[QUIZ] Knight's Travails (#27)

[Ruby Quiz]

The three rules of Ruby Quiz:

1. Please do not post any solutions or spoiler discussion for this quiz until
48 hours have passed from the time on this message.

2. Support Ruby Quiz by submitting ideas as often as you can:

http://www.rubyquiz.com/

3. Enjoy!

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

by Jason Bailey

Given a standard 8 x 8 chessboard where each position is indicated in algebraic
notation (with the lower left corner being a1), design a script that accepts two
or more arguments.

The first argument indicates the starting position of the knight. The second
argument indicates the ending position of the knight. Any additional arguments
indicate positions that are forbidden to the knight.

Return an array indicating the shortest path that the knight must travel to get
to the end position without landing on one of the forbidden squares. If there is
no valid path to the destination return nil.

example 1:
a8, b7, b6

could return
[ c7 , b5 , d6 , b7 ]

Example 2:
a8 , g6 , b6 , c7

would return
nil

Note: That in the majority of cases it would be possible to have more then one
valid path.

 

[Phil Tomson]

The three rules of Ruby Quiz:

1. Please do not post any solutions or spoiler discussion for this quiz until
48 hours have passed from the time on this message.

2. Support Ruby Quiz by submitting ideas as often as you can:

http://www.rubyquiz.com/

3. Enjoy!

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

by Jason Bailey

Given a standard 8 x 8 chessboard where each position is indicated in algebraic
notation (with the lower left corner being a1), design a script that
accepts two
or more arguments.

The first argument indicates the starting position of the knight. The second
argument indicates the ending position of the knight. Any additional arguments
indicate positions that are forbidden to the knight.

Sounds fun... <sigh>I just wish I had the time...</sigh>

Phil

 

[Carlos]

Return an array indicating the shortest path that the knight must travel to get
to the end position without landing on one of the forbidden squares. If there is
no valid path to the destination return nil.

Here is a solution. I'm having a hard time trying to explain it in English,
I hope the code is clear enough (it's very simple).

Thanks.

# code:

class String
def to_coords
return [self[0] - ?a, self[1] - ?1]
end
end

class Array
def to_algebraic
return (self[0] + ?a).chr + (self[1] + ?1).chr
end
end


def where_can_jump_from (here, visited)
col, row = here
[
[col+2, row+1], [col+2, row-1], [col+1, row-2], [col-1, row-2],
[col-2, row-1], [col-2, row+1], [col-1, row+2], [col+1, row+2]
].select { |c,r|
r >= 0 && r < 8 && c >= 0 && c < 8 && !visited[c][r]
}
end


def knight_path (start_pos, finish_pos, forbidden)
visited = Array.new(8) { Array.new(8) }
forbidden.each do |col,row| visited[col][row] = true end

# special cases:
# shortest path: no movement at all
return [] if start_pos == finish_pos
# impossible task:
return nil if forbidden.include? finish_pos

# setup...
paths = [[start_pos]]
visited[start_pos[0]][start_pos[1]] = true

while !paths.empty?
# pp paths.map {|p| p.map {|c| c.to_algebraic } }
new_paths = []
paths.each do |path|
where_next = where_can_jump_from(path.last, visited)
where_next.each do |coord|
newpath = path.dup << coord
if coord == finish_pos
# clear first cell (start position)
newpath.shift
return newpath
end
c, r = coord
visited[c][r] = true
new_paths.push newpath
end
end
paths = new_paths
end

return nil
end

start_pos = ARGV.shift.to_coords
finish_pos = ARGV.shift.to_coords
forbidden = ARGV.map {|arg| arg.to_coords }

result = knight_path start_pos, finish_pos, forbidden

if (result)
result.map! { |coord| coord.to_algebraic }
puts "[ " + result.join(" , ") + " ]"
else
p nil
end

 

[Timothy Byrd]

Here's mine - still very C-like...

module Knight

Moves = [
[-2,-1], [-2, 1], [2,-1], [2, 1],
[-1,-2], [-1, 2], [1,-2], [1, 2]
]

def self.possible_moves(pos)
result = []
mv = 'a1'

Moves.each {|delta|
(0..1).each { |i| mv = pos + delta }
if (?a..?h).include?(mv[0]) && (?1..?8).include?(mv[1])
result.push(mv.clone)
end
}
result
end

def self.find_path_recurse(pStart, pEnd, forbidden, max_moves = 64)

