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# Copyright (c) 2014 The Bitcoin Core developers
# Distributed under the MIT software license, see the accompanying
# file COPYING or http://www.opensource.org/licenses/mit-license.php.
#
# Helpful routines for regression testing
#
# Add python-bitcoinrpc to module search path:
import os
import sys
from decimal import Decimal, ROUND_DOWN
import json
estimatefee / estimatepriority RPC methods New RPC methods: return an estimate of the fee (or priority) a transaction needs to be likely to confirm in a given number of blocks. Mike Hearn created the first version of this method for estimating fees. It works as follows: For transactions that took 1 to N (I picked N=25) blocks to confirm, keep N buckets with at most 100 entries in each recording the fees-per-kilobyte paid by those transactions. (separate buckets are kept for transactions that confirmed because they are high-priority) The buckets are filled as blocks are found, and are saved/restored in a new fee_estiamtes.dat file in the data directory. A few variations on Mike's initial scheme: To estimate the fee needed for a transaction to confirm in X buckets, all of the samples in all of the buckets are used and a median of all of the data is used to make the estimate. For example, imagine 25 buckets each containing the full 100 entries. Those 2,500 samples are sorted, and the estimate of the fee needed to confirm in the very next block is the 50'th-highest-fee-entry in that sorted list; the estimate of the fee needed to confirm in the next two blocks is the 150'th-highest-fee-entry, etc. That algorithm has the nice property that estimates of how much fee you need to pay to get confirmed in block N will always be greater than or equal to the estimate for block N+1. It would clearly be wrong to say "pay 11 uBTC and you'll get confirmed in 3 blocks, but pay 12 uBTC and it will take LONGER". A single block will not contribute more than 10 entries to any one bucket, so a single miner and a large block cannot overwhelm the estimates.
11 years ago
import random
import shutil
import subprocess
import time
import re
from . import coverage
from .authproxy import AuthServiceProxy, JSONRPCException
COVERAGE_DIR = None
def enable_coverage(dirname):
"""Maintain a log of which RPC calls are made during testing."""
global COVERAGE_DIR
COVERAGE_DIR = dirname
def get_rpc_proxy(url, node_number, timeout=None):
"""
Args:
url (str): URL of the RPC server to call
node_number (int): the node number (or id) that this calls to
Kwargs:
timeout (int): HTTP timeout in seconds
Returns:
AuthServiceProxy. convenience object for making RPC calls.
"""
proxy_kwargs = {}
if timeout is not None:
proxy_kwargs['timeout'] = timeout
proxy = AuthServiceProxy(url, **proxy_kwargs)
proxy.url = url # store URL on proxy for info
coverage_logfile = coverage.get_filename(
COVERAGE_DIR, node_number) if COVERAGE_DIR else None
return coverage.AuthServiceProxyWrapper(proxy, coverage_logfile)
def p2p_port(n):
return 11000 + n + os.getpid()%999
def rpc_port(n):
return 12000 + n + os.getpid()%999
def check_json_precision():
"""Make sure json library being used does not lose precision converting BTC values"""
n = Decimal("20000000.00000003")
satoshis = int(json.loads(json.dumps(float(n)))*1.0e8)
if satoshis != 2000000000000003:
raise RuntimeError("JSON encode/decode loses precision")
def count_bytes(hex_string):
return len(bytearray.fromhex(hex_string))
def sync_blocks(rpc_connections, wait=1):
"""
Wait until everybody has the same block count
"""
while True:
counts = [ x.getblockcount() for x in rpc_connections ]
if counts == [ counts[0] ]*len(counts):
break
time.sleep(wait)
def sync_mempools(rpc_connections, wait=1):
"""
Wait until everybody has the same transactions in their memory
pools
"""
while True:
pool = set(rpc_connections[0].getrawmempool())
num_match = 1
for i in range(1, len(rpc_connections)):
if set(rpc_connections[i].getrawmempool()) == pool:
num_match = num_match+1
if num_match == len(rpc_connections):
break
time.sleep(wait)
bitcoind_processes = {}
def initialize_datadir(dirname, n):
datadir = os.path.join(dirname, "node"+str(n))
if not os.path.isdir(datadir):
os.makedirs(datadir)
with open(os.path.join(datadir, "bitcoin.conf"), 'w') as f:
f.write("regtest=1\n");
f.write("rpcuser=rt\n");
f.write("rpcpassword=rt\n");
f.write("port="+str(p2p_port(n))+"\n");
f.write("rpcport="+str(rpc_port(n))+"\n");
estimatefee / estimatepriority RPC methods New RPC methods: return an estimate of the fee (or priority) a transaction needs to be likely to confirm in a given number of blocks. Mike Hearn created the first version of this method for estimating fees. It works as follows: For transactions that took 1 to N (I picked N=25) blocks to confirm, keep N buckets with at most 100 entries in each recording the fees-per-kilobyte paid by those transactions. (separate buckets are kept for transactions that confirmed because they are high-priority) The buckets are filled as blocks are found, and are saved/restored in a new fee_estiamtes.dat file in the data directory. A few variations on Mike's initial scheme: To estimate the fee needed for a transaction to confirm in X buckets, all of the samples in all of the buckets are used and a median of all of the data is used to make the estimate. For example, imagine 25 buckets each containing the full 100 entries. Those 2,500 samples are sorted, and the estimate of the fee needed to confirm in the very next block is the 50'th-highest-fee-entry in that sorted list; the estimate of the fee needed to confirm in the next two blocks is the 150'th-highest-fee-entry, etc. That algorithm has the nice property that estimates of how much fee you need to pay to get confirmed in block N will always be greater than or equal to the estimate for block N+1. It would clearly be wrong to say "pay 11 uBTC and you'll get confirmed in 3 blocks, but pay 12 uBTC and it will take LONGER". A single block will not contribute more than 10 entries to any one bucket, so a single miner and a large block cannot overwhelm the estimates.
