 3) ^ (b << 1));\n", |
| m, o); |
| } |
| else if ((m != o-1) && (m != n-1) && (o != m-1) && (o != n-1)) |
| { |
| final->used = 3; |
| sprintf(final->line[0], " ub4 newval = val & 0x%lx;\n", |
| (((ub4)1<<m)^((ub4)1<<n)^((ub4)1<<o))); |
| if (o==0) /* 0x00,0x01,0x04,0x10 */ |
| { |
| sprintf(final->line[1], " ub4 b = -newval;\n"); |
| } |
| else /* 0x00,0x04,0x09,0x10 */ |
| { |
| sprintf(final->line[1], " ub4 b = -(newval >> %ld);\n", o); |
| } |
| if (m==0) /* 0x00,0x04,0x09,0x10 */ |
| { |
| sprintf(final->line[2], |
| " ub4 rsl = ((newval ^ (newval>>%ld) ^ b) & 3);\n", n-1); |
| } |
| else /* 0x00,0x03,0x04,0x10 */ |
| { |
| sprintf(final->line[2], |
| " ub4 rsl = (((newval>>%ld) ^ (newval>>%ld) ^ b) & 3);\n", |
| m, n-1); |
| } |
| } |
| else if (o == m-1) |
| { |
| if (o==0) /* 0x02,0x03,0x0a,0x10 */ |
| { |
| sprintf(final->line[0], " ub4 b = (val<<1) & 2;\n"); |
| } |
| else if (o==1) /* 0x00,0x02,0x04,0x10 */ |
| { |
| sprintf(final->line[0], " ub4 b = val & 2;\n"); |
| } |
| else /* 0x00,0x04,0x08,0x20 */ |
| { |
| sprintf(final->line[0], " ub4 b = (val>>%ld) & 2;\n", o-1); |
| } |
| |
| if (o==0) /* 0x02,0x03,0x0a,0x10 */ |
| { |
| sprintf(final->line[1], |
| " ub4 rsl = ((val & 3) ^ ((val>>%ld) & 1) ^ b);\n", |
| n); |
| } |
| else /* 0x00,0x02,0x04,0x10 */ |
| { |
| sprintf(final->line[1], |
| " ub4 rsl = (((val>>%ld) & 3) ^ ((val>>%ld) & 1) ^ b);\n", |
| o, n); |
| } |
| } |
| else /* h4ax: 10 instructions, but not reached */ |
| { |
| sprintf(final->line[1], |
| " ub4 rsl = (((val>>%ld) & 1) ^ ((val>>%ld) & 2) ^ b ^ (b<<1));\n", |
| m, n-1); |
| } |
| |
| return; |
| } |
| |
| /* five instructions, multiply bit o by 0, covered before the big guns */ |
| w = ((a>>m)&1)^(a>>(n-1)&2); |
| x = ((b>>m)&1)^(b>>(n-1)&2); |
| y = ((c>>m)&1)^(c>>(n-1)&2); |
| z = ((d>>m)&1)^(d>>(n-1)&2); |
| if (testfour(w,x,y,z)) |
| { /* h4v, not reached */ |
| sprintf(final->line[0], |
| " ub4 rsl = (((val>>%ld) & 1) ^ ((val>>%ld) & 2));\n", m, n-1); |
| return; |
| } |
| } |
| |
| printf("fatal error: bug in hexfour!\n"); |
| exit(SUCCESS); |
| return; |
| } |
| |
| |
| /* test if a_k is distinct and in range for all keys */ |
| static int testeight(keys, badmask) |
| key *keys; /* keys being hashed */ |
| ub1 badmask; /* used for minimal perfect hashing */ |
| { |
| ub1 mask = badmask; |
| key *mykey; |
| |
| for (mykey=keys; mykey; mykey=mykey->next_k) |
| { |
| if (bit(mask, 1<<mykey->a_k)) return FALSE; |
| bis(mask, 1<<mykey->a_k); |
| } |
| return TRUE; |
| } |
| |
| |
| |
| /* |
| * Try to find a perfect hash when there are five to eight keys. |
| * |
| * We can't deterministically find a perfect hash, but there's a reasonable |
| * chance we'll get lucky. Give it a shot. Return TRUE if we succeed. |
| */ |
| static int hexeight(keys, nkeys, final, form) |
| key *keys; |
| ub4 nkeys; |
| gencode *final; |
| hashform *form; |
| { |
| key *mykey; /* walk through the keys */ |
| ub4 i,j,k; |
| ub1 badmask; |
| |
| printf("hexeight\n"); |
| |
| /* what hash values should never be used? */ |
| badmask = 0; |
