# Solving Tatham's Puzzles - Signpost (backtracking)

I've started writing solvers for games of *Simon Tatham's Portable Puzzle Collection*.
Those Problems can usually be reduced to some NP-complete problem. This is quite
a fun exercise.

We will be continuing with the puzzle "Signpost", which is a variant of Hamiltonian paths in a directed graph where, for some fields or nodes, the position within the path is given.

The goal of the puzzle is to connect the Node "1" to some node, which is then
node "2" in the direction the arrow points to. All numbers need to be connected
to their successor **until all nodes have a single unique number** from "1" to (in
this example) "25".

We will be solving this problem in two different ways, both with the help
of the programming language Python. But first of all we'll
need to convert a puzzle into a machine readable format. Luckily, Signpost
has an easy and accessible format. The problem instance from the image is the instance
`5x5:1cceefcfggeeccghcac3e12hch10ah25a`. It starts off with the **size of the
problem** and then a list of **letters that show the directions of the arrow** where
`a` is north or 12 o'clock and all other letters go clockwise with `h` being
north-west. A letter can be prefixed by a **number which is the hint** for that node.
That node's number is fixed. The cells or nodes are **defined line by line from the
top left to the bottom right**.

Problems don't need to start with 1 or end with the last number. The start and the end of the path can be anywhere but puzzles generated with the ends at opposite corner look nicer.

First we define what our data looks like. We implement a simple form of **directed graph**
that we define as a **set of nodes and their successors**. From that we derive following
definition:

```
class Node(object):
hint = None
following = frozenset()
def __init__(self, direction):
self.direction = direction
```

We don't want to use any of that code in a library or something something so we don't do anything fancy here and we use the simplest class definition that comes to mind.

Now we go on to parsing the problem. The parsing function should get the problem
as a string and return a **3-tuple containing width and height** of the problem
as well as a **mapping of positions within the grid to ``Node`` instances**.

Onto actually parsing. We can split at the colon and then regular expressions to the rescue.
The size definition has the form `^(?P<size_x>\d+)x(?P<size_y>\d+)$` while
a single node definition has the form `(?P<defn>\d*[a-h])`. As we can see,
the digits are optional but we need to greedily accept them. **All non-overlapping
occurences in order** will be our cell definitions.

In code we, again, do nothing too too fancy:

```
def parse(puzzle_str):
directions = {
'a': (0,-1), 'b': (1,-1), 'c': (1,0), 'd': (1,1),
'e': (0,1), 'f': (-1,1), 'g': (-1,0),'h': (-1,-1)
}
size, _, definition = puzzle_str.partition(':')
r_size = re.compile(r"^(?P<size_x>\d+)x(?P<size_y>\d+)$")
r_defn = re.compile(r"(?P<defn>\d*[a-h])")
size_t = tuple(map(int, r_size.match(size).groups()))
w, h = size_t
nodes = {}
for n, m in enumerate(r_defn.finditer(definition)):
pos = (n % w, n // w)
nodes[pos] = Node(directions[m.group(0)[-1:]])
hint = m.group(0)[:-1]
if hint:
nodes[pos].hint = int(hint)
return w, h, nodes
```

And we also need to fill in the **successor Nodes**. This is no problem at all.

```
w, h, nodes = parse(sys.argv[1])
from itertools import product
for x, y in product(range(w), range(h)):
this_node = nodes[(x,y)]
dx, dy = this_node.direction
x, y = x + dx, y + dy
while x >= 0 and x < w and y >= 0 and y < h:
this_node.following = this_node.following | frozenset([nodes[x,y]])
x, y = x + dx, y + dy
```

And now on to actually solving this thing. The idea is, that we slowly **build
a list of Nodes that is our solution**. The first element is the Node with the
hint "1" and we iterate through all nodes that follow that Node and do not have a hint
that is other than "2", and so on. And of course, the Node that we add to the
solution is not allowed to have **already been part of the solution** and if there
is a node with that hint, it should be this exact one. **If we find
such a conflict** in our solution, we **reject that solution** and return to an
earlier partial solution.

Instead of using recursion, we build **a queue of partial solutions** until we get
a complete solution. This is functionally identical to restarting a solver function
with partial solutions of increasing size. In that case, the call stack manages
the backtracking. But we'll do it with a queue this time. We build the queue so
that every element in the list is either

- a partial solution (beginning from 1) without conflicts
- a partial solution (beginning from 1) where the last element is in some conflict
- a complete solution

For **case 1**, we remove the partial solution that has the length n and **add all
partial solutions** of length n+1 by iterating through all successor Nodes that
have not yet been part of the partial solution. For
**case 2**, we **reject** that partial solution. For **case 3**, we immediately return that
solution and are **finished**.

```
def solve_dhc(nodes):
# get the Node that has hint 1
first = next(iter(n for n in nodes.values() if n.hint == 1))
# get the set of all hints, so it is easier to check for conflicts
hints = frozenset(n.hint for n in nodes.values() if n.hint is not None)
from collections import deque
q = deque([[first]]) # initializing queue
while q:
curr = q.pop()
idx = len(curr)
if (idx in hints or curr[-1].hint is not None) and curr[-1].hint != idx:
continue # case 2
if idx == len(nodes):
return curr # case 3
for _next in curr[-1].following: # case 1
if _next not in curr:
q.append(curr + [_next])
```

**This algorithm terminates** because we always remove one element from the queue of
some length n and possibly add a large but finite amount of elements of length
n+1 and the algorithm terminates when we have an element in the queue of some
possibly large but finite length.

Sidenote: if you use `popleft` instead of `pop` you get a breadth-first search
instead the depth-first search, that is implemented here. Add to the right
and pop from the right for depth-first and add to the right and pop from the left
for breadth-first. Since every proper solution has the exact same length/depth
(that we know and never go further anyways),
**searching depth-first is strictly better than breadth-first**.

And given a solution, we still have to print it. That's the easy part.

```
dhc = solve_dhc(nodes)
fill = len(str(w*h))
for l in range(h):
for c in range(w):
node = nodes[(c,l)]
print(str(dhc.index(node) + 1).rjust(fill), end=" ")
print("")
```

And there's the **solution** for our puzzle.

```
$ python3 signpost_dhc.py 5x5:1cceefcfggeeccghcac3e12hch10ah25a
1 20 9 2 21
23 14 13 22 24
15 5 7 6 8
18 19 11 3 12
16 17 10 4 25
```

There is code linked to this article. Get it here!

**This post is part of the series "Tatham's Puzzles":**

- 31.01.2016 – Solving Tatham's Puzzles - Pattern
- 21.03.2016 – Solving Tatham's Puzzles - Signpost (backtracking)

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