#|HDLBits 系列（38）值得一看的状态机设计题目 #|HDLBits

目录
背景
原题复现
我的方案
【#|HDLBits 系列（38）值得一看的状态机设计题目】状态转移图
我的设计
更新方案
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背景这是这个系列中的一个状态机的题目，但是相比于给了你完整状态转移图之类的题目，这个题目还是稍微有点难的，我实在不知道该怎么给这个博客起个什么名字？
我在线等一个简单的方式去解决今天的问题，而如题所说，我用最无能的方式来解决这个问题，但简单的方式一定存在。
2019/12/16更新
今天一个帅兄弟给了我一个答案，很巧妙，这里十分感谢。我把它放到后面来给出。
原题复现原题复现：

Consider a finite state machine with inputs s and w. Assume that the FSM begins in a reset state called A, as depicted below. The FSM remains in state A as long as s = 0, and it moves to state B when s = 1. Once in state B the FSM examines the value of the input w in the next three clock cycles. If w = 1 in exactly two of these clock cycles, then the FSM has to set an output z to 1 in the following clock cycle. Otherwise z has to be 0. The FSM continues checking w for the next three clock cycles, and so on. The timing diagram below illustrates the required values of z for different values of w.
Use as few states as possible. Note that the s input is used only in state A, so you need to consider just the w input.

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我的方案状态转移图我的解决方案，有点无奈，多加了一些状态：

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我的设计根据此状态转移图给出我的设计：

module top_module ( input clk, input reset,// Synchronous reset input s, input w, output z ); localparam A = 0, B = 1, S0 = 3, S1 = 4, S2 = 5, S3 = 6, S4 = 7, S5 = 8, S6 = 9, S7 = 10, S8 = 11, S9 = 12, S10 = 13, S11 = 14; reg [3:0] state, next_state; always@(*) begin case(state) A: begin if(s) next_state = B; else next_state = A; end B: begin if(w) next_state = S1; else next_state = S0; end S0: begin if(w) next_state = S2; else next_state = S4; end S1: begin if(w) next_state = S9; else next_state = S6; end S2: begin if(w) next_state = S3; else next_state = S5; end S3: begin if(w) next_state = S1; else next_state = S0; end S4: begin next_state = S5; end S5: begin if(w) next_state = S1; else next_state = S0; end S6: begin if(w) next_state = S7; else next_state = S8; end S7: begin if(w) next_state = S1; else next_state = S0; end S8: begin if(w) next_state = S1; else next_state = S0; end S9: begin if(w) next_state = S11; else next_state = S10; end S10: begin if(w) next_state = S1; else next_state = S0; end S11: begin if(w) next_state = S1; else next_state = S0; end default: begin next_state = A; endendcaseend always@(posedge clk) begin if(reset) state <= 0; else state <= next_state; end assign z = (state == S10 || state == S7 || state == S3) ? 1 : 0; endmodule

测试成功。
更新方案今天群里的大佬给了一种简单的方法，状态转移图确实简单了，但是理解起来呢？
我之前用的方案是在B之后的三个周期内，列举w的值，取值情况有8种，然后添加更多的状态去解决这个问题，不得不说状态转移图看起来复杂很多，也绝对不是推荐的方案。
我等待的确实是今天的这个方案，通过计数，不需要添加更多的状态，这也是我一开始就想用的方法，只是计数的时序当时没有搞定，今天很感谢，群里的一位兄弟。
直接给出设计，通过代码就应该能看出来设计的思路：

module top_module ( input clk, input reset,// Synchronous reset input s, input w, output z ); parameter A = 1'b0, B = 1'b1; reg current_state; reg next_state; always@(posedge clk)begin if(reset)begin current_state <= A; end else begin current_state <= next_state; end endalways@(*)begin case(current_state) A:begin next_state = s ? B : A; end B:begin next_state = B; end endcase endreg w_reg1; reg w_reg2; always@(posedge clk)begin if(reset)begin w_reg1 <= 1'b0; w_reg2 <= 1'b0; end else if(next_state == B)begin w_reg1 <= w; w_reg2 <= w_reg1; end else begin w_reg1 <= 1'b0; w_reg2 <= 1'b0; end endalways@(posedge clk)begin if(reset)begin z <= 1'b0; end else if(next_state == B && counter == 2'd0)begin if(~w & w_reg1 & w_reg2 | w & ~w_reg1 & w_reg2 | w & w_reg1 & ~w_reg2)begin z <= 1'b1; end else begin z <= 1'b0; end end else begin z <= 1'b0; end endreg [1:0] counter; always@(posedge clk)begin if(reset)begin counter <= 2'd0; end else if(counter == 2'd2)begin counter <= 2'd0; end else if(next_state == B)begin counter <= counter + 1'b1; end endendmodule

巧妙之处在于将输入w延迟两拍之后进行判断，如果有两个w为1，则在下一个周期将输出z置位.
有的朋友，也许在状态机的设计中，习惯将第三段使用组合逻辑来实现，这个题目的第三段也可以使用组合逻辑，但是呢？确实也没有必要，因为状态机的第三段本身既可以使用组合逻辑，也可以使用时序逻辑，如果使用时序逻辑。
下面给出组合逻辑的方案：

module top_module ( input clk, input reset,// Synchronous reset input s, input w, output reg z ); parameter A = 1'b0, B = 1'b1; reg current_state; reg next_state; always@(posedge clk)begin if(reset)begin current_state <= A; end else begin current_state <= next_state; end endalways@(*)begin case(current_state) A:begin next_state = s ? B : A; end B:begin next_state = B; end endcase endreg w_reg1; reg w_reg2; always@(posedge clk)begin if(reset)begin w_reg1 <= 1'b0; w_reg2 <= 1'b0; end else if(next_state == B)begin w_reg1 <= w; w_reg2 <= w_reg1; end else begin w_reg1 <= 1'b0; w_reg2 <= 1'b0; end endreg z_mid; always@(*)begin if(reset)begin z_mid <= 1'b0; end else if(current_state == B && counter == 2'd0)begin if(~w & w_reg1 & w_reg2 | w & ~w_reg1 & w_reg2 | w & w_reg1 & ~w_reg2)begin z_mid <= 1'b1; end else begin z_mid <= 1'b0; end end else begin z_mid <= 1'b0; end endalways@(posedge clk)begin if(reset)begin z <= 1'b0; end else begin z <= z_mid; end endreg [1:0] counter; always@(posedge clk)begin if(reset)begin counter <= 2'd0; end else if(counter == 2'd2)begin counter <= 2'd0; end else if(next_state == B)begin counter <= counter + 1'b1; end endendmodule

只需要将第三段改为组合逻辑，但是输出需要延迟一拍，为什么？看时序图。
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