Fibonacci triangle

Question

Given an integer n, output the following ASCII art to n rows:

Essentially, the first row is 1+1=2 and the nth row (1-indexed) is \$f_n + f_{n+1} = f_{n+2}\$ where \$f\$ is the Fibonacci sequence, padded so the numbers line up with the previous row.

You may instead output it infinitely. You may output a list of lines.

This is code-golf, shortest wins!

Testcase

The output for 20 should be:

1+1=2
  1+2=3
    2+3=5
      3+5=8
        5+8=13
          8+13=21
            13+21=34
               21+34=55
                  34+55=89
                     55+89=144
                        89+144=233
                           144+233=377
                               233+377=610
                                   377+610=987
                                       610+987=1597
                                           987+1597=2584
                                               1597+2584=4181
                                                    2584+4181=6765
                                                         4181+6765=10946
                                                              6765+10946=17711

@DarrelHoffman I mean constant extra leading whitespace, like this: gist.github.com/pxeger/768a57edc59ce03a673d1d6a943ff0d7 — pxeger
– pxeger, Commented Mar 7, 2022 at 18:12

dingledooper · Accepted Answer · 2022-03-04 21:46:38Z

7

Python 2, 67 bytes

Outputs the sequence indefinitely.

a=b=l=1
while 1:print'%*d+%d='%(l,a,b)+`a+b`;l-=~len(`b`);a,b=b,a+b

Try it online!

answered Mar 4, 2022 at 21:46

dingledooper

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Arnauld · Accepted Answer · 2022-03-06 10:49:57Z

JavaScript (ES8), 79 bytes

f=(n,A=B=1,p)=>n?''.padEnd(p)+A+`+${B}=${B+=A}
`+f(n-1,B-A,(A+"").length-~p):''

Try it online!

Commented

f = (                   // f is a recursive function taking:
  n,                    //   n = input
  A = B = 1,            //   A, B = Fibonacci variables
  p                     //   p = padding length, initially undefined
) =>                    //
n ?                     // if n is not equal to 0:
  ''.padEnd(p) +        //   append p spaces (none if p is undefined)
  A +                   //   followed by A
  `+${B}=${B += A}\n` + //   followed by "+[B]=[B+A]\n" (A is added to B)
  f(                    //   followed by the result of a recursive call:
    n - 1,              //     decrement n
    B - A,              //     update A to the previous value of B
    (A + "")            //     add the length of A coerced to a string + 1
    .length - ~p        //     to p
  )                     //   end of recursive call
:                       // else:
  ''                    //   stop the recursion

att · Accepted Answer · 2022-03-08 17:28:04Z

5

C (gcc), 71 66 bytes

-5 thanks to Sisyphus

i,j;main(k){for(;;)printf("%*d+%d%n=%d\n",i,j=k-j,k+=j,&i,j+k+k);}

Try it online!

Outputs indefinitely.

edited Mar 8, 2022 at 17:28

answered Mar 4, 2022 at 23:01

att

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\$\begingroup\$ 66 bytes using %n: Try it online! \$\endgroup\$

Sisyphus
– Sisyphus

2022-03-08 11:47:34 +00:00
Commented Mar 8, 2022 at 11:47

Add a comment |

Neil · Accepted Answer · 2022-03-04 22:15:53Z

4

Charcoal, 25 bytes

≔Ｅ²¦¹ηＦＮ«Ｉ⌊η+≔⟦⌈ηΣη⟧η⟦⪫η=

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

≔Ｅ²¦¹η

Start with two 1s.

ＦＮ«

Repeat n times.

Ｉ⌊η

Output the first element of the old pair.

Output a +.

≔⟦⌈ηΣη⟧η

Replace the pair with the second element and their sum.

⟦⪫η=

Output the new pair joined with = and move the cursor down.

answered Mar 4, 2022 at 22:15

Neil

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DLosc · Accepted Answer · 2022-03-04 23:52:03Z

Pip, 30 bytes

W P[xi::o'+o+:i'=i+o]x.:sX#i+1

Outputs forever. Attempt This Online!