11 years ago
return datadir
def initialize_chain(test_dir):
"""
Create (or copy from cache) a 200-block-long chain and
4 wallets.
bitcoind and bitcoin-cli must be in search path.
"""
if (not os.path.isdir(os.path.join("cache","node0"))
or not os.path.isdir(os.path.join("cache","node1"))
or not os.path.isdir(os.path.join("cache","node2"))
or not os.path.isdir(os.path.join("cache","node3"))):
#find and delete old cache directories if any exist
for i in range(4):
if os.path.isdir(os.path.join("cache","node"+str(i))):
shutil.rmtree(os.path.join("cache","node"+str(i)))
devnull = open(os.devnull, "w")
# Create cache directories, run bitcoinds:
for i in range(4):
estimatefee / estimatepriority RPC methods New RPC methods: return an estimate of the fee (or priority) a transaction needs to be likely to confirm in a given number of blocks. Mike Hearn created the first version of this method for estimating fees. It works as follows: For transactions that took 1 to N (I picked N=25) blocks to confirm, keep N buckets with at most 100 entries in each recording the fees-per-kilobyte paid by those transactions. (separate buckets are kept for transactions that confirmed because they are high-priority) The buckets are filled as blocks are found, and are saved/restored in a new fee_estiamtes.dat file in the data directory. A few variations on Mike's initial scheme: To estimate the fee needed for a transaction to confirm in X buckets, all of the samples in all of the buckets are used and a median of all of the data is used to make the estimate. For example, imagine 25 buckets each containing the full 100 entries. Those 2,500 samples are sorted, and the estimate of the fee needed to confirm in the very next block is the 50'th-highest-fee-entry in that sorted list; the estimate of the fee needed to confirm in the next two blocks is the 150'th-highest-fee-entry, etc. That algorithm has the nice property that estimates of how much fee you need to pay to get confirmed in block N will always be greater than or equal to the estimate for block N+1. It would clearly be wrong to say "pay 11 uBTC and you'll get confirmed in 3 blocks, but pay 12 uBTC and it will take LONGER". A single block will not contribute more than 10 entries to any one bucket, so a single miner and a large block cannot overwhelm the estimates.
11 years ago
datadir=initialize_datadir("cache", i)
args = [ os.getenv("BITCOIND", "bitcoind"), "-keypool=1", "-datadir="+datadir, "-discover=0" ]
if i > 0:
args.append("-connect=127.0.0.1:"+str(p2p_port(0)))
bitcoind_processes[i] = subprocess.Popen(args)
if os.getenv("PYTHON_DEBUG", ""):
print "initialize_chain: bitcoind started, calling bitcoin-cli -rpcwait getblockcount"
subprocess.check_call([ os.getenv("BITCOINCLI", "bitcoin-cli"), "-datadir="+datadir,
"-rpcwait", "getblockcount"], stdout=devnull)
if os.getenv("PYTHON_DEBUG", ""):
print "initialize_chain: bitcoin-cli -rpcwait getblockcount completed"
devnull.close()
rpcs = []
for i in range(4):
try:
url = "http://rt:rt@127.0.0.1:%d" % (rpc_port(i),)
rpcs.append(get_rpc_proxy(url, i))
except:
sys.stderr.write("Error connecting to "+url+"\n")
sys.exit(1)
