Triangle Tripler

Question

Simpler version of this

Task

You must create a program that outputs a triangle of a given size N.

A triangle is made of 3 identical triangles
A triangle of size N should be made of N lines
A triangle's line starts from 1 star, and increases by 2 stars each line

Test Cases:

1 ->

 *
* *

3 ->

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

5 ->

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

10 ->


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

Scoring:

This is code-golf so shortest code size wins!

Trailing whitespace is allowed

Credits to Ninsuo for the puzzle

Yes, it's not as nearly duplicate as I initially thought. Anyway, we usually link related challenges for the benefit of readers — Luis Mendo, Feb 11, 2022 at 13:46
Can we use a different character than * to create the triangles? And any reason why the input is always a triangular number (which 5 isn't as mentioned by @Giuseppe), since any positive integer works for the intended output? — Kevin Cruijssen, Feb 11, 2022 at 14:13

Lynn · Accepted Answer · 2022-02-12 14:00:33Z

9

Python 3, 74 bytes

lambda n:[print(i//n*f"{i%n*'**'+'*':^{8*n>>i//n}}")for i in range(n,3*n)]

Try it online!

Deera Wijesundara saved 2 bytes by suggesting print-based I/O instead of "\n".join.

Explanation

For each i in range(n,3*n), we center i%n*2+1 asterisks in 8*n>>i//n spaces and repeat this string i//n times.

`i`	`i%n`	`i%n''+''`	`8*n>>i//n`	`i//n`
n	0	`*`	4n	1
n+1	1	`***`	4n	1
n+2	2	`*****`	4n	1
…	…	…	…	…
2n	0	`*`	2n	2
2n+1	1	`***`	2n	2
2n+2	2	`*****`	2n	2
…	…	…	…	…

i%n*'**'+'*' is a byte shorter than (i%n*2+1)*'*'.
f"{s:^{n}}" is an icky but short way to write (s).center(n).

edited Feb 12, 2022 at 14:00

answered Feb 12, 2022 at 0:21

Lynn

68.1k11 gold badges130 silver badges279 bronze badges

\$\begingroup\$ You can add print and make it 2 bytes shorter lambda n:[print(i//n*f"{i%n*'**'+'*':^{8*n>>i//n}}")for i in range(n,3*n)] \$\endgroup\$
– Deera Wijesundara
Feb 12, 2022 at 6:00
\$\begingroup\$ you can add both io but it'll add 6 more bytes n=int(input());[print(i//n*f"{i%n*'**'+'*':^{8*n>>i//n}}")for i in range(n,3*n)] \$\endgroup\$
– Deera Wijesundara
Feb 12, 2022 at 6:04

Add a comment |

hakr14 · Accepted Answer · 2022-02-12 03:26:32Z

5

Canvas, 11 bytes

*×［］：｛│］∔／│

Try it here!

Explanation:

*x[]:{|]+/| | Full code (replaced with ASCII)
------------+-----------------------------------------
*           | Push '*'
 x          | Repeat <input> times horizontally
  []        | List of prefixes
    :       | Duplicate
     { ]    | Map over each element
      |     | Palindromize horizontally with 1 overlap
        +   | Add the two lists together
         /  | Anti-diagonalize
          | | Palindromize horizontally with 1 overlap

answered Feb 12, 2022 at 3:26

hakr14

4,9641 gold badge10 silver badges18 bronze badges

1

\$\begingroup\$ Anti-diagonalize and palindromize are quite nice primitives. \$\endgroup\$
– Jonah
Feb 12, 2022 at 14:51
\$\begingroup\$ @Jonah Canvas is literally designed for ASCII art, so taking advantage of common ASCII art features like diagonals and symmetry is something of a must... really comes in handy for these types of challenges! \$\endgroup\$
– hakr14
Feb 12, 2022 at 18:05

Add a comment |

Neil · Accepted Answer · 2022-02-11 21:11:26Z

5

Charcoal, 12 bytes

ＮθＦ²«Ｇ⌊θ*↖‖Ｏ

Try it online! Link is to verbose version of code. Explanation:

Ｎθ

Input the size.

