# Forum: Ruby Number Spiral (#109)

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on 2007-01-19 16:30
```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!

Suggestion:  A [QUIZ] in the subject of emails about the problem helps
everyone
message,
if you can.

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

by Bob Showalter

(Taken from the puzzle by William Wu at
http://www.ocf.berkeley.edu/~wwu/riddles/cs.shtml)

[Editor's Note:  This was also a code golf problem a few months back:
http://codegolf.com/oblongular-number-spirals  --JEG2]

Write a Ruby program to print out a "spiral" of numbers that fill a NxN
square.
Your program will take a single argument to specify the dimensions of
the square
(1 or higher). The number zero represents the center of the spiral, and
the
succeeding integers spiral out in a clockwise (or counterclockwise; your
choice)
direction from the center until the square is filled.

Your program should write the output line by line, without using an
array to
build up the data first.

Here's the output for an 8x8 spiral:

56   57   58   59   60   61   62   63

55   30   31   32   33   34   35   36

54   29   12   13   14   15   16   37

53   28   11    2    3    4   17   38

52   27   10    1    0    5   18   39

51   26    9    8    7    6   19   40

50   25   24   23   22   21   20   41

49   48   47   46   45   44   43   42```
on 2007-01-19 16:30
```On 12/01/07, Ruby Quiz <james@grayproductions.net> wrote:
> Write a Ruby program to print out a "spiral" of numbers that fill a NxN square.
> Your program will take a single argument to specify the dimensions of the square
> (1 or higher). The number zero represents the center of the spiral, and the
> succeeding integers spiral out in a clockwise (or counterclockwise; your choice)
> direction from the center until the square is filled.
>
> Your program should write the output line by line, without using an array to
> build up the data first.

Ok, I like a challenge and my maths needed some dusting down.

Here is my solution, after a fair bit of scribbling on paper to work
out the formula for sprial_value_at(x,y).

I found it a bit frustrating that ranges in Ruby can only be
ascending; but soon found that we have #downto, which achieves the
desired result.

Thanks to Bob & James for setting this quiz.

Marcel

#! /usr/bin/env ruby
#
# Marcel Ward   <wardies ^a-t^ gmaildotcom>
# Sunday, 14 January 2007
# Solution for Ruby Quiz number 109 - Number Spiral

# Prints a clockwise spiral, starting with zero at the centre (0,0).
# Note, here x increases to the east and y increases to the north.
def spiral(size)
# maximum -ve/+ve reach from the centre point "0" at (0,0)
neg_reach = -pos_reach = size/2
# we miss out the bottom/left sides for even-sized spirals
neg_reach += 1 if size % 2 == 0

# Compute width to allocate a cell based on the max value printed
cell_width = (size**2 - 1).to_s.size + 3

pos_reach.downto(neg_reach) do
|y|
spiral_line((neg_reach..pos_reach), y, cell_width)
end
end

def spiral_line(x_range, y, cell_width)
x_range.each do
|x|
print spiral_value_at(x, y).to_s.center(cell_width)
end
puts
end

# calculate the value in the spiral at location (x,y)
def spiral_value_at(x, y)
if x + y > 0  # top/right side
if x > y    # right side
4 * x**2 - x - y
else        # top side
4 * y**2 - 3 * y + x
end
else          # bottom/left side
if x < y    # left side
4 * x**2 - 3 * x + y
else        # bottom side
4 * y**2 - y - x
end
end
end

spiral(10)```
on 2007-01-19 16:30
```* Ruby Quiz, 01/12/2007 03:29 PM:
> The number zero represents the center of the spiral

Let me suggest that you address the issue of the centre being
ill-defined under the condition that the number of columns/rows is
even. Given that altogether four positions meet the condition leaving
this question open will result in a multitude of different outputs
that are solutions of the task. In my opinion a programming quiz
should use a well-posed problem so that the different programs can be
compared directly. Especially if the problem can be solved by
constructing a clever mathematical formula which my intuition suggest
to be the case with the number spiral problem (I did not yet try to
verify this).

Jupp```
on 2007-01-19 16:30
```Dear Ruby Quiz,

this isn't really a solution to the quiz 109 because it violates
some (if not all) of the rules. But as James noted there was a
code golf problem very similar to this quiz and here is my
solution to that.
(see http://codegolf.com/oblongular-number-spirals for detailed
description of the code golf problem)

