 # Topic: Difficulty (Read 9901 times)

sr. member    Activity: 440
Merit: 250 October 27, 2010, 05:22:48 AM
nullvoid offline?
hero member     Activity: 574
Merit: 507  August 04, 2010, 09:45:46 AM
Thanks ArtForz! ^_^

Code:
import struct

def SetCompact(nCompact):
"""convert bc compact uint to number"""
nSize = (nCompact >> 24) & 0xFF
tbuf = "\x00\x00\x00" + chr(nSize)
if nSize >= 1:
tbuf += chr((nCompact >> 16) & 0xFF)
if nSize >= 2:
tbuf += chr((nCompact >> 8) & 0xFF)
if nSize >= 3:
tbuf += chr((nCompact >> 0) & 0xFF)
tbuf += "\x00" * (nSize - 3)
return mpi2num(tbuf)

Code:
def uint256_from_compact(c):
nbytes = (c >> 24) & 0xFF
v = (c & 0xFFFFFFL) << (8 * (nbytes - 3))
return v

I confirmed again that both of these produce same output when using 0x1d00ffff
sr. member    Activity: 406
Merit: 253 August 04, 2010, 09:21:22 AM
this is my quick and dirty reference implementation for python, tested against bitcoins implementation for -256...256 and with about 4000 random numbers varying in size from 8-256 bits.
Writing a faster and more pythonic version should be pretty easy, but I can't be arsed right now...
Code:
import struct

def mpi2num(m):
"""convert MPI string to number"""
datasize = struct.unpack(">I", m[0:4])
r = 0
if datasize:
neg_flag = bool(ord(m) & 0x80)
r = ord(m) & 0x7F
for i in xrange(1, datasize):
r <<= 8
r += ord(m[4+i])
if neg_flag:
r = -r
return r

def num2mpi(n):
"""convert number to MPI string"""
if n == 0:
return struct.pack(">I", 0)
r = ""
neg_flag = bool(n < 0)
n = abs(n)
while n:
r = chr(n & 0xFF) + r
n >>= 8
if ord(r) & 0x80:
r = chr(0) + r
if neg_flag:
r = chr(ord(r) | 0x80) + r[1:]
datasize = len(r)
return struct.pack(">I", datasize) + r

def GetCompact(n):
"""convert number to bc compact uint"""
mpi = num2mpi(n)
nSize = len(mpi) - 4
nCompact = (nSize & 0xFF) << 24
if nSize >= 1:
nCompact |= (ord(mpi) << 16)
if nSize >= 2:
nCompact |= (ord(mpi) << 8)
if nSize >= 3:
nCompact |= (ord(mpi) << 0)
return nCompact

def SetCompact(nCompact):
"""convert bc compact uint to number"""
nSize = (nCompact >> 24) & 0xFF
tbuf = "\x00\x00\x00" + chr(nSize)
if nSize >= 1:
tbuf += chr((nCompact >> 16) & 0xFF)
if nSize >= 2:
tbuf += chr((nCompact >> 8) & 0xFF)
if nSize >= 3:
tbuf += chr((nCompact >> 0) & 0xFF)
tbuf += "\x00" * (nSize - 3)
return mpi2num(tbuf)
hero member     Activity: 574
Merit: 507  August 04, 2010, 04:06:57 AM
If C++ SetCompact(unsigned int nCompact) function is
Code:
CBigNum& SetCompact(unsigned int nCompact) {
unsigned int nSize = nCompact >> 24;
std::vector vch(4 + nSize);
vch = nSize;
if (nSize >= 1) vch = (nCompact >> 16) & 0xff;
if (nSize >= 2) vch = (nCompact >> 8) & 0xff;
if (nSize >= 3) vch = (nCompact >> 0) & 0xff;
BN_mpi2bn(&vch, vch.size(), this);
return *this;
}

With Python equivalent of
Code:
def uint256_from_compact(c):
nbytes = (c >> 24) & 0xFF
v = (c & 0xFFFFFFL) << (8 * (nbytes - 3))
return v

And PHP equvalent of
Code:
function uint256_from_compact(\$c) {
\$nbytes = (\$c >> 24) & 0xFF;
return bcmul(\$c & 0xFFFFFF,bcpow(2,8 * (\$nbytes - 3)));
}

then for C++ Getcompact() function
Code:
unsigned int GetCompact() const {
unsigned int nSize = BN_bn2mpi(this, NULL);
std::vector vch(nSize);
nSize -= 4;
BN_bn2mpi(this, &vch);
unsigned int nCompact = nSize << 24;
if (nSize >= 1) nCompact |= (vch << 16);
if (nSize >= 2) nCompact |= (vch << 8);
if (nSize >= 3) nCompact |= (vch << 0);
return nCompact;
}

how can I produce simplistic code for python/php to produce equivalent function?

