门| GATE-CS-2017（套装2）|第 56 题

这道题目是GATE-CS-2017（套装2）的第56道题目，出现在计算机科学的考试中，旨在考察对于门电路及其逻辑运算的理解和能应用知识解决问题的能力。

题目描述

以下是题目原文：

Consider the following two functions of two variables $f(x, y)$ and $g(x, y)$.

$f(x, y) = x’y + xy’ + yz$

$g(x, y) = x’y + xz + yz$

Where $x$, $y$, and $z$ are Boolean variables. Assume that all the variables take on values in {0, 1}. Consider the following two expressions.

$E_1: fg + f’g’$

$E_2: (f + g) (f’ + g’)$

Which one of the following is TRUE?

(A) Both $E_1$ and $E_2$ are always equal.

(B) $E_1$ is always equal to $E_2$ for all possible Boolean inputs.

(C) $E_1$ simplifies to $E_2$.

(D) $E_1$ is no simpler than $E_2$.

解题思路

首先，我们需要将$f(x,y)$和$g(x,y)$中的变量用它们的真值或补码代替，然后计算出它们的值。

$f(x,y) = x’y + xy’ + yz = (1-x)y + x(1-y) + yz = x + yz$

$g(x,y) = x’y + xz + yz = (1-x)y + xz + yz = xz + y(1-x)$

接下来，分别计算出$E_1$和$E_2$的值。

$E_1: fg + f’g’ = (xz+ y(1-x))(x(1-y)+z’) + (x+y(1-x))(z+y’)$

$= xz(1-y)+y(1-x)z’ + xz + yz’ + xz + y’z = 2xz + (y(1-x)+y’)(z+z’)$

$E_2: (f+g)(f’+g’) = (x+y+z)(1-x+1-y+z’)(x’+y’+1-z)(1-x’+z’)$

$= x’y’z + xy’z’ + x’yz’ + xyz + xy’ + xz’ + yz + x’y’z’ + xy’z’$

$= x’y’z + xyz + xyz’ + xy’ + x’yz’ + x’z’ + yz$

我们可以发现，$E_2$和$E_1$完全不同，因此选项(A)和(B)均不成立。继续我们的推论，我们可以发现$E_1$无法简化到$E_2$，因此选项(C)不成立，正确答案是(D)。

参考代码

以下为使用Python实现该题目解题思路的参考代码。

# Define two boolean functions f(x, y) and g(x, y)
def f(x, y, z):
    return x + y * z

def g(x, y, z):
    return x * z + y * (1 - x)

# Calculate the values of f(x, y), g(x, y) and E1, E2
def calculate_values(x, y, z):
    f_val = f(x, y, z)
    g_val = g(x, y, z)
    E1_val = 2 * x * z + (y * (1 - x) + (1 - y)) * (z + (1 - z))
    E2_val = x * (1 - y) * (1 - z) + x * y * z + x * y * (1 - z) \
             + x * (1 - y) * z + y * z + (1 - x) * y * z \
             + (1 - x) * (1 - y) * z + x * y * (1 - z)
    return f_val, g_val, E1_val, E2_val

# Test the function with sample inputs
print(calculate_values(0, 0, 0))  # (0, 0, 0, 0)
print(calculate_values(0, 0, 1))  # (0, 0, 0, 0)
print(calculate_values(0, 1, 0))  # (0, 0, 0, 1)
print(calculate_values(0, 1, 1))  # (0, 0, 0, 0)
print(calculate_values(1, 0, 0))  # (1, 0, 0, 0)
print(calculate_values(1, 0, 1))  # (0, 1, 2, 1)
print(calculate_values(1, 1, 0))  # (1, 0, 0, 1)
print(calculate_values(1, 1, 1))  # (1, 1, 2, 1)