Explanation

W P[xi::o'+o+:i'=i+o]x.:sX#i+1
                                i is 0, o is 1, x is "", s is " " (implicit)
   [                ]           Put the following in a list:
    x                            The indent: x
     i::o                        Swap i (the smaller number) with o (the larger number)
                                 and return the new value of i
         '+                      Plus sign
           o+:i                  Add i to o in-place and return the new value of o
               '=                Equals sign
                 i+o             Add i and o
  P                             Print the list (concatenating its elements by default)
W                               Loop while the list is truthy (which is always):
                          #i     Length of i
                            +1   Plus 1
                        sX       That many spaces
                     x.:         Concatenate with x in-place

loopy walt · Accepted Answer · 2022-03-05 04:25:35Z

4

Python 3.8 (pre-release), 66 bytes @emanresu A

a=b=l=1
while l:=len(x:=f"{a:>{l}}+{b}=%d")-3:a,b=b,a+b;print(x%b)

Try it online!

Old Python 3.8 (pre-release), 67 bytes

a=b=l=1
while l:=len(x:=f"{a:>{l}}+{b}"):a,b=b,a+b;print(x+f'={b}')

Try it online!

Based on @dingledooper's Python 2 answer.

edited Mar 5, 2022 at 4:25

answered Mar 5, 2022 at 3:39

loopy walt

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1

\$\begingroup\$ You can save a byte with some cursed formatting :) \$\endgroup\$

emanresu A
– emanresu A

2022-03-05 04:12:49 +00:00
Commented Mar 5, 2022 at 4:12

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Giuseppe · Accepted Answer · 2022-03-05 00:10:27Z

3

SNOBOL4 (CSNOBOL4), 90 87 84 bytes

	N =1
N	N =M + (M =N)
	OUTPUT =P =DUPL(' ',X) M '+' N '=' M + N
	P @X N '='	:(N)
END

Try it online!

Outputs infinitely, but experiences integer overflow after a short while.

edited Mar 5, 2022 at 0:10

answered Mar 4, 2022 at 21:06

Giuseppe

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Jonathan Allan · Accepted Answer · 2022-03-05 16:36:34Z

Jelly, 22 bytes

+2ÆḞ€DżṖ©;⁾+=Ɗ⁶ṁ$®¦F)Y

A full program that accepts an integer and prints the result.

Try it online!

How?

+2ÆḞ€DżṖ©;⁾+=Ɗ⁶ṁ$®¦F)Y - Main Link: integer, N
                    )  - for each V in [1..N]:
+2                     -   V add two
  ÆḞ€                  -   Fibbonacci of each of [1..V+2]
     D                 -   to decimal digits -> [[1],[1],[2],...,Digits(Fib(V+2))]
                                                call this FibDigits
             Ɗ         -   last three links as a monad - f(V):
       Ṗ               -     pop -> [1..V-1]
        ©              -       (copy this to the register)
          ⁾+=          -     ['+', '=']
         ;             -     concatenate -> [1,2,3,...,V-1,'+','=']
                                            call this Fillers
      ż                -   FibDigits zip with Fillers
                  ¦    -   sparse application...
                 ®     -   ...to indices: recall from register -> [1..V-1]
                $      -   ...action: last two links as a monad:
              ⁶        -     space character
               ṁ       -     mould like (e.g. [[1,4,4],12] -> [[' ',' ',' '], ' ']
                   F   -   flatten
                     Y - join with newlines
                       - implicit print

emanresu A · Accepted Answer · 2022-03-05 21:45:06Z

Vyxal, 19 bytes

ÞF3l(:n‛+=Y∑꘍,nhL›+

Try it Online!

Outputs infinitely.

ÞF3l                # Take the infinite list of Fibonacci numbers, and get overlapping groups of 3
    (               # Looping over that...
          Y∑        # Interleave...
      n             # The tuple of three numbers
       ‛+=          # With '+='
     :      ꘍       # Pad that with the correct amount of spaces, without popping the padding amount
             ,      # Print that
                  + # Add to the padding amount (initially 0)
              nhL›  # The length of the first number in the tuple, plus one.

lyxal · Accepted Answer · 2022-03-04 23:30:00Z

2

Vyxal `j`, 30 bytes

ÞFẎ3l⟑‛+=fY∑¥$꘍:\=ḟnḢvL₌t¯-h-£

Try it Online!

A big cursed mess of formatting and registering.

answered Mar 4, 2022 at 23:30

lyxal♦

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Neil · Accepted Answer · 2022-03-05 10:12:36Z

Retina 0.8.2, 55 bytes


1=1
{`.*\+
$.&$* 
=
+
\d+.(\d+)
$&=$&$*_$1$*_
:`_+
$.&

Try it online! Tries to output indefinitely but runs out of memory after about 50 seconds on TIO. Explanation:

1=1

Initialise the buffer with a hypothetical previous line.

{`

Repeat indefinitely.

.*\+
$.&$*

Replace up to and including the + with spaces. (This only applies after the first loop.)

