Robbers: Crack the regex - Make a snake

Question

This is the robber's thread. The cop's thread is here.

---

A snake matrix is a square matrix that follows this pattern:

3-by-3:

1  2  3
6  5  4
7  8  9

and 4-by-4:

1  2  3  4
8  7  6  5
9  10 11 12
16 15 14 13

Your task is to write a code that takes an input n and creates such a matrix, in the same language as a cop post, and with a code that matches the cop's regex. The output format of your code must match the output format of the cop's code.

Please leave a comment under the Cop's post to indicate that you have cracked it.

Winning criterion:

The winner will be the user that has cracked the most submissions. In case of a tie, then there will be multiple winners.

Answer 1

Jelly, 9 bytes, cracks @Dennis' answer

²sµUFḤ$¦G

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Should be correct now; I had to rather rethink what I was doing.

Explanation

The hardest part of this problem is getting a list of the even indices inside the ¦ (which applies an operation at specific indices). Unlike most operations, which use the input to the whole program as a default for the second operand, ¦ uses the most recently seen value as a default (because it conceptually has two left operands, rather than a left and a right operand like most things that take in two values).

However, we can observe that we have a list of numbers, including all integers from 1 to half the input, already in the current value at the time. As such, flattening and doubling it gives us a list of even numbers, including all the even indices (also some other even numbers but we don't care about those). It's even possible to spend one µ solely to avoid a parsing ambiguity and still come in within 9 characters.

²sµUFḤ$¦G
 s         Split
²          {the list of numbers from 1 to} the square of {the input}
  µ        into a number of pieces equal to {the input};
   U       then reverse
       ¦   the elements at indexes
    F      obtained by flattening the split list
      $    and
     Ḥ     doubling each element in the flattened list;
        G  finally, format it into a grid.         

Answer 2

05AB1E, Emigna

This was my first time using 05AB1E. Got it with a little bit of help. That was fun. :)

UXFXLNX*+N2BSR1k_iR}ˆ

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Explanation:

U                       Assign input to X
 XF                     For N in range(0,X):
   XL                   Push list [1 .. X]
     NX*+               Add N*X to each element of the list
         N2BSR1k_       Super clever way to get N%2:
                            not bool(reversed(str(bin(N))).index('1')) (N%2 is odd)
                 iR}    if N%2 is odd, then reverse the list
                    ˆ   Add row to global array
                        Implicit print

I actually found this similar program on my own, but the output format is different:

UXFXLNX*+N2BSR1k_iR}=

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[1, 2, 3, 4]
[8, 7, 6, 5]
[9, 10, 11, 12]
[16, 15, 14, 13]

See the edit history for my previous two attempts.

Answer 3

Python 2, Dennis

This is a fun token golfing problem.

while ord>eval:1;int,eval=int<1and(int,eval+1)or(-input(),1);i=int;ord=i*i;print'%*s'%(-i,(eval+~i+~-eval%-i*~1,eval)[eval/i&1])+'\nt'[:-1<eval%int],

Regex verification

Answer 4

Ohm, Nick Clifford

My first time attempting Ohm.
Really nice language that I look forward to use again :)

²@┼σ▓_^è?R

Explanation

²             # square input
 @            # range [1 ... input^2]
  ┼σ          # split in input sized chunks
    ▓         # perform the following block on each element of the array
     _        # current element
      ^è?     # if index is odd
         R    # reverse

My first attempt that didn't work as adding an array and an int is not possible:

@┼MDR┼+;W

My second attempt that didn't match the regex:

²@┼σ▓_^MR

Answer 5

05AB1E, Emigna (2nd submission)

First time working with 05AB1E.

VYLUYFYXDˆ+RU

Try it online! | Regex verification

Explanation

VYLUYFYXDˆ+RU  # Implicit input
V              # Save input to Y
 YL            # Push [1 .. Y]
   U           # Save list to X
    YF         # Repeat Y times:
      YX       # Push Y, then X
        Dˆ     # Add X into the global array (could've used X here instead)
          +    # Push X + Y
           R   # Reverse top of stack
            U  # Save updated list to X
               # Implicit loop end
               # Implicit global array print if stack is empty

Answer 6

CJam, Lynn

esmpmpmeimtmemqmememqicelic
esmpmpmeimememqmlmtmemoc
esmpmpmeimememqmtmtmtmtmeic
esmpmpmeimememqmtmtmtmtmeic
esmpmpmeimeiscic
esmpmpmeimemeimfsic
esmpmpmeisciscimqmtmemeic
esmpmpmeiscimlmqmqmemeic
esmpmpmeimemomqmqmemeic
esmpmpmeisciscimfsimqic
esmpmpmeimeiscic
esmpmpmeisciscimfsimqic
esmpmpmeimemomqmemqmemtmemoc
esmpmpmeiscic
esmpmpmeimemomeimqmeic
esmpmpmeimemeimqmlmtmeic
esmpmpmeimtmtmqmemtmtmeic
esmpmpmeimemomqmqmtmeic
esmpmpmeimemqmqmemeic
esmpmpmeiscimlmqmqmemeic
esmpmpmeiscimqmtmtmtmqmemeic
esmpmpmeimeimemtmqmemeic
esmpmpmeimeiscimlmlmtmlmtic
esmpmpmeimemeimqmlmtmeic
~~

All linefeeds are for cosmetic purposes and can be removed without affecting the program.

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Explanation

After Lynn removed {|} from the list of allowed characters, I had to try something new. It turns out we can still construct arbitrary strings and evaluate them as code.

First, we need to get some value onto the stack. The only available built-ins that push something without popping something else first (and without reading the input) are es, ea and et. I'm sure you could start from all of these one way or another, but I went with es which pushes the current timestamp. Since I didn't want to make any assumptions about its actual value, I test its primality with mp (which gives 0 and 1) and test that value's primality again to ensure I've got a 0 on the stack. A 1 will be more useful, so we compute exp(0) with me and turn it into an integer with i. So all the numbers start with:

esmpmpmei

Now we've got a whole bunch of unary maths operators to work with:

i    int(x) (floor for positive numbers, ceiling for negative)
me   exp(x)
ml   ln(x)
mq   sqrt(x)
mo   round(x)
mt   tan(x)

We can also combine a few built-ins for more elaborate functions of x:

sci     Extract first digit of x and add 48 (convert to string, convert
        to character, convert to integer).
ceui    Convert to character, convert to upper case, convert to integer.
celi    Convert to character, convert to lower case, convert to integer.
mfsi    Get a sorted list of prime factors of x and concatenate them into 
        a new number.
mfseei  Get a sorted list of prime factors, interleave it with 1,2,3,..., and
        concatenate the result into a new number.

