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A high performance data logging and graphing system for time series data.
/*
* Copyright 2010 Tom Gibara
*
* Licensed under the Apache License, Version 2.0 (the "License");
* you may not use this file except in compliance with the License.
* You may obtain a copy of the License at
*
* http://www.apache.org/licenses/LICENSE-2.0
* Unless required by applicable law or agreed to in writing, software
* distributed under the License is distributed on an "AS IS" BASIS,
* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
* See the License for the specific language governing permissions and
* limitations under the License.
*
*/
package com.tomgibara.crinch.hashing;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.Comparator;
/**
*
* A "minimal perfect hash" for Strings. After construction with an array of
* n unique non-null strings, an instance of this class will return a
* unique hash value h (0 <= h < n) for any string s in the
* array. A negative has value will typically be returned for a string that is
* not in the array.
*
*
*
* However, the supplied array is not retained. This means that the
* implementation cannot necessarily confirm that a string is not in the
* supplied array. Where this implementation cannot distinguish that a string is
* not in the array, a 'valid' hash value may be returned. Under no
* circumstances will a hash value be returned that is greater than or equal to
* n.
*
*
*
* IMPORTANT NOTE: The array of strings supplied to the
* constructor will be mutated: it is re-ordered so that
* hash(a[i]) == i. Application code must generally use this
* information to map hash values back onto the appropriate string value.
*
*
*
* NOTE: Good performance of this algorithm is predicated on
* string hash values being cached by the String class. Experience
* indicates that is is a good assumption.
*
*
*
* @author Tom Gibara
*/
public class PerfectStringHash implements Hash {
// statics
/**
* Comparator used to order the supplied string array. Hashcodes take
* priority, we will do a binary search on those. Otherwise, lengths take
* priority over character ordering because the hash algorithm prefers to
* compare lengths, it's faster.
*/
private static final Comparator comparator = new Comparator() {
@Override
public int compare(String s1, String s2) {
final int h1 = s1.hashCode();
final int h2 = s2.hashCode();
if (h1 == h2) {
final int d = s1.length() - s2.length();
return d == 0 ? s1.compareTo(s2) : d;
}
return h1 < h2 ? -1 : 1;
}
};
/**
* Builds a (typically v. small) decision tree for distinguishing strings
* that share the same hash value.
*
* @param values
* the string values to distinguish
* @param start
* the index from which the values should be read
* @param length
* the number of string values that need to be distinguished
* @param pivots
* the array that will hold our decision nodes
* @param pivotIndex
* the index at which the tree should be written
*/
private static void generatePivots(String[] values, int start, int length, int[] pivots, int pivotIndex) {
final int capacity = Integer.highestOneBit(length - 1) << 1;
final int depth = Integer.numberOfTrailingZeros(capacity);
pivots[ pivotIndex << 1 ] = depth;
pivots[(pivotIndex << 1) + 1] = length;
pivotIndex++;
//build the array
for (int i = 0; i < depth; i++) {
int step = capacity >> i;
for (int j = (1 << (depth-i-1)) - 1; j < capacity; j += step) {
final int part;
final int comp;
if (j >= length - 1) {
part = Integer.MIN_VALUE;
comp = 0;
} else {
final String v1 = values[start + j];
final String v2 = values[start + j + 1];
final int l1 = v1.length();
final int l2 = v2.length();
if (l1 == l2) {
int tPart = -1;
int tComp = -1;
for (int k = 0; k < l1; k++) {
final char c1 = v1.charAt(k);
final char c2 = v2.charAt(k);
if (c1 == c2) continue;
if (c1 < c2) { //must occur at some point because we have already checked that the two strings are unequal
tPart = k;
tComp = c1;
} else {
//shouldn't be possible - we've sorted the strings to avoid this case
throw new IllegalStateException();
}
break;
}
//check if we've been passed a duplicated value
if (tPart == -1) throw new IllegalArgumentException("duplicate value: " + v1);
part = tPart;
comp = tComp;
} else {
part = -1;
comp = l1;
}
}
pivots[ pivotIndex<<1 ] = part;
pivots[(pivotIndex<<1) + 1] = comp;
pivotIndex++;
}
}
}
// fields
/**
* The hashcodes of the supplied strings.
