Ok from the example you provided (which I converted into a funky table on a bizarre whim!) we have the tree as constructed directly by huffman encoding. The idea of this encoding as you know is to have each leaf at such a position that minimizes the overall depth x weight of the tree. We want to convert that tree into an equivelent in those terms but with the additional constraint that they are ordered left to right by weight.

The fact that a huffman tree is constructed so this branch goes left and that branch right is completely spurious so long as the relative weighting between the nodes is maintained. Also the problem is not to generate a huffman tree. You get that as input. The problem is to find its equivelent tree.

It has occurred to me that this not really golf but rather a brainteaser and for that maybe I should change the title. Also I will post my solution if you think I should.

Step	Trees
0	a:10	b:10	c:12	d:12	e:21	f:22
1	c:12	d:12	a:10 b:10 20	e:21	f:22
2	a:10 b:10 20	e:21	f:22	c:12 d:12 24
3	f:22	c:12 d:12 24	a:10 b:10 20 e:21 41
4	a:10 b:10 20 e:21 41	f:22 c:12 d:12 24 46
5	a:10 b:10 20 e:21 41 f:22 c:12 d:12 24 46 87

Which is this tree:

                             ():87
                               |
                +--------------+--------+
                |                       |
              ():41                   ():46
        +-------+-------+         +-----+--------+
        |               |         |              |
      ():20            e:21      f:22          ():24
   +----+----+                              +----+----+
   |         |                              |         |
  a:10      b:10                           c:12      d:12
[download]

                             ():87
                               |
                +--------------+--------+
                |                       |
              ():44                   ():43
        +-------+-------+         +-----+----+
        |               |         |          |
      ():20           ():24      e:21       f:22
   +----+----+     +----+----+
   |         |     |         |
  a:10      b:10  c:12      d:12
[download]

Symbol	Weight	D*W	Huffman	Fixed
a	10	30	000	000
b	10	30	001	001
c	12	36	110	010
d	12	36	111	011
e	21	42	01	10
f	22	44	10	11

Yves
--
You are not ready to use symrefs unless you already know why they are bad. -- tadmc (CLPM)

Note that the property by which you are defining the Huffman encoding is not entirely obvious from the algorithm, and was not stated in your original node.

Certainly I had no idea what you meant. I had a distinct impression that we were to come up with another tree which could be produced by the same algorithm as the first. I think it would have really helped to have some complex inputs and complex outputs. Or to have a program which could be used to validate things.

Otherwise I guarantee that, for instance, someone will try to get the bonus and (for instance) use a naive recursive algorithm get the wrong result on the case where a-r are weighted 1 each, and S and T get 3. The ouput should be:

a 1 00000
b 1 00001
c 1 00010
d 1 00011
e 1 00100
f 1 00101
g 1 00110
h 1 00111
i 1 01000
j 1 01001
k 1 01010
l 1 01011
m 1 01100
n 1 01101
o 1 01110
p 1 01111
q 1 1000
r 1 1001
S 3 101
T 3 11
[download]

However a recursive approach will have a hard time figuring out where is globally best to take the penalties of incorrect weights.