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C# So how does this hexadecimal stuff work?

I'm doing some entry level programming challenges at codefights.com and I came across the following question. The link is to a blog that has the answer, but it includes the question in it as well. If only it had an explanation...
https://codefightssolver.wordpress.com/2016/10/19/swap-adjacent-bits/
My concern is with the line of code (it is the only line of code) below.
return (((n & 0x2AAAAAAA) >> 1) | ((n & 0x15555555) << 1)) ;
Specifically, I'm struggling to find some decent info on how the "0x2AAAAAAA" and "0x15555555" work, so I have a few dumb questions. I know they represent binary values of 10101010... and 01010101... respectively.
1. I've messed around some and found out that the number of 5s and As corresponds loosely and as far as I can tell to bit size, but how?
2. Why As? Why 5s?
3. Why the 2 and the 1 before the As and 5s?
4. Anything else I should know about this? Does anyone know a cool blog post or website that explains some of this in more detail?

0x2AAAAAAA is 00101010101010101010101010101010 in 32 bits binary,
0x15555555 is ‭00010101010101010101010101010101‬ in 32 bits binary.
Note that the problem specifies Constraints: 0 ≤ n < 2^30. For this reason the highest two bits can be 00.
The two hex numbers have been "built" starting from their binary representation, that has a particular property (that we will see in the next paragraph).
Now... We can say that, given the constraint, x & 0x2AAAAAAA will return the even bits of x (if we count the bits as first, second, third... the second bit is even), while x & 0x15555555 will return the odd bits of x. By using << 1 and >> 1 you move them of one step. By using | (or) you re-merge them.

0x2AAAAAAA is used to get 30 bits, which is the constraint.
Constraints:
0 ≤ n < 2^30.
0x15555555 also represent 30 bits with bits opposite of other number.
I would start with binary number (101010101010101010101010101010) in the calculator and select hex using programmer calculator to show the number in hex.
you can also use 0b101010101010101010101010101010 too, if you like, depending on language.

Get random double (floating point) value from random byte array between 0 and 1 in C#?

Assume I have an array of bytes which are truly random (e.g. captured from an entropy source).
byte[] myTrulyRandomBytes = MyEntropyHardwareEngine.GetBytes(8);
Now, I want to get a random double precision floating point value, but between the values of 0 and positive 1 (like the Random.NextDouble() function performs).
Simply passing an array of 8 random bytes into BitConverter.ToDouble() can yield strange results, but most importantly, the results will almost never be less than 1.
I am fine with bit-manipulation, but the formatting of floating point numbers has always been mysterious to me. I tried many combinations of bits to apply randomness to and always ended up finding the numbers were either just over 1, always VERY close to 0, or very large.
Can someone explain which bits should be made random in a double in order to make it random within the range 0 and 1?