# Are we there yet?
#
return [pEnd.clone] if pStart == pEnd

# You can't get there from here!
#
return nil if forbidden.include?(pEnd) || max_moves <= 0

moves = possible_moves(pStart)
moves.delete_if {|x| forbidden.include?(x)}

return [pEnd.clone] if moves.include?(pEnd)

best_solution = nil
moves_left = max_moves - 1
cant_go = forbidden.clone.concat(moves)

moves.each do |m|
s = find_path_recurse(m, pEnd, cant_go, moves_left)
next if !s

s.insert(0, m)
if !best_solution || s.size < best_solution.size
# From now on only interested in solutions that
# are at least two moves shorter
#
moves_left = s.size - 2
best_solution = s
end
end

best_solution
end


def self.find_path(pStart, pEnd, forbidden)
forbidden = [] if !forbidden
if forbidden.empty?
puts "From #{pStart} to #{pEnd}"
else
puts "From #{pStart} to #{pEnd} excluding
[#{forbidden.join(', ')}]"
end
s = find_path_recurse(pStart, pEnd, forbidden, 64)
if s
puts s.join(', ')
else
puts nil
end
puts ''
end
end

if ARGV[1]
Knight.find_path(ARGV[0], ARGV[1], ARGV[2..-1])
else
Knight.find_path('a8', 'b7', ['b6'])
Knight.find_path('a8', 'g6', ['b6', 'c7'])
end

 

[Dominik Bathon]

Here is my solution:
It uses a ChessPos class to validate the given positions and to simplify
the moves.
The find_path method uses Dijkstra's Shortest Path Algorithm in a
simplified form.

Btw. does anybody know if this behavior is defined somewhere (adding
elements to an Array while iterating over it):
irb(main):001:0> (a=[0]).each { |i| a << i+1 if i < 10 }; a
=> [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]


The Code:

class ChessPos
attr_reader os

def initialize(str)
unless str.size==2 && (?a..?h).include?(str[0]) &&
(?1..?8).include?(str[1])
raise "#{str} is not a valid chess position"
end
@pos=str
end

def move(x, y)
ChessPos.new((pos[0]+x).chr+(pos[1]+y).chr)
end

def hash; pos.hash; end
def eql?(other); pos.eql?(other.pos) rescue false; end
alias :== :eql?
end

def all_knight_moves_from(pos)
[-2, -1, 1, 2].each { |x|
yt=3-x.abs
[-yt, yt].each { |y|
np=(pos.move(x, y) rescue nil)
yield np if np
}
}
end

def find_path(start, endp, forbidden={})
# simplified dijkstra
# all weights are equal -> no sorting
return [] if start==endp
pre=forbidden.merge({start=>nil})
(front=[start]).each { |pos|
all_knight_moves_from(pos) { |m|
unless pre.has_key? m # if not visited before
pre[m]=pos
front << m
if (endp==m) # path found
path=[endp]
path.unshift(pos) until start==(pos=pre[path[0]])
return path
end
end
}
}
nil
end

def main(s, e, *forb)
forbidden={}
forb.each { |f| forbidden[ChessPos.new(f)]=nil } # all keys are
forbidden
if path=find_path(ChessPos.new(s), ChessPos.new(e), forbidden)
puts "[ #{path.collect { |p| p.pos }.join(", ")} ]"
else
puts nil
end
end

main(*ARGV) rescue puts $!.message

 

[linus sellberg]

Here is my solution. It could use a fair bit of refactoring, but I'm too
tired now to do it

start_node, end_node,*forbidden = ARGV
start_node = [start_node[0] - 97, Integer(start_node[1,1]) - 1]
end_node = [end_node[0] - 97, Integer(end_node[1,1]) - 1]

success = false

Moves = [[1,2],[-1,2],[1,-2],[-1,-2],
[2,1],[-2,1],[2,-1],[-2,-1]]

board = Array.new(8) { Array.new(8) }

forbidden.each{|el|
board[el[0] - 97][Integer(el[1,1]) - 1] = :forbidden
}

board[start_node[0]][start_node[1]] = :start
queue = [start_node]

queue.each{ |i,j|
#create some moves
Moves.collect {|k,l|
[i+k, j+l]
}.reject{|k,l|
#remove the impossible and already used moves
k < 0 || l < 0 || k > 7 || l > 7 || (board[k][l])
}.collect{|k,l|
#checks if done, end looping or equeue the move.
if [k,l] == end_node
success = true
queue = []
else
queue << [k,l]
end
#mark the node
board[k][l] = [i,j]#node
}
}

if success
#traverse backwards from the end node
path = [end_node]
path.inject([]){|acc,node|
unless node == start_node
path << board[node[0]][node[1]]
acc << node
end
}

path.reverse!


