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This put up is partly brought on by a reveal on Hacker Info which acknowledged:

• The three examples used to designate recursion, towers of Hanoi,
fibonacci, and binary search are not essentially the most convincing ones.
All will be solved with iteration.

Others pointed out that (in thought) any recursive algorithm can
be expressed as an iterative one, and vice versa, however I spoke back
announcing:

• Academic examples are now and again ever “real world”, and the relaxation
“real world” is typically too advanced to map the underlying
principles obvious … I could write a follow-up put up to
this with a pleasant real-world (smartly, Pure Maths calculation)
instance … presumably that you just must well per chance derive that extra enjoyable.

So here’s that put up.

Some time ago I wrote a put up on Thinking About Recursion, and additional
not too long ago I wrote a couple of deliver touching on the deliver of the
sequence of Vertices Required For Cycles. As it happens, the code for
that deliver is heavily recursive, and affords a large instance
of the yelp of recursion to resolve a real deliver.

So I believed I’d demonstrate you.

## The cycle deliver

The deliver in quiz is this:

 Given a run integer \$C\$, what’s the minimum sequence of vertices wished for a graph to grasp exactly \$C\$ cycles?

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

However how are values computed?

## Counting cycles

On this case we can follow the frequent definition from
Graph Theory, thru which we enable neither loops nor plenty of
edges between any pair of vertices. This explicit deliver
affords an very most involving instance of why the so-known as “Easy Graph” is
the frequent convention: if we allowed loops then we could well per chance presumably even possess
any sequence of cycles we wanted honest with a single vertex.

The final technique is to search at all graphs on \$N\$ vertices and
for every, depend what number of cycles there are. Since we will be
procuring for the minimum sequence of vertices for a given sequence of
cycles we can ignore graphs which can per chance per chance be not associated, and we can
ignore graphs which possess a vertex of stage 1. So we search at all
the non-isomorphic associated graphs of minimum stage 2.

However at its heart, for a given graph we would prefer to depend what number of
cycles it has. Right here is the algorithm.

In actuality the explicit illustration of the graph have to not
be a deliver, and I’ve used this explicit illustration because
then the code reads moderately naturally. Well, to me, anyway.
I could well per chance presumably also abstract it away additional, however this code feels moderately
recount.

We account for routine “count_cycles” which takes a graph in the secure
of a dictionary mimicking a arena of edges. If {u,v} is an edge in
the graph, then both graph_edges[(u,v)] or graph_edges[(v,u)]
is arena to 1. There are extra atmosphere qualified info structures for this
algorithm, however for the functions of describing it we will follow
this one.

Routine “count_cycles” is a recursive routine.

If there are no edges then the answer is 0 … there are no cycles.

```
def count_cycles( graph_edges ):

if not graph_edges:
return 0
```

In every other case we can catch an edge. In thought we could well per chance presumably also derive some
suave heuristic for selecting the brink, however in practice we simply
obtain the principle edge from the checklist. Even so, we call out to a
routine to map the selection:

```
selected_edge = choose_edge( graph_edges )
```

Every cycle both does embody this edge, or does not embody
this edge. So we obtain away it from the graph:

```
del graph_edges[ selected_edge ]
```

Then depend the cycles in the graph with out that edge. To map that we
simply call this routine – here’s the principle instance of recursion.

```
cycles_without = count_cycles( graph_edges )
```

Cycles with this edge

So now now we possess to depend what number of cycles there are in conjunction with this
chosen edge. We search at this draw here on the factual, and our
chosen edge is the one in purple becoming a member of the vertices \$u\$ and \$v\$,
shown in blue.

Any cycle that comprises this edge \$(u,v)\$ will then encompass the
edge, and a path from \$u\$ to \$v\$. So having earlier eliminated the
edge \$(u,v)\$ we can now depend the sequence of cycles by counting the
paths from \$u\$ to \$v\$ on this reduced graph.

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

We then put the brink assist, and return the total:

```
graph_edges[ selected_edge ] = 1

return cycles_without + cycles_with
```

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

## Counting paths

So now now we possess to write the code to depend the paths from \$u\$ to \$v\$:

```
def count_paths( graph_edges, u, v ):
```

If \$u\$ and \$v\$ are the identical vertex then now we possess one (degenerate)
path:

Paths from \$u\$ to \$v\$

In every other case our path must yelp likely the most purple edges shown here on
the draw. Further, since paths (and cycles) can’t repeat a vertex,
once one of those purple edges is used, the opposite ones can’t be used
because in every other case we would be returning to vertex \$u\$. Which is
forbidden.

So we can obtain away the purple edges, and for every of the purple vertices
we then depend the sequence of paths from that to \$v\$.

So let’s obtain our checklist of purple edges, and obtain away them:

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

Then we obtain our checklist of purple vertices.

```
red_vertices = [ e for e in purple_edges
if e==u ]
+ [ e for e in purple_edges
if e==u ]
```

The total sequence of paths from \$u\$ to \$v\$ is the total sequence of
paths from the purple vertices to \$v\$. So we obtain that total by a
recursive call to this identical routine:

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

Within the discontinuance, we put all those edges assist, and return our result:

```
for edge in purple_edges:
graph_edges[ edge ] = 1
```

```
def count_paths( graph_edges, u, v ):

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

def count_cycles( graph_edges ):

if not 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 here is the code total. The routine “count_cycles” calls
itself recursively, and additionally calls “count_paths”. In its turn
the routine “count_paths” calls itself recursively, and we’re
finished.

With any recursive routine we must consistently search info from: How will we all know
the recursion bottoms out?

The answer on this case is that in every call the sequence of
edges decreases, is a total number, and can not lope beneath zero.
On myth of that, in every case the resolution tree has to discontinuance.

And the code will not be undoubtedly that massive, nor that advanced. Right here’s the
strength of recursion when accurately utilized.

## Discontinuance Notes

Right here’s not intended to be especially atmosphere qualified code, neither is it
particularly magnificent Python code. That’s not the explanation (on this
case). The supreme version changed into once extra atmosphere qualified, however less obvious due
to the modifications made for that efficiency.

Even so, I hope the algorithm, and the intention the recursion works, is
obvious.