题目：
For the following Karnaugh map, give the circuit implementation using one 4-to-1 multiplexer and as many 2-to-1 multiplexers as required, but using as few as possible. You are not allowed to use any other logic gate and you must use a and b as the multiplexer selector inputs, as shown on the 4-to-1 multiplexer below.
You are implementing just the portion labelled top_module, such that the entire circuit (including the 4-to-1 mux) implements the K-map.

解题：

module top_module (
    input c,
    input d,
    output [3:0] mux_in
); 

    assign mux_in[0]=({c,d}==2'b00)?0:1;
    assign mux_in[1]=0;
    assign mux_in[3]=({c,d}==2'b11)?1:0;
    assign mux_in[2]=({c,d}==2'b00|{c,d}==2'b10)?1:0;

endmodule

结果正确：

标准答案：

module top_module (
	input c,
	input d,
	output [3:0] mux_in
);
	
	// After splitting the truth table into four columns,
	// the rest of this question involves implementing logic functions
	// using only multiplexers (no other gates).
	// I will use the conditional operator for each 2-to-1 mux: (s ? a : b)
	assign mux_in[0] = c ? 1 : d;          // 1 mux:   c|d
	assign mux_in[1] = 0;                  // No muxes:  0
	assign mux_in[2] = d ? 0 : 1;          // 1 mux:    ~d
	assign mux_in[3] = c ? d : 0;          // 1 mux:   c&d
	
endmodule