Java中的哈夫曼树

时间:2013-05-20 21:06:15

标签: java huffman-code

我的哈夫曼树代码有问题。在main方法中,我输入了一串符号,我还输入了一个包含符号频率的整数数组。它应该打印出每个符号及其霍夫曼代码,但我认为它错了......

以下是代码:

 package huffman;

import java.util.*;

abstract class HuffmanTree implements Comparable<HuffmanTree> {
    public final int frequency; // the frequency of this tree
    public HuffmanTree(int freq) { frequency = freq; }

    // compares on the frequency
    public int compareTo(HuffmanTree tree) {
        return frequency - tree.frequency;
    }
}

class HuffmanLeaf extends HuffmanTree {
    public final char value; // the character this leaf represents

    public HuffmanLeaf(int freq, char val) {
        super(freq);
        value = val;
    }
}

class HuffmanNode extends HuffmanTree {
    public final HuffmanTree left, right; // subtrees

    public HuffmanNode(HuffmanTree l, HuffmanTree r) {
        super(l.frequency + r.frequency);
        left = l;
        right = r;
    }
}

public class Huffman {
    // input is an array of frequencies, indexed by character code
    public static HuffmanTree buildTree(int[] charFreqs, char[] test2) {
        PriorityQueue<HuffmanTree> trees = new PriorityQueue<HuffmanTree>();
        // initially, we have a forest of leaves
        // one for each non-empty character
        for (int i = 0; i < charFreqs.length; i++)
            if (charFreqs[i] > 0)
                trees.offer(new HuffmanLeaf(charFreqs[i], test2[i]));

        assert trees.size() > 0;
        // loop until there is only one tree left
        while (trees.size() > 1) {
            // two trees with least frequency
            HuffmanTree a = trees.poll();
            HuffmanTree b = trees.poll();

            // put into new node and re-insert into queue
            trees.offer(new HuffmanNode(a, b));
        }
        return trees.poll();
    }

    public static void printCodes(HuffmanTree tree, StringBuffer prefix) {
        assert tree != null;
        if (tree instanceof HuffmanLeaf) {
            HuffmanLeaf leaf = (HuffmanLeaf)tree;

            // print out character, frequency, and code for this leaf (which is just the prefix)
            System.out.println(leaf.value + "\t" + leaf.frequency + "\t" + prefix);

        } else if (tree instanceof HuffmanNode) {
            HuffmanNode node = (HuffmanNode)tree;

            // traverse left
            prefix.append('0');
            printCodes(node.left, prefix);
            prefix.deleteCharAt(prefix.length()-1);

            // traverse right
            prefix.append('1');
            printCodes(node.right, prefix);
            prefix.deleteCharAt(prefix.length()-1);
        }
    }

    public static void main(String[] args) {
        //Symbols:
        String str = "12345678"; 
        char[] test2 = str.toCharArray();
        //Frequency (of the symbols above):
        int[] charFreqs = {36,18,12,9,7,6,5,4};


        // build tree
        HuffmanTree tree = buildTree(charFreqs,test2);

        // print out results
        System.out.println("SYMBOL\tFREQ\tHUFFMAN CODE");
        printCodes(tree, new StringBuffer());
    }
}

我得到的输出是:

SYMBOL  FREQ    HUFFMAN CODE
1           36          0
3           12          100
6           6           1010
5           7           1011
2           18          110
4           9           1110
8           4           11110
7           5           11111

这很奇怪,例如符号7应该是:11110和符号8应该是:11111

你能帮我吗?

4 个答案:

答案 0 :(得分:1)

位模式的分配与代码的最优性无关。你的任务将工作得很好。它没什么奇怪的。你本可以对2:110,3:100或4:1110,5:1011表示担忧,但这些也很好。

对代码强加命令的唯一原因是减少将代码从压缩器传送到解压缩器所需的位数。您可以发送每个符号的代码长度,而不是发送代码,只要代码的构造在两侧都是相同的长度。

在这种情况下,方法通常是将数字顺序的代码分配给符号的排序列表。那么你确实会让符号7的代码“值”低于符号8,如果这是它们的分配顺序。

对于您的示例,这样的规范代码将是:

1: 1 - 0
2: 3 - 100
3: 3 - 101
4: 4 - 1100
5: 4 - 1101
6: 4 - 1110
7: 5 - 11110
8: 5 - 11111

您只需获取长度并在相同长度内对符号进行排序。然后分配以0开头并递增的代码,在您逐步增加长度时将位添加到末尾。

请注意,这是一个不常见的示例,其中符号顺序也是频率顺序。通常情况并非如此。

答案 1 :(得分:0)

只需添加一个0来理解结束位。 (超过3位读数)

1 36 0 3 12 100 6 6 1010 5 7 1011'0 2 18 110 4 9 1110 8 4 11110 7 5 11111'0

答案 2 :(得分:0)

回答答案评论中的问题:

  嘿马克,谢谢你的帮助,但我真的不明白你是如何得到这些代码的?我是否必须在代码中进行大量更改?&lt;

Mark只是指霍夫曼编码的目标,即为每个符号找到最有效的深度(比特数),以便所有符号的整体编码(所有符号的频率[符号] * codeLenght [符号])是最小化。

因此,实际上您需要做的就是确定每个符号的叶子深度(级别)。现在按每个符号的深度对其进行排序,然后开始计算它们。

现在你只需要计算模式了。很简单:

Example:
2x2: 00, 01  (next is 10)
4x3: 10 + (00, 01, 10) = 1000, 1001, 1010 (next is 1011)
5x3: 1011 + (0, 1, 0 + 10) = 10110, 10111, 10110 + 10 = 11000 (next would be 11001)...

