Play Caesar's Cipher with Letters only

Question

Description

The Caesar cipher is a cipher, where every letter in the alphabet will be rotated by a secret number.
If the rotation is \$7\$, then a -> h, g -> n, v -> c and z -> g.

Today, we're playing Caesar's Cipher with ASCII chars, (0-127 inclusive). So, with rotation \$7\$, b -> i, Z -> a, x -> DEL (127),

But, even if the encoder shifted the chars around the ASCII table, you know, that the original string consisted of only the lowercase, uppercase alphabet, and space.

Task

You're given a string of ASCII code points, your task is to print all of the possible original strings if the encoder only used the lowercase and the upper case alphabet (plus space) in the original string.

Rules

The program must take the ASCII char codes of the encoded string as the input
The encoded string is not constricted on lowercase, uppercase, and space, it can (because it's ASCII shifted) include ASCII chars
The output must be all possible original strings (as a list, separated by newlines, etc.)
Default loopholes apply
Lowercase, uppercase, and space: abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ
This is code-golf, so the shortest answer wins!

Examples

[In]: [91, 111, 112, 122, 39, 112, 122, 39, 104, 39, 123, 108, 122, 123, 39, 116, 108, 122, 122, 104, 110, 108]
[Out]: This is a test message

[In]: [43, 35, 49, 49, 31, 37, 35]
MESSAGE
NFTTBHF
OGUUCIG
PHVVDJH
QIWWEKI
RJXXFLJ
SKYYGMK
TLZZHNL
message
nfttbhf
oguucig
phvvdjh
qiwweki
rjxxflj
skyygmk
tlzzhnl

[In]: [103, 123, 120, 51, 97, 124, 1, 120, 7, 120, 120, 1, 7, 123, 51, 85, 12, 7, 120]
[Out]: The Nineteenth Byte
[In]: [78, 82, 78, 78]
[Out]:
NRNN
OSOO
PTPP
QUQQ
RVRR
SWSS
TXTT
UYUU
VZVV
aeaa
bfbb
cgcc
dhdd
eiee
fjff
gkgg
hlhh
imii
jnjj
kokk
lpll
mqmm
nrnn
osoo
ptpp
quqq
rvrr
swss
txtt
uyuu
vzvv
AEAA
BFBB
CGCC
DHDD
EIEE
FJFF
GKGG
HLHH
IMII
JNJJ
KOKK
LPLL
MQMM

\$\begingroup\$ Suggests adding testcases with NULL bytes in it. \$\endgroup\$
– tsh
Apr 1, 2022 at 9:09 — tsh, Apr 1, 2022 at 9:09

Jonathan Allan · Accepted Answer · 2022-03-31 18:11:19Z

Jelly, 16 bytes

+ⱮØ⁷%Ø⁷ỌfƑƇØẠ;⁶¤

A monadic Link that accepts a list of code-points and yields a list of lists of characters.

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How?

+ⱮØ⁷%Ø⁷ỌfƑƇØẠ;⁶¤ - Link: list of code-points C
  Ø⁷             - 128
 Ɱ               - map (across v in [1..128]) with:
+                -   C add v (vectorises)
    %Ø⁷          - modulo 128
       Ọ         - convert to characters (list of potentials, P)
               ¤ - nilad followed by link(s) as a nilad:
           ØẠ    -   alphabetic characters
              ⁶  -   space character
             ;   -   concatenate (list of characters "A..Za..z ", A)
          Ƈ      - filter keep those p in P for which:
         Ƒ       -   is invariant under:
        f        -     p filter keep only those in A

pajonk · Accepted Answer · 2022-03-31 19:22:10Z

4

R, 82 bytes

\(s)for(i in 0:127)if(all((x=(s+i)%%128)%in%c(32,97:122,65:90)))show(intToUtf8(x))

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With the use of grepl instead of filtering char codes:

R, 84 bytes

\(s)for(i in 0:127)if(!grepl("[^ a-z]",v<-intToUtf8(x<-(s+i)%%128),T)&all(x))show(v)

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intToUtf8 has some weird behaviour with the input of 0 (it omits that char altogether) - the workaround here is using all to filter that out.

answered Mar 31, 2022 at 19:22

pajonk

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Sisyphus · Accepted Answer · 2022-03-31 23:16:02Z

3

Perl 5 `-na`, 45 bytes

y//\1-\x7f\0/c,/[^\pL ]/||say for(pack'C*',@F)x128

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The program works as is, but replace \0, \1 and \x7f with their literal equivalents for the claimed score.

