Question

我们有二进制数，我们需要从给定数字生成2 1的组合。如果给出这样的2 1的组合，我们应该能够产生下一个组合。

实施例： -

Given vector   : 10101111 Given combination : 10100000 output       : 10001000
Given vector   : 10101111 Given combination : 10001000 output       : 10000100
Given vector   : 10101111 Given combination : 10000010 output       : 10000001
Given vector   : 10101111 Given combination : 10000001 output       : 00101000
Given vector   : 10101111 Given combination : 00101000 output       : 00100100

编辑：一旦第二个1到达给定二进制数中的最后一个，则第一个1递增（在二进制数中设置为下一个'1'，第二个'1'在第一个'1'之后变为'1' （如在例如4））

这是在硬件中完成的，所以它不应该在计算上复杂。我们如何用VHDL设计这个模块。

Answer 1

以下是一些可以完成工作的异步代码：

library IEEE;
use IEEE.STD_LOGIC_1164.ALL;

entity nex2ones is
    Port ( vector : in   STD_LOGIC_VECTOR (1 to 8);
           combo1 : in   STD_LOGIC_VECTOR (1 to 8);
           combo2 : out  STD_LOGIC_VECTOR (1 to 8);
           error  : out  STD_LOGIC);
end nex2ones;

architecture Behavioral of nex2ones is

type int_array_8 is array (1 to 8) of integer range 0 to 8;

begin

process (vector,combo1)

variable ones_ixs        : int_array_8; 
variable first_combo1_ix : integer range 0 to 8 := 0;
variable second_combo1_ix: integer range 0 to 8 := 0;
variable first_combo1_k  : integer range 0 to 9 := 0;
variable second_combo1_k : integer range 0 to 9 := 0;
variable k               : integer range 1 to 9;

begin

ones_ixs := (others => 0); -- indices of 1s in vector
combo2   <= (others => '0');
k := 1;
first_combo1_ix  := 0;
second_combo1_ix := 0;
first_combo1_k   := 0;  -- corresponding ptr to ones_ixs
second_combo1_k  := 0;
error            <= '0';

for j in 1 to 8 loop

  if combo1(j) = '1' then
      if first_combo1_ix = 0 then
        first_combo1_ix := j;   
        first_combo1_k := k;
      else
        second_combo1_ix := j;
        second_combo1_k := k;
      end if;
  end if;

  if vector(j) = '1' then
    ones_ixs(k) := j; 
    k := k + 1;
  end if;

end loop;  

if k > 1 then k := k - 1; end if; -- point to last nonzero index

if (first_combo1_ix = 0 or second_combo1_ix = 0)
  --or (first_combo1_ix = ones_ixs(k-1) and second_combo1_ix = ones_ixs(k)) 
  or (k < 2) then 
  error <= '1';
else -- no error proceed
  if second_combo1_ix = ones_ixs(k) then              -- can't slide 2nd anymore
    if (second_combo1_k - first_combo1_k) > 1 then    -- is 1st movable
        combo2(ones_ixs(first_combo1_k + 1)) <= '1';    -- move 1st
        if (second_combo1_k - first_combo1_k) > 2 then  -- is 2nd movable
          combo2(ones_ixs(first_combo1_k + 2)) <= '1'; -- move 2nd
        else
          combo2(ones_ixs(second_combo1_k)) <= '1';     -- leave 2nd be
        end if;
      else
        error <= '1'; -- no mas
      end if;
  else
    combo2(ones_ixs(first_combo1_k)) <= '1';      -- leave 1st be
      combo2(ones_ixs(second_combo1_k + 1)) <= '1'; -- next
  end if;
end if;

end process;

end Behavioral;

Testbench输出：

   ps       vector   combo1   combo2        
                                   error      
      0     00000000 00000000 00000000 1 
 100000     10101111 10100000 10001000 0 
 200000     10101111 10001000 10000100 0 
 300000     10101111 10000010 10000001 0 
 400000     10101111 10000001 00101000 0 
 500000     10101111 00101000 00100100 0 
 600000     10101111 00100100 00100010 0 
 700000     10101111 00000011 00000000 1 
 800000     11001110 00000110 00000000 1 
 900000     10001110 00001010 00000110 0 
1000000     11001110 00001010 00000110 0

在二进制数中找到2 1的组合

1 个答案: