currently im working on a solution for a prime-number calculator/checker. The algorythm is already working and verry efficient (0,359 seconds for the first 9012330 primes). Here is a part of the upper region where everything is declared:
const uint anz = 50000000;
uint a = 3, b = 4, c = 3, d = 13, e = 12, f = 13, g = 28, h = 32;
bool[,] prim = new bool[8, anz / 10];
uint max = 3 * (uint)(anz / (Math.Log(anz) - 1.08366));
uint[] p = new uint[max];
Now I wanted to go to the next level and use ulong's instead of uint's to cover a larger area (you can see that already), where i tapped into my problem: the bool-array.
Like everybody should know, bool's have the length of a byte what takes a lot of memory when creating the array... So I'm searching for a more resource-friendly way to do that.
My first idea was a bit-array -> not byte! <- to save the bool's, but haven't figured out how to do that by now. So if someone ever did something like this, I would appreciate any kind of tips and solutions. Thanks in advance :)
You can use BitArray collection:
http://msdn.microsoft.com/en-us/library/system.collections.bitarray(v=vs.110).aspx
MSDN Description:
Manages a compact array of bit values, which are represented as Booleans, where true indicates that the bit is on (1) and false indicates the bit is off (0).
You can (and should) use well tested and well known libraries.
But if you're looking to learn something (as it seems to be the case) you can do it yourself.
Another reason you may want to use a custom bit array is to use the hard drive to store the array, which comes in handy when calculating primes. To do this you'd need to further split addr, for example lowest 3 bits for the mask, next 28 bits for 256MB of in-memory storage, and from there on - a file name for a buffer file.
Yet another reason for custom bit array is to compress the memory use when specifically searching for primes. After all more than half of your bits will be 'false' because the numbers corresponding to them would be even, so in fact you can both speed up your calculation AND reduce memory requirements if you don't even store the even bits. You can do that by changing the way addr is interpreted. Further more you can also exclude numbers divisible by 3 (only 2 out of every 6 numbers has a chance of being prime) thus reducing memory requirements by 60% compared to plain bit array.
Notice the use of shift and logical operators to make the code a bit more efficient.
byte mask = (byte)(1 << (int)(addr & 7)); for example can be written as
byte mask = (byte)(1 << (int)(addr % 8));
and addr >> 3 can be written as addr / 8
Testing shift/logical operators vs division shows 2.6s vs 4.8s in favor of shift/logical for 200000000 operations.
Here's the code:
void Main()
{
var barr = new BitArray(10);
barr[4] = true;
Console.WriteLine("Is it "+barr[4]);
Console.WriteLine("Is it Not "+barr[5]);
}
public class BitArray{
private readonly byte[] _buffer;
public bool this[long addr]{
get{
byte mask = (byte)(1 << (int)(addr & 7));
byte val = _buffer[(int)(addr >> 3)];
bool bit = (val & mask) == mask;
return bit;
}
set{
byte mask = (byte) ((value ? 1:0) << (int)(addr & 7));
int offs = (int)addr >> 3;
_buffer[offs] = (byte)(_buffer[offs] | mask);
}
}
public BitArray(long size){
_buffer = new byte[size/8 + 1]; // define a byte buffer sized to hold 8 bools per byte. The spare +1 is to avoid dealing with rounding.
}
}
Related
So I've been trying to add bits of a value to a MemoryStream but the issue is I have no idea how. I've seen that it's used for performance when it comes to networking.
I know I want a function that takes the bit value and how many bits it takes to store that value. So for instance, to store the value 3 I would need to allocate 2 bits 0000 0000 0000 0011. I would essentially pack the bits into a byte array and then add that byte array to the MemoryStream
var ms = new MemoryStream();
ms.WriteByte(1);
ms.WriteByte(1);
ms.WriteByte(1);
ms.WriteByte(1);
ms.WriteByte(1);
WriteBits(2, 3);
WriteBits(1, 1);
void WriteBits(int numbBits, int value)
{
/* Convert the "value" to a byte or bytes and add it to the MemoryStream */
}
How do I properly implement this?
Java Example
public void writeBits(int numBits, int value) {
int bytePos = bitPosition >> 3;
int bitOffset = 8 - (bitPosition & 7);
bitPosition += numBits;
for (; numBits > bitOffset; bitOffset = 8) {
buffer[bytePos] &= ~bitMaskOut[bitOffset]; // mask out the desired area
buffer[bytePos++] |= (value >> (numBits - bitOffset))
& bitMaskOut[bitOffset];
numBits -= bitOffset;
}
if (numBits == bitOffset) {
buffer[bytePos] &= ~bitMaskOut[bitOffset];
buffer[bytePos] |= value & bitMaskOut[bitOffset];
} else {
buffer[bytePos] &= ~(bitMaskOut[numBits] << (bitOffset - numBits));
buffer[bytePos] |= (value & bitMaskOut[numBits]) << (bitOffset - numBits);
}
}
So I've been trying to add bits of a value to a MemoryStream
You don't, MemoryStream only handles bytes.