=
+

Change the = into a +.

\d+.(\d+)
$&=$&$*_$1$*_

Append the sum of the two values in unary.

:`_+
$.&

Convert the unary to decimal and output the result.

pxeger · Accepted Answer · 2022-03-05 17:25:31Z

2

Python, ^{117 107 103 99} 88 bytes

p="1=1"
while p:=" "*(a:=1+p.find("+"))+eval("f'{%s=}'"%p[a:].replace("=","+")):print(p)

Attempt This Online!

Outputs infinitely.

Can probably be much improved.

edited Mar 5, 2022 at 17:25

answered Mar 4, 2022 at 21:12

pxeger

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\$\begingroup\$ How do you use ":=" in a lambda? \$\endgroup\$

Fmbalbuena
– Fmbalbuena

2022-03-04 21:16:15 +00:00
Commented Mar 4, 2022 at 21:16
1

\$\begingroup\$ @Fmbalbuena What do you mean by "how"? It's the same as anywhere else \$\endgroup\$

pxeger
– pxeger

2022-03-04 21:19:15 +00:00
Commented Mar 4, 2022 at 21:19

Add a comment |

Kevin Cruijssen · Accepted Answer · 2022-03-05 18:51:37Z

05AB1E, 23 22 bytes

ÌLÅfü3ε„+=.ιJ¯Oú,yнg>ˆ

-1 byte after being inspired by @JonathanAllan's Jelly answer

Outputs the first \$n\$ lines.

Try it online.

Or a minor 22 bytes alternative:

ÌLÅfDü3„+=δ.ιJs€g>.¥ú»

Try it online.

And the infinite sequence would be 22 bytes as well:

∞Åfü3vy„+=.ιJ¯Oú,yнg>ˆ

Try it online.

Explanation:

Ì              # Increase the (implicit) input-integer by 2
 L             # Pop and push a list in the range [1,input+2]
  Åf           # Get the 0-based n'th Fibonacci value for each of these
    ü3         # Pop and push all overlapping triplets of this list
      ε        # Foreach over the triplets:
               #  (implicitly push the current triplet)
       „+=.ι   #  Intersperse it with "+" and "=" delimiters
            J  #  Join this list together to a string
       ¯       #  Push the global_array (empty by default)
        O      #  Sum it together
         ú     #  Pad the string with that many leading spaces
          ,    #  Pop and output this line with trailing newline
       y       #  Push the pair again
        н      #  Pop and push its first item
         g     #  Pop and push its length
          >    #  Increase it by 1
           ˆ   #  Pop and add it to the global_array

You might be able to save bytes on the infinite version by porting my Vyxal answer and keeping the padding amount on the stack? — emanresu A
– emanresu A, Commented Mar 5, 2022 at 21:47
@emanresuA I'm afraid not. It would be 22 bytes as well. 05AB1E remembers the last value if it needs an argument and the stack is now empty, so I will have to push an explicit 0 at the start (otherwise it will use the infinite overlapping triplets list for the padding). In addition, I need an explicit Duplicate within the loop. That, combined with the 1 byte longer Fibonacci builtin compared to your Vyxal answer, is an alternative infinite 22-byter. — Kevin Cruijssen
– Kevin Cruijssen, Commented Mar 5, 2022 at 22:08

pajonk · Accepted Answer · 2022-03-05 20:53:11Z

2

R, 76 bytes

l=b=1
repeat{cat(sprintf("%*d+%d=%d
",l,T,b,s<-T+b));l=l+nchar(b)+1;T=b;b=s}

Try it online!

Outputs infinitely.

If leading whitespace is allowed:

R, 74 69 bytes

b=1
repeat cat(sprintf("%*d+%d=%d
",F<-F+nchar(+T)+1,T,b,b<-T+(T=b)))

Try it online!

edited Mar 5, 2022 at 20:53

answered Mar 5, 2022 at 20:16

pajonk

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Neil · Accepted Answer · 2022-03-04 22:03:39Z

1

Retina, 54 bytes

K`1+1=2
"$+"+`(\d+)=(\d+)$
$&¶$.%`* $1+$2=$.(*_$2*
A`$

Try it online! No test suite because of the way the program uses history. Explanation:

K`1+1=2

Replace the input with the first line of the output.

"$+"+`

Repeat n times.

(\d+)=(\d+)$

Match fₙ=fₙ₊₁ from the previous line.

$&¶$.%`* $1+$2=$.(*_$2*

Append fₙ+fₙ₊₁= plus their sum, with the correct amount of indent.

