I enjoyed this problem. It’s fun to play with and gives you a little
room to
get creative with solving techniques.
As many people pointed out, this is pretty much the Knight’s Tour
problem with
some unusual Knight jumps. Most solutions will solve either problem, if
you
change their idea of neighbor squares. The good news about that is that
we can
take advantage of the shortcuts people use to solve that problem.
The most straightforward solution to this problem is I can dream up is:
1. Pick a random starting square
2. Make a list of all the possible moves from the current square
3. If there are no moves, undo the last move and try the next choice
from that square
4. Otherwise, make the first move in the list
5. Goto step 2 until the grid is full
That’s a boring brute force approach, which is the computer science term
for
“try everything until something works.” When it gets stuck somewhere,
it
backtracks and tries another jump. It may, at times, need to backtrack
several
steps to get back to a place where it could try another move.
This approach has one major plus and one major minus. First, the good
news: it
will find a solution. The bad news: eventually. Because it has to
check every
possibility, it can take a good long while to find a path that works.
The
bigger the board gets, the more places it has to check. The waits get
longer
and longer.
Now, when solving the Knight’s Tour there is a common shortcut that
often leads
to a solution much faster. Luckily it works here too. The idea is
that, you
should visit the squares with the least choices first. Those are the
places you
are likely to run out of options, so getting them out of the way while
you still
have plenty of open squares increases your chances of making it around
the
board. This is called the JC Warnsdorff heuristic, because JC
Warnsdorff made
the suggestion all the way back in 1823.
The downside of the JC Warnsdorff heuristic is that is doesn’t always
work.
Depending on your starting position and which squares you visit first,
it can
paint itself into a corner. The problem is more common on certain board
sizes,
but it does happen. The upside is, it’s so darn quick (because it
doesn’t
backtrack), you can do multiple searches and still be quicker than the
brute
force approach. If one attempt fails, just try again. Most solutions
used this
approach.
I found one more corner to cut. When the quiz mentioned circular
solutions, it
made me realize you could cheat a bit, if you had one. If you can go
from the
end of the line back to the beginning (the definition of a circular
solution),
you can start anywhere on that path and follow it for the entire length
to get
all the numbers out. In truth, these are all the same solution (in my
opinion),
but the numbers move around so they look different. It’s also lightning
quick
to shift all the numbers by some offset. Sadly it can be pretty slow to
find a
circular solution in the first place. It’s a trade off.
Another technique, subdividing the grid and piecing together multiple
smaller
solutions, was used be Elliot T… See the later postings in the
quiz thread
for a good discussion of that approach.
OK, let’s get to the code already!
I’m going to so show my solution, just because it takes both shortcuts
and I’m
very familiar with how it works. I do not think my solution came out
the
cleanest though, so definitely go through the others to find some pretty
code.
Here’s the beginning of my solution:
#!/usr/bin/env ruby -w
require "enumerator"
class PenAndPaperGame
def self.circular_solutions
@circular ||= if File.exist?("circular_solutions.dump")
File.open("circular_solutions.dump") { |file| Marshal.load(file)
}
else
Array.new
end
end
def initialize(size)
@size = size
@largest = @size * @size
@grid = Array.new(@largest)
end
# ...
I pull in enumerator here, because I’m addicted. Can’t wait until that
library
is in the core.
The class method here is my persistent memory for circular solutions.
It reads
the data file, if it exists, and creates an Array of circular solutions
at
various sizes. If we don’t have the file, a new Array is used. The ||=
caching
operator is used for the assignment so we only look the value up once.
The constructor is trivial and almost every solution had one just like
it. We
record the size, figure out what the largest number needs to be, and
build the
grid Array.
The Array was a dumb choice on my part. It somehow made me feel more
manly to
use a one dimensional Array and deal with the two dimensional indexing.
However, looking at David T.s super clear 45 line solution that uses
normal
nested Arrays just made me feel dumb.
Here’s the cheat solver I described earlier:
# ...
def solve
if self.class.circular_solutions[@size].nil?
solve_manually
else
@grid = self.class.circular_solutions[@size]
offset = @grid[rand(@grid.size)]
@grid.map! { |n| (n + offset) % @largest + 1 }
to_s
end
end
# ...
If we haven’t yet found a circular solution for this size, a handoff is
made to
solve_manually(). If we do have one, the else clause is a full
(cheating)
solution. We set the grid to the complete solution, choose a random
value from
the grid (similar to picking a starting square), adjust all the values
in the
grid by the chosen starting point, and print the results.