----------------------------------------------------------------
s,f=1,proc{|x,y|y<1?[]:[[*s...s+=x]]+f[y-1,x].reverse.transpose}
puts f[w=gets(' ').to_i,gets.to_i].map{|i|['%3i']*w*' '%i}
----------------------------------------------------------------

It draws a number spiral, starting with '1' in the upper left
corner and the highest number in the middle, it also features

Yes, you will get some score at the codegolf site if you repost
this solution there - but nowadays you will only get to Rank 9
with this solution and of course you will start to feel ill and
you won't be able to sleep for days and other nasty things might
happen if you do so.

If someone can derive an even shorter solution from this i would
be very interested to see it (the best ruby solution today has 7
bytes less)

cheers

Simon```
on 2007-01-19 16:30
```Here's my solution to the quiz.  I used a recursive solution.  An
odd-sized spiral is the smaller-by-one even-sized spiral with a number
added to the end of each row, and a new row across the bottom.  And an
even-sized spiral is the smaller-by-one odd-sized spiral with a number
added to the beginning of each row and a new row across the top.

I decided to make my solution use relatively few lines of code.
However in doing that I added some inefficiencies, where a value might
be calculated multiple times in a loop/iterator when it would have been
better to calculate it once before the loop/iterator, and use that
stored value in the loop/iterator.

With respect to not building the solution in an array and then
displaying the array, I read that to mean not creating a
two-dimensional array (i.e., array of arrays) in which to build the
entire spiral.  I assemble each line of output in an array before
displaying that line, but each line is displayed before any subsequent
lines are calculated.  The solution could be adapted to avoid even the
one-dimensional array.

Eric
----------------
Interested in on-site, hands-on Ruby training?  At www.LearnRuby.com
you can read previous students' reviews!

================

def odd_spiral(size, row)
if row < size - 1 : even_spiral(size - 1, row) << (size - 1)**2 + row
else (0...size).collect { |n| size**2 - 1 - n }
end
end

def even_spiral(size, row)
if row == 0 : (0...size).collect { |n| size**2 - size + n }
else odd_spiral(size - 1, row - 1).unshift(size**2 - size - row)
end
end

size = (ARGV[0] || 8).to_i
(0...size).each do |row|
puts ((size % 2 == 0 ? even_spiral(size, row) : odd_spiral(size,
row)).
map { |n| n.to_s.rjust((size**2 - 1).to_s.length) }.join("
"))
end```
on 2007-01-19 16:30
```Quiz #109 even leaves it up to the implementer whether to go clockwise
or counter-clockwise.  So it's designed to have some flexibility in it.
In fact, it's typical of the Ruby Quiz to define a core problem and
leave many of the details up to the individual implementers.  And given
that the Quiz is designed to enhance understanding and appreciation of
Ruby (and not to be an exam or contest) that seems reasonable.

the quiz statement does use an even number (8) to define the size.  And
as you can see from the sample output, the 0 is located just to the
right and just below the exact center.  So you could reasonably use
that to nail down that detail.

Eric

================

Hands-on Ruby training at your location is available from
www.LearnRuby.com .  Read reviews from actual students there.```
on 2007-01-19 16:31
```My second attempt/solution...  Slightly different in that it does a
counter-clockwise spiral, but basically follows a similar idea as my
previous solution, though I think this looks nicer.

N  = ARGV[0].to_i
FW = (N ** 2 - 1).to_s.size + 2

def fmt(x)
" " * (FW - x.to_s.size) + x.to_s
end

def o(n, r, c)
x = (n - 1) ** 2
if    c == 0      then x + r
elsif r == n - 1  then x + r + c
else                   e(n - 1, r, c - 1)
end
end

def e(n, r, c)
x = (n ** 2) - 1
if    r == 0      then x - c
elsif c == n - 1  then x - c - r
else                   o(n - 1, r - 1, c)
end
end

def spiral(n)
(0...n).map do |r|
if (n % 2).zero?  # even
(0...n).map { |c| fmt(e(n, r, c)) }
else
(0...n).map { |c| fmt(o(n, r, c)) }
end.join
end.join("\n")
end

puts spiral(N)```
on 2007-01-19 16:31
```#!/usr/bin/env ruby
# Script to print an N by N spiral as shown in the following example:
#
#  56   57   58   59   60   61   62   63
#
#  55   30   31   32   33   34   35   36
#
#  54   29   12   13   14   15   16   37
#
#  53   28   11    2    3    4   17   38
#
#  52   27   10    1    0    5   18   39
#
#  51   26    9    8    7    6   19   40
#
#  50   25   24   23   22   21   20   41
#
#  49   48   47   46   45   44   43   42
#
# Let item with value 0 be at x, y coordinate (0, 0).  Consider the
# spiral to be rings of numbers.  For the numbers 1 through 8 make
# up ring level 1, and numbers 9 through 24 make up ring level 2.
# To figure out the value at a particular x, y position, note that
# the first value at any level is (2 * level - 1) ** 2 and use that
# value to count up or down to the coordinate.

