UNDIRECTED HAMCYCLE to HAMPATH reduction

Question

I'll define the problems

UHAMPATH

Input: A undirected graph G and 2 nodes, s and t

Question: Is there a hamiltonian path from s to t in G?

UHAMCYCLE

Input: A undirected graph G

Question: Is there a hamiltonian cycle in G?

$$UHAMCYCLE \leq_p UHAMPATH$$

my reduction is as followed $(G) \to (G', s, t)$

function(G)

    for each e = (u, v) in E(G)
        G' = G
        add nodes u' and v' to G'
        add edges (u', u) and (v, v')
        s = u'
        t = v'
        if there is a hamilton path from s to t:
            return (G', s, t)

Basically if there is a hamilton cycle in G then some edge in $G$ will form a hamilton path in $G'$.

An example to further illustrate the reduction:

 Graph G

(a) ---- --(d)
 | \        |
 |   \      |
 |    \     |
 |     \    | 
 |      \   | 
 |       \  |
 |         \|      |
(b) ------ (c)

Has a clear Hamilton cycle (a, b, c, d). 

If we choose edges (a, d)

 Graph G'

(s)--(a) ---- --(d)
      | \        |
      |   \      |
      |    \     |
      |     \    | 
      |      \   | 
      |       \  |
      |         \|      
     (b) ------ (c)--(t)

Doesn't have a hamilton path from s to t. However if you choose any other edge such as (a, d) then

(s)--(a) ---- --(d)--(t)
      | \        |
      |   \      |
      |    \     |
      |     \    | 
      |      \   | 
      |       \  |
      |         \|      
     (b) ------ (c)

$(s \to a \to b \to c \to d \to t)$. Holds.

---

I'm confused whether or not I can use this line:

if there is a hamilton path from s to t:
    return (G', s, t)

Checking to see if a graph has a hamilton path is NP complete but since we are looking to reduce it to it I think we can?

Answer 1

No. Since you are performing a many-one reduction, you have no capabilities beyond the model you are using for the reduction, in this case a poly-time Turing machine. Queries to the problem you are reducing to are only allowed if you are performing a Turing reduction, but that is a different notion and (strictly) more powerful than a many-one reduction (and is not what is used when establishing NP-completeness).

Of course, the line would be allowed if you managed to prove HAMPATH is solvable in polynomial time. I don't think you will be able to.