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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.
,.(0&<#":)"+9-+/~|i:8
Thanks to FUZxxl for the "+ trick (I don't think I've ever used u"v before, heh).
i:8 "steps" vector: _8 _7 _6 ... _1 0 1 ... 7 8
| magnitude
+/~ outer product using +
9- inverts the diamond so that 9 is in the center
( )"+ for each digit:
# copy
0&< if positive then 1 else 0
": copies of the string representation of the digit
(in other words: filter out the strictly positive
digits, implicitly padding with spaces)
,. ravel each item of the result of the above
(necessary because the result after `#` turns each
scalar digit into a vector string)
#(loop[[r & s](range 18)h 1](print(apply str(repeat(if(< r 8)(- 8 r)(- r 8))\ )))(doseq[m(concat(range 1 h)(range h 0 -1))](print m))(println)(if s(recur s((if(< r 8)inc dec)h))))
-12 bytes by changing the outer doseq to a loop, which allowed me to get rid of the atom (yay).
A double "for-loop". The outer loop (loop) goes over each row, while the inner loop (doseq) goes over each number in the row, which is in the range (concat (range 1 n) (range n 0 -1)), where n is the highest number in the row.
(defn diamond []
(let [spaces #(apply str (repeat % " "))] ; Shortcut function that produces % many spaces
(loop [[row-n & r-rows] (range 18) ; Deconstruct the row number from the range
high-n 1] ; Keep track of the highest number that should appear in the row
(let [top? (< row-n 8) ; Are we on the top of the diamond?
f (if top? inc dec) ; Decided if we should increment or decrement
n-spaces (if top? (- 8 row-n) (- row-n 8))] ; Calculate how many prefix-spaces to print
(print (spaces n-spaces)) ; Print prefix-spaces
(doseq [m (concat (range 1 high-n) (range high-n 0 -1))] ; Loop over the row of numbers
(print m)) ; Print the number
(println)
(if r-rows
(recur r-rows (f high-n)))))))
Due to a bug in the logic in my first attempt (accidentally inserting the prefix-spaces between each number instead), I managed to get this:
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 5 6 5 4 3 2 1
1 2 3 4 5 6 7 6 5 4 3 2 1
1 2 3 4 5 6 7 8 7 6 5 4 3 2 1
12345678987654321
1 2 3 4 5 6 7 8 9 10 9 8 7 6 5 4 3 2 1
1 2 3 4 5 6 7 8 9 8 7 6 5 4 3 2 1
1 2 3 4 5 6 7 8 7 6 5 4 3 2 1
1 2 3 4 5 6 7 6 5 4 3 2 1
1 2 3 4 5 6 5 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
Not even correct ignoring the obvious bug, but it looked cool.
A⍪1↓⊖A←A,0 1↓⌽A←⌽↑⌽¨⍴∘(1↓⎕D)¨⍳9
If spaces separating the numbers are allowed (as in the Mathematica entry), it can be shortened to 28 26:
A⍪1↓⊖A←A,0 1↓⌽A←⌽↑⌽∘⍕∘⍳¨⍳9
Explanation:
⍳9: a list of the numbers 1 to 91↓⎕D: ⎕D is the string '0123456789', 1↓ removes the first element⍴∘(1↓⎕D)¨⍳9: for each element N of ⍳9, take the first N elements from 1↓⎕D. This gives a list: ["1", "12", "123", ... "123456789"] as strings⌽¨: reverse each element of this list. ["1", "21", "321"...]
(Short program:)
⍳¨⍳9: the list of 1 to N, for N [1..9]. This gives a list [[1], [1,2], [1,2,3] ... [1,2,3,4,5,6,7,8,9]] as numbers.⌽∘⍕∘: the reverse of string representation of each of these lists. ["1", "2 1"...]A←⌽↑: makes a matrix from the list of lists, padding on the right with spaces, and then reverse that. This gives the upper quadrant of the diamond. It is stored in A.A←A,0 1↑⌽A: A, with the reverse of A minus its first column attached to the right. This gives the upper half of the rectangle. This is then stored in A again.A⍪1↓⊖A: ⊖A is A mirrored vertically (giving the lower half), 1↓ removes the top row of the lower half and A⍪ is the upper half on top of 1↓⊖A.