Ｆ²«

Repeat twice.

Ｇ⌊θ*

Draw half of a triangle. The first time, this draws the left half of the bottom left triangle; the second time, the left half of the top triangle.

↖

Move to the next triangle. (See below as to why this moves left instead of right. This command has no effect on the second iteration.)

‖Ｏ

Reflect to complete the triangle. On the first iteration, this also reflects the current position to where it needs to be. On the second iteration, this also reflects the bottom left triangle to create the bottom right triangle.

answered Feb 11, 2022 at 21:11

Neil

166k12 gold badges71 silver badges273 bronze badges

\$\begingroup\$ how do code in charcoal so well? like how do know what to use in the verbose lang of charcoal \$\endgroup\$
– DialFrost
Feb 12, 2022 at 0:25
\$\begingroup\$ @DialFrost Practice, mostly (after all I've written over 1,000 programs in Charcoal) but obviously the Wiki was very helpful for getting started. \$\endgroup\$
– Neil
Feb 12, 2022 at 0:52
\$\begingroup\$ what wiki? the github page? (i wanna learn a golf lang soon lol) \$\endgroup\$
– DialFrost
Feb 12, 2022 at 0:53
\$\begingroup\$ @DialFrost Yeah, the wiki on the github. \$\endgroup\$
– Neil
Feb 12, 2022 at 0:54

Add a comment |

Kevin Cruijssen · Accepted Answer · 2022-02-11 14:11:12Z

05AB1E, 20 16 bytes

·ÅÉ'*×DāRúí.º«.c

Try it online or verify the test cases in the range \$[1,10]\$.

Explanation:

           #  e.g. input=3
·          # Double the (implicit) input-integer
           #  STACK: 6
 ÅÉ        # Pop and push a list of all odd numbers below this
           #  STACK: [1,3,5]
   '*×    '# Map each to that many "*" as strings
           #  STACK: ["*","***","*****"]
D          # Duplicate this list
           #  STACK: ["*","***","*****"],["*","***","*****"]
 ā         # Push a list in the range [1,length] (without popping)
           # (which is basically a list in the range [1,input])
           #  STACK: ["*","***","*****"],["*","***","*****"],[1,2,3]
  R        # Reverse it to range [input,1]
           #  STACK: ["*","***","*****"],["*","***","*****"],[3,2,1]
   ú       # Pad the strings of the top copy with that many leading spaces
           #  STACK: ["*","***","*****"],["   *","  ***"," *****"]
    í      # Reverse each string in the list so the spaces are trailing
           #  STACK: ["*","***","*****"],["*   ","***  ","***** "]
     .º    # Overlapping mirror each string
           #  STACK: ["*","***","*****"],["*     *","***   ***","***** *****"]
       «   # Merge the two lists of lines together
           #  STACK: ["*","***","*****","*     *","***   ***","***** *****"]
        .c # Join each line by newlines, and centralize it
           #  STACK: "     *\n    ***\n   *****\n  *     *\n ***   ***\n***** *****"
           # (after which the result is output implicitly)

Axuary · Accepted Answer · 2022-02-11 15:42:44Z

4

Excel, 94 bytes

=LET(x,SEQUENCE(A1*2),a,REPT(" ",A1*2-x),b,REPT("*",MOD(x-1,A1)*2+1),a&b&REPT(a&" "&a&b,x>A1))

Link to Spreasheet

answered Feb 11, 2022 at 15:42

Axuary

3,4065 silver badges11 bronze badges

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Jonah · Accepted Answer · 2022-02-12 14:44:19Z

J, ³⁹ 37 bytes

'* '{~](](=|.)/@,:{.)(>i.@-+/|@i:)@+:

Try it online!

^{-2 thanks to Lynn!}

Includes leading whitespace that can be removed with a few bytes.

