Output a Magical 8 Trapezium

Question

Your task is to output a Magical 8 Trapezium:

        1 × 8 + 1 = 9
       12 × 8 + 2 = 98
      123 × 8 + 3 = 987
     1234 × 8 + 4 = 9876
    12345 × 8 + 5 = 98765
   123456 × 8 + 6 = 987654
  1234567 × 8 + 7 = 9876543
 12345678 × 8 + 8 = 98765432
123456789 × 8 + 9 = 987654321

• Output in your chosen language in the fewest bytes possible.
• Note the number of spaces at the start of each line to maintain the trapezium shape.
• Trailing spaces are allowed.
• You can use × or the letter x - whichever you prefer.

Answer 1

Python 2, 59 bytes

a=i=1
exec"print'%9d x 8 +'%a,i,'=',a*8+i;i+=1;a=a*10+i;"*9

The numbers a and i the equation a * 8 + i are generated arithmetically. Each line, i is incremented, and a has the next digit appended via a=a*10+i. For example, if a=12345, i=5, then i becomes 6, so the new a is 12345*10 + 6 which is 123456.

Storing these as numbers rather than strings lets us compute the RHS as given by the equation a*8+i, which is shorter than string reversing.

Answer 2

V, 37 bytes

i¸ 1 X 8 + 1 = 98ñYp|Eylp^Xf+$ylp

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This contains unprintable, so here is a hexdump:

00000000: 69c2 b820 3120 5820 3820 2b20 3120 3d20  i.. 1 X 8 + 1 = 
00000010: 391b 38c3 b159 707c 4579 6c70 015e 5866  9.8..Yp|Eylp.^Xf
00000020: 2b01 2479 6c70 18                        +.$ylp.

Answer 3

Pyth, 32 bytes

VS9ss[*dK-9NSN" x 8 + "N" = "r9K

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VS9ss[*dK-9NSN" x 8 + "N" = "r9K
VS9                                  # For N in 1..9
   s                                 # Join without delimiter
    s[                               # Reduce the array on + (flattens)
      *dK-9N                         # - Space, repeated K=(9-N) times
            SN                       # - The string sequence 1..N
              " x 8 + "              # - This string literal
                       N             # - N itself
                        " = "        # - This string literal
                             r9K     # - The string sequence 9..K

---

Thanks to @FryAmTheEggman for saving 2 bytes. Thanks to @KennyLau for saving 3 bytes.

Answer 4

05AB1E, 32 31 30 28 bytes

Code:

TG9N-ð×NLJðN"x8+ÿ="€ðJžmN£J,

Uses the CP-1252 encoding. Try it online!.

Answer 5

PHP, 105 89 60 57 bytes

my first golf try here (thanks to manatwork & user55641)

for(;$i++<9;)printf("%9s x 8 + $i = %s
",$s.=$i,$s*8+$i);

59

for(;$i++<9;)printf("%9s x 8 + $i = %s
",$s.=$i,$t.=10-$i);

89 (my own try)

for(;@++$i<=9;){printf("%9s x 8 + %s = %s\n",join(range(1,$i)),$i,join(range(9,10-$i)));}

105 (first)

for($j='123456789';@$j[$i++];){printf("%9s x 8 + %s = %s\n",substr($j,0,$i),$i,strrev(substr($j,-$i)));}

Answer 6

CJam, 39 38 36 bytes

Thanks to Optimizer for saving 2 bytes.

9{)_,:)9Se[" x 8 + "@S'=S9_,fm4$<N}/

Test it here.

Same byte count:

9{)_,:)9Se[]"x8+"+:\'=9_,f-Y$<]S*n}/

This requires the latest version, available on Try it online!

Answer 7

Python 2, 87 84 78 75 bytes

s="123456789"
n=1
exec'print"%9s"%s[:n],"x 8 + %s ="%n,s[::-1][:n];n+=1;'*9

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A previous version uses some string magic.

