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POMapping a list to a Huffman Tree whilst preserving relative order
 

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  1. POMapping a list to a Huffman Tree whilst preserving relative order
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    <p>I'm having an issue with a search algorithm over a Huffman tree: for a given probability distribution I need the Huffman tree to be identical regardless of permutations of the input data.</p> <p>Here is a picture of what's happening vs what I want: </p> <p><img src="https://i.stack.imgur.com/t4G7u.jpg" alt="Expectation vs reality"></p> <p>Basically I want to know if it's possible to preserve the relative order of the items from the list to the tree. If not, why is that so?</p> <p>For reference, I'm using the Huffman tree to generate sub groups according to a division of probability, so that I can run the search() procedure below. Notice that the data in the merge() sub-routine is combined, along with the weight. The codewords themselves aren't as important as the tree (which should preserve the relative order).</p> <p>For example if I generate the following Huffman codes:</p> <pre><code>probabilities = [0.30, 0.25, 0.20, 0.15, 0.10] items = ['a','b','c','d','e'] items = zip(items, probabilities) t = encode(items) d,l = hi.search(t) print(d) </code></pre> <p>Using the following Class:</p> <pre><code>class Node(object): left = None right = None weight = None data = None code = None def __init__(self, w,d): self.weight = w self.data = d def set_children(self, ln, rn): self.left = ln self.right = rn def __repr__(self): return "[%s,%s,(%s),(%s)]" %(self.data,self.code,self.left,self.right) def __cmp__(self, a): return cmp(self.weight, a.weight) def merge(self, other): total_freq = self.weight + other.weight new_data = self.data + other.data return Node(total_freq,new_data) def index(self, node): return node.weight def encode(symbfreq): pdb.set_trace() tree = [Node(sym,wt) for wt,sym in symbfreq] heapify(tree) while len(tree)&gt;1: lo, hi = heappop(tree), heappop(tree) n = lo.merge(hi) n.set_children(lo, hi) heappush(tree, n) tree = tree[0] def assign_code(node, code): if node is not None: node.code = code if isinstance(node, Node): assign_code(node.left, code+'0') assign_code(node.right, code+'1') assign_code(tree, '') return tree </code></pre> <p>I get: </p> <pre><code>'a'-&gt;11 'b'-&gt;01 'c'-&gt;00 'd'-&gt;101 'e'-&gt;100 </code></pre> <p>However, an assumption I've made in the search algorithm is that more probable items get pushed toward the left: that is I need 'a' to have the '00' codeword - and this should always be the case regardless of any permutation of the 'abcde' sequence. An example output is:</p> <pre><code>codewords = {'a':'00', 'b':'01', 'c':'10', 'd':'110', 'e':111'} </code></pre> <p>(N.b even though the codeword for 'c' is a suffix for 'd' this is ok). </p> <p>For completeness, here is the search algorithm:</p> <pre><code>def search(tree): print(tree) pdb.set_trace() current = tree.left other = tree.right loops = 0 while current: loops+=1 print(current) if current.data != 0 and current is not None and other is not None: previous = current current = current.left other = previous.right else: previous = other current = other.left other = other.right return previous, loops </code></pre> <p>It works by searching for the 'leftmost' 1 in a group of 0s and 1s - the Huffman tree has to put more probable items on the left. For example if I use the probabilities above and the input:</p> <pre><code>items = [1,0,1,0,0] </code></pre> <p>Then the index of the item returned by the algorithm is 2 - which isn't what should be returned (0 should, as it's leftmost).</p>
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    1. USTom Kealy
    1. USTom Kealy
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    1. COThe tree you drew for what you're getting is not what you're getting. E.g. the tree on the left shows 111 for `d`, but you are getting 101 for `d`.
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      1. USMark Adler
    2. CO@MarkAdler, sorry yes. I've been staring at this for 3 days, and can't tell up from down at the moment. Will you forgive a little slip up?
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      1. USTom Kealy
    3. COYour sample code has `t = hi.encode(items)` early on, but `hi` isn't defined. What is it?
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      1. USTim Peters
 

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