基于FPGA的并行排序算法

1.前言

        并行排序法的总体思路就是,把每个数都跟其他数比较一下,大于则得一分,然后根据分数大小将输入的数排序。 相较于其他排序法,并行排序只需四个周期就能给他排出来,不过会占用很多资源,用面积换速度

        八个16位数据并行排序RTL图如下:

基于FPGA的并行排序算法_第1张图片

        可以看出占用了很多资源(大约600多个ALM)。

  • 开发环境:Quartus Prime Standard 18.1
  • 操作系统:Windows 10 Pro 1809

2.第一个时钟周期

        例如待排序的有n个数,把每个数都跟其他数比较一下,大于则得一分,这样就会产生从0到(n-1)的分数。但如果有两个一样的数,这俩数的得分就会一样。所以需要设定一个规则来处理相同数的问题。我这里采用的方法是,原始顺序靠前的相同数得一分,原始顺序靠后的相同数不得分,即原始顺序考前的数与原始顺序靠后的数比较是采用大于等于号,而原始顺序靠后的数与原始顺序考前的数比较时采用大于号。代码如下:

always@(posedge clk or negedge rst_n)
begin
	if(!rst_n)
		begin
			a <= 7'b000_0000;
			b <= 7'b000_0000;
			c <= 7'b000_0000;
			d <= 7'b000_0000;
			e <= 7'b000_0000;
			f <= 7'b000_0000;
			g <= 7'b000_0000;
			h <= 7'b000_0000;

			add_start <= 0;
		end
		
		else
			begin
				if(in0 >= in1) a[0] <= 1'b1; //in0和所有除自己外的其他数据比较,大于则标志置1,否则为0
				else a[0] <= 1'b0;
				if(in0 >= in2) a[1] <= 1'b1;
				else a[1] <= 1'b0;
				if(in0 >= in3) a[2] <= 1'b1;
				else a[2] <= 1'b0;
				if(in0 >= in4) a[3] <= 1'b1;
				else a[3] <= 1'b0;
				if(in0 >= in5) a[4] <= 1'b1;
				else a[4] <= 1'b0;
				if(in0 >= in6) a[5] <= 1'b1;
				else a[5] <= 1'b0;
				if(in0 >= in7) a[6] <= 1'b1;
				else a[6] <= 1'b0;

						
				if(in1 > in0) b[0] <= 1'b1;//in1和所有除自己外的数据比较,大于标志位置1,否则为0
				else b[0] <= 1'b0;
				if(in1 >= in2) b[1] <= 1'b1;
				else b[1] <= 1'b0;
				if(in1 >= in3) b[2] <= 1'b1;
				else b[2] <= 1'b0;
				if(in1 >= in4) b[3] <= 1'b1;
				else b[3] <= 1'b0;
				if(in1 >= in5) b[4] <= 1'b1;
				else b[4] <= 1'b0;
				if(in1 >= in6) b[5] <= 1'b1;
				else b[5] <= 1'b0;
				if(in1 >= in7) b[6] <= 1'b1;
				else b[6] <= 1'b0;

						
				if(in2 > in0) c[0] <= 1'b1;
				else c[0] <= 1'b0;
				if(in2 > in1) c[1] <= 1'b1;
				else c[1] <= 1'b0;
				if(in2 >= in3) c[2] <= 1'b1;
				else c[2] <= 1'b0;
				if(in2 >= in4) c[3] <= 1'b1;
				else c[3] <= 1'b0;
				if(in2 >= in5) c[4] <= 1'b1;
				else c[4] <= 1'b0;
				if(in2 >= in6) c[5] <= 1'b1;
				else c[5] <= 1'b0;
				if(in2 >= in7) c[6] <= 1'b1;
				else c[6] <= 1'b0;

						
				if(in3 > in0) d[0] <= 1'b1;
				else d[0] <= 1'b0;
				if(in3 > in1) d[1] <= 1'b1;
				else d[1] <= 1'b0;
				if(in3 > in2) d[2] <= 1'b1;
				else d[2] <= 1'b0;
				if(in3 >= in4) d[3] <= 1'b1;
				else d[3] <= 1'b0;
				if(in3 >= in5) d[4] <= 1'b1;
				else d[4] <= 1'b0;
				if(in3 >= in6) d[5] <= 1'b1;
				else d[5] <= 1'b0;
				if(in3 >= in7) d[6] <= 1'b1;
				else d[6] <= 1'b0;

						
				if(in4 > in0) e[0] <= 1'b1;
				else e[0] <= 1'b0;
				if(in4 > in1) e[1] <= 1'b1;
				else e[1] <= 1'b0;
				if(in4 > in2) e[2] <= 1'b1;
				else e[2] <= 1'b0;
				if(in4 > in3) e[3] <= 1'b1;
				else e[3] <= 1'b0;
				if(in4 >= in5) e[4] <= 1'b1;
				else e[4] <= 1'b0;
				if(in4 >= in6) e[5] <= 1'b1;
				else e[5] <= 1'b0;
				if(in4 >= in7) e[6] <= 1'b1;
				else e[6] <= 1'b0;

						
				if(in5 > in0) f[0] <= 1'b1;
				else f[0] <= 1'b0;
				if(in5 > in1) f[1] <= 1'b1;
				else f[1] <= 1'b0;
				if(in5 > in2) f[2] <= 1'b1;
				else f[2] <= 1'b0;
				if(in5 > in3) f[3] <= 1'b1;
				else f[3] <= 1'b0;
				if(in5 > in4) f[4] <= 1'b1;
				else f[4] <= 1'b0;
				if(in5 >= in6) f[5] <= 1'b1;
				else f[5] <= 1'b0;
				if(in5 >= in7) f[6] <= 1'b1;
				else f[6] <= 1'b0;