| if (form->perfect == MINIMAL_HP) |
| { |
| for (i=nkeys; i<8; ++i) |
| bis(badmask,(1<<i)); |
| } |
| |
| /* one instruction */ |
| for (mykey=keys; mykey; mykey=mykey->next_k) |
| mykey->a_k = mykey->hash_k & 7; |
| if (testeight(keys, badmask)) |
| { /* h8a */ |
| final->used = 1; |
| sprintf(final->line[0], " ub4 rsl = (val & 7);\n"); |
| return TRUE; |
| } |
| |
| /* two instructions */ |
| for (i=final->lowbit; i<=final->highbit-2; ++i) |
| { |
| for (mykey=keys; mykey; mykey=mykey->next_k) |
| mykey->a_k = (mykey->hash_k >> i) & 7; |
| if (testeight(keys, badmask)) |
| { /* h8b */ |
| final->used = 1; |
| sprintf(final->line[0], " ub4 rsl = ((val >> %ld) & 7);\n", i); |
| return TRUE; |
| } |
| } |
| |
| /* four instructions */ |
| for (i=final->lowbit; i<=final->highbit; ++i) |
| { |
| for (j=i+1; j<=final->highbit; ++j) |
| { |
| for (mykey=keys; mykey; mykey=mykey->next_k) |
| mykey->a_k = ((mykey->hash_k >> i)+(mykey->hash_k >> j)) & 7; |
| if (testeight(keys, badmask)) |
| { |
| final->used = 1; |
| if (i == 0) /* h8c */ |
| sprintf(final->line[0], |
| " ub4 rsl = ((val + (val >> %ld)) & 7);\n", j); |
| else /* h8d */ |
| sprintf(final->line[0], |
| " ub4 rsl = (((val >> %ld) + (val >> %ld)) & 7);\n", i, j); |
| return TRUE; |
| } |
| |
| for (mykey=keys; mykey; mykey=mykey->next_k) |
| mykey->a_k = ((mykey->hash_k >> i)^(mykey->hash_k >> j)) & 7; |
| if (testeight(keys, badmask)) |
| { |
| final->used = 1; |
| if (i == 0) /* h8e */ |
| sprintf(final->line[0], |
| " ub4 rsl = ((val ^ (val >> %ld)) & 7);\n", j); |
| else /* h8f */ |
| sprintf(final->line[0], |
| " ub4 rsl = (((val >> %ld) ^ (val >> %ld)) & 7);\n", i, j); |
| |
| return TRUE; |
| } |
| |
| for (mykey=keys; mykey; mykey=mykey->next_k) |
| mykey->a_k = ((mykey->hash_k >> i)-(mykey->hash_k >> j)) & 7; |
| if (testeight(keys, badmask)) |
| { |
| final->used = 1; |
| if (i == 0) /* h8g */ |
| sprintf(final->line[0], |
| " ub4 rsl = ((val - (val >> %ld)) & 7);\n", j); |
| else /* h8h */ |
| sprintf(final->line[0], |
| " ub4 rsl = (((val >> %ld) - (val >> %ld)) & 7);\n", i, j); |
| |
| return TRUE; |
| } |
| } |
| } |
| |
| |
| /* six instructions */ |
| for (i=final->lowbit; i<=final->highbit; ++i) |
| { |
| for (j=i+1; j<=final->highbit; ++j) |
| { |
| for (k=j+1; k<=final->highbit; ++k) |
| { |
| for (mykey=keys; mykey; mykey=mykey->next_k) |
| mykey->a_k = ((mykey->hash_k >> i) + |
| (mykey->hash_k >> j) + |
| (mykey->hash_k >> k)) & 7; |
| if (testeight(keys, badmask)) |
| { /* h8i */ |
| final->used = 1; |
| sprintf(final->line[0], |
| " ub4 rsl = (((val >> %ld) + (val >> %ld) + (val >> %ld)) & 7);\n", |
| i, j, k); |
| return TRUE; |
| } |
| } |
| } |
| } |
| |
| |
| return FALSE; |
| } |
| |
| |
| |
| /* |
| * Guns aren't enough. Bring out the Bomb. Use tab[]. |
| * This finds the initial (a,b) when we need to use tab[]. |
| * |
| * We need to produce a different (a,b) every time this is called. Try all |
| * reasonable cases, fastest first. |
| * |
| * The initial mix (which this determines) can be filled into final starting |