Print@@#&/@(Sum[k~DiamondMatrix~17,{k,0,8}]/.0->" ")
With 3 bytes saved thanks to Kelly Lowder.
Analysis
The principal part of the code, Sum[DiamondMatrix[k, 17], {k, 0, 8}], can be checked on WolframAlpha.
The following shows the underlying logic of the approach, on a smaller scale.
a = 0~DiamondMatrix~5;
b = 1~DiamondMatrix~5;
c = 2~DiamondMatrix~5;
d = a + b + c;
e = d /. 0 -> "";
Grid /@ {a, b, c, d, e}
f = Table[# - Abs@k, {k, -8, 8}] &; f[f[9]] /. n_ /; n < 1 -> "" // Grid
\$\endgroup\$
– DavidC
Mar 22 '13 at 17:56
Table[9 - ManhattanDistance[{9, 10}, {j, k}], {j, 18}, {k, 18}] /. n_ /; n < 1 -> "" // Grid
\$\endgroup\$
– DavidC
Mar 22 '13 at 17:57
Grid@#@#@9&[Table[#-Abs@k,{k,-8,8}]&]/.n_/;n<1->"" 50 chars.
\$\endgroup\$
– chyanog
Mar 23 '13 at 4:26
ArrayPlot[Sum[k~DiamondMatrix~17, {k, 0, 8}], AspectRatio -> 2]
\$\endgroup\$
– DavidC
Nov 23 '13 at 15:53
Not clever:
s=str(111111111**2)
for i in map(int,s):print'%8s'%s[:i-1]+s[-i:]
'L'.
\$\endgroup\$
– Steven Rumbalski
Oct 15 '12 at 14:51
v;main(i){for(;i<307;putchar(i++%18?v>8?32:57-v:10))v=abs(i%18-9)+abs(i/18-8);}
for n in`111111111**2`:print`int('1'*int(n))**2`.center(17)
Abuses backticks and repunits.
in keyword can be removed, just like you did with print keyboard.
\$\endgroup\$
– Konrad Borowski
Oct 23 '12 at 17:02
L in the middle seven lines of output.
\$\endgroup\$
– Steven Rumbalski
Mar 28 '13 at 14:56
Another GolfScript solution
17,{8-abs." "*10@-,1>.-1%1>n}%
Thank you to @PeterTaylor for another char.
Previos versions:
17,{8-abs" "*9,{)+}/9<.-1%1>+}%n*
17,{8-abs" "*9,{)+}/9<.-1%1>n}%
17,{8-abs." "*10@-,1>.-1%1>n}%
\$\endgroup\$
– Peter Taylor
Mar 28 '13 at 16:21
Mathematica 55 50 45 41 38
(10^{9-Abs@Range[-8,8]}-1)^2/81//Grid
Grid[(10^Array[{9}-Abs[#-9]&,17]-1)^2/81]
My first entry on Codegolf!
for(l=n=1;l<18;n-=2*(++l>9)-1,console.log(s+z)){for(x=n,s="";x<9;x++)z=s+=" ";for(x=v=1;x<2*n;v-=2*(++x>n)-1)s+=v}If this can be shortened any further, please comment :)
<?for($a=-8;$a<9;$a++){for($b=-8;$b<9;){$c=abs($a)+abs($b++);echo$c>8?" ":9-$c;}echo"\n";}
Calculates and prints the Manhattan distance of the position from the centre. Prints a space if it's less than 1.
An anonymous user suggested the following improvement (84 characters):
<?for($a=-8;$a<9;$a++,print~õ)for($b=-8;$b<9;print$c>8?~ß:9-$c)$c=abs($a)+abs($b++);
<? in the bytecount. I've made some other improvements too.