Using these, we can obtain any number in 0 <= x < 128 (and many others) in less than 10 steps from 1. I'm sure a much smaller subset of these commands would also suffice. I've written a small Mathematica program, to determine all of these snippets (it's not very readable, sorry):

codes = SortBy[
  Select[Nest[Select[DeleteDuplicatesBy[SortBy[Join @@ (Through[{
               List,
               If[1 <= # < 50, {Exp@#, #2 <> "me"}, Nothing] &,
               If[# >= 1, {Log@#, #2 <> "ml"}, Nothing] &,
               If[# > 1, {Sqrt@#, #2 <> "mq"}, Nothing] &,
               {If[# > 0, Floor@#, Ceiling@#], #2 <> "i"} &,
               {Floor[# + 1/2], #2 <> "mo"} &,
               {Tan@#, #2 <> "mt"} &,               
               If[NumberQ@# && # >= 0, {First@
                   ToCharacterCode@ToString@#, #2 <> "sci"}, 
                 Nothing] &,
               If[IntegerQ@# && 
                  32 < # < 65536, {First@
                   ToCharacterCode@
                    ToUpperCase@FromCharacterCode@#, #2 <> "ceui"}, 
                 Nothing] &,
               If[IntegerQ@# && 
                  32 < # < 65536, {First@
                   ToCharacterCode@
                    ToLowerCase@FromCharacterCode@#, #2 <> "celi"}, 
                 Nothing] &,
               If[IntegerQ@# && # > 0, ## & @@ {
                   {FromDigits[
                    "" <> (ToString /@ (f = 
                    Join @@ Table @@@ FactorInteger@#))], #2 <> 
                    "mfsi"},
                   {FromDigits[
                    "" <> (ToString /@ 
                    MapIndexed[## & @@ {#2[[1]] - 1, #} &, f])], #2 <>
                     "mfeesi"}
                   }, Nothing] &
               }@##] &) @@@ #, StringLength@*Last], 
       First], #[[1]] < 1000000 &] &, {{1, "esmpmpmei"}}, 9], 
   IntegerQ@# && 0 <= # < 128 &@*First], First]

With that, we can simply push an arbitrary list of character codes, converting each of them to a character with c afterwards. Once we've pushed the entire code we want to execute, we push 95 (]). We eval that one with ~ to wrap all the others in a string, and then we eval that string with ~.

The actual code run at the end of the program is again:

ri__2#,:)/2/[1W]f.%:~<p

See my previous solution for an explanation.

Answer 7

Python 3, TuukkaX

Sorry, the regex you used was too easy to trivialize. No 0, # or ? No problem!

I might've misinterpreted the example output but it's still pretty easy to tweak since I have 45 spare characters left over

n=int(input())
print(str(n)+":")
x=1-1
exec("print([*range(1+x*n,1+n*-~x)][::(1,-1)[x%2]]);x+=1;"*n)
"no-op:a_string_that_doesnt_actually_matter"

Answer 8

R, MickyT

lets_make_a_snake<-function(n)`for`(i,1:n,cat(i*n+1-`if`(i%%2,n:1,1:n),"\n"))

Tests cases:

> lets_make_a_snake(4)
1 2 3 4 
8 7 6 5 
9 10 11 12 
16 15 14 13 
> lets_make_a_snake(7)
1 2 3 4 5 6 7 
14 13 12 11 10 9 8 
15 16 17 18 19 20 21 
28 27 26 25 24 23 22 
29 30 31 32 33 34 35 
42 41 40 39 38 37 36 
43 44 45 46 47 48 49 

And regex confirmation: https://regex101.com/r/OB8ZIM/1

I also had:

invisible(sapply(1:(n=scan()),function(i)cat(i*n+1-`if`(i%%2,n:1,1:n),"\n")))

which gives the same output and match the same regex.

Answer 9

Röda, fergusq

{|i|;a=1;while[a<=i]do;b=a*i-i+1;c=[];while[b<=a*i]do;c+=b;b++;done;{d=[];x=0;while[x<i]do;d+=c[i-x-1];x++;done[d]}if[a%2<1]else{[c]};a++;done;r="||||||"}

This is an anonymous function that matches this PCRE regex: ^{(\|[^\/#\s]*){8}$.

Try it online!

Answer 10

Bash, @Marcos M

sort -n <(seq 1 $((n * n)) | xargs -n$n | sed -n 'p;n'; seq $((n * n)) 1 | xargs -n$n | sort -n | sed -n 'n;p')

Prettified:

sort -n <(               # feed the stdout of this subshell into stdin for `sort -n`
    seq 1 $((n * n)) |   #   generate 1 2 3 ... n²
        xargs -n$n |     #     add line break every n numbers
        sed -n 'p;n';    #     print only odd lines
    seq $((n * n)) 1 |   #   generate n² ... 3 2 1
        xargs -n$n |     #     add line break every n numbers
        sort -n |        #     reverse all lines (keeping each row decreasing)
        sed -n 'n;p'     #     print only even lines
)

The first part of the subcommand will generate 1 2 3 4, 9 10 11 12, and the second part will generate 8 7 6 5, 16 15 14 13. The outer sort -n will properly mix them together to form the snake pattern.

I used the trick in https://superuser.com/a/101760 to print the odd and even lines. Thank you Marcos, really a fun one.

