// kelondroHashtable.java
// ------------------
// part of the Kelondro Database
// (C) by Michael Peter Christen; mc@anomic.de
// first published on http://www.anomic.de
// Frankfurt, Germany, 2005
// last major change: 21.06.2005
//
// This program 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 2 of the License, or
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/*
we implement a hashtable based on folded binary trees
each hight in these binary trees represents one step of rehasing
the re-hashing is realised by extending the number of relevant bits in the given hash
We construct the binary tree as follows
- there exists no root node
- at height-1 are 2 nodes, and can be accessed by using only the least significant bit of the hash
- at height-2 are 4 nodes, addresses by (hash & 3) - mapping the 2 lsb of the hash
- at height-3 are 8 nodes, addresses by (hash & 7)
- .. and so on.
The number of nodes N(k) that are needed for a tree of height-k is
N(k) = 2**k + N(k-1) = 2**(k + 1) - 2 [where k > 0]
We fold this tree by putting all heights of the tree in a sequence
Computation of the position (the index) of a node:
given:
hash h, with k significant bits (representing a height-k): h|k
then the position of a node node(h,k) is
node(h,k) = N(k - 1) + h|k [where k > 0]
We use these nodes to sequentially store a hash h at position node(h, 1), and
if that fails on node(h, 2), node(h, 3) and so on.
This is highly inefficient for the most heights k = 1, ..., (?)
The 'inefficient-border' depends on the number of elements that we want to store.
We therefore introduce an offset o which is the number of bits that are not used
at the beginning of (re-)hashing. But even if these o re-hasing steps are not done,
all bits of the hash are relevant.
Now the number of nodes N(k) that are needed is computed by N(k,o):
N(k,o) = N(k) - N(o) = 2**(k + 1) - 2**(o + 1) [where k > o, o >= 0]
When o=0 then this is equivalent to N(k).
The node-formula must be adopted as well
node(h,k,o) = N(k - 1, o) + h|k [where k > o, o >= 0]
So if you set an offset o, this leads to a minimum number of nodes
at level k=o+1: node(0,o + 1,o) = N(o, o) = 0 (position of the first entry)
Computatiion of the maxlen 'maxk', the maximum height of the tree for a given
number of maximum entries 'maxsize' in the hashtable:
maxk shall be computed in such a way, that N(k,o) <= maxsize, for any o or k
This means paricualary, that
node(h,k,o) < maxsize
for h|k we must consider the worst case:
h|k (by maxk) = 2**k - 1
therefore
node(h,maxk,o) < maxsize
N(maxk - 1, o) + h|maxk < maxsize [where maxk > o, o >= 0]
2**maxk - 2**(o + 1) + 2**maxk - 1 < maxsize [where maxk > o, o >= 0]
2**maxk - 2**(o + 1) + 2**maxk < maxsize + 1 [where maxk > o, o >= 0]
2**maxk + 2**maxk < maxsize + 2**(o + 1) + 1 [where maxk > o, o >= 0]
2**(maxk+1) < maxsize + 2**(o + 1) + 1 [where maxk > o, o >= 0]
maxk < log2(maxsize + 2**(o + 1) + 1) [where maxk > o, o >= 0]
setting maxk to
maxk = log2(maxsize)
will make this relation true in any case, even if maxk = log2(maxsize) + 1
would also be correct in some cases
Now we can use the following formula to create the folded binary hash tree:
node(h,k,o) = 2**k - 2**(o + 1) + h|k
to compute the node index and
maxk = log2(maxsize)
to compute the upper limit of re-hashing
*/
package de.anomic.kelondro;
import java.io.File;
import java.io.IOException;
public class kelondroHashtable {
private kelondroFixedWidthArray hashArray;
protected int offset;
protected int maxk;
private int maxrehash;
private kelondroRow.Entry dummyRow;
private static final byte[] dummyKey = kelondroBase64Order.enhancedCoder.encodeLong(0, 5).getBytes();
public kelondroHashtable(File file, kelondroRow rowdef, int offset, int maxsize, int maxrehash) throws IOException {
// this creates a new hashtable
// the key element is not part of the columns array
// this is unlike the kelondroTree, where the key is part of a row
// the offset is a number of bits that is omitted in the folded tree hierarchy
// a good number for offset is 8
// the maxsize number is the maximum number of elements in the hashtable
// this number is needed to omit grow of the table in case of re-hashing
// the maxsize is re-computed to a virtual folding height and will result in a tablesize
// less than the given maxsize. The actual maxsize can be retrieved by maxsize()
boolean fileExisted = file.exists();
this.hashArray = new kelondroFixedWidthArray(file, extCol(rowdef), 6);
if (fileExisted) {
this.offset = hashArray.geti(0);
this.maxk = hashArray.geti(1);
this.maxrehash = hashArray.geti(2);
} else {
this.offset = offset;
this.maxk = kelondroMSetTools.log2a(maxsize); // equal to |log2(maxsize)| + 1
if (this.maxk >= kelondroMSetTools.log2a(maxsize + power2(offset + 1) + 1) - 1) this.maxk--;
this.maxrehash = maxrehash;
dummyRow = this.hashArray.row().newEntry();
dummyRow.setCol(0, dummyKey);
//for (int i = 0; i < hashArray.row().columns(); i++)
hashArray.seti(0, this.offset);
hashArray.seti(1, this.maxk);
hashArray.seti(2, this.maxrehash);
}
}
private kelondroRow extCol(kelondroRow rowdef) {
kelondroColumn[] newCol = new kelondroColumn[rowdef.columns() + 1];
newCol[0] = new kelondroColumn("Cardinal key-4 {b256}");
for (int i = 0; i < rowdef.columns(); i++) newCol[i + 1] = rowdef.column(i);
return new kelondroRow(newCol, rowdef.objectOrder, rowdef.primaryKey);
}
public static int power2(int x) {
int p = 1;
while (x > 0) {p = p << 1; x--;}
return p;
}
public synchronized byte[][] get(int key) throws IOException {
Object[] search = search(new Hash(key));
if (search[1] == null) return null;
byte[][] row = (byte[][]) search[1];
byte[][] result = new byte[row.length - 1][];
System.arraycopy(row, 1, result, 0, row.length - 1);
return result;
}
public synchronized kelondroRow.Entry put(int key, kelondroRow.Entry rowentry) throws IOException {
Hash hash = new Hash(key);
// find row
Object[] search = search(hash);
kelondroRow.Entry oldhkrow;
int rowNumber = ((Integer) search[0]).intValue();
if (search[1] == null) {
oldhkrow = null;
} else {
oldhkrow = (kelondroRow.Entry) search[1];
}
// make space
while (rowNumber >= hashArray.size()) hashArray.overwrite(hashArray.size(), dummyRow);
// write row
kelondroRow.Entry newhkrow = hashArray.row().newEntry();
newhkrow.setCol(0, hash.key());
newhkrow.setCol(1, rowentry.bytes());
hashArray.overwrite(rowNumber, newhkrow);
return hashArray.row().newEntry(oldhkrow.getColBytes(1));
}
private Object[] search(Hash hash) throws IOException {
kelondroRow.Entry hkrow;
int rowKey;
int rowNumber;
do {
rowNumber = hash.node();
if (rowNumber >= hashArray.size()) return new Object[]{new Integer(rowNumber), null};
hkrow = hashArray.get(rowNumber);
rowKey = (int) hkrow.getColLong(0);
if (rowKey == 0) return new Object[]{new Integer(rowNumber), null};
hash.rehash();
} while (rowKey != hash.key());
return new Object[]{new Integer(rowNumber), hkrow};
}
private class Hash {
int key;
int hash;
int depth;
public Hash(int key) {
this.key = key;
this.hash = key;
this.depth = offset + 1;
}
public int key() {
return key;
}
private int hash() {
return hash & (power2(depth) - 1); // apply mask
}
public int depth() {
return depth;
}
public void rehash() {
depth++;
if (depth > maxk) {
depth = offset + 1;
hash = (int) ((5 * (long) hash - 7) / 3 + 13);
}
}
public int node() {
// node(h,k,o) = 2**k - 2**(o + 1) + h|k
return power2(depth) - power2(offset + 1) + hash();
}
}
}