双向搜索以选择最佳储蓄率

时间:2018-10-31 17:43:17

标签: python search binary-search bisection

您好,我希望获得有关此问题的帮助,这是MIT OCW计算机科学和Python课程中的问题之一。我知道人们也问过类似的问题,并且发现了有用的帖子,例如Bisection search code doesnt work,但我仍然被困住了!

我已经为这个问题苦苦挣扎了很多天,并试图以不同的方式来解决这个问题,但是却以各种方式失败了。如果可能的话,有人能暗示我要去哪里,而不是告诉我答案。我想通过一点帮助自己解决这个问题。

供参考,问题是C部分,在这里:https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-0001-introduction-to-computer-science-and-programming-in-python-fall-2016/assignments/MIT6_0001F16_ps1.pdf

在努力奋斗的过程中,我已将这项任务分解为一个总体目标,然后分解为解决问题的步骤。

目标:尝试找到最佳的储蓄率,以在36个月内实现首付100万美元房屋的首付。##解决问题的步骤:
1)猜测储蓄率,即0和1000的平均值
2)计算36个月后的增长情况
3a)如果在36个月内达到的金额超过100万美元的25%,那么较低的储蓄率应该是新的猜测
... max = guesss(旧的猜测)和min = 0并更新猜测(高和低的平均值)
...使用新的猜测来运行步骤2中的计算
3b)如果金额在36个月内未达到100万美元的25%,那么更高的储蓄率应该是新的猜测
... min = guess(旧猜测)并更新猜测(高低平均) ...使用新的猜测来运行步骤2中的计算
3c)如果金额在第36个月的截止日期内达到100万美元的25%,则退出并记录储蓄率作为最佳猜测。
为简单起见:假设没有利息,并且假设工资保持不变

所以这是我目前正在努力解决此问题的代码。 (这导致“猜测”变量趋向于0,然后无限循环)

total_cost=1000000 #cost of house
portion_down_payment=0.25 #fraction of cost needed for downpayment on house
downpayment=total_cost*portion_down_payment

starting_annual_salary=float(input("Enter the starting salary: "))
low=0
high=1000
bisect_steps=0
month=1 #set as 1 because the first calculation will occur in month 1 
guess=(low+high)//2
current_savings=0

def calSavings(current_savings,monthly_salary,guess,month):
    while month<37:
        monthly_savings=monthly_salary*(guess/1000)
        current_savings+=monthly_savings
        month+=1
    return(current_savings) 

current_savings=calSavings(current_savings,monthly_salary,guess,1)
while True:
    current_savings=calSavings(current_savings,monthly_salary,guess,1)
    if current_savings>downpayment and month<=35: #if amount reached goes over 25% of $1m within 36 months
        #a lower savings rate should be the new guess
        high=guess #max=guess (old guess) and min=0 and update the guess
        bisect_steps+=1
        guess=(low+high)//2
        print("The guess was too high, so the new lower guess is",guess," This is bisect step",bisect_steps)
        continue #send new guess up to beginning of while loop to calculate 
    elif current_savings<downpayment and month>=36: #if amount does not reach 25% of $1m within 36 months
        low=guess
        bisect_steps=+1
        guess=(low+high)//2
        print("The guess was too low, so the new higher guess is",guess," This is bisect step",bisect_steps)
        continue #send new guess up to beginning of while loop to calculate 
    elif current_savings>=downpayment and month==36: #if amount reaches 25% of $1m in the 36th months then quit
        # record the savings rate as the best guess
        print("The best savings rate is ",guess/100,"%, the amount saved was ",current_savings," in ",month," months")
        break #break out of while loop

我知道其他人也曾问过类似的问题(我已经看过这些答案了,但仍未解决我的问题),但我想获得的更多答案是 HOW 来解决此问题。

2 个答案:

答案 0 :(得分:1)

更新

循环没有停止的原因是您没有给它足够的时间。您忘记的是您正在处理decimal类型。将==decimal值一起使用总是很危险的。 decimal类型(默认情况下)精确到28位,这意味着您正在尝试找到一个非常这个问题的近似值,因为只有在正确的28位小数时,{ {1}}评估为(current_savings>downpayment or current_savings<downpayment)调用您的退出条件。

基本上,导致您遇到问题的问题是,即使您最终获得的估价为$ 1,000,000.0000000001,python仍说这不等于$ 1,000,000.0000000000,因此它一直持续到得到下一个0,然后才加上另一个零等等。这将持续非常长的时间,在极少数情况下,由于并非所有十进制数字都可以存储为二进制数字(1,000,000不在这种情况下),因此可能永远不会停止。