Now that we’ve looked at my super cheat, let’s get to a real solution:
# ...
def solve_manually
x, y = rand(@size), rand(@size)
count = mark(x, y)
loop do
to = jumps(x, y)
return self.class.new(@size).solve_manually if to.empty?
scores = rate_jumps(to)
low = scores.min
next_jump = to.enum_for(:each_with_index).select do |jump|
scores[jump.last] == low
end.sort_by { rand }.first.first
count = mark(*(next_jump + [count]))
x, y = next_jump
if count > @largest
if circular?
self.class.circular_solutions[@size] = @grid
File.open("circular_solutions.dump", "w") do |file|
Marshal.dump(self.class.circular_solutions, file)
end
return to_s
else
puts "Found this solution:"
puts to_s
puts "Continuing search for a circular solution..."
return self.class.new(@size).solve_manually
end
end
end
end
# ...
This method looks big, but hopefully it’s fairly high level and thus
easy enough
to read. First, we select a random starting x and y and mark() that
square.
Then we dive into the main solution loop.
In the loop, we pull all possible jumps() from this square and make sure
we have
at least one choice. If we don’t, we got stuck and can’t find a
solution so we
just build a new solver, trigger the search for another attempt, and
return the
results of that.
If we did get some jumps, we need to score them, based on how many moves
we
would have from there. The call to rate_jumps() does this. Then we
pluck out
the low score as our target move. The complicated ball of iterators
right after
that is just a lazy way to select a random move from all the choices
matching
the lowest score, which gets slotted into next_jump.
With a jump selected we mark the new square and move.
Before we loop(), we check to see if that was the final mark. When it
is, we
have a solution, but we want to check if it’s a circular solution we
could reuse
for cheating. If it is circular, we add it to the collection and update
our
storage file. Then we show the solution. When it’s not circular, we go
ahead
and show it, but trigger another search to see if we can hunt down a
circular
solution.
Here’s the printing code:
# ...
def to_s
width = @largest.to_s.size
border = " -" + (["-" * width] * @size).join("-") + "- \n"
border +
@grid.enum_for(:each_slice, @size).inject(String.new) do |grid, row|
grid + "| " + row.map { |n| n.to_s.center(width) }.join(" ") + "
|\n"
end +
border
end
# ...
Nothing too tricky here. We find a cell size and build a border. Then
we print
a border, each row, and another border. The complicated row iteration
is just
another sign that I used the wrong Array.
Here are the missing helper methods:
# ...
private
def at(x, y)
x + y * @size
end
def mark(current_x, current_y, mark = 1)
@grid[at(current_x, current_y)] = mark
mark + 1
end
def jumps(from_x, from_y, grid = @grid)
[ [-3, 0],
[3, 0],
[0, -3],
[0, 3],
[2, 2],
[-2, 2],
[2, -2],
[-2, -2] ].map do |jump|
[from_x + jump.first, from_y + jump.last]
end.select do |jump|
jump.all? { |to| (0...@size).include? to } and
grid[at(*jump)].nil?
end
end
def rate_jumps(choices)
choices.map { |jump| jumps(*jump).size }
end
def circular?
grid = @grid.dup
grid[grid.index(@largest)] = nil
x, y = grid.index(1).divmod(@size).reverse
not jumps(x, y, grid).empty?
end
end
# ...
The at() method is used to turn x and y coordinates into an index for
the one
dimensional grid I am using. mark() will place a number in the
indicated square
and return the next mark that should be placed. (This was another odd
choice.
I have no idea why I didn’t use an instance variable here.)
The jumps() method uses offset to locate all possible moves from the
current
location. It then filters those by removing any out-of-bound indices
and any
squares that aren’t nil. rate_jumps() is just a thin shell over jumps
to count
the moves available at each step.
The circular?() check duplicates the solved grid, knocks the last move
back to
nil, locates the starting square, and checks to see if the starting
square now
has one jump (to the only nil square).
Using those pieces, here’s the application code:
# ...
if __FILE__ == $PROGRAM_NAME
size = ARGV.first && ARGV.first =~ /\A-s(?:ize)?\Z/ ? ARGV.last.to_i
: 5
puts PenAndPaperGame.new(size).solve
end
That just reads the size parameter or sets the default to 5 and kicks
off the
solver process.
A big thank you to all those who were so creative with their solutions
this time
around. You guys taught me how to finish off my own solution.
Tomorrow we will play around with interactively defining Ruby methods…