110
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Given a number N, how can I print out a Christmas tree of height N using the least number of code characters? N is assumed constrained to a minimum value of 3, and a maximum value of 30 (bounds and error checking are not necessary). N is given as the one and only command line argument to your program or script.

All languages appreciated, if you see a language already implemented and you can make it shorter, edit if possible - comment otherwise and hope someone cleans up the mess. Include newlines and White Spaces for clarity, but don't include them in the character count.

A Christmas tree is generated as such, with its "trunk" consisting of only a centered "*"

N = 3:

   *
  ***
 *****
   *

N = 4:

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

N = 5:

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

N defines the height of the branches not including the one line trunk.

Merry Christmas PPCG!

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116 Answers 116

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PHP (133 relevant characters):

function xmastree($h) {
    for($i=0;$i<$h;++$i)
        echo str_repeat(' ',$h-$i-1).str_repeat('*',2*$i+1)."\n";
    echo str_repeat(' ',$h-1)."*\n";
}
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1
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Language: Pike

101 Relevant characters

int main (int c, array a) {
    int n=(int)a[1], i,l;
    for(;i<=n; l = ++i < n ? i : 0)
        write(" " *(n-l) + "*" * (l*2+1) +"\n");
}
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1
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Rhino Javascript shell: 117 chars minified

t=['*'];
for(i=1;i<arguments[0];++i)
{
  s='*'+t[i-1]+'*';
  for(j in t) 
    t[j]=' '+t[j];
  t[i]=s;
}
t[i]=t[0];print(t.join('\n'));

minified:

t=['*'];for(i=1;i<arguments[0];++i){s='*'+t[i-1]+'*';for(j in t) t[j]=' '+t[j];t[i]=s;}t[i]=t[0];print(t.join('\n'));

results:

c:\>java -jar C:\appl\Java\rhino1_7R1\js.jar c:/tmp/Xtree.js 10
         *
        ***
       *****
      *******
     *********
    ***********
   *************
  ***************
 *****************
*******************
         *
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1
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Language: Php, Char count: 110 (3 relevant spaces)

<? function x($n,$a,$t){return $n?str_repeat(' ',$n).$a.x($n-1,"*$a"," $t"):$t;}echo x($argv[1],"\n","*\n");

A bit of php recursion to reduce the count of chars to 110.

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1
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FreePascal:

program xmastree;

{$mode objfpc}{$H+}

uses
  {$IFDEF UNIX}{$IFDEF UseCThreads}
  cthreads,
  {$ENDIF}{$ENDIF}
  Classes
  { you can add units after this };

var x,y,h:integer;

{$IFDEF WINDOWS}{$R xmastree.rc}{$ENDIF}

procedure printRow(sp,st:integer);
var i:integer;
begin
    for i := 1 to sp do begin
    write(' ');
  end;
    for x := 1 to st do begin
    write('*');
  end;
    for x := 1 to sp do begin
    write(' ');
  end;
    writeln();
end;

begin
    val(ParamStr(1),h);
  for y := 1 to h do begin
    printRow(h-y,(y-1)*2+1);
  end;
  printRow(h-1,1);
end.

Output for xmastree.exe 9

        *
       ***
      *****
     *******
    *********
   ***********
  *************
 ***************
*****************
        *
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1
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Language: C, Char count: 116

I realized I could improve on my original design:

main(int c,char**v){char l[99],i=0;for(c=atoi(1[v]);i<c;printf("%*s%.*s\n",c,l,i++,l))l[i]=42;printf("%*c\n",c,42);}

Different approach (119 characters):

s[99],w,i=0;p(n){printf("%*.*s\n",w+n,n*2+1,s);}main(int c,char**v){w=atoi(v[1]);for(memset(s,42,99);i<w;p(i++));p(0);}

Old version (123 characters):

main(int c,char**v){char*l=calloc(c=atoi(v[1]),2),i=0;for(;i<c;printf("%*s%.*s\n",c,l,i++,l))l[i]=42;printf("%*c\n",c,42);}

(One byte can be saved by putting char *l=... in the for loop. That makes it non-standard, however (though gcc still accepts it).)

