I am working with bitcoin source code want to set initial difficulty to 1 (I changed bdiff,nBits field). So I need to change pdiff as well. according to :
difficulty = difficulty_1_target / current_target (target is a 256
bit number)
difficulty_1_target can be different for various ways to measure
difficulty. Traditionally, it represents a hash where the leading 32
bits are zero and the rest are one (this is known as "pool difficulty"
or "pdiff"). The Bitcoin protocol represents targets as a custom
floating point type with limited precision; as a result, Bitcoin
clients often approximate difficulty based on this (this is known as
"bdiff").
Anyone knows where pdiff is stored ? Is it hard coded ?
I found the solution! It's not exactly a pdiff field in the code but there is a function in blockchain.cpp :
double GetDifficulty(const CBlockIndex* blockindex)
{
if (blockindex == nullptr)
{
return 1.0;
}
int nShift = (blockindex->nBits >> 24) & 0xff;
double dDiff =
(double)0x0000ffff / (double)(blockindex->nBits & 0x00ffffff);
while (nShift < 29)
{
dDiff *= 256.0;
nShift++;
}
while (nShift > 29)
{
dDiff /= 256.0;
nShift--;
}
return dDiff;
}
for bitcoin initial nBits is equal to 0x1d00ffff so dDiff field above becomes 1 and nshift is equal to 1D. For my private version I set nBits to 0x1f0fffff and should calculate dDiff like
double dDiff =(double)0x000ffff / (double)(blockindex->nBits & 0x00ffffff);
and nShift field for me is 0x1f so I changed while conditions to while(nShift < 31) andwhile (nShift > 31). by running command bitcoin-cli getdifficulty I got 1 as initial difficulty.
Related
I have some questions regarding the flash memory with a dspic33ep512mu810.
I'm aware of how it should be done:
set all the register for address, latches, etc. Then do the sequence to start the write procedure or call the builtins function.
But I find that there is some small difference between what I'm experiencing and what is in the DOC.
when writing the flash in WORD mode. In the DOC it is pretty straightforward. Following is the example code in the DOC
int varWord1L = 0xXXXX;
int varWord1H = 0x00XX;
int varWord2L = 0xXXXX;
int varWord2H = 0x00XX;
int TargetWriteAddressL; // bits<15:0>
int TargetWriteAddressH; // bits<22:16>
NVMCON = 0x4001; // Set WREN and word program mode
TBLPAG = 0xFA; // write latch upper address
NVMADR = TargetWriteAddressL; // set target write address
NVMADRU = TargetWriteAddressH;
__builtin_tblwtl(0,varWord1L); // load write latches
__builtin_tblwth(0,varWord1H);
__builtin_tblwtl(0x2,varWord2L);
__builtin_tblwth(0x2,varWord2H);
__builtin_disi(5); // Disable interrupts for NVM unlock sequence
__builtin_write_NVM(); // initiate write
while(NVMCONbits.WR == 1);
But that code doesn't work depending on the address where I want to write. I found a fix to write one WORD but I can't write 2 WORD where I want. I store everything in the aux memory so the upper address(NVMADRU) is always 0x7F for me. The NVMADR is the address I can change. What I'm seeing is that if the address where I want to write modulo 4 is not 0 then I have to put my value in the 2 last latches, otherwise I have to put the value in the first latches.
If address modulo 4 is not zero, it doesn't work like the doc code(above). The value that will be at the address will be what is in the second set of latches.
I fixed it for writing only one word at a time like this:
if(Address % 4)
{
__builtin_tblwtl(0, 0xFFFF);
__builtin_tblwth(0, 0x00FF);
__builtin_tblwtl(2, ValueL);
__builtin_tblwth(2, ValueH);
}
else
{
__builtin_tblwtl(0, ValueL);
__builtin_tblwth(0, ValueH);
__builtin_tblwtl(2, 0xFFFF);
__builtin_tblwth(2, 0x00FF);
}
I want to know why I'm seeing this behavior?
2)I also want to write a full row.
That also doesn't seem to work for me and I don't know why because I'm doing what is in the DOC.