Not super golfy but the concept is fun:

how

Let's take n=3 as our example:

(>i.@-+/|@i:)@+: Create a single triangle double the requested input size:

0 0 0 0 0 0 1 0 0 0 0 0 0
0 0 0 0 0 1 1 1 0 0 0 0 0
0 0 0 0 1 1 1 1 1 0 0 0 0
0 0 0 1 1 1 1 1 1 1 0 0 0
0 0 1 1 1 1 1 1 1 1 1 0 0
0 1 1 1 1 1 1 1 1 1 1 1 0

](..........{.) Take the first n lines of that:

0 0 0 0 0 0 1 0 0 0 0 0 0
0 0 0 0 0 1 1 1 0 0 0 0 0
0 0 0 0 1 1 1 1 1 0 0 0 0

].......,: And append it to the big triangle, auto-filling the extra rows with zeros:

0 0 0 0 0 0 1 0 0 0 0 0 0
0 0 0 0 0 1 1 1 0 0 0 0 0
0 0 0 0 1 1 1 1 1 0 0 0 0
0 0 0 1 1 1 1 1 1 1 0 0 0
0 0 1 1 1 1 1 1 1 1 1 0 0
0 1 1 1 1 1 1 1 1 1 1 1 0

0 0 0 0 0 0 1 0 0 0 0 0 0
0 0 0 0 0 1 1 1 0 0 0 0 0
0 0 0 0 1 1 1 1 1 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0

(=|.)/@ Reverse rows of the little triangle and check where the planes are still equal:

0 0 0 0 0 0 1 0 0 0 0 0 0
0 0 0 0 0 1 1 1 0 0 0 0 0
0 0 0 0 1 1 1 1 1 0 0 0 0
0 0 0 1 1 1 1 1 1 1 0 0 0
0 0 1 1 1 1 1 1 1 1 1 0 0
0 1 1 1 1 1 1 1 1 1 1 1 0

     = elementwise?

0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 1 1 1 1 1 0 0 0 0
0 0 0 0 0 1 1 1 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0 0 0 0

           v

1 1 1 1 1 1 0 1 1 1 1 1 1
1 1 1 1 1 0 0 0 1 1 1 1 1
1 1 1 1 0 0 0 0 0 1 1 1 1
1 1 1 0 1 1 1 1 1 0 1 1 1
1 1 0 0 0 1 1 1 0 0 0 1 1
1 0 0 0 0 0 1 0 0 0 0 0 1

'* '{~ Convert to asterisks and spaces:

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

You can replace +|. by =|., then get rid of 2| and invert ' *' to '* '. — Lynn, Feb 12, 2022 at 14:14

rebane2001 · Accepted Answer · 2022-02-11 14:33:24Z

3

Python 3.8 (pre-release), 114 bytes

def f(n,x=1):[print([n*(p:=" "),(t:=(s:=p*(~-n-i))+"*"+"**"*i+s)+p][x]+t)for i in range(0if x and f(n,0)else 0+n)]

Try it online!

Less-golfed version

def f(n, drawingTopPart=True):
    for i in range(n):
        space = " "
        spacing = space*(n-i-1)
        stars = spacing + '*' + '**' * i + spacing
        print(n*space + stars) if drawingTopPart else print(stars + space + stars)
    if drawingTopPart:
        f(n, False)

answered Feb 11, 2022 at 14:33

rebane2001

3071 silver badge4 bronze badges

\$\begingroup\$ Is “0+n” necessary? \$\endgroup\$
– ophact
Feb 11, 2022 at 15:11
\$\begingroup\$ @ophact I couldn't think of any other way to do a singular conditional recursive call \$\endgroup\$
– rebane2001
Feb 11, 2022 at 17:19
\$\begingroup\$ You can do 0if x and f(n,0)else 0+n=>x and f(n,0)or n, since the function never returns anything, it's value is None which is falsy, which means that n is always the value of the expression \$\endgroup\$
– AnttiP
Feb 11, 2022 at 20:19

Add a comment |

Level River St · Accepted Answer · 2022-02-11 19:55:20Z

3

Ruby, 70 bytes

->n{a=[]
n.times{|i|puts" "*n+b=(?**i-=~i).center(n*2)
a<<b*2}
puts a}

Try it online!

