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Recursion Revisited

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This post is partly resulted in by a comment[0] on Hacker Data[1] which mentioned:

  • The three examples susceptible to show conceal recursion, towers of Hanoi,
    fibonacci, and binary search are no longer potentially the most convincing ones.
    All could moreover be solved with iteration.

Others pointed out that (in theory) any recursive algorithm can
be expressed as an iterative one, and vice versa, but I answered[2]
pronouncing:

  • Tutorial examples are no longer incessantly ever “staunch world”, and one thing
    “staunch world” is incessantly too complex to kind the underlying
    solutions sure … I will write a follow-up post to
    this with a good staunch-world (properly, Pure Maths calculation)
    example … maybe it’s essential to well gain that extra gratifying.

So right here’s that post.

A whereas ago I wrote a post on Pondering About Recursion, and further
no longer too long ago I wrote a pair of scenario referring to the scenario of the
option of Vertices Required For Cycles. As it happens, the code for
that scenario is closely recursive, and offers a huge example
of utilizing recursion to resolve a staunch scenario.

So I believed I’d show conceal you.

The cycle scenario

The scenario in ask is this:

Given a obvious integer $C$, what’s the minimum option of
vertices wished for a graph to possess exactly $C$ cycles?

The characteristic $C{rightarrow}N$ is now within the OEIS:

However how are values computed?

Counting cycles

On this case we’re going to follow the long-established definition from
Graph Theory, wherein we permit neither loops nor multiple
edges between any pair of vertices. This particular scenario
provides an correct example of why the so-known as “Easy Graph” is
the long-established convention: if we allowed loops then we could well maintain
any option of cycles we wished factual with a single vertex.

The general approach is to secret agent at all graphs on $N$ vertices and
for every body, count how many cycles there are. Since we’re going to be
hunting for the minimum option of vertices for a given option of
cycles we’re going to ignore graphs that are no longer linked, and we’re going to
ignore graphs that maintain a vertex of level 1. So we secret agent at all
the non-isomorphic linked graphs of minimum level 2.

However at its coronary heart, for a given graph we maintain now to count how many
cycles it has. Right here is the algorithm.

Basically the particular illustration of the graph must no longer
be a scenario, and I’ve vulnerable this particular illustration on fable of
then the code reads moderately naturally. Effectively, to me, anyway.
I could abstract it away extra, but this code feels moderately
direct.

We define routine “count_cycles” which takes a graph within the kill
of a dictionary mimicking a situation of edges. If {u,v} is an edge in
the graph, then both graph_edges[(u,v)] or graph_edges[(v,u)]
is situation to 1. There are extra efficient records structures for this
algorithm, but for the capabilities of describing it we will follow
this one.

Routine “count_cycles” is a recursive routine.

If there usually are no longer any edges then the reply is 0 … there usually are no longer any cycles.

 
def count_cycles( graph_edges ):

    if no longer graph_edges:
        return 0

Otherwise we’re going to earn an edge. In theory we could gain some
artful heuristic for picking the threshold, but in practice we simply
hold the first edge from the checklist. Even so, we name out to a
routine to kind the option:

 
    selected_edge = choose_edge( graph_edges )

Every cycle both does encompass this edge, or does no longer encompass
this edge. So we employ away it from the graph:

 
    del graph_edges[ selected_edge ]

Then count the cycles within the graph without that edge. To affect that we
simply name this routine – right here’s the first instance of recursion.

 
    cycles_without = count_cycles( graph_edges )

https://www.solipsys.co.uk/images/CyclesIncludingThisEdge.png



Cycles with this edge

So now we maintain now to count how many cycles there are at the side of this
chosen edge. We secret agent at this map right here on the appropriate, and our
chosen edge is the one in red joining the vertices $u$ and $v$,
proven in blue.