I tried a simple write row code and at the end I just read back the first 3 or 4 element that I wrote to see if it works:
NVMCON = 0x4002; //set for row programming
TBLPAG = 0x00FA; //set address for the write latches
NVMADRU = 0x007F; //upper address of the aux memory
NVMADR = 0xE7FA;
int latchoffset;
latchoffset = 0;
__builtin_tblwtl(latchoffset, 0);
__builtin_tblwth(latchoffset, 0); //current = 0, available = 1
latchoffset+=2;
__builtin_tblwtl(latchoffset, 1);
__builtin_tblwth(latchoffset, 1); //current = 0, available = 1
latchoffset+=2;
.
. all the way to 127(I know I could have done it in a loop)
.
__builtin_tblwtl(latchoffset, 127);
__builtin_tblwth(latchoffset, 127);
INTCON2bits.GIE = 0; //stop interrupt
__builtin_write_NVM();
while(NVMCONbits.WR == 1);
INTCON2bits.GIE = 1; //start interrupt
int testaddress;
testaddress = 0xE7FA;
status = NVMemReadIntH(testaddress);
status = NVMemReadIntL(testaddress);
testaddress += 2;
status = NVMemReadIntH(testaddress);
status = NVMemReadIntL(testaddress);
testaddress += 2;
status = NVMemReadIntH(testaddress);
status = NVMemReadIntL(testaddress);
testaddress += 2;
status = NVMemReadIntH(testaddress);
status = NVMemReadIntL(testaddress);
What I see is that the value that is stored in the address 0xE7FA is 125, in 0xE7FC is 126 and in 0xE7FE is 127. And the rest are all 0xFFFF.
Why is it taking only the last 3 latches and write them in the first 3 address?
Thanks in advance for your help people.
The dsPIC33 program memory space is treated as 24 bits wide, it is
more appropriate to think of each address of the program memory as a
lower and upper word, with the upper byte of the upper word being
unimplemented
(dsPIC33EPXXX datasheet)
There is a phantom byte every two program words.
Your code
if(Address % 4)
{
__builtin_tblwtl(0, 0xFFFF);
__builtin_tblwth(0, 0x00FF);
__builtin_tblwtl(2, ValueL);
__builtin_tblwth(2, ValueH);
}
else
{
__builtin_tblwtl(0, ValueL);
__builtin_tblwth(0, ValueH);
__builtin_tblwtl(2, 0xFFFF);
__builtin_tblwth(2, 0x00FF);
}
...will be fine for writing a bootloader if generating values from a valid Intel HEX file, but doesn't make it simple for storing data structures because the phantom byte is not taken into account.
If you create a uint32_t variable and look at the compiled HEX file, you'll notice that it in fact uses up the least significant words of two 24-bit program words. I.e. the 32-bit value is placed into a 64-bit range but only 48-bits out of the 64-bits are programmable, the others are phantom bytes (or zeros). Leaving three bytes per address modulo of 4 that are actually programmable.
What I tend to do if writing data is to keep everything 32-bit aligned and do the same as the compiler does.
Writing:
UINT32 value = ....;
:
__builtin_tblwtl(0, value.word.word_L); // least significant word of 32-bit value placed here
__builtin_tblwth(0, 0x00); // phantom byte + unused byte
__builtin_tblwtl(2, value.word.word_H); // most significant word of 32-bit value placed here
__builtin_tblwth(2, 0x00); // phantom byte + unused byte
Reading:
UINT32 *value
:
value->word.word_L = __builtin_tblrdl(offset);
value->word.word_H = __builtin_tblrdl(offset+2);
UINT32 structure:
typedef union _UINT32 {
uint32_t val32;
struct {
uint16_t word_L;
uint16_t word_H;
} word;
uint8_t bytes[4];
} UINT32;
I have a Data object in swift that is an array of Int16 objects. For some reason using ".pcmFormatInt16" did not work for the format of my AVAudioPCMBuffer and yielded no sound, or a memory error. Eventually, I was able to get white noise/static to play from the speakers by converting the Int16 to a float and putting that onto both channels of my AVAudioPCMBuffer. I have a feeling that I am getting close to the answer, because whenever I speak into the microphone I hear a different frequency of static. I think the issue is that I am not converting the converted Int16 into the buffer floatChannelData.