R=range(1,10)
for n in R:print" "*(9-n)+`R`[1:n*3:3]+" x 8 + %d = "%n+`R`[-2:27-3*n:-3]

Casting range(1,10) to a string gives [1, 2, 3, 4, 5, 6, 7, 8, 9], and this is nice since every number is only a single digit. So getting the string 123456789 from this is simple with `range(1,10)`[1::3]. The reversed range is `range(1,10)`[-2::-3]. Then, to get only as far as I want each iteration, I slice it off at either 3*n, or at 3*(9-n) (27-3*n) for the reversed digits.

Answer 8

Ruby, 77 73 65 60 bytes

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Major revamps from @manatwork

Another overhaul from @xsot

a=i=0;9.times{puts"%9d x 8 + %d = %d"%[a=a*10+i+=1,i,a*8+i]}

Answer 9

Java 10, 151 133 130 129 126 110 bytes

v->{String p="\n",r="";for(int n=123456789,i=9;i>0;n/=10,p+=" ")r=p+n+" x 8 + "+i+" = "+(n*8+i--)+r;return r;}

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Explanation:

v->{                   // Method with empty unused parameter and String return-type
  String p="\n",       //  Prefix-String, starting at a newline
         r="";         //  Result-String, starting empty
  for(int n=123456789, //  Multiply-number, starting at 123456789
      i=9;i>0          //  Loop `i` in the range [9, 0):
      ;                //    After every iteration:
       n/=10,          //     Remove the last digit from the integer
       p+=" ")         //     Append a space after the prefix
    r=...+r;           //   Prepend the following to the result-String:
      p                //    The prefix-String
      +n               //    Followed by the integer
      +" x 8 + "       //    Followed by the literal String " x 8 + "
      +i               //    Followed by the loop-index `i`
      +" = "           //    Followed by the literal String " = "
      +(n*8+i--)       //    Followed by the result of that equation
  return r;}           //  Return the result-String

Answer 10

Perl, 49 bytes

printf"%9s x 8 + $_ = %s
",$@.=$_,$_+8*$@for 1..9

Usage

perl -e 'printf"%9s x 8 + $_ = %s
",$@.=$_,$_+8*$@for 1..9'

Answer 11

C, 74 bytes

d(i,n){for(i=n=1;i<10;n=++i+n*10)printf("%9d x 8 + %d = %d\n",n,i,n*8+i);}

Answer 12

Batch, 117 bytes

@echo off
set a=         12345678987654321
for /l %%i in (1,1,9)do call echo %%a:~%%i,9%% x 8 + %%i = %%a:~17,%%i%%

Yes, that is 16 % signs on one line; that's Batch for you!

Answer 13

C#, 113 bytes

void f(){for(int n=1,i=1;i<10;n=10*n+ ++i)Console.WriteLine(new string(' ',9-i)+n+" x "+"8 + "+i+" = "+(n*8+i));}

if you have anyway to improve this solution feel free to share.

Answer 14

APL (Dyalog Unicode), 61 52 39 bytes^SBCS

↑(⍳9)((¯9↑↑),' x 8 +',⊣,'= ',↑∘⌽)¨⊂1↓⎕D

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-9 bytes by using the 10⊥ trick to parse the number, instead of a reduction. Thanks to @Adám for -13!

Explanation:

↑    ((¯9↑↑),' x 8 +',⊣,'= ',↑∘⌽)¨⊂1↓⎕D
                                     ⎕D  ⍝ Numbers from 0 to 9
                                   1↓    ⍝ Drop the 0
 (⍳9)(                          )¨⊂      ⍝ Do 9 times, N=current
                             ↑∘⌽         ⍝ Reverse the string (9..1) and cut off N elements
                      ⊣                  ⍝ N itself
      (   ↑)                             ⍝ Drop N elements off the 1..9 string...
      (¯9↑ )                             ⍝ ...then pad it back with spaces
            ,' x 8 +', ,'= ',            ⍝ Join with a few constant strings
↑                                        ⍝ Format

Answer 15

PowerShell v2+, 85 64 58 57 52 bytes

8..0|%{" "*$_+-join(1..++$i+" x 8 + $i = "+9..++$_)}

Loops from 8 to 0 8..0|%{...} via the range operator. Each iteration, we output a string concatenation consisting of (the appropriate number of spaces " "*$_), plus a -joined string of (a range from 1 to a pre-incremented helper number ++$i, plus the middle bit " x 8 + $i = ", plus the final range from 9 to the current number $_ pre-incremented).