\$\endgroup\$
– RedClover
May 26 '18 at 19:58
for(i=9;--i+9;console.log(s))for(j=9;j;s=j--^9?k>0?k+s+k:" "+s:k+"")k=i<0?j+i:j-i
-7
\$\endgroup\$
– Christiaan Westerbeek
Jul 8 '20 at 22:23
Thanks to @DJMcMayhem for saving a ton of bytes!
My first Vim answer, so exciting!
i12345678987654321<ESC>qqYP9|xxI <ESC>YGpHq7@q
I tried to write the numbers via a recording, but it is much longer
i123 ... 321<ESC> Write this in insert mode and enter normal mode
qq Start recording into register q
YP Yank this entire line and Paste above
9| Go to the 9th column
xx Delete character under cursor twice
I <ESC> Go to the beginning of the line and insert a space and enter normal mode
Y Yank this entire line
G Go to the last line
p Paste in the line below
H Go to the first line
q End recording
7@q Repeat this 7 times
EDIT:
I used H instead of gg and saved 1 byte
ma and change `ai<space> to I<space>.
\$\endgroup\$
– DJMcMayhem
Dec 4 '16 at 20:20
(defun x(n)(if(= n 0)1(+(expt 10 n)(x(1- n)))))(dotimes(n 17)(format t"~17:@<~d~>~%"(expt(x(- 8(abs(- n 8))))2)))
First I noticed that the elements of the diamond could be expressed like so:
1 = 1 ^ 2
121 = 11 ^ 2
12321 = 111 ^ 2
etc.
x recursively calculates the base (1, 11, 111, etc), which is squared, and then printed centered by format. To make the numbers go up to the highest term and back down again I used (- 8 (abs (- n 8))) to avoid a second loop
1..8+9..1|%{' '*(9-$_)+[int64]($x='1'*$_)*$x}
1..9+8..1|%{' '*(9-$_)+[int64]($x='1'*$_)*$x}
Thanks to Strigoides for the hint to use 1^2,11^2,111^2...
Shaved some characters by:
- Eliminating
$w.- Nested the definition of
$xin place of its first use.- Took some clues from Rynant's solution:
- Combined the integer arrays with
+instead of,which allows elimination of the parenthesis around the arrays and a layer of nesting in the loops.- Used
9-$_to calculate the length of spaces needed, instead of more complicated maths and object methods. This also eliminated the need for$y.
Explanation:
1..8+9..1 or 1..9+8..1 generates an array of integers ascending from 1 to 9 then descending back to 1.
|%{...} pipes the integer array into a ForEach-Object loop via the built-in alias %.
' '*(9-$_)+ subtracts the current integer from 9, then creates a string of that many spaces at the start of the output for this line.
[int64]($x='1'*$_)*$x defines $x as a string of 1s as long as the current integer is large. Then it's converted to int64 (required to properly output 1111111112 without using E notation) and squared.
(⍉⊢⍪1↓⊖)⍣2⌽↑,⍨\1↓⎕d
⎕d are the digits '0123456789'
1↓ drop the first ('0')
,⍨\ swapped catenate scan, i.e. the reversed prefixes '1' '21' '321' ... '987654321'
↑ mix into a matrix padded with spaces:
1
21
321
...
987654321
⌽ reverse the matrix horizontally
(...)⍣2 do this twice:
⍉⊢⍪1↓⊖ the transposition (⍉) of the matrix itself (⊢) concatenated vertically (⍪) with the vertically inverted matrix (⊖) without its first row (1↓)
(( {#2-1}| {8})($3${#3+1}|1)(${#4+1}$4|)\n){9}(( $6| )(${$7/10}|${$3/10})(${$8-1+1/$8*#7-1/$8}|7){#7-1}\n){8}
The solution is divided into two parts, which generate the upper half of the diamond (including the center line) and the lower half, respectively. The upper half consists of group 1, repeated nine times (...\n){9}; the lower half consists of group 5, repeated eight times (...\n){8}.
