A Pollux problem to solve

 

Someone has encoded a phrase using the Pollux cipher and told you that 2,3 are Dots, 5,6 are Dashes and 8,9 are spaces (x).  What does it say?

 

12059811012278473374491805946698143393935026296198313

 

0455866718756946591628223037761517666963203

Background on Solving Pollux

The Pollux cipher works by first converting the text into Morse code which is written as a series of dots (●), dashes (–), and spaces.  To make it more convenient to solve, we typically represent the spaces as an ื. A single space is used at the end of a Morse code letter and a pair of spaces is used at the end of a word.

 

The person encoding the text then decides with digits will stand for dots/dashes/spaces with no restriction on that choice.  For example, all the spaces could be represented by a 2, all of the dots by a 1 and all the other digits stand for a dash.  Given the mapping of the digits, the Morse code is translated to the cipher text by picking a digit for the dash/dot/space.  Since more than one digit can stand for a dash/dot/space, the encoding can choose whatever digit they would like. 

 

Decoding a Pollux applies the process in reverse.  It starts by mapping the known digits to their corresponding dot/dash/space and looking for complete Morse code characters.  A complete Morse code character is one where an uninterrupted series of dots/dashes are delimited by a space.  For example: ●●●ื at the beginning represents the very familiar letter S (three dots).  Finding ื●●ื in the middle would represent the letter I (two dots).  However, if we had ื● ื (with an unmapped digit after the dot), we wouldn’t know what the plain text is until we figured out the mapping for the digit.

 

With that in mind, the strategy for solving a Pollux consists of a set of steps:

1)     Build a table of the possibilities for the digits.

0

1

2

3

4

5

6

7

8

9

2)     Fill in the table with the known mappings and then just put ●–ื for everything else since we don’t know what they map to.

3)     Underneath the digits of the cipher, fill in the known mappings with the corresponding Morse code character (●, –, ื).

4)     Solve.. As digits are eliminated, removed them from the possibility table and fill in known mappings under the cipher text.  One special case that makes it easier to solve.  If you eliminate ื as a possibility, leaving ● or –, filling in the corresponding cipher spot with Solving A Pollux Cipher? makes it easier to find places where a ื belongs.

 

Some good solving rules that help quickly solve a Pollux

1)     The first character will never be an ื.  If the cipher digit at the start could map to an ื, you can eliminate that choice.

2)     There will never be three spaces (ืืื) in a row. Hence if you find a cipher digit that is tripled, you know that it can’t map to a ื.

3)     Also looking for three spaces, if you have digits that already map to ื and either have a doubled digit next to it which is unknown or ืื next to an unknown, you can eliminate ื from that unknown.

4)     No Morse letter is more than 4 dots/dashes and all numbers are exactly 5 dots/dashes.  If there is a sequence of 6 characters with an unknown and all the remainder are known to be a dot/dash (●–?) then you know that the unknown must be a ื.

5)     Not all sequences of 4 dots/dashes are legal Morse characters.  (●●––, ●–●–, –––●, and ––––).  If you have a pattern that would map to it, you know that you can eliminate it.

How to solve

Since we are told the mapping of 235689 ciphertext, we can build the following table:

0

1

2

3

4

5

6

7

8

9

●–ื

●–ื

●

●

●–ื

–

–

●–ื

ื

ื

Based on that information we can map the cipher text as:

12059811012278473374491805946698143393935026296198313

 ● –ืื    ●● ื  ●●   ื ื –ื ––ืื  ●●ื●ื●– ●–●ื– ืื● ●

   /                   /     E          /

0455866718756946591628223037761517666963203

  ––ื––  ื ––ื ––ื –●ื●●● ●  – –  –––ื–●● ●

 

 

The first Morse code character can never be an ื,

0

1

2

3

4

5

6

7

8

9

●–ื

●–

●

●

●–ื

–

–

●–ื

ื

ื

Based on that information we can map the cipher text as:

12059811012278473374491805946698143393935026296198313

?● –ืื?? ?●● ื  ●●   ื?ื –ื ––ืื? ●●ื●ื●– ●–●ื–?ืื●?●

    /                         /      E          /

0455866718756946591628223037761517666963203

  ––ื––  ?ื ––ื ––ื?–●ื●●● ● –?–? –––ื–●● ●

 

 

At this point in time, 4 ciphertext characters still need to be mapped. Looking at the ciphertext, we see the sequence 449 which would result in three ืs in a row if 4 were an ื.

