10
\$\begingroup\$

This is the Robber post. The Cop post is here.


Your task is to take an integer input N and output the Nth digit in the sequence OEIS A002942.

The sequence consists of the square numbers written backwards:

1, 4, 9, 61, 52, 63, 94, 46, 18, 1, 121, 441, ...

Note that leading zeros are trimmed away (100 becomes 1, not 001). Concatenating this into a string (or one long number gives):

1496152639446181121441

You shall output the Nth digit in this string/number. You may choose to take N as 0-indexed or 1-indexed (please state which one you choose).

Test cases (1-indexed):

N = 5,      ==> 1
N = 17,     ==> 1   <- Important test case! It's not zero.
N = 20,     ==> 4
N = 78,     ==> 0
N = 100,    ==> 4
N = 274164, ==> 1

Your code should work for numbers up to N = 2^15 (unless your language can't handles 32 bit integers by default, in which case N can be lower).


Robbers:

You should try to crack the Cops' posts.

Your code must be in the same language as the Cop post, and have a Levenshtein distance exactly equal to the distance given by the cop. Your code cannot be longer than the original solution (but it can be the same size).

You may check the Levenshtein distance here!

The winner will be the robber that cracked the most posts.

\$\endgroup\$
3
  • \$\begingroup\$ Wait... so if the robber's result doesn't have to be the same as the original intended program... Can't the cop just write one program and make up a distance...? \$\endgroup\$ Nov 2, 2017 at 15:55
  • \$\begingroup\$ Well, the cops must provide the alternative code in order to mark the submission as safe and be eligible for the win. I've clarified in the cop post. :) \$\endgroup\$ Nov 2, 2017 at 17:03
  • \$\begingroup\$ I've never tried a cop-and-robber challenge. All of this was very confusing to me hah! \$\endgroup\$ Nov 2, 2017 at 17:11

13 Answers 13

4
\$\begingroup\$

Haskell, Laikoni

((show.(*1).read.reverse.show.(^2)=<<[1..])!!)

Try it online!

The (*1) was necessary for type checking.

\$\endgroup\$
0
3
\$\begingroup\$

JavaScript, Arnauld

/*ZZ*/m=>[...Array(m+1).keys()].map(eval(atob("eD0+K1suLi4iIit4KnhdLnJldmVyc2VgYC5qb2luYGA="))).join``[m]
\$\endgroup\$
2
  • 1
    \$\begingroup\$ @Arnauld OK, I think Array(m+1) fixed it. \$\endgroup\$
    – Lynn
    Nov 3, 2017 at 9:09
  • \$\begingroup\$ It does. FWIW, I've added the intended solution to my post. \$\endgroup\$
    – Arnauld
    Nov 3, 2017 at 9:30
2
\$\begingroup\$

cQuents 0, Stephen

":\r$$

Try it online! I have no idea how this code works, but it still worked after removing the *.

\$\endgroup\$
2
\$\begingroup\$

Wolfram Language (Mathematica), Jenny_mathy

(Join@@Table[k@FromDigits@Reverse@(k=IntegerDigits)[i*i],{i,10^4}])[[#]]&

Try it online!

Alternative version also at distance 43:

(Join@@IntegerDigits@IntegerReverse[Range@#^2])[[1#]]&

Try it online!

\$\endgroup\$
0
2
\$\begingroup\$

6502 Machine Code (C64), Felix Palmen

I tested this with all the questions test cases and quite a few extras (like 2^15... that took awhile), and it appears to work the same as the original with LD = 1.

00 C0 20 FD AE A0 00 99 5B 00 C8 20 73 00 90 F7 99 5B 00 A2 0B CA 88 30 09 B9
5B 00 29 0F 95 5B 10 F3 A9 00 95 5B CA 10 F9 A9 01 A0 03 99 69 00 88 10 FA A0
20 A2 76 18 B5 E6 90 02 09 10 4A 95 E6 E8 10 F4 A2 03 76 69 CA 10 FB 88 F0 11
A2 09 B5 5C C9 08 30 04 E9 03 95 5C CA 10 F3 30 D6 A2 03 B5 69 95 57 CA 10 F9
A9 01 85 FB A2 03 A9 00 95 FB CA D0 FB A2 03 B5 FB 95 22 95 26 CA 10 F7 A9 00
A2 03 95 69 CA 10 FB A0 20 A2 02 46 25 76 22 CA 10 FB 90 0C A2 7C 18 B5 AA 75
ED 95 ED E8 10 F7 A2 7D 06 26 36 AA E8 10 FB 88 10 DD A0 0B A9 00 99 5A 00 88
D0 FA A0 20 A2 09 B5 5C C9 05 30 04 69 02 95 5C CA 10 F3 06 69 A2 FD 36 6D E8
D0 FB A2 09 B5 5C 2A C9 10 29 0F 95 5C CA 10 F4 88 D0 D7 E0 0A F0 05 E8 B5 5B
F0 F7 09 30 99 5B 00 C8 E8 E0 0B F0 04 B5 5B 90 F1 88 B9 5B 00 C9 30 F0 F8 A2
7C 18 B5 DB E9 00 95 DB E8 10 F7 90 14 88 30 05 B9 5B 00 D0 EA A2 7C F6 7F D0
03 E8 10 F9 4C 73 C0 B9 5B 00 4C D2 FF