Here is my code:
for ch in 0..<2 {
for i in 0..<audio.count {
var val = Float( Int16(audio[i]) ) / Float(Int16.max)
if( val > 1 ){
val = 1;
}
if( val < -1 ){
val = -1;
}
self.buffer.floatChannelData![ch][i+self.bufferCount] = val
self.bufferCount+=1
}
}
self.audioFilePlayer.scheduleBuffer(self.buffer, at:nil, options: .interruptsAtLoop, completionHandler: {
print("played sum")
self.bufferCount=0
})
a typical multi-channel PCM buffer has the channels interleaving on a per sample basis although, not being familiar with swift audio, I find it refreshing to see here channels given a dimension on the buffer datastructure
... a flag goes up when I see your guard checks clamping val > 1 set to val = 1 etc. ... elsewhere that is not needed as those boundary checks are moot as the data nicely falls into place as is
... my guess is your input audio[] is signed int 16 because of your val > 1 and val < -1 ? if true then dividing by max int float is wrong as you would be loosing half your dynamic range ...
I suggest you look closely at your
var val = Float( Int16(audio[i]) ) / Float(Int16.max)
lets examine range of your ints in audio[]
2^16 == 65536 // if unsigned then values range from 0 to (2^16 - 1) which is 0 to 65535
2^15 == 32768 // if signed then values would range from -32768 to (2^15 - 1) which is -32768 to 32767
Please tell is whether input buffer audio[] is signed or not ... sometimes its helpful to identify the max_seen and min_seen values of your input data ... do this and tell us the value of max and min of your input audio[]
Now lets focus on your desired output buffer self.buffer.floatChannelData ... since you are saying its 16 bit float ... what is the valid range here ? -1 < valid_value < 1 ?
We can continue once you tell us answers to these basic questions
My program uses alot of memory and Processing power, I can only search up to 6000, is there any way to reduce the amount of memory this uses? This will really help with future programming endevours as it will be nice to know how to work with memory smartly.
ArrayList<Integer> factor = new ArrayList<Integer>();
ArrayList<Integer> non = new ArrayList<Integer>();
ArrayList<Integer> prime = new ArrayList<Integer>();
Scanner sc = new Scanner(System.in);
System.out.println("Please enter how high we want to search");
long startTime = System.nanoTime();
int max = sc.nextInt();
int number = 2;
while (number < max)
{
for (int i=0;i<prime.size();i++)
{
int value = prime.get(i);
if (number % value == 0)
{
factor.add(value);
}
else
{
non.add(value);
}
}
if(factor.isEmpty())
{
prime.add(number);
}
else
{
composite.add(number);
}
factor.clear();
number++;
}
int howMany=prime.size();
System.out.printf("The are "+howMany+" prime numbers up to " +max + " and they are: " +prime );
System.out.println();
}
You do not say what language you are using, so this answer will be general.
To store primes up to 6,000 you only need about 3,000 bits which is less than 380 bytes. Your basic solution is the Sieve of Eratosthenes and the fact that 2 is the only even prime. You set up the sieve to handle only odd numbers, which halves the storage needed. Since the sieve only holds prime or not prime for each odd number, the storage can be reduced to a single bit for each number.
Once you have set up your sieve, there are many sites including this one which have instructions in different languages, you just need to retrieve the prime/not prime value from the sieve for the numbers in your range. Here is the pseudocode for checking if a number is prime, assuming the sieve has already been set up:
boolean function isPrime(number)
// Low numbers
if (number < 2)
return false
endif
// Even numbers
if (number is even)
return number == 2
endif
// Odd numbers >= 3
return sieve[(number - 1) / 2] == 1
end function
Low numbers are not prime. 2 is the only even prime; all other even numbers are not prime. The prime flag for the odd number 2n+1 is stored at bit n in the sieve. This assumes that the language you are using allows bit level access, something like a BitSet in Java.
When yo generate a barcode with Barcode4j as an image, you could obtain the human readable text, too, for instance:
EAN13 barcode example
In this picture we can see that the human readable text is: 1000000012026
In this example the barcode has been generated with the code 100000001202 and the number 6 is the check digit added by Barcode4j generator.
So, my question is: Is possible obtain the check digit of an EAN13 generated barcode with Barcode4j? Because I know how to render this as a image, but I don't know how to obtain the human readable text, as a plain text.
