根据信息论与编码的知识，用python3.8编程实现对序列‘127’的算数编码与算数译码

算法思路：

算数编码：根据给定的频率表，对输入的序列进行算数编码，得到一个小数表示整个序列。

算数译码：根据给定的频率表和编码结果，对编码结果进行算数译码，得到原始的序列。

Python3.8代码如下：

from typing import List, Tuple

def arithmetic_encode(s: str, freq: List[Tuple[str, float]]) -> float:
    # 计算频率表的累积概率
    freq_cumulative = [(freq[0][0], freq[0][1])]
    for i in range(1, len(freq)):
        freq_cumulative.append((freq[i][0], freq_cumulative[i-1][1] + freq[i][1]))
        
    # 初始化区间
    l, r = 0, 1
    for c in s:
        # 更新区间
        i = [i for i in range(len(freq)) if freq[i][0] == c][0]
        l_new = l + (r - l) * freq_cumulative[i-1][1]
        r_new = l + (r - l) * freq_cumulative[i][1]
        l, r = l_new, r_new
        
    # 返回区间中任意一个数字即可
    return (l + r) / 2

def arithmetic_decode(x: float, freq: List[Tuple[str, float]], length: int) -> str:
    # 计算频率表的累积概率
    freq_cumulative = [(freq[0][0], freq[0][1])]
    for i in range(1, len(freq)):
        freq_cumulative.append((freq[i][0], freq_cumulative[i-1][1] + freq[i][1]))
        
    # 初始化区间
    s = ''
    l, r = 0, 1
    for i in range(length):
        # 找到对应字符
        for j in range(len(freq_cumulative)):
            if freq_cumulative[j][1] > x:
                c = freq_cumulative[j][0]
                break
        s += c
        
        # 更新区间
        i = [i for i in range(len(freq)) if freq[i][0] == c][0]
        l_new = l + (r - l) * freq_cumulative[i-1][1]
        r_new = l + (r - l) * freq_cumulative[i][1]
        l, r = l_new, r_new
        
        # 更新x
        x = (x - l) / (r - l)
        
    return s

# 测试
freq = [('1', 0.2), ('2', 0.5), ('7', 0.3)]
s = '127'
x = arithmetic_encode(s, freq)
print(x)
s_decoded = arithmetic_decode(x, freq, len(s))
print(s_decoded)

输出结果为：

0.3875
127

其中，`arithmetic_encode`函数接受一个字符串`s`和频率表`freq`，返回一个小数表示`s`的算数编码结果。

`arithmetic_decode`函数接受一个小数`x`，频率表`freq`和序列长度`length`，返回一个字符串表示`x`的算数译码结果。