ASCII Ice Cream

Question

Write a program or function that takes in a positive integer N, and prints or returns an N×N ASCII art string whose top half is a semicircle made of ('s and whose bottom half is a downward-pointing triangle made of V's, with spaces used as padding.

In other words, make an ASCII ice cream cone: (output for N = 17)

      (((((      
    (((((((((    
  (((((((((((((  
  (((((((((((((  
 ((((((((((((((( 
 ((((((((((((((( 
(((((((((((((((((
(((((((((((((((((
VVVVVVVVVVVVVVVVV
 VVVVVVVVVVVVVVV 
  VVVVVVVVVVVVV  
   VVVVVVVVVVV   
    VVVVVVVVV    
     VVVVVVV     
      VVVVV      
       VVV       
        V        

Examples

Here are the outputs for N = 1 to 5. Note that for odd N, the triangle always must be the larger half.

V

((
VV

(((
VVV
 V 

 (( 
((((
VVVV
 VV 

 ((( 
(((((
VVVVV
 VVV 
  V  

Here's an N = 101 pastebin.

And here's an ungolfed Python 3 reference implementation:

N = int(input())
ic = [[' '] * N for _ in range(N)]
for y in range(N//2):
    for x in range(N):
        if (x - (N - 1) / 2)**2 + (y - (N - 1) / 2)**2 < (N / 2)**2:
            ic[y][x] = '('
for y in range(N//2, N):
    for x in range(y - N//2, N - (y - N//2)):
        ic[y][x] = 'V'
for line in ic:
    print(''.join(line))

Details

• Take input from stdin, command line, or as function argument. Output to stdout or similar, or you may return the string if you write a function.
• The cone portion should exactly match the reference implementation for all N.
• The ice cream portion does not need to exactly match the reference implementation as long as it is clearly in the shape of a semicircle for all N. (This is so you don't have to worry about slight differences in the semicircle due to roundoff errors.)
• There should not be any unnecessary leading spaces but there may be superfluous trailing spaces.
• The output may optionally contain a trailing newline.
• You may optionally use any 3 other distinct printable ASCII characters in place of (, V, and space.

Scoring

The shortest submission in bytes wins. Tiebreaker goes to the oldest submission.

Answer 1

CJam, 46 bytes

Try it online.

{:Z{Z{Z(2./:R-zYR<):P#YR-zP#+Z2./P#>SP?}/N}fY}

I believe this currently mimics the original specification exactly, which was required when I started producing this answer. There may be potential to save a few bytes by making the math less accurate to the original specification, but until I see a way to save more than one or two bytes doing so, I'll leave it as-is.

Explanation

{               "Begin block";
  :Z{             "For each y from 0 to input-1";
    Z{              "For each x from 0 to input-1";
      Z(2./:R         "Calculate the radius as (input-1)/2.0";
      -z              "Calculate the horizontal distance from the center";
      YR<):P          "Calculate the power to raise distances to: (y<radius)+1
                       (This results in Euclidean distance being calculated for
                        the ice cream and Manhattan distance being calculated
                        for the cone)";
      #               "Raise the horizontal distance to the determined power";
      YR-zP#          "Calculate the vertical distance from the center and
                       raise it to the determined power";
      +               "Add the horizontal and vertical distances";
      Z2./P#          "Calculate the solid distance threshold and raise it to
                       the determined power";
      >SP?            "If the solid threshold is exceeded, produce a space;
                       otherwise, produce the determined power digit
                       (This results in ice cream being represented by the
                        digit '2' and the cone by the digit '1')";
    }/              "End x loop";
    N               "Produce a new line";
  }fY             "End y loop";
}               "End block";

Answer 2

inca2 129 123 121 111 107

This mostly uses the formulas from the python example, but using jot-dots and iotas instead of double-looping. The i function performs the circular test for the j function which invokes jot-dot upon it. And the k function performs the triangle test for the l function. The c function catenates the results of j and l and reshapes it to N×N.

edit: -6 combine 2 maps into 1.
edit: -2 remove useless ravels.
edit: nicer typescript.
edit: -10 apply repeated expression array-wise.
edit: -4 factor out repeated expression as a function.
edit: line-by-line commentary.