Prints the single triangle as it iterates while building up an array of double triangles.

Then dumps a printout of the array at the end.

answered Feb 11, 2022 at 19:55

Level River St

25.7k4 gold badges36 silver badges104 bronze badges

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Arnauld · Accepted Answer · 2022-02-12 17:22:29Z

JavaScript (ES6), 80 bytes

f=(n,x=y=h=n*2)=>y?` *
`[--x+h?x>=y-h&x<=h-y^y<=n&x*x<y*y:(x=h,y--,2)]+f(n,x):''

Try it online!

Patterns

Coordinates:

\$y\$ goes from \$2n\$ at the top to \$1\$ at the bottom
\$x\$ goes from \$2n-1\$ on the left to \$-2n+1\$ on the right

The examples below are given with \$n=3\$.

x >= y - h        x <= h - y
               
XXXXXX-----       -----XXXXXX       -----X-----
XXXXXXX----       ----XXXXXXX       ----XXX----
XXXXXXXX---  AND  ---XXXXXXXX  -->  ---XXXXX---
XXXXXXXXX--       --XXXXXXXXX       --XXXXXXX--
XXXXXXXXXX-       -XXXXXXXXXX       -XXXXXXXXX-
XXXXXXXXXXX       XXXXXXXXXXX       XXXXXXXXXXX

y <= n            x² < y²
               
-----------       XXXXXXXXXXX       -----------
-----------       -XXXXXXXXX-       -----------
-----------  AND  --XXXXXXX--  -->  -----------
XXXXXXXXXXX       ---XXXXX---       ---XXXXX---
XXXXXXXXXXX       ----XXX----       ----XXX----
XXXXXXXXXXX       -----X-----       -----X-----


-----X-----       -----------       -----X-----
----XXX----       -----------       ----XXX----
---XXXXX---  XOR  -----------  -->  ---XXXXX---
--XXXXXXX--       ---XXXXX---       --X-----X--
-XXXXXXXXX-       ----XXX----       -XXX---XXX-
XXXXXXXXXXX       -----X-----       XXXXX-XXXXX

loopy walt · Accepted Answer · 2022-02-11 16:01:47Z

2

Python NumPy, 95 bytes

lambda n:where((t:=tril(1>tri(4*n)[::-1])[2*n:])^(t[::-1]&t[:,n:n+1]),*'* ')
from numpy import*

Attempt This Online!

Old Python NumPy, 99 bytes

lambda n:[r:=tri(n),z:=0*r,l:=1-r[::-1]]and where(bmat("z l r z;l r l r"),*'* ')
from numpy import*

Attempt This Online!

Returns a numpy array of "*"s and " "s.

edited Feb 11, 2022 at 16:01

answered Feb 11, 2022 at 14:16

loopy walt

16k2 gold badges10 silver badges65 bronze badges

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amelies · Accepted Answer · 2022-02-17 11:21:27Z

2

Julia 1.0, 86 bytes

n/k=1:n.|>j->((s=" "^(k-j))*" "*"*"^(2j-1)*s)^(1+(n==k))
~n=n./[2n,n].|>l->println.(l)

Try it online!

Golfing David Scholz's answer (cannot comment under other people solutions yet!)

Basically got rid of join by nesting the println call. println also gives you trailing \n for free. Finally I removed i iteration index as it can be expressed in terms of j

answered Feb 17, 2022 at 11:21

amelies

6513 silver badges7 bronze badges

1

\$\begingroup\$ This is really clever! (I'll upvote in 11 hours, I've used mine all :p) \$\endgroup\$
– David Scholz
Feb 17, 2022 at 12:42
\$\begingroup\$ aha thanks! :D also, the other thing to point out is that △ counts for 3 bytes, simply replacing it saved 6 bytes \$\endgroup\$
– amelies
Feb 17, 2022 at 13:00
\$\begingroup\$ Yes indeed. I didn't know that one could remove the brackets for two arguments as well! \$\endgroup\$
– David Scholz
Feb 17, 2022 at 16:35
2

