Output/print this block of text:
1234567890
2468013579
3691470258
4815926037
5049382716
6172839405
7306295184
8520741963
9753108642
0987654321
Acceptable formats include:
However, list of integers is not acceptable because the last line is not an integer.
This is code-golf. Shortest answer in bytes wins. Standard loopholes apply.
Mod[1##,11]~Mod~10&~Array~{10,10}
Try it online! (Using Mathics.)
The cell at 1-based index (x,y) has value ((x*y) % 11) % 10
Saved a byte thanks to Luis. I keep forgetting the & is a shortcut for duplicating and transposing.
10:&*11\10\
Using @Martin's algorithm: x*y % 11 % 10
Explanation:
10 % Pust 10 to the stack. Stack: 1
: % 1-based range. Stack: [1 2 3 ... 10]
& % Duplicate range. Stack: [1 2 3 ... 10],[1 2 3 ... 10]
% Transpose last range. Stack [1 2 3 ... 10],[1;2;3 ...10]
* % Multiply with broadcasting. Stack: [1 2 3 ...;2 4 6...] (10-by-10 array)
11 % Push 11 to the stack. Stack [1 2 3 ...;2 4 6 ...], 11
\ % Modulus.
10 % Push 10 to the stack.
\ % Modulus
% Implicit display
Same bytecount:
10t:&*11\w\
t!* by&*
\$\endgroup\$
Jun 26, 2017 at 10:13
10|11|∘.×⍨⍳10
A port of my Mathematica answer.
∘.×⍨ ⍝ Multiplication table of...
⍳10 ⍝ The list from 1 to 10.
11| ⍝ mod 11.
10| ⍝ mod 10.
Byte count assumes ISO 8859-1 encoding.
10$*
1
,1$`
,1+
$_¶
(?<=(¶?.+)+)1
$#1$*
1{10}1?
,(1*)
$.1
Another implementation of the ... % 11 % 10 algorithm. The fun part of doing it with a regex is that we can take care of both modulo computations at once.
10$*
Initialise the string to ten 1s.
1
,1$`
Replace each of those with a comma, a one, and the prefix in front of that one. This gives ,1,11,...,1111111111, i.e. a unary range.
,1+
$_¶
Now replace each of the range elements with the entire string followed by a linefeed. This gives us a 10x10 grid of unary numbers indicating the current column.
(?<=(¶?.+)+)1
$#1$*
Match each 1 and determine which row it's on by repeating group one that many times. Replace the 1 with that many 1s. This multiplies the values in each row by the row's 1-based index.
1{10}1?
Now let's do mod 11, mod 10 in one step. To do mod 11, we'd normally just remove all 1{11} from the string to be left with the remainders. And then we'd remove 1{10} after that. But if we just remove ten 1s plus another if possible, the regex engine's greediness will do mod 11 for us as long as possible, and if not, then it'll try at least mod 10.
,(1*)
$.1
Finally, we just convert each number to decimal by replacing it with its length.
l=[1..10]
f=[[x*i`mod`11`mod`10|i<-l]|x<-l]
_=>[...1e9+''].map((_,a,b)=>b.map((_,c)=>-~a*++c%11%10))
Saved 4 bytes thanks to Shaggy and 8 bytes thanks to Arnauld.
_=>[...a=Array(10)].map((_,x)=>[...a].map((_,y)=>(x+1)*++y%11%10)). You save me 4 bytes, I save you 4 bytes :)
\$\endgroup\$
map() and 3 more bytes by using 1e9+'' instead of Array(10). That leads to _=>[...1e9+''].map((_,x,a)=>a.map((_,y)=>-~x*++y%11%10)).
\$\endgroup\$
1e9 trick. I didn't know that one. I did think about using the third argument, but for some reason I didn't use it.
\$\endgroup\$
Turns out this was my 200 (undeleted) answer here :)
Looks like this is the same formula Martin spotted.
Aõ
£®*X%B%A
Test it (-R flag for visualisation purposes only)
Aõ :Generate an array of integers from 1 to 10, inclusive.
£ :Map over each element in the array, returning...
® :Another map of the same array, which...
*X :Multiplies the current element of the inner function by the current element of the outer function...
%B :Modulus 11...
%A :Modulus 10.
:Implicit output of resulting 2D array
-R flag
\$\endgroup\$
o->{String r="";for(int x=0,y;++x<11;r+="\n")for(y=0;++y<11;r+=x*y%11%10);return r;}
Uses the same algorithm as @MartinEnder's Mathematica answer: 1-indexed x*y%11%10.
Explanation:
o->{ // Unused Object parameter and String return-type
String r=""; // Result-String
for(int x=0,y;++x<11; // Loop (1) from 1 to 11 (exclusive)
r+="\n") // And append a new-line after every iteration
for(y=0;++y<11; // Inner loop (2) from 1 to 11 (exclusive)
r+=x*y%11%10 // And append the result-String with `x*y%11%10`
); // End of inner loop (2)
// End of loop (1) (implicit / single-line body)
return r; // Return result-String
} // End of method
-6 bytes thanks to offcialaimm.