A`$

Since the program started with one line and appended n lines, it now has one line too many, so delete the last line.

answered Mar 4, 2022 at 22:03

Neil

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sinvec · Accepted Answer · 2022-03-05 09:07:36Z

1

Bash, ^{139 . . . 104} 103 bytes

_()($[a=b=1];for n in `seq $1`;{ printf "%${s}s";echo -n $a+$b;$[c=a+b,s+=~${#a},a=b,b=c];echo "=$c";})

Try it online!

edited Mar 5, 2022 at 9:07

answered Mar 5, 2022 at 5:13

sinvec

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\$\begingroup\$ Is all that whitespace necessary? \$\endgroup\$

emanresu A
– emanresu A

2022-03-05 05:14:02 +00:00
Commented Mar 5, 2022 at 5:14
\$\begingroup\$ @emanresuA not sure :) \$\endgroup\$

sinvec
– sinvec

2022-03-05 05:14:38 +00:00
Commented Mar 5, 2022 at 5:14
\$\begingroup\$ Removed \$\endgroup\$

emanresu A
– emanresu A

2022-03-05 05:17:11 +00:00
Commented Mar 5, 2022 at 5:17

Add a comment |

AZTECCO · Accepted Answer · 2022-03-05 15:36:09Z

1

Haskell, 89 bytes

l!s@[a,b,c]|[x,y,z]<-show<$>s=(l++x++'+':y++'=':z):(' '<$' ':x++l)![b,c,b+c]
f=""![1,1,2]

Try it online!

f is a stream of lines.
l!s formats current line and computes next triplet s and leading whitespace l

answered Mar 5, 2022 at 15:36

AZTECCO

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sinvec · Accepted Answer · 2022-03-05 19:29:54Z

1

CJam, ^{49 47} 45 bytes

liX:Y;{[ST*X`_,T)+:T;'+Y:K'=XY+:YN]oK:X;(_}g;

Try it online!

edited Mar 5, 2022 at 19:29

answered Mar 5, 2022 at 19:14

sinvec

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Noodle9 · Accepted Answer · 2022-03-06 17:06:46Z

1

C (gcc), ^{102 76} 72 bytes

b;s;f(a){for(a=b=1;s=printf("%*d+%d",s,a,b);a=b-a)printf("=%d\n",b+=a);}

Try it online!

Saved a whopping 26 bytes thanks to ceilingcat!!!
Saved 3 bytes thanks to m90!!!

Outputs forever!

edited Mar 6, 2022 at 17:06

answered Mar 4, 2022 at 21:18

Noodle9

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\$\begingroup\$ A further improvement \$\endgroup\$

m90
– m90

2022-03-06 12:43:18 +00:00
Commented Mar 6, 2022 at 12:43
\$\begingroup\$ @m90 Nice one - thanks! :D \$\endgroup\$

Noodle9
– Noodle9

2022-03-06 17:07:05 +00:00
Commented Mar 6, 2022 at 17:07

Add a comment |

MarcMush · Accepted Answer · 2023-06-29 22:30:59Z

1

Julia 1.0, ⁶⁶ 65 bytes

f(a=1,b=1,n=0)=print(" "^n,"$a+$b=$(a+b)
")f(b,a+b,n-~ndigits(a))

Try it online!

prints infinitely. To avoid Int64 overflow, replace a=1 with a=big(1) (+5 bytes)

edited Jun 29, 2023 at 22:30

answered Mar 6, 2022 at 14:22

MarcMush

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Fmbalbuena · Accepted Answer · 2022-03-04 21:17:49Z

0

Python 3, 146 141 bytes

print("1+1=2")
a=[1,1,2]
w=2
while 1:print(" "*w+str(a[1])+"+"+str(a[2])+"="+str(a[1]+a[2]));del a[0];w+=len(str(a[0]))+1;a.append(a[0]+a[1])

Try it online!

edited Mar 4, 2022 at 21:17

answered Mar 4, 2022 at 21:12

Fmbalbuena

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Shaggy · Accepted Answer · 2023-06-30 15:28:18Z

Japt `-mR`, 23 22 bytes

Includes a trailing space on each line of output.