\$\begingroup\$ you can even do it for three arguments tio \$\endgroup\$
– amelies
Feb 17, 2022 at 18:47

Add a comment |

ophact · Accepted Answer · 2022-02-11 14:57:53Z

1

Python 3.8 (pre-release), 96 bytes

lambda n:'\n'.join(' '*(d:=2*n-1-i)+(c:='*'*(1+i%n*2))+('  '*d+' '+c)*(i>=n)for i in range(n+n))

Try it online!

answered Feb 11, 2022 at 14:57

ophact

3,1441 gold badge5 silver badges24 bronze badges

\$\begingroup\$ 90 bytes \$\endgroup\$
– AnttiP
Feb 11, 2022 at 18:34

Add a comment |

Neil · Accepted Answer · 2022-02-12 01:13:28Z

Retina 0.8.2, 54 bytes

.+
$**
.
$._$*_$.'$* $`*$`$.'$* ¶
_+
$&$%'¶$%' 
_
 
O`

Try it online! Link includes test cases. Explanation:

.+
$**

Convert the input to a string of asterisks.

.
$._$*_$.'$* $`*$`$.'$* ¶

Create a triangle of asterisks indented by underscores. $._$*_ is the string of asterisks replaced by underscores. $.'$* is the space padding on either side of the triangle. $`*$` is the triangle.

_+
$&$%'¶$%'

Duplicate each line. $&$%' ends up keeping the first line the same, while the $%' causes the second copies of the lines to have an extra copy of the triangle, but without the indent.

Convert the underscores back to spaces.

O`

Sort the lines into order. This puts the lines with the most spaces first, thus grouping the top triangle's lines back together (and also the remaining lines of the bottom double triangle of course).

Graham · Accepted Answer · 2022-02-12 12:42:05Z

1

APL+WIN, 41 bytes

Prompts for n

(⌽0 1↓m),m←⊃(-⍳2×n)↑¨(m,¯1+2×m←⍳n←⎕)⍴¨'*'

Try it online! Thanks to Dyalog Classic

answered Feb 12, 2022 at 12:42

Graham

5,52614 silver badges11 bronze badges

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math scat · Accepted Answer · 2022-02-13 12:48:36Z

Vyxal `C`, 35 bytes

λhd›×*;→ẋė:£ƛ←†;¥ƛ←†:?d‹nhd-ð*JȮJ;J

Try it Online!

Explanation

λhd›×*;→  Simple Lambda
λ         Start lambda
 h        Get head
  d›      Double and add 1
    ×*    Repeat it "*" time
      ;→  Close lambda and set it to var.

ẋė:£ƛ←†;  Top triangle
ẋė        Make an empty list with length n and generate tuple (index, item)
  :£      Duplicate tuple and set it to the register
    ƛ     Map through tuple of (index, item)
     ←†;  Get var, call it and close lambda

¥ƛ←†:?d‹nhd-ð*JȮJ;J  Bottom left ans right triangle
¥ƛ                   Get register (index, item) and start a mapping lambda
  ←†:                Get var, call it and duplicate
     ?d‹             Double n and increment by one
        hd-          Get head, double and decrement by one
           ð*        Repeat space that many times
             J       Join the left triangle row with spaces
              ȮJ     Get second to last item (left triangle row) and join
                ;J   Join the top triangle with the bottom triangles

C flag centers everything with spaces.

David Scholz · Accepted Answer · 2022-02-17 03:22:04Z

Julia, 129 bytes

l="\n"
△(n,k)=((i,j)->(" "^(k-j+1)*"*"^i*" "^(k-j))^(n==k ? 2 : 1)).(1:2:2n,1:n)
~n=print(join(△(n,2n),l),l,join(△(n,n),l))

Try it online!

I had a hard time with this one, any optimizations are highly appreciated!

The function △ takes two arguments n,k whereby n is the size of the triangle. The parameter k determines the number of whitespaces for each row of the triangle.