So for instance, to store the value 3 I would need to allocate 2 bits
This would only be true if the range of values you want to store is [0, 3]. If you want the possibility of storing any larger value you need more bits.
How do I properly implement this?
you would need to implement your own bit-stream. The java example looks like it has a byte[] buffer, and a bitPosition. You would need to implement this. The bif-fiddeling code looks like it should work just about the same in c#. Once you have a byte[] it is trivial to write this out to whatever stream you want, and usually possible to send directly over the network.
I've seen that it's used for performance when it comes to networking
I think there is a significant misunderstanding here. While you could manually manipulate individual bits, in most cases it would just be a waste of (development) time.
In general, a better way to get good performance is to use existing, well optimized and designed libraries. And there are a variety of serialization libraries that converts objects to byte-streams for you. An example would be protobuf (.net), this actually encodes numbers with a variable number of bytes.
If you still need smaller data it is usually more efficient to use some form of compression. The old classic deflate usually gives good compromise between size and performance, while algorithms like lz4 prioritizes speed over compression ratio.
I had exactly the same problem and wrote an entire BitStream library which can handle any reads and writes of an arbitrary number of bits to a MemoryStream (and any other stream, too). The library is open-source, MIT-licensed and fast (https://github.com/martinweihrauch/BitStream).
Writing bits to a MemoryStream.
These are the steps to write a value to a certain number of bits to a specific position in the MemoryStream:
Have a Stream available, e. g. a MemoryStream(), to which you want to write.
Connect this Stream to a new Bitstream
using SharpBitStream;
uint[] testDataUnsigned = { 5, 62, 17, 50, 33 };
var ms = new MemoryStream();
var bs = new BitStream(ms);
Now, you can start writing to the BitStream like this:
foreach(var bits in testDataUnsigned)
{
bs.WriteUnsigned(6, (ulong)bits);
}
Writing can be done as above by only providing the bitlength and the value, but you of course also have full controll of exactly where to write the bits like so:
bs.WriteUnsigned(3, 2, 4, 5);
// Overloaded signature of WriteUnsigned:
// public void WriteUnsigned(long offsetByteStream, int offsetBit, int bitLength, ulong value)
// For signed numbers (e. g. -17), use
// bs.WriteSigned(3, 2, 4, -5);
This means, you can control that you write to the 4th byte (3, because starting at 0) in the underlying byte Stream,
starting from the the 3rd (=2) position of the byte with a length of 6 bits and the value 5 (=0b0101);
Reading works similarly:
Just read the next 6 bits, wherever your byte and bit position is (e. g. for loops, etc):
ulong number = bs.ReadUnsigned(6);
// For Signed, use
// long number = bs.ReadSigned(6);
Read a specific position, in this example read 4 bits from 3rd byte in Stream (2= 3rd position), starting with bit #0:
ulong number = bs.ReadUnsigned(2, 0, 4);
// For signed, use
// long number = bs.ReadSigned(2, 0, 4);
Note: The bit offset is always counting from 0 from the left-most position.
I am trying to find a simple algorithm that reverses the bits of a number up to N number of bits. For example:
For N = 2:
01 -> 10
11 -> 11
For N = 3:
001 -> 100
011 -> 110
101 -> 101
The only things i keep finding is how to bit reverse a full byte but thats only going to work for N = 8 and thats not always what i need.
Does any one know an algorithm that can do this bitwise operation? I need to do many of them for an FFT so i'm looking for something that can be very optimised too.
It is the C# implementation of bitwise reverse operation:
public uint Reverse(uint a, int length)
{
uint b = 0b_0;
for (int i = 0; i < length; i++)
{
b = (b << 1) | (a & 0b_1);
a = a >> 1;
}
return b;
}
The code above first shifts the output value to the left and adds the bit in the smallest position of the input to the output and then shifts the input to right. and repeats it until all bits finished. Here are some samples:
uint a = 0b_1100;
uint b = Reverse(a, 4); //should be 0b_0011;
And
uint a = 0b_100;
uint b = Reverse(a, 3); //should be 0b_001;
This implementation's time complexity is O(N) which N is the length of the input.
Edit in Dotnet fiddle
Here's a small look-up table solution that's good for (2<=N<=32).