# Create a 200-block-long chain; each of the 4 nodes
# gets 25 mature blocks and 25 immature.
# blocks are created with timestamps 10 minutes apart, starting
# at 1 Jan 2014
block_time = 1388534400
for i in range(2):
for peer in range(4):
for j in range(25):
set_node_times(rpcs, block_time)
rpcs[peer].generate(1)
block_time += 10*60
# Must sync before next peer starts generating blocks
sync_blocks(rpcs)
# Shut them down, and clean up cache directories:
stop_nodes(rpcs)
wait_bitcoinds()
for i in range(4):
os.remove(log_filename("cache", i, "debug.log"))
os.remove(log_filename("cache", i, "db.log"))
os.remove(log_filename("cache", i, "peers.dat"))
os.remove(log_filename("cache", i, "fee_estimates.dat"))
for i in range(4):
from_dir = os.path.join("cache", "node"+str(i))
to_dir = os.path.join(test_dir, "node"+str(i))
shutil.copytree(from_dir, to_dir)
initialize_datadir(test_dir, i) # Overwrite port/rpcport in bitcoin.conf
def initialize_chain_clean(test_dir, num_nodes):
"""
Create an empty blockchain and num_nodes wallets.
Useful if a test case wants complete control over initialization.
"""
for i in range(num_nodes):
datadir=initialize_datadir(test_dir, i)
def _rpchost_to_args(rpchost):
'''Convert optional IP:port spec to rpcconnect/rpcport args'''
if rpchost is None:
return []
match = re.match('(\[[0-9a-fA-f:]+\]|[^:]+)(?::([0-9]+))?$', rpchost)
if not match:
raise ValueError('Invalid RPC host spec ' + rpchost)
rpcconnect = match.group(1)
rpcport = match.group(2)
if rpcconnect.startswith('['): # remove IPv6 [...] wrapping
rpcconnect = rpcconnect[1:-1]
rv = ['-rpcconnect=' + rpcconnect]
if rpcport:
rv += ['-rpcport=' + rpcport]
return rv
def start_node(i, dirname, extra_args=None, rpchost=None, timewait=None, binary=None):
estimatefee / estimatepriority RPC methods New RPC methods: return an estimate of the fee (or priority) a transaction needs to be likely to confirm in a given number of blocks. Mike Hearn created the first version of this method for estimating fees. It works as follows: For transactions that took 1 to N (I picked N=25) blocks to confirm, keep N buckets with at most 100 entries in each recording the fees-per-kilobyte paid by those transactions. (separate buckets are kept for transactions that confirmed because they are high-priority) The buckets are filled as blocks are found, and are saved/restored in a new fee_estiamtes.dat file in the data directory. A few variations on Mike's initial scheme: To estimate the fee needed for a transaction to confirm in X buckets, all of the samples in all of the buckets are used and a median of all of the data is used to make the estimate. For example, imagine 25 buckets each containing the full 100 entries. Those 2,500 samples are sorted, and the estimate of the fee needed to confirm in the very next block is the 50'th-highest-fee-entry in that sorted list; the estimate of the fee needed to confirm in the next two blocks is the 150'th-highest-fee-entry, etc. That algorithm has the nice property that estimates of how much fee you need to pay to get confirmed in block N will always be greater than or equal to the estimate for block N+1. It would clearly be wrong to say "pay 11 uBTC and you'll get confirmed in 3 blocks, but pay 12 uBTC and it will take LONGER". A single block will not contribute more than 10 entries to any one bucket, so a single miner and a large block cannot overwhelm the estimates.
11 years ago
"""
Start a bitcoind and return RPC connection to it
"""
datadir = os.path.join(dirname, "node"+str(i))
if binary is None:
binary = os.getenv("BITCOIND", "bitcoind")
args = [ binary, "-datadir="+datadir, "-keypool=1", "-discover=0", "-rest" ]
estimatefee / estimatepriority RPC methods New RPC methods: return an estimate of the fee (or priority) a transaction needs to be likely to confirm in a given number of blocks. Mike Hearn created the first version of this method for estimating fees. It works as follows: For transactions that took 1 to N (I picked N=25) blocks to confirm, keep N buckets with at most 100 entries in each recording the fees-per-kilobyte paid by those transactions. (separate buckets are kept for transactions that confirmed because they are high-priority) The buckets are filled as blocks are found, and are saved/restored in a new fee_estiamtes.dat file in the data directory. A few variations on Mike's initial scheme: To estimate the fee needed for a transaction to confirm in X buckets, all of the samples in all of the buckets are used and a median of all of the data is used to make the estimate. For example, imagine 25 buckets each containing the full 100 entries. Those 2,500 samples are sorted, and the estimate of the fee needed to confirm in the very next block is the 50'th-highest-fee-entry in that sorted list; the estimate of the fee needed to confirm in the next two blocks is the 150'th-highest-fee-entry, etc. That algorithm has the nice property that estimates of how much fee you need to pay to get confirmed in block N will always be greater than or equal to the estimate for block N+1. It would clearly be wrong to say "pay 11 uBTC and you'll get confirmed in 3 blocks, but pay 12 uBTC and it will take LONGER". A single block will not contribute more than 10 entries to any one bucket, so a single miner and a large block cannot overwhelm the estimates.