Online demo, usage: sys49152,n where n is the 0-indexed input.

\$\endgroup\$
1
  • \$\begingroup\$ This is correct because you found some utterly useless code I wasn't aware of and the change is in this code :o \$\endgroup\$ Nov 10, 2017 at 8:27
1
\$\begingroup\$

Lua, Katenkyo

i=1s=""while(#s<...+0)do s=s..((i*i)..""):reverse():gsub("(0+)(%d+)$","%2")i=i+1 end
print(s:sub(...,...))

Try it online!

I don't know Lua, but this was a simple one, just replaced a space with a newline.

\$\endgroup\$
1
  • \$\begingroup\$ Hum, didn't think about that, the original was about replacing (0+)(%d+)$ with (0+)(%d+), so it was about regex ^^' \$\endgroup\$
    – Katenkyo
    Nov 2, 2017 at 14:32
1
\$\begingroup\$

Python 3, HyperNeutrino

lambda o:"".join(str(p*p+2*p+1)[::~0].lstrip("0")for p in range(o+1))[o]

Try it online!

\$\endgroup\$
10
  • \$\begingroup\$ Ninja'd in 9 sec :P \$\endgroup\$
    – Uriel
    Nov 2, 2017 at 14:19
  • \$\begingroup\$ huh interesting, was not the one I had but nice :) \$\endgroup\$
    – hyper-neutrino
    Nov 2, 2017 at 14:19
  • \$\begingroup\$ You could also do '' instead of "" \$\endgroup\$
    – Uriel
    Nov 2, 2017 at 14:19
  • \$\begingroup\$ @Uriel Yeah, I know a ton of alternatives. \$\endgroup\$
    – Mr. Xcoder
    Nov 2, 2017 at 14:20
  • \$\begingroup\$ An alternative crack would be lambda i:"".join(str( (-~k)**2)[::-1]for k in range(i+1))[i]. \$\endgroup\$ Nov 2, 2017 at 14:24
1
\$\begingroup\$

Python 2, dylnan

d=lambda y:y if y%10>0 else d(y/10)
lambda n:''.join([str(d(x*x))[::-1]for x in range(1,n+1)])[n-1]#fix

Try it online!

Note: this cop submission was bugged and didn't work for inputs lower than 5. While I was at it I built this solution which has the correct Levenshtein distance AND fixes the bug.

\$\endgroup\$
1
\$\begingroup\$

Perl 5, (-p) Xcali

Updated after comment, Levenshtein Distance between

a$j.=int reverse$_**2for 1..$_;$_--;say$j=~s/.{$_}(.).*/$1/r

and

p$_=substr w.(join"",map{whyddwzz;0|reverse$_**2}1..$_),$_,1

is 55

Try it online

\$\endgroup\$
2
  • \$\begingroup\$ Given that M5.010 is "free", I don't think it should count here. I'm not really sure how to count the -a versus -p flags. The two solutions I came up with both used the same flags. I would think the flag would just be tacked onto the front without a space, but I'm willing to be swayed by others on that. \$\endgroup\$
    – Xcali
    Nov 3, 2017 at 13:04
  • \$\begingroup\$ updated my answer \$\endgroup\$ Nov 3, 2017 at 16:05
1
\$\begingroup\$

Java 8, Kevin Cruijssen

/*!FooBarFooBarFoo!*/N->{String R="",T=R;for(int I=1,J;N+2>R.length();I++){for(T="",J=(I*I+"").length();0<J;T+=(I*I+"").charAt(--J));R+=new Long(T)+"";}return R.charAt(N);}

Try it online!

