Move string distance function to utils and format error message

libdevcore/StringUtils.cpp

@@ -0,0 +1,77 @@
+/*
+	This file is part of solidity.
+
+	solidity is free software: you can redistribute it and/or modify
+	it under the terms of the GNU General Public License as published by
+	the Free Software Foundation, either version 3 of the License, or
+	(at your option) any later version.
+
+	solidity is distributed in the hope that it will be useful,
+	but WITHOUT ANY WARRANTY; without even the implied warranty of
+	MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
+	GNU General Public License for more details.
+
+	You should have received a copy of the GNU General Public License
+	along with solidity.  If not, see <http://www.gnu.org/licenses/>.
+*/
+/** @file StringUtils.h
+ * @author Balajiganapathi S <balajiganapathi.s@gmail.com>
+ * @date 2017
+ *
+ * String routines
+ */
+
+#include "StringUtils.h"
+#include <algorithm>
+#include <string>
+#include <vector>
+
+using namespace std;
+using namespace dev;
+
+namespace dev
+{
+
+bool stringWithinDistance(string const& _name1, string const& _name2, size_t _maxDistance)
+{
+	if (_name1 == _name2)
+		return true;
+
+	size_t n1 = _name1.size(), n2 = _name2.size();
+	vector<vector<size_t>> dp(n1 + 1, vector<size_t>(n2 + 1));
+
+	// In this dp formulation of Damerau–Levenshtein distance we are assuming that the strings are 1-based to make base case storage easier.
+	// So index accesser to _name1 and _name2 have to be adjusted accordingly
+	for (size_t i1 = 0; i1 <= n1; ++i1)
+	{
+		for (size_t i2 = 0; i2 <= n2; ++i2)
+		{
+			if (min(i1, i2) == 0)
+				// Base case
+				dp[i1][i2] = max(i1, i2);
+			else
+			{
+				dp[i1][i2] = min(dp[i1 - 1][i2] + 1, dp[i1][i2 - 1] + 1);
+				// Deletion and insertion
+				if (_name1[i1 - 1] == _name2[i2 - 1])
+					// Same chars, can skip
+					dp[i1][i2] = min(dp[i1][i2], dp[i1 - 1][i2 - 1]);
+				else
+					// Different chars so try substitution
+					dp[i1][i2] = min(dp[i1][i2], dp[i1 - 1][i2 - 1] + 1);
+
+				if (i1 > 1 && i2 > 1 && _name1[i1 - 1] == _name2[i2 - 2] && _name1[i1 - 2] == _name2[i2 - 1])
+					// Try transposing
+					dp[i1][i2] = min(dp[i1][i2], dp[i1 - 2][i2 - 2] + 1);
+			}
+		}
+	}
+
+	size_t distance = dp[n1][n2];
+
+	// if distance is not greater than _maxDistance, and distance is strictly less than length of both names,
+	// they can be considered similar this is to avoid irrelevant suggestions
+	return distance <=  _maxDistance && distance < n1 && distance < n2;
+}
+
+}