renames

benchmarks/microbench/src/ssort/ssort.cc

@@ -0,0 +1,111 @@
+// This is the Skew algorithm's reference implementation.
+
+#include <benchmark.h>
+
+static int N;
+
+inline bool leq(int a1, int a2,   int b1, int b2) { // lexic. order for pairs
+  return(a1 < b1 || (a1 == b1 && a2 <= b2));
+}                                                   // and triples
+inline bool leq(int a1, int a2, int a3,   int b1, int b2, int b3) {
+  return(a1 < b1 || (a1 == b1 && leq(a2,a3, b2,b3)));
+}
+// stably sort a[0..n-1] to b[0..n-1] with keys in 0..K from r
+static void radixPass(int* a, int* b, int* r, int n, int K)
+{ // count occurrences
+  int* c = (int*)bench_alloc(sizeof(int)*(K+1));
+  for (int i = 0;  i <= K;  i++) c[i] = 0;         // reset counters
+  for (int i = 0;  i < n;  i++) c[r[a[i]]]++;    // count occurences
+  for (int i = 0, sum = 0;  i <= K;  i++) { // exclusive prefix sums
+     int t = c[i];  c[i] = sum;  sum += t;
+  }
+  for (int i = 0;  i < n;  i++) b[c[r[a[i]]]++] = a[i];      // sort
+}
+
+// find the suffix array SA of s[0..n-1] in {1..K}^n
+// require s[n]=s[n+1]=s[n+2]=0, n>=2
+void suffixArray(int* s, int* SA, int n, int K) {
+  int n0=(n+2)/3, n1=(n+1)/3, n2=n/3, n02=n0+n2;
+  int* s12  = (int*)bench_alloc(sizeof(int)*(n02+3));  s12[n02]= s12[n02+1]= s12[n02+2]=0;
+  int* SA12 = (int*)bench_alloc(sizeof(int)*(n02+3)); SA12[n02]=SA12[n02+1]=SA12[n02+2]=0;
+  int* s0   = (int*)bench_alloc(sizeof(int)*n0);
+  int* SA0  = (int*)bench_alloc(sizeof(int)*n0);
+
+  // generate positions of mod 1 and mod  2 suffixes
+  // the "+(n0-n1)" adds a dummy mod 1 suffix if n%3 == 1
+  for (int i=0, j=0;  i < n+(n0-n1);  i++) if (i%3 != 0) s12[j++] = i;
+
+  // lsb radix sort the mod 1 and mod 2 triples
+  radixPass(s12 , SA12, s+2, n02, K);
+  radixPass(SA12, s12 , s+1, n02, K);
+  radixPass(s12 , SA12, s  , n02, K);
+
+  // find lexicographic names of triples
+  int name = 0, c0 = -1, c1 = -1, c2 = -1;
+  for (int i = 0;  i < n02;  i++) {
+    if (s[SA12[i]] != c0 || s[SA12[i]+1] != c1 || s[SA12[i]+2] != c2) {
+      name++;  c0 = s[SA12[i]];  c1 = s[SA12[i]+1];  c2 = s[SA12[i]+2];
+    }
+    if (SA12[i] % 3 == 1) { s12[SA12[i]/3]      = name; } // left half
+    else                  { s12[SA12[i]/3 + n0] = name; } // right half
+  }
+
+  // recurse if names are not yet unique
+  if (name < n02) {
+    suffixArray(s12, SA12, n02, name);
+    // store unique names in s12 using the suffix array
+    for (int i = 0;  i < n02;  i++) s12[SA12[i]] = i + 1;
+  } else // generate the suffix array of s12 directly
+    for (int i = 0;  i < n02;  i++) SA12[s12[i] - 1] = i;
+
+  // stably sort the mod 0 suffixes from SA12 by their first character
+  for (int i=0, j=0;  i < n02;  i++) if (SA12[i] < n0) s0[j++] = 3*SA12[i];
+  radixPass(s0, SA0, s, n0, K);
+
+  // merge sorted SA0 suffixes and sorted SA12 suffixes
+  for (int p=0,  t=n0-n1,  k=0;  k < n;  k++) {
+#define GetI() (SA12[t] < n0 ? SA12[t] * 3 + 1 : (SA12[t] - n0) * 3 + 2)
+    int i = GetI(); // pos of current offset 12 suffix
+    int j = SA0[p]; // pos of current offset 0  suffix
+    if (SA12[t] < n0 ?
+        leq(s[i],       s12[SA12[t] + n0], s[j],       s12[j/3]) :
+        leq(s[i],s[i+1],s12[SA12[t]-n0+1], s[j],s[j+1],s12[j/3+n0]))
+    { // suffix from SA12 is smaller
+      SA[k] = i;  t++;
+      if (t == n02) { // done --- only SA0 suffixes left
+        for (k++;  p < n0;  p++, k++) SA[k] = SA0[p];
+      }
+    } else {
+      SA[k] = j;  p++;
+      if (p == n0)  { // done --- only SA12 suffixes left
+        for (k++;  t < n02;  t++, k++) SA[k] = GetI();
+      }
+    }
+  }
+}
+
+extern "C" {
+
+static int *s, *sa;
+
+void bench_ssort_prepare() {
+  N = setting->size;
+  bench_srand(1);
+  s = (int*)bench_alloc(sizeof(int)*(N+10));
+  sa = (int*)bench_alloc(sizeof(int)*(N+10));
+
+  for (int i = 0; i < N; i ++) {
+    s[i] = bench_rand() % 26;
+  }
+}
+
+void bench_ssort_run() {
+  suffixArray(s, sa, N, 26);
+}
+
+int bench_ssort_validate() {
+  return checksum(sa, sa + N) == setting->checksum;
+}
+
+}
+