logarithm

This earlier question addresses some of the factors that might cause an algorithm to have O(log n) complexity. What would cause an algorithm to have time complexity O(log log n)? O(log log n) terms can show up in a variety of different places, but there are typically two main routes that will arrive at this runtime. Shrinking by a Square Root As mentioned in the answer to the linked question, a common way for an algorithm to have time complexity O(log n) is for that algorithm to work by repeatedly cut the size of the input down by some constant factor on each iteration. If this is the case,

I need to rewrite the log function (base 2 or base 10 doesn't matter which) without using float type, but I need to get the precision of few decimal digits after the decimal point. ( like a float * 100 to get 2 decimals inside integer type eg: if the 1.4352 would be the result, my function should return something like 143 ( int type) and I know that the last 2 numbers are the decimals. I found over the stackoverflow some methods like: How can I compute a base 2 logarithm without using the built-in math functions in C#? but all of them returns int precision ( avoiding the decimals ). I have no

How can I calculate the logarithm of a BigDecimal? Does anyone know of any algorithms I can use? My googling so far has come up with the (useless) idea of just converting to a double and using Math.log. I will provide the precision of the answer required. edit: any base will do. If it's easier in base x, I'll do that. Peter Java Number Cruncher: The Java Programmer's Guide to Numerical Computing provides a solution using Newton's Method . Source code from the book is available here . The following has been taken from chapter 12.5 Big Decmial Functions (p330 & p331): /** * Compute the natural

In the C++ standard libraries I found only a floating point log method. Now I use log to find the level of an index in a binary tree ( floor(2log(index)) ). Code (C++): int targetlevel = int(log(index)/log(2)); I am afraid that for some of the edge elements (the elements with value 2^n) log will return n-1.999999999999 instead of n.0. Is this fear correct? How can I modify my statement so that it always will return a correct answer? You can use this method instead: int targetlevel = 0; while (index >>= 1) ++targetlevel; Note: this will modify index. If you need it unchanged, create another