与Heron of Alexandria合作开发一个程序

Question

鉴于avg_guess中的.5 - 我是否需要所有浮动（arg）规范？

# Heron of Alexandria method for approximating
# the square root of a number

def heron(num, guess, tolerance):
    print "Your guess squared:",guess**2
    if guess**2 != num:
        print "When squared, the guess:", guess, "is",abs(float(num) - float(guess)**2), "away from", num, "\n"
        if abs(float(num) - float(guess)**2) > float(tolerance):
            avg_guess = 0.5 * (float(guess) + (float(num) / float(guess)))
            return heron(num, avg_guess, tolerance)
        print "Given your tolerance, this is Heron's best guess:", guess, "\n"
    else:
        print guess, "is correct!\n"

我注意到一定的容差，这个函数将返回5.0作为答案......不确定为什么？

heron(25, 3, .00000001)

从＆＃34;返回5.0;否则：＆＃34;条件，但

heron(25, 3, .0000001)

终止于初始嵌套的容差＆＃34; if：＆＃34;条件，印刷： 鉴于你的宽容，这是Heron的最佳猜测：5.00000000233。