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1
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VBScript, 106 characters

n = WScript.Arguments(0)
For i = 1 To n
  WScript.Echo Space(n-i+1) & String(2*i-1, "*")
Next
WScript.Echo Space(n) & "*"

Usage and output example:

> cscript christmastree.vbs 7 //nologo
       *
      ***
     *****
    *******
   *********
  ***********
 *************
       *
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1
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Golfscript - 27 chars

~:i,0+{.i\-(' '*\.)+'*'*n}%

~:i  # eval the command line arg, store in i  
,0+  # create a list [0..i-1] add 0 to the end  
{}%  # map this block over the list  
.    # make a copy of the list element 
i\-  # subtract the list element from i 
(    # decrease by one more  
' '* # multiply the result by ' '
\    # swap, so the list element is back on top of stack
.)+  # duplicate, add one, add the two numbers together
'*'* # multiply the result by '*'
n    # put a newline here
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1
  • \$\begingroup\$ nice, i did mine before seeing yours, was about to refactor it to look like this \$\endgroup\$
    – Claudiu
    Jun 10, 2010 at 6:29
1
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Q, 54

{-1(-:)[((x+(!)x),x)]$(((&)((!)(2*x))mod 2),1)#\:"*";}

example:

q){-1(-:)[((x+(!)x),x)]$(((&)((!)(2*x))mod 2),1)#\:"*";} 10
         *
        ***
       *****
      *******
     *********
    ***********
   *************
  ***************
 *****************
*******************
         *
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1
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This is my first experience with code golf. So advices are welcome :)

C: 170 chars:

int main(){
    int n,m=1,x;
    scanf("%d",&n);x=2*n-1;
    for(;m<x*n+n+2;m++)
        printf("%c",m==x*n|m==x*n+n+1?42:m==x*n+1?10:m>x*n?32:!(m%(2*n-1))?10:!(m%x<n-m/x)&m%x<m/x+n+1?42:32);
}
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1
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It's in PHP.

<center><?$t=$_GET[1]*2-$i=1;while($i<=$t){echo str_repeat('*',$i).'<br>';$i+=2;}echo'*'?><center>

Total characters(with spaces):98 Total characters(with no spaces):97 Bytes:98

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1
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I did not see a solution using R. The code below may not be efficient, but it seems to work (170 characters with spaces if all code placed on one line):

for(i in 1:4){ cat( paste( paste(rep(' ', (3-(i-1))), collapse=''), 
                           paste(rep('*', (2*i-1)),   collapse=''), collapse=''),  
               sep='\n'); if(i==4) cat('    *    ', sep='\n')}


    *
   ***
  *****
 *******
    *
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1
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Javascript, 89

i=4
c=t=""
while(i){o=c,j=(i--+1)*2-3
while(j--)o+="*"
t=o+"\n"+t
c+=" "}console.log(t+o)

Javascript 81

o=s=''
for(a=3;a--;){l=s+=' '
for(k=a*2+1;k--;)l+='*'
o=l+'\n'+o}console.log(o+l)
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1
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><>: 136 chars

This one's not going to win, but because Christmas is coming again, I thought fish needed a Christmas tree too! (pun intended)

1-:&1$>:> :1(?!v~$:> :1(?!vv
        ^-1o" "<   ^-1o"*"<>~2+$1-:0(?!v~~&ao>" "o1-v
      ^                              oa<;o"*"^!?(1: <

It can be run as follows with the standard interpreter:

./fish.py christmas.fish -v 5

... for N=5. (If the file is called christmas.fish :P) In this way, one prepopulates the stack with the value 5. This is the closest I could get to a command-line argument.

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1
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JavaScript, 110 characters (2 relevant spaces)

function p(l)
{
    o=''
    for(c=0; c<=n+l; c++)
      o += c < n - l ? ' ' : '*'
    print(o)
}

n = parseInt(arguments[0])

for(l = 0; l < n; l++)
  p(l)
p(0)

Ran using spidermonkey. $ smjs christmas_tree.js 4

For Javascript Console 105

function p(l){o='';for(c=0;c<=n+l;)o+=c++<n-l?' ':'*';console.log(o)}n=+prompt();for(l=0;l<n;)p(l++);p(0)
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1
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Scala, 73 chars

val n=args(0).toInt
(1 to n):+1 foreach(i=>println(" "*(n-i)+"*"*(2*i-1)))

run as

golf.scala 10
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2
  • \$\begingroup\$ Wait, does that display the trunk? \$\endgroup\$
    – Johannes Kuhn
    Dec 12, 2013 at 19:56
  • \$\begingroup\$ yes, :+1 would do that \$\endgroup\$
    – prashanth
    Dec 12, 2013 at 20:05
1
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Japt, 11 bytes

õ_ç* êÃp* û

Try it online!