*/
private final int[] hashes;
/**
* Stores two ints for every string, an offset into the pivot array (-1 if
* not necessary) and the depth of the decision tree that is rooted there.
*/
private final int[] offsets;
/**
* Stores two ints for every decision, the index at which a character
* comparison needs to be made, followed by the character value to be
* compared against; or -1 to indicate a length comparison, followed by the
* length to be compared against.
*/
private final int[] pivots;
/**
* Cache a range object which indicates the range of hash values generated.
*/
private final HashRange range;
/**
* Constructs a minimal perfect string hashing over the supplied strings.
*
* @param values
* an array of unique non-null strings that will be reordered
* such that hash(values[i]) == i.
*/
public PerfectStringHash(final String values[]) {
final int length = values.length;
if (length == 0) throw new IllegalArgumentException("No values supplied");
final int[] hashes = new int[length];
final int[] offsets = new int[2 * length];
final int[] runLengths = new int[length];
//sort values so that we can assume ordering by hashcode, length and char[]
Arrays.sort(values, comparator);
//pull the hashcodes into an array for analysis
for (int i = 0; i < length; i++) hashes[i] = values[i].hashCode();
//test for unique hashes
int offset = 0;
if (length > 1) {
int previousHash = hashes[0];
int runLength = 1;
for (int i = 1; i <= length; i++) {
int currentHash = i == length ? ~previousHash : hashes[i];
if (currentHash == previousHash) {
runLength++;
} else {
if (runLength > 1) {
final int firstIndex = i - runLength;
for (int j = i - 1; j >= firstIndex; j--) {
runLengths[j] = runLength;
//offset points to the first node in decision tree
offsets[ j<<1 ] = offset;
//adjustment is number of indices to first duplicate
offsets[(j<<1) + 1] = j - firstIndex;
}
//extra one for recording depth
offset += (Integer.highestOneBit(runLength - 1) << 1);
runLength = 1;
} else {
runLengths[i-1] = 1;
offsets[(i-1)<<1] = -1;
}
}
previousHash = currentHash;
}
}
//shortcut for when all hashes are unique
if (offset == 0) {
this.hashes = hashes;
this.offsets = null;
this.pivots = null;
this.range = new HashRange(0, length - 1);
return;
}
//build the decision trees
final int[] pivots = new int[offset * 2];
for (int i = 0; i < length;) {
final int runLength = runLengths[i];
if (runLength > 1) generatePivots(values, i, runLength, pivots, (int) offsets[i << 1]);
i += runLength;
}
//setup our state
this.pivots = pivots;
this.offsets = offsets;
this.hashes = hashes;
this.range = new HashRange(0, length - 1);
}
// hash generator methods
@Override
public HashRange getRange() {
return range;
}
@Override
public BigInteger hashAsBigInt(String value) {
return BigInteger.valueOf(hash(value));
}
//TODO decide whether to throw an IAE if -1 is returned from hash
@Override
public int hashAsInt(String value) {
return hash(value);
}
@Override
public long hashAsLong(String value) {
return hash(value);
}
/**
* Generates a hashcode for the supplied string.
*
* @param value
* any string, not null
* @return a minimal hashcode for the supplied string, or -1
*/
private int hash(String value) {
final int h = value.hashCode();
final int index = Arrays.binarySearch(hashes, h);
final int[] pivots = this.pivots;
if (pivots == null || index < 0) return index;
final int offset = offsets[index << 1];
if (offset == -1) return index;
final int depth = pivots[(offset << 1) ];
final int count = pivots[(offset << 1) + 1];
int i = 0;
for (int d = 0; d < depth; d++) {
final int t = (offset + (1 << d) + i) << 1;
final int part = pivots[t ];
final int comp = pivots[t + 1];
final boolean right;
if (part == Integer.MIN_VALUE) { //easy case - no right value
right = false;
} else if (part == -1) { //compare length
right = value.length() > comp;
} else { //lengths are equal, compare character
right = value.charAt(part) > (char) comp;
}
i <<= 1;
if (right) i++;
}
return i >= count ? -1 : index + i - offsets[(index << 1) + 1];
}
}
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