class Spiral

def initialize(size)
@size = size
@center = size/2
end

# returns the value for a given row and column of output
def position_value(row, col)
x, y = coordinate = coordinate_for(row, col)
level = [x.abs, y.abs].max
if x < level && y > -level
# return number for top left portion of ring
first_number(level) +
steps_between(first_coordinate(level), coordinate)
else
last_number(level) -
steps_between(last_coordinate(level), coordinate)
end
end

def maximum_value
@size * @size - 1
end

def first_number(level)
(2 * level - 1) ** 2
end

def last_number(level)
first_number(level + 1) - 1
end

def first_coordinate(level)
[-level, -level + 1]
end

def last_coordinate(level)
[-level, -level]
end

def coordinate_for(row, col)
[col - @center, @center - row]
end

def steps_between(point1, point2)
(point1[0] - point2[0]).abs + (point1[1] - point2[1]).abs
end
end

if __FILE__ == \$0
size = ARGV[0].to_i
spiral = Spiral.new(size)
width = spiral.maximum_value.to_s.length + 3
(0...size).each do |row|
(0...size).each do |col|
print spiral.position_value(row, col).to_s.rjust(width)
end
print "\n\n"
end
end```
on 2007-01-19 16:31
```Here is my solution. I used a recursive printing routine to handle the
insides of the middle rows.

Ben

class NumberSpiral
def initialize(n)
@size = n
@format = "%#{(n*n - 1).to_s.length+1}d"
if n % 2 == 0
@top_row = proc{|x| (x*(x-1)).upto(x*x-1) {|i| print_num(i) } }
@bottom_row = proc{|x| ((x-1)*(x-1)).downto((x-1)*(x-2)) {|i|
print_num(i) } }
@middle_first = proc{|x,row| print_num(x*(x-1)-row) }
@middle_last = proc{|x,row| print_num((x-2)*(x-2)-1+row) }
else
@top_row = proc{|x| ((x-1)*(x-2)).upto((x-1)*(x-1)) {|i|
print_num(i) } }
@bottom_row = proc{|x| (x*x-1).downto(x*(x-1)) {|i| print_num(i)
} }
@middle_first = proc{|x,row| print_num((x-1)*(x-2)-row) }
@middle_last = proc{|x,row| print_num((x-1)*(x-1)+row) }
end
end

def print_num(i)
printf @format, i
end

def print_row(size, row)
if row == 0
@top_row.call(size)
elsif row == size - 1
@bottom_row.call(size)
else
@middle_first.call(size, row)
print_row(size-2, row-1)
@middle_last.call(size, row)
end
end

def print_clockwise
@size.times {|i| print_row(@size, i) ; puts ; puts if i < @size-1 }
end
end

if ARGV.size == 0 or not ARGV[0] =~ /^\d+\$/
puts "Usage:  #\$0 N"
puts "Output:  Prints a \"spiral\" of numbers that fill a NxN
square."
else
NumberSpiral.new(ARGV[0].to_i).print_clockwise
end```
on 2007-01-19 16:31
```n = ARGV[0].to_i
square = Array.new(n+2) { Array.new(n+2) }

# boundaries
(n+1).times {|i|
square[i][0] = square[i][n+1] = square[0][i] = square[n+1][i] = 0
}

dirs = [[1, 0], [0, -1], [-1, 0], [0, 1]]

