[superprismatic]

A certain quotation containing 26

letters is written below a normal

alphabet and a third line is generated

by writing below each letter in the

second line the letter which is below

it in the first line. If, for

example, A in the first line has a B

below it in the second line, any A in

the second line will have a B below it

in the third line. Thus: "Give me

liberty or give me death" will lead

to the sequence:

    LBDMYMTBIMVMTRVLBDMYMEMGMI

[/code]


Another 26-letter quotation treated

in the same way yields the third line:

[code]

    YIYVNEASLROEEEEUAIYVAUAUAL

What does it say? SUPERPRISMATIC EXPANDING THE EXAMPLE:

    ABCDEFGHIJKLMNOPQRSTUVWXYZ

    GIVEMELIBERTYORGIVEMEDEATH

    LBDMYMTBIMVMTRVLBDMYMEMGMI

[/code]


SUPERPRISMATIC'S REQUEST:

If you solve this, please describe

how you did it.  My method is not

very slick, so I'd like to see a

better one -- perhaps one using a

result from Group Theory, e.g.

[Guest]

ABCDEFGHIJKLMNOPQRSTUVWXYZ

EVERYCLOUDHASASILVERLINING

YIYVNEASLROEEEEUAIYVAUAUAL

Or...

Every cloud has a silver lining. The font sizing doesn't come out right.

I can't honestly say how I did it. Sort of guess and test, I guess. Actually, I'm not even sure if that's right. I'll have to go through it.

If I think of how I got it...I'll tell someone.

Edited August 12, 2009 by AnapesticTetrameter

[bushindo]

ABCDEFGHIJKLMNOPQRSTUVWXYZ

EVERYCLOUDHASASILVERLINING

YIYVNEASLROEEEEUAIYVAUAUAL

Or...

Every cloud has a silver lining. The font sizing doesn't come out right.

I can't honestly say how I did it. Sort of guess and test, I guess. Actually, I'm not even sure if that's right. I'll have to go through it.

If I think of how I got it...I'll tell someone.

This is the approach using group theory. Many thanks to profmmv for introducing me to the basic elements of cycles in this

The puzzle is equivalent to applying a permutation S to the alphabet, call it A, twice. The permutation S does not need to be 1-to-1, that is, two letters from the alphabet can go to the same letter. Let the cypher code be called C. The problem, in these notation, is,

S*S( A ) = C. Find S( A ).

Let the letters in the alphabet be replaced with 1-26 (1=A, 2 =B, ... , 26=Z). The data that we have is the following

       A  C

 [1,]  1 25

 [2,]  2  9

 [3,]  3 25

 [4,]  4 22

 [5,]  5 14

 [6,]  6  5

 [7,]  7  1

 [8,]  8 19

 [9,]  9 12

[10,] 10 18

[11,] 11 15

[12,] 12  5

[13,] 13  5

[14,] 14  5

[15,] 15  5

[16,] 16 21

[17,] 17  1

[18,] 18  9

[19,] 19 25

[20,] 20 22

[21,] 21  1

[22,] 22 21

[23,] 23  1

[24,] 24 21

[25,] 25  1

[26,] 26 12

By going through data above, we can construct the following graphs for the cycles of S*S

Now, we need to construct the permutation S such that S*S looks like the graph above. The first thing we need to notice is that there is a cycle in S such that

1 -> a -> 25 -> b -> 1

Since there are two only sub-cycles in S*S (1->25->1 and 5->14->5), they have to form a 4-length cycle in S. So, we now have 2 choices for this 4-length cycle, (1 -> 5 -> 25 -> 14 -> 1) or (1 -> 14 -> 25 -> 5 -> 1). Let's pick the first cycle and follow through with the analysis,

The number 3 in S*S points to 25. Therefore 3-> a -> 25 - > 14 in S. Since 5 is the only number in S*S pointing to 14, the sequence is 3-> 5 -> 25 - > 14.

By looking at sequence 10->18->9->12->5 in the bottom cycle in S*S, the next observation is that in S, there is a long sequence

10 -> a -> 18 -> b -> 9 -> c -> 12 -> d -> 5 -> 25.

Observation of the top cycle in the graph of S*S above indicates that d must be 1, and (a, b, c) is a sequence of length 3. So the sequence above is either

10 -> 4 -> 18 -> 22 -> 9 -> 21 -> 12 -> 1 -> 5 -> 25 or

10 -> 20 -> 18 -> 22 -> 9 -> 21 -> 12 -> 1 -> 5 -> 25.

By going through similar reasoning, we can find the appropriate places for 11, 8, 15, 19, 2, 24, 16, 26. The incomplete constructed cycles of S now looks like

Where the red dash under the 4 and 20 means that they could be interchangable. The blue numbers on the right hand side means that they aren't added to the graph yet because there are several possibilities. However, this constructed of S is big enough that if we apply it to the plaintext A, we can get a reasonable guess of S(A) and complete the graph, thus solving the puzzle.

Edited August 16, 2009 by bushindo