最后一部分显示了如果元素数量大于两个组之间的可用比特差异会发生什么。它只是添加到前缀。

这样就创建了一个使用最少空间的霍夫曼代码。由于这只是一棵树,你也可以从11111开始并删除1并获得另一个在位数方面同样有效的代码系统。

可以添加的一件事是,有一些修改可以将比较中的1(或0)的数量增加到0(或1),因此您有另一种机会压缩用于压缩消息的位模式。


摘要:按频率树中的深度对符号进行排序。通过添加(减去)一个来构造代码。下一个免费代码是下一组的开始前缀。对于每个新成员,只需添加(减去)。首先,在代码中填写0(1)。

为了存储这样的树,记住对同一个symbole组使用相同的算法,你只需要存储以下信息:

组数n,n * {+ bitDepth,符号数i,s1,s2 ... si}。由于存储n,+ bitDepth,符号数本身就是压缩对象,你可以使用可变位格式甚至发出一个霍夫曼树,因为你会发现几乎不均匀的分布,这是你看到压缩发生的主要原因与霍夫曼树。

答案 3 :(得分:0)

发布基于我基于Princeton.EDU版本创建的Java中的霍夫曼树的完整实现。作为一个学术示例,普林斯顿软件包的版本是可以的,但在上下文之外实际上并不能使用。

下面的示例可以通过提供一个字符串作为输入并获取一个字节数组作为输出来使用。

如果您想将输出编码为人类可读的格式,请使用标准的Java Base64编码器(在空间方面效率很低)或检查我发布的here

的HumanByte类。
import java.io.ByteArrayInputStream;
import java.io.ByteArrayOutputStream;
import java.util.PriorityQueue;

/**************************************************************************************************************
 * Used edu.princeton.cs version of huffman algorithm and made it easily usable outside of example context
 * replaced proprietary priority queue by Java standard one
 * created binary output/input that has minimal features needed
 *
 * If human readable encoding of huffman encoded string is desired then look at HumanByte class
 * that was posted here
 * 
 * https://stackoverflow.com/questions/4141317/how-to-convert-byte-array-into-human-readable-format/58332208#58332208
 *
 * @author Stan Sokolov
 * 10/9/19
 **************************************************************************************************************/
public class BetterHuffman {
    private static final int R = 256;
    private static final char PARENT = '\u0000';
    private static final char EMPTY = '\u0001';
    private static final int UNDEFINED = -1;


    /**************************************************************************************************************
     *   Compress string in bytes
     **************************************************************************************************************/
    public static byte[] compress(final String s) {

        final HuffmanOut binaryOut = new HuffmanOut();
        final char[] input = s.toCharArray();
        final int[] freq = new int[R];

        for (char anInput1 : input) {
            ++freq[anInput1];
        }

        final BetterHuffman.Node root = buildTrie(freq);
        final String[] st = new String[R];
        buildCode(st, root, "");

        writeTrie(root, binaryOut);
        binaryOut.write(input.length);


        for (char anInput : input) {
            final String code = st[anInput];
            for (char ch : code.toCharArray()) {
                binaryOut.writeBit(ch == '1');
            }
        }

        return binaryOut.value();
    }

    /**************************************************************************************************************
     *   build huffman tree
     **************************************************************************************************************/
    private static BetterHuffman.Node buildTrie(int[] freq) {
        final PriorityQueue<BetterHuffman.Node> pq2 = new PriorityQueue<>();

        for (char i = 0; i < R; ++i) {
            if (freq[i] > 0) {
                //pq.insert(new BetterHuffman.Node(i, freq[i], null, null));
                pq2.add(new BetterHuffman.Node(i, freq[i], null, null));
            }
        }

        if (pq2.size() == 1) {//if entire string is just one char repeated
            if (freq[0] == 0) {//empty string
                pq2.add(new BetterHuffman.Node(PARENT, 0, null, null));
            } else {
                pq2.add(new BetterHuffman.Node(EMPTY, 0, null, null));
            }
        } else
            while (pq2.size() > 1) {
                final BetterHuffman.Node left = pq2.poll();
                final BetterHuffman.Node right = pq2.poll();
                //aggregate two nodes into one by summing frequency
                final BetterHuffman.Node parent = new BetterHuffman.Node(PARENT, left.freq + right.freq, left, right);
                pq2.add(parent);
            }