11 years ago
if extra_args is not None: args.extend(extra_args)
bitcoind_processes[i] = subprocess.Popen(args)
devnull = open(os.devnull, "w")
if os.getenv("PYTHON_DEBUG", ""):
print "start_node: bitcoind started, calling bitcoin-cli -rpcwait getblockcount"
subprocess.check_call([ os.getenv("BITCOINCLI", "bitcoin-cli"), "-datadir="+datadir] +
estimatefee / estimatepriority RPC methods New RPC methods: return an estimate of the fee (or priority) a transaction needs to be likely to confirm in a given number of blocks. Mike Hearn created the first version of this method for estimating fees. It works as follows: For transactions that took 1 to N (I picked N=25) blocks to confirm, keep N buckets with at most 100 entries in each recording the fees-per-kilobyte paid by those transactions. (separate buckets are kept for transactions that confirmed because they are high-priority) The buckets are filled as blocks are found, and are saved/restored in a new fee_estiamtes.dat file in the data directory. A few variations on Mike's initial scheme: To estimate the fee needed for a transaction to confirm in X buckets, all of the samples in all of the buckets are used and a median of all of the data is used to make the estimate. For example, imagine 25 buckets each containing the full 100 entries. Those 2,500 samples are sorted, and the estimate of the fee needed to confirm in the very next block is the 50'th-highest-fee-entry in that sorted list; the estimate of the fee needed to confirm in the next two blocks is the 150'th-highest-fee-entry, etc. That algorithm has the nice property that estimates of how much fee you need to pay to get confirmed in block N will always be greater than or equal to the estimate for block N+1. It would clearly be wrong to say "pay 11 uBTC and you'll get confirmed in 3 blocks, but pay 12 uBTC and it will take LONGER". A single block will not contribute more than 10 entries to any one bucket, so a single miner and a large block cannot overwhelm the estimates.
11 years ago
_rpchost_to_args(rpchost) +
["-rpcwait", "getblockcount"], stdout=devnull)
if os.getenv("PYTHON_DEBUG", ""):
print "start_node: calling bitcoin-cli -rpcwait getblockcount returned"
devnull.close()
estimatefee / estimatepriority RPC methods New RPC methods: return an estimate of the fee (or priority) a transaction needs to be likely to confirm in a given number of blocks. Mike Hearn created the first version of this method for estimating fees. It works as follows: For transactions that took 1 to N (I picked N=25) blocks to confirm, keep N buckets with at most 100 entries in each recording the fees-per-kilobyte paid by those transactions. (separate buckets are kept for transactions that confirmed because they are high-priority) The buckets are filled as blocks are found, and are saved/restored in a new fee_estiamtes.dat file in the data directory. A few variations on Mike's initial scheme: To estimate the fee needed for a transaction to confirm in X buckets, all of the samples in all of the buckets are used and a median of all of the data is used to make the estimate. For example, imagine 25 buckets each containing the full 100 entries. Those 2,500 samples are sorted, and the estimate of the fee needed to confirm in the very next block is the 50'th-highest-fee-entry in that sorted list; the estimate of the fee needed to confirm in the next two blocks is the 150'th-highest-fee-entry, etc. That algorithm has the nice property that estimates of how much fee you need to pay to get confirmed in block N will always be greater than or equal to the estimate for block N+1. It would clearly be wrong to say "pay 11 uBTC and you'll get confirmed in 3 blocks, but pay 12 uBTC and it will take LONGER". A single block will not contribute more than 10 entries to any one bucket, so a single miner and a large block cannot overwhelm the estimates.