Answer 11

Javascript, Tom

f=(a,i=1,j=0)=>Array(a).fill(0).map(b=>Array(a).fill(0).map(a=>i++)).map(a=>j++%2?a.reverse():a).map(a=>a)

console.log(f(4))

Answer 12

Python 3, @TuukkaX

n=int(input());j=0;exec("print([(j-+i%n-n++2*n-0,j+i%n+1)[1&int(i/n)//1^(0x1)//1]*(2*(i%n)*0+2222222//2222222)for i in range(j,j+n)]);j+=n;"*n)

• Verify: https://regex101.com/r/GI1Zvh/1

---

Slightly analyzing the cop's regex shows a fixed template:

________________________"___________i%n____2*n-____i%n__________i/n)//1_____)//1___2*(i%n)____^^^^^^^^^^^^^^^^for i in range(j,____])______"*n)

where _ is any character except [ '"#] and ^ is any of [int()2/].

The "*n) at the end clearly shows an eval("..."*n) or exec("..."*n) is going on, so we just need to make sure the "..." prints the j-th row.

The for i in range(j, is too close to the end of the string, hinting list comprehension without any if. So we need to construct the i-th column using those i%n, 2*n stuff.

n = int(input())
j=0
exec("""print([
    (
        j - +i%n - n ++ 2*n - 0,    # equivalent to (n + j - i%n) for the decreasing rows
        j + i%n + 1                 # equivalent to (j + i%n + 1 == i + 1) for the increasing rows
    )[1 & int(i/n)//1 ^ (0x1)//1]   # int(i/n)   ->    get row number 0, 1, 2, 3, ...; 
                                    # 1 & int(i/n)//1    ->    0 for increasing rows, 1 for decreasing rows, 
                                    # 1 & int(i/n)//1 ^ (0x1)//1    ->   flip 0 and 1
    * (2*(i%n)*0+2222222//2222222)  # multiply by the constant 1.
    for i in range(j,j+n)
]); j+=n; "*n)

Answer 13

dc, Mitchell Spector

This was my first entry to a cops and robbers challenge, and I had a lot of fun. The regex needed to be matched was simple, ^[^# !]{59}$, which basically turned my job into a golfing one, without using those 3 characters. Initially I had difficulties getting below 60 bytes, but I cracked it in the end.

?sN0[AP]sP[ddlN~_2*lN+1-r2%*+1+n32P1+dlN%0=PdlNd*>L]dsLxqqq

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Explanation:

My code uses one loop with N ² iterations, keeping a zero based counter (1D), and calculates what number needs to be printed based on the corresponding matrix row and column (r, c) coordinates.

Example of what I mean, if N = 4:

 0  1  2  3            (0,0) (0,1) (0,2) (0,3)             1  2  3  4
 4  5  6  7     ->     (1,0) (1,1) (1,2) (1,3)     ->      8  7  6  5
 8  9 10 11            (2,0) (2,1) (2,2) (2,3)             9 10 11 12
12 13 14 15            (3,0) (3,1) (3,2) (3,3)            16 15 14 13

It looks complicated, but the intermediary step is helpful. Plus, I've tried using 2 loops from the start, but I ended up above the regex character limit. Number generation at each iteration (zero based):

• if r % 2 = 0 (normal row), n = (r * N) + c = counter
• if r % 2 = 1 (reversed row), n = ((r + 1) * N) - c - 1 = counter + N - (2 * c) - 1

Or all at once, as one based numbering: n = counter + ((N - (2 * c) - 1) * (r % 2)); n++

?sN0             # read input, save as N, initialize iteration counter
[AP]sP           # macro 'P' that prints a newline (ASCII code 10 = A)
[                # start loop
    ddlN~        # push N, calculate row and column coordinates:
                 #r = int(counter / N), c = counter % N, '~' calculates both
    _2*lN+1-     # c is on top, so this does: N - (2 * c) - 1
    r2%*+        # now r is on top, do: (r % 2) * (previous result) + counter
    1+n32P       # do: n++, print space (ASCII code 32)
    1+           # increment counter
    dlN%0=P      # call macro 'P' every Nth printed number
    dlNd*>L      # if: N * N > counter, repeat loop
]dsLx            # this saves the loop to macro 'L', then executes it
qqq              # my script was shorter, so I added a bunch of quit commands to
                 #fit the regex limit. Use of comments ('#') was prohibited.

Answer 14

PowerShell, ConnorLSW

Crack

$mySnakeIndex=1;$seq=1..$args[0];$seq|%{$rowNum=$seq|%{($mySnakeIndex++)};if(!($_%2)){[array]::Reverse($rowNum)};$rowNum-join" "}

I started with a smaller solution to the issue and padded my variable names to get the regex to match. Trying to find a use for the colon I suppose was the hardest part to wrap my head around.

$a=1..$args[0];$i=1;$a|%{$r=$a|%{($i++)};if(!($_%2)){[array]::Reverse($r)};$r-join" "}

Explanation

# Initialize a counter that starts at one.
$mySnakeIndex=1
# Save the integer array from 1 to the input value. 
$seq=1..$args[0]
# For each row of the output...
$seq|%{
    # Build the integer array for this row sequentially
    $rowNum=$seq|%{
        # Increase the integer index while sending it down the pipeline
        ($mySnakeIndex++)}
        # Check if this is and odd row. If so reverse the integer array.
        if(!($_%2)){[array]::Reverse($rowNum)}
        # Take this row and join all the numbers with spaces.
        $rowNum-join" "

Answer 15

CJam, Lynn

Something like this:

ri
{s}seu~~ci{zs}seu~~c{a}seu~~|~{w}seu~~z{w}seu~~sc~c{z}seu~~{w}seu~~sc~c{w}seu~~z{w}seu~~sc~c~

