pieguy

In statistics, "average" often means "expected value". No time frame is specified (although you could consider it an infinite time frame). With a small sample size your actual average might not be 10 minutes, but as your sample size grows, it will tend toward 10 minutes.

Let me say it again in a different way. My resolution of the paradox is that the prisoner incorrectly negates "x will happen" as "x won't happen" instead of negating "x definitely will happen" as "x might not happen". Thus,the prisoner cannot conclude that the hanging won't be a surprise. The prisoner can only conclude that the hanging might not be a surprise.

> he believes he wouldn't be hanged at all

> he assumes what the judge said is true

The judge says he will be hanged. Your comments are inconsistent.

He cannot logically conclude that. He can only conclude that either he won't be hanged or his hanging might not be a surprise.

No, it would not have been a surprise if he were hanged on Friday. The prisoner disproves the existence of a strategy for the judge that guarantees surprise, but does not disprove the existence of a strategy that gives the possibility of surprise.

Here's my take. The prisoner considers the statement "I will be hanged, and it will be a surprise", and after "proving" it false, takes its negation "Either I won't be hanged, or it won't be a surprise". But this is not correct. The prisoner actually disproves the statement "I will be hanged, and it will be a surprise no matter which day it happens". When you negate this statement, you obtain "Either I won't be hanged, or there is a day that it would not be a surprise", which is consistent with the outcome.

I don't think multi-algorithm is taken into account for multi-algorithm coins. Several coins have multiple independent mining algorithms each with their own difficulty, and need to be attacked simultaneously.

   (c(mod p)) (mod 2) = (p * q + 2 * r + m (mod p)) (mod 2) = 2r + m (mod 2) = m

This breaks if 2*r > p. Even if you choose r to be small during encryption, the r values accumulate with each homomorphic operation and will eventually be too big. The only restriction stated is that r is "from a different interval than the private key one". This should be made more clear.

I used to answer secret questions with bogus answers that I deemed unguessable. Then I discovered that when my bank asks me the questions back it does multiple choice, displaying the answer I gave along with 4 other possible options! Sometimes my answer would not be shown and the correct answer is "none of the above", but otherwise my answer sticks out like a sore thumb.

On my machine this produces

  (STDIN)= 5BAA61E4C9B93F3F0682250B6CF8331B7EE68FD8

Which is obviously not what you want. So I changed it to:

  echo -n "$password" | openssl sha1 -binary | xxd -p -u

There's a decent chance the passwords you found aren't the actual passwords either, considering how easy it is to generate collisions for the hash function.

"(long)pow(a, len_s - (i+1))" is both inefficient and inaccurate. You should store the current power in a variable which you multiply by 'a' and reduce by 'm' on each iteration.

Also, your double hashing scheme has a bug. You identified that it's problematic if hash_b returns 0, but adding 1 to the result doesn't actually solve anything, because now you have the same problem if hash_b returns num_buckets-1.

Not sure how you came up with 5 cents but I get $1.36: 4000 dollars/btc * 10^-8 btc/satoshi * 150 satoshi/byte * 226 bytes/transaction = $1.356 per transaction