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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.
Inspired by @BradC's answer
DECLARE @ INT=8a:PRINT
space(abs(@))+stuff('12345678987654321',9-abs(@),abs(@)*2,'')SET
@-=1IF~@<8GOTO a
(defn s[n](concat(range 1 n)(range n 0 -1)))(print(apply str(mapcat #(concat(repeat(- 9 %)" ")(s %)'("\n"))(s 9))))
Shortened from 127 to 121 to 115.
interface A{static void main(String[]v){for(long m=0,a=0;++a<18;System.out.printf("%"+(a>9?26-a:8+a)+"d%n",m*m))m=a>9?m/10:m*10+1;}}
This is a different way to do it than my first answer for Java. It is using integer arithmetics. It is even shorter, maybe the shortest possible.
Formatted version of this code:
interface A {
static void main(String[] v) {
for (long m = 0, a = 0; ++a < 18; System.out.printf("%" + (a > 9 ? 26 - a : 8 + a) + "d%n", m * m))
m = a > 9 ? m / 10 : m * 10 + 1;
}
}
Or more readable:
class A {
public static void main(String[] v) {
for (long m = 0, a = 0; ++a < 18; ) {
long width = a > 9 ? 26 - a : 8 + a;
m = a > 9 ? m / 10 : m * 10 + 1;
System.out.printf("%" + width + "d%n", m * m);
}
}
}
k=[]
for t in range(10):
l=str(111111111**2)[:t]
k.append(l.rjust(9)+l[:-1][::-1])
print("\n".join(k+k[:-1][::-1]))
interface A{static void main(String[]a){f(i->{var o=System.out;f(j->o.print(i+j>8?" ":""+(9-i-j)));o.println();});}static void f(java.util.function.IntConsumer c){java.util.stream.IntStream.range(-8,9).map(Math::abs).forEach(c);}}
Readable:
interface A {
static void main(String[] a) {
f(i -> {
var o = System.out;
f(j -> o.print(i + j > 8 ? " " : "" + (9 - i - j)));
o.println();
});
}
static void f(java.util.function.IntConsumer c) {
java.util.stream.IntStream.range(-8, 9).map(Math::abs).forEach(c);
}
}
9╒ñmÆ╒ñyFΩn
Explanation:
9╒ # Push a list in the range [1,9]
ñ # Palindromize it: [1,2,3,4,5,6,7,8,9,8,7,6,5,4,3,2,1]
m # Map over each inner integer,
Æ # using the following five character as inner code-block:
╒ # Convert the integer to a list in the range [1,n]
ñ # Palindromize it similar as before
y # Join the list together to a string
Ω # Prepend/append potential leading/trailing spaces (centralize)
F # to make the string length 17
n # After the map: join the strings in the list by newlines
# (after which the entire stack is output implicitly as result)
-5 bytes thanks to @ceilingcat
#define f(x,y);for(j=0;j++<(i<9?x:y);)printf(
j;D(i){for(i=0;++i<19;puts("")){f(9-i,i-9)" ")f(i,18-i)"%d",j)f(i-1,17-i)"%d",i<9?i-j:18-i-j);}}
This code first prints a certain amount of spaces and then the numbers.
i,j;D(){for(;++i<19;){for(j=0;++j<19;i<10?(i+j>9&&j-i<9?printf("%d",j<9?i+j-9:9-j+i):printf(" ")):i-j<9&&i+j<27?printf("%d",j<9?j-i+9:27-i-j):printf(" "));puts("");}}
This code always chooses whether to print a space or a number, depending on the values of i and j.