Though working answers have been given, I'll give an other one, that looks worse but isn't:
long asLong = BitConverter.ToInt64(myTrulyRandomBytes, 0);
double number = (double)(asLong & long.MaxValue) / long.MaxValue;
The issue with casting from an ulong to double is that it's not directly supported by hardware, so it compiles to this:
vxorps xmm0,xmm0,xmm0
vcvtsi2sd xmm0,xmm0,rcx ; interpret ulong as long and convert it to double
test rcx,rcx ; add fixup if it was "negative"
jge 000000000000001D
vaddsd xmm0,xmm0,mmword ptr [00000060h]
vdivsd xmm0,xmm0,mmword ptr [00000068h]
Whereas with my suggestion it will compile more nicely:
vxorps xmm0,xmm0,xmm0
vcvtsi2sd xmm0,xmm0,rcx
vdivsd xmm0,xmm0,mmword ptr [00000060h]
Both tested with the x64 JIT in .NET 4, but this applies in general, there just isn't a nice way to convert an ulong to a double.
Don't worry about the bit of entropy being lost: there are only 262 doubles between 0.0 and 1.0 in the first place, and most of the smaller doubles cannot be chosen so the number of possible results is even less.
Note that this as well as the presented ulong examples can result in exactly 1.0 and distribute the values with slightly differing gaps between adjacent results because they don't divide by a power of two. You can change them exclude 1.0 and get a slightly more uniform spacing (but see the first plot below, there is a bunch of different gaps, but this way it is very regular) like this:
long asLong = BitConverter.ToInt64(myTrulyRandomBytes, 0);
double number = (double)(asLong & long.MaxValue) / ((double)long.MaxValue + 1);
As a really nice bonus, you can now change the division to a multiplication (powers of two usually have inverses)
long asLong = BitConverter.ToInt64(myTrulyRandomBytes, 0);
double number = (double)(asLong & long.MaxValue) * 1.08420217248550443400745280086994171142578125E-19;
Same idea for ulong, if you really want to use that.
Since you also seemed interested specifically in how to do it with double-bits trickery, I can show that too.
Because of the whole significand/exponent deal, it can't really be done in a super direct way (just reinterpreting the bits and that's it), mainly because choosing the exponent uniformly spells trouble (with a uniform exponent, the numbers are necessarily clumped preferentially near 0 since most exponents are there).
But if the exponent is fixed, it's easy to make a double that's uniform in that region. That cannot be 0 to 1 because that spans a lot of exponents, but it can be 1 to 2 and then we can subtract 1.
So first mask away the bits that won't be part of the significand:
x &= (1L << 52) - 1;
Put in the exponent (1.0 - 2.0 range, excluding 2)
x |= 0x3ff0000000000000;
Reinterpret and adjust for the offset of 1:
return BitConverter.Int64BitsToDouble(x) - 1;
Should be pretty fast, too. An unfortunate side effect is that this time it really does cost a bit of entropy, because there are only 52 but there could have been 53. This way always leaves the least significant bit zero (the implicit bit steals a bit).
There were some concerns about the distributions, which I will address now.
The approach of choosing a random (u)long and dividing it by the maximum value clearly has a uniformly chosen (u)long, and what happens after that is actually interesting. The result can justifiably be called a uniform distribution, but if you look at it as a discrete distribution (which it actually is) it looks (qualitatively) like this: (all examples for minifloats)
Ignore the "thicker" lines and wider gaps, that's just the histogram being funny. These plots used division by a power of two, so there is no spacing problem in reality, it's only plotted strangely.
Top is what happens when you use too many bits, as happens when dividing a complete (u)long by its max value. This gives the lower floats a better resolution, but lots of different (u)longs get mapped onto the same float in the higher regions. That's not necessarily a bad thing, if you "zoom out" the density is the same everywhere.
The bottom is what happens when the resolution is limited to the worst case (0.5 to 1.0 region) everywhere, which you can do by limiting the number of bits first and then doing the "scale the integer" deal. My second suggesting with the bit hacks does not achieve this, it's limited to half that resolution.
For what it's worth, NextDouble in System.Random scales a non-negative int into the 0.0 .. 1.0 range. The resolution of that is obviously a lot lower than it could be. It also uses an int that cannot be int.MaxValue and therefore scales by approximately 1/(231-1) (cannot be represented by a double, so slightly rounded), so there are actually 33 slightly different gaps between adjacent possible results, though the majority of the gaps is the same distance.
Since int.MaxValue is small compared to what can be brute-forced these days, you can easily generate all possible results of NextDouble and examine them, for example I ran this:
const double scale = 4.6566128752458E-10;
double prev = 0;
Dictionary<long, int> hist = new Dictionary<long, int>();
for (int i = 0; i < int.MaxValue; i++)
{
long bits = BitConverter.DoubleToInt64Bits(i * scale - prev);
if (!hist.ContainsKey(bits))
hist[bits] = 1;
else
hist[bits]++;
prev = i * scale;
if ((i & 0xFFFFFF) == 0)
Console.WriteLine("{0:0.00}%", 100.0 * i / int.MaxValue);
}

This is easier than you think; its all about scaling (also true when going from a 0-1 range to some other range).
Basically, if you know that you have 64 truly random bits (8 bytes) then just do this:
double zeroToOneDouble = (double)(BitConverter.ToUInt64(bytes) / (decimal)ulong.MaxValue);
The trouble with this kind of algorithm comes when your "random" bits aren't actually uniformally random. That's when you need a specialized algorithm, such as a Mersenne Twister.

I don't know wether it's the best solution for this, but it should do the job:
ulong asLong = BitConverter.ToUInt64(myTrulyRandomBytes, 0);
double number = (double)asLong / ulong.MaxValue;
All I'm doing is converting the byte array to a ulong which is then divided by it's max value, so that the result is between 0 and 1.

To make sure the long value is within the range from 0 to 1, you can apply the following mask:
long longValue = BitConverter.ToInt64(myTrulyRandomBytes, 0);
longValue &= 0x3fefffffffffffff;
The resulting value is guaranteed to lay in the range [0, 1).
Remark. The 0x3fefffffffffffff value is very-very close to 1 and will be printed as 1, but it is really a bit less than 1.
If you want to make the generated values greater, you could set a number higher bits of an exponent to 1. For instance:
longValue |= 0x03c00000000000000;
Summarizing: example on dotnetfiddle.

If you care about the quality of the random numbers generated, be very suspicious of the answers that have appeared so far.
Those answers that use Int64BitsToDouble directly will definitely have problems with NaNs and infinities. For example, 0x7ff0000000000001, a perfectly good random bit pattern, converts to NaN (and so do thousands of others).
Those that try to convert to a ulong and then scale, or convert to a double after ensuring that various bit-pattern constraints are met, won't have NaN problems, but they are very likely to have distributional problems. Representable floating point numbers are not distributed uniformly over (0, 1), so any scheme that randomly picks among all representable values will not produce values with the required uniformity.
To be safe, just use ToInt32 and use that int as a seed for Random. (To be extra safe, reject 0.) This won't be as fast as the other schemes, but it will be much safer. A lot of research and effort has gone into making RNGs good in ways that are not immediately obvious.

Simple piece of code to print the bits out for you.
for (double i = 0; i < 1.0; i+=0.05)
{
var doubleToInt64Bits = BitConverter.DoubleToInt64Bits(i);
Console.WriteLine("{0}:\t{1}", i, Convert.ToString(doubleToInt64Bits, 2));
}
0.05: 11111110101001100110011001100110011001100110011001100110011010
0.1: 11111110111001100110011001100110011001100110011001100110011010
0.15: 11111111000011001100110011001100110011001100110011001100110100
0.2: 11111111001001100110011001100110011001100110011001100110011010
0.25: 11111111010000000000000000000000000000000000000000000000000000
0.3: 11111111010011001100110011001100110011001100110011001100110011
0.35: 11111111010110011001100110011001100110011001100110011001100110
0.4: 11111111011001100110011001100110011001100110011001100110011001
0.45: 11111111011100110011001100110011001100110011001100110011001100
0.5: 11111111011111111111111111111111111111111111111111111111111111
0.55: 11111111100001100110011001100110011001100110011001100110011001
0.6: 11111111100011001100110011001100110011001100110011001100110011
0.65: 11111111100100110011001100110011001100110011001100110011001101
0.7: 11111111100110011001100110011001100110011001100110011001100111
0.75: 11111111101000000000000000000000000000000000000000000000000001
0.8: 11111111101001100110011001100110011001100110011001100110011011
0.85: 11111111101011001100110011001100110011001100110011001100110101
0.9: 11111111101100110011001100110011001100110011001100110011001111
0.95: 11111111101110011001100110011001100110011001100110011001101001

Math.Ceiling() on expression when dividing and multiplying by same number

I have a little trouble with Math.Ceiling() in C#.
I call it on the result of a division by a number followed by a multiplication by the same number, e.g. 20 000 / 184 * 184. I would expect that result to be 20 000 but it is 20 001. Are there any possible ways how to avoid this behavior when trying round up value?
Thank you in advance

When running the code you supplied we have the following
twentyThousand/oneEightyFour * oneEightyFour
The answer is 20000.000000000000000000000001
Hence when you do the ceiling we have 20001.
By the following article I think the result is due to in inaccuracy introduced when performing the division , this yields 108.69565217391304347826086957 and as Jon stated
As a very broad rule of thumb, if you end up seeing a very long string representation (ie most of the 28/29 digits are non-zero) then chances are you've got some inaccuracy along the way.
http://csharpindepth.com/Articles/General/Decimal.aspx