For N==8, I think everyone agrees that a 256 byte array lookup table is the way to go. Similarly, for N from 2 to 7, you could create 4, 8, ... 128 lookup byte arrays.
For N==16, you could flip each byte and then reorder the two bytes. Similarly, for N==24, you could flip each byte and then reorder things (which would leave the middle one flipped but in the same position). It should be obvious how N==32 would work.
For N==9, think of it as three 3-bit numbers (flip each of them, reorder them and then do some masking and shifting to get them in the right position). For N==10, it's two 5-bit numbers. For N==11, it's two 5-bit numbers on either side of a center bit that doesn't change. The same for N==13 (two 6-bit numbers around an unchanging center bit). For a prime like N==23, it would be a pair of 8- bit numbers around a center 7-bit number.
For the odd numbers between 24 and 32 it gets more complicated. You probably need to consider five separate numbers. Consider N==29, that could be four 7-bit numbers around an unchanging center bit. For N==31, it would be a center bit surround by a pair of 8-bit numbers and a pair of 7-bit numbers.
That said, that's a ton of complicated logic. It would be a bear to test. It might be faster than #MuhammadVakili's bit shifting solution (it certainly would be for N<=8), but it might not. I suggest you go with his solution.
Using string manipulation?
static void Main(string[] args)
{
uint number = 269;
int numBits = 4;
string strBinary = Convert.ToString(number, 2).PadLeft(32, '0');
Console.WriteLine($"{number}");
Console.WriteLine($"{strBinary}");
string strBitsReversed = new string(strBinary.Substring(strBinary.Length - numBits, numBits).ToCharArray().Reverse().ToArray());
string strBinaryModified = strBinary.Substring(0, strBinary.Length - numBits) + strBitsReversed;
uint numberModified = Convert.ToUInt32(strBinaryModified, 2);
Console.WriteLine($"{strBinaryModified}");
Console.WriteLine($"{numberModified}");
Console.Write("Press Enter to Quit.");
Console.ReadLine();
}
Output:
269
00000000000000000000000100001101
00000000000000000000000100001011
267
Press Enter to Quit.
I am using a Microsoft example for interprocess communication. In the example there are two methods for reading/writing a string to/from a stream. The code sends the length of the string being streamed in the data. I need similar code, but need to make some modifications. An explanation of the highlighted lines would be helpful.
In WriteString(), they taken the length of the byte array being written and divide it by 256. The opposite is done in ReadString(), but an explanation as to why 256 is used would be great. Then, it writes another byte by taking the length and & it with 255. I don't understand the reasoning for this either. I'm thinking it shifts the value, but I don't really understand why this is needed. And then in ReadString() it does a += to the length by reading a byte. An explanation of this would be really helpful. I'm new to streaming and just want to understand exactly what is happening and why.
public string ReadString()
{
int len = 0;
// the next two lines
len = ioStream.ReadByte() * 256;
len += ioStream.ReadByte();
byte[] inBuffer = new byte[len];
ioStream.Read(inBuffer, 0, len);
return streamEncoding.GetString(inBuffer);
}
public int WriteString(string outString)
{
byte[] outBuffer = streamEncoding.GetBytes(outString);
int len = outBuffer.Length;
if (len > UInt16.MaxValue)
{
len = (int)UInt16.MaxValue;
}
// the next to lines
ioStream.WriteByte((byte)(len / 256));
ioStream.WriteByte((byte)(len & 255));
ioStream.Write(outBuffer, 0, len);
ioStream.Flush();
return outBuffer.Length + 2;
}
This code is bad, find a new tutorial.
The 256 stuff is there to convert the low 2 bytes of the length integer to bytes in order to serialize/deserialize them. This is not how it's normally done. Use BinaryReader/Writer or code not based on multiplication but on binary and and shift.
Dividing by 256 is equivalent to x >> 8. Also this only works with positive integers. x & 255 is used to take the lowest byte. This could simply be (byte)x. Sometimes people write x % 256 for this which not idiomatic and has problems with signedness.
Good code would be new byte[] { (byte)(x >> 8), (byte)(x >> 0) } and x = bytes[1] << 8 | bytes[0]. Much simpler, faster and idiomatic. I like writing >> 0 which does nothing for the sake of symmetry. It's optimized away. This might seem ridiculous for 16 bit ints but with longer ints there are 4 or 8 components and having one of them slightly off seems like needless inconsistency.
ioStream.Read(inBuffer, 0, len);
is a bug because it assumes the read completes in one chunk. Need a loop or again BinaryReader.
if (len > UInt16.MaxValue)
{
len = (int)UInt16.MaxValue;
}
I'm going to use this opportunity to warn anyone who reads this: Microsoft .NET sample code often is of extremely poor quality. Read it with great skepticism.