eval s+={1..9}';<<<${(l:8:)s%?}`rev<<<$s`;'>f;{<f;tac f}|uniq
eval s+={1..9}';<<<${(l:8:)s%?}`rev<<<$s`;'>f;{<f;tac f}|uniq
eval {1..9}' ;' # evaluate this 9 times
s+= ; # append the number to s
<<< # print
${ s } # s
%? # with the last character removed
(l:8:) # padded to 8 spaces
`rev<<<$s` # then s, reversed
>f; # all output to the file f
<f; # print f
tac f # print f in reverse
{ }|uniq # remove the duplicated line in the middle
```
for i=-8,8 do x=9-math.abs(i)n=math.ceil(10^x-1)//9print((" "):rep(9-x)..n*n)end
Ap<Fun<If<A<1>,Cat<Ap<Fun<Cat<Ap<Fun<If<A<1>,Cat<Ap<A<0>,Sub<A<1>,T>>,SP>,LF>>,Sub<I<9>,A<1>>>,Ap<Fun<If<lt<A<1>,A<2>>,Cat<Cat<A<1>,Ap<A<0>,Add<A<1>,T>,A<2>>>,A<1>>,A<1>>>,T,A<1>>>>,If<lt<A<1>,I<9>>,A<1>,Sub<I<18>,A<1>>>>,Ap<A<0>,Sub<A<1>,T>>>,LF>>,I<17>>
Has leading and trailing whitespace -- sorry! tch has no concept of "empty string" (at least not that I've found)
I doubt the ungolfed version would answer any questions (probably the opposite). I like to think of TCH as a lisp-like language with no functions, loops, or variables, so all the logic is done through recursive lambdas.
The basic structure:
lambda: (run lower lambdas with int 1,2,3...,17,18, join output)
lambda: (convert input, run lower lambdas, join output)
lambda: (int -> string of spaces & newline)
lambda: (int -> string of numbers)
b=Math.abs,e="";for(i=0;i<17;i++){for(j=0;j<17;j++){z=9-(b(i-8)+b(j-8));e+=z>0?z:" "}e+="\n"}console.log(e)
#define W printf("%*lld\n",i,1ll*j*j)
g(i,j){i<17&&g(W,j*10+1);W;}D(){g(9,1);}
main=mapM putStrLn[(' '<$[n..8])++(s n>>=show)|n<-s 9]
s n=[1..n-1]++[n,n-1..1]
=CONCAT(LET(x,ABS(9-ROW(1:17)),y,9-ABS(9-COLUMN(A:R)),IFS(y>x,y-x,y," ",1,"
")))
C, 17 bytes»÷rṡF»S¦øṁ½ƛ÷⋎;øm
»÷rṡF»S - 123456789 as string
¦ - Take all the prefixes
øṁ - Vertical mirror
½ƛ÷⋎; - Halve and split, then join while removing the duplicate ends.
øm - Horizontal palindromize.
<C flag> - Center and join list by newlines
<implicit output> - Print
say$"x abs.(1x(9-abs))**2for-8..8
Using 11 * 11 = 121, 111 * 111 = 12321, 1111 * 1111 = 1234321, ...
3 bytes saved thanks to naffetS.
-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);}}
var r={(1..<$0)+(1...$0).reversed()};r(9).map{(0...9-$0).map{_ in print(terminator:" ")};r($0).map{print($0,terminator:"")};print()}
-1 byte thanks to Donat!
Not the shortest JS solution, but not the longest either!
for(i=9;--i+9;)print(("".padEnd(a=i<0?-i:i)+"1".repeat(9-a)**2).replace(0,1))
i goes from 8 to -8.
We generate the correct number of spaces for each line by using the absolute value of i.
We take advantage of the fact that:
1 * 1 = 1
11 * 11 = 121
111 * 111 = 12321
1111 * 1111 = 1234321
(and so on)
So we generate the correct number of 1 by repeating it 9 minus the absolute value of i.