Regards,
Miguel.
Thanks to Barcode4j plugin, you can calculate the checksum with the barcode format you need. In Java 7, you can calculate the checkSum as this way:
private String calculateCodeWithcheckSum(String codigo){
EAN13Bean generator = new EAN13Bean();
UPCEANLogicImpl impl = generator.createLogicImpl();
codigo += impl.calcChecksum(codigo);
return codigo;
}
First, you need the EAN13 barcode format, so you can get the class that the plugin provides you, and call its only method: createLogicImpl().
This method is used to give you a class of type UPCEANLogicImpl.
This is the class you need, because you can find in it the method to calculate the checkSum. So, you only have to call the method calcChecksum giving your code (100000001202), and will give you the checkSum value (6).
You can check it in the next web: http://www.gs1.org/check-digit-calculator
Adding your code and the checkSum value will give you the value you need (1000000012026)
In case you what to compute the check digit without importing a java library here's the code:
public static String ean13CheckDigit(String barcode) {
int s = 0;
for (int i = 0; i < 12; i++) {
int c = Character.getNumericValue(barcode.charAt(i));
s += c * ( i%2 == 0? 1: 3);
}
s = (10 - s % 10) % 10;
barcode += s;
return barcode;
}
This question already has answers here:
What is a simple example of floating point/rounding error?
(9 answers)
Why is Lua arithmetic is not equal to itself? [duplicate]
(1 answer)
Closed 9 years ago.
Today morning I found a bug on my Lua Script, wich seem very weird. How can this evaluation fail this way? Examples can be tested in here
First example:
if( math.abs(29.7 - 30) <= 0.3 ) then
result = 1
else
result = 0
end
print("result = "..result )
-->> result = 0
Second example:
if( 0.3 <= 0.3 ) then
result = 1
else
result = 0
end
print("result = "..result )
-->> result = 1
Third example
if( math.abs(29.7-30) == 0.3 )then
print("Lua says: "..math.abs(29.7-30).." == 0.3")
else
print("Lua says: "..math.abs(29.7-30).." ~= 0.3")
end
-->> Lua says: 0.3 ~= 0.3 WHAT?
I am really confuse, and I would like to understand this to avoid similiar bugs in the future. Thanks
You are being hit by the fact that Lua uses (IEEE 754) 64-bit double-precision floating point numbers.
Look at the following examples
> print(0.3 == 0.3)
true
> print(0.3 <= 0.3)
true
> print(0.3 >= 0.3)
true
The actual value of 0.3 in memory is:
> print(string.format("%1.64f",math.abs(-0.3)))
0.2999999999999999888977697537484345957636833190917968750000000000
Now look at you example:
> print(math.abs(29.7-30) == 0.3)
false
> print(math.abs(29.7-30) >= 0.3)
true
> print(math.abs(29.7-30) <= 0.3)
false
The actual value of 29.7-30 is:
> print(string.format("%1.64f",29.7-30))
-0.3000000000000007105427357601001858711242675781250000000000000000
The actual value of math.abs(29.7-30) is:
> print(string.format("%1.64f", math.abs(29.7-30))
0.3000000000000007105427357601001858711242675781250000000000000000
And just for fun the value of math.abs(-0.3) is:
> print(string.format("%1.64f", math.abs(-0.3)))
0.2999999999999999888977697537484345957636833190917968750000000000
There are two solutions to you problem, the first is read What Every Computer Scientist Should Know About Floating-Point Arithmetic, and understand it :-). The second solution is to configure Lua to use another type for numbers, see Values and Types for hints.
Edit
I just thought of another way of "solving" the problem, but it is a bit of a hack, and not guarantied to always work. You can use fixed point numbers in lua by first converting the float to a string with a fixed precision.
In your case that would look something like:
a = string.format("%1.1f", math.abs(29.7 - 30))
print(a == "0.3")
or a bit more robust:
a = string.format("%1.1f", math.abs(29.7 - 30))
print(a == string.format("%1.1f", 0.3))
However you must make sure that you use a precision that is both adequate and the same for all you comparisons.
As we know, float point has a precision problem
Refer: http://lua-users.org/wiki/FloatingPoint
a = 1
if a < 1 then print("<1") end
Will never print "<1". Not unless you actually change a