Print this diamond

Question

This question has been spreading like a virus in my office. There are quite a variety of approaches:

Print the following:

        1
       121
      12321
     1234321
    123454321
   12345654321
  1234567654321
 123456787654321
12345678987654321
 123456787654321
  1234567654321
   12345654321
    123454321
     1234321
      12321
       121
        1

Answers are scored in characters with fewer characters being better.

Answer 1

Javascript, 103 96 93 89 88 87

f=(i=[...Array(17)].map((x,i)=>i<8?i:i-2*(i-8)))=>i.map(x=>i.map(y=>x+y>7?(x+y-7):' ').join``).join`\n`

I haven't looked at the other answers up until this point, and just started out with this spreadsheet. Just saw that this answer isn't too bad of a golf compared to the other Javascript answers.

I tried to be clever coming up with something that generates a particular array (i) that I use for the x and y-axis. But it turns out, spelling out that array is shorter.

f=(i=[0,1,2,3,4,5,6,7,8,7,6,5,4,3,2,1,0])=>i.map(x=>i.map(y=>x+y>7?(x+y-7):' ').join``).join`\n`

Update. Spelling out the array differently, but I'm sure that I can golf this further

f=(i=[...'01234567876543210'])=>i.map(x=>i.map(y=>1*x+1*y>7?(1*x+1*y-7):' ').join``).join`\n`

Last update for tonight:

f=(i=[...'01234567876543210'])=>i.map(x=>i.map(y=>+x+1*y>7?+x+1*y-7:' ').join``).join`\n`

Getting rid of one more byte:

f=(i=[...'0'+10*11111111**2])=>i.map(x=>i.map(y=>+x+1*y>7?+x+1*y-7:' ').join``).join`\n`

And one more:

f=(i=[0,...11111111**2+'0'])=>i.map(x=>i.map(y=>+x+1*y>7?+x+1*y-7:' ').join``).join`\n`

Answer 2

Duocentehexaquinquagesimal, 58 bytes

(Úv×H.F°ǝ=ÉÈNŸ=~ʒ€
₁˨Dûl`N₆rÇX³ΣJd‘×∞*Åv€rØ(0՜֤ΔNā—St€ô

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Answer 3

FreeBASIC, 100 characters

Dim As Byte i,j,k
for i=-8 to 8 : for j=-8 to 9
k=abs(i)+abs(j)
if k > 8 then print " "; : else : ? chr(57-k);
next
? 
next 
Sleep
End

Answer 4

Perl 5, 33 bytes

say$"x abs.(1x(9-abs))**2for-8..8

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Using 11 * 11 = 121, 111 * 111 = 12321, 1111 * 1111 = 1234321, ...

3 bytes saved thanks to naffetS.

Answer 5

Go, 239 196 187 185 bytes

import(."fmt";."strings")
func g(i int){P:=Print;P(Repeat(" ",(9-i)))
for j:=1;j<=i;j++{P(j)}
for k:=i-1;k>0;k--{P(k)}
P("\n")}
func f(){
for i:=1;i<10;i++{g(i)}
for i:=8;i>0;i--{g(i)}}

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My first attempt at golfing in Go. This can definitely be shortened, but I don't know how.

Commented

import (."fmt"; ."strings")        // Import modules for later
func g(i int) {                    // Helper function g taking an int i
P := Print                         //  Useful function for later
P(Repeat(" ", (9 - i)))            //  Repeat a space 9-i times and print
for j := 1; j <= i; j++ { P(j) }   //  Print each j for j in [1 .. i]
for k := i-1; k > 0; k-- { P(k) }  //  Print each k for k in [i .. 1]
P("\n") }                          //  Print a newline
func f(){                          // Function boilerplate
for i := 1; i < 10; i++ { g(i) }   //  Call g for each i in [1 .. 9]
for i := 8; i > 0; i-- { g(i) }}   //  Call g for each i in [8 .. 1]

Answer 6

Java 8 (OpenJDK 8), 157 ¹⁶¹ bytes

-4 bytes thanks to @ceilingcat

Tried to use only 1 for instead of 2 for two-dimensional printing, but couldn't beat the best Java answers.

interface M{static void main(String[]b){for(int r,s,a=0;a<289;System.out.print((r<1?s<1?" ":"":r)+(s>7?"\n":"")))r=9-Math.abs(s=a%17-8)-Math.abs(8-a++/17);}}

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