# spiral inwards from a corner
x, y, i, d = 1, 1, n*n - 1, 0

while i >= 0 do
square[x][y] = i

# move to the next square in line
x += dirs[d][0]
y += dirs[d][1]
if square[x][y]
# if it is already full, backtrack
x -= dirs[d][0]
y -= dirs[d][1]
# change direction
d = (d - 1) % 4
# and move to the new next square in line
x += dirs[d][0]
y += dirs[d][1]
end
i -= 1
end

# remove the boundaries
square.shift; square.pop
square.map {|i| i.shift; i.pop}

puts square.map {|i| i.map {|j| "%02s" % j}.join(" ")}```
on 2007-01-19 16:31
```My first attempt... A recursive solution recognizing than a spiral of
even dimension can be formed by a top row, a left column, and an odd
spiral. Likewise, an odd spiral is a smaller even spiral with a right
column and bottom row.

The functions erow and orow reflect the even/odd-ness of the spiral,
not the row.

DIM = ARGV[0].to_i
FLD = (DIM ** 2 - 1).to_s.size + 2

def fmt(x)
" " * (FLD - x.to_s.size) + x.to_s
end

def orow(n, i)
m = n ** 2
x = m - n

if i == n - 1
(1..n).inject("") { |o, v| o + fmt(m - v) }
else
erow(n - 1, i) + fmt(x - n + i + 1)
end
end

def erow(n, i)
m = n ** 2
x = m - n

if i == 0
(0...n).inject("") { |o, v| o + fmt(x + v) }
else
fmt(x - i) + orow(n - 1, i - 1)
end
end

def spiral(n)
if (n % 2).zero?
n.times { |i| puts erow(n, i) }
else
n.times { |i| puts orow(n, i) }
end
end

spiral(ARGV[0].to_i)```
on 2007-01-19 16:31
```Hi all!
This is my first partecipation to Ruby Quiz. I developed a pretty messy
solution for clockwise (ck) solution. When I started to tackle the
counter-ck solution I started messing around with lambdas everywhere,
but eventually I found out that I just could reverse each line of a ck
solution to have the correct output. Also I didnt code the ck/cck
picking part, so you need to change it manually in initialize
My approach is still of the kind "over-use all the power of the
language" to crack the solution instead of a more reccomendable
mathematical one. But there is time to it.
I tried to comment my code extensively, maybe to understand it you need
to go through the example spiral output and check what it does ...
I loved all the really compact solutions that have been posted so far,
keep them coming!
This community just rocks!
Take care you all!
Francesco Levorato aka flevour

#! /usr/bin/env ruby
#
# Francesco Levorato aka flevour   <flevour ^a-t^ gmaildotcom>
# Sunday, 14 January 2007
# Solution for Ruby Quiz number 109 - Number Spiral

class Array
def decrease_all
self.map! { |x| x = x - 1}
end
def increase_all
self.map! { |x| x = x + 1}
end
def enqueue(x)
self.insert(0, x)
end

# sort of a hackish method to remove unwanted numbers from @left and
@right
# i haven't figured out a valid reason to explain why i need to
remove these values
# but otherwise things won't work and I haven't time to think more on
the topic
def delete_invalid
self.map! {|x| (x > 1) ? x : nil}
self.compact!
end
end

class NumberSpiral
# this solution addresses clockwise from center to outside filling
method
# my approach is based on the observation that each row of the matrix
is composed
# of 3 parts: 0 or more columns, a series of consecutive numbers, 0
or more columns
def initialize(n, direction = :ck)
@n = n
@dim = @n*@n
# left contains the first part of a row
# right contains the third part of a row
# in a 8x8: if the row is 54,29,12,13,14,15,16,37
# left: [54, 29], right: [37]
@left = []
@right = []
# just wanted to try out this block thingie Ruby is so famous about
@format = Proc.new { |x| print sprintf("%3s", x.to_s + " ") }
@direction = direction # :ck or :cck
end

# the 3 following methods, h,l,d are were the funniest part of the
quiz: finding
# the relationships intercurring between special elements of the
spiral.
# they are used to build only the first (N/2 + 1) rows, as the other
ones are
# built according only to the data structures @left and @right

# to explain these 3 methods, define the following function
# pivot(row): returns the number at given row just before the start
of the second part
#             of the row (the part containing the consecutive
numbers)