One big trick here is we leverage the "left-preference" for typecasting, which allows us to "add" arrays together inside the -join parens, meaning we use only one -join operator.

Example

PS C:\Tools\Scripts\golfing> .\magical-8-trapezium.ps1
        1 x 8 + 1 = 9
       12 x 8 + 2 = 98
      123 x 8 + 3 = 987
     1234 x 8 + 4 = 9876
    12345 x 8 + 5 = 98765
   123456 x 8 + 6 = 987654
  1234567 x 8 + 7 = 9876543
 12345678 x 8 + 8 = 98765432
123456789 x 8 + 9 = 987654321

Answer 16

Haskell, 92 bytes

s=(show=<<)
[1..9]>>= \x->([x..8]>>" ")++s[1..x]++" x 8 + "++s[x]++" = "++s[9,8..10-x]++"\n"

How it works:

s=(show=<<)                   -- helper function that turns a list of numbers into
                              -- a string without delimiters, e.g. [1,2] -> "12"

[1..9]>>=                     -- for each number 1 to 9
     ([x..8]>>" ")            -- take length of [x..8] copies of a space
     s[1..x]                  -- the digits from 1 to x
     " x 8 + "                -- a string literal
     s[x]                     -- the digit of x
     " = "                    -- another string literal
     s[9,8..10-x]             -- the digits from 9 down to 10-x
     "\n"                     -- an a newline

Answer 17

Pyke, 30 29 bytes

9Fd*~utj+9<\x8\+9i-\=ji>_dJ)X

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9F                         )  -  for i in range(9):
  d*                          -       " " * i
        +                     -      ^ + V
       j                      -       j = V
    ~ut                       -        "123456789"
         9<                   -     ^[:9]
           \x8\+9i-\=         -    [^, "x", 8, "+", (9-i), "=", V]
                        _     -     reversed(V)
                     ji>      -      j[i:]
                         dJ   -   " ".join(^)
                            X - print(reversed(^))

Answer 18

Retina, 66 bytes

Byte count assumes ISO 8859-1 encoding. The leading linefeed is significant.

123456789!9 = 987654321
+`^((.)+)\B.!.(.+).
 $1!$2$3¶$&
!
 x 8 + 

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Answer 19

MATL, 38 36 35 bytes

9:"9@-Z"@:!V' x 8 + '@VO61O58@:-v!D

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Answer 20

J, 51 bytes

(|."1|.\p),.' x 8 + ',"1]p,.' = ',"1]\|.p=:u:49+i.9

Creates the string 123456789 and then operates on prefixes and suffixes of it to create the output.

Usage

   (|."1|.\p),.' x 8 + ',"1]p,.' = ',"1]\|.p=:u:49+i.9
        1 x 8 + 1 = 9        
       12 x 8 + 2 = 98       
      123 x 8 + 3 = 987      
     1234 x 8 + 4 = 9876     
    12345 x 8 + 5 = 98765    
   123456 x 8 + 6 = 987654   
  1234567 x 8 + 7 = 9876543  
 12345678 x 8 + 8 = 98765432 
123456789 x 8 + 9 = 987654321

Answer 21

JavaScript ES6 (88)

Taking advantage of the new repeat method, backticks and templating...

i=10;for(y="";--i;)console.log(`${" ".repeat(i)+(y+=(x=10-i))} x 8 + ${x} = ${y*8+x}\n`)

Answer 22

R, 107 103 bytes

a=1;for(i in 2:10){cat(rep("",11-i),paste(a,"x",8,"+",(i-1),"=",strtoi(a)*8+(i-1)),"\n");a=paste0(a,i)}

Ungolfed :

a=1

for(i in 2:10)
    cat(rep("",11-i),paste(a,"x",8,"+",(i-1),"=",strtoi(a)*8+(i-1)),"\n")
    a=paste0(a,i)

Result :

        1 x 8 + 1 = 9 
       12 x 8 + 2 = 98 
      123 x 8 + 3 = 987 
     1234 x 8 + 4 = 9876 
    12345 x 8 + 5 = 98765 
   123456 x 8 + 6 = 987654   
  1234567 x 8 + 7 = 9876543 
 12345678 x 8 + 8 = 98765432 
123456789 x 8 + 9 = 987654321

Answer 23

05AB1E, 24 bytes

9Lε©LJ'x8'+®'=T®L-Jðý}.c

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Uses a newer version than the challenge, which is now allowed.