path.each{|node|
i,j = *node
print (i+97).chr
puts j + 1
}
else
puts "no path found"
end

 

[linus sellberg]

#traverse backwards from the end node
path = [end_node]
path.inject([]){|acc,node|
unless node == start_node
path << board[node[0]][node[1]]
acc << node
end
}

I wonder what I thought here, this should be an #each instead, I never
use the accumulated value

 

[Carlos]

However, even with the code part, I think it could be more rubyist, and
so, prettier to read. Any comments on it are welcome.

(!) Your code is clear as water. I think yours is the most readable solution
so far. (And linus' the best thought.)

 

[Matthew D Moss]

Here's my solution... This actually turned out to be relatively easy
for me, both in algorithm and programming. Simple but fun problem.

#!/usr/bin/env ruby

# Helper class
class Tile
attr_reader :x, :y
protected :x, :y

def initialize(x, y)
@x, @y = x, y
end

def Tile.named(s)
Tile.new(s.downcase[0] - 'a'[0], s.downcase[1] - '1'[0])
end

def valid?
(0...8) === @x and (0...8) === @y
end

def to_s
to_str
end

def to_str
%w(a b c d e f g h)[@x] + %w(1 2 3 4 5 6 7 8)[@y] if valid?
end

def ==(c)
@x == c.x and @y == c.y
end

def adjacent?(c)
dx = (@x - c.x).abs
dy = (@y - c.y).abs
valid? and c.valid? and (dx == 1 && dy == 2 or dx == 2 && dy == 1)
end
end


def knights_trip(start, finish, *forbidden)
# First, build big bucket o' tiles.
board = (0...64).collect { |n| Tile.new(n % 8, n / 8) }

# Second, pull out forbidden tiles.
board.reject! { |t| forbidden.include?(t) }

# Third, prepare a hash, where layer 0 is just the start.
# Remove start from the board.
x = 0
flood = { x => [start] }
board.delete(start)

# Fourth, perform a "flood fill" step, finding all board tiles
# adjacent to the previous step.
until flood[x].empty? or flood[x].include?(finish) do
x += 1
flood[x] = flood[x-1].inject([]) do |mem, obj|
mem.concat(board.find_all { |t| t.adjacent?(obj) })
end

# Remove those found from the board.
board.reject! { |t| flood[x].include?(t) }
end

# Finally, determine if we found a way to the finish and, if so,
# build a path.
if not flood[x].empty?
# We found a way. Time to build the path. This is built
# backwards, so finish goes in first.
path = [finish]

# Since we got to finish in X steps, we know there must be
# at least one adjancent to finish at X-1 steps, and so on.
until x == 0
x -= 1

# Find in flood[x] a tile adjacent to the head of our
# path. Doesn't matter which one. Make it the new head
# of our path.
jumps = flood[x].find_all { |t| t.adjacent?(path.first) }
path[0,0] = jumps.sort_by { rand }.first
end

# Tada!
path
end
end


# main
args = ARGV.collect { |arg| Tile.named(arg) }
if args.any? { |c| not c.valid? }
puts "Invalid argument(s)!"
else
trip = knights_trip(*args)
if trip
puts "Knight's trip: " + trip.join(", ")
else
puts "No route available!"
end
end

 

[Kero]

# Ruby Quiz #27, Knight's Travails
#
# Author: Kero van Gelder
# License: LGPL, see http://www.gnu.org/licenses/lgpl.html
#
# Given: Chess board, start_pos, finish_pos and forbidden squares
# Find: a shortest route from start to finish, for a knight, without using the
# forbidden squares.
#
# Observations:
# - shortest path requires Dijkstra, but all distances are 1, so we do not need
# a priority queue, we can use a regular queue
# - breadth first search like this (dynamic programming style, too) keeps
# pointers (steps) towards the point where you start the algorithm, so we
# have to start at the finish. Quite normal for Dijkstra, now that I think of
# it...
#
# Apologies for:
# - not checking the input (ignoring commas happily, no spaces)
# - the use of @board and @q which are pseudo-global variables
# - not returning an array, but printing the result (hey, you left the
# *content* of the array undescribed; mine is like [[?b, 7], [?c, 5]],
# perhaps you need ["b7", "c5"] ?) Printing is with spaces before the commas,
# too? How tasteless