{s}seu~~ci{z}seu~~{s}seu~~c{a}seu~~|~{w}seu~~z{w}seu~~sc~c
{s}seu~~ci{z}seu~~{s}seu~~c{a}seu~~|~{w}seu~~z{w}seu~~sc~c
{s}seu~~ci{z}seu~~{s}seu~~c{a}seu~~|~{w}seu~~z{w}seu~~sc~c
{s}seu~~c{a}seu~~|
{s}seu~~c{c}seu~~|
{t}seu~~sc{a}seu~~|
{s}seu~~c{a}seu~~|{w}seu~~z{w}seu~~sc~c
{s}seu~~sc{fb}seu~~||
{s}seu~~sc{i}seu~~|
{s}seu~~sc{fb}seu~~||
{s}seu~~ci{z}seu~~{s}seu~~c{a}seu~~|~{w}seu~~z{w}seu~~sc~c{z}seu~~{w}seu~~sc~c{w}seu~~z{w}seu~~sc~c
{a}seu~~scs
{w}seu~~
{s}seu~~ci{z}seu~~{s}seu~~c{a}seu~~|~{z}seu~~{w}seu~~sc~c
{fb}s{b}s{w}seu~~sc~
{s}seu~~sc{ee}seu~~||
{s}seu~~sc{z}seu~~|{w}seu~~{w}seu~~sc~{w}seu~~{w}seu~~sc~
{t}seu~~sc{a}seu~~|
{~}s{}s{w}seu~~sc~
{t}seu~~sc{c}seu~~|

{s}seu~~ci{z}seu~~{s}seu~~c{a}seu~~|~{z}seu~~{w}seu~~sc~c~
s~
p

All whitespace is for... "readability"... and can be omitted to comply with Lynn's regex.

Try it online!

Explanation

The regex requires that we solve problem using only:

• Lower-case letters.
• {}, which can be used to create blocks.
• |, mostly used for bitwise OR.
• ~, "eval" and bitwise NOT (also "dump array", but I'm not going to use it).

Since we have, ~, if we can construct arbitrary strings, we can run arbitrary code. However, at first it's not obvious how to do that.

The first piece of the puzzle is that blocks are unevaluated bits of code, which can turn into strings with s. So {abc}s gives us "{abc}". Next, we can use eu to convert these strings to upper case.

{abc}seu  e# gives "{ABC}"

The benefit of this is that upper-case letters are pre-initialised variables, so we can get a lot of constant values by creating such a string, and eval'ing it twice (once to turn the string back into a block and once to execute that block). We can't get all the letters, because some, like x aren't valid commands (so CJam will refuse to parse a block containing them). We can't use f as is, because it needs to be followed by another command, but we can use fb and then OR the two values together. Likewise, we can use ee instead of e. With that, we can get the numbers 0, -1, 3, and 10 to 19. The -1 is convenient, because if we turn it into a string ("-1") then into a character ('-) and then eval it, we can get either subtraction or set difference. Like I said, we can't get X (for 1), but we can take the absolute value of -1 with z.

We can also use s to get a string containing a space, and use c to turn that into a space character:

{s}seu~~c

This is convenient, because from there we can get lots of useful commands in the lower ASCII range by ORing the space with various numbers. To get some of the characters above code point 48, we use the character '0 as the basis instead:

{t}seu~~si

This is already enough to construct arbitrary strings, because we can get '+ (addition and string concatenation) from the following snippet:

{s}seu~~c{b}seu~~|

And we have a literal 1 so we can just push space characters, increment them to the value we need and then concatenate them all together, but that's a bit boring and the code would become massive.

Instead, I've generated [ and ] and evalled them, so that all the characters I push in between are wrapped in a string automatically. That's these two lines:

{s}seu~~ci{zs}seu~~c{a}seu~~|~{w}seu~~z{w}seu~~sc~c{z}seu~~{w}seu~~sc~c{w}seu~~z{w}seu~~sc~c~

...

{s}seu~~ci{z}seu~~{s}seu~~c{a}seu~~|~{z}seu~~{w}seu~~sc~c~

And finally, we'll need f and ~ in the string we're generating. While those are valid characters already, we don't have string literals or character literals, so we'd have to generate these as well and building up larger code points from the space is a bit annoying. Instead, I've made use of the set subtraction here, but subtracting two blocks (to get rid of the {}):

{fb}s{b}s{w}seu~~sc~
...
{~}s{}s{w}seu~~sc~

That's pretty much all there is to it. We eval [. We push all the characters, obtained by various computations from the few built-in constants we have, |, - (via eval) and + (via eval). We eval ]. We flatten the entire thing into a string, because at some point I added some strings or numbers into the list. We eval our arbitrary string with ~.

The ri...p are part of the actual final program, but I've extracted them because they don't need encoding.

Finally, this is the program we're actually running:

ri___*,:)/2/[1-1]f.%:~<p

ri      e# Read input and convert to integer.
__      e# Make two copies.
_*      e# Square the last copy.
,       e# Turn into range [0 1 ... n^2-1].
:)      e# Increment each to get [1 2 ... n^2].
/       e# Split into chunks of length n, creating a square.
2/      e# Split into pairs of lines.
[1-1]   e# Push [1 -1].
f.%     e# Use this to reverse the second line in each pair. If n was odd,
        e# this will pair a -1 with the last line.
:~      e# Flatten the pairs back into the square.
<       e# Truncate to n lines to get rid of that extraneous -1 for odd inputs.
p       e# Pretty-print.

Answer 16

tinylisp, @DLosc

(v(c(h(q(d)))(c(h(q(f)))(q((c(q(n))(q((g(v(h(q(n))))(s(v(h(q(n))))(v(h(q(1)))))())))))))))(v(c(h(q(d)))(c(h(q(mod)))(q((c(c(h(q(x)))(q(y)))(q((i(l(v(h(q(x))))(v(h(q(y)))))x(mod(s(v(h(q(x))))(v(h(q(y)))))y))))))))))(v(c(h(q(d)))(c(h(q(range)))(q((c(c(h(q(x)))(c(h(q(y)))(c(h(q(z)))(q(w)))))(q((i(l(times(v(h(q(z))))(v(h(q(x))))(v(h(q(0)))))(times(v(h(q(z))))(v(h(q(y))))(v(h(q(0))))))(range(v(h(q(x))))(s(v(h(q(y))))(v(h(q(z)))))z(c(s(v(h(q(y))))(v(h(q(z)))))w))w)))))))))(v(c(h(q(d)))(c(h(q(times)))(q((c(c(h(q(x)))(c(h(q(y)))(q(acc))))(q((i(l(v(h(q(x))))(v(h(q(0)))))(times(s(v(h(q(0))))(v(h(q(x)))))(s(v(h(q(0))))(v(h(q(y)))))acc)(i(e(v(h(q(x))))(v(h(q(0)))))acc(times(s(v(h(q(x))))(v(h(q(1)))))y(a(v(h(q(y))))(v(h(q(acc))))))))))))))))(v(c(h(q(d)))(c(h(q(g)))(q((c(c(h(q(n)))(c(h(q(w)))(q(r))))(q((i(l(v(h(q(w))))(v(h(q(0)))))r(g(v(h(q(n))))(s(v(h(q(w))))(v(h(q(1)))))(c(i(e(v(h(q(0))))(mod(v(h(q(w))))(v(h(q(2))))))(range(a(v(h(q(1))))(times(v(h(q(w))))(v(h(q(n))))(v(h(q(0))))))(a(a(v(h(q(1))))(times(v(h(q(w))))(v(h(q(n))))(v(h(q(0))))))n)1())(range(a(times(v(h(q(w))))(v(h(q(n))))(v(h(q(0)))))n)(times(v(h(q(w))))(v(h(q(n))))(v(h(q(0)))))(s(v(h(q(0))))(v(h(q(1)))))()))r)))))))))))