Very simple approach. 128 times, translate \0-\x7f to \1-\x7f\0, thus shifting each char by one, and check whether it meets the criteria.

answered Mar 31, 2022 at 23:16

Sisyphus

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mossymountain · Accepted Answer · 2022-04-03 17:00:20Z

Based on ophact's answer. (Sadly I don't have the reputation to comment this instead of duplicating it like this)

Python 3.8+, 97 bytes

lambda l:[b for i in range(128)if(b:=bytes(i+c&127for c in l))*all(c==32or 64<c&95<91for c in b)]

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Since both A-Z and a-z are permitted, we can ignore the "case bit" 0x20 (c | 0x20 or c & 0x5f) to get a single range for the letters. There's no need to worry about the out of bounds 0x80.
As we can write 0x5b (b'Z'[0]+1) as 91, while 0x7b (b'z'[0]+1) could only be as short as 123 (and 0x20==32 is as expensive as 0x5f==95), we'll choose the uppercase letters' range.
bytes([1,2,3]) is, for all purposes I can think of "just read-only [1,2,3]" while already being something string-like we can return.

Python 3.8+, 105 bytes

If it has to output str instead for some reason, we need to add a .decode():

lambda l:[w.decode()for i in range(128)if(w:=bytes(c+i&127for c in l))*all(64<c&95<91or c==32for c in w)]

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~~Is it required to be a lambda? Just the business part would be 86 bytes.~~

b for i in range(128)if(b:=bytes(i+c&127for c in l))*all(c==32or 64<c&95<91for c in b)

You would get the same output by giving the input in the variable l and writing something like

sys.stdout.buffer.write(b'\n'.join(
-paste here-
))

Welcome to the site. The last two solutions there are what we call "snippets" and they are generally disallowed. You can leave them in though since you do have a valid answer. — Wheat Wizard, Apr 1, 2022 at 21:46
@WheatWizard How could returning str be ok while returning bytes wouldn't? They are the same thing for ASCII in most encodings. Python 3 writes str to stdout by converting it to bytes first. — mossymountain, Apr 3, 2022 at 17:24
It probably is valid. I would guess, but I don't know. You can ask on meta if it is really a concern. — Wheat Wizard, Apr 3, 2022 at 20:13

Arnauld · Accepted Answer · 2022-03-31 16:53:30Z

2

JavaScript (Node.js), 85 bytes

a=>[...Array(128)].flatMap(_=>/[^a-z ]/i.test(s=Buffer(a=a.map(c=>c+1&127))+'')?[]:s)

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edited Mar 31, 2022 at 16:53

answered Mar 31, 2022 at 16:47

Arnauld

189k20 gold badges171 silver badges631 bronze badges

\$\begingroup\$ Why do you need global regex? \$\endgroup\$
– l4m2
Mar 31, 2022 at 16:50
\$\begingroup\$ @l4m2 Actually, I don't need to match the whole string either. :-p \$\endgroup\$
– Arnauld
Mar 31, 2022 at 16:54

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score 2 · Accepted Answer · 2022-04-01 00:11:31Z

2

PHP 4 (123 chars)

The function f is accepting an encoded string.

function f($s){for($i=128;$i-=print chop($s,'A..Za..z ')?'':"$s\n";)for($j=strlen($s);$j--;)$s[$j]=chr(1+ord($s[$j])%128);}

edited Apr 1, 2022 at 0:11

answered Mar 31, 2022 at 20:48

user111777

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Kevin Cruijssen · Accepted Answer · 2022-04-01 14:02:51Z

05AB1E, 18 16 bytes

₅ÝIδ+žy%êçʒðKDáQ

Outputs as a list of list of characters.

Try it online or verify all test cases.

Explanation:

₅Ý        # Push a list in the range [0,255]
          # (255 is the lowest single-byte value above 127)
   δ      # Map over this list of integers:
  I +     #  Add the integer to each value in the input-list
     žy%  # Modulo-128
        ê # Sorted uniquify this list of lists
ç         # Convert each inner-most integer to a character with this codepoint
 ʒ        # Filter this list of lists of characters by:
  ðK      #  Remove all spaces
    D     #  Duplicate it
     á    #  Only keep all ASCII letters
      Q   #  Check if both lists are the same
          # (after which the filtered list is output implicitly)

Seggan · Accepted Answer · 2022-04-03 17:38:14Z

2

Vyxal, 17 15 bytes

₇ƛ⁰+₇%C;U'kBṄ↔⁼

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-2 thanks to @lyxal

Explanation is for my old code:

₇ƛ⁰+₇%Cṅ;U'ðkB+↔= # This comment line needs some love
₇                 # 128, implicitly cast to `0..128`
 ƛ      ;         # Map to...
  ⁰+              # The input plus the offset
    ₇%            # Modulo 128
      Cṅ          # Converted into a string
         U        # Remove duplicates
          '       # Filter by...
                = # Is the original string equal to...
               ↔  # The string with the following chars removed
           ðkB+   # " ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz"

edited Apr 3, 2022 at 17:38

answered Mar 31, 2022 at 18:40

Seggan

5,27121 silver badges47 bronze badges

\$\begingroup\$ Try it Online! for 16 bytes. \$\endgroup\$
– lyxal
Apr 3, 2022 at 9:37
1

\$\begingroup\$ Try it Online! for 15, outputting as a list of lists. \$\endgroup\$
– lyxal
Apr 3, 2022 at 11:12

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ophact · Accepted Answer · 2022-03-31 16:06:30Z

1

Python 3.8 (pre-release), 117 bytes

lambda l:[''.join(map(chr,w))for i in range(128)if(w:=[c+i&127for c in l])*all(64<c<91or 96<c<123or c==32for c in w)]

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Yep... that long.

Nothing much to say here; simply loops through the numbers from 0-127 and keeps the word found by shifting by that many units if all characters in such a word are part of the required alphabet.

answered Mar 31, 2022 at 16:06

ophact

3,1441 gold badge5 silver badges24 bronze badges

\$\begingroup\$ -7 \$\endgroup\$
– loopy walt
Mar 31, 2022 at 23:54
\$\begingroup\$ @loopywalt not sure I really understand that one, python bytes stuff is beyond my knowledge... but I could add it in soon when I'm at a computer \$\endgroup\$
– ophact
Apr 1, 2022 at 5:06

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ovs · Accepted Answer · 2022-03-31 16:26:31Z

1

BQN, 39 bytes^SBCS

Returns a matrix with one possible original per row.

128⊸|{@+𝔽𝕩+⌜˜(∊/⊣)˝𝔽𝕩-˜⌜32∾⥊65‿97+⌜↕26}

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This computes the possible shifts for each character, takes the intersection, and shifts the input by all remaining numbers.
Generating all shifts and then only keeping alphabetic ones is 2 bytes longer:

@+·(∧´˘∊⟜(32∾⥊65‿97+⌜↕26))⊸/128|(↕128)+⌜⊢

Run online!

edited Mar 31, 2022 at 16:26

answered Mar 31, 2022 at 16:03

ovs

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l4m2 · Accepted Answer · 2022-03-31 19:29:47Z

1

JavaScript (Node.js), 78 bytes

x=>Buffer(128).map(_=>/[^ a-z]/i.test(x=x.map(y=>y+1&127))||console.log(''+x))

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JavaScript (Node.js), 83 bytes

a=>[...Array(128)].flatMap(_=>/[^a-z ]/i.test(a=Buffer(a.map(c=>c+1&127)))?[]:a+'')

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Handleing part already near Arnauld enough, so I place it here

edited Mar 31, 2022 at 19:29

answered Mar 31, 2022 at 16:48

l4m2

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1

\$\begingroup\$ You can't assume that the console function is input \$\endgroup\$
– ophact
Mar 31, 2022 at 17:23
\$\begingroup\$ @ophact I remember so but it turns out that it's not well agreed \$\endgroup\$
– l4m2
Mar 31, 2022 at 19:00

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Neil · Accepted Answer · 2022-03-31 19:48:37Z

Charcoal, 24 bytes

ΦＥ¹²⁸⭆θ℅﹪⁺ιλ¹²⁸⬤ι№⁺⁺αβ λ

Try it online! Link is to verbose version of code. Explanation:

  ¹²⁸                       Literal integer `128`
 Ｅ                          Map over implicit range
      θ                     Input array
     ⭆                      Map over elements and join
          ι                 Outer value
         ⁺                  Plus
           λ                Inner value
        ﹪                   Modulo
            ¹²⁸             Literal integer `128`
       ℅                    Convert from ASCII code
Φ                           Filtered where
                ι           Current string
               ⬤            All characters satisfy
                 №          (Non-zero) Count of
                       λ    Current character in
                    α       Uppercase letters
                   ⁺        Concatenated with
                     β      Lowercase letters
                  ⁺         Concatenated with space
                            Implicitly print each solution on its own line

Kjetil S · Accepted Answer · 2022-03-31 21:36:32Z

1

Perl 5, 63 bytes

sub{grep!/[^a-z ]/i,map{//;join'',map chr($_+$'&127),@_}0..127}

Try it online!

answered Mar 31, 2022 at 21:36

Kjetil S

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chunes · Accepted Answer · 2022-03-31 23:14:36Z

1

Factor, 84 bytes

[ 128 iota [ '[ _ + 128 mod ] ""map-as ] with [ R/ [a-z ]+/i matches? ] map-filter ]