# given a row number
# returns the distance from the pivot to the first "spiral wall"
below it
# subtracts 1 not to overlap with l(x) results
# in a 8x8: given row 7 returns length from 54 down to 50
def h(x)
2*x - @n - 1
end

# given a row number
# returns the width of the next horizontal segment going from pivot
toward
# the center of the spiral
# in a 8x8: given row 6 returns length from 25 to 20
def l(x)
2 * ( x + 1 ) - @n
end

# given a row number, returns the difference between the pivot and
the number
# just at its right
# in a 8x8: given 7 returns the difference between 55 and 30
def d(x)
2 * ( l(x) + h(x) ) - 1
end

def print_me
row = @n
start = @dim - @n
# prints first row
print_row(consecutive_numbers(start))
print "\n"

# prepare for loop
pivot = start - 1
@left << pivot

# prints the top rows, it stops after printing the row containing
the zero
while(pivot >= 0) do
row = row - 1
pivot = pivot - d(row)

# gets middle consecutive numbers
middle = consecutive_numbers(pivot)
print_row(middle)

@left << pivot
@left.decrease_all

@right.enqueue(middle.last) # last number of consecutive series
will be in the right part in next iteration
@right.increase_all

pivot = @left.last
print "\n"
end

@left.delete_invalid
@right.delete_invalid
row = row -1

# prints the remainder of the spiral
while(row > 0) do
from= @left.pop

middle = consecutive_numbers(from, :down)
last_printed = middle.last
print_row(middle)

@right.delete_at(0)

@left.decrease_all
@right.increase_all
row = row - 1
print "\n"
end
end

def consecutive_numbers(n, go = :up)
array = []
(@n - @left.size - @right.size).times do
array << n
if go == :up
n = n + 1
else # go == :down
n = n - 1
end
end
array
end

def print_row(middle)
if @direction == :ck
(@left + middle + @right).each(&@format)
else
(@left + middle + @right).reverse.each(&@format)
end
end
end

if ARGV[0]
NumberSpiral.new(ARGV[0].to_i).print_me
else
puts "Call me: #{\$0} <matrix_dim>\n"
end```
on 2007-01-19 16:32
```Simon Kröger wrote:
> s,f=1,proc{|x,y|y<1?[]:[[*s...s+=x]]+f[y-1,x].reverse.transpose}
> you won't be able to sleep for days and other nasty things might
> happen if you do so.
>
> If someone can derive an even shorter solution from this i would
> be very interested to see it (the best ruby solution today has 7
> bytes less)

A reduction:

["stuff"]+[[4,5,6],[:x,:y,:z]].reverse.transpose
==>["stuff", [:x, 4], [:y, 5], [:z, 6]]
["stuff"]+(a,b=[[4,5,6],[:x,:y,:z]];b.zip a)
==>["stuff", [:x, 4], [:y, 5], [:z, 6]]```
on 2007-01-19 17:16
```Simon Kröger wrote:
> s,f=1,proc{|x,y|y<1?[]:[[*s...s+=x]]+f[y-1,x].reverse.transpose}
This is more obfuscated than

s=1
f=proc{|x,y|y<1?[]:[[*s...s+=x]]+f[y-1,x].reverse.transpose}

and is no shorter.

> puts f[w=gets(' ').to_i,gets.to_i].map{|i|['%3i']*w*' '%i}
> ----------------------------------------------------------------
>
> It draws a number spiral, starting with '1' in the upper left
> corner and the highest number in the middle, it also features
> spirals that are not quadratic.
>
> Yes, you will get some score at the codegolf site if you repost
> this solution there

If the site accepted this, then it wasn't tested thoroughly
enough.  '%3i' gives every number-spiral a column-width
of 3;  the column-width should equal the width of the
largest number.```
on 2007-01-20 15:40
```William James wrote:
>> s,f=1,proc{|x,y|y<1?[]:[[*s...s+=x]]+f[y-1,x].reverse.transpose}
>
> This is more obfuscated than
>
> s=1
> f=proc{|x,y|y<1?[]:[[*s...s+=x]]+f[y-1,x].reverse.transpose}
>
> and is no shorter.

depends on your line end character(s) :)

> If the site accepted this, then it wasn't tested thoroughly
> enough.  '%3i' gives every number-spiral a column-width
> of 3;  the column-width should equal the width of the
> largest number.

Well, consider it cheating, most spirals have a column width of
3 - so you may have to post it twice to have it accepted.

cheers

Simon```
on 2007-01-20 21:45
```Simon Kröger wrote:
> depends on your line end character(s) :)
Only if you're under windoze and you neglect to remove
doing this, then you are really hurting your score.