Try it online!

This defines a function f that returns the answer. It also prints the names of the functions I defined to stdout, but I assume that doesn't matter, since [we're allowed to choose our output stream] for code golf, at least. If it's a big deal, I think I can tweak it to not print those. How did I do it? I started with something fairly standard:

(d f (q ((n)
  (g n (s n 1) ()))))

(d mod (q((x y)
  (i (l x y) x
    (mod (s x y) y)))))

(d range (q((x y z w)
  (i (l (times z x 0) (times z y 0))
    (range x (s y z) z (c (s y z) w))
    w))))

(d times (q ((x y acc)
  (i (l x 0) (times (s 0 x) (s 0 y) acc)
  (i (e x 0) acc
    (times (s x 1) y (a y acc)))))))

(d g (q ((n w r)
  (i (l w 0) r
    (g n (s w 1)
       (c (i (e 0 (mod w 2))
             (range (a 1 (times w n 0)) (a (a 1 (times w n 0)) n) 1 ())
             (range (a (times w n 0) n) (times w n 0) (s 0 1) ()))
          r))))))

Then, I observed that one can transform functions definitions like this:

(d mod (q ((x y) (body))))

becomes

(v(c(h(q(d)))(c(h(q(mod)))(q((c(c(h(q(x)))(q(y)))(q((body)))))))))

And function calls like this:

(a x y)

becomes

(a(v(h(q(x))))y)

I used this recursive Vim macro, stored in register q, to do the second one (I have jk mapped to <Esc>): f s(v(h(q(jkea))))jk@q.

These two transformations were sufficient to eliminate all spaces.

Answer 17

Swift, @James Webster

{n in for j in 0..<n{print((1...n).map{j%2==0 ?$0+j*n:j*n+n+1-$0})}} as(CShort)->()

Verification: https://regex101.com/r/7ukrM0/1

Answer 18

PHP, @JörgHülsermann

<?=(function($n,$snake){foreach(array_chunk(range(1,$n*$n),$n)as$i=>$a){if($i%2)$a=array_reverse($a);echo"\n",join('',array_map(function($e){return(sprintf("%3d",$e));},$a));}})($argv[1],'-=-=-=-=-=-=-=-=-=-=-=-=-=-o~')?>

• Verify: https://regex101.com/r/81pUrB/1
• TIO: https://tio.run/#aNvF0

---

221 bytes is too long (thus the snake), and the lack of whitespace can be easily worked-around.

Prettified:

<?=
(
    function($n, $snake) {
        foreach (array_chunk(range(1, $n*$n), $n) as $i => $a) {
            if($i % 2)
                $a = array_reverse($a);
            echo "\n", join('', array_map(function($e) {
                return (sprintf("%3d", $e));
            }, $a));
        }
    }
)($argv[1], '-=-=-=-=-=-=-=-=-=-=-=-=-=-o~')
?>

Answer 19

Jelly, length 12, @JonathanAllan

Ḷ-*m@"s@²$G

Try it online!

How it works

Ḷ-*m@"s@²$G  Main link. Argument: n

Ḷ            Unlength; yield [0, ..., n-1].
 -*          Yield [(-1)**0, ..., (-1)**(n-1)].
         $   Combine the two links to the left into a chain.
        ²    Yield n².
      s@     Split [1, ..., n²] into chunks of length n.
   m@"       Take the lists to the right modulo the units to the left, 1 being
             normal order and -1 being reversed.

Answer 20

Jelly, 12 bytes, cracks @JonathanAllan's second answer

²sµ;@/€FḤ$¦G

Try it online!

Explanation

This is almost the same as my other answer. I just made two changes:

First, I changed U ("reverse each element") to Ṛ€ ("reverse" "each element"). That doesn't help by itself, because Ṛ is also banned.

Then, I changed the Ṛ ("reverse") to ;@/ (/ "fold by" ; "concatenating" @ "in the opposite order to the original list"). That avoids all the banned characters, giving a valid solution.

I assume the next step would be to start banning array-manipulation quicks, in addition to the atoms.

Answer 21

Jelly, length 13, @JonathanAllan

1r-*Nm@"s@²$G

Try it online!

How it works

1r-*Nm@"s@²$G  Main link. Argument: n

1r             Range 1; yield [1, ..., n].
 -*            Yield [(-1)**1, ..., (-1)**n].
   N           Negate each unit.
           $   Combine the two links to the left into a chain.
          ²    Yield n².
        s@     Split [1, ..., n²] into chunks of length n.
     m@"       Take the lists to the right modulo the units to the left, 1 being
               normal order and -1 being reversed.

Answer 22

Jelly, length 14, @JonathanAllan

²sµðạ"J×2$$¦GµL.xị"ḅ1µ$

Try it online!

Answer 23

Scala, @Soapy

def g(n:Int) = {
    var vec = Vector.fill(0)(Vector.fill(0)(1))
    for (p <- 1 to n) {
        var vec2 = Vector.fill(0)(1)
        for (q <- (p-1)*n+1 to p*n) {
            vec2 = vec2 ++ Vector(q)
        }
        if (p%2==1) vec = vec ++ Vector(vec2)
        else vec = vec ++ Vector(vec2.reverse)

    }
    println(vec)
}

Haven't touched Scala in awhile, it was fun to revisit. Unfortunately, this solution misses out on a lot of Scala's cool features.