| * at line[1]. val is set and a,b are declared. The final hash (at line[7]) |
| * is a^tab[b] or a^scramble[tab[b]]. |
| * |
| * The code will probably look like this, minus some stuff: |
| * val += CONSTANT; |
| * val ^= (val<<16); |
| * val += (val>>8); |
| * val ^= (val<<4); |
| * b = (val >> l) & 7; |
| * a = (val + (val<<m)) >> 29; |
| * return a^scramble[tab[b]]; |
| * Note that *a* and tab[b] will be computed in parallel by most modern chips. |
| * |
| * final->i is the current state of the state machine. |
| * final->j and final->k are counters in the loops the states simulate. |
| */ |
| static void hexn(keys, salt, alen, blen, final) |
| key *keys; |
| ub4 salt; |
| ub4 alen; |
| ub4 blen; |
| gencode *final; |
| { |
| key *mykey; |
| ub4 highbit = final->highbit; |
| ub4 lowbit = final->lowbit; |
| ub4 alog = mylog2(alen); |
| ub4 blog = mylog2(blen); |
| |
| for (;;) |
| { |
| switch(final->i) |
| { |
| case 1: |
| /* a = val>>30; b=val&3 */ |
| for (mykey=keys; mykey; mykey=mykey->next_k) |
| { |
| mykey->a_k = (mykey->hash_k << (UB4BITS-(highbit+1)))>>(UB4BITS-alog); |
| mykey->b_k = (mykey->hash_k >> lowbit) & (blen-1); |
| } |
| if (lowbit == 0) /* hna */ |
| sprintf(final->line[5], " b = (val & 0x%lx);\n", |
| blen-1); |
| else /* hnb */ |
| sprintf(final->line[5], " b = ((val >> %ld) & 0x%lx);\n", |
| lowbit, blen-1); |
| if (highbit+1 == UB4BITS) /* hnc */ |
| sprintf(final->line[6], " a = (val >> %ld);\n", |
| UB4BITS-alog); |
| else /* hnd */ |
| sprintf(final->line[6], " a = ((val << %ld ) >> %ld);\n", |
| UB4BITS-(highbit+1), UB4BITS-alog); |
| |
| ++final->i; |
| return; |
| |
| case 2: |
| /* a = val&3; b=val>>30 */ |
| for (mykey=keys; mykey; mykey=mykey->next_k) |
| { |
| mykey->a_k = (mykey->hash_k >> lowbit) & (alen-1); |
| mykey->b_k = (mykey->hash_k << (UB4BITS-(highbit+1)))>>(UB4BITS-blog); |
| } |
| if (highbit+1 == UB4BITS) /* hne */ |
| sprintf(final->line[5], " b = (val >> %ld);\n", |
| UB4BITS-blog); |
| else /* hnf */ |
| sprintf(final->line[5], " b = ((val << %ld ) >> %ld);\n", |
| UB4BITS-(highbit+1), UB4BITS-blog); |
| if (lowbit == 0) /* hng */ |
| sprintf(final->line[6], " a = (val & 0x%lx);\n", |
| alen-1); |
| else /* hnh */ |
| sprintf(final->line[6], " a = ((val >> %ld) & 0x%lx);\n", |
| lowbit, alen-1); |
| |
| ++final->i; |
| return; |
| |
| case 3: |
| /* |
| * cases 3,4,5: |
| * for (k=lowbit; k<=highbit; ++k) |
| * for (j=lowbit; j<=highbit; ++j) |
| * b = (val>>j)&3; |
| * a = (val<<k)>>30; |
| */ |
| final->k = lowbit; |
| final->j = lowbit; |
| ++final->i; |
| break; |
| |
| case 4: |
| if (!(final->j < highbit)) |
| { |
| ++final->i; |
| break; |
| } |
| for (mykey=keys; mykey; mykey=mykey->next_k) |
| { |
| mykey->b_k = (mykey->hash_k >> (final->j)) & (blen-1); |
| mykey->a_k = (mykey->hash_k << (UB4BITS-final->k-1)) >> (UB4BITS-alog); |
| } |
| if (final->j == 0) /* hni */ |
| sprintf(final->line[5], " b = val & 0x%lx;\n", |
| blen-1); |
| else if (blog+final->j == UB4BITS) /* hnja */ |
| sprintf(final->line[5], " b = val >> %ld;\n", |
| final->j); |
| else |
