Draw an hourglass

Question

Again inspired by a task for Programming 101 here's another challenge.

Input:

• A positive integer n >= 3. (has to be odd)

Output:

• n lines of asterisks, where the first line has n asterisks and every new line has two asterisks less than the line before. Until hitting 1 asterisk. From there every new line has two asterisks more than the line before until getting back to n asterisks. Spaces or something like spaces have to used to align the asterisks so that it really will look like an hourglass.

General rules:

• Trailing newlines are allowed but do not have to be used.
• indentation is a must.
• This is code-golf, so shortest answer in bytes wins.
• Since the course is taught in C++, I'm eager to see solutions in C++.

Test case (n=5):

*****
 ***
  *
 ***
*****

Answer 1

Haskell, 84 bytes

f n|l<-div n 2,k<-[-l..l]=putStr$unlines[[" *"!!(fromEnum$abs x<=abs y)|x<-k]|y<-k]

Answer 2

C (gcc), 80 74 bytes

Thank ceilingcat for 6 bytes

-Du(x,k)=for(x=~n;x+=2,x<n;putchar(k))

i;j;f(n){u(i,10)u(j,i*i<j*j?32:42);}

Try it online!

Answer 3

Jelly, 15 bytes

HŻżḂƇṚ$⁾ *xⱮŒḄY

Try it online!

I'm surprised there wasn't a Jelly answer yet. Jelly is not good at strings.

Explanation

HŻżḂƇṚ$⁾ *xⱮŒḄY   Main monadic link
H                 Halve: n/2
 Ż                Range from 0 to floor(n/2)
  ż               Zip with
      $           (
                    Range from 1 to n
    Ƈ               Filter by
   Ḃ                  parity [to get the odd numbers]
     Ṛ              Reverse
      $           )
           Ɱ      For each pair of numbers
          x         repeat
       ⁾ *            the characters ' ' and '*' that many times
            ŒḄ    Bounce (palindromize)
              Y   Join with newlines

Answer 4

Java 7, 156 bytes

Fairly simple. Keeps track of lines with n, stars with j, spaces with s, and direction with d. I really just wanted a non-recursive Java answer on the board, but it doesn't hurt that it's also a bit shorter :)

String f(int n){String o="";int j=n,s=0,i,d=0;for(;n-->0;o+="\n"){for(i=0;i++<s;)o+=" ";for(i=0;i++<j;)o+="*";d+=j<2?1:0;j+=d<1?-2:2;s+=d<1?1:-1;}return o;}

With line breaks:

String f(int n){
    String o="";
    int j=n,s=0,i,d=0;
    for(;n-->0;o+="\n"){
        for(i=0;i++<s;)
            o+=" ";
        for(i=0;i++<j;)
            o+="*";
        d+=j<2?1:0;
        j+=d<1?-2:2;
        s+=d<1?1:-1;
    }
    return o;
}

Answer 5

Vyxal CRM, 5 bytes

y×ẋṘ∞

Try it Online!

Newer Vyxal 2 version with a ton of flags but hey 5 bytes

Answer 6

Groovy, 66 Bytes

{n->((n..1)+(2..n)).each{if(it%2>0){println(("*"*it).center(n))}}}

Try it: https://groovyconsole.appspot.com/script/5145735624392704

Explained:

((n..1)+(2..n)) - Reverse palindromize to n [n,..,1,..,n]

.each{if(it%2>0){...} - Iterate through odd elements.

println(("*"*it).center(n)) - Center n stars and print each on newline.

Answer 7

PHP, 191 bytes

$b=[];for($i=$a=$argv[1]+1;$i>0;$i--){$i--;if($i<=1){$c=str_pad("*",$a," ",2)."\n";break;}$b[]=str_pad(str_repeat("*",$i),$a," ",2)."\n";}echo implode("",$b).$c.implode("",array_reverse($b));

Run like php -f golf_hourglass.php 15

# php -f golf_hourglass.php 15
***************
 *************
  ***********
   *********
    *******
     *****
      ***
       *
      ***
     *****
    *******
   *********
  ***********
 *************
***************

The idea behind it is to create the top half (the part before the single *), then just echo the top part twice, but the second time in reverse order.

Answer 8

Pyke, 22 19 bytes

F-ed*ih\**+)2%'X_OX

Try it here!