Try it out here

Regex Confirmation

Answer 24

QBasic (QB64), @DLosc

Note that since the . does not match \n (U+000A, LF), the newline here is a \r (U+000D, CR).

INPUT N:ZERO=N-N:ONE=N/N:FOR I=ZERO TO N-ONE:FOR J=ONE TO N:IF ZERO=I MOD(ONE+ONE)THEN PRINT I*N+J;ELSE PRINT I*N+N-J+ONE;REM
NEXT:PRINT:NEXT

Verify:

>>> re.match('^([A-Z]+.)+$', 'INPUT N:ZERO=N-N:ONE=N/N:FOR I=ZERO TO N-ONE:FOR J=ONE TO N:IF ZERO=I MOD(ONE+ONE)THEN PRINT I*N+J;ELSE PRINT I*N+N-J+ONE;REM\rNEXT:PRINT:NEXT')
<_sre.SRE_Match object; span=(0, 141), match='INPUT N:ZERO=N-N:ONE=N/N:FOR I=ZERO TO N-ONE:FOR >

---

The main difficulty is how to insert a word after the ;. Thankfully, QB64 treats CR as a newline while Python's regex doesn't, so we could slip a REM\r here. Out of the five allowed regex flavors,

• Perl accepts "\r" =~ /./
• JavaScript rejects !/./.test('\r') (\n, \r, \u2028, \u2029 are all line separators)
• Python accepts re.match('.', '\r') (only \n is considered a line break)
• Golang accepts regexp.Match(".", "\r")
• Ruby accepts /./ =~ "\r"

So this crack is fine as long as we don't mention JavaScript...

Answer 25

Python 3, kennytm

def S(N,A=1,K=1,E=list,r=range):print(E(r(A,A+N)[::K]))or(A>N*N-N)or(((S(N,A+N,-K))))

Regex verification

Answer 26

><>, torcado

!v &0_!
_<>~ao1+>_v_
?______;__>:&:&=?;::2%:}+&:&*{3+0$.
?!v1+:n' 'o:&:&%_
?!v:n' 'o1-:&:&%_

Answer 27

C, @Yimin Rong

main(int c,char**p){int n=atoi(*++p),i=n-n,t,o=c;for(--o;i<n;++i)for(t=o;t<=n;++t)printf("%-*d%c",n-o,i%c?i*n+n+o-t:i*n+t,t%n?' ':'\n');}

The program cannot contain numbers, but we can get numbers via:

1. c, commonly known as "argc", which is always 2.
2. + and - are available, so we can create 0 with n-n, and create 1 with o=c;--o.

Answer 28

PHP, @Ionut Botizan

In the moment I have no better idea to crack the original solution.

Supports n<=15

It is the first time that I use getopt. Maybe not the best idea to use options as input.

start from the command line like this

php hack.php -a=4

Original Regex

Level 1:

^<[^'"\d{vV;<$]+$

Very nice combination of letters. Feel free to vote up the cops thread.

It blocks me functions like - strrev - array_reverse - get_defined_vars

https://regex101.com/r/5rGTnw/2

Level 2:

^<[^'"\d{v;<$_~|&A-Z]+$

https://regex101.com/r/XtVl9G/1

Solution

 <?php
    error_reporting(~E_NOTICE)&
    define(A,a.chr(E_COMPILE_ERROR-E_NOTICE+E_WARNING))
    &define(B,getopt(A,[])[a])&print_r(array_chunk(
    array_slice(
    array_merge(
    range(E_ERROR,B)
    ,range(E_WARNING*B,E_ERROR+B)
    ,range(E_WARNING*B+E_ERROR,(E_WARNING+E_ERROR)*B)
    ,range(E_PARSE*B,+E_ERROR+(E_WARNING+E_ERROR)*B)
    ,range(E_PARSE*B+E_ERROR,(E_PARSE+E_ERROR)*B)
    ,range((E_PARSE+E_WARNING)*B,+E_ERROR+(E_PARSE+E_ERROR)*B)
    ,range((E_PARSE+E_WARNING)*B+E_ERROR,(E_NOTICE-E_ERROR)*B)
    ,range(E_NOTICE*B,+E_ERROR+(E_NOTICE-E_ERROR)*B)
    ,range(E_NOTICE*B+E_ERROR,(E_NOTICE+E_ERROR)*B)
    ,range((E_NOTICE+E_WARNING)*B,E_ERROR+(E_NOTICE+E_ERROR)*B)
    ,range((E_NOTICE+E_WARNING)*B+E_ERROR,(E_NOTICE+E_WARNING+E_ERROR)*B)
    ,range((E_NOTICE+E_PARSE)*B,E_ERROR+(E_NOTICE+E_WARNING+E_ERROR)*B)
    ,range((E_NOTICE+E_PARSE)*B+E_ERROR,(E_NOTICE+E_PARSE+E_ERROR)*B)
    ,range((E_CORE_ERROR-E_WARNING)*B,E_ERROR+(E_NOTICE+E_PARSE+E_ERROR)*B)
    ,range((E_CORE_ERROR-E_WARNING)*B+E_ERROR,(E_CORE_ERROR-E_ERROR)*B)
    )
    ,B-B,B*B
    ),B)
    )
    ?>

Level 2:

<?php
define(aa,a.chr(ord(strtoupper(a))-ord(h)+ord(a)))and
define(bb,getopt(aa,[])[a])and
define(us,chr(ord(a)-true-true))and
(prin.t.(us).r)(
(arra.y.(us).chunk)(
(arra.y.(us).slice)(
(arra.y.(us).merge)(
range((ord(b)-ord(a)),bb)
,range((ord(c)-ord(a))*bb,(ord(b)-ord(a))+bb)
,range((ord(c)-ord(a))*bb+(ord(b)-ord(a)),((ord(c)-ord(a))+(ord(b)-ord(a)))*bb)
,range((ord(e)-ord(a))*bb,+(ord(b)-ord(a))+((ord(c)-ord(a))+(ord(b)-ord(a)))*bb)
,range((ord(e)-ord(a))*bb+(ord(b)-ord(a)),((ord(e)-ord(a))+(ord(b)-ord(a)))*bb)
,range(((ord(e)-ord(a))+(ord(c)-ord(a)))*bb,+(ord(b)-ord(a))+((ord(e)-ord(a))+(ord(b)-ord(a)))*bb)
,range(((ord(e)-ord(a))+(ord(c)-ord(a)))*bb+(ord(b)-ord(a)),((ord(e)-ord(a))-(ord(b)-ord(a)))*bb)
,range((ord(e)-ord(a))*bb,+(ord(b)-ord(a))+((ord(e)-ord(a))-(ord(b)-ord(a)))*bb)
,range((ord(e)-ord(a))*bb+(ord(b)-ord(a)),((ord(e)-ord(a))+(ord(b)-ord(a)))*bb)
,range(((ord(e)-ord(a))+(ord(c)-ord(a)))*bb,(ord(b)-ord(a))+((ord(e)-ord(a))+(ord(b)-ord(a)))*bb)
,range(((ord(e)-ord(a))+(ord(c)-ord(a)))*bb+(ord(b)-ord(a)),((ord(e)-ord(a))+(ord(c)-ord(a))+(ord(b)-ord(a)))*bb)
,range(((ord(e)-ord(a))+(ord(e)-ord(a)))*bb,(ord(b)-ord(a))+((ord(e)-ord(a))+(ord(c)-ord(a))+(ord(b)-ord(a)))*bb)
,range(((ord(e)-ord(a))+(ord(e)-ord(a)))*bb+(ord(b)-ord(a)),((ord(e)-ord(a))+(ord(e)-ord(a))+(ord(b)-ord(a)))*bb)
,range(((ord(q)-ord(a))-(ord(c)-ord(a)))*bb,(ord(b)-ord(a))+((ord(e)-ord(a))+(ord(e)-ord(a))+(ord(b)-ord(a)))*bb)
,range(((ord(q)-ord(a))-(ord(c)-ord(a)))*bb+(ord(b)-ord(a)),((ord(q)-ord(a))-(ord(b)-ord(a)))*bb)
)
,bb-bb,bb*bb
),bb)
)
?>

Answer 29

Ruby, @Value Ink

->n{(1..n).map{|r|x=(r*n-n+1..r*n).to_a;if(r.modulo(2)==1)then(x)else(x.reverse)end}}#1-2-3-4-5-6--

[(-=Z-~]* means "I can write anything I like :)"

Answer 30

PHP, @Ionut Botizan

Once upon a time I remember that PHP has a goto

https://regex101.com/r/m5kNRd/1/

double assigments are alternatives for example Newline under Mac and Linux

solution with tab

<?="";
$p="M"&";";
$cr="M"&"-";
$cr="Z"&"*";
$zero=$one=$line=$text="";
$zero++;$zero--;
$a=$zero;
$s="$zero"&"$";
$one=++$one;
$g=$argv{$one};
$dir=$one;
a:++$a;
$line=$dir?"$line$a":"$a$line";
$line=$a%$g?$dir?"$line$p":"$p$line":"$line$cr";
$dir=$a%$g?$dir:$one-$dir;
$text=$a%$g?$text:"$text$line";
$line=$a%$g?$line:"";
echo$a-$g*$g?"":$text;
$a-$g*$g?:exit;
goto a;

First I have tought that I must use dynamic spaces

<?="";
$cr="M"&"-";
$cr="Z"&"*";
$zero=$one=$line=$text="";
$zero++;$zero--;
$a=$zero;
$s="$zero"&"$";
$one=++$one;
$five=$one+$one+$one+$one+$one;
$ten=$five+$five;
${"v$zero"}="$s$s$s$s$s$s$s";
${"v$one"}="$s$s$s$s$s$s";
${"v$ten"}="$s$s$s$s$s";
$g=$argc-$one;
$g=$argv{$one};
$space=$zero;
$p=${"v$space"};
$dir=$one;
a:++$a;
$a%$ten?:$space++;
$p=${"v$space"}??$p;
$line=$dir?"$line$a":"$a$line";
$line=$a%$g?$dir?"$line$p":"$p$line":"$line$cr";
$dir=$a%$g?$dir:$one-$dir;
$text=$a%$g?$text:"$text$line";
$line=$a%$g?$line:"";
echo$a-$g*$g?"":$text;
$a-$g*$g?:exit;
goto a;

Answer 31

tinylisp, @DLosc

A very straightforward solution, and totally unoptimized :)

(d p(q((m n)(s m(s(s 1 1)n)))))(d j(q((f g)(i(l f g)(c f(j(p f 1) g))()))))(d r(q((f g)(i(l f g)(c(s g 1)(r f(s g 1)))()))))(d k(q((m o f g n)(i(l n g)()(c(m f(p f n))(k o m(p f n)(p g 1) n))))))(d f(q((n)(k j r 1 1 n))))

Call as (disp (f 4)).

• (p m n) computes m + n using subtraction s (m + n == m - ((1 - 1) - n))
• (j f g) generates (f f+1 f+2 ... g-1)
• (r f g) generates (g-1 g-2 g-3 ... f)
• (k m o f g n) generate one row of the snake matrix, and then recurve itself for the next row, until n rows are created. The arguments m, o are substituted by j/r to generate increasing or decreasing rows. The arguments f, g are running indices to know which row we are on.
• (f n) calls (k j r 1 1 n) to start the generation.

Answer 32

PHP, @Jörg Hülsermann

$n=next($argv);$a=array();for($i=0;$i<$n;++$i){$f=$i*$n;$g=$f+++$n;$a{$i}=$i&1?range($g,$f):range($f,$g);}echo(serialize($a));;

In PHP, range(1, 4) gives [1, 2, 3, 4] while range(4, 1) gives [4, 3, 2, 1]. o_O.

Answer 33

Racket Intermediate Student with Lambda, @Brian McCutchon

(λ(n)(foldr(λ(p q)(cons((λ(x)(if(even? p)(range(+ x 1)(+ x n 1)1)(range(+ x n)x -1)))(* p n))q))'()(range 0 n 1)))

(λ can be replaced by lambda, both are allowed in the regex.)