						
				if(in6 > in0) g[0] <= 1'b1;
				else g[0] <= 1'b0;
				if(in6 > in1) g[1] <= 1'b1;
				else g[1] <= 1'b0;
				if(in6 > in2) g[2] <= 1'b1;
				else g[2] <= 1'b0;
				if(in6 > in3) g[3] <= 1'b1;
				else g[3] <= 1'b0;
				if(in6 > in4) g[4] <= 1'b1;
				else g[4] <= 1'b0;
				if(in6 > in5) g[5] <= 1'b1;
				else g[5] <= 1'b0;
				if(in6 >= in7) g[6] <= 1'b1;
				else g[6] <= 1'b0;

						
				if(in7 > in0) h[0] <= 1'b1;
				else h[0] <= 1'b0;
				if(in7 > in1) h[1] <= 1'b1;
				else h[1] <= 1'b0;
				if(in7 > in2) h[2] <= 1'b1;
				else h[2] <= 1'b0;
				if(in7 > in3) h[3] <= 1'b1;
				else h[3] <= 1'b0;
				if(in7 > in4) h[4] <= 1'b1;
				else h[4] <= 1'b0;
				if(in7 > in5) h[5] <= 1'b1;
				else h[5] <= 1'b0;
				if(in7 > in6) h[6] <= 1'b1;
				else h[6] <= 1'b0;
				
						
				add_start <= 1; //开始相加标志
			end
end

        不过这样只能解决两个相同数的排序,三个就尴尬了……

        最后比较完成会有一个开始相加的标志位置1。


3.第二个时钟周期

        把上一周期每个数的得分总数分别求出来,装进寄存器。代码如下:

always@(posedge clk or negedge rst_n)
begin
	if(!rst_n)
		begin
			mid0 <= 7'b000_0000;
			mid1 <= 7'b000_0000;
			mid2 <= 7'b000_0000;
			mid3 <= 7'b000_0000;
			mid4 <= 7'b000_0000;
			mid5 <= 7'b000_0000;
			mid6 <= 7'b000_0000;
			mid7 <= 7'b000_0000;
		end
		
	else if(add_start == 1)
		begin
			mid0 <= a[0] + a[1] + a[2] + a[3] + a[4] + a[5] + a[6]; //标志位相加,所得结果就是其所在位置
			mid1 <= b[0] + b[1] + b[2] + b[3] + b[4] + b[5] + b[6];
			mid2 <= c[0] + c[1] + c[2] + c[3] + c[4] + c[5] + c[6];
			mid3 <= d[0] + d[1] + d[2] + d[3] + d[4] + d[5] + d[6];
			mid4 <= e[0] + e[1] + e[2] + e[3] + e[4] + e[5] + e[6];
			mid5 <= f[0] + f[1] + f[2] + f[3] + f[4] + f[5] + f[6];
			mid6 <= g[0] + g[1] + g[2] + g[3] + g[4] + g[5] + g[6];
			mid7 <= h[0] + h[1] + h[2] + h[3] + h[4] + h[5] + h[6];
		end
		
	assignm_start <= 1;//相加结束,赋值开始标志
end

4.第三个时钟周期

        将输入的数按照得分高低依次装进输出暂存寄存器,代码如下:

always@(posedge clk or negedge rst_n)
begin
	if(!rst_n)
		begin
			out_temp[0]  <= 16'b00000000_00000000;
			out_temp[1]  <= 16'b00000000_00000000;
			out_temp[2]  <= 16'b00000000_00000000;
			out_temp[3]  <= 16'b00000000_00000000;
			out_temp[4]  <= 16'b00000000_00000000;
			out_temp[5]  <= 16'b00000000_00000000;
			out_temp[6]  <= 16'b00000000_00000000;
			out_temp[7]  <= 16'b00000000_00000000;
			out_start <= 0;
		end
			
	else if(assignm_start == 1)
		begin
			out_temp[mid0] <= in0;
			out_temp[mid1] <= in1;
			out_temp[mid2] <= in2;
			out_temp[mid3] <= in3;
			out_temp[mid4] <= in4;
			out_temp[mid5] <= in5;
			out_temp[mid6] <= in6;
			out_temp[mid7] <= in7;

					
			out_start <= 1;//赋值结束,输出开始标志位
		end
end

5.第四个时钟周期

        按照输出暂存寄存器的顺序把数值输出,代码如下:

always@(posedge clk or negedge rst_n)
begin
	if(!rst_n)
		begin
			out0 <= 16'b00000000_00000000;
			out1 <= 16'b00000000_00000000;
			out2 <= 16'b00000000_00000000;
			out3 <= 16'b00000000_00000000;
			out4 <= 16'b00000000_00000000;
			out5 <= 16'b00000000_00000000;
			out6 <= 16'b00000000_00000000;
			out7 <= 16'b00000000_00000000;
			
			complete <= 0;
		end
		
	else if(out_start == 1)
		begin
			out0 <= out_temp[0];
			out1 <= out_temp[1];
			out2 <= out_temp[2];
			out3 <= out_temp[3];
			out4 <= out_temp[4];
			out5 <= out_temp[5];
			out6 <= out_temp[6];
			out7 <= out_temp[7];
			
			complete <= 1;

		end	
			
end

6.仿真

        用Modelsim仿真一下。

基于FPGA的并行排序算法_第2张图片

        很明显经过四个周期后输入数据从小到大并行输出,而且相同的两个数也能都被输出。


        告辞。


原文地址:基于FPGA的并行排序算法
源代码:https://github.com/Verdvana/Sort_Paralell

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