The von Koch conjecture

Question

You may know the mathematician von Koch by his famous snowflake. However he has more interesting computer science problems up his sleeves. Indeed, let's take a look at this conjecture:

Given a tree with n nodes (thus n-1 edges). Find a way to enumerate the nodes from 1 to n and, accordingly, the edges from 1 to n-1 in such a way, that for each edge k the difference of its node numbers equals to k. The conjecture is that this is always possible.

Here's an example to make it perfectly clear :

YOUR TASK

Your code will take as input a tree, you can take the format you want but for the test cases I will provide the tree by their arcs and the list of their nodes.

For example this is the input for the tree in the picture :

[a,b,c,d,e,f,g]
d -> a
a -> b
a -> g
b -> c
b -> e
e -> f

Your code must return the tree with nodes and edges numbered. You can return a more graphical output but I will provide this kind of output for the test cases :

[a7,b3,c6,d1,e5,f4,g2]
d -> a 6
a -> b 4
a -> g 5
b -> c 3
b -> e 2
e -> f 1

TEST CASES

[a,b,c,d,e,f,g]             [a7,b3,c6,d1,e5,f4,g2]
d -> a                      d -> a 6
a -> b                      a -> b 4
a -> g             =>       a -> g 5
b -> c                      b -> c 3
b -> e                      b -> e 2
e -> f                      e -> f 1


[a,b,c,d]                   [a4,b1,c3,d2]
a -> b                      a -> b 3
b -> c            =>        b -> c 2
b -> d                      b -> d 1


[a,b,c,d,e]                [a2,b3,c1,d4,e5]
a -> b                      a -> b 1
b -> c                      b -> c 2
c -> d             =>       c -> d 3
c -> e                      c -> e 4

This is code-golf this the shortest answer in bytes win!

Note : This is stronger than the Ringel-Kotzig conjecture, which states every tree has a graceful labeling. Since in the Koch conjecture it is not possible to skip integers for the labeling contrary to the graceful labeling in the Ringel-Kotzig conjecture. Graceful labeling has been asked before here.

Answer 1

Jelly, 30 bytes

JŒ!³,$€
ǵy⁴VIAµ€Q⁼$€TðḢịø³JŒ!

Try it online! (Use GṄ³çG as footer to make output prettier. )

Inputs similar to example, e.g. abcdef and [d,a],[a,b],[a,g],[b,c],[b,e],[e,f]

Outputs the list e.g. a,b,c,d,e,f in order.

Note: My program produces different values than the test cases as there are several possibilities which are all valid.

Explanation

JŒ!³,$€                - helper function, generates all possible numberings, input is e.g. 'abcd'
J                      - range(len(input)). e.g. [1,2,3,4]
 Œ!                    - all permutations of the range.
   ³,$                 - pair the input with ... 
      €                - each permutation. Sample element e.g. ['abcd',[3,1,2,4]]

ǵy⁴VIAµ€Q⁼$€TðḢịø³JŒ! - main dyadic link, input is e.g. 'abcd' and '[a,b],[b,c],[b,d]'
 µy                    - use a numbering as an element-wise mapping e.g. 'abcd'->[3,1,2,4]
   ⁴                   - apply this to the list of edges. e.g. '[3,1],[1,2],[1,4]'
    V                  - turn this into an internal list.
     IAµ€              - find absolute difference on each edge
         Q⁼            - Is this invariant under deduplication? Returns 1 if the numbering is valid; 0 otherwise.
Ç          $€          - apply this to all possible numberings
             Tð        - return the indices of all valid numberings
               Ḣ       - choose the first one and
                ị      - get the element corresponding to its index in 
                 ø³JŒ! - all possible numberings 

Save 1 byte by showing all possible solutions:

JŒ!³,$€
ǵy⁴VIAµ€Q⁼$€Tðịø³JŒ!

Try it online! (Use GṄ³çG⁷³G as a footer to make output prettier)

Use converter to copy-paste test case into an input list.

Answer 2

Ruby, 108 bytes

lamba function, accepts an array of 2-element arrays containing the edges (where each edge is expressed as a pair of numbers corresponding to the relevant notes.)

->a{[*1..1+n=a.size].permutation.map{|i|k=a.map{|j|(i[j[0]-1]-i[j[1]-1]).abs}
(k&k).size==n&&(return[i,k])}}

Ungolfed in test program

f=->a{                                    #Accept an array of n tuples (where n is the number of EDGES in this case)
  [*1..1+n=a.size].permutation.map{|i|    #Generate a range 1..n+1 to label the nodes, convert to array, make an array of all permutations and iterate through it.
    k=a.map{|j|(i[j[0]-1]-i[j[1]-1]).abs} #Iterate through a, build an array k of differences between nodes per current permutation, as a trial edge labelling.
    (k&k).size==n&&(return[i,k])          #Intersect k with itself to remove duplicates. If all elements are unique the size will still equal n so
  }                                       #return a 2 element array [list of nodes, list of edges]
}

p f[[[4,1],[1,2],[1,7],[2,3],[2,5],[5,6]]]

p f[[[1,2],[2,3],[2,4]]]

p f[[[1,2],[2,3],[3,4],[2,5]]]

Output

output is a 2 element array, containing:

the new node numbering

the edge numbering.

For example the first edge of the first example [4,1] is between nodes 6 and 1 under the new node numbering and is therefore edge 6-1=5.

[[1, 5, 2, 6, 3, 4, 7], [5, 4, 6, 3, 2, 1]]
[[1, 4, 2, 3], [3, 2, 1]]
[[1, 5, 3, 4, 2], [4, 2, 1, 3]]

There are in fact multiple solutons for each test case. the return stops the function once the first one is found.

Answer 3

Python3, 237 bytes:

def f(d,e,C=[]):
 if len(C)==len(d)-1:yield d
 for u,v in e:
  K={*range(1,len(d)+1)}-{*d.values()}
  for x in[K,[d[u]]][d[u]!=0]:
   for y in[K,[d[v]]][d[v]!=0]:
    if abs(x-y)==max(C+[0])+1:yield from f({**d,u:x,v:y},e,C+[abs(x-y)]) 

Try it online!