The first part of each group handles the spaces. (I've replaced space with underscore in the explanations for better visibility.)
Upper-half spaces:
( {#2-1}| {8})
( ) Group 2:
_{#2-1} Space, repeated one less time than the previous length of group 2
| Or, if this is the first row and there is no previous value of group 2...
_{8} Space, repeated eight times
Lower-half spaces:
( $6| )
( ) Group 6:
_$6 Space, concatenated to the previous value of group 6
| Or, if this is the first row and there is no previous value of group 6...
_ A single space
Now for the diamond itself. It can be divided into four quadrants, and I used three different approaches to generate them:
We generate the left half (including the center) of each row with group 3, and the right half with group 4. For both halves, observe that we can generate the next row by concatenating a digit to the previous row:
1234 321
| |
v v
12345 4321
So we can use a similar approach to what we did for the bottom-half spaces. The only difference is that the character we're concatenating at each step needs to be calculated.
Top left quadrant:
($3${#3+1}|1)
( ) Group 3:
$3 Previous value of group 3
${ } concatenated to the result of...
#3+1 Previous length of group 3 plus 1
| Or, if this is the first row and there is no previous value of group 3...
1 The digit 1
Top right quadrant:
(${#4+1}$4|)
( ) Group 4:
${ } The result of...
#4+1 Previous length of group 4 plus 1
$4 concatenated to the previous value of group 4
| Or, if this is the first row and there is no previous value of group 4...
Empty string
We generate the left half (including the center) of each row with group 7. Observe that each row is obtained by removing the rightmost digit from the previous row:
12345
|
v
1234
If we treat the entire thing as an integer, we can perform this transformation easily by dividing it by ten:
(${$7/10}|${$3/10})
( ) Group 7:
${ } The result of...
$7/10 Previous value of group 7, divided by 10
| Or, if there is no previous value of group 7...
${ } The result of...
$3/10 Group 3 (the final upper-left quadrant row) divided by 10
There isn't an easy way to remove a digit from the left side of a previous match, so for the right half of each bottom row, we resort to generating one digit at a time with group 8. Here are the first three rows that we need to generate:
7654321
654321
54321
Each digit is one less than the previous digit, except that we have to "reset" when we hit 1. The reset value is always one less than the length of the left half of the row, which we have access to as group 7.
How can we tell when we've hit 1? I used an integer-division trick I first developed for programming in Acc!!: if N is a positive integer, then 1/N will be one iff N = 1 and zero if N > 1. Multiplying this quantity by (length of group 7) - 1 gives us the reset we need.
(${$8-1+1/$8*#7-1/$8}|7){#7-1}
( ) Group 8 (generates one digit at a time):
${ } The result of...
$8-1 Previous value of group 8, minus 1
+ #7 Plus the length of group 7
1/$8* if the previous value of group 8 equals 1
-1/$8 Minus 1 if the previous value of group 8 equals 1
| Or, if there is no previous value of group 8...
7 The digit 7
{ } Generate this many digits on each row:
#7-1 One less than the length of group 7
For the records:
s=c(1:9,8:1);for(i in s)cat(rep(" ",9-i),s[0:i],s[(i-1):0],"\n",sep="")
message(rep(" ",9-i),s[c(1:i,i:1-1)])
\$\endgroup\$
– flodel
Apr 11 '15 at 17:14
for(i in s<-c(1:9,8:1))... to save a byte
\$\endgroup\$
– Giuseppe
Dec 14 '17 at 17:10
-1'(::;1_|:)@\:((|!9)#'" "),'$i*i:"J"$(1+!9)#'"1";
Old method:
-1',/(::;1_|:)@\:((|!9)#\:" "),',/'+(::;1_'|:')@\:i#\:,/$i:1+!9;
CJam is a lot newer than this challenge, so this answer is not eligible for being accepted. This was a neat little Saturday evening challenge, though. ;)
8S*9,:)+9*9/2%{_W%1>+z}2*N*
The idea is to form the upper left quadrant first. Here is how that works:
First, form the string " 123456789", using 8S*9,:)+. This string is 17 characters long. Now we repeat the string 9 times, and then split it into substrings of length 9 with 9/. The mismatch between 9 and 17 will offset every other row one character to the left. Printing each substring on its own line we get:
1
23456789
12
3456789
123
456789
1234
56789
12345
6789
123456
789
1234567
89
12345678
9
123456789
So if we just drop every other row (which conveniently works by doing 2%), we obtain one quadrant as desired:
1
12
123
1234
12345
123456
1234567
12345678
123456789
Finally, we mirror this twice, transposing the grid in between to ensure that the two mirroring operations go along different axes. The mirroring itself is just
_ "Duplicate all rows.";
W% "Reverse their order.";
1> "Discard the first row (the centre row).";
+ "Add the other rows.";
Lastly, we just join all lines with newlines, with N*.