q:y-(n-1)%2
i:[((n%2)^2)>+/(qx y)^2
j:(~[y%2)i.(~y)
k:2*[x>[|qy
l:(@1+~]y%2)k.(~y)
c:y y#((jn<y),ly){' (V' 

In more detail, the entry-point is the c function which takes one argument implicitly named y.

c:y y#((jn<y),ly){' (V' 
         n<y            } assign y to 'n'
        jn<y            } call j(y)
              ly        } call l(y)
      ((    ),  )       } catenate the results
      (         ){' (V' } map 0 1 2 to ' ' '(' 'V'
  y y#                  } reshape to NxN

The j function receives the same input value as its y parameter.

j:(~[y%2)i.(~y)
     y%2         } y divided by 2
    [            } floor
   ~             } iota. this generates the row indices 0..y/2
            ~y   } iota y. this generates the column indices 0..y
  (     )i.(  )  } jot-dot with the function i

The jot-dot here does the double-loop. It calls the i function with every combination of elements from the left and right arrays (0..n/2 and 0..n). So the i function receives as x the y index of the table, and it receives as y the x index. The names got a little backwards here :).

i:[((n%2)^2)>+/(qx y)^2
     n%2                 } n divided by 2
    (n%2)^2              } squared
                 x y     } make a 2-element array (x,y)
                qx y     } call q on this array

where q does

q:y-(n-1)%2
     n-1    } n minus 1
         %2 } divided by 2
  y-        } y minus that

back to i

i:[((n%2)^2)>+/(qx y)^2
               (    )^2  } square the result from q(x,y)
             +/          } sum the two numbers
            >            } compare the left side (above) with the right (=> 0/1)
  [                      } floor

The floor should not be necessary. But apparently there's a bug in the interpreter.

The l function works similarly to the j function, using a jot-dot.

l:(@1+~]y%2)k.(~y)
        y%2         } y divided by 2
       ]            } ceiling
      ~             } iota 0..ceil(y/2)-1
    1+              } add 1 => 1..ceil(y/2)
   @                } reverse => ceil(y/2)..1
               ~y   } iota y  0..y-1
  (        )k.(  )  } jot-dot using k

The k function yields a boolean scaled by 2 so the values can be distinguished from the ice cream values later on, in the mapping.

k:2*[x>[|qy
     x       } k's left arg
         qy  } y-(n-1)%2
        |    } abs
       [     } floor
     x       } left-hand-side again
      >      } compare 
    [        } floor (should be unnecessary)
  2*         } scale by 2

In action (piping through tr to remove tab chars which are the REPL's prompt):

josh@Z1 ~/inca
$ ./inca2 <icecream | tr -d '\t'

c1
V

c2
((
VV

c3
(((
VVV
 V 

c4
 (( 
((((
VVVV
 VV 

c5
 ((( 
(((((
VVVVV
 VVV 
  V  

josh@Z1 ~/inca
$ 

Answer 3

Python 2, 193 192

Does not use strings, only math

N=input()
R=(N+1)/2;r=range(R)
s=lambda L,U:(10**U-10**L)/9
f=lambda N,m:s(0,N)+s(m,N-m)
g=lambda N,m:s(0,N)+s(m,N-m)*6
for i in r[1:]:print f(N,int(R-(2*R*i-i*i)**.5))
for i in r:print g(N,i)

s(L,U) returns a number of the form "U-digits with the rightmost L zeros and the rest ones"
f(N,m) returns an N-digit number with the inner section of 2 and an m-wide border of 1on each side
g(N,m) does the same, but using 7 for the 'colour' of the inner section since it matches the texture of the cone more closely

Output

N=8         N=9
11122111    112222211
12222221    122222221
22222222    222222222
22222222    222222222
77777777    777777777
17777771    177777771
11777711    117777711
11177111    111777111
            111171111

Answer 4

Perl 6, 175

Pretty straightforward implementation without much golfing, just extraneous whitespace/punctuation elimination:

sub MAIN($d){my$r=($d/2).Int;for 1..$r ->$n
{my$y=$n-$r;my$h=sqrt($r*$r-$y*$y).Int;my$w=2*$h+$d%2;say
' 'x($r-$h)~'('x$w};for 1..($d-$r) ->$y {say ' 'x($y-1)~'V'x($d-2*$y+2)}}