Answer 24

Canvas, 20 bytes

９｛Ｒ⤢x∙8∙+¹=¹◂±ｍ） ＊］ｒ

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Answer 25

VBA (Excel), 51 bytes

Using Immediate Window

For z=1To 9:a=a &z:?Spc(9-z)a" x 8 +"z"="a*8+z:Next

Answer 26

JavaScript (ES6), 99 bytes

_=>[...Array(9)].map((n,i)=>`${n="        123456789".substr(i,9)} x 8 + ${++i} = ${n*8+i}`).join`\n`
_=>".........".replace(/./g,(n,i)=>`${n="        123456789".substr(i,9)} x 8 + ${++i) = ${n*8+i}\n`)

Where \n represents a literal newline character. The second version outputs a trailing newline. I came up with a formula for the numbers ('1'.repeat(9-i)+0+i)/9 but the padding was easier to do this way.

Answer 27

Brainfuck, 232 bytes

++++++++[->+>+>+++++[->+>+>+>+>+>+++<<<<<<]>->>+>++>+++<<<<<<<<]>++>+>>>+++>>>---<<<<<<[-[-<<+>>>>.<<]>+[-<+>>>>+.<<<]>.>>>>>.<<<<<.>>>.<<<.>.<.>>.<<.>>>>.<<<<.<<<<[->>>+>>>+<<<<<<]>>[-<<+>>>>>>.-<<<<]>[->>>-<<<<+>]<<<[->>>+<<<]>.>]

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Can be golfed much further...

Answer 28

Javascript (using external library) (143 bytes)

n=>_.Range(1,9).WriteLine(v=>_.Range(0,10-v).Write("",x=>" ")+_.Range(1,v).Write("")+" x 8 + " + v + " = "+_.Range(10-v,v).Reverse().Write(""))

Link to lib: https://github.com/mvegh1/Enumerable/

Explanation of code: Create range 1 to 9, and for each value, write a line corresponding to the complex predicate. The predicate is passed the current integer value, and creates a range spanning 10-currentValue elements, in order to create that many spaces. Those spaces are concatenated with the formula part of the line, and then that is concatenated with the tailend of the range matching the number of elements as the frontend, in reverse order.

Note: In the image, the first line is off by one space because the console added a quotation mark since the return value is a string. The actual value is formatted correctly

Answer 29

dc, 79 bytes

1[32P]sS[d0<S1-d0<R]sR[dZ9r-lRxrdn[ x 8 + ]PdZn[ = ]Pdd8*rZ+prd10*dZ+dZA>M]dsMx

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Somewhat better than I expected. Started with a solution that juggled stack and an incremental register before realizing I only need to keep the left-most number around - since the value to be added is the number of digits the left-most number has, this can be determined via Z and the result can just be calculated according to the formula.

[32P]sS[d0<S1-d0<R]sR feels wasteful, but I couldn't come up with a much better way to handle the indentation. The main macro, [dZ9r-lRxrdn[ x 8 + ]PdZn[ = ]Pdd8*rZ+prd10*dZ+dZA>M]dsMx should be fairly self-explanatory based on my above description. The initial 1 just seeds the whole thing.

Answer 30

Pyth - 30 bytes

First attempt, working on optimization.