++++++++[->+>++++>+>++++++<<<<]>>>++<<+[-[-<+>>.<]<[->+<]>>>>+>[->+<<.+>]>+[-<+<.->>]<<<.<<]>>>>-[<<<<+[-<+>>.<]<[->+<]>>>>>[->+<<+.>]>-[-<+<-.>>]<<-<.>>]
[spaceMem, spaceCount, space, lf, "0", numCount, numMem]
++++++++[->+>++++>+>++++++<<<<]>>>++<<+
[ for spaceCount
-[- for spaceCount minus 1
<+ inc spaceMem
>>. print space
< go to spaceCount
]
<[->+<]> fetch spaceCount from spaceMem
>>>+ inc number
>[- for numCount
>+ inc numMem
<<.+ print and inc number
> go to numCount
]
>+[- for numMem plus 1
<+ inc numCount
<.- print and decrement number
>> go to numMem
]
<<<. print lf
<< go to spaceCount
]
>>>>-[ for numCount
<<<<+[- for spaceCount plus 1
<+ inc spaceMem
>>. print space
< go to spaceCount
]
<[->+<]> fetch spaceCount from spaceMem
>>> inc number
>[- for numCount
>+ inc numMem
<<+. print and inc number
> go to numCount
]
>-[- for numMem minus 1
<+ inc numCount
<-. print and decrement number
>> go to numMem
]
<<- decrement number
<. print lf
>> go to numCount
]
f=->x{[*1..x]+[*1...x].reverse};puts f[9].map{|i|(f[i]*'').center 17}
(Thanks, Patrick!)
Improvements welcome. :)
f=->x{[*1..x]+[*1...x].reverse};puts f[9].map{|i|(f[i]*'').center 17}
\$\endgroup\$
– Patrick Oscity
Nov 5 '12 at 21:11
[*-8..8].map{|i|puts' '*i.abs+"#{eval [?1*(9-i.abs)]*2*?*}"}
\$\endgroup\$
– G B
Nov 25 '16 at 8:05
(-8..8).map{i=_1.abs;puts' '*i+"#{eval(?1*(9-i))**2}"}. It's the same solution optimized.
\$\endgroup\$
– Siim Liiser
Oct 2 '20 at 12:22
Assuming that this is meant as a code-golf challenge, here's a basic GolfScript solution:
9,.);\-1%+:a{a{1$+7-.0>\" "if}%\;n}%
adding +1 for -E which is required for say
say$"x(9-$_).(1x$_)**2for 1..9,reverse 1..8
edit: shortened a bit
perl -E is the same number of characters as perl -e. :)
\$\endgroup\$
– Mark Reed
Aug 23 '20 at 13:58
for i in map(int,str(int('1'*9)**2)):print' '*(9-i),int('1'*i)**2
I=int; to your code and replacing all subsequent instances of int with I
\$\endgroup\$
– Cyoce
Nov 27 '16 at 2:51
int is used, and it's used 3 times, so it saves 6 chars at a cost of 6 chars.
