Kevacoin source tree
You can not select more than 25 topics Topics must start with a letter or number, can include dashes ('-') and can be up to 35 characters long.

526 lines
20 KiB

#!/usr/bin/env python3
# Copyright (c) 2014-2016 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."""
from base64 import b64encode
from binascii import hexlify, unhexlify
from decimal import Decimal, ROUND_DOWN
import json
import logging
import os
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 re
import time
from . import coverage
from .authproxy import AuthServiceProxy, JSONRPCException
logger = logging.getLogger("TestFramework.utils")
# Assert functions
##################
def assert_fee_amount(fee, tx_size, fee_per_kB):
"""Assert the fee was in range"""
target_fee = tx_size * fee_per_kB / 1000
if fee < target_fee:
raise AssertionError("Fee of %s BTC too low! (Should be %s BTC)" % (str(fee), str(target_fee)))
# allow the wallet's estimation to be at most 2 bytes off
if fee > (tx_size + 2) * fee_per_kB / 1000:
raise AssertionError("Fee of %s BTC too high! (Should be %s BTC)" % (str(fee), str(target_fee)))
def assert_equal(thing1, thing2, *args):
if thing1 != thing2 or any(thing1 != arg for arg in args):
raise AssertionError("not(%s)" % " == ".join(str(arg) for arg in (thing1, thing2) + args))
def assert_greater_than(thing1, thing2):
if thing1 <= thing2:
raise AssertionError("%s <= %s" % (str(thing1), str(thing2)))
def assert_greater_than_or_equal(thing1, thing2):
if thing1 < thing2:
raise AssertionError("%s < %s" % (str(thing1), str(thing2)))
def assert_raises(exc, fun, *args, **kwds):
assert_raises_message(exc, None, fun, *args, **kwds)
def assert_raises_message(exc, message, fun, *args, **kwds):
try:
fun(*args, **kwds)
except exc as e:
if message is not None and message not in e.error['message']:
raise AssertionError("Expected substring not found:" + e.error['message'])
except Exception as e:
raise AssertionError("Unexpected exception raised: " + type(e).__name__)
else:
raise AssertionError("No exception raised")
def assert_raises_jsonrpc(code, message, fun, *args, **kwds):
"""Run an RPC and verify that a specific JSONRPC exception code and message is raised.
Calls function `fun` with arguments `args` and `kwds`. Catches a JSONRPCException
and verifies that the error code and message are as expected. Throws AssertionError if
no JSONRPCException was raised or if the error code/message are not as expected.
Args:
code (int), optional: the error code returned by the RPC call (defined
in src/rpc/protocol.h). Set to None if checking the error code is not required.
message (string), optional: [a substring of] the error string returned by the
RPC call. Set to None if checking the error string is not required.
fun (function): the function to call. This should be the name of an RPC.
args*: positional arguments for the function.
kwds**: named arguments for the function.
"""
try:
fun(*args, **kwds)
except JSONRPCException as e:
# JSONRPCException was thrown as expected. Check the code and message values are correct.
if (code is not None) and (code != e.error["code"]):
raise AssertionError("Unexpected JSONRPC error code %i" % e.error["code"])
if (message is not None) and (message not in e.error['message']):
raise AssertionError("Expected substring not found:" + e.error['message'])
except Exception as e:
raise AssertionError("Unexpected exception raised: " + type(e).__name__)
else:
raise AssertionError("No exception raised")
def assert_is_hex_string(string):
try:
int(string, 16)
except Exception as e:
raise AssertionError(
"Couldn't interpret %r as hexadecimal; raised: %s" % (string, e))
def assert_is_hash_string(string, length=64):
if not isinstance(string, str):
raise AssertionError("Expected a string, got type %r" % type(string))
elif length and len(string) != length:
raise AssertionError(
"String of length %d expected; got %d" % (length, len(string)))
elif not re.match('[abcdef0-9]+$', string):
raise AssertionError(
"String %r contains invalid characters for a hash." % string)
def assert_array_result(object_array, to_match, expected, should_not_find=False):
"""
Pass in array of JSON objects, a dictionary with key/value pairs
to match against, and another dictionary with expected key/value
pairs.