My intention is to use the function in this php snippet
Code:
function GetNextWorkRequired(\$block, \$bits, \$nActualTimespan) {
\$nTargetTimespan = 60 * 60 * 24 * 14; // 2 weeks
if (\$nActualTimespan < \$nTargetTimespan / 4) { \$nActualTimespan = \$nTargetTimespan / 4; }
if (\$nActualTimespan > \$nTargetTimespan * 4) { \$nActualTimespan = \$nTargetTimespan * 4; }
\$bits = uint256_from_compact(\$bits);
\$bits = bcmul(\$bits,\$nActualTimespan);
\$bits = bcdiv(\$bits,\$nTargetTimespan);
\$bits = uint256_to_compact(\$bits); // <-- Need to translate C++ code for GetCompact()
return \$bits;
}
founder Activity: 364
Merit: 5240 July 28, 2010, 08:16:23 PM
You were looking at the wrong code.  Here's the code that applies:

Code:
bool CBlock::CheckBlock() const
{
...
// Check timestamp
if (nTime > GetAdjustedTime() + 2 * 60 * 60)
return error("CheckBlock() : block timestamp too far in the future");
...

bool CBlock::AcceptBlock()
{
...
// Check timestamp against prev
if (nTime <= pindexPrev->GetMedianTimePast())
return error("AcceptBlock() : block's timestamp is too early");

The timestamp is limited to up to 2 hours in the future.  It can be earlier than the previous block, but it must be greater than the median of the last 11 blocks.  The reason for doing it that way is so the time can get corrected in the next block if the previous block had the time too far in the future, like what happened.

hero member     Activity: 574
Merit: 507  July 27, 2010, 08:44:28 PM
It appears that block 70718 was found two hours in the future and followed with block 70719 being found about two hours in the past (from the future date of previous block).

Code:
-7017 seconds to find block 70719
8153 seconds to find block 70718
49 seconds to find block 70717
524 seconds to find block 70716

Code:
block timestamp
70715 1280275621
70716 1280276145
70717 1280276194
70718 1280284347
70719 1280277330

This doesn't seem to have any effect on things so much other than providing a kind of inaccurate depiction of statistical results every now and then (when a block is claimed to be from the future or the past), however what would happen if either the first or second (whichever one triggers new difficulty being established) of a new set of 2016 blocks is generated  to have a timestamp 2+ weeks in the past or the future?  If in the past, I presume according to the code in main.cpp: GetNextWorkRequired function that the lowest accepted interval is 3.5 days.  And if in the future I presume according to the code the lowest accepted interval is 56 days.
hero member     Activity: 574
Merit: 507  July 27, 2010, 07:31:31 PM
Actually, calculating with another digit comes out to 244.213223092375323881335701184896212407930361873708113773326734402748260502489 1554339282075169582811726687822061777635447902413461818393548574922122914786456 6654051064077556842993973623686279076537890477160907158869253835628402573582928 178404029229380767999068 and thus could be rounded up to the value you suggested Insti.
sr. member    Activity: 294
Merit: 252 Firstbits: 1duzy July 27, 2010, 10:45:03 AM
I'm just thankful the difficulty isn't 244.213223092375323881335701184896212407930361873708113773326734402748260502489 1554339282075169582811726687822061777635447902413461818393548574922122914786456 6654051064077556842993973623686279076537890477160907158869253835628402573582928 17840402922938076799907

Just imagine how hard it would be to win a block then!!!!!
hero member     Activity: 574
Merit: 507  July 27, 2010, 09:52:04 AM
Hehe, thanks. I could provide more accurate precision, but I think that is good enough.
full member   Activity: 183
Merit: 100  July 27, 2010, 09:32:42 AM
nice precision! latest difficulty = 244.213223092375323881335701184896212407930361873708113773326734402748260502489 1554339282075169582811726687822061777635447902413461818393548574922122914786456 6654051064077556842993973623686279076537890477160907158869253835628402573582928 17840402922938076799906

hero member     Activity: 574
Merit: 507  July 27, 2010, 05:11:51 AM
http://nullvoid.org/bitcoin/difficultiez.php

Code:
require_once 'jsonRPCClient.php';
\$data=new jsonRPCClient('http://127.0.0.1:8332');
\$blockcount = \$data->getblockcount();
\$now = date("U");
\$blockfile = "blockdata";
\$data = file(\$blockfile, FILE_IGNORE_NEW_LINES | FILE_SKIP_EMPTY_LINES); array_pop(\$data);
?>

`	function humantime(\$secs) {		if (\$secs<0) return false;		\$m = (int)(\$secs / 60); \$s = \$secs % 60; \$s = (\$s <= 9) ? "0\$s" : \$s;		\$h = (int)(\$m / 60); \$m = \$m % 60; \$m = (\$m <= 9) ? "0\$m" : \$m;		\$d = (int)(\$h / 24); \$h = \$h % 24; \$h = (\$h <= 9) ? "0\$h" : \$h;		\$d = (\$d <= 9) ? "0\$d" : \$d;		return \$d."d \$h:\$m:\$s";	}	// Converted from ArtForz's python code http://bitcointalk.org/index.php?topic=464.msg4080#msg4080	function uint256_from_compact(\$c) {		\$nbytes = (\$c >> 24) & 0xFF;		return bcmul(\$c & 0xFFFFFF,bcpow(2,8 * (\$nbytes - 3)));	}		\$tblock = 0;	\$nTargetTimespan = 60 * 60 * 24 * 14;	bcscale(256);	foreach (\$data as \$line) {		\$blocks = strtok(\$line, " ");		\$date = strtok(" ");		\$avghash = strtok(" ");		\$bits = strtok(" ");		if (\$blocks == 0 || \$blocks == \$tblock) {			\$tblock = \$blocks + 2016;			\$blocknum = str_repeat(" ", 6 - strlen(\$blocks)).\$blocks;			echo "Block \$blocknum was generated at \$date (".date("r", \$date).")";			if (\$blocks != 0) {				\$intervalnum = \$date - \$lastdate;				\$interval = str_repeat(" ", 9 - strlen(\$intervalnum)).\$intervalnum;				echo "    \$interval seconds interval (".humantime(\$intervalnum).")";				echo "   Difficulty: ".bcdiv(bcdiv(bcmul(uint256_from_compact(0x1D00FFFF),1000), uint256_from_compact(\$bits)), 1000);				\$lastdate = \$date;			}			echo "";		}	}?>  `