11 years ago
url = "http://rt:rt@%s:%d" % (rpchost or '127.0.0.1', rpc_port(i))
proxy = get_rpc_proxy(url, i, timeout=timewait)
if COVERAGE_DIR:
coverage.write_all_rpc_commands(COVERAGE_DIR, proxy)
return proxy
estimatefee / estimatepriority RPC methods New RPC methods: return an estimate of the fee (or priority) a transaction needs to be likely to confirm in a given number of blocks. Mike Hearn created the first version of this method for estimating fees. It works as follows: For transactions that took 1 to N (I picked N=25) blocks to confirm, keep N buckets with at most 100 entries in each recording the fees-per-kilobyte paid by those transactions. (separate buckets are kept for transactions that confirmed because they are high-priority) The buckets are filled as blocks are found, and are saved/restored in a new fee_estiamtes.dat file in the data directory. A few variations on Mike's initial scheme: To estimate the fee needed for a transaction to confirm in X buckets, all of the samples in all of the buckets are used and a median of all of the data is used to make the estimate. For example, imagine 25 buckets each containing the full 100 entries. Those 2,500 samples are sorted, and the estimate of the fee needed to confirm in the very next block is the 50'th-highest-fee-entry in that sorted list; the estimate of the fee needed to confirm in the next two blocks is the 150'th-highest-fee-entry, etc. That algorithm has the nice property that estimates of how much fee you need to pay to get confirmed in block N will always be greater than or equal to the estimate for block N+1. It would clearly be wrong to say "pay 11 uBTC and you'll get confirmed in 3 blocks, but pay 12 uBTC and it will take LONGER". A single block will not contribute more than 10 entries to any one bucket, so a single miner and a large block cannot overwhelm the estimates.
11 years ago
def start_nodes(num_nodes, dirname, extra_args=None, rpchost=None, binary=None):
estimatefee / estimatepriority RPC methods New RPC methods: return an estimate of the fee (or priority) a transaction needs to be likely to confirm in a given number of blocks. Mike Hearn created the first version of this method for estimating fees. It works as follows: For transactions that took 1 to N (I picked N=25) blocks to confirm, keep N buckets with at most 100 entries in each recording the fees-per-kilobyte paid by those transactions. (separate buckets are kept for transactions that confirmed because they are high-priority) The buckets are filled as blocks are found, and are saved/restored in a new fee_estiamtes.dat file in the data directory. A few variations on Mike's initial scheme: To estimate the fee needed for a transaction to confirm in X buckets, all of the samples in all of the buckets are used and a median of all of the data is used to make the estimate. For example, imagine 25 buckets each containing the full 100 entries. Those 2,500 samples are sorted, and the estimate of the fee needed to confirm in the very next block is the 50'th-highest-fee-entry in that sorted list; the estimate of the fee needed to confirm in the next two blocks is the 150'th-highest-fee-entry, etc. That algorithm has the nice property that estimates of how much fee you need to pay to get confirmed in block N will always be greater than or equal to the estimate for block N+1. It would clearly be wrong to say "pay 11 uBTC and you'll get confirmed in 3 blocks, but pay 12 uBTC and it will take LONGER". A single block will not contribute more than 10 entries to any one bucket, so a single miner and a large block cannot overwhelm the estimates.
11 years ago
"""
Start multiple bitcoinds, return RPC connections to them
"""
if extra_args is None: extra_args = [ None for i in range(num_nodes) ]
if binary is None: binary = [ None for i in range(num_nodes) ]
return [ start_node(i, dirname, extra_args[i], rpchost, binary=binary[i]) for i in range(num_nodes) ]
def log_filename(dirname, n_node, logname):
return os.path.join(dirname, "node"+str(n_node), "regtest", logname)
def stop_node(node, i):
node.stop()
bitcoind_processes[i].wait()
del bitcoind_processes[i]
def stop_nodes(nodes):
for node in nodes:
node.stop()
del nodes[:] # Emptying array closes connections as a side effect
def set_node_times(nodes, t):
for node in nodes:
node.setmocktime(t)
def wait_bitcoinds():
# Wait for all bitcoinds to cleanly exit
for bitcoind in bitcoind_processes.values():
bitcoind.wait()
bitcoind_processes.clear()
def connect_nodes(from_connection, node_num):
ip_port = "127.0.0.1:"+str(p2p_port(node_num))
from_connection.addnode(ip_port, "onetry")
# poll until version handshake complete to avoid race conditions
# with transaction relaying
while any(peer['version'] == 0 for peer in from_connection.getpeerinfo()):
time.sleep(0.1)
def connect_nodes_bi(nodes, a, b):
connect_nodes(nodes[a], b)
connect_nodes(nodes[b], a)
estimatefee / estimatepriority RPC methods New RPC methods: return an estimate of the fee (or priority) a transaction needs to be likely to confirm in a given number of blocks. Mike Hearn created the first version of this method for estimating fees. It works as follows: For transactions that took 1 to N (I picked N=25) blocks to confirm, keep N buckets with at most 100 entries in each recording the fees-per-kilobyte paid by those transactions. (separate buckets are kept for transactions that confirmed because they are high-priority) The buckets are filled as blocks are found, and are saved/restored in a new fee_estiamtes.dat file in the data directory. A few variations on Mike's initial scheme: To estimate the fee needed for a transaction to confirm in X buckets, all of the samples in all of the buckets are used and a median of all of the data is used to make the estimate. For example, imagine 25 buckets each containing the full 100 entries. Those 2,500 samples are sorted, and the estimate of the fee needed to confirm in the very next block is the 50'th-highest-fee-entry in that sorted list; the estimate of the fee needed to confirm in the next two blocks is the 150'th-highest-fee-entry, etc. That algorithm has the nice property that estimates of how much fee you need to pay to get confirmed in block N will always be greater than or equal to the estimate for block N+1. It would clearly be wrong to say "pay 11 uBTC and you'll get confirmed in 3 blocks, but pay 12 uBTC and it will take LONGER". A single block will not contribute more than 10 entries to any one bucket, so a single miner and a large block cannot overwhelm the estimates.
