Add the rt linux 4.1.3-rt3 as base

Import the rt linux 4.1.3-rt3 as OPNFV kvm base.

It's from git://git.kernel.org/pub/scm/linux/kernel/git/rt/linux-rt-devel.git linux-4.1.y-rt and
the base is:

commit 0917f823c59692d751951bf5ea699a2d1e2f26a2
Author: Sebastian Andrzej Siewior <bigeasy@linutronix.de>
Date:   Sat Jul 25 12:13:34 2015 +0200

    Prepare v4.1.3-rt3

    Signed-off-by: Sebastian Andrzej Siewior <bigeasy@linutronix.de>

We lose all the git history this way and it's not good. We
should apply another opnfv project repo in future.

Change-Id: I87543d81c9df70d99c5001fbdf646b202c19f423
Signed-off-by: Yunhong Jiang <yunhong.jiang@intel.com>

1 files changed, 84 insertions, 0 deletions

diff --git a/kernel/tools/perf/util/levenshtein.c b/kernel/tools/perf/util/levenshtein.c
new file mode 100644
index 000000000..e521d1516
--- /dev/null
+++ b/kernel/tools/perf/util/levenshtein.c
@@ -0,0 +1,84 @@
+#include "cache.h"
+#include "levenshtein.h"
+
+/*
+ * This function implements the Damerau-Levenshtein algorithm to
+ * calculate a distance between strings.
+ *
+ * Basically, it says how many letters need to be swapped, substituted,
+ * deleted from, or added to string1, at least, to get string2.
+ *
+ * The idea is to build a distance matrix for the substrings of both
+ * strings.  To avoid a large space complexity, only the last three rows
+ * are kept in memory (if swaps had the same or higher cost as one deletion
+ * plus one insertion, only two rows would be needed).
+ *
+ * At any stage, "i + 1" denotes the length of the current substring of
+ * string1 that the distance is calculated for.
+ *
+ * row2 holds the current row, row1 the previous row (i.e. for the substring
+ * of string1 of length "i"), and row0 the row before that.
+ *
+ * In other words, at the start of the big loop, row2[j + 1] contains the
+ * Damerau-Levenshtein distance between the substring of string1 of length
+ * "i" and the substring of string2 of length "j + 1".
+ *
+ * All the big loop does is determine the partial minimum-cost paths.
+ *
+ * It does so by calculating the costs of the path ending in characters
+ * i (in string1) and j (in string2), respectively, given that the last
+ * operation is a substition, a swap, a deletion, or an insertion.
+ *
+ * This implementation allows the costs to be weighted:
+ *
+ * - w (as in "sWap")
+ * - s (as in "Substitution")
+ * - a (for insertion, AKA "Add")
+ * - d (as in "Deletion")
+ *
+ * Note that this algorithm calculates a distance _iff_ d == a.
+ */
+int levenshtein(const char *string1, const char *string2,
+		int w, int s, int a, int d)
+{
+	int len1 = strlen(string1), len2 = strlen(string2);
+	int *row0 = malloc(sizeof(int) * (len2 + 1));
+	int *row1 = malloc(sizeof(int) * (len2 + 1));
+	int *row2 = malloc(sizeof(int) * (len2 + 1));
+	int i, j;
+
+	for (j = 0; j <= len2; j++)
+		row1[j] = j * a;
+	for (i = 0; i < len1; i++) {
+		int *dummy;
+
+		row2[0] = (i + 1) * d;
+		for (j = 0; j < len2; j++) {
+			/* substitution */
+			row2[j + 1] = row1[j] + s * (string1[i] != string2[j]);
+			/* swap */
+			if (i > 0 && j > 0 && string1[i - 1] == string2[j] &&
+					string1[i] == string2[j - 1] &&
+					row2[j + 1] > row0[j - 1] + w)
+				row2[j + 1] = row0[j - 1] + w;
+			/* deletion */
+			if (row2[j + 1] > row1[j + 1] + d)
+				row2[j + 1] = row1[j + 1] + d;
+			/* insertion */
+			if (row2[j + 1] > row2[j] + a)
+				row2[j + 1] = row2[j] + a;
+		}
+
+		dummy = row0;
+		row0 = row1;
+		row1 = row2;
+		row2 = dummy;
+	}
+
+	i = row1[len2];
+	free(row0);
+	free(row1);
+	free(row2);
+
+	return i;
+}