If the should_not_find flag is true, to_match should not be found
in object_array
"""
if should_not_find:
assert_equal(expected, {})
num_matched = 0
for item in object_array:
all_match = True
for key, value in to_match.items():
if item[key] != value:
all_match = False
if not all_match:
continue
elif should_not_find:
num_matched = num_matched + 1
for key, value in expected.items():
if item[key] != value:
raise AssertionError("%s : expected %s=%s" % (str(item), str(key), str(value)))
num_matched = num_matched + 1
if num_matched == 0 and not should_not_find:
raise AssertionError("No objects matched %s" % (str(to_match)))
if num_matched > 0 and should_not_find:
raise AssertionError("Objects were found %s" % (str(to_match)))
# Utility functions
###################
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 bytes_to_hex_str(byte_str):
return hexlify(byte_str).decode('ascii')
def hex_str_to_bytes(hex_str):
return unhexlify(hex_str.encode('ascii'))
def str_to_b64str(string):
return b64encode(string.encode('utf-8')).decode('ascii')
def satoshi_round(amount):
return Decimal(amount).quantize(Decimal('0.00000001'), rounding=ROUND_DOWN)
def wait_until(predicate, *, attempts=float('inf'), timeout=float('inf'), lock=None):
if attempts == float('inf') and timeout == float('inf'):
timeout = 60
attempt = 0
timeout += time.time()
while attempt < attempts and time.time() < timeout:
if lock:
with lock:
if predicate():
return
else:
if predicate():
return
attempt += 1
time.sleep(0.05)
# Print the cause of the timeout
assert_greater_than(attempts, attempt)
assert_greater_than(timeout, time.time())
raise RuntimeError('Unreachable')
# RPC/P2P connection constants and functions
############################################
# The maximum number of nodes a single test can spawn
MAX_NODES = 8
# Don't assign rpc or p2p ports lower than this
PORT_MIN = 11000
# The number of ports to "reserve" for p2p and rpc, each
PORT_RANGE = 5000
class PortSeed:
# Must be initialized with a unique integer for each process
n = None
def get_rpc_proxy(url, node_number, timeout=None, coveragedir=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(
coveragedir, node_number) if coveragedir else None
return coverage.AuthServiceProxyWrapper(proxy, coverage_logfile)
def p2p_port(n):
assert(n <= MAX_NODES)
return PORT_MIN + n + (MAX_NODES * PortSeed.n) % (PORT_RANGE - 1 - MAX_NODES)
def rpc_port(n):
return PORT_MIN + PORT_RANGE + n + (MAX_NODES * PortSeed.n) % (PORT_RANGE - 1 - MAX_NODES)
def rpc_url(datadir, i, rpchost=None):
rpc_u, rpc_p = get_auth_cookie(datadir)
host = '127.0.0.1'
port = rpc_port(i)
if rpchost:
parts = rpchost.split(':')
if len(parts) == 2:
host, port = parts
else:
host = rpchost
return "http://%s:%s@%s:%d" % (rpc_u, rpc_p, host, int(port))
# Node functions
################
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', encoding='utf8') as f:
f.write("regtest=1\n")
f.write("port=" + str(p2p_port(n)) + "\n")
f.write("rpcport=" + str(rpc_port(n)) + "\n")
f.write("listenonion=0\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 get_datadir_path(dirname, n):
return os.path.join(dirname, "node" + str(n))
def get_auth_cookie(datadir):
user = None
password = None
if os.path.isfile(os.path.join(datadir, "bitcoin.conf")):
with open(os.path.join(datadir, "bitcoin.conf"), 'r', encoding='utf8') as f:
for line in f:
if line.startswith("rpcuser="):
assert user is None # Ensure that there is only one rpcuser line
user = line.split("=")[1].strip("\n")
if line.startswith("rpcpassword="):
assert password is None # Ensure that there is only one rpcpassword line
password = line.split("=")[1].strip("\n")
if os.path.isfile(os.path.join(datadir, "regtest", ".cookie")):
with open(os.path.join(datadir, "regtest", ".cookie"), 'r') as f:
userpass = f.read()
split_userpass = userpass.split(':')
user = split_userpass[0]
password = split_userpass[1]
if user is None or password is None:
raise ValueError("No RPC credentials")
return user, password
def log_filename(dirname, n_node, logname):
return os.path.join(dirname, "node" + str(n_node), "regtest", logname)
def get_bip9_status(node, key):
info = node.getblockchaininfo()
return info['bip9_softforks'][key]
def set_node_times(nodes, t):
for node in nodes:
node.setmocktime(t)
def disconnect_nodes(from_connection, node_num):
for peer_id in [peer['id'] for peer in from_connection.getpeerinfo() if "testnode%d" % node_num in peer['subver']]:
from_connection.disconnectnode(nodeid=peer_id)
for _ in range(50):
if [peer['id'] for peer in from_connection.getpeerinfo() if "testnode%d" % node_num in peer['subver']] == []:
break
time.sleep(0.1)
else:
raise AssertionError("timed out waiting for disconnect")
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)
def sync_blocks(rpc_connections, *, wait=1, timeout=60):
"""
Wait until everybody has the same tip.
sync_blocks needs to be called with an rpc_connections set that has least
one node already synced to the latest, stable tip, otherwise there's a
chance it might return before all nodes are stably synced.
