This is the function that I use for this calculation: public static int binlog( int bits ) // returns 0 for bits=0 { int log = 0; if( ( bits & 0xffff0000 ) != 0 ) { bits >>>= 16; log = 16; } if( bits >= 256 ) { bits >>>= 8; log += … Read more
Your code is doing a log of a number that is less than or equal to zero. That’s mathematically undefined, so Python’s log function raises an exception. Here’s an example: >>> from math import log >>> log(-1) Traceback (most recent call last): File “<pyshell#59>”, line 1, in <module> log(-1) ValueError: math domain error Without knowing … Read more
“Change of Base” Formula / Identity The numerical value for logarithm to the base 10 can be calculated with the following identity. Since Math.log(x) in JavaScript returns the natural logarithm of x (same as ln(x)), for base 10 you can divide by Math.log(10) (same as ln(10)): function log10(val) { return Math.log(val) / Math.LN10; } Math.LN10 … Read more
I have to agree that it’s pretty weird the first time you see an O(log n) algorithm… where on earth does that logarithm come from? However, it turns out that there’s several different ways that you can get a log term to show up in big-O notation. Here are a few: Repeatedly dividing by a … Read more
A histogram is a poor-man’s density estimate. Note that in your call to hist() using default arguments, you get frequencies not probabilities — add ,prob=TRUE to the call if you want probabilities. As for the log axis problem, don’t use ‘x’ if you do not want the x-axis transformed: plot(mydata_hist$count, log=”y”, type=”h”, lwd=10, lend=2) gets … Read more
I finally found some time to do some experiments in order to understand the difference between them. Here’s what I discovered: log only allows positive values, and lets you choose how to handle negative ones (mask or clip). symlog means symmetrical log, and allows positive and negative values. symlog allows to set a range around … Read more
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 … Read more
If you are on a recent-ish x86 or x86-64 platform (and you probably are), use the bsr instruction which will return the position of the highest set bit in an unsigned integer. It turns out that this is exactly the same as log2(). Here is a short C or C++ function that invokes bsr using … Read more
You need to use fixed point precision/arithmetics/math for this. It means you use integer type variables but some of the bits are after the decimal point. for example let assume 8 decimal bits so operations are done like this: a = number1*256 b = number2*256 c=a+b // + c=a-b // – c=(a*b)>>8 // * c=(a/b)<<8 … Read more
The usual strategy is based on the identity log(a*b) = log(a) + log(b), or in this case log2( 2^exponent * mantissa) ) = log2( 2^exponent ) + log2(mantissa). Or simplifying, exponent + log2(mantissa). The mantissa has a very limited range, 1.0 to 2.0, so a polynomial for log2(mantissa) only has to fit over that very … Read more