Output Rhombus Sequence without Nested Loops [duplicate]

Question

The following is the rhombus sequence.

1
121
12321
1234321
123454321
12345654321
1234567654321
123456787654321
12345678987654321

Your task is to output this, taking no input and using standard output methods (STDOUT, list of numbers or list of strings, etc.).

There's a hitch. You may only use up to one loop - no nested loops. Recursion counts as a loop, so you can either recurse or loop, not both. Non-standard languages will be dealt with on a case-by-case basis.

This is code-golf, so shortest program in bytes wins. This is kolmorgorov-complexity, so you may not take any input whatsoever.

Answer 1

Mathematica, 23 bytes

((10^#-1)/9)^2&~Array~9

Try it online!

Answer 2

05AB1E, 7 bytes

Only one for-loop here

TG1N×n,

Uses the 05AB1E encoding. Try it online!

Explanation:

TG        # For N in range(1, 10):
  1N×     # Push a string of N 1's
     n    # Square that number
      ,   # Pop and print with a newline

Answer 3

Python 2, 32 31 bytes

^{-1 byte thanks to officialaimm}

o=0
exec"o=o*10+1;print o*o;"*9

Try it online!

Answer 4

Jelly, 5 bytes

9ŒḄ€Y

Try it online!

This only uses one mapping loop (€) since ŒḄ (palindromize) is a builtin already. Explanation:

9ŒḄ€Y
9     9
   €  Map (our looping construct)
 ŒḄ     Palindromize (make range)
    Y Join by newlines.

Answer 5

Python 2, 42 40 38 35 bytes

-2 thanks to @officialaimm

I ported Jenny_mathy's Mathematica answer to Python, because the arithmetic approach is much shorter:

i=1;exec"print(10**i/9)**2;i+=1;"*9

Try it online!

Exaplanation

• i=1 - Initializes a variable i to 1.

• exec"..."*9 - Executes ... 9 times.

• print(10**i/9)**2 - Prints 10 ⁱ and divides the result by 9, which then gets sqaured.

• i+=1 - Increment i and execute again.

---

Python 2, 37 bytes

for i in range(9):print(10**-~i/9)**2

Try it online!

---

Python 2, 63 bytes

This is the initial version.

l='123456789'
for i in l:s=l.index(i);print l[:s]+l[:s+1][::-1]

Try it online!

Answer 6

SOGL V0.12, 8 7 bytes

9∫Δø∑ΓO

Try it Here!

Explanation:

9∫       do 9 times, pushing 1-indexed counter
  Δ        get 1-indexed (last inclusive) range
   ø∑      join together
     Γ     palindromise
      O    output it

Answer 7

Dyalog APL, 20 bytes

⍪(' '~⍨∘⍕⍳,1↓⌽∘⍳)¨⍳9

Try it online!

How?

¨⍳9 - for each in range 1..9

    ⍳, - produce the range of this number

    1↓⌽∘⍳ - appended to itself reversed without the last item

    ⍕ - format to string

    ' '~⍨ - remove spaces

⍪ - output vertically

17 bytes, doesn't work because of precision issues

⌽⍕×⍨⍪(10⊥1⍴⍨⊢)¨⍳9

This one uses the i ones squared identity, but replaces the first one on the last line with a 0, because the decoding rounds it up.

Answer 8

Brachylog, 15 bytes

9⟦₁{⟦₁⟨kc↔⟩cẉ}ᵐ

Try it online!

Yeah this doesn't feel much declarative...

Answer 9

C#, 101 bytes

using System.Linq;_=>new int[9].Select((n,i)=>"123456789".Substring(0,i+1)+"87654321".Substring(8-i))

Try it online!

---

Or the loop approach for 111 bytes:

_=>{var a=new string[9];for(int i=0;i<9;)a[i]="123456789".Substring(0,++i)+"87654321".Substring(9-i);return a;}

Try it online!

Answer 10

Vim + coreutils, 42 bytes

i1⏎12<Esc>qqYp$ylp<C-a>q6@qggqqjYpv$!rev⏎kJd2lq7@q

Try it online!

