在Python中使用递归实现二分法

Question

我在一个玩具示例中实现了带有递归的Bisection算法和Python递归示例，目的是找到我应该每月支付的最低每月分期付款，为期12个月，以便在12个月结束时我的余额为零或稍少。

假设我在期初的余额为320000，年利率为0.2。 正确答案是29157.09。

我的代码由于无法收敛而遇到异常。

代码如下（我已插入打印语句以方便调试）

balance = 320000
annualInterestRate = 0.2

Balance = balance

Annual_interest_rate = annualInterestRate



Monthly_interest_rate = (Annual_interest_rate) / 12.0
Monthly_payment_lower_bound = Balance / 12
Monthly_payment_upper_bound = (Balance * (1 + Monthly_interest_rate)) / 12.0

def bisect(Balance, Annual_interest_rate, a, b ):



    Previous_balance = Balance

    for i in range(12):



            Monthly_unpaid_balance = (Previous_balance) - a
            Updated_balance_each_month = (Monthly_unpaid_balance) + (Monthly_interest_rate * Monthly_unpaid_balance)
            Previous_balance =  Updated_balance_each_month

    f_a = Previous_balance

    for i in range(12):



            Monthly_unpaid_balance = (Previous_balance) - b
            Updated_balance_each_month = (Monthly_unpaid_balance) + (Monthly_interest_rate * Monthly_unpaid_balance)
            Previous_balance =  Updated_balance_each_month

    f_b = Previous_balance

    c = (a + b)/2


    for i in range(12):



            Monthly_unpaid_balance = (Previous_balance) - c
            Updated_balance_each_month = (Monthly_unpaid_balance) + (Monthly_interest_rate * Monthly_unpaid_balance)
            Previous_balance =  Updated_balance_each_month

    f_c = Previous_balance

    print('a is {0}, b is {1}, c = {2}, f_a = {3}, f_b = {4}, f_c = {5}'.format(a, b, c, f_a, f_b, f_c))

    if abs(f_c) <=0.01:
        return(c)

    elif f_c * f_a >0:

        a =c

        print('a is {0}, b is {1}, c = {2}, f_a = {3}, f_b = {4}, f_c = {5}'.format(a, b, c, f_a, f_b, f_c))

        return(bisect(Balance, Annual_interest_rate, a,  b ))

    else:

        b = c

        print('a is {0}, b is {1}, c = {2}, f_a = {3}, f_b = {4}, f_c = {5}'.format(a, b, c, f_a, f_b, f_c))

        return(bisect(Balance, Annual_interest_rate, a,  b ))

当我运行该函数时，我得到以下打印输出：

bisect(Balance, Annual_interest_rate, Monthly_payment_lower_bound, Monthly_payment_upper_bound )
a is 26666.666666666668, b is 27111.11111111111, c = 26888.88888888889, f_a = 33328.98239049623, f_b = -322183.0368628964, f_c = -752717.255677314
a is 26666.666666666668, b is 26888.88888888889, c = 26888.88888888889, f_a = 33328.98239049623, f_b = -322183.0368628964, f_c = -752717.255677314
a is 26666.666666666668, b is 26888.88888888889, c = 26777.77777777778, f_a = 33328.98239049623, f_b = -319209.06882307003, f_c = -747603.8415428438
a is 26666.666666666668, b is 26777.77777777778, c = 26777.77777777778, f_a = 33328.98239049623, f_b = -319209.06882307003, f_c = -747603.8415428438
a is 26666.666666666668, b is 26777.77777777778, c = 26722.222222222226, f_a = 33328.98239049623, f_b = -317722.08480315685, f_c = -745047.1344756087
a is 26666.666666666668, b is 26722.222222222226, c = 26722.222222222226, f_a = 33328.98239049623, f_b = -317722.08480315685, f_c = -745047.1344756087

Answer 1

以下代码使用了前端函数，因此用户不必处理诸如a和b类的“魔术”参数。 它将年利率重新调整为月利率，并通过将递归输入缩放100来将计算转换为整数，然后在返回之前重新缩放解决方案。

def calculate_payment(balance, interest):
    balance = int(100 * balance)
    return bisect(balance, interest / 12.0, 0, balance) * 0.01

def bisect(balance, monthly_interest, low, high):
    payment = round((high + low) / 2)
    remaining_balance = balance
    for _ in range(12):
        remaining_balance -= payment
        remaining_balance += int(monthly_interest * remaining_balance)
    if remaining_balance > 0:
        return bisect(balance, monthly_interest, payment, high)
    elif remaining_balance <= -12:
        return bisect(balance, monthly_interest, low, payment)
    else:
        return payment

print(calculate_payment(320000.00, 0.2))