0

1

2

3

4

5

6

7

8

9

●–ื

●–

●

●

●–

–

–

●–ื

ื

ื

Based on that information we can map the cipher text as:

12059811012278473374491805946698143393935026296198313

?● –ืื?? ?●● ื? ●● ??ื?ื –ื?––ืื??●●ื●ื●– ●–●ื–?ืื●?●

    /                         /     E           /

0455866718756946591628223037761517666963203

 ?––ื–– ?ื ––ื?––ื?–●ื●●● ●  –?–? –––ื–●● ●

 

 

At this point in time, 4 ciphertext characters still need to be mapped. Based on the sequence 350262 with 0 possibly being one of ●–ื, only ื results in a legal Morse code character, so we can mark 0 as being ื.

0

1

2

3

4

5

6

7

8

9

ื

●–

●

●

●–

–

–

●–ื

ื

ื

Based on that information we can map the cipher text as:

12059811012278473374491805946698143393935026296198313

?●ื–ืื??ื?●● ื? ●● ??ื?ืื–ื?––ืื??●●ื●ื●–ื●–●ื–?ืื●?●

   T/                  / T    /      E A  R     /

0455866718756946591628223037761517666963203

ื?––ื–– ?ื ––ื?––ื?–●ื●●●ื●  –?–? –––ื–●●ื●

                      S              D   E

 

At this point in time, 3 ciphertext characters still need to be mapped. Since 1 can still map to ●– we simply try them and look at the first word or two to see if it makes sense. Trying ● for 1 gives us a chunk: EARN S. Trying – for 1 gives us a chunk: EARM R. Which means we know that 1 must map to ●

0

1

2

3

4

5

6

7

8

9

ื

●

●

●

●–

–

–

●–ื

ื

ื

Based on that information we can map the cipher text as:

12059811012278473374491805946698143393935026296198313

●●ื–ืื●●ื●●● ื? ●● ??ื●ืื–ื?––ืื●?●●ื●ื●–ื●–●ื–●ืื●●●

I  T/ I               E/ T    /      E A  R   N / S

0455866718756946591628223037761517666963203

ื?––ื–– ●ื ––ื?––ื●–●ื●●●ื●  –●–● –––ื–●●ื●

                  R   S               D   E

 

At this point in time, 2 ciphertext characters still need to be mapped. Based on the sequence 37761517666 with 7 possibly being one of ●–ื, only ื results in a legal Morse code character, so we can mark 7 as being ื.

0

1

2

3

4

5

6

7

8

9

ื

●

●

●

●–

–

–

ื

ื

ื

Based on that information we can map the cipher text as:

12059811012278473374491805946698143393935026296198313

●●ื–ืื●●ื●●●ืื?ื●●ื??ื●ืื–ื?––ืื●?●●ื●ื●–ื●–●ื–●ืื●●●

I  T/ I  S  /   I     E/ T    /      E A  R   N / S

0455866718756946591628223037761517666963203

ื?––ื––ื●ืื––ื?––ื●–●ื●●●ื●ืื–●–●ื–––ื–●●ื●

     M  E/ M      R   S   E/ C    O   D   E

 

At this point in time, 1 ciphertext characters still need to be mapped. Since 4 can still map to ●– we simply try them and look at the first word or two to see if it makes sense. Trying ● for 4 gives us a chunk: IT IS EIIE TW HEARN SWME MWRSE COD. Trying – for 4 gives us a chunk: IT IS TIME TO LEARN SOME MORSE CODE. Which means we know that 4 must map to –

0

1

2

3

4

5

6

7

8

9

ื

●

●

●

–

–

–

ื

ื

ื

 

Based on that information we can map the cipher text as:

12059811012278473374491805946698143393935026296198313

●●ื–ืื●●ื●●●ืื–ื●●ื––ื●ืื–ื–––ืื●–●●ื●ื●–ื●–●ื–●ืื●●●

I  T/ I  S  / T I  M  E/ T O  / L    E A  R   N / S

0455866718756946591628223037761517666963203

ื–––ื––ื●ืื––ื–––ื●–●ื●●●ื●ืื–●–●ื–––ื–●●ื●

 O   M  E/ M  O   R   S   E/ C    O   D   E

 

Now that we have mapped all the ciphertext characters, the decoded Morse code is the answer:

IT IS TIME TO LEARN SOME MORSE CODE