VS9+*dK-9Njd[jkSN\×8\+N\=jkr9K

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Answer 31

Forth (gforth), 76 bytes

: f 0 10 1 do cr 10 * i + dup 9 .r ."  x 8 + "i . ." = "dup 8 * i + . loop ;

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Explanation

Start a value at 0. Starts a loop from 1 to 9. Each iteration:

• Multiply value by 10 and add the loop index
• output this value right-aligned
• output " x 8 + "
• output the loop index
• output " = "
• calculate results and output

Code Explanation

: f                  \ start word definition
  0                  \ set up counter/value
  10 1 do            \ start a counted loop from 1 to 9
    cr               \ output a newline
    10 * i +         \ calculate the next value for the first term
    dup 9 .r         \ make a copy and then output in a right-aligned space of 9 characters
    ."  x 8 + "      \ output " x 8 +"
    i .              \ output the loop index
    ." = "           \ output " ="
    dup 8 * i +      \ perform the actual calculation
    .                \ output result
  loop               \ end the counted loop
;                    \ end the word definition

Answer 32

Perl 6, 58 bytes

my $a;printf "%9s x 8 + $_ = %s\n",$a~=$_,$a.flip for 1..9

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Answer 33

k (77 bytes)

Could probably be shortened a bit more

-1@'((|t)#\:" "),'(t#\:n),'" x 8 + ",/:n,'" = ",/:(t:"I"$'n)#\:|n:"123456789";

Example:

k)-1@'((|t)#\:" "),'(t#\:n),'" x 8 + ",/:n,'" = ",/:(t:"I"$'n)#\:|n:"123456789";
         1 x 8 + 1 = 9
        12 x 8 + 2 = 98
       123 x 8 + 3 = 987
      1234 x 8 + 4 = 9876
     12345 x 8 + 5 = 98765
    123456 x 8 + 6 = 987654
   1234567 x 8 + 7 = 9876543
  12345678 x 8 + 8 = 98765432
 123456789 x 8 + 9 = 987654321

Answer 34

golflua, 56 characters

p=""~@i=1,9p=p..i; w(S.q("%9s x 8 + %d = "..p*8+i,p,i))$

Sample run:

bash-4.3$ golflua -e 'p=""~@i=1,9p=p..i; w(S.q("%9s x 8 + %d = "..p*8+i,p,i))$'
        1 x 8 + 1 = 9
       12 x 8 + 2 = 98
      123 x 8 + 3 = 987
     1234 x 8 + 4 = 9876
    12345 x 8 + 5 = 98765
   123456 x 8 + 6 = 987654
  1234567 x 8 + 7 = 9876543
 12345678 x 8 + 8 = 98765432
123456789 x 8 + 9 = 987654321

Answer 35

Pascal, 156 127 103 bytes

Trivial solution, using the Format library func. Shaved some more bytes by using only integers and no strings. Reduced in size thanks to @manatwork (for teaching me some basic stuff about writeln).

var i,c:int64;begin i:=0;for c in[1..9]do begin i:=i*10+c;WriteLn(i:9,' x 8 + ',c,' = ',i*8+c);end;end.

Ungolfed:

var
  i, c: int64;
begin
  i := 0;
  for c in [1..9] do
  begin
    i := i*10+c;
    WriteLn(i:9, ' x 8 + ', c, ' = ', i*8+c);
  end;
end.

Answer 36

C++, 196 bytes

Ungolfed:

#include <iostream>

using namespace std;

int main()
{
    int i=1;
    while (i<10){
        for (int j=1;j<10-i;j=j+ 1){
            cout << " ";
        }
        for (int j=1;j<=i;j++){
            cout << j;
        }
        cout << " x 8 + 1 = ";

        for(int m=1;m<=i;m++){
                cout << 10 - m;
        }

        cout << endl;
        i = i + 1;
   }
}

Run example:

sh-4.3$ main                                                                                                                                                    
        1 x 8 + 1 = 9                                                                                                                                           
       12 x 8 + 1 = 98                                                                                                                                          
      123 x 8 + 1 = 987                                                                                                                                         
     1234 x 8 + 1 = 9876                                                                                                                                        
    12345 x 8 + 1 = 98765                                                                                                                                       
   123456 x 8 + 1 = 987654                                                                                                                                      
  1234567 x 8 + 1 = 9876543                                                                                                                                     
 12345678 x 8 + 1 = 98765432                                                                                                                                    
123456789 x 8 + 1 = 987654321 

Answer 37

///, 149 bytes

/:/ x 8 + //-/ = //0/1234//_/9876//|/  /||||1:1-9
||| 12:2-98
|||123:3-987
|| 0:4-_
||05:5-_5
| 056:6-_54
|0567:7-_543
 05678:8-_5432
056789:9-_54321

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