Since I can't get TIO to restrict to htdp-intermediate-lambda and I don't want to download a new language, I will just assume it is fine as long as I only use functions given on that page:
λ foldr cons if even? range + *.

Since TIO does not include the words "list" in the output (it just produces '((1 2 3 4) (8 7 6 5)…), if the above crack is rejected, consider this quick fix:

(λ(n)(cons 'list(foldr(λ(p q)(cons(cons 'list((λ(x)(if(even? p)(range(+ x 1)(+ x n 1)1)(range(+ x n)x -1)))(* p n)))q))'()(range 0 n 1))))

Answer 34

Mathematica, @Martin Ender (non-competing)

((#//Range)+i#-#)~SortBy~(-(-#)^i&)~Table~{i,#} &

Answer 35

QBasic, @DLosc

DEFSTR S:CLS:INPUT N:ONE=N\N:TWO=ONE+ONE:FOUR=TWO+TWO:SELECT CASE N:CASE IS<FOUR:FW=ONE:CASE IS<FOUR*TWO+TWO:FW=TWO:CASE ELSE:FW=ONE+TWO:END SELECT:READ S:FOR I=ONE TO N:FOR J=ONE TO N:COL=J*FW+J-FW:LOCATE I+ONE,COL:G=I MOD TWO:F=J*TWO-ONE-N:V=I*N-J+ONE+G*F:PRINT V:LOCATE I+ONE,COL:PRINT S:NEXT:LOCATE I+ONE,N*FW+N+ONE:PRINT S:NEXT:DATA|

---

Prettified (s/:/\n/g):

DEFSTR S
CLS
INPUT N
ONE = N\N
TWO = ONE+ONE
FOUR = TWO+TWO
SELECT CASE N
    CASE IS < FOUR: FW = ONE
    CASE IS < FOUR*TWO+TWO: FW = TWO
    CASE ELSE: FW = ONE+TWO
END SELECT
READ S
FOR I=ONE TO N
    FOR J=ONE TO N
        COL = J*FW+J-FW
        LOCATE I+ONE, COL
        G = I MOD TWO
        F = J*TWO-ONE-N
        V = I*N-J+ONE+G*F
        PRINT V
        LOCATE I+ONE, COL
        PRINT S
    NEXT
    LOCATE I+ONE, N*FW+N+ONE
    PRINT S
NEXT
DATA|

Use LOCATE to force the output position. Use DATA/READ to obtain "|" without using functions.

Answer 36

Racket Intermediate Student with Lambda, @Brian McCutchon

(λ (n)(((λ (f) (f f))(λ(s)(λ (i a b) (cond [(= i(* n n))'()] [#t(cons(a 1 i)((s s) (+ n i)b a))]))))0((λ (g) (g g))(λ(u)(λ (j i) (cond [(> j n)'()][#t(cons(+ j i)((u u) (+ 1 j)i))]))))((λ(g) (g g))(λ(d) (λ(j i) (cond [(> j n)'()][#t(cons(+(- n j)1 i)((d d) (+ 1 j)i))]))))))

Uses this trick to perform recursion without define/letrec/let*, otherwise a pretty standard algorithm. Some spaces cannot be eliminated due to the regex restriction.

Answer 37

Jelly, 15 bytes, @ais523

s@²UJḤ$¦Ȯṛ“”
Dzs

Try it online!

Answer 38

Jelly, 16 bytes, @ais523

s@²UJḤ$¦Ȯṛ“”
Çœ?²s

Try it online!

Answer 39

05AB1E, mbomb007

Unlikely to be the intended solution :)

Crack

UXF**XLNNVX*F>}YFR}ˆ

Explanation

U                      # store input in X
 XF                    # X times do:
   **                  # multiply nothing twice
     XL                # range [1 ... input]
       N               # push current iteration number
        NV             # store current iteration number in Y
          X*           # multiply current iteration by input
            F>}        # increment the range that many times
               YFR}    # iteration number times, reverse the range
                   ˆ   # add it to global array
                       # implicitly print global array after program end

Answer 40

Go, @Kristoffer Sall-Storgaard

package main

import . "fmt"

const (
    zero = iota
    one
    two
)

func inc(p int) int {
    var bit = one
    for {
        if (p & bit) == zero {
            return p | bit
        }
        p ^= bit
        bit *= two
    }
}

func dec(p int) int {
    var bit = one
    for {
        if (p & bit) == bit {
            return p ^ bit
        }
        p |= bit
        bit *= two
    }
}

func main() {
    var n int
    var i int
    var row = one
    var col = zero
    Scan(&n)
    print("[")
    for n*n > i {
        i = inc(i)
        if col == zero {
            print("[")
        }
        if row%two == one {
            print(i)
        } else {
            var m = row * n
            var c = col
            for c > zero {
                m = dec(m)
                c = dec(c)
            }
            print(m)
        }
        col = inc(col)
        if col == n {
            print("]")
            col = zero
            row = inc(row)
        }
        if i != n*n {
            print(" ")
        }
    }
    print("]")
}

---

The regex means we only use one character from + to <, i.e.

+,-./:;<0123456789

The "." is needed for fmt.Scanf anyway. Although the character set is restricted, with unlimited length we could still write very complex programs.

• The lack of numbers is trivial to workaround, just use iota to define constants.
• The lack of < is also simple, just use > instead.
• The lack of + and - are trickier. Here I reimplemented inc(x) == x+1 and dec(x) == x-1 using bit operations. Perhaps there is an easier way but this isn't a code-golf
• The lack of , means we cannot use functions with two or more arguments, so make() and append() are out. There might be other ways to construct an array, but I find printing the output as if it is an array is simpler.

Answer 41

C#, @Eklipz

static int[,]F(int n){var a=new int[n,n];for(int r=n-n,z=r;r<n;++r)for(var c=z;c<n;++c)a[r,c]=n*r+((r&-~z)!=z?n-c:c-~z);;;;;;;;;;;;;;;;;;;;;;;return a;}

Uses standard tricks like z := n - n == 0 and -~z == 1 to produce the missing numbers.