那么,我们该如何解决呢?有两种选择。最简单的方法是忽略美分,仅将比较值转换为False,这将确保接受任何不足一美元的值。其他选项是创建一系列可接受的答案。举例来说,在这36个月中,我想节省100万美元,但这不太可能实现。因此,我将以$ 1,000,000.00-$ 1,000,010.00范围内的任何金额结算(例如)。这样,我们可以确保任何过高的猜测都将被拒绝,并且仅接受非常有限的猜测。

不管走哪条路线,通常最好将无限循环的退出条件放在顶部,这样可以保证始终对它进行评估。

我的建议是编写一个这样的函数,并使用该函数退出循环(放在顶部):

int

这将认为0.00009(以及所有小于该数字的小数)等于0.0000。

原始

首先,请注意,您所做的不是二等分,而是二分查找。

由于这个问题,您永远不会在主循环中更改month的值。这意味着,def are_decimals_equal(a, b): accuracy = 0.0001 return abs(a-b) < accuracy 的计算结果始终为False时,一旦current_savings>downpayment的计算结果为False,您的程序便会进入无限循环,因为它没有任何条件可以求值为True。

据我所见,if / elif语句中条件的第二部分是不必要的,您的calSavings将始终计算出36个月的节省价值,永远不会多,永远不会少。因此,如果您从if / elif语句中删除该条件,则程序最终将停止,并且此时应确定正确的答案。

最后,您看到month>=36作为输出的原因是最后的划分。如果执行0,则将看到它是一个整数,100也是一个整数,因此该除法将导致某些值,例如print(typeof(guess)),该值将被截断为0.3123。将您的输出更改为0,它将消失。

答案 1 :(得分:0)

我希望我可以在这里为自己的问题提供答案-尽管这不是一个完美的答案。

代码产生的结果似乎是合理的。

total_cost=1000000 #cost of house

portion_down_payment=0.25 #fraction of cost needed for downpayment on house
downpayment=total_cost*portion_down_payment

starting_annual_salary=float(input("Enter the starting salary: "))
monthly_salary=starting_annual_salary/12
low=0
high=1000
binary=0
month=1 #set as 1 because the first calculation will occur in month 1
guess=(low+high)//2
current_savings=0
tolerance=500

def calSavings(current_savings,monthly_salary,guess,month):
    while month<37:
        monthly_savings=int(monthly_salary*(guess/1000))
        current_savings+=monthly_savings
        month+=1
    return(current_savings)

current_savings=calSavings(current_savings,monthly_salary,guess,1)

while True:
    if abs(current_savings-downpayment)<=tolerance: #if the difference between the current savings and downpayment is less than $500
        # record the savings rate as the best guess
        print("The best savings rate is ",guess/10,"%, the amount saved was $",current_savings," in 36 months")
        break #break out of while loop
    elif (current_savings-downpayment)>tolerance: #if amount reached goes over 25% of $1m within 36 months
        #a lower savings rate should be the new guess
        high=guess #high=guess (old guess) and low=low (stays same) and update the guess
        binary=binary+1
        guess=(low+high)//2
        print("The guess was too high, so the new lower savings rate is",guess/10,"%. This is binary-search step",binary)
        current_savings=calSavings(0,monthly_salary,guess,1)
        continue #send new guess up to beginning of while loop to check if conditionals
    elif (downpayment-current_savings)>tolerance: #if amount does not come to within tolerance amount of 25% of $1m within 36 months
        low=guess #to make the guess higher, make low=guess (old guess) and high stay the same
        binary=binary+1
        guess=(low+high)//2
        print("guess is ",guess)
        if guess>=990: #check if the savings rate guess is getting too high
            print("Your wages are too low. You can't save up enough")
            break #exit the while loop because conditions will never be met
        print("The guess was too low, so the new higher savings rate is",guess/10,"%. This is binary-search step",binary)
        current_savings=calSavings(0,monthly_salary,guess,1)
        continue #send new guess up to beginning of while loop to check over the conditionals

可接受的答案的公差在$ 500以内,但是如果我将其降低到$ 50,我将再次陷入看似无限的循环,猜测和低点最终都相同。我很高兴我已经取得了一些明显的进步,但是对不能再忍受它而无法降低容忍度感到困惑。

顺便说一句,我不想​​不想忽略尼克关于将变量转换为浮点数的评论,但是我解释了为什么我在评论中使用整数-看起来正确吗?