By the way, in the ASCII Art contest I got the size
down to 76.  I would pay to see how "flagitious" did it
in 71 bytes!

> >
> > If the site accepted this, then it wasn't tested thoroughly
> > enough.  '%3i' gives every number-spiral a column-width
> > of 3;  the column-width should equal the width of the
> > largest number.
>
> Well, consider it cheating, most spirals have a column width of
> 3 - so you may have to post it twice to have it accepted.

The site really should do more thorough testing.```
on 2007-09-25 22:31
```Here's the solution I came up with before submitting this idea:

# spiral.rb
# RubyQuiz #109
# Bob Showalter

class Integer

def odd?
self % 2 == 1
end

end

class Spiral

# order must be > 0
def initialize(order)
raise ArgumentError, "order must be > 0" unless order.to_i > 0
@order = order
end

# writes the spiral to stdout
def output
puts "\n"
0.upto(@order - 1) do |r|
row_for(@order, r)
puts "\n\n"
end
end

private

# emits row r for spiral of order p
def row_for(p, r)
if p <= 1
cell(0)
elsif p.odd?
if r == p - 1
row(p)
else
row_for(p - 1, r)
col(p, r)
end
else
if r == 0
row(p)
else
col(p, r)
row_for(p - 1, r - 1)
end
end
end

# emits the full row (top or bottom) for spiral of order p
def row(p)
x = p * (p - 1)
y = x + p - 1
x.upto(y) {|i| cell(p.odd? ? x - i + y : i) }
end

# emits the single column cell for row r of spiral of order p
def col(p, r)
x = p * (p - 1)
r = p - r - 1 if p.odd?
cell(x - r)
end

# emits a single cell
def cell(i)
printf ' %3d ', i
end

end

n = (ARGV.first || 3).to_i
Spiral.new(n).output```
on 2007-09-25 22:32
```Here's a modified version of my previous solution that uses no arrays
other than ARGV.  For the general approach, see the discussion in my
previous posting.  And please note that it's written to favor brevity
over clarity.

Eric

----------------
On-site, hands-on Ruby training is available from www.LearnRuby.com !

================

def odd_spiral(size, row, col)
if row == size - 1 : size**2 - 1 - col
elsif col == size - 1 : (size - 1)**2 + row
else even_spiral(size - 1, row, col)
end
end

def even_spiral(size, row, col)
if row == 0 : size**2 - size + col
elsif col == 0 : size**2 - size - row
else odd_spiral(size - 1, row - 1, col - 1)
end
end

size = (ARGV[0] || 8).to_i
(0...size).each do |row|
(0...size).each do |col|
v = size % 2 == 0 ? even_spiral(size, row, col) : odd_spiral(size,
row, col)
print v.to_s.rjust((size**2 - 1).to_s.length), ' '
end
puts
end```
on 2007-09-25 22:33
```Simon Kröger wrote:
> s,f=1,proc{|x,y|y<1?[]:[[*s...s+=x]]+f[y-1,x].reverse.transpose}
> you won't be able to sleep for days and other nasty things might
> happen if you do so.
>
> If someone can derive an even shorter solution from this i would
> be very interested to see it (the best ruby solution today has 7
> bytes less)
>
> cheers
>
> Simon

I can't get this to work.

E:\Ruby>ruby try.rb
4 4
try.rb:2:in `%': too few arguments. (ArgumentError)
from try.rb:2
from try.rb:2:in `map'
from try.rb:2```
on 2007-09-25 22:34
```On 1/15/07, Matthew Moss <matthew.moss.coder@gmail.com> wrote:
> My second attempt/solution...  Slightly different in that it does a
> counter-clockwise spiral, but basically follows a similar idea as my
> previous solution, though I think this looks nicer.