As light pointed out in the comments, you shouldn't be getting 20001 at all.
20000 / 184 would yield 108. Which then would give you 19872 when multiplied by 184.
Somewhere you are doing something other than what you posted. Where is Math.Ceiling() even called?
I will say, if the numbers are hard coded, you can put a decimal in the code and it will treat it as such. If you are using variables that represent numbers, be sure they are formatted as some floating point type (decimal,double,float) depending on the accuracy needed.
Console.WriteLine(20000 / 184 * 184); // 19872
Console.WriteLine((20000.0 / 184.0 * 184.0)); // 20000

Are there any possible ways how to avoid this behavior when trying round up value?
In this particular case you can avoid the problem by multiplying first, then dividing:
result = (20000m * 184m) / 184m ;
Since the precision is lost in the division, multiplying first prevents that imprecision form getting exaggerated when you multiply.

Increment forever and you get -2147483648?

For a clever and complicated reason that I don't really want to explain (because it involves making a timer in an extremely ugly and hacky way), I wrote some C# code sort of like this:
int i = 0;
while (i >= 0) i++; //Should increment forever
Console.Write(i);
I expected the program to hang forever or crash or something, but, to my surprise, after waiting for about 20 seconds or so, I get this ouput:
-2147483648
Well, programming has taught me many things, but I still cannot grasp why continually incrementing a number causes it to eventually be negative...what's going on here?

In C#, the built-in integers are represented by a sequence of bit values of a predefined length. For the basic int datatype that length is 32 bits. Since 32 bits can only represent 4,294,967,296 different possible values (since that is 2^32), clearly your code will not loop forever with continually increasing values.
Since int can hold both positive and negative numbers, the sign of the number must be encoded somehow. This is done with first bit. If the first bit is 1, then the number is negative.
Here are the int values laid out on a number-line in hexadecimal and decimal:
Hexadecimal Decimal
----------- -----------
0x80000000 -2147483648
0x80000001 -2147483647
0x80000002 -2147483646
... ...
0xFFFFFFFE -2
0xFFFFFFFF -1
0x00000000 0
0x00000001 1
0x00000002 2
... ...
0x7FFFFFFE 2147483646
0x7FFFFFFF 2147483647
As you can see from this chart, the bits that represent the smallest possible value are what you would get by adding one to the largest possible value, while ignoring the interpretation of the sign bit. When a signed number is added in this way, it is called "integer overflow". Whether or not an integer overflow is allowed or treated as an error is configurable with the checked and unchecked statements in C#. The default is unchecked, which is why no error occured, but you got that crazy small number in your program.
This representation is called 2's Complement.

The value is overflowing the positive range of 32 bit integer storage going to 0xFFFFFFFF which is -2147483648 in decimal. This means you overflow at 31 bit integers.
It's been pointed out else where that if you use an unsigned int you'll get different behaviour as the 32nd bit isn't being used to store the sign of of the number.

What you are experiencing is Integer Overflow.
In computer programming, an integer overflow occurs when an arithmetic operation attempts to create a numeric value that is larger than can be represented within the available storage space. For instance, adding 1 to the largest value that can be represented constitutes an integer overflow. The most common result in these cases is for the least significant representable bits of the result to be stored (the result is said to wrap).

int is a signed integer. Once past the max value, it starts from the min value (large negative) and marches towards 0.
Try again with uint and see what is different.

Try it like this:
int i = 0;
while (i >= 0)
checked{ i++; } //Should increment forever
Console.Write(i);
And explain the results

What the others have been saying. If you want something that can go on forever (and I wont remark on why you would need something of this sort), use the BigInteger class in the System.Numerics namespace (.NET 4+). You can do the comparison to an arbitrarily large number.

It has a lot to do with how positive numbers and negative numbers are really stored in memory (at bit level).
If you're interested, check this video: Programming Paradigms at 12:25 and onwards. Pretty interesting and you will understand why your code behaves the way it does.

This happens because when the variable "i" reaches the maximum int limit, the next value will be a negative one.