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answered Mar 31, 2022 at 23:14

chunes

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Graham · Accepted Answer · 2022-04-01 10:05:38Z

1

APL+WIN, 74 bytes

Prompts for code points

⎕av[((⍴n)=+/m∊(i,53)⍴32,∊(⊂65 97)+¨⍳26)⌿m←i|((i,⍴n)⍴n)+⍉((⍴n←⎕),i)⍴⍳i←128]

Unfortunately Dyalog APL Classic's atomic vector does not line up with ASCII code points whereas APL+WIN does so I cannot demonstrate via TIO as I would do normally.

answered Apr 1, 2022 at 10:05

Graham

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Lince Assassino · Accepted Answer · 2022-04-01 11:41:33Z

Behaviour, 78 bytes

u=&(127*&(f=a;s=c*&(f+a)%127~(s<~&[a==32(a>64a<91)(a>96a<123)])@s*ascii>&a+b))

test with:

@u:c={43 35 49 49 31 37 35}
@u:c={91 111 112 122 39 112 122 39 104 39 123 108 122 123 39 116 108 122 122 104 110 108}

ungolfed and commented:

uncypher = &(
  127*&(  // do the following sequence 127 times
    off = a  // get offset from iterator
    s = c*&(off+a)%128  // rotate each character from /c by the offset
    ~(s<~&[  // stop sequence if the previous result has any invalid character
      (a==32)
      (a>64 a<91)
      (a>96 a<123)
    ])
    @(s*ascii)>&a+b // convert ascii num to char, reduce to string and print
  )
)

ErikF · Accepted Answer · 2022-04-02 19:55:53Z

1

C (gcc), 159 136 bytes

-23 bytes thanks to ceilingcat

As NUL is a valid character, also takes the counted string length. The function first iterates through each shifted string to verify that all the decoded characters are valid (and bails early if not), then prints the shifted string if it is still valid.

#define v;for(r=0;a&&r<l;r++
a,h,r,c;f(s,l)char*s;{for(h=128;a=h--;a&&puts("")){v)a=isalpha(c=s[r]+h&127)|c==32 v)putchar(s[r]+h&127);}}

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edited Apr 2, 2022 at 19:55

answered Mar 31, 2022 at 20:03

ErikF

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Jordan · Accepted Answer · 2022-10-10 20:04:38Z

1

Ruby, 66 bytes

->a{(1..127).map{|n|a.map{((_1+n)%128).chr}*""}.grep /^[a-z ]+$/i}

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answered Oct 10, 2022 at 20:04

Jordan

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Stack Exchange Network

Play Caesar's Cipher with Letters only

Description

Task

Rules

Examples

18 Answers 18

Jelly, 16 bytes

How?

R, 82 bytes

R, 84 bytes

Perl 5 `-na`, 45 bytes

Python 3.8+, 97 bytes

Python 3.8+, 105 bytes

JavaScript (Node.js), 85 bytes

PHP 4 (123 chars)

05AB1E, 18 16 bytes

Vyxal, 17 15 bytes

Python 3.8 (pre-release), 117 bytes

BQN, 39 bytes^SBCS

JavaScript (Node.js), 78 bytes

JavaScript (Node.js), 83 bytes

Charcoal, 24 bytes

Perl 5, 63 bytes

Factor, 84 bytes

APL+WIN, 74 bytes

Behaviour, 78 bytes

C (gcc), 159 136 bytes

Ruby, 66 bytes

Your Answer

Not the answer you're looking for? Browse other questions tagged
code-golf
string
puzzle-solver
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or ask your own question.

Hot Network Questions

Play Caesar's Cipher with Letters only

Description

Task

Rules

Examples

18 Answers 18

Jelly, 16 bytes

How?

R, 82 bytes

R, 84 bytes

Perl 5 -na, 45 bytes

Python 3.8+, 97 bytes

Python 3.8+, 105 bytes

JavaScript (Node.js), 85 bytes

PHP 4 (123 chars)

05AB1E, 18 16 bytes

Vyxal, 17 15 bytes

Python 3.8 (pre-release), 117 bytes

BQN, 39 bytesSBCS

JavaScript (Node.js), 78 bytes

JavaScript (Node.js), 83 bytes

Charcoal, 24 bytes

Perl 5, 63 bytes

Factor, 84 bytes

APL+WIN, 74 bytes

Behaviour, 78 bytes

C (gcc), 159 136 bytes

Ruby, 66 bytes

Your Answer

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

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Hot Network Questions

Perl 5 `-na`, 45 bytes

BQN, 39 bytes^SBCS

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