Spiral matrix
You are encouraged to solve this task according to the task description, using any language you may know.
Produce a spiral array. A spiral array is a square arrangement of the first N2 natural numbers, where the numbers increase sequentially as you go around the edges of the array spiralling inwards.
For example, given 5, produce this array:
0 1 2 3 4 15 16 17 18 5 14 23 24 19 6 13 22 21 20 7 12 11 10 9 8
Ada
<ada>-- Spiral Square with Ada.Text_Io; use Ada.Text_Io; with Ada.Integer_Text_Io; use Ada.Integer_Text_Io; with Ada.Numerics.Elementary_Functions; use Ada.Numerics.Elementary_Functions;
procedure Spiral_Square is
type Array_Type is array(Positive range <>, Positive range <>) of Natural;
function Spiral (N : Positive) return Array_Type is
Result : Array_Type(1..N, 1..N);
Row : Natural := 1;
Col : Natural := 1;
Max_Row : Natural := N;
Max_Col : Natural := N;
Min_Row : Natural := 1;
Min_Col : Natural := 1;
begin
for I in 0..N**2 - 1 loop
Result(Row, Col) := I;
if Row = Min_Row then
Col := Col + 1;
if Col > Max_Col then
Col := Max_Col;
Row := Row + 1;
end if;
elsif Col = Max_Col then
Row := Row + 1;
if Row > Max_Row then
Row := Max_Row;
Col := Col - 1;
end if;
elsif Row = Max_Row then
Col := Col - 1;
if Col < Min_Col then
Col := Min_Col;
Row := Row - 1;
end if;
elsif Col = Min_Col then
Row := Row - 1;
if Row = Min_Row then -- Reduce spiral
Min_Row := Min_Row + 1;
Max_Row := Max_Row - 1;
Row := Min_Row;
Min_Col := Min_Col + 1;
Max_Col := Max_Col - 1;
Col := Min_Col;
end if;
end if;
end loop;
return Result;
end Spiral;
procedure Print(Item : Array_Type) is
Num_Digits : constant Float := Log(X => Float(Item'Length(1)**2), Base => 10.0);
Spacing : constant Positive := Integer(Num_Digits) + 2;
begin
for I in Item'range(1) loop
for J in Item'range(2) loop
Put(Item => Item(I,J), Width => Spacing);
end loop;
New_Line;
end loop;
end Print;
begin
Print(Spiral(5));
end Spiral_Square; </Ada> The following is a variant using a different algorithm (which can also be used recursively): <Ada>
function Spiral (N : Positive) return Array_Type is
Result : Array_Type (1..N, 1..N);
Left : Positive := 1;
Right : Positive := N;
Top : Positive := 1;
Bottom : Positive := N;
Index : Natural := 0;
begin
while Left < Right loop
for I in Left..Right - 1 loop
Result (Top, I) := Index;
Index := Index + 1;
end loop;
for J in Top..Bottom - 1 loop
Result (J, Right) := Index;
Index := Index + 1;
end loop;
for I in reverse Left + 1..Right loop
Result (Bottom, I) := Index;
Index := Index + 1;
end loop;
for J in reverse Top + 1..Bottom loop
Result (J, Left) := Index;
Index := Index + 1;
end loop;
Left := Left + 1;
Right := Right - 1;
Top := Top + 1;
Bottom := Bottom - 1;
end loop;
Result (Top, Left) := Index;
return Result;
end Spiral;
</Ada>
Haskell
Solution based on the J hints:
grade xs = map snd. sort $ zip xs [0..] values n = cycle [1,n,-1,-n] counts n = (n:).concatMap (ap (:) return) $ [n-1,n-2..1] reshape n = unfoldr (\xs -> if null xs then Nothing else Just (splitAt n xs)) spiral n = reshape n . grade. scanl1 (+). concat $ zipWith replicate (counts n) (values n)
J
This function is the result of some beautiful insights:
spiral =. ,~ $ [: /: }.@(2 # >:@i.@-) +/\@# <:@+: $ (, -)@(1&,) spiral 5 0 1 2 3 4 15 16 17 18 5 14 23 24 19 6 13 22 21 20 7 12 11 10 9 8
Would you like some hints that will allow you to reimplement it in another language?
These inward spiralling arrays are known as "involutes"; we can also generate outward-spiraling "evolutes", and we can start or end the spiral at any corner, and go in either direction (clockwise or counterclockwise). See the first link (to JSoftware.com).
Python
<python> def spiral(n):
dx,dy = 1,0 # Starting increments
x,y = 0,0 # Starting location
i = 0 # starting value
myarray = [[None]* n for j in range(n)]
while i < n**2:
myarray[x][y] = i
i += 1
nx,ny = x+dx, y+dy
if 0<=nx<n and 0<=ny<n and myarray[nx][ny] == None:
x,y = nx,ny
continue
else:
dx,dy = (-dy,dx) if dy else (dy,dx)
x,y = x+dx, y+dy
return myarray
def printspiral(myarray):
n = range(len(myarray))
for y in n:
for x in n:
print "%2i" % myarray[x][y],
print
printspiral(spiral(5)) </python> Sample output:
0 1 2 3 4 15 16 17 18 5 14 23 24 19 6 13 22 21 20 7 12 11 10 9 8
Recursive Solution
<python> def spiral_part(x,y,n):
if x==-1 and y==0:
return -1
if y==(x+1) and x<(n/2):
return spiral_part(x-1, y-1, n-1) + 4*(n-y)
if x<(n-y) and y<=x:
return spiral_part(y-1, y, n) + (x-y) + 1
if x>=(n-y) and y<=x:
return spiral_part(x, y-1, n) + 1
if x>=(n-y) and y>x:
return spiral_part(x+1, y, n) + 1
if x<(n-y) and y>x:
return spiral_part(x, y-1, n) - 1
def spiral(n):
array = [[None]*n for j in range(n)]
for x in range(n):
for y in range(n):
array[x][y] = spiral_part(x,y,n)
return array
</python>
Adding a cache for the spiral_part function it could be quite efficient.
Another way, based on preparing lists ahead
<Python> def spiral(n):
dat = [[None] * n for i in range(n)]
le = [[i + 1, i + 1] for i in reversed(range(n))]
le = sum(le, [])[1:] # for n = 5 le will be [5, 4, 4, 3, 3, 2, 2, 1, 1]
dxdy = [[1, 0], [0, 1], [-1, 0], [0, -1]] * ((len(le) + 4) / 4) # long enough
x, y, val = -1, 0, -1
for steps, (dx, dy) in zip(le, dxdy):
x, y, val = x + dx, y + dy, val + 1
for j in range(steps):
dat[y][x] = val
if j != range(steps)[-1]:
x, y, val = x + dx, y + dy, val + 1
return dat
for row in spiral(5): # calc spiral and print it
print ' '.join('%3s' % x for x in row)
</Python>