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Got it! Here is my solution in Python: <pre><code>a = 5402147 b = 54321 n = 10000001 def psi(x): return (a * x + b) % n inverse1000 = 9990001 max_k = (n-1) / 1000 + 1 def first_input(y): global last_input, i, possible_k last_input = y possible_k = [set(range(max_k))] i = 0 def add_input(y): global last_input, i c = inverse1000 * (b + a * last_input - y) % n sk0 = set() sk1 = set() for k0 in possible_k[i]: ak0 = a * k0 % n for k1 in range(max_k): if (k1 - ak0) % n == c: sk0.add(k0) sk1.add(k1) #print "found a solution" last_input = y possible_k[i] = possible_k[i] & sk0 possible_k.append(sk1) i += 1 if len(possible_k[i-1]) == 0 or len(possible_k[i]) == 0: print "Wrong machine" return if len(possible_k[i]) == 1: x = y + 1000 * possible_k[i].copy().pop() for j in range(10): x = psi(x) print x % 1000, print return print "Not enough observations" </code></pre> It could probably be optimized (and cleaned), but since it runs in less than 30sec on my 3 years old laptop I probably won't bother making it faster... The program doesn't accept exactly the same input as requested, here is how to use it: <pre><code>>>> first_input(767) >>> add_input(308) Not enough observations >>> add_input(284) 577 428 402 291 252 544 735 545 771 34 >>> first_input(78) >>> add_input(880) Not enough observations >>> add_input(53) 698 235 762 18 98 703 456 676 621 291 >>> add_input(698) 235 762 18 98 703 456 676 621 291 488 >>> add_input(235) 762 18 98 703 456 676 621 291 488 332 >>> first_input(862) >>> add_input(452) Not enough observations >>> add_input(303) Wrong machine >>> add_input(558) Wrong machine </code></pre> As you can see, usually 3 observations are enough to determine the future outcomes. Since it's a pain to write math stuff in a text editor I took a picture of my <strike>demonstration</strike> explanation: <img src="https://i.stack.imgur.com/5vFDJ.jpg" alt="hand written "demonstration"">
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