I am trying to find a way to remove a bit from an integer. The solution must not use string operations.
For example, I have the number 27, which is 11011 in binary.
I want to remove the third bit so it leaves me with 1011.
Or we have 182 (10110110), remove the 6th bit so the result is 1110110 (which is 118). I am trying to think of the algorithm how to do that, but so far no luck, and I can't find useful information on the internet.
I know how to use bitwise operators and how to extract or manipulate bits in integers (change values, exchange values etc), but I don't know how to 'remove' a certain bit.
I am not looking for code, just the logic of the operation. If anyone could help me, that would be awesome!
Regards,
Toni
No problem, just decompose the number into the "upper part" and the "lower part", and put them together without the middle bit that now disappeared.
Not tested:
uint upper = x & 0xFFFFFFF0;
uint lower = x & 7;
return (upper >> 1) | lower;
More generally: (also not tested)
uint upper = x & (0xFFFFFFFE << n);
uint lower = x & ((1u << n) - 1);
return (upper >> 1) | lower;
In order to do this you need two bit masks and a shift.
The first bit mask gives you the portion of the number above bit n, exclusive of the n-th bit. The mask is constructed as follows:
var top = ~((1U<<(n+1))-1); // 1111 1111 1000 000, 0xFF80
The second bit mask gives you the portion of the number below bit n, exclusive of the n-th bit:
var bottom = (1U<<n)-1; // 0000 0000 0011 1111, 0x003F
Comments above show the values for your second example (i.e. n == 6)
With the two masks in hand, you can construct the result as follows:
var res = ((original & top)>>1) | (original & bottom);
Demo.
You could use the following approach:
int value = 27;
string binary = Convert.ToString(value, 2);
binary = binary.Remove(binary.Length-3-1,1); //Remove the exact bit, 3rd in this case
int newValue = Convert.ToInt32(binary, 2);
Console.WriteLine(newValue);
Hope it helps!
int Place = 7;
int TheInt = 182;
string binary = Convert.ToString(TheInt, 2);
MessageBox.Show(binary.Remove(binary.Length - Place, 1));
Here is a version that needs slightly fewer operations than the solution by harold:
x ^ (((x >> 1) ^ x) & (0xffffffff << n));
The idea is that below n, bits are xored with zero, leaving them unchanged, while from n and above the two x xored cancel each other out, leaving x >> 1.
int a = 27;//int= 4byte equal to 32 bit
string binary = "";
for (int i = 0; i < 32; i++)
{
if ((a&1)==0)//if a's least significant bit is 0 ,add 0 to str
{
binary = "0" + binary;
}
else//if a's least significant bit is 1 ,add 1 to str
{
binary = "1" + binary;
}
a = a >> 1;//shift the bits left to right and delete lsb
//we are doing it for 32 times because integer have 32 bit.
}
Console.WriteLine("Integer to Binary= "+binary);
//Now you can operate the string(binary) however you want.
binary = binary.Remove(binary.Length-4,1);//remove 4st bit from str
I'm trying to debug some bit shifting operations and I need to visualize the bits as they exist before and after a Bit-Shifting operation.
I read from this answer that I may need to handle backfill from the shifting, but I'm not sure what that means.
I think that by asking this question (how do I print the bits in a int) I can figure out what the backfill is, and perhaps some other questions I have.
Here is my sample code so far.
static string GetBits(int num)
{
StringBuilder sb = new StringBuilder();
uint bits = (uint)num;
while (bits!=0)
{
bits >>= 1;
isBitSet = // somehow do an | operation on the first bit.
// I'm unsure if it's possible to handle different data types here
// or if unsafe code and a PTR is needed
if (isBitSet)
sb.Append("1");
else
sb.Append("0");
}
}
Convert.ToString(56,2).PadLeft(8,'0') returns "00111000"
This is for a byte, works for int also, just increase the numbers
To test if the last bit is set you could use:
isBitSet = ((bits & 1) == 1);
But you should do so before shifting right (not after), otherwise you's missing the first bit:
isBitSet = ((bits & 1) == 1);
bits = bits >> 1;
But a better option would be to use the static methods of the BitConverter class to get the actual bytes used to represent the number in memory into a byte array. The advantage (or disadvantage depending on your needs) of this method is that this reflects the endianness of the machine running the code.
byte[] bytes = BitConverter.GetBytes(num);
int bitPos = 0;
while(bitPos < 8 * bytes.Length)
{
int byteIndex = bitPos / 8;
int offset = bitPos % 8;
bool isSet = (bytes[byteIndex] & (1 << offset)) != 0;
// isSet = [True] if the bit at bitPos is set, false otherwise
bitPos++;
}