But unfortunately, JS can't handle properly the biggest calculation: 111111111 at power 2 gives 12345678987654320, so we also have to replace this single 0 with a 1, which costs 14 bytes :(
l=9,n;p(i){for(i=18;i;putchar(i?n>l?48+n-l:32:10))n=9<--i?18-i:i;}
main(){p(l--);l&&p(main());l++;}
for i=1 to 9:?string(27-i*3,32)l;:for j=1 to i:?j;:next:for j=i-1 to 1 step -1:?j;:next:?:next:for i=8 to 1 step -1:?string(27-i*3,32)l;:for j=1 to i:?j;:next:for j=i-1 to 1 step -1:?j;:next:?:next
if vba didn't automatically add a space for the + sign it doesn't print, perhaps I could avoid the 27-i*3 construct for making it look right
public class A{public static void main(String[]v){
for(int a=-8,b,c;a<9;a++){
for(b=-8;b<9;){c=Math.abs(a)+Math.abs(b++);
System.out.print(c>8?" ":9-c);}System.out.println();}}}
correct formatting of this solution (352 chars):
public class A {
public void main(String[] args) {
for (int a = -8; a < 9; a++) {
for (int b = -8; b < 9;) {
int c = Math.abs(a) + Math.abs(b++);
System.out.print(c > 8 ? " " : 9 - c);
}
System.out.println("");
}
}
}
String[] args can be String[]v, and System.out.println("") can be System.out.println(). Similarly, int b= -8; b < 9; can be int b=-8;b<9;.
\$\endgroup\$
for: for(int b=-8,c;b<9;){c=Math.….
\$\endgroup\$
Oct 24, 2012 at 6:41
=MAX(A2,B1,B3,C2)-1 (19 chars)0; (2 chars)8.9 in J10 (3 char)Note: Some might call step 2 cheating as it creates 19 to 23 characters in every cell for a total of over 6000 chars. If you really want to count it that way, you would be better off not putting the formula into those squares that are to remain blank. In that case, you can use a 9 in J10 and you don't need the custom formatting. The total character count would then be just over 3000.
FOR I=-8TO 8A=ABS(I)Q=VAL("1"*(9-A))?" "*A;Q*Q
NEXT
LOCATE 16,8?1
I used the fact that 11*11=121, 111*111=12321, etc. Unfortunately, 12345678987654321 can't be stored as a 64 bit float, so I had to add the last 1 separately, adding 14 bytes.
(do((b 1)(l'(1 2 3 4 5 6 7 8 9))(i 1(+ i b)))((= i 0))(format t"~17:@<~v{~a~}~v{~a~}~>~%"i l i(reverse(subseq l 0(1- i))))(if(> i 8)(setf b -1)))
It is worse than the other Common Lisp solution, but it works and I think it is different enough to be posted.
i is incremented up to 9 and then decremented to 0 (when i=9 sign of b is changed effectively changing addition to subtraction - this avoids second loop.
In each line I print numbers: `123...i(i-1)...1 using loops of format function (for first loop I use list '(1 2 3 4 5 6 7 8 9) and for decrementing loop I use reversed subsequence of this list. Text is then centered.
I didn't notice that numbers are squares of 11..11. As far as this solution is concerned the diamond could be made out of letters or (! @ # ...) (for that you would need to change ~a to ~c in format function.
proc P i {puts [format %[expr 8+$i]s [expr [string repe 1 $i]**2]]}
time {P [incr i]} 9
time {P [incr i -1]} 8
i=8;exec'print`int("1"*(9-abs(i)))**2`.center(18);i-=1;'*17
\$\endgroup\$
Sep 20, 2018 at 18:10
i=8;exec'j=abs(i);print" "*j+`int("1"*(9-j))**2`;i-=1;'*17
\$\endgroup\$
Sep 20, 2018 at 18:13
Anonymous VBE immediate window function that takes no input and outputs to the VBE Immediate Window
For i=-8To 8:j=Abs(i):r=Mid(987654321,j+1):?Spc(j)StrReverse(r)Mid(r,2):Next
For i=-9To 8:For j=-8To 9:k=Abs(i)+Abs(j):l=l &IIf(k>8," ",9-k):Next:?l:l="":Next
Anonymous function that takes no input and outputs to STDOUT.
For i=-8To 8
For j=-8To 9
k=Abs(i)+Abs(j)
If k>8Then?" ";
Else?Chr$(57-k);
Fi
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