Code is taking too much time

Question

I wrote code to arrange numbers after taking user input. The ordering requires that the sum of adjacent numbers is prime. Up until 10 as an input code is working fine. If I

Answer 1

For posterity ;-), here's one more based on finding a Hamiltonian path. It's Python3 code. As written, it stops upon finding the first path, but can easily be changed to generate all paths. On my box, it finds a solution for all n in 1 through 900 inclusive in about one minute total. For n somewhat larger than 900, it exceeds the maximum recursion depth.

The prime generator (psieve()) is vast overkill for this particular problem, but I had it handy and didn't feel like writing another ;-)

The path finder (ham()) is a recursive backtracking search, using what's often (but not always) a very effective ordering heuristic: of all the vertices adjacent to the last vertex in the path so far, look first at those with the fewest remaining exits. For example, this is "the usual" heuristic applied to solving Knights Tour problems. In that context, it often finds a tour with no backtracking needed at all. Your problem appears to be a little tougher than that.

def psieve():
    import itertools
    yield from (2, 3, 5, 7)
    D = {}
    ps = psieve()
    next(ps)
    p = next(ps)
    assert p == 3
    psq = p*p
    for i in itertools.count(9, 2):
        if i in D:      # composite
            step = D.pop(i)
        elif i < psq:   # prime
            yield i
            continue
        else:           # composite, = p*p
            assert i == psq
            step = 2*p
            p = next(ps)
            psq = p*p
        i += step
        while i in D:
            i += step
        D[i] = step

def build_graph(n):
    primes = set()
    for p in psieve():
        if p > 2*n:
            break
        else:
            primes.add(p)

    np1 = n+1
    adj = [set() for i in range(np1)]
    for i in range(1, np1):
        for j in range(i+1, np1):
            if i+j in primes:
                adj[i].add(j)
                adj[j].add(i)
    return set(range(1, np1)), adj

def ham(nodes, adj):
    class EarlyExit(Exception):
        pass

    def inner(index):
        if index == n:
            raise EarlyExit
        avail = adj[result[index-1]] if index else nodes
        for i in sorted(avail, key=lambda j: len(adj[j])):
            # Remove vertex i from the graph.  If this isolates
            # more than 1 vertex, no path is possible.
            result[index] = i
            nodes.remove(i)
            nisolated = 0
            for j in adj[i]:
                adj[j].remove(i)
                if not adj[j]:
                    nisolated += 1
                    if nisolated > 1:
                        break
            if nisolated < 2:
                inner(index + 1)
            nodes.add(i)
            for j in adj[i]:
                adj[j].add(i)

    n = len(nodes)
    result = [None] * n
    try:
        inner(0)
    except EarlyExit:
        return result

def solve(n):
    nodes, adj = build_graph(n)
    return ham(nodes, adj)