#|HDLBits 系列（37）此系列关于独热码的题目的疑问（） #|HDLBits

目录
背景
我的做法
第一题
第二题
第三题
解决办法
第一题
第二题
第三题
推荐
背景目前为止，关于状态机独热码的题目，几乎没一个题目能做对，这令我疑惑？是不是题目的答案有问题？在此请大家一试？（已解决，谢谢）
我的做法第一题
第一题（点击蓝色字体进入题目链接做答）
本人答案：

module top_module ( input [6:1] y, input w, output Y2, output Y4); localparam A = 6'b0000_01, B = 6'b0000_10, C = 6'b0001_00, D = 6'b0010_00, E = 6'b0100_00, F = 6'b1000_00; reg [6:1] next_state; assign Y2 = next_state[2]; assign Y4 = next_state[4]; always@(*) begin next_state = A; case(y) A: begin if(w) next_state = A; else next_state = B; end B: begin if(w) next_state = D; else next_state = C; end C: begin if(w) next_state = D; else next_state = E; end D: begin if(w) next_state = A; else next_state = F; end E: begin if(w) next_state = D; else next_state = E; end F: begin if(w) next_state = D; else next_state = C; end endcase endendmodule

第二题
第二题
本人答案：

module top_module ( input [5:0] y, input w, output Y1, output Y3 ); localparam A = 6'b0000_01, B = 6'b0000_10, C = 6'b0001_00, D = 6'b0010_00, E = 6'b0100_00, F = 6'b1000_00; reg [5:0] next_state; assign Y1 = next_state[1]; assign Y3 = next_state[3]; always@(*) begin case(y) A: begin if(~w) next_state = A; else next_state = B; end B: begin if(~w) next_state = D; else next_state = C; end C: begin if(~w) next_state = D; else next_state = E; end D: begin if(~w) next_state = A; else next_state = F; end E: begin if(~w) next_state = D; else next_state = E; end F: begin if(~w) next_state = D; else next_state = C; end default: begin next_state = A; end endcase endendmodule

第三题
第三题
https://blog.csdn.net/Reborn_Lee/article/details/103428895
本人答案：

module top_module( input in, input [9:0] state, output [9:0] next_state, output out1, output out2); localparam S0 = 10'b0000_0000_01, S1 = 10'b0000_0000_10, S2 = 10'b0000_0001_00, S3 = 10'b0000_0010_00,S4 = 10'b0000_0100_00; localparam S5 = 10'b0000_1000_00, S6 = 10'b0001_0000_00, S7 = 10'b0010_0000_00, S8 = 10'b0100_0000_00,S9 = 10'b1000_0000_00; always@(*) begin case(state) S0: begin if(in) next_state = S1; else next_state = S0; end S1: begin if(in) next_state = S2; else next_state = S0; end S2: begin if(in) next_state = S3; else next_state = S0; end S3: begin if(in) next_state = S4; else next_state = S0; end S4: begin if(in) next_state = S5; else next_state = S0; end S5: begin if(in) next_state = S6; else next_state = S8; end S6: begin if(in) next_state = S7; else next_state = S9; end S7: begin if(in) next_state = S7; else next_state = S0; end S8: begin if(in) next_state = S1; else next_state = S0; end S9: begin if(in) next_state = S1; else next_state = S0; end default: begin next_state = S0; endendcase endassign out1 = (state == S8 | state == S9) ? 1 : 0; assign out2 = (state == S9 | state == S7) ? 1 : 0; endmodule

如有大神知道，还望告知，谢谢。
（崩溃中）
解决办法更新：
群里的一个同学给我解答了，道理是有的，但是这个网站上有关独热码的题目真的没必要深究了，上面的设计本身就是没有问题的，仅仅为了正确的答案，给出Success的答案：
第一题
第一题：

module top_module ( input [6:1] y, input w, output Y2, output Y4); assign Y2 = y[1]&&(~w); //assign Y4 = y[2]&&w || y[3]&&w || y[5]&&w || y[6]&&w; assign Y4= w&&(y[2] || y[3] || y[5] || y[6]); endmodule

第二题
第二题：

module top_module ( input [5:0] y, input w, output Y1, output Y3 ); assign Y1 = y[0]&& w; assign Y3= ~w && (y[1] || y[2] || y[4] || y[5]); endmodule

第三题
第三题：

module top_module( input in, input [9:0] state, output [9:0] next_state, output out1, output out2); assign next_state[0] = ~in & (state[0] | state[1] | state[2] | state[3] | state[4] | state[7] | state[8] | state[9]); assign next_state[1] = in & (state[0] | state[8] | state[9]); assign next_state[2] = in & state[1]; assign next_state[3] = in & state[2]; assign next_state[4] = in & state[3]; assign next_state[5] = in & state[4]; assign next_state[6] = in & state[5]; assign next_state[7] = in & (state[6] | state[7]); assign next_state[8] = ~in & state[5]; assign next_state[9] = ~in & state[6]; assign out1 = state[8] | state[9]; assign out2 = state[7] | state[9]; endmodule

推荐思想见：
https://blog.csdn.net/Reborn_Lee/article/details/103338804
https://blog.csdn.net/Reborn_Lee/article/details/103339538
最后感谢群里的小伙伴，进群见：
https://blog.csdn.net/Reborn_Lee/article/details/99715080