文章目录

• 简单支付验证
• • 练习1
  • • 代码实现
    • 测试
  • 练习2
  • • 代码实现
    • 测试
  • 练习3
  • • 代码实现
    • 测试
  • 练习4
  • • 代码实现
    • 运行结果
  • 练习5
  • • 代码实现
    • 运行结果
  • 练习6
  • • 代码实现
    • 运行结果
  • 练习7
  • • 代码实现
    • 运行结果

《区块链编程》第十一章

简单支付验证

练习1

p180

代码实现

# -*- coding: utf-8 -*-
# @Author: 从化北(喵星人)
# @Date:   2022-01-16 10:43:44
# @Last Modified by:   从化北
# @Last Modified time: 2022-01-16 11:00:29
from helper import hash256


def merkle_parent(hash1, hash2):
    '''Takes the binary hashes and calculates the hash256'''
    return hash256(hash1 + hash2)


hash0 = bytes.fromhex('c117ea8ec828342f4dfb0ad6bd140e03a50720ece40169ee38b\
dc15d9eb64cf5')
hash1 = bytes.fromhex('c131474164b412e3406696da1ee20ab0fc9bf41c8f05fa8ceea\
7a08d672d7cc5')
print(merkle_parent(hash0, hash1).hex())

测试

8b30c5ba100f6f2e5ad1e2a742e5020491240f8eb514fe97c713c31718ad7ecd
[Finished in 318ms]

练习2

p181

代码实现

# -*- coding: utf-8 -*-
# @Author: 从化北(喵星人)
# @Date:   2022-01-16 10:53:22
# @Last Modified by:   从化北
# @Last Modified time: 2022-01-16 11:06:46

from helper import merkle_parent


def merkel_parent_level(hashes):
    '''Takes a list of binary hashes and returns a list that's half
    the length'''
    if len(hashes) == 1:
        raise RuntimeError('Cannot take a parent level with only 1 item')
    if len(hashes) % 2 == 1:
        hashes.append(hashes[-1])
    parent_level = []
    for i in range(0, len(hashes), 2):
        parent = merkle_parent(hashes[i], hashes[i + 1])
        parent_level.append(parent)
    return parent_level


hex_hashes = [
    'c117ea8ec828342f4dfb0ad6bd140e03a50720ece40169ee38bdc15d9eb64cf5',
    'c131474164b412e3406696da1ee20ab0fc9bf41c8f05fa8ceea7a08d672d7cc5',
    'f391da6ecfeed1814efae39e7fcb3838ae0b02c02ae7d0a5848a66947c0727b0',
    '3d238a92a94532b946c90e19c49351c763696cff3db400485b813aecb8a13181',
    '10092f2633be5f3ce349bf9ddbde36caa3dd10dfa0ec8106bce23acbff637dae',
]
hashes = [bytes.fromhex(h) for h in hex_hashes]
level = merkel_parent_level(hashes)
for h in level:
    print(h.hex())

测试

8b30c5ba100f6f2e5ad1e2a742e5020491240f8eb514fe97c713c31718ad7ecd
7f4e6f9e224e20fda0ae4c44114237f97cd35aca38d83081c9bfd41feb907800
3ecf6115380c77e8aae56660f5634982ee897351ba906a6837d15ebc3a225df0
[Finished in 323ms]

练习3

p182

代码实现

# -*- coding: utf-8 -*-
# @Author: 从化北(喵星人)
# @Date:   2022-01-16 11:07:52
# @Last Modified by:   从化北
# @Last Modified time: 2022-01-16 11:14:24
from helper import merkle_parent_level


def merkle_root(hashes):
    '''Takes a list of binary hashes and returns the merkle root
    '''
    # current level starts as hashes
    current_level = hashes
    # loop until there's exactly 1 element
    while len(current_level) > 1:
        # current level becomes the merkle parent level
        current_level = merkle_parent_level(current_level)
    # return the 1st item of the current level
    return current_level[0]


hex_hashes = [
    'c117ea8ec828342f4dfb0ad6bd140e03a50720ece40169ee38bdc15d9eb64cf5',
    'c131474164b412e3406696da1ee20ab0fc9bf41c8f05fa8ceea7a08d672d7cc5',
    'f391da6ecfeed1814efae39e7fcb3838ae0b02c02ae7d0a5848a66947c0727b0',
    '3d238a92a94532b946c90e19c49351c763696cff3db400485b813aecb8a13181',
    '10092f2633be5f3ce349bf9ddbde36caa3dd10dfa0ec8106bce23acbff637dae',
    '7d37b3d54fa6a64869084bfd2e831309118b9e833610e6228adacdbd1b4ba161',
    '8118a77e542892fe15ae3fc771a4abfd2f5d5d5997544c3487ac36b5c85170fc',
    'dff6879848c2c9b62fe652720b8df5272093acfaa45a43cdb3696fe2466a3877',
    'b825c0745f46ac58f7d3759e6dc535a1fec7820377f24d4c2c6ad2cc55c0cb59',
    '95513952a04bd8992721e9b7e2937f1c04ba31e0469fbe615a78197f68f52b7c',
    '2e6d722e5e4dbdf2447ddecc9f7dabb8e299bae921c99ad5b0184cd9eb8e5908',
    'b13a750047bc0bdceb2473e5fe488c2596d7a7124b4e716fdd29b046ef99bbf0',
]
hashes = [bytes.fromhex(x) for x in hex_hashes]
root = merkle_root(hashes)
print(root.hex())

