gavinksong

Yeah, it's a little vague. I'm rusty on math, but I don't think there are any continuous and differentiable functions that are zero everywhere except for a specific region. If it doesn't have to be differentiable, you could just construct a composite function with linear parts or maybe stitch together bezier curves. If it doesn't have to be continuous except for the region between $1 and $40, you could just use something like f(x | a, b, c, d) = a*e^[-(x - b)^c / d] and solve for a, b, c, d.

Haha, this is wonderful. BMAD, you must mean 0.7 waist-to-hip ratio and not the other way around.

Nice job! As you said, our fellow brain-denizens had all the bits and pieces (especially DeGe and Logophobic, also Molly Mae). Well done on putting them all together! There IS one last piece of the puzzle required to make n = 2 billion run in a reasonable amount of time. I didn't know this. If this is true, this is indeed an alternative solution.

Nice job! As you said, our fellow brain-denizens had all the bits and pieces (especially DeGe and Logophobic, also Molly Mae). Well done on putting them all together! There IS one last piece of the puzzle required to make n = 2 billion run in a reasonable amount of time.

I appreciate the amount of thought you've put into these responses. Unfortunately, I do believe all three of your original assumptions are correct. Again, I don't know the solution to this problem, so I can't be sure. The basis for my opinion is the following line from the OP: I am thus led to believe that "memory" refers to binary memory. This implies that assumption (1) is true as long as each record is a fixed number of bits and that assumption (2) is true, since instead of arbitrarily precise "memory cells" we have bits. It follows that your proposal for relaxing assumption (2) is likely not intended by the OP. I think you were on the right track with using the maze itself as a source of boundless memory. The problem certainly seems impossible, but I assure you that this has been a signature of BWOC puzzles. I truly believe that there is a way to solve the problem as we currently understand it. I suspect that it's necessary to somehow leverage the infinite amount of time we have. The OP states: "probability 1" suggests that we can converge on it over infinite time. If we could design an algorithm that every now and then has some probability of returning the correct answer with no chance of being incorrect, then over an infinite timeline, the probability of success would converge to 1.

I will confirm that what you observed is true. You need to be clever.

Hmm. Maybe this is trivial, but I have a counterexample.

Let me explain my code from before. The results it gives suggests that the surgeon's claim is correct.

I wrote some short Haskell code to verify the surgeon's claim for all worms of a given length. I have yet to think of an actual proof.

Where did the bin come from?

Actually, DeGe, your first method is approaching one of the solutions, but it isn't quite there because it still uses irrational numbers in the final calculation.

Hey, thanks for giving this puzzle some attention. Let me rephrase the problem so that it is less ambiguous. There will still be multiple solutions. How can you determine the last three digits of (3 + √5)n before the decimal using only integer operations?

Wow, I really didn't think anybody would get this! I'm actually super impressed. ))) Good job!

Yeah, I think post #6 should be the Best Answer.

I'd like to know the answer

Incidentally, this explanation provides a much simpler way to arrive at the solution to your aha​ problem, bonanova. Edit - I just realized that there was an equally elegant words-only solution further along the aha thread. And it has a slightly different angle than the explanation we have above. Man, this is what I love about math.

Say we want to simulate an N sided die.

I'm gonna go out on a limb here, but based partially from a trend I have noticed and the formula that bonanova cited...

Suppose there are M gold fish and K silver fish and W red herrings in a lake. The gold and silver fish are caught and eaten one at a time until only two colors of fish remain in the lake. One of the silver fish is named George. One of the red herrings is named Harold. Find the probability George is not eaten. I am finding this problem very hard. :\

But I wonder if there is a better way.

You, good sir, have just blown my mind. I have not seen this type of solution. In my mind, you are a true winner. But there is still a problem lurking in your blind side. You would have to calculate either √5 or phi to an absurd amount of precision in order to multiply them by such a large integer. You would probably either run out of space or time. The trick is to exclude irrational numbers entirely from the final calculation - or at least, that's what the trick is for one solution.

Brilliant!