Uses Martin's algorithm which I have no understanding of how he came up with it so fast. o0
r=range(1,11)
print[[x*y%11%10for y in r]for x in r]
r=range(1,11) saves 6 bytes
\$\endgroup\$
Jun 26, 2017 at 11:16
e instead of % T
\$\endgroup\$
Jun 26, 2017 at 12:11
1:10%o%1:10%%11%%10
The least "R"-looking bit of R code I have ever written. Uses the same algorithm as Martin Ender's answer (and almost all the other answers as well). x %o% y is the same as outer(x, y).
Ｆχ«ＦχＩ﹪﹪×⁺¹ι⁺¹κ¹¹χ⸿
Uses Martin's formula.
»s and while you can use ω instead of ”” you can save a whole bunch of bytes by using ⸿ as this then becomes Ｆχ«ＦχＩ﹪﹪×⁺¹ι⁺¹κ¹¹χ⸿. (Before I knew about ⸿ I would have suggested Ｊ⁰ι which would still have saved a number of bytes.)
\$\endgroup\$
⸿ is the reverse operator, what does it do at the end of your code without arguments? Is it documented?
\$\endgroup\$
⮌ is the Reverse operator, ⸿ is the move cursor to start of next line character (like ¶ but can be in a separate string).
\$\endgroup\$
f(x,y){for(x=0;x++<10;puts(""))for(y=0;y++<10;putchar(x*y%11%10+48));}
[|?[|?a*b%11%z';
This, of course, uses Martin's Method. It translates to this QBasic code.
[| FOR A = 1 to 10 ([ starts a FOR loop, | delimits the list of arguments;
a FOR loop with 0 args loops from 1 to 10 by default with increment 1.
? PRINT a newline
[| Start a second FOR loop from 1-10, iterator b
? PRINT
a*b%11%z the result of Martin's formula.
'; and suppress newlines/tabs/spaces
_=>{var r="";for(int x=0,y;++x<11;r+="\n")for(y=0;++y<11;r+=x*y%11%10);return r;}
Same algorithm as most of the other answers and essentially the C# port of @Kevins Java answer.
12345678902468013579369147025848159260375049382716
.*
$&$&
O$^50>`.
.{10}
$&¶
10,{){\)*11%10%}+10,%}%`
-13 thanks to a clever trick Martin Ender suggested.
{ --> ;, } --> `), you can at least drop the first [.
\$\endgroup\$
Jun 26, 2017 at 16:28
{){\)*11%10%}+10,/n}10,/
\$\endgroup\$
Jun 26, 2017 at 16:32
/. ;)
\$\endgroup\$
Jun 26, 2017 at 16:35
int blk + -> {int space contents-of-blk}.
\$\endgroup\$
Jun 26, 2017 at 16:36
+ trick...although I altered your code a bit
\$\endgroup\$
Jun 26, 2017 at 16:43
10$*T
M!&`T+
m`$
;110$*T10$*
1
09876543210
m`(?<=^\1;\1{0,9}(T+))T
C
(?<!C.{108})\S
TS F~u0+*i>i%
TS - [1, 2, 3, 4, 5, 6, 7, 8, 9]
F~u0+*i>i% - for i in ^:
~u0+ - "01234567890"
* - ^ * i
i> - ^[i:]
i% - ^[::i]
1":(10|11|*&>:)"0/~@(i.@9:)
Uses the same trick as in Martin Ender's Mathematica answer.
1un@i/
/10<@i/01234567890/jl10<qnc0a^t>jtl%n>
A (fairly) straightforward implementation of Rod's Python answer.
1un !initialize register n to 1!
@i/<nl>/ !insert a newline!
10< !loop for 10 rows!
@i/01234567890/ !insert the mysterious string of digits!
j !move point to start of buffer!
l !move forward past the newline!
10< !loop for 10 digits on a line!
qnc !move point forward by n characters!
0a^t !print the character at point!
> !end inner loop!
j !move point to start of buffer!
t !print (empty) line!
l !move to start of digit string!
%n !increment register n (for next line)!
> !end outer loop!
Using <ESC>-terminated inserts and a control character for the ^T command would save another three five bytes, at the expense of readability.
Using Martin's mod-11/mod-10 formula actually makes it longer at 43 bytes using controls for ^A and ^T, mostly because TECO doesn't have a mod operator.
0ur10<%run10<qn-10"g-11%n'qn\r0a^Tqr%n>^a
^A>
Mod 11 is done in an ongoing fashion by incrementing the number in qn by -11 whenever it exceeds 10. The qn\r0a^T sequence inserts the number in the editing buffer as decimal digits, reverses past the last digit, retrieves it from the buffer and types it, essentially doing mod-10.
I expected it to be shorter. Oh, well.