I have not found a way to get rid of the join operator, which joins the array of strings with \n as a separator. This costs me 8 bytes already.

MarcMush · Accepted Answer · 2022-02-17 20:05:34Z

1

Julia 1.0, 58 bytes

!n=1:2n.|>i->((s=' '^(2n-i))*'*'^2(~-i%n)*"* $s")^-~+(i>n)

Try it online!

output is a list of strings

edited Feb 17, 2022 at 20:05

answered Feb 17, 2022 at 20:00

MarcMush

6,26511 silver badges16 bronze badges

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alephalpha · Accepted Answer · 2022-02-23 11:19:36Z

1

Pari/GP, 79 bytes

n->for(i=1,m=2*n,print(Strchr([32+10*(i<=n||i+abs(j)>m&&i>-j)|j<-[1-m..i-1]])))

Try it online!

edited Feb 23, 2022 at 11:19

answered Feb 11, 2022 at 14:25

alephalpha

46.7k7 gold badges61 silver badges172 bronze badges

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naffetS · Accepted Answer · 2022-05-29 02:41:50Z

Vyxal `C`, 12 bytes

d~∷×*øĊDZvṄJ

Try it Online!

How?

d~∷×*øĊDZvṄJøĊ⁋
d               # Double the (implicit) input
 ~∷             # Filter-keep for only odd numbers (the number is implicitly converted into a range [1,n])
   ×*           # Repeat an asterisk that many times for each
     øĊ         # Pad each with spaces to center
       D        # Triplicate
        ZvṄ     # Zip the top two copies together, and join by spaces
           J    # Join these two lists together
                # C flag centers and joins by newlines

Stack Exchange Network

Triangle Tripler

Task

Test Cases:

Scoring:

19 Answers 19

Python 3, 74 bytes

Explanation

Canvas, 11 bytes

Explanation:

Charcoal, 12 bytes

05AB1E, 20 16 bytes

Excel, 94 bytes

J, ³⁹ 37 bytes

how

Python 3.8 (pre-release), 114 bytes

Less-golfed version

Ruby, 70 bytes

JavaScript (ES6), 80 bytes

Patterns

Python NumPy, 95 bytes

Old Python NumPy, 99 bytes

Julia 1.0, 86 bytes

Python 3.8 (pre-release), 96 bytes

Retina 0.8.2, 54 bytes

APL+WIN, 41 bytes

Vyxal `C`, 35 bytes

Explanation

Julia, 129 bytes

Julia 1.0, 58 bytes

Pari/GP, 79 bytes

Vyxal `C`, 12 bytes

How?

Your Answer

Not the answer you're looking for? Browse other questions tagged
code-golf
string
ascii-art
or ask your own question.

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Hot Network Questions

Triangle Tripler

Task

Test Cases:

Scoring:

19 Answers 19

Python 3, 74 bytes

Explanation

Canvas, 11 bytes

Explanation:

Charcoal, 12 bytes

05AB1E, 20 16 bytes

Excel, 94 bytes

J, 39 37 bytes

how

Python 3.8 (pre-release), 114 bytes

Less-golfed version

Ruby, 70 bytes

JavaScript (ES6), 80 bytes

Patterns

Python NumPy, 95 bytes

Old Python NumPy, 99 bytes

Julia 1.0, 86 bytes

Python 3.8 (pre-release), 96 bytes

Retina 0.8.2, 54 bytes

APL+WIN, 41 bytes

Vyxal C, 35 bytes

Explanation

Julia, 129 bytes

Julia 1.0, 58 bytes

Pari/GP, 79 bytes

Vyxal C, 12 bytes

How?

Your Answer

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Not the answer you're looking for? Browse other questions tagged code-golfstringascii-art or ask your own question.

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Hot Network Questions

J, ³⁹ 37 bytes

Vyxal `C`, 35 bytes

Vyxal `C`, 12 bytes

Not the answer you're looking for? Browse other questions tagged
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string
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or ask your own question.