Looking at the website, it seems like Eric I and I came up with
similar solutions, but I'll say he wins, because his solution looks
better, arrived first, and uses rjust.  =)```
on 2007-09-25 22:34
```Here is my solution:

n = (ARGV[0] || 8).to_i
(0...n).each do |row|
lev = (row-n/2).abs
m = [2*lev+1,n].min
p = (n-m+1)/2
(0...p).each do |col|
s = (n/2-col)*2
s = s*(s-1)-(row-col)
printf "%2d ",s
end
delta = n/2<=>row
s = lev*2
s *= (s-delta)
s += m-1 if delta<0
m.times do
printf "%2d ",s
s += delta
end
(0...n-p-m).each do |col|
s = (lev+col+1)*2
s = s*(s+1)-(p+m+col-row)
printf "%2d ",s
end
puts
end```
on 2007-09-25 22:40
```Here is one that uses an empirical algorithm but with an array (Saw the
constraint a little late) . Uses :clock or :counter parameter to print
either spiral.

class Array
def cnext
@p ||= -1
@p += 1
@p = 0 if @p == self.length
self[@p]
end
end

class ClockState
def initialize
@seq = [:left, :up, :right, :down]
@count = 1
@count_state = 0
@times = 0
@val = @seq.cnext
end

def next
if @count == @count_state
@val = @seq.cnext
@count_state = 0
@times += 1
if @times == 2
@count += 1
@times = 0
end
end
@count_state += 1
@val
end
end

class Spiral
def initialize(dim)
@m = []
dim.times do
@m << Array.new(dim, 0)
end
@x = dim/2
@y = dim/2
@val = 0
@sz = dim
end

def left
@x -= 1
end

def up
@y -= 1
end

def right
@x += 1
end

def down
@y +=1
end

def make_spiral dir_hash={:dir=>:clock}
c = ClockState.new
while ((@x < @sz) && (@y < @sz))
if dir_hash[:dir]==:counter
@m[@x][@y] = @val
elsif dir_hash[:dir]==:clock
@m[@y][@x] = @val
else
raise "Legal values are :clock and :counter"
end
self.send(c.next)
@val += 1
end
end

def print_spiral
fmt_sz = (@sz*@sz).to_s.length + 2
for i in 0...@sz do
print "\n"
for j in 0...@sz do
printf("%#{fmt_sz}d", @m[i][j])
end
end
end
end

s = Spiral.new(20)
s.make_spiral :dir=>:clock
s.print_spiral

#---```
on 2007-09-25 22:40
```My first submission to Ruby Quiz:

n = ARGV[0].to_i

# pass this method two coordinates relative to the center of the spiral
def spiral(x, y)
max_xy = [x,y].collect{|num| num.abs}.max
offset = (max_xy * 2 - 1)**2 - 1

if -(x) == max_xy and x != y
y + offset + max_xy
elsif y == max_xy
x + offset +  (3 * max_xy)
elsif x == max_xy
-y + offset + (5 * max_xy)
elsif -(y) == max_xy
-x + offset + (7 * max_xy)
end
end

for row in 0..(n - 1)
# the ease of writing one-liners in ruby lends itself to abuse...
puts (0..(n - 1)).map{|col| spiral(col - (n / 2), (n / 2) -
row).to_s.rjust(4) }.join
end```
on 2007-09-25 22:41
```William James wrote:
> > ----------------------------------------------------------------
> > with this solution and of course you will start to feel ill and
>
> I can't get this to work.
>
> E:\Ruby>ruby try.rb
> 4 4
> try.rb:2:in `%': too few arguments. (ArgumentError)
>         from try.rb:2
>         from try.rb:2:in `map'
>         from try.rb:2

irb(main):006:0> s=1; x=5; [*s...s+x]
=> [1, 2, 3, 4, 5]
irb(main):007:0> s=1; x=5; [*s...s+=x]
=> []
irb(main):009:0> s=1; x=5; s...s+=x
=> 6...6```
on 2007-09-25 22:43
```On 1/15/07, Martin DeMello <martindemello@gmail.com> wrote:
> n = ARGV[0].to_i
> square = Array.new(n+2) { Array.new(n+2) }

oops - didn't read the question carefully enough. ignore.

m.```
on 2007-09-25 22:43
```>> I can't get this to work.

Hmm.

> => []
> irb(main):009:0> s=1; x=5; s...s+=x
> => 6...6

Interesting:

C:\development>ruby -v -e "s=1; x=5; p s...s+=x"
ruby 1.8.5 (2006-08-25) [i386-mswin32]
1...6

cheers

Simon```
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