Ungolfed/Explanation

i1⏎12<Esc>                                          " start with lines 1,12 in the buffer
          qq           q6@q                         " record macro and run it 6 times:
            Yp                                      "  - duplicate line
              $ylp                                  "  - duplicate last character
                  <C-a>                             "  - increment it
                           gg                       " go to the beginning (buffer is now: 1,12,123,...,12..89)
                             qq               q7@q  " record macro and run it 6 times:
                               jYp                  "  - go to line below and duplicate it
                                  v$!rev⏎           "  - mark it and reverse it
                                         kJ         "  - join the line above with the current one
                                           d2l      "  - remove the space and character that are too much

Answer 11

Haskell, 39 32 bytes

No for loops/recursion at all, makes use of list comprehension:

[[1..x]++[x-1,x-2..1]|x<-[1..9]]

Try it online!

Answer 12

Scala, 69 bytes

There is only one "loop", when I iterate over (1 to 9). But you could consider that there are loops in reverse and Range instanciations.

(1 to 9).map(x=>println(((1 to x)++(1 to x-1).reverse).mkString("")))

Try it online!

Answer 13

AHK, 55 bytes

a=1
Loop,9{ 
b:=a**2
Send %b%`n
a:=a 1
}
Send {VK08 2}1

This was too stupid of a result for me to not post. AHK messes rounds the last output to 12345678987654320 so we backspace twice and add a 1.

Answer 14

Pyth, 11 bytes

jm^s*d\12S9

Try it online!

---

How does this work?

 m       S9         - Map over [1...9]
  ^s    2           - The square of the integer formed by:
    *d\1            - the string with n consecutive 1s.
j                   - Join by newlines

Answer 15

Google Sheets, 66 bytes

=ArrayFormula(Join("
",REPT(1,ROW(A1:A8))^2,"12345678987654321"))

No loops, just a matrix. It would only be 46 bytes but Sheets only has 15 decimal points of precision so the last number is displayed as 12345678987654300 or 1.23457E+1621 so I had to add it manually.

Answer 16

R, 18 bytes

cumsum(10^(0:8))^2

Returns the values.

Try it online!

Due to floating point precision issues, the last number is 12345678987654320 which is equal to 12345678987654321 in R, as the footer shows. In the header, it sets the options to get the numbers to print in non-scientific notation.

Answer 17

Kotlin, 46 bytes

(1..9).map{"1".repeat(it).toLong()}.map{it*it}

Answer 18

QBIC, 17 bytes

[9|?((z^a-1)/9)^2

From oeis.

Answer 19

Swift, 76 bytes

import Foundation
for i in 1...9{print(Int(pow((pow(10,Double(i))-1)/9,2)))}

Try it online!

If printing as an array of lines is allowed, here is 74 byte approach:

import Foundation
print((1...9).map{Int(pow((pow(10,Double($0))-1)/9,2))})

Try it online!

Answer 20

Perl 5, 22 bytes

say((1x$_)**2)for 1..9

Try it online!

Using Jenny_mathy's formula to save a lot of bytes.

Answer 21

PHP, 38 37 bytes

for(;$i<9**8;)echo(($i.='1')**2)."
";

with warnings

PHP, 42 41 bytes

for($i=0;$i<9**8;)echo(($i.='1')**2)."
";

without warnings

Answer 22

Javascript 66 bytes

Returns an array of strings representing the numbers in the sequence (necessary because 12345678987654321 is too large to be accurately represented as an int in Javascript)

o=>Array(9).fill``.map(x=>(x+(n=n*10+1)*n).replace(0,1),n=‌​0)

Explanation

o=>                               // o is unused input (saves 1 byte vs ())
    Array(9)                      // create an array of length 9
        .fill``                   // fill it with empty strings
                                  // we must fill it with something, because map skips
                                  // empty entries
        .map(                     // loop over the array and return a new array with the
                                  // values returned by the callback
            x=>(                  // x is the element in the array
                (x+(n=n*10+1)*n)  // make n 1, 11, 111, 1111, etc, and get the string
                                  // value of n^2
                    .replace(0,1) // replace 0's with 1's (necessary for
                                  // 12345678987654321, because
                                  // 111111111^2 = 12345678987654320 in javascript)
            ),
            n=0                   // n^2 is the rhombus number at each stage
        )