11 years ago
def find_output(node, txid, amount):
"""
Return index to output of txid with value amount
Raises exception if there is none.
"""
txdata = node.getrawtransaction(txid, 1)
for i in range(len(txdata["vout"])):
if txdata["vout"][i]["value"] == amount:
return i
raise RuntimeError("find_output txid %s : %s not found"%(txid,str(amount)))
def gather_inputs(from_node, amount_needed, confirmations_required=1):
estimatefee / estimatepriority RPC methods New RPC methods: return an estimate of the fee (or priority) a transaction needs to be likely to confirm in a given number of blocks. Mike Hearn created the first version of this method for estimating fees. It works as follows: For transactions that took 1 to N (I picked N=25) blocks to confirm, keep N buckets with at most 100 entries in each recording the fees-per-kilobyte paid by those transactions. (separate buckets are kept for transactions that confirmed because they are high-priority) The buckets are filled as blocks are found, and are saved/restored in a new fee_estiamtes.dat file in the data directory. A few variations on Mike's initial scheme: To estimate the fee needed for a transaction to confirm in X buckets, all of the samples in all of the buckets are used and a median of all of the data is used to make the estimate. For example, imagine 25 buckets each containing the full 100 entries. Those 2,500 samples are sorted, and the estimate of the fee needed to confirm in the very next block is the 50'th-highest-fee-entry in that sorted list; the estimate of the fee needed to confirm in the next two blocks is the 150'th-highest-fee-entry, etc. That algorithm has the nice property that estimates of how much fee you need to pay to get confirmed in block N will always be greater than or equal to the estimate for block N+1. It would clearly be wrong to say "pay 11 uBTC and you'll get confirmed in 3 blocks, but pay 12 uBTC and it will take LONGER". A single block will not contribute more than 10 entries to any one bucket, so a single miner and a large block cannot overwhelm the estimates.
11 years ago
"""
Return a random set of unspent txouts that are enough to pay amount_needed
"""
assert(confirmations_required >=0)
utxo = from_node.listunspent(confirmations_required)
estimatefee / estimatepriority RPC methods New RPC methods: return an estimate of the fee (or priority) a transaction needs to be likely to confirm in a given number of blocks. Mike Hearn created the first version of this method for estimating fees. It works as follows: For transactions that took 1 to N (I picked N=25) blocks to confirm, keep N buckets with at most 100 entries in each recording the fees-per-kilobyte paid by those transactions. (separate buckets are kept for transactions that confirmed because they are high-priority) The buckets are filled as blocks are found, and are saved/restored in a new fee_estiamtes.dat file in the data directory. A few variations on Mike's initial scheme: To estimate the fee needed for a transaction to confirm in X buckets, all of the samples in all of the buckets are used and a median of all of the data is used to make the estimate. For example, imagine 25 buckets each containing the full 100 entries. Those 2,500 samples are sorted, and the estimate of the fee needed to confirm in the very next block is the 50'th-highest-fee-entry in that sorted list; the estimate of the fee needed to confirm in the next two blocks is the 150'th-highest-fee-entry, etc. That algorithm has the nice property that estimates of how much fee you need to pay to get confirmed in block N will always be greater than or equal to the estimate for block N+1. It would clearly be wrong to say "pay 11 uBTC and you'll get confirmed in 3 blocks, but pay 12 uBTC and it will take LONGER". A single block will not contribute more than 10 entries to any one bucket, so a single miner and a large block cannot overwhelm the estimates.