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1
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Japt -R, 11 bytes

ÇÑÄ î*Ãp* û

Try it

Ç               :Map the range [0,input)
 Ñ              :  Multiply by 2
  Ä             :  Add 1
    î*          :  Repeat * that many times
      Ã         :End map
       p*       :Push a *
          û     :Centre pad each element with spaces to the length of the longest
                :Implicitly join with newlines
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1
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Lua, 100 99 bytes

p,r,n=io.write,('').rep,arg[1]for i=0,n do p(r(' ',n-i+1),r('*',i*2+1),'\n')end p(r(' ',n+1),'*\n')

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Another Lua version, a little shorter.

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0
1
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PHP, 79 bytes

while($argn)echo$r[]=str_pad("
",1+$argn--).str_pad("",++$n*2-1,"*");echo$r[0];

Run as pipe with -nR or try it online.

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1
  • \$\begingroup\$ Save 2 bytes using variable variables: Try it online! \$\endgroup\$
    – Night2
    Oct 13, 2019 at 13:50
1
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Zsh, 77 64 bytes

Since all these old questions are now open, I think the oldest q. on codegolf.se needs a zsh answer!
Try it Online!

s=*;for ((;i<$1;))printf %$[$1+i++]s\\n $s&&s+=**
printf %$1s \*

Hat-tip to Alastair from Boxing Day 2008
NB: Putting the printf stuff in a function is no advantage...
NB2: original for 77 bytes.. saved 12 bytes by eliminating quotes, braces, trailing newline

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1
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Charcoal, 12 bytes

Ｇ<Ｎ*‖↑¶Ｍ⊖θ←*

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explanation

Ｇ<Ｎ*             #Draw a triangle size input N with character "*"
      ‖↑           #Mirror the triangle so it is the right way round
        ¶          #Newline
         Ｍ⊖θ←    #move left input-1 
               *   #"*" character
                   #the result is implicitly printed
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2
  • \$\begingroup\$ Nice answer! I took the liberty of creating a TIO link so others can see it in action, if you'd liked to edit it into your post: tio.run/##ASkA1v9jaGFyY29hbP//… \$\endgroup\$
    – Malivil
    Dec 19, 2019 at 18:25
  • \$\begingroup\$ my bad @Malivil! I meant to add one, but forgot, thank you :) \$\endgroup\$
    – mabel
    Dec 19, 2019 at 18:33
1
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Keg, 42 38 32 23 bytes

:&1ɧ1⑷:⑻$-⑬*$⑵;`*`*⑸(
'

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-9 bytes due to generating odd numbers inside of the map

Answer History

32 Bytes

¿&1(⑻|:2+)1^⑷:;½ℤ⑻$-⑬*^`*`*⑸(,
,

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-6 bytes due to using a mapping approach

38 Bytes

:&®l1(©l|⑬©l*,:`*`*,
,2+©l;®l)&⑬*,`*`,

-4 bytes due to ⑬ operator (space string)

Funny story: I almost said this was 40 bytes before I realised that I forgot how to count.

Try it online!

I might as well place up the Keg Christmas tree for the year seeing as how it's almost December!

Explained

:&®l    #Store the amount of lines in variable l and the register
1   #Push the number of stars per row
(©l|    #Repeat variable l times
    ⑬©l*,   #Print that amount of spaces
    :`*`*,  #Print that amount of stars

    ,   #Print a literal newline
    2+  #Increment the number of stars
    ©l;®l   #Update the amount of spaces to print
)
&⑬*,`*`,    #Print the bottom row
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1
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Java, 180 chars

interface a{static void main(String[]a){for(int N=new Byte(a[0]),i=-1;i++<N;p('*',i%N*2),p('\n',0))p(' ',N-i%N-1);}static<T>void p(T c,int i){System.out.print(c);if(i-->0)p(c,i);}}

I believe this is a new record for Java for this question!

interface a
{
    static void main(String[]a){
        for(int N=new Byte(a[0]),i=-1;i++<N;p('*',i%N*2),p('\n',0))p(' ',N-i%N-1);
    }
    static<T>void p(T c,int i){
        System.out.print(c);
        if(i-->0)p(c,i);
    }
}

This program uses a recursive method to get rid of unnecessary for loops and print calls. Everything in main is run in one for loop.

