# Proving that Hamiltonian Cycle is reducible to a travelling problem?

I chanced upon the following question online:

A company has two trucks, and must deliver a number of parcels to a number of addresses. They want both drivers to be home at the end of the day. This gives the following decision problem.

Instance: Set V of locations, with for each pair of locations
v, w ∈ V , a distance d(v, w) ∈ N, a starting location s ∈ V ,
and an integer K.


Question: Are there two cycles, that both start in s, such that every location in V is on at least one of the two cycles, and both cycles have length at most K?

The solution provided uses Hamiltonian Cycle to prove the parcel problem is NP-Complete.

Given any instance of Hamiltonian Cycle with n vertices, construct the following special instance of the above problem.

• Copy the graph and make the distance d(v, w) equal to 2 if there is an edge in the graph, and 4, otherwise.
• Label one of these vertices as s. Add one vertex 0 such that d(s, 0) = d(0, s) = n and d(v, 0) = d(0, v) = 2n + 1 for all v != s. The threshold K is equal to 2n.

The explanation provided is as follows:

The only thing we have to do is to keep the other driver busy, which is easily taken care of by including an additional vertex with distance K/2 from s and distance K + 1 to all other addresses.

I tried constructing a simple graph of N = 3 whereby each vertex is connected by an edge of distance 2. Then I introduced a new node which is distance 4 away from the source vertex(i.e. starting point), and distance 7 from every other vertex.

But I am unable to understand how this construction fits into the original decisional problem. What exactly does the solution mean by "keeping the other driver busy?"