11 years ago
random.shuffle(utxo)
inputs = []
total_in = Decimal("0.00000000")
while total_in < amount_needed and len(utxo) > 0:
t = utxo.pop()
total_in += t["amount"]
inputs.append({ "txid" : t["txid"], "vout" : t["vout"], "address" : t["address"] } )
if total_in < amount_needed:
raise RuntimeError("Insufficient funds: need %d, have %d"%(amount_needed, total_in))
estimatefee / estimatepriority RPC methods New RPC methods: return an estimate of the fee (or priority) a transaction needs to be likely to confirm in a given number of blocks. Mike Hearn created the first version of this method for estimating fees. It works as follows: For transactions that took 1 to N (I picked N=25) blocks to confirm, keep N buckets with at most 100 entries in each recording the fees-per-kilobyte paid by those transactions. (separate buckets are kept for transactions that confirmed because they are high-priority) The buckets are filled as blocks are found, and are saved/restored in a new fee_estiamtes.dat file in the data directory. A few variations on Mike's initial scheme: To estimate the fee needed for a transaction to confirm in X buckets, all of the samples in all of the buckets are used and a median of all of the data is used to make the estimate. For example, imagine 25 buckets each containing the full 100 entries. Those 2,500 samples are sorted, and the estimate of the fee needed to confirm in the very next block is the 50'th-highest-fee-entry in that sorted list; the estimate of the fee needed to confirm in the next two blocks is the 150'th-highest-fee-entry, etc. That algorithm has the nice property that estimates of how much fee you need to pay to get confirmed in block N will always be greater than or equal to the estimate for block N+1. It would clearly be wrong to say "pay 11 uBTC and you'll get confirmed in 3 blocks, but pay 12 uBTC and it will take LONGER". A single block will not contribute more than 10 entries to any one bucket, so a single miner and a large block cannot overwhelm the estimates.
11 years ago
return (total_in, inputs)
def make_change(from_node, amount_in, amount_out, fee):
"""
Create change output(s), return them
"""
outputs = {}
amount = amount_out+fee
change = amount_in - amount
if change > amount*2:
# Create an extra change output to break up big inputs
change_address = from_node.getnewaddress()
# Split change in two, being careful of rounding:
outputs[change_address] = Decimal(change/2).quantize(Decimal('0.00000001'), rounding=ROUND_DOWN)
change = amount_in - amount - outputs[change_address]
estimatefee / estimatepriority RPC methods New RPC methods: return an estimate of the fee (or priority) a transaction needs to be likely to confirm in a given number of blocks. Mike Hearn created the first version of this method for estimating fees. It works as follows: For transactions that took 1 to N (I picked N=25) blocks to confirm, keep N buckets with at most 100 entries in each recording the fees-per-kilobyte paid by those transactions. (separate buckets are kept for transactions that confirmed because they are high-priority) The buckets are filled as blocks are found, and are saved/restored in a new fee_estiamtes.dat file in the data directory. A few variations on Mike's initial scheme: To estimate the fee needed for a transaction to confirm in X buckets, all of the samples in all of the buckets are used and a median of all of the data is used to make the estimate. For example, imagine 25 buckets each containing the full 100 entries. Those 2,500 samples are sorted, and the estimate of the fee needed to confirm in the very next block is the 50'th-highest-fee-entry in that sorted list; the estimate of the fee needed to confirm in the next two blocks is the 150'th-highest-fee-entry, etc. That algorithm has the nice property that estimates of how much fee you need to pay to get confirmed in block N will always be greater than or equal to the estimate for block N+1. It would clearly be wrong to say "pay 11 uBTC and you'll get confirmed in 3 blocks, but pay 12 uBTC and it will take LONGER". A single block will not contribute more than 10 entries to any one bucket, so a single miner and a large block cannot overwhelm the estimates.
11 years ago
if change > 0:
outputs[from_node.getnewaddress()] = change
estimatefee / estimatepriority RPC methods New RPC methods: return an estimate of the fee (or priority) a transaction needs to be likely to confirm in a given number of blocks. Mike Hearn created the first version of this method for estimating fees. It works as follows: For transactions that took 1 to N (I picked N=25) blocks to confirm, keep N buckets with at most 100 entries in each recording the fees-per-kilobyte paid by those transactions. (separate buckets are kept for transactions that confirmed because they are high-priority) The buckets are filled as blocks are found, and are saved/restored in a new fee_estiamtes.dat file in the data directory. A few variations on Mike's initial scheme: To estimate the fee needed for a transaction to confirm in X buckets, all of the samples in all of the buckets are used and a median of all of the data is used to make the estimate. For example, imagine 25 buckets each containing the full 100 entries. Those 2,500 samples are sorted, and the estimate of the fee needed to confirm in the very next block is the 50'th-highest-fee-entry in that sorted list; the estimate of the fee needed to confirm in the next two blocks is the 150'th-highest-fee-entry, etc. That algorithm has the nice property that estimates of how much fee you need to pay to get confirmed in block N will always be greater than or equal to the estimate for block N+1. It would clearly be wrong to say "pay 11 uBTC and you'll get confirmed in 3 blocks, but pay 12 uBTC and it will take LONGER". A single block will not contribute more than 10 entries to any one bucket, so a single miner and a large block cannot overwhelm the estimates.