| sprintf(final->line[5], " b = (val >> %ld) & 0x%lx;\n", /* hnj */ |
| final->j, blen-1); |
| if (UB4BITS-final->k-1 == 0) /* hnk */ |
| sprintf(final->line[6], " a = (val >> %ld);\n", |
| UB4BITS-alog); |
| else /* hnl */ |
| sprintf(final->line[6], " a = ((val << %ld) >> %ld);\n", |
| UB4BITS-final->k-1, UB4BITS-alog); |
| while (++final->j < highbit) |
| { |
| if (((final->diffbits>>(final->j)) & (blen-1)) > 2) |
| break; |
| } |
| return; |
| |
| case 5: |
| while (++final->k < highbit) |
| { |
| if ((((final->diffbits<<(UB4BITS-final->k-1))>>alog) & (alen-1)) > 0) |
| break; |
| } |
| if (!(final->k < highbit)) |
| { |
| ++final->i; |
| break; |
| } |
| final->j = lowbit; |
| final->i = 4; |
| break; |
| |
| |
| case 6: |
| /* |
| * cases 6,7,8: |
| * for (k=0; k<UB4BITS-alog; ++k) |
| * for (j=0; j<UB4BITS-blog; ++j) |
| * val = val+f(salt); |
| * val ^= (val >> 16); |
| * val += (val << 8); |
| * val ^= (val >> 4); |
| * b = (val >> j) & 3; |
| * a = (val + (val << k)) >> 30; |
| */ |
| final->k = 0; |
| final->j = 0; |
| ++final->i; |
| break; |
| |
| case 7: |
| /* Just do something that will surely work */ |
| { |
| ub4 addk = 0x9e3779b9*salt; |
| |
| if (!(final->j <= UB4BITS-blog)) |
| { |
| ++final->i; |
| break; |
| } |
| for (mykey=keys; mykey; mykey=mykey->next_k) |
| { |
| ub4 val = mykey->hash_k + addk; |
| if (final->highbit+1 - final->lowbit > 16) |
| val ^= (val >> 16); |
| if (final->highbit+1 - final->lowbit > 8) |
| val += (val << 8); |
| val ^= (val >> 4); |
| mykey->b_k = (val >> final->j) & (blen-1); |
| if (final->k == 0) |
| mykey->a_k = val >> (UB4BITS-alog); |
| else |
| mykey->a_k = (val + (val << final->k)) >> (UB4BITS-alog); |
| } |
| sprintf(final->line[1], " val += 0x%lx;\n", addk); |
| if (final->highbit+1 - final->lowbit > 16) /* hnm */ |
| sprintf(final->line[2], " val ^= (val >> 16);\n"); |
| if (final->highbit+1 - final->lowbit > 8) /* hnn */ |
| sprintf(final->line[3], " val += (val << 8);\n"); |
| sprintf(final->line[4], " val ^= (val >> 4);\n"); |
| if (final->j == 0) /* hno: don't know how to reach this */ |
| sprintf(final->line[5], " b = val & 0x%lx;\n", blen-1); |
| else /* hnp */ |
| sprintf(final->line[5], " b = (val >> %ld) & 0x%lx;\n", |
| final->j, blen-1); |
| if (final->k == 0) /* hnq */ |
| sprintf(final->line[6], " a = val >> %ld;\n", UB4BITS-alog); |
| else /* hnr */ |
| sprintf(final->line[6], " a = (val + (val << %ld)) >> %ld;\n", |
| final->k, UB4BITS-alog); |
| |
| ++final->j; |
| return; |
| } |
| |
| case 8: |
| ++final->k; |
| if (!(final->k <= UB4BITS-alog)) |
| { |
| ++final->i; |
| break; |
| } |
| final->j = 0; |
| final->i = 7; |
| break; |
| |
| case 9: |
| final->i = 6; |
| break; |
| } |
| } |
| } |
| |
| |
| |
| /* find the highest and lowest bit where any key differs */ |
| static void setlow(keys, final) |
| key *keys; |
| gencode *final; |
| { |
| ub4 lowbit; |
| ub4 highbit; |
| ub4 i; |
| key *mykey; |
| ub4 firstkey; |
| |
| /* mark the interesting bits in final->mask */ |
| final->diffbits = (ub4)0; |
| if (keys) firstkey = keys->hash_k; |