I hope this does not sound like smart-ass advice, because its well meant, and not meant to be snarky.
What you are asking is for us to describe that which is pretty fundamental behaviour for integer datatypes.
There is a reason why datatypes are covered in the 1st year of any computer science course, its really very fundamental to understanding how and where things can go wrong (you can probably already see how the behaviour above if unexpected causes unexpected behaviour i.e. a bug in your application).
My advice is get hold of the reading material for 1st year computer science + Knuth's seminal work "The art of computer pragramming" and for ~ $500 you will have everything you need to become a great programmer, much cheaper than a whole Uni course ;-)

Get number of digits in an unsigned long integer c#

I'm trying to determine the number of digits in a c# ulong number, i'm trying to do so using some math logic rather than using ToString().Length. I have not benchmarked the 2 approaches but have seen other posts about using System.Math.Floor(System.Math.Log10(number)) + 1 to determine the number of digits.
Seems to work fine until i transition from 999999999999997 to 999999999999998 at which point, it i start getting an incorrect count.
Has anyone encountered this issue before ?
I have seen similar posts with a Java emphasis # Why log(1000)/log(10) isn't the same as log10(1000)? and also a post # How to get the separate digits of an int number? which indicates how i could possibly achieve the same using the % operator but with a lot more code
Here is the code i used to simulate this
Action<ulong> displayInfo = number =>
Console.WriteLine("{0,-20} {1,-20} {2,-20} {3,-20} {4,-20}",
number,
number.ToString().Length,
System.Math.Log10(number),
System.Math.Floor(System.Math.Log10(number)),
System.Math.Floor(System.Math.Log10(number)) + 1);
Array.ForEach(new ulong[] {
9U,
99U,
999U,
9999U,
99999U,
999999U,
9999999U,
99999999U,
999999999U,
9999999999U,
99999999999U,
999999999999U,
9999999999999U,
99999999999999U,
999999999999999U,
9999999999999999U,
99999999999999999U,
999999999999999999U,
9999999999999999999U}, displayInfo);
Array.ForEach(new ulong[] {
1U,
19U,
199U,
1999U,
19999U,
199999U,
1999999U,
19999999U,
199999999U,
1999999999U,
19999999999U,
199999999999U,
1999999999999U,
19999999999999U,
199999999999999U,
1999999999999999U,
19999999999999999U,
199999999999999999U,
1999999999999999999U
}, displayInfo);
Thanks in advance
Pat

log10 is going to involve floating point conversion - hence the rounding error. The error is pretty small for a double, but is a big deal for an exact integer!
Excluding the .ToString() method and a floating point method, then yes I think you are going to have to use an iterative method but I would use an integer divide rather than a modulo.
Integer divide by 10. Is the result>0? If so iterate around. If not, stop.
The number of digits is the number of iterations required.
Eg. 5 -> 0; 1 iteration = 1 digit.
1234 -> 123 -> 12 -> 1 -> 0; 4 iterations = 4 digits.

I would use ToString().Length unless you know this is going to be called millions of times.
"premature optimization is the root of all evil" - Donald Knuth

From the documentation:
By default, a Double value contains 15
decimal digits of precision, although
a maximum of 17 digits is maintained
internally.
I suspect that you're running into precision limits. Your value of 999,999,999,999,998 probably is at the limit of precision. And since the ulong has to be converted to double before calling Math.Log10, you see this error.

Other answers have posted why this happens.
Here is an example of a fairly quick way to determine the "length" of an integer (some cases excluded). This by itself is not very interesting -- but I include it here because using this method in conjunction with Log10 can get the accuracy "perfect" for the entire range of an unsigned long without requiring a second log invocation.
// the lookup would only be generated once
// and could be a hard-coded array literal
ulong[] lookup = Enumerable.Range(0, 20)
.Select((n) => (ulong)Math.Pow(10, n)).ToArray();
ulong x = 999;
int i = 0;
for (; i < lookup.Length; i++) {
if (lookup[i] > x) {
break;
}
}
// i is length of x "in a base-10 string"
// does not work with "0" or negative numbers
This lookup-table approach can be easily converted to any base. This method should be faster than the iterative divide-by-base approach but profiling is left as an exercise to the reader. (A direct if-then branch broken into "groups" is likely quicker yet, but that's way too much repetitive typing for my tastes.)
Happy coding.