"""
# Use getblockcount() instead of waitforblockheight() to determine the
# initial max height because the two RPCs look at different internal global
# variables (chainActive vs latestBlock) and the former gets updated
# earlier.
maxheight = max(x.getblockcount() for x in rpc_connections)
start_time = cur_time = time.time()
while cur_time <= start_time + timeout:
tips = [r.waitforblockheight(maxheight, int(wait * 1000)) for r in rpc_connections]
if all(t["height"] == maxheight for t in tips):
if all(t["hash"] == tips[0]["hash"] for t in tips):
return
raise AssertionError("Block sync failed, mismatched block hashes:{}".format(
"".join("\n {!r}".format(tip) for tip in tips)))
cur_time = time.time()
raise AssertionError("Block sync to height {} timed out:{}".format(
maxheight, "".join("\n {!r}".format(tip) for tip in tips)))
def sync_chain(rpc_connections, *, wait=1, timeout=60):
"""
Wait until everybody has the same best block
"""
while timeout > 0:
best_hash = [x.getbestblockhash() for x in rpc_connections]
if best_hash == [best_hash[0]] * len(best_hash):
return
time.sleep(wait)
timeout -= wait
raise AssertionError("Chain sync failed: Best block hashes don't match")
def sync_mempools(rpc_connections, *, wait=1, timeout=60):
"""
Wait until everybody has the same transactions in their memory
pools
"""
while timeout > 0:
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):
return
time.sleep(wait)
timeout -= wait
raise AssertionError("Mempool sync failed")
# Transaction/Block functions
#############################
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"]})
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 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
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
change = amount_in - amount
if change > amount * 2:
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 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 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)
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
(total_in, inputs) = gather_inputs(from_node, amount + fee)
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
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)
# Helper to create at least "count" utxos
# Pass in a fee that is sufficient for relay and mining new transactions.
def create_confirmed_utxos(fee, node, count):
to_generate = int(0.5 * count) + 101
while to_generate > 0:
node.generate(min(25, to_generate))
to_generate -= 25
utxos = node.listunspent()
iterations = count - len(utxos)
addr1 = node.getnewaddress()
addr2 = node.getnewaddress()
if iterations <= 0:
return utxos
for i in range(iterations):
t = utxos.pop()
inputs = []
inputs.append({"txid": t["txid"], "vout": t["vout"]})
outputs = {}
send_value = t['amount'] - fee
outputs[addr1] = satoshi_round(send_value / 2)
outputs[addr2] = satoshi_round(send_value / 2)
raw_tx = node.createrawtransaction(inputs, outputs)
signed_tx = node.signrawtransaction(raw_tx)["hex"]
node.sendrawtransaction(signed_tx)
while (node.getmempoolinfo()['size'] > 0):
node.generate(1)
utxos = node.listunspent()
assert(len(utxos) >= count)
return utxos
# Create large OP_RETURN txouts that can be appended to a transaction
# to make it large (helper for constructing large transactions).
def gen_return_txouts():
# Some pre-processing to create a bunch of OP_RETURN txouts to insert into transactions we create
# So we have big transactions (and therefore can't fit very many into each block)
# create one script_pubkey
script_pubkey = "6a4d0200" # OP_RETURN OP_PUSH2 512 bytes
for i in range(512):
script_pubkey = script_pubkey + "01"
# concatenate 128 txouts of above script_pubkey which we'll insert before the txout for change
txouts = "81"
for k in range(128):
# add txout value
txouts = txouts + "0000000000000000"
# add length of script_pubkey
txouts = txouts + "fd0402"
# add script_pubkey
txouts = txouts + script_pubkey
return txouts
def create_tx(node, coinbase, to_address, amount):
inputs = [{"txid": coinbase, "vout": 0}]
outputs = {to_address: amount}
rawtx = node.createrawtransaction(inputs, outputs)
signresult = node.signrawtransaction(rawtx)
assert_equal(signresult["complete"], True)
return signresult["hex"]
# Create a spend of each passed-in utxo, splicing in "txouts" to each raw
# transaction to make it large. See gen_return_txouts() above.
def create_lots_of_big_transactions(node, txouts, utxos, num, fee):
addr = node.getnewaddress()
txids = []
for _ in range(num):
t = utxos.pop()
inputs = [{"txid": t["txid"], "vout": t["vout"]}]
outputs = {}
change = t['amount'] - fee
outputs[addr] = satoshi_round(change)
rawtx = node.createrawtransaction(inputs, outputs)
newtx = rawtx[0:92]
newtx = newtx + txouts
newtx = newtx + rawtx[94:]
signresult = node.signrawtransaction(newtx, None, None, "NONE")
txid = node.sendrawtransaction(signresult["hex"], True)
txids.append(txid)
return txids
def mine_large_block(node, utxos=None):
# generate a 66k transaction,
# and 14 of them is close to the 1MB block limit
num = 14
txouts = gen_return_txouts()
utxos = utxos if utxos is not None else []
if len(utxos) < num:
utxos.clear()
utxos.extend(node.listunspent())
fee = 100 * node.getnetworkinfo()["relayfee"]
create_lots_of_big_transactions(node, txouts, utxos, num, fee=fee)
node.generate(1)