        

测试

acbcab8bcc1af95d8d563b77d24c3d19b18f1486383d75a5085c4e86c86beed6
[Finished in 399ms]

练习4

p183

代码实现

# -*- coding: utf-8 -*-
# @Author: 从化北(喵星人)
# @Date:   2022-01-16 11:20:26
# @Last Modified by:   从化北
# @Last Modified time: 2022-01-16 11:20:26
class Block:
...
    def validate_merkle_root(self):
        '''Gets the merkle root of the tx_hashes and checks that it's
        the same as the merkle root of this block.
        '''
        # reverse each item in self.tx_hashes
        hashes = [h[::-1] for h in self.tx_hashes]
        # compute the Merkle Root and reverse
        root = merkle_root(hashes)
        # return whether self.merkle_root is the same
        return root[::-1] == self.merkle_root

运行结果

无

练习5

p187

代码实现

# -*- coding: utf-8 -*-
# @Author: 从化北(喵星人)
# @Date:   2022-01-16 11:24:09
# @Last Modified by:   从化北
# @Last Modified time: 2022-01-16 11:26:48
import math
total = 27
max_depth = math.ceil(math.log(total, 2))
merkle_tree = []
for depth in range(max_depth + 1):
    num_itmes = math.ceil(total / 2 ** (max_depth - depth))
    level_hashes = [None] * num_itmes
    merkle_tree.append(level_hashes)
for level in merkle_tree:
    print(level)

运行结果

[None]
[None, None]
[None, None, None, None]
[None, None, None, None, None, None, None]
[None, None, None, None, None, None, None, None, None, None, None, None, None, None]
[None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None]
[Finished in 397ms]

练习6

p193

代码实现

# -*- coding: utf-8 -*-
# @Author: 从化北(喵星人)
# @Date:   2022-01-16 11:38:42
# @Last Modified by:   从化北
# @Last Modified time: 2022-01-16 11:39:00


class MerkleTree:
...
    @classmethod
    def parse(cls, s):
        '''Takes a byte stream and parses a merkle block. Returns a Merkle Block object'''
        # version - 4 bytes, Little-Endian integer
        version = little_endian_to_int(s.read(4))
        # prev_block - 32 bytes, Little-Endian (use [::-1])
        prev_block = s.read(32)[::-1]
        # merkle_root - 32 bytes, Little-Endian (use [::-1])
        merkle_root = s.read(32)[::-1]
        # timestamp - 4 bytes, Little-Endian integer
        timestamp = little_endian_to_int(s.read(4))
        # bits - 4 bytes
        bits = s.read(4)
        # nonce - 4 bytes
        nonce = s.read(4)
        # total transactions in block - 4 bytes, Little-Endian integer
        total = little_endian_to_int(s.read(4))
        # number of transaction hashes - varint
        num_hashes = read_varint(s)
        hashes = []
        # each transaction is 32 bytes, Little-Endian
        for _ in range(num_hashes):
            hashes.append(s.read(32)[::-1])
        # length of flags field - varint
        flags_length = read_varint(s)
        # read the flags field
        flags = s.read(flags_length)
        # initialize class
        return cls(version, prev_block, merkle_root, timestamp, bits, nonce, total, hashes, flags)

运行结果

无

练习7

p197

代码实现

# -*- coding: utf-8 -*-
# @Author: 从化北(喵星人)
# @Date:   2022-01-16 11:45:26
# @Last Modified by:   从化北
# @Last Modified time: 2022-01-16 11:45:26

class MerkleBlock:
    ...
    def is_valid(self):
        '''Verifies whether the merkle tree information validates to the merkle root'''
        # convert the flags field to a bit field
        flag_bits = bytes_to_bit_field(self.flags)
        # reverse self.hashes for the merkle root calculation
        hashes = [h[::-1] for h in self.hashes]
        # initialize the merkle tree
        merkle_tree = MerkleTree(self.total)
        # populate the tree with flag bits and hashes
        merkle_tree.populate_tree(flag_bits, hashes)
        # check if the computed root reversed is the same as the merkle root
        return merkle_tree.root()[::-1] == self.merkle_root

运行结果

无