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2
  • 1
    \$\begingroup\$ The whitespace is there to make it easier to read. The 180 char count is for the necessary characters only. \$\endgroup\$
    – Ethan Gallagher
    Apr 30, 2020 at 2:37
  • 1
    \$\begingroup\$ Nice. I see that you have already included the reduced version as well. That would have been what I suggested, and it is generally best to include both versions. \$\endgroup\$
    – Wheat Wizard ♦
    Apr 30, 2020 at 3:39
1
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MAWP, 45 bytes

=M@![![32;1-]`M[42;1-]`M2+=M10;1-]`[32;1-]42;

Try it!

Takes a while for N bigger than 10

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1
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Rockstar, 107 bytes

listen to N
X's0
while N-X
let X be+1
let Y be N-X
let Z be X*2-1
say " "*Y+"*"*Z

let N be-1
say " "*N+"*"

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

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2
  • \$\begingroup\$ A good idea would be to append the name of a related rock song to each of your rockstar programs. \$\endgroup\$
    – Razetime
    Sep 23, 2020 at 16:56
  • \$\begingroup\$ Not a bad idea, @Razetime, I might just do that - I already use song names for all my challenge titles. Of course, if I was anyway skilled at such things, I'd actually write a song in Rockstar related to the challenge! \$\endgroup\$
    – Shaggy
    Sep 23, 2020 at 16:59
1
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sed 4.2.2, 62 bytes, -r flag

s/0\B/ /g
:
H
s/ (0+)/0\10/
//b
:a
s/0(0+)0/ \1/
ta
H
x
y/0/*/

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

Try it online!

With comments:

#Replace all zeroes except the last by a space
s/0\B/ /g
#Unnamed branch label
:
#Append to hold buffer
H
#Put zeroes around a stretch of zeroes
s/ (0+)/0\10/
#Jump to unnamed label if there are any spaces left
//b
#Create a label a, and recursively remove initial/final zeroes
:a
s/0(0+)0/ \1/
ta
#Append to hold buffer
H
#Exchange hold/pattern buffers
x
#Replace 0s with asterisks
y/0/*/
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1
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Ruby 2.7, 50 bytes

->h{t=(0...h).map{" "*(h-_1)+?**(_1*2+1)};t<<t[0]}

Try it online!

TIO uses an older version of Ruby, whereas in Ruby 2.7, we've numbered parameters, which saves a byte.

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1
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Pip, 18 bytes

@PPZsXa-_.'*X_M\,a

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I don't know how pip recognizes this lambda, but it somehow does.

-3 from DLosc.

Explanation

YPPZsXa-_.'*X_M\,ay@0 a → input
              M\,a    map inclusive range of 0 - a, loop var '_'
    sXa-_             repeat space (input - _) times
         .'*X_        repeat * _ times
         .            join them:               
  PZ                  palindromize the result: ' ***' → ' ***** '
YP                    print each item of the array with newline
                      and assign the array to variable y
                  y@0 return first element of y (always a single *)
                      implicitly print it

@y cannot be used here, since it changes \,a in the first statement.

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1
  • \$\begingroup\$ Nice! Here's 18 bytes, taking advantage of the fact that P returns its operand. \$\endgroup\$
    – DLosc
    Oct 12, 2021 at 5:31
1
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Clojure, 214 bytes

(defn p[n](let[c(map #(inc(* % 2))(range n))l(map #(apply str %)(partition 2(interleave(map #(apply str(repeat(/ % 2)" "))(reverse c))(map #(apply str(repeat % "*"))c))))](doseq[a l](println a))(println(first l))))

Try it online!

Ungolfed:

(defn print-xmas-tree-of-height [n]
  (let [asterisk-counts       (map #(inc (* % 2)) (range n))
        asterisk-strings      (map #(apply str (repeat % "*")) asterisk-counts)
        leading-spaces        (map #(apply str (repeat (/ % 2) " ")) (reverse asterisk-counts))
        lines                 (map #(apply str %) (partition 2 (interleave leading-spaces asterisk-strings)))
        ]
    (doseq [aLine  lines]
      (println aLine))
    (println (first lines))))
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2
  • \$\begingroup\$ I think you can make this a lambda instead of a named function \$\endgroup\$
    – Seggan
    Dec 1, 2022 at 21:29
  • \$\begingroup\$ @Seggan You can't use #() because you can't nest them, but you could do (fn[n](...)) \$\endgroup\$
    – naffetS
    Dec 2, 2022 at 2:50
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