11 years ago
return outputs
def send_zeropri_transaction(from_node, to_node, amount, fee):
"""
Create&broadcast a zero-priority transaction.
Returns (txid, hex-encoded-txdata)
Ensures transaction is zero-priority by first creating a send-to-self,
then using its output
estimatefee / estimatepriority RPC methods New RPC methods: return an estimate of the fee (or priority) a transaction needs to be likely to confirm in a given number of blocks. Mike Hearn created the first version of this method for estimating fees. It works as follows: For transactions that took 1 to N (I picked N=25) blocks to confirm, keep N buckets with at most 100 entries in each recording the fees-per-kilobyte paid by those transactions. (separate buckets are kept for transactions that confirmed because they are high-priority) The buckets are filled as blocks are found, and are saved/restored in a new fee_estiamtes.dat file in the data directory. A few variations on Mike's initial scheme: To estimate the fee needed for a transaction to confirm in X buckets, all of the samples in all of the buckets are used and a median of all of the data is used to make the estimate. For example, imagine 25 buckets each containing the full 100 entries. Those 2,500 samples are sorted, and the estimate of the fee needed to confirm in the very next block is the 50'th-highest-fee-entry in that sorted list; the estimate of the fee needed to confirm in the next two blocks is the 150'th-highest-fee-entry, etc. That algorithm has the nice property that estimates of how much fee you need to pay to get confirmed in block N will always be greater than or equal to the estimate for block N+1. It would clearly be wrong to say "pay 11 uBTC and you'll get confirmed in 3 blocks, but pay 12 uBTC and it will take LONGER". A single block will not contribute more than 10 entries to any one bucket, so a single miner and a large block cannot overwhelm the estimates.
11 years ago
"""
# Create a send-to-self with confirmed inputs:
self_address = from_node.getnewaddress()
(total_in, inputs) = gather_inputs(from_node, amount+fee*2)
outputs = make_change(from_node, total_in, amount+fee, fee)
outputs[self_address] = float(amount+fee)
self_rawtx = from_node.createrawtransaction(inputs, outputs)
self_signresult = from_node.signrawtransaction(self_rawtx)
self_txid = from_node.sendrawtransaction(self_signresult["hex"], True)
vout = find_output(from_node, self_txid, amount+fee)
# Now immediately spend the output to create a 1-input, 1-output
# zero-priority transaction:
inputs = [ { "txid" : self_txid, "vout" : vout } ]
outputs = { to_node.getnewaddress() : float(amount) }
rawtx = from_node.createrawtransaction(inputs, outputs)
signresult = from_node.signrawtransaction(rawtx)
txid = from_node.sendrawtransaction(signresult["hex"], True)
return (txid, signresult["hex"])
def random_zeropri_transaction(nodes, amount, min_fee, fee_increment, fee_variants):
"""
Create a random zero-priority transaction.
Returns (txid, hex-encoded-transaction-data, fee)
"""
from_node = random.choice(nodes)
to_node = random.choice(nodes)
fee = min_fee + fee_increment*random.randint(0,fee_variants)
(txid, txhex) = send_zeropri_transaction(from_node, to_node, amount, fee)
return (txid, txhex, fee)
def random_transaction(nodes, amount, min_fee, fee_increment, fee_variants):
"""
Create a random transaction.
Returns (txid, hex-encoded-transaction-data, fee)
"""
from_node = random.choice(nodes)
to_node = random.choice(nodes)
fee = min_fee + fee_increment*random.randint(0,fee_variants)
(total_in, inputs) = gather_inputs(from_node, amount+fee)
outputs = make_change(from_node, total_in, amount, fee)
outputs[to_node.getnewaddress()] = float(amount)
rawtx = from_node.createrawtransaction(inputs, outputs)
signresult = from_node.signrawtransaction(rawtx)
txid = from_node.sendrawtransaction(signresult["hex"], True)
return (txid, signresult["hex"], fee)
def assert_equal(thing1, thing2):
if thing1 != thing2:
raise AssertionError("%s != %s"%(str(thing1),str(thing2)))
def assert_greater_than(thing1, thing2):
if thing1 <= thing2:
raise AssertionError("%s <= %s"%(str(thing1),str(thing2)))
def assert_raises(exc, fun, *args, **kwds):
try:
fun(*args, **kwds)
except exc:
pass
except Exception as e:
raise AssertionError("Unexpected exception raised: "+type(e).__name__)
else:
raise AssertionError("No exception raised")
def satoshi_round(amount):
return Decimal(amount).quantize(Decimal('0.00000001'), rounding=ROUND_DOWN)