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start:
2 4 // B = A % 8
1 5 // B = B ^ 5
7 5 // C = A >> B
1 6 // B = B ^ 6
4 1 // B = B ^ C
5 5 // print B % 8
0 3 // A = A >> 3
3 0 // If A != 0 goto start
for (A = a; A != 0; A >> 3) {
B1 = (A % 8) ^ 5;
C = A >> B1;
B2 = (B1 ^ 6) ^ C;
print(B2 % 8);
}
Goal: Find A such that one iteration prints x and leaves A = a
Going backwards:
A >> 3 = a -> A = 8a + Y
B2 = x
B1 = (A % 8) ^ 5 = Y^5
x = (B1 ^ 6) ^ (A >> B1) = (Y^5 ^ 6) ^ (A >> Y^5)
= (Y ^ 3) ^ (A >> Y^5)
-> x ^ (Y^3) = A >> Y^5
-> A = ((a^Y^3) << Y^5) + Z for some Z < Y^5 < 8
At the last iteartion, x = 0 and a = 0 so:
-> A = Y
-> x = (Y^3) ^ (Y >> Y^5)
Y = 0 => x = 3
Y = 1 => x = 2 ^ (1 >> 4) = 2
Y = 2 => x = 1 ^ (2 >> 7) = 1
Y = 3 => x = 0 ^ (3 >> 2) = 0
=> A = Y = 3
Second to last, x = 3 and a = 3, so
-> A = 24 + Y
3 = (A - 24) ^ 3 ^ ((A-24) >> (A-24)^5)
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