        //this will be the root node that would have total length of input as frequency
        return pq2.poll();
    }

    /**************************************************************************************************************
     *   write tree into byte output
     **************************************************************************************************************/
    private static void writeTrie(final BetterHuffman.Node x, final HuffmanOut binaryOut) {
        if (x.isLeaf()) {//if this is a node representing symbol in alphabet
            binaryOut.writeBit(true);
            binaryOut.writeByte((int) x.ch);
        } else {
            binaryOut.writeBit(false); //this is an aggregate node used for branching
            writeTrie(x.left, binaryOut);
            writeTrie(x.right, binaryOut);
        }
    }

    /**************************************************************************************************************
     *   make substitutes for incoming words
     **************************************************************************************************************/
    private static void buildCode(final String[] st, final BetterHuffman.Node x, final String s) {
        if (!x.isLeaf()) {
            buildCode(st, x.left, s + '0');
            buildCode(st, x.right, s + '1');
        } else {
            st[x.ch] = s;
        }
    }

    /**************************************************************************************************************
     *   Return uncompressed string
     **************************************************************************************************************/
    public static String expand(final byte[] value) {

        final HuffmanIn binaryIn = new HuffmanIn(value);
        final StringBuilder out = new StringBuilder();

        final BetterHuffman.Node root = readTrie(binaryIn);
        final int length = binaryIn.readInt();

        for (int i = 0; i < length; ++i) {
            BetterHuffman.Node x = root;

            while (!x.isLeaf()) {
                boolean bit = binaryIn.readBoolean();
                if (bit) {
                    x = x.right;
                } else {
                    x = x.left;
                }
            }

            out.append(x.ch);
        }


        return out.toString();
    }

    /**************************************************************************************************************
     *   get tree from bytes
     **************************************************************************************************************/
    private static BetterHuffman.Node readTrie(final HuffmanIn binaryIn) {
        boolean isLeaf = binaryIn.readBoolean();
        if (isLeaf) {
            char ch = binaryIn.readChar();
            return new BetterHuffman.Node(ch, UNDEFINED, null, null);
        } else {
            return new BetterHuffman.Node(PARENT, UNDEFINED, readTrie(binaryIn), readTrie(binaryIn));
        }
    }


    /**************************************************************************************************************
     *   Simple implementation of node
     **************************************************************************************************************/
    private static class Node implements Comparable<Node> {
        private final char ch;
        private final int freq;
        private final Node left;
        private final Node right;

        Node(char ch, int freq, Node left, Node right) {
            this.ch = ch;
            this.freq = freq;
            this.left = left;
            this.right = right;
        }

        private boolean isLeaf() {
            return left == null && right == null;
        }

        @Override
        public int compareTo(Node that) {
            return this.freq - that.freq;
        }
    }

    /**************************************************************************************************************
     *   class to read bits from stream
     **************************************************************************************************************/
    private static class HuffmanIn {

        private final ByteArrayInputStream in;
        private int buffer;
        private byte n;

        HuffmanIn(final byte[] input) {
            in = new ByteArrayInputStream(input);
            fillBuffer();
        }

        private void fillBuffer() {
            buffer = in.read();
            n = 8;
        }

        boolean readBoolean() {
            boolean bit = (buffer >> --n & 1) == 1;
            if (n == 0) {
                fillBuffer();
            }
            return bit;
        }

        char readChar() {
            int x = buffer <<= 8 - n;
            if (n == 8) {
                fillBuffer();
            } else {
                byte oldN = n;
                fillBuffer();
                n = oldN;
                x |= buffer >>> n;
            }
            return (char) (x & 255);
        }

        int readInt() {
            int x = 0;
            for (int i = 0; i < 4; ++i) {
                char c = readChar();
                x <<= 8;
                x |= c;
            }
            return x;
        }
    }

    /**************************************************************************************************************
     *   Output
     **************************************************************************************************************/
    private static class HuffmanOut {

        private ByteArrayOutputStream out = new ByteArrayOutputStream();
        private int buffer;
        private byte n;

        /**************************************************************************************************************
         * @return what was compressed so far in a human readable (no funny characters) format
         **************************************************************************************************************/
        public byte[] value() {
            clearBuffer();
            return out.toByteArray();
        }

        void writeBit(final boolean bit) {
            buffer = (buffer <<= 1) | (bit ? 1 : 0);
            if (++n == 8) {
                clearBuffer();
            }
        }

        void writeByte(final int x) {
            for (int i = 0; i < 8; ++i) {
                writeBit((x >>> 8 - i - 1 & 1) == 1);
            }
        }

        void clearBuffer() {
            if (n != 0) {
                out.write(buffer <<= 8 - n);
                n = 0;
                buffer = 0;
            }
        }


        /**************************************************************************************************************
         *   write all 4 bytes of int
         **************************************************************************************************************/
        void write(final int x) {
            for (int i = 3; i >= 0; i--)
                writeByte(x >>> (i * 8) & 255);//write 4 bytes of int
        }


    }

}