| for (mykey=keys; mykey!=(key *)0; mykey=mykey->next_k) |
| final->diffbits |= (firstkey ^ mykey->hash_k); |
| |
| /* find the lowest interesting bit */ |
| for (i=0; i<UB4BITS; ++i) |
| if (final->diffbits & (((ub4)1)<<i)) |
| break; |
| final->lowbit = i; |
| |
| /* find the highest interesting bit */ |
| for (i=UB4BITS; --i; ) |
| if (final->diffbits & (((ub4)1)<<i)) |
| break; |
| final->highbit = i; |
| } |
| |
| /* |
| * Initialize (a,b) when keys are integers. |
| * |
| * Normally there's an initial hash which produces a number. That hash takes |
| * an initializer. Changing the initializer causes the initial hash to |
| * produce a different (uniformly distributed) number without any extra work. |
| * |
| * Well, here we start with a number. There's no initial hash. Any mixing |
| * costs extra work. So we go through a lot of special cases to minimize the |
| * mixing needed to get distinct (a,b). For small sets of keys, it's often |
| * fastest to skip the final hash and produce the perfect hash from the number |
| * directly. |
| * |
| * The target user for this is switch statement optimization. The common case |
| * is 3 to 16 keys, and instruction counts matter. The competition is a |
| * binary tree of branches. |
| * |
| * Return TRUE if we found a perfect hash and no more work is needed. |
| * Return FALSE if we just did an initial hash and more work is needed. |
| */ |
| int inithex(keys, nkeys, alen, blen, smax, salt, final, form) |
| key *keys; /* list of all keys */ |
| ub4 nkeys; /* number of keys to hash */ |
| ub4 alen; /* (a,b) has a in 0..alen-1, a power of 2 */ |
| ub4 blen; /* (a,b) has b in 0..blen-1, a power of 2 */ |
| ub4 smax; /* maximum range of computable hash values */ |
| ub4 salt; /* used to initialize the hash function */ |
| gencode *final; /* output, code for the final hash */ |
| hashform *form; /* user directives */ |
| { |
| setlow(keys, final); |
| |
| switch (nkeys) |
| { |
| case 1: |
| hexone(keys, final); |
| return TRUE; |
| case 2: |
| hextwo(keys, final); |
| return TRUE; |
| case 3: |
| hexthree(keys, final, form); |
| return TRUE; |
| case 4: |
| hexfour(keys, final); |
| return TRUE; |
| case 5: case 6: case 7: case 8: |
| if (salt == 1 && /* first time through */ |
| hexeight(keys, nkeys, final, form)) /* get lucky, don't need tab[] ? */ |
| return TRUE; |
| /* fall through */ |
| default: |
| if (salt == 1) |
| { |
| final->used = 8; |
| final->i = 1; |
| final->j = final->k = final->l = final->m = final->n = final->o = 0; |
| sprintf(final->line[0], " ub4 a, b, rsl;\n"); |
| sprintf(final->line[1], "\n"); |
| sprintf(final->line[2], "\n"); |
| sprintf(final->line[3], "\n"); |
| sprintf(final->line[4], "\n"); |
| sprintf(final->line[5], "\n"); |
| sprintf(final->line[6], "\n"); |
| if (blen < USE_SCRAMBLE) |
| { /* hns */ |
| sprintf(final->line[7], " rsl = (a^tab[b]);\n"); |
| } |
| else |
| { /* hnt */ |
| sprintf(final->line[7], " rsl = (a^scramble[tab[b]]);\n"); |
| } |
| } |
| hexn(keys, salt, alen, blen, final); |
| return FALSE; |
| } |
| } |