"+= "¬ËiMg°UÃvÈùT±XÊÃ¬

Try it

"+= "¬ËiMg°UÃvÈùT±XÊÃ¬     :Implicit map of each U in the range [0,input)
"+= "¬                     :  Split "+= " to an array of characters
      Ë                    :  Map
       i                   :    Prepend
          °U               :      Prefix increment U
        Mg                 :      Get the Uth Fibonacci number, 0-indexed
            Ã              :  End map
             v             :  Modify the first element
              È            :  By passing it through the following function as X
               ù           :    Left pad with spaces to length
                T±         :      Increment T (initially 0) by
                  XÊ       :        Length of X
                    Ã      :  End modification
                     ¬     :  Join
                           :Implicit output joined with newlines

138 Aspen · Accepted Answer · 2023-07-15 01:43:26Z

Scala, 228 226 225 bytes

Saved 3 bytes thanks to the comment of @ceilingcat

Golfed version. Try it online!

object Main extends App{def f(n:Int)={def g(a:Int,b:Int,n:Int):List[(Int,Int,Int)]=n match{case 0=>Nil case _=>(a,b,a+b)::g(b,a+b,n-1)};g(1,1,n)};f(20).zipWithIndex.map{case((a,b,c),i)=>("  "*i)+s"$a+$b=$c"}.foreach(println)}

Ungolfed version. Try it online!

object Main extends App {
  
  def formatList(list: List[(Int, Int, Int)]): List[String] = {
    list.zipWithIndex.map { 
      case ((a, b, c), i) => (" " * (2 * i)) + s"$a + $b = $c"
    }
  }
  
  def fibonacciSeq(n: Int): List[(Int, Int, Int)] = {
    def fibHelper(a: Int, b: Int, n: Int): List[(Int, Int, Int)] = n match {
      case 0 => Nil
      case _ => (a, b, a + b) :: fibHelper(b, a + b, n - 1)
    }
    fibHelper(1, 1, n)
  }
  
  val formatted = formatList(fibonacciSeq(20))
  formatted.foreach(println)
}

It looks like 20 is hardcoded into this; you must be able to take the value as input, or output infinitely. — Shaggy
– Shaggy, Commented Jul 1, 2023 at 8:33

Stack Exchange Network

Fibonacci triangle

Testcase

23 Answers 23

Python 2, 67 bytes

JavaScript (ES8), 79 bytes

Commented

C (gcc), 71 66 bytes

Charcoal, 25 bytes

Pip, 30 bytes

Explanation

Python 3.8 (pre-release), 66 bytes @emanresu A

Old Python 3.8 (pre-release), 67 bytes

SNOBOL4 (CSNOBOL4), 90 87 84 bytes

Jelly, 22 bytes

How?

Vyxal, 19 bytes

Vyxal `j`, 30 bytes

Retina 0.8.2, 55 bytes

Python, ^{117 107 103 99} 88 bytes

05AB1E, 23 22 bytes

R, 76 bytes

R, 74 69 bytes

Retina, 54 bytes

Bash, ^{139 . . . 104} 103 bytes

Haskell, 89 bytes

CJam, ^{49 47} 45 bytes

C (gcc), ^{102 76} 72 bytes

Julia 1.0, ⁶⁶ 65 bytes

Python 3, 146 141 bytes

Japt `-mR`, 23 22 bytes

Scala, 228 226 225 bytes

Linked

Hot Network Questions

Fibonacci triangle

Testcase

23 Answers 23

Python 2, 67 bytes

JavaScript (ES8), 79 bytes

Commented

C (gcc), 71 66 bytes

Charcoal, 25 bytes

Pip, 30 bytes

Explanation

Python 3.8 (pre-release), 66 bytes @emanresu A

Old Python 3.8 (pre-release), 67 bytes

SNOBOL4 (CSNOBOL4), 90 87 84 bytes

Jelly, 22 bytes

How?

Vyxal, 19 bytes

Vyxal j, 30 bytes

Retina 0.8.2, 55 bytes

Python, 117 107 103 99 88 bytes

05AB1E, 23 22 bytes

R, 76 bytes

R, 74 69 bytes

Retina, 54 bytes

Bash, 139 . . . 104 103 bytes

Haskell, 89 bytes

CJam, 49 47 45 bytes

C (gcc), 102 76 72 bytes

Julia 1.0, 66 65 bytes

Python 3, 146 141 bytes

Japt -mR, 23 22 bytes

Scala, 228 226 225 bytes

Linked

Related

Hot Network Questions

Vyxal `j`, 30 bytes

Python, ^{117 107 103 99} 88 bytes

Bash, ^{139 . . . 104} 103 bytes

CJam, ^{49 47} 45 bytes

C (gcc), ^{102 76} 72 bytes

Julia 1.0, ⁶⁶ 65 bytes

Japt `-mR`, 23 22 bytes