F          )        -    for i in range(input)
 -                  -        Q-i
  e                 -       floor(^/2)
   d*               -      ^*" "
          +         -     ^+V
     ih             -       i+1
       \**          -      ^*"*"
            2%      -   ^[::2]
              'X_   - splat(^),
                       reversed(^)
                 OX - splat(^[:-1])

Answer 9

C, 117 bytes

void p(c,n){while(n--)putchar(c);}void h(n){for(int i=n;i>=-n;i-=i==1?4:2){p(32,(n-abs(i))/2);p(42,abs(i));p(10,1);}}

Ungolfed

void printNum(c, n) {
  while (n--)
    putchar(c);
}

void hourGlass(n) {
  for (int i = n; i >= -n; i-=i==1?4:2) {
    printNum(32, (n - abs(i)) / 2);
    printNum(42, abs(i));
    printNum(10, 1);
  }
}

Answer 10

PHP, 158 bytes

(-2 if I can omit the start tag)

<?$v=$argv[1]/2;while($c++<$v){$a="";$b=$v+.5;while(--$b)$a.=$b<$c?" ":"*";$z[]=strrev($a)."*$a";}$c=array_pop($z);$a=join($z,"
");echo"$a
$c
".strrev($a);

A similar approach to the one @hd took but uses manual concatenation instead of built-in string functions with overly long names. I wish I could call $z->pop()

Answer 11

Pyth, 22 bytes

VQ+*d-J/Q2aJN*\*hatQyN

Try online.

Lame attempt at explaining:

VQ+*d-J/Q2aJN*\*hatQyN   The whole thing.
VQ                       For N in [0,1,2, ... ,Q-1]; Q takes the input.
      J/Q2               Integer division Q/2, store in J
          aJN            Absolute difference between N and J
     -                   Subtract the above numbers...
   *d                    ...and repeat whitespace that many times
  +                      Concatenate that with output from the next block
                 atQyN   Absolute difference between Q-1 and 2*N
             *\*h        Repeat an asterisk the above number of times + 1

Interestingly I matched TheBikingViking's score on the same language using a different approach... O:

Answer 12

Racket 177 bytes

(let p((n N)(d #t))(for((i(/(- N n)2)))(display" "))(for((i n))(display #\*))(displayln"")
(cond[(and(not d)(= n N))][(= n 1)(set! d #f)(p(+ 2 n)d)][d(p(- n 2)d)][(p(+ 2 n)d)]))

Ungolfed:

(define (f N)
  (let loop ((n N)
             (d #t))                      ; descending
    (for ((i (/(- N n)2)))
          (display " "))
    (for ((i n))
      (display #\*))
    (displayln "")
    (cond
      [(and (not d) (= n N))]             ; end
      [(= n 1) (set! d #f)
               (loop (+ 2 n) d)]
      [d (loop (- n 2) d)]
      [(loop (+ 2 n) d)]))
  )

Testing:

(f 9)

Output:

*********
 *******
  *****
   ***
    *
   ***
  *****
 *******
*********
#t

Answer 13

Batch, 191 154 bytes

@echo off
set s=*
set l=for /l %%i in (2,1,%1)do call:
%l%s
%l%s
%l%u
%l%l
:l
echo %s%
:s
set s=%s:~1%**
exit/b
:u
echo %s%
set s= %s:~0,-2%

The u subroutine generates the top half of the hourglass by prepending a space and removing two trailing stars each time. The l subroutine generates the bottom half of the hourglass by removing a space and appending two trailing stars each time. The code falls through into the l subroutine at the end to save bytes. The second half of the l subroutine is also used to build up the initial string, although it has to be called twice as many times as the string doesn't have its spaces yet.

Edit: Saved 37 bytes by using an idea by @ConorO'Brien.

Answer 14

CJam, 26 bytes

ri:A,_W%.e<{_S*A@2*-'**N}/

Try it online!

ri:A     e# read integer, save as A
,_W%     e# make range and mirror (e.g.[0,1,2,3,4],[4,3,2,1,0])
.e<      e# take minimums (e.g.[0,1,2,1,0])
{        e# for each minimum "m"
  _S*    e#   push m spaces
  A@2*-  e#   calculate A-2m (rotates to spare m)
  '**    e#   push A-2m asterisks
  N      e#   push a newline
}/       e# end, prints stack

Answer 15

Swift 3, 139 130

let c={String(repeating:$0,count:$1)};func d(_ n:Int,_ i:Int=0)->String{let b=c(" ",i)+c("*",n)+"\n";return n>1 ?b+d(n-2,i+1)+b:b}

On-line test

let c={(s,i)in(0..<i).reduce(""){(b,_)in b+s}};func d(_ n:Int,_ i:Int=0)->String{let b=c(" ",i)+c("*",n)+"\n";return n>1 ?b+d(n-2,i+1)+b:b}

Answer 16

T-SQL, 93 bytes

This will also provide an answer when an even number is used - the bottom row will be missing.