Change log

  • Replaced .replaceAll() with new Long().
  • Removed the test for perfect squares. Now using perfect squares directly.
  • Updated all variable names to upper-case.
  • Rewritten the inequalities.
  • And finally added a 21-byte leading comment to reach the correct LD.
\$\endgroup\$
1
  • 1
    \$\begingroup\$ Oh nice. That's completely different than I had in mind, but nice that you got to 92 LD anyway. The solution I had in mind was: n->{String r="";for(int i=1;r.length()<=n+1;r+=new Long(new StringBuffer(i*i+++"").reverse()+""));return r.charAt(n);} (118 bytes, 92 LD compared to my other answer.) \$\endgroup\$ Nov 4, 2017 at 11:09
1
\$\begingroup\$

Octave, Stewie Griffin

@(n)regexprep(fliplr(num2str((1:n)'.^2))'(:)',' +0*','')(n)%abcdefghijk

Try it online!

I was actually attempting my own Octave answer and spotted the existing one. Mine was already significantly shorter so adding a comment at the end was sufficient to get to the required distance of 63.

\$\endgroup\$
1
  • \$\begingroup\$ Well done :-) I had a loop with input() and everything that goes with it... \$\endgroup\$ Nov 4, 2017 at 19:25
1
\$\begingroup\$

PHP, Jo.

<?for($j++;strlen($p)<$argv[1];$j++)$p.=(int)strrev($j**2);echo($p[$argv[1]+2-3]);

Try it online!

(I was planning to switch the inequality in order to get even larger LD...)

\$\endgroup\$
0
\$\begingroup\$

6502 Machine Code (C64), Felix Palmen

May also be a "simple" crack, but it appears to work as the original.
Having the LD = 1 is just so tempting to try to crack it (sorry, Felix). :)

00 C0 20 FD AE A0 00 99 5B 00 C8 20 73 00 90 F7 99 5B 00 A2 0B CA 98 88 30 09
B9 5B 00 29 0F 95 5B 10 F2 95 5B CA 10 FB A0 20 A2 76 18 B5 E6 90 02 09 10 4A
95 E6 E8 10 F4 A2 03 76 69 CA 10 FB 88 F0 11 A2 09 B5 5C C9 08 30 04 EB 03 95
5C CA 10 F3 30 D6 A2 03 B5 69 95 57 CA 10 F9 A9 01 85 FB A2 03 A9 00 95 FB CA
D0 FB A2 03 B5 FB 95 22 95 26 CA 10 F7 A9 00 A2 03 95 69 CA 10 FB A0 20 A2 02
46 25 76 22 CA 10 FB 90 0C A2 7C 18 B5 AA 75 ED 95 ED E8 10 F7 A2 7D 06 26 36
AA E8 10 FB 88 10 DD A2 0B A9 00 95 5A CA D0 FB A0 20 A2 09 B5 5C C9 05 30 04
69 02 95 5C CA 10 F3 06 69 A2 FD 36 6D E8 D0 FB A2 09 B5 5C 2A C9 10 29 0F 95
5C CA 10 F4 88 D0 D7 E8 B5 5B F0 FB 09 30 99 5B 00 C8 E8 E0 0B F0 04 B5 5B 90
F1 88 B9 5B 00 C9 30 F0 F8 A2 7C 18 B5 DB E9 00 95 DB E8 10 F7 90 14 88 30 05
B9 5B 00 D0 EA A2 7C F6 7F D0 03 E8 10 F9 4C 68 C0 B9 5B 00 4C D2 FF

Online demo, usage: sys49152,n where n is the 0-indexed input.

\$\endgroup\$
4
  • \$\begingroup\$ I'm not sure whether I have to accept this. It replaces E9 (a subtract command) by EB which is undefined in 6502 machine code, but happens to do the same on NMOS 6502 and 6510 chips. This program would for example crash on the C64 DTV1. But it's unlikely to find a real C64 that doesn't execute it correctly, so it could be considered a valid crack? I might ask for opinions on meta.... \$\endgroup\$ Nov 15, 2017 at 10:25
  • \$\begingroup\$ I'd be interested in input here on meta. \$\endgroup\$ Nov 15, 2017 at 10:55
  • \$\begingroup\$ @FelixPalmen I'll take this answer down soon. \$\endgroup\$
    – Jo.
    Nov 16, 2017 at 2:24
  • \$\begingroup\$ keep it, see also my comment on meta. The community clearly expressed the opinion that it's valid. It's my fault not to require only documented 6502 code, and I'll keep that in mind for the future. \$\endgroup\$ Nov 16, 2017 at 16:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.