--- title: 【探索】利用后缀表达式解方程 urlname: Solving-equations-using-postfix-notation index_img: https://api.limour.top/randomImg?d=2024-01-19 22:41:03 date: 2020-12-12 06:41:03 tags: 探索 excerpt: 这段Python代码实现了使用后缀表达式解方程的功能。它包括了一个栈类(Stack)和一个多项式类(Polynomial)。通过中缀表达式转后缀表达式的方式处理方程,最终实现了解一元一次方程的功能。代码包含了多个方法,如多项式的加法、减法、乘法、除法等操作。在解方程时,它先进行表达式的简化和转换,然后输出方程的解或者判断方程是否是一元一次方程。 --- ```python import re from fractions import Fraction class Stack(list): def isEmpty(self): return self == [] def peek(self): if self == []: return None else: return self[-1] def size(self): return len(self) push = list.append def pop(self): if self == []: return None else: return super().pop() class Polynomial(list): def __init__(self, value): for item in value: self.append(Fraction(item)) def add(self, value): if len(self) < len(value): self += [Fraction(0)]*(len(value)-len(self)) for i in range(len(value)): self[i] += value[i] def sub(self, value): if len(self) < len(value): self += [Fraction(0)]*(len(value)-len(self)) for i in range(len(value)): self[i] -= value[i] def mul(self, value): tmp = self.copy() size = len(self) self.clear() self += [Fraction(0)]*(size+len(value)-1) for i,item in enumerate(value): for j in range(size): self[i+j] += tmp[j]*item def divn(self, n): if type(n) is Polynomial: for i in range(len(self)): self[i] /= n[0] else: _n = Fraction(n) for i in range(len(self)): self[i] /= _n def __str__(self): if self == []: return '0' elif len(self) == 1: return str(self[0]) elif len(self) == 2: return f'({self[1]})x + {self[0]}' else: pass def get_Formula(equation): return equation.replace(' ','').split('=') _ep = re.compile(r'([\+\-\*/()][^\+\-\*/()]+)') _op = { '+': lambda x,y:x.add(y), '-': lambda x,y:x.sub(y), '*': lambda x,y:x.mul(y), '/': lambda x,y:x.divn(y) } def _middle2behind(Fma, res, s, e): sta = Stack() _s = s while s < e: if Fma[s] in '+-': if sta.isEmpty(): sta.push(Fma[s]) s += 1 elif sta.peek() in '*/': while sta: res.push(sta.pop()) sta.push(Fma[s]) s += 1 elif sta.peek() in '+-': res.push(sta.pop()) sta.push(Fma[s]) s += 1 elif Fma[s] in '*/': if sta.isEmpty(): sta.push(Fma[s]) s += 1 elif sta.peek() in '+-': sta.push(Fma[s]) s += 1 elif sta.peek() in '*/': res.push(sta.pop()) sta.push(Fma[s]) s += 1 elif Fma[s] == '(': s += 1 d = _middle2behind(Fma, res, s, e) s += d elif Fma[s] == ')': s += 1 break else: res.push(Fma[s]) s += 1 while sta: res.push(sta.pop()) return s-_s def middle2behind(Formula): if Formula.startswith('-'): Formula = '0' + Formula expr = _ep.findall(Formula.replace('(-','(0-')) res = Stack() _middle2behind(expr, res, 0, len(expr)) return res def str2polynomial(_str): if _str.endswith('x'): if _str == 'x': return Polynomial((0, 1)) return Polynomial((0, _str.rstrip('x'))) else: return Polynomial((_str,)) def not_eval(Formula): expr = middle2behind(Formula) #print(expr) sta = Stack() for item in expr: if item in _op: y = sta.pop() x = sta.pop() _op[item](x, y) sta.push(x) else: sta.push(str2polynomial(item)) #print(sta) return sta.pop() _x = re.compile(r'[a-zA-Z]+') def solve_eq(equation): print(f'equation is \t\t{equation}') xname = set(_x.findall(equation)) if len(xname) != 1: #print(xname) print(f'别逗,{equation}是一元一次方程吗?') return xname = xname.pop() Fma = get_Formula(_x.sub('x', equation)) if len(Fma) != 2: print('{equation}不是标准方程!') return Fma = [not_eval(expr) for expr in Fma] print(f'Simplification is \t{Fma[0]} = {Fma[1]}') tmp = Fma[0] tmp.sub(Fma[1]) print(f'Transposition is \t{tmp} = 0') if len(tmp) < 2 or tmp[1] == 0: print(f'{xname} 无解') else: x = -tmp[0]/tmp[1] print(f'{xname} = x = {x}') ```