Golfed:

DECLARE @ INT=5

,@i INT=0z:PRINT SPACE(@/2-abs(@/2-@i))+REPLICATE('*',abs(@/2-@i)*2+1)SET
@i+=1IF @i<@ GOTO z

Ungolfed:

DECLARE @ INT=5

,@i INT=0
z:
PRINT SPACE(@/2-abs(@/2-@i))+REPLICATE('*',abs(@/2-@i)*2+1)
SET @i+=1
IF @i<@ GOTO z

A bit longer alternative solution without looping

USE master
DECLARE @ INT=5
SELECT top(@)SPACE(@/2-abs(@/2-number))+REPLICATE('*',abs(@/2-number)*2+1)
FROM spt_values
WHERE'P'=type

Fiddle

Answer 17

Actually, 34 bytes

;╗R`1&`░`'**╜kd#"{:^%d}"%f`M;pXR+i

Try it online!

Explanation:

;╗R`1&`░`'**╜kd#"{:^%d}"%f`M;pXR+i
;╗                                  store a copy of n in register 0
  R                                 range(1, n+1) ([1, n])
   `1&`░                            filter: keep only odd values
        `'**╜kd#"{:^%d}"%f`M        for each value:
         '**                          push a string containing that many asterisks
            ╜                         push n
             kd#                      list from stack, dequeue, listify (make both stack elements singleton lists)
                "{:^%d}"%             format string to center string in n-wide field
                         f            do the formatting
                            ;pXR+   duplicate list, discard the extra "*" element, reverse, and append original list (vertical mirroring)
                                 i  flatten and implicitly print

Answer 18

Common Lisp, 101 bytes

(dotimes(i(1+(set'x(read))))(format t"~v:@<~v{*~}~>~&"x(set'b(abs(- x(* i 2))))1)(if(= 1 b)(incf i)))

Answer 19

SOGL V0.12, 12 bytes

.↔∫ **}⁰№§╬3

Try it Here!

Answer 20

Excel VBA, ₆₈ 67 Bytes

Anonymous VBE immediate window function that takes input from [A1] and outputs to the VBE immediate window

n=[Int(A1/2)]:For i=-n To n:j=abs(i):?Spc(n-j)String(2*j+1,42):Next

Answer 21

Ruby, 66 bytes

->n{(-n..n).each{|x|x.abs>1?(puts"*"*x.abs):(puts"*"*x if x>0)}}

Probably a shorter way of doing this, but this is the most straight forward one i came up with.

Answer 22

Japt -R, 11 bytes

1õU2_ç*Ãê û

Try it

                :Implicit input of integer U
1õU2            :Range [U,1] in increments of 2
    _           :Map each Z
     ç*         :  Repeat "*" Z times
       Ã        :End map
        ê       :Palindromise
          û     :Centre pad each element with spaces to the length of the longest
                :Implicitly join with newlines and output

Answer 23

Japt, 11 bytes

õ_î*ÃÔë ê û

Try it online

Answer 24

Python 3, 79 bytes

def f(i):[print(' '*(i//2-abs(j-i//2))+'*'*2*abs(j-i//2)+'*')for j in range(i)]

Answer 25

Oracle SQL, 119 bytes

SQL> select lpad(' ',ceil(n/2)-x-1,' ')||rpad('*',x*2+1,'*')from(select n,abs(ceil(n/2)-rownum)x from t connect by level<=n)
  2  /

LPAD('',CEIL(N/2)-X-1,'')||RPAD('*',X*2+1,'*')
--------------------------------------------------------------------------------
*******
 *****
  ***
   *
  ***
 *****
*******

7 rows selected.

Answer 26

Perl 5 -a, 51 bytes

$_%2&&say$"x(("@F"-abs)/2),'*'x abs for-$_..0,2..$_

Try it online!

Answer 27

sed 4.2.2, 54 42 bytes

:
H
s/00/ /
t
:a
s/ 0/000/
H
//ba
g
y/0/*/

Takes input in unary (e.g. 3 is 000), allowed as per meta consensus. Try it online!

Code with comments:

#Create an unnamed branch label (sed 4.2 only)
:
#Append pattern buffer to hold buffer
H
#Replace first two zeroes with a space
s/00/ /
#If we did the previous replacement, branch back to the unnamed label
t
#Create a label a
:a
#Replace last space with two zeroes
s/ 0/000/
#Append pattern buffer to the hold buffer
H
#Check if there is still something to replace (// refers to the previous pattern)
#If so, branch to label a
//ba
#Replace the pattern buffer with the hold buffer
g
#Replace zeroes with asterisks
y/0/*/

-12 bytes thanks to Xcali

Answer 28

Rockstar, 141 bytes

listen to N
let X be N--2
F takes X
let Y be N-X
let Y be/2
say " "*Y+"*"*X

while X-1
let X be-2
F taking X

while N-X
let X be+2
F taking X

Try it here (Code will need to be pasted in)