# Input helper
def decode_pos(str)
[str[0], str[1,1].to_i]
end

# Used in breadth first search
def try_pos(c, d, rc, rd)
(c >= ?a and c <= ?h) or return
(d >= 1 and d <= 8) or return
unless @board[c][d]
@board[c][d] = [rc, rd]
@q << [c, d]
end
end

start_pos, finish_pos, *forbidden = *ARGV
@board = {}
(?a..?h).each { |c| @board[c] = [] }
forbidden.each { |pos|
c, d = decode_pos(pos)
@board[c][d] = :forbidden
}

fc, fd = decode_pos(finish_pos)
@board[fc][fd] = :finish
@q = [[fc, fd]]
sc, sd = decode_pos(start_pos)

while not @q.empty?
c, d = *@q.shift
break if c == sc and d == sd
try_pos(c-2, d-1, c, d)
try_pos(c-2, d+1, c, d)
try_pos(c-1, d-2, c, d)
try_pos(c-1, d+2, c, d)
try_pos(c+1, d-2, c, d)
try_pos(c+1, d+2, c, d)
try_pos(c+2, d-1, c, d)
try_pos(c+2, d+1, c, d)
end

# It is a good defensive programming habit to look whether you actually found a
# solution (and don't check q.empty? as I did, 'coz the queue will be empty
# when the search looked at all 64 squares for a route from e.g. a8 to h1).
if @board[sc][sd]
route = []
rc, rd = sc, sd
while rc != fc or rd != fd
next_pos = @board[rc][rd]
route << "#{next_pos[0].chr}#{next_pos[1]}"
rc, rd = *next_pos
end
puts "[ #{route.join" , "} ]"
else
puts nil
end

 

[Timothy Byrd]

Btw. does anybody know if this behavior is defined

somewhere (adding elements to an Array while
iterating over it):
irb(main):001:0> (a=[0]).each { |i| a << i+1 if i < 10 }; a
=> [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

If you just want to accumualte an array, here is a snippet from the
Programming Ruby HTML book:

(1..10).to_a » [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

-- Timothy
just realized I should have added start_pos to my initial forbidden
array so the algorithm wouldn't try to double back.

 

[linus sellberg]

# - breadth first search like this (dynamic programming style, too) keeps
# pointers (steps) towards the point where you start the algorithm, so we
# have to start at the finish. Quite normal for Dijkstra, now that I think of

Oh, I wish I thought of that, my solution has to reverse the path
afterwards

 

[Dominik Bathon]

Btw. does anybody know if this behavior is defined
somewhere (adding elements to an Array while
iterating over it):
irb(main):001:0> (a=[0]).each { |i| a << i+1 if i < 10 }; a
=> [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

If you just want to accumualte an array, here is a snippet from the
Programming Ruby HTML book:

(1..10).to_a » [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

Thanks for the reply, but that wasn't the intetion.

While coding the solution, I first had something like this in my find_path
method:

front=[start]
until front.empty?
pos=front.shift
all_moves(pos) { |m|
if ...
front << m
end
}
end

I experimented a bit and changed it to the following:

(front=[start]).each { |pos|
all_moves(pos) { |m|
if ...
front << m
end
}
}

It basicly does the same. And I wondered if this is "defined" behavior, or
if it is just coincidence, that it works (adding elements to front, while
iterating over it).

Dominik

 

[Timothy Byrd]

And here is an iterative, breadth-first search that can be used in
place of my find_path_recurse().


module Knight

def self.find_path_bf(pStart, pEnd, forbidden)

# Are we there yet?
#
return [pEnd.clone] if pStart == pEnd

# You can't get there from here!
#
return nil if forbidden.include?(pEnd)

position_list = Hash.new
forbidden.each {|f| position_list[f] = 'forbidden' }
position_list[pStart] = []

moves_to_check = [pStart]

until moves_to_check.empty? do
pos = moves_to_check.shift
possible_moves(pos).each do |m|
next if position_list[m]
position_list[m] = position_list[pos] + [m]
return position_list[m] if m == pEnd
moves_to_check.push(m)
end
end

nil
end
end

-- Timothy

 

[Brian Schröder]

A very late entry for the knigh's travail. The solution can be found at:

http://ruby.brian-schroeder.de/quiz/knights-travail/

It's just the standard dijkstra.

regards,

Brian