Alphacipher Solution
====================

Solved by Adrian Dale 19/12/2013

From the example here:
http://www.puzzler.com/Puzzles-encyclopedia/Alphacipher.htm
http://www.puzzler.com/Puzzles-encyclopedia/Puzzle-samples/Alphacipher.pdf

ANGELOU = 45
ATWOOD = 55
BALZAC = 59
BRAINE = 47
CONRAD = 59
EVELYN = 63
FORESTER = 58
GASKELL = 47
GOGOL = 41
HAMSUN = 53
HELLER = 35
JEROME = 53
KAFKA = 74
LESSING = 45
NESBIT = 45
PARKER = 58
POTTER = 47
PROUST = 42
QUENEAU = 56
RANSOME = 45
RENAULT = 40
SALINGER = 55
SHUTE = 37
SYMONS = 50
WALTON = 49

From NESBIT = LESSING, cancel letters to get BT = LSG
From BRAINE + 8 = SALINGER, cancel to get B + 8 = LSG

Therefore BT = B + 8, hence T = 8

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From POTTER = PROUST + 5, cancel letters to get TE = US + 5
Replace T with 8 to get E = US - 3 (1)

From SHUTE = 37, replace T with 8 to get SHUE = 29
Now from eqn (1) replace US in SHUE with E + 3 to get HEE + 3 = 29, hence HEE = 26 (2)

Replace HEE (from eqn (2) in HELLER = 35 to get 26 + LLR = 35, hence LLR = 9 (3)

From ANGELOU = GOGOL + 4, cancel letters to get ANEU = GO + 4 (4)
From QUENEAU = 56, replace ANEU to get QUE GO + 4 = 56, hence QUEGO = 52 (5)

From QUENEAU = RENAULT + 16, cancel letters to get QUE = RL + 24 (6)
Add L to each side of result to give QUEL = LLR + 24
Replace LLR with value 9 (eqn (3))to give QUEL = 33 (7)

Subtract QUEL = 33 from QUEGO = 52 to give GO - L = 19, hence L = GO - 19 (8)

Replace the L in GOGOL = 41 to give GOGOGO - 19 = 41
Therefore GOGOGO = 60, hence GO = 20 (9), therefore L = 1

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From LLR = 9 (3), we now know R = 7

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From PARKER = POTTER + 11, cancelling letters and substituting known
values we get AK = O + 20  (10)

From GASKELL = 47, replace AK with O + 20 (10), and known value for L
to get GS + O + 20 + E + 2 = 47, which gives GSOE = 25 (11)
Replace GO = 20 (eqn 9) in (11) to give 20 + SE = 25, hence SE = 5 (12)

Replace SE, and known value of T in SHUTE = 37 to get HU = 24 (13)

Replace HU in HAMSUN = 24 to get AMSN = 29 (14)

From BRAINE = RANSOME + 2, cancel letters and replace knowns to get BI = SOM + 2 (15)
Using (15), replace BI, and known value for T in NESBIT = 45 to get NESSOM = 35 (16)

From SYMONS = NESSOM + 15, cancel letters to get Y = E + 15 (17)

From (1) E = US - 3 and rearranged (17) E = Y - 15, get US - 3 = Y - 15,
hence Y = US + 12 (18)

From (1) E = US - 3, add S to both sides to get SE = USS - 3, replace SE (12), 
hence USS = 8 (19)

Add A to both sides of SYMONS = 50, to get AMSN + SYO = 50 + A
Replace known value of AMSN to get SYO = A + 21 (25)
Replace Y with eqn (18) to get SUS + 12 + O = A + 21
Replace USS (eqn 19) to get O = A + 1 (20)

From (10) and (20) we get AK - 20 = A + 1, so K - 20 = 1, hence K = 21

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From GASKELL = 47, replace known values for K, L and SE (12) to get GA = 19 (21)

From NESBIT + 10 = SALINGER, cancel letters and replace values to get B + 18 = GA + 8 (22)

Replace GA (21) in (22) to get B + 18 = 27, hence B = 9

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Replace known values SE (12) GA (21) L R in SALINGER = 55 to get IN = 23 (23)

From BRAINE = 47 cancel letters and replace known values B, R, IN (23) to get AE = 8 (27)

From eqn (20) O = A + 1 and (27) O - 1 + E = 8, hence OE = 9 (24)

Replace known values OE (24), T, R in POTTER = 47, hence P = 15

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Replace known values OE (24), SE (12), T, R in FORESTER = 58 to get
F + 9 + 7 + 5 + 8 + 7 = 58, hence F = 22

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From KAFKA = 74, replacing known value of K and F gives  64 + AA = 74,
therefore AA = 10, hence A = 5

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From eqn (20) O = A + 1, hence O = 6

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From eqn (9) GO = 20, hence G = 14

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From eqn (24) OE = 9, hence E = 3

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From eqn (12) SE = 5, hence S = 2

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From eqn (16) Y = E + 15, hence Y = 18

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From PROUST = 42, replacing known letters gives 15 + 7 + 6 + U + 2 + 8 = 42,
hence U = 4

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From SHUTE = 37, replacing known letters gives 2 + H + 4 + 8 + 3 = 37,
hence H = 20

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From ANGELOU = 45, replacing known letters gives 5 + N + 14 + 3 + 1 + 6 + 4 = 45,
hence N = 12

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From SYMONS = 50, replacing known letters gives 2 + 18 + M + 6 + 12 + 2 = 50,
hence M = 10

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From SALINGER = 55, replacing known letters gives 2 + 5 + 1 + I + 12 + 14 + 3 + 7 = 55,
hence I = 11

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From WALTON = 49, replacing known letters gives W + 5 + 1 + 8 + 6 + 12 = 49,
hence W = 17

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From ATWOOD = 55, replacing known letters gives 5 + 8 + 17 + 6 + 6 + D = 55,
hence D = 13

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From CONRAD = 59, replacing known letters gives C + 6 + 12 + 7 + 5 + 13 = 59,
hence C = 16

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From BALZAC = 59, replacing known letters gives 9 + 5 + 1 + Z + 5 + 16 = 59,
hence Z = 23

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From JEROME = 53, replacing known letters gives J + 3 + 7 + 6 + 10 + 3 = 53,
hence J = 24

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From QUENEAU = 56, replacing known letters gives Q + 4 + 3 + 12 + 3 + 5 + 4 = 56,
hence Q = 25

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From EVELYN = 63, replacing known letters gives 3 + V + 3 + 1 + 18 + 12 = 63,
hence V = 26

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Since all letters are now allocated, X must be the value which has not yet been taken,
hence X = 19