\$\endgroup\$
– cardboard_box
Nov 27 '16 at 4:05
87:
q()(printf %$[9+$1%9]s\\n $[$2*$2];[ 7 -lt $1 ]||(q $[$1+1] ${2}1;q $[$1+9] $2))
q 0 1
As it usually goes, changing from iterative to recursive solution helps us win additional bytes.
Meaning of parameters to q:
$1 How much to remove from 8 to get the number of spaces in the beginning. Note value modulo 9 counts here (actual value is also a hint to quit recursion).
$2 The current chain of 1s to be squared and output by printf.
The modus operandi is:
(if not yet at 12..9..21 - the recursive step)
2.1. output the next sequence (here: 111111 > $2 , output 12345654321
2.2. output the sequence once again (123454321).
In the step 2.2 , we pass (indent value + 9) instead of indent value however, so that the algoritm "knows" we are printing the row for the second time. Without this, the [ 7 -lt $1 ] would be false, causing us to retrigger the recursive step 1. This would never finish then.
The recursion goes like this:
q 0 1: 1
q 1 11: 121
q 2 111: 12321
q 3 1111: 1234321
q 4 11111: 123454321
q 5 111111: 12345654321
q 6 1111111: 1234567654321
q 7 11111111: 123456787654321
q 8 111111111: 12345678987654321
q 16 11111111: 123456787654321
q 15 1111111: 1234567654321
q 14 111111: 12345654321
q 13 11111: 123454321
q 12 1111: 1234321
q 11 111: 12321
q 10 11: 121
q 9 1: 1
p()(printf "%$[8+i]s\n" $[k*k])
k=;for i in `seq 9`;do k+=1;p;done;for i in `seq 8 -1 1`;do k=${k:1};p;done;
"k+=1" is much cheaper as k=$[10*k+1] , and for k being a string of ones it's the same. Same goes for ${k:1} and $[k/10] .
126:
p() (printf "%$[$1+i]s\n" $[k*k];)
k=1;for i in `seq 8`;do p 8;k=$[10*k+1];done;for i in `seq 8 -1 0`;do p 9;k=$[k/10];done;
I guess there may be even shorter solution, but weather is glorious, I can't stand sitting in front of computer any more :).
9:v:<,+55<v5*88<v-\9:$_68v
> v> ^>3p2vpv -1<! *
, 1^ 2p45*3+9<4: ,: +
g -^_75g94+4pg7^! +^ ,<
1 : ^ `0 :-1$_:68*^$
^1_$:55+\-0\>:#$1-#$:_^
It could definitely be golfed more, but it's my first Funge program and my head already hurts. Had a lot of fun, though
My first code golf :)
a="";function b(c){a+=" ".repeat(10-c);for(i=1;i<c;i++)a+=i;for(i=2;i<c;i++)a+=c-i;a+="\n";}for(i=2;i<11;i++)b(i);for(i=9;i>1;i--)b(i);document.write("<pre>"+a+"</pre>");
var str = "";
function row(line) {
str += " ".repeat(10 - line);
for (var i = 1; i < line; i++) {
str += i;
}
for (var i = 2; i < line; i++) {
str += line - i;
}
str += "\n";
}
for (var line = 2; line < 11; line++) {
row(line);
}
for (var line = 9; line > 1; line--) {
row(line);
}
document.write("<pre>" + str + "</pre>");
x=abs(-8:8);m=x+x';m(m>8)=25;[57-m,'']
To make it work in MATLAB too, you'd need to write x=ndgrid(abs(-8:8));m=x+x';m(m>8)=25;[57-m,''] or x=meshgrid(abs(-8:8));m=x+x';m(m>8)=25;[57-m,'']
9┅…ṫ¦€|ṫṣ
9┅ Push [1 2 3 4 5 6 7 8 9]
… Prefixes; push [[1] [1 2] [1 2 3] ... [1 2 3 4 5 6 7 8 9]]
ṫ¦ Palindromize each row: e.g. [1 2 3 4] -> [1 2 3 4 3 2 1]
€| Centre-align the rows, padding with spaces
ṫ Palindromize the rows
ṣ Join with newlines