Any cycle that contains this edge $(u,v)$ will then encompass the
edge, and a route from $u$ to $v$. So having earlier eradicated the
edge $(u,v)$ we’re going to now count the option of cycles by counting the
paths from $u$ to $v$ on this diminished graph.

 
    u, v = vertices_of_edge( selected_edge )
    cycles_with = count_paths( graph_edges, u, v )

We then place the threshold aid, and return the whole:

 
    graph_edges[ selected_edge ] = 1

    return cycles_without + cycles_with

That’s how we count the cycles, and all that is left is easy ideas to count
the paths from $u$ to $v$.

Counting paths

So now we maintain now to put in writing the code to count the paths from $u$ to $v$:

 
def count_paths( graph_edges, u, v ):

If $u$ and $v$ are the the same vertex then we maintain now one (degenerate)
route:


https://www.solipsys.co.uk/images/PathsFromU2V.png



Paths from $u$ to $v$

Otherwise our route must employ one amongst the red edges proven right here on
the map. Extra, since paths (and cycles) can no longer repeat a vertex,
once one amongst these red edges is vulnerable, the a sort of ones can no longer be vulnerable
on fable of otherwise we could be returning to vertex $u$. Which is
forbidden.

So we’re going to employ away the red edges, and for every of the red vertices
we then count the option of paths from that to $v$.

So let’s get our checklist of red edges, and employ away them:

 
    purple_edges = [ e for e in graph_edges
                         if e[0]==u or e[1]==u
                     ]
    for edge in purple_edges:
        del graph_edges[ edge ]

Then we get our checklist of red vertices.

 
    red_vertices = [ e[1] for e in purple_edges
                              if e[0]==u ]
                 + [ e[0] for e in purple_edges
                              if e[1]==u ]

The whole option of paths from $u$ to $v$ is the whole option of
paths from the red vertices to $v$. So we employ that whole by a
recursive name to this same routine:

 
    whole = 0
    for red_vertex in red_vertices:
        whole += count_paths( graph_edges,
                              red_vertex, v
                              )

At reliable, we place all these edges aid, and return our consequence:

 
    for edge in purple_edges:
        graph_edges[ edge ] = 1
    return whole

 
def count_paths( graph_edges, u, v ):

    if u == v: return 1
    purple_edges = [ e for e in graph_edges
                        if e[0]==u or e[1]==u
                    ]
    for edge in purple_edges:
        del graph_edges[ edge ]
    red_vertices =
        [ e[1] for e in purple_edges if e[0]==u ] +
        [ e[0] for e in purple_edges if e[1]==u ]
    whole = 0
    for red_vertex in red_vertices:
        whole += count_paths( graph_edges,
                              red_vertex, v )
    for edge in purple_edges:
        graph_edges[ edge ] = 1
    return whole

def count_cycles( graph_edges ):

    if no longer graph_edges:
        return 0

    selected_edge = choose_edge( graph_edges )

    del graph_edges[ selected_edge ]
    cycles_without = count_cycles( graph_edges )
    u, v = vertices_of_edge( selected_edge )
    cycles_with = count_paths( graph_edges, u, v )
    graph_edges[ selected_edge ] = 1

    return cycles_without + cycles_with

Abstract

So right here is the code whole. The routine “count_cycles” calls
itself recursively, and moreover calls “count_paths”. In its flip
the routine “count_paths” calls itself recursively, and we’re
finished.

With any recursive routine we can maintain to aloof incessantly query: How will each person is conscious of
the recursion bottoms out?

The reply on this case is that in every name the option of
edges decreases, is a whole quantity, and could no longer streak under zero.
Thanks to that, in every case the option tree has to cease.

And the code is no longer that giant, nor that advanced. Right here is the
energy of recursion when appropriately utilized.

Dwell Notes

Right here is no longer intended to be especially efficient code, nor is it
in particular correct Python code. That’s no longer the diagram (on this
case). The reliable version became extra efficient, but much less sure due
to the adjustments made for that efficiency.

Even so, I am hoping the algorithm, and the procedure the recursion works, is
sure.

References and links

[0] The comment: https://news.ycombinator.com/merchandise?id=24744814

[1] Hacker Data: https://news.ycombinator.com/news

[2] My reply: https://news.ycombinator.com/merchandise?id=24744962

My attributable to Neil Walker for some valuable solutions and solutions.




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