puzzle

I am trying to figure an agreement protocol on an unreliable channel. Basically two parties (A and B) have to agree to do something, so it's the two generals' problem . Since there is no bullet-proof solution, I am trying to do this. A continuously sends a message with sequence 1 When B receives seq 1 it continuously replies with sequence 2 At this point A receives seq 2 so it starts sending seq 3 ... My question. When can the two parties conclude that they can take the action ? Obviously I can't set a condition: " do it after receiving 10 messages " since the last sender won't have any

We recently encountered an interesting problem at work where we discovered duplicate user submitted data in our database. We realized that the Levenshtein distance between most of this data was simply the difference between the 2 strings in question. That indicates that if we simply add characters from one string into the other then we end up with the same string, and for most things this seems like the best way for us to account for items that are duplicate. We also want to account for typos. So we started to think about on average how often do people make typos online per word, and try to

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center . Closed 6 years ago . I wrote following code. class String { private final java.lang.String s; public String(java.lang.String s){ this.s = s; } public java.lang.String toString(){ return s; } public static void main(String[] args) { String s = new String("Hello world"); System

Possible Duplicate: Why is this statement not working in java x ^= y ^= x ^= y; Sample code int a=3; int b=4; a^=(b^=(a^=b)); In c++ it swaps variables, but in java we get a=0, b=4 why? By writing your swap all in one statement, you are relying on side effects of the inner a^=b expression relative to the outer a^=(...) expression. Your Java and C++ compilers are doing things differently. In order to do the xor swap properly, you have to use at least two statements: a ^= b; a ^= (b ^= a); However, the best way to swap variables is to do it the mundane way with a temporary variable, and let the

I've been learning Ruby, so I thought I'd try my hand at some of the project Euler puzzles. Embarrassingly, I only made it to problem 4... Problem 4 goes as follows: A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99. Find the largest palindrome made from the product of two 3-digit numbers. So I figured I would loop down from 999 to 100 in a nested for loop and do a test for the palindrome and then break out of the loops when I found the first one (which should be the largest one): final=nil range = 100...1000

From the question " Is partitioning easier than sorting? ": Suppose I have a list of items and an equivalence relation on them, and comparing two items takes constant time. I want to return a partition of the items, e.g. a list of linked lists, each containing all equivalent items. One way of doing this is to extend the equivalence to an ordering on the items and order them (with a sorting algorithm); then all equivalent items will be adjacent. (Keep in mind the distinction between equality and equivalence .) Clearly the equivalence relation must be considered when designing the ordering