code problem?

Question

In a certain code 'ANTICIPATE' is written as 'LRHUZDULHM' and 'CONSTIPATION' is written as 'MALUHURDWRHW' and 'DIE' is written as 'ZYH', 'PIE' is written as 'HMZ' ' the following can be a code for 'ANACONDA'?


.
1]LRDRWLRY
2]URDRWUYU
3]LLRDWRYL
4]LRDDLLYW



PLEASE EXPLAIN the answer.

Answer 1

James had the right idea...but...I don't see how U can be A, when there is only one A in CONSTIPATION, but there are two Us in MALUHURDWRHW.

So, following the same idea as he had:

DIE and PIE share 2 letters in common (IE).
ZYH and HMZ share 2 letters in common (ZH).
Therefore, D → Y, P → M, and either (I → Z and E → H), or (I → H and E → Z).

ANTICIPATE has: 2 As, 2 Is, 2 Ts, 1 N, 1 C, 1 P, 1 T, and 1 E.
LRHUZDULHM has: 2 Ls, 2 Us, 2 Hs, 1 R, 1 Z, 1 D, 1 H, and 1 M.

Since we know that I has to be either Z or H, and there are 2 Is in ANTICIPATE, but only 1 Z in LRHUZDULHM, we can conclude that I → H, and therefore E → Z.

So now we have: D → Y, P → M, I → H, E → Z. {A, T} maps somehow to {L, U}, and {N, C} maps somehow to {R, D}.

CONSTIPATION has: 2 Os, 2 Ns, 2 Ts, 1 C, 1 S, and 1 A. (We've already taken care of I and P.)
MALUHURDWRHW has: 2 Us, 2 Rs, 2 Ws, 1 A, 1 L, and 1 D. (We've already taken care of H and M.)

Since there are 2 Ns and 2 Rs and 1 C and 1 D, we have N → R and C → D. Since there are 2 Ts and 2 Us and 1 A and 1 L, we have T → U and A → L. By process of elimination, O → W and S → A.

Altogether we have:

A → L
C → D
D → Y
E → Z
I → H
N → R
O → W
P → M
S → A
T → U

So ANACONDA must be an arrangement of LRLDWRYL. 4 is out because it contains 2 Ds. 2 is out because it contains Us. 1 is out because it contains only 2 Ls.

Therefore the answer must be 3...and the letters do match up, so I'm pretty sure that's right.

Answer 2

DIE and PIE both share 2 letters in common, as do their encrypted counterparts. ANTICIPATE has 2 A's, T's, & I's in it, but 1 N, C, P and E. LRHUZDULHM has 2 L's, U's & H's in it, but 1 Z, M, D and R in. this indicates a likely scrambling of letters.

comparing the words...
I = H, thus
A = U

ANACONDA has 3 A's, 2 N's and 1 C, O and D. each of the code words has the same pattern of letters, however as A = U and there must be 3 U's in the code word, number 2 is correct as it is the only one with 3 U's.