Alphabetic Soup

[superprismatic]

I have taken the standard A-Z alphabet and permuted it

in such a way that the permuted alphabet starts with

a long English word. The object of this puzzle is to

determine my permuted alphabet.

For clues, I give you five sets of pairs of letters.

Each set has the property that every pair of letters within

that set are the same distance apart, in the same direction,

in my permuted alphabet. Furthermore, each set uses a

distance/direction pair which is not used in any other set.

For example, if my permuted alphabet were

ACEGIKLNPRTVXBDFHJLNPRTVXZ

Two of my sets may be

{AG,LR,HN,XC,ZE,KP} and {BA,DC,FE,HG,JI}

in which case, the first set represents pairs of letters

where the second of the pair is three away to the right

of the first letter of the pair (AG has G three to the

right of A; XC has C three to the right of X) or,

equivalently, 23 away to the left of the first letter.

The second set contains pairs which contain pairs of

letters which are 13 apart from each other. Notice that

the alphabet is considered cyclic and counting is done

'around the corner'.

The five sets are:

Set 1: {EF,HA,IH,QV,TB,WZ,ZN}

Set 2: {JA,KB,QE,WJ,ZM,HO,UQ,BR,RU,CV}

Set 3: {AK,CH,DV,FW,JX,NG,QN,TJ,UI,VO,WP,YB,ZR}

Set 4: {LC,UD,NF,QH,BN,JR,RS,ST,GW}

Set 5: {CD,GS,HV,KY,TW}

Can you find my permuted alphabet which starts with a

long English word?

[curr3nt]

copyright?

[curr3nt]

because I think I found 12 combinations of letters that would fit the puzzle. Six unique sets with each one mirrored.

For the given distances of each set here are the six unique combinations. The longest word I found was in the last set backwards.

_1 _2 _3 _4 _5

14 10 23 _1 13 : O K I V A U D P M G W B N F Y Q H X L C J R S T Z E

16 _4 17 _3 13 : M X O G L K W C I B J V N R A F S U Y T D Q Z P H E

18 24 11 _5 13 : J Q W U O R H B D K S X N P I T L F M V Z C Y G A E

20 18 _5 _7 13 : Y V L P S B O Q A C M T N K H U J G Z F I X D R W E

22 12 25 _9 13 : I U M B Y X J T O V D G N Q L R Z K A P W F H C S E

24 _6 19 11 13 : L B A T H G I R Y P O C N U Z X W V S Q M K J F D E

edit - getting formatting right is a puzzle of it's own...

Edited June 18, 2012 by curr3nt

[superprismatic]

because I think I found 12 combinations of letters that would fit the puzzle. Six unique sets with each one mirrored.

For the given distances of each set here are the six unique combinations. The longest word I found was in the last set backwards.

_1 _2 _3 _4 _5

14 10 23 _1 13 : O K I V A U D P M G W B N F Y Q H X L C J R S T Z E

16 _4 17 _3 13 : M X O G L K W C I B J V N R A F S U Y T D Q Z P H E

18 24 11 _5 13 : J Q W U O R H B D K S X N P I T L F M V Z C Y G A E

20 18 _5 _7 13 : Y V L P S B O Q A C M T N K H U J G Z F I X D R W E

22 12 25 _9 13 : I U M B Y X J T O V D G N Q L R Z K A P W F H C S E

24 _6 19 11 13 : L B A T H G I R Y P O C N U Z X W V S Q M K J F D E

edit - getting formatting right is a puzzle of it's own...

Yes, you've got it!

The 12 are basically all the same because decimations of the alphabet by integers relatively prime to 26 (there are 12 of them) preserves the fact that all sets contain pairs which all have the same distances, and the fact that all sets represent different distances. The purpose of the long word was to help nail down the correct alphabet. It started with the word UNCOPYRIGHTABLE and was followed by the remaining letters in alphabetical order.