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I recently posted a solution for the numbers game on the Countdown TV show using a recursive technique. For those unfarmiliar with the rules…

  • A fixed set of numbers (usually 6) are chosen at random
  • A target value is set
  • The player must reach the target value using some or all of the numbers from the list of 6 random numbers using only the operators +, *, - and /. The use of brackets is also permitted

I have attached a new program that uses the itertools library and for loops. As with the last code the print statement

  • "Eval..." means the program has used BODMAS (e.g: 1+3*10 = 31 as 10*3 = 30 and 30+1 = 31) whereas
  • "Running sum..." produces a running sum of the totals (1+3*10 = 40 as 1+3 = 4, 4*10 = 40)
import itertools
import re
def unpack(method):
    string = method
    special = ["*","+","-","/"]
    list_sum = []
    list_special = []
    numbers = (re.findall(r"[\w']+", string))
    for char in string: 
        if char in special: 
            list_special.append(char)

    for index in range (len(numbers)-1):
        to_eval = numbers[index] + list_special[index] + numbers[index+1]
        list_sum.append(f'{to_eval} = {eval(to_eval)}')
        numbers[index+1] = str(eval(to_eval))

    return list_sum

def plus (x,y): return x+y, "+"
def minus (x,y): return x-y, "-"
def times (x,y): return x*y, "*"
def divide (x,y): return x/y, "/"

def solve():
    for numbers in list_sols:
        total = 0
        running = ""
        list_to_call = [order_ops1,order_ops2,order_ops3,order_ops4,order_ops5,order_ops6]
        for operator in list_to_call[len(numbers)-2]:
            zipped = itertools.zip_longest(numbers, operator)
            to_calculate = (list(zipped))
            running = ""
            total = 0
            for (index,item) in enumerate(to_calculate):
                if index == 0:
                    total += item[0]
                    next_operator = item[1]
                    running = str(item[0])
                else:
                    total, symbol = next_operator(total,item[0])
                    running += symbol + str(item[0])
                    if index < len(numbers)-1: next_operator = item[1]
            if eval(running)==target:
                print(f"Eval: {running}")
                break
            if total == target:
                print (f'Running sum: {unpack(running)}')
                break


def main():
    global list_sols
    global order_ops1, order_ops2, order_ops3, order_ops4, order_ops5, order_ops6
    global target 

    numbers = [100,50,75,25,3,6]
    target = 952

    list_sols = []
    for number in itertools.chain.from_iterable(itertools.permutations(numbers,i) for i in range(2,7)):
        list_sols.append(list(number))

    order_ops1,order_ops2,order_ops3,order_ops4,order_ops5,order_ops6 = [],[],[],[],[],[]
    for number in itertools.chain.from_iterable(itertools.product([plus,minus,times,divide],repeat=i) for i in range(1,6)):
        if len(number) ==1: order_ops1.append(list(number))
        if len(number) ==2: order_ops2.append(list(number))
        if len(number) ==3: order_ops3.append(list(number))
        if len(number) ==4: order_ops4.append(list(number))
        if len(number) ==5: order_ops5.append(list(number))
        if len(number) ==6: order_ops6.append(list(number))


    if target in numbers:
        print (target)
        return 
    else:
        solve()
        return 

if __name__=="__main__":
    main()
    exit_key = input("Press any key to exit...")


```
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    \$\begingroup\$ Why do you keep putting "countdown" in your question titles? What does that mean? \$\endgroup\$ Commented May 21, 2019 at 17:26
  • \$\begingroup\$ Its the name of a British TV game show (en.wikipedia.org/wiki/Countdown_(game_show)). I have tried solving the game show in different ways and thought it'd be nice to share the code with people in case they are working on a similar solution \$\endgroup\$
    – MrJoe
    Commented May 21, 2019 at 17:34

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