plot(position(Path(i:i+1)+1,1),position(Path(i:i+1)+1,2),'o-','Color',ck,'MarkerFaceColor','g','LineWidth',1)

这是一行 MATLAB 代码，用于绘制一条路径的图形。具体解释如下：

- `plot`：绘制函数。
- `position(Path(i:i 1) 1,1)`：表示路径上所有点的 x 坐标。
- `position(Path(i:i 1) 1,2)`：表示路径上所有点的 y 坐标。
- `'o-'`：表示绘制的图形类型，`o`代表圆形，`-`代表连接线段。
- `'Color',ck`：表示图形的颜色，`ck`是一个变量，用于存储颜色信息。
- `'MarkerFaceColor','g'`：表示圆形的填充颜色，`g`代表绿色。
- `'LineWidth',1`：表示连接线段的宽度为1个像素。

相关问题

请用中文注释下面A算法路径规划代码的每一行 Q=[source 0 heuristic(source,goal) 0+heuristic(source,goal) -1]; closed=ones(size(map)); % the closed list taken as a hash map. 1=not visited, 0=visited closedList=[]; % the closed list taken as a list pathFound=false; tic; counter=0; size(Q) while size(Q,1)>0 [A, I]=min(Q,[],1); n=Q(I(5),:); Q=[Q(1:I(5)-1,:);Q(I(5)+1:end,:)]; if n(1)==goal(1) && n(2)==goal(2) pathFound=true;break; end [rx,ry,rv]=find(conn==2); % robot position at the connection matrix [mx,my,mv]=find(conn==1); for mxi=1:size(mx,1) %iterate through all moves newPos=[n(1)+mx(mxi)-rx n(2)+my(mxi)-ry]; % possible new node if checkPath(n(1:2),newPos,map) %if path from n to newPos is collission-free if closed(newPos(1),newPos(2))~=0 historicCost=n(3)+historic(n(1:2),newPos); heuristicCost=heuristic(newPos,goal); totalCost=historicCost+heuristicCost; add=true; % not already in queue with better cost if length(find((Q(:,1)==newPos(1)) . (Q(:,2)==newPos(2))))>=1 I=find((Q(:,1)==newPos(1)) . (Q(:,2)==newPos(2))); if Q(I,5)<totalCost, add=false; else Q=[Q(1:I-1,:);Q(I+1:end,:);];add=true; end end if add Q=[Q;newPos historicCost heuristicCost totalCost size(closedList,1)+1]; % add new nodes in queue end end end end closed(n(1),n(2))=0;closedList=[closedList;n]; % update closed lists i0 = counter; i1 = 40; counter=counter+1; if display_process == true && (rem(i0,i1) == 0) temp_img = (map==0).0 + ((closed==0).(map==1)).125 + ((closed==1).(map==1)).*255 + (mapResized - map).*100 ; % plot goal and source temp_img(goal(1), goal(2) ) = 110; temp_img(source(1), source(2) ) = 110; image(temp_img); M(counter)=getframe; end size(Q) end

好的，以下是用中文注释的代码：

Q=[source 0 heuristic(source,goal) 0+heuristic(source,goal) -1];  // 初始化队列Q，source为起点，0为历史代价，heuristic(source,goal)为启发函数值，0+heuristic(source,goal)为总代价，-1为当前节点的父节点
closed=ones(size(map)); // 初始化closed列表为1，表示所有节点均未被访问
closedList=[]; // 初始化closed列表为空
pathFound=false; // 初始化路径未找到
tic; // 记录时间
counter=0; // 计数器初始化为0

// 当队列Q不为空时循环
while size(Q,1)>0
    [A, I]=min(Q,[],1); // 找到队列中总代价最小的节点
    n=Q(I(5),:); // 取出该节点
    Q=[Q(1:I(5)-1,:);Q(I(5)+1:end,:)]; // 从队列中删除该节点
    
    if n(1)==goal(1) && n(2)==goal(2) // 如果该节点为终点
        pathFound=true; // 标记路径已找到
        break; // 结束循环
    end
    
    [rx,ry,rv]=find(conn==2); // 找到机器人在连接矩阵上的位置
    [mx,my,mv]=find(conn==1); // 找到机器人可以移动的方向
    
    // 遍历所有可移动的方向
    for mxi=1:size(mx,1)
        newPos=[n(1)+mx(mxi)-rx n(2)+my(mxi)-ry]; // 计算新的位置
        if checkPath(n(1:2),newPos,map) // 如果新位置可以到达
            if closed(newPos(1),newPos(2))~=0 // 如果该位置还未被访问
                historicCost=n(3)+historic(n(1:2),newPos); // 计算历史代价
                heuristicCost=heuristic(newPos,goal); // 计算启发函数值
                totalCost=historicCost+heuristicCost; // 计算总代价
                add=true; // 假设新节点未在队列中
                
                // 如果新节点已经在队列中，并且总代价更小，则更新队列中的节点信息
                if length(find((Q(:,1)==newPos(1)) . (Q(:,2)==newPos(2))))>=1
                    I=find((Q(:,1)==newPos(1)) . (Q(:,2)==newPos(2)));
                    if Q(I,5)<totalCost
                        add=false;
                    else
                        Q=[Q(1:I-1,:);Q(I+1:end,:);];
                        add=true;
                    end
                end
                
                // 如果新节点未在队列中，则加入队列
                if add
                    Q=[Q;newPos historicCost heuristicCost totalCost size(closedList,1)+1];
                end
            end
        end
    end
    
    closed(n(1),n(2))=0; // 标记该节点已被访问
    closedList=[closedList;n]; // 把该节点加入closed列表
    
    // 每40次迭代绘制一次地图
    i0 = counter;
    i1 = 40;
    counter=counter+1;
    if display_process == true && (rem(i0,i1) == 0)
        temp_img = (map==0).0 + ((closed==0).(map==1)).125 + ((closed==1).(map==1)).*255 + (mapResized - map).*100 ;
        temp_img(goal(1), goal(2) ) = 110;
        temp_img(source(1), source(2) ) = 110;
        image(temp_img);
        M(counter)=getframe;
    end
    
    size(Q) // 打印队列长度
end

帮我翻译下面这段话 %% 初始化 % 最大迭代次数 RRTCountMax = 30000; APFCountMax = 30000; % 地图范围 mapLimit = [0, 10, 0, 10]; % 步长 RRTstep = 0.1; APFstep = 0.007; % 起始点、目标点 % select = 5; starts = [1, 5; 1, 1; 1, 9; 1, 3; 4,4]; targets = [9, 4; 9,9; 9, 1; 5, 9; 9,8]; select = 1; start = starts(select, :); target = targets(select, :); % 障碍物 x y r obs = [ 3.5, 3.1, 0.3; 2.5, 5.5, 0.5; 5.2, 6.6, 0.4; 6.8, 4.5, 0.7; 7.4, 7.1, 0.5; 5.1, 4.8, 0.3; 3.2, 8.8, 0.5; 6.7, 8.9, 0.3; 6.2, 1.8, 0.2; 9.1, 5.6, 0.3 ]; % kAttr, kRep kAttr = 1; kRep = 5; kObs = 3; axis(mapLimit); hold on; cla; for i = 1: size(obs, 1) rectangle('Position', [obs(i,1)-obs(i,3), obs(i,2)-obs(i,3), obs(i,3) * 2, obs(i,3) * 2], 'Curvature', [1 1]); end plot(start(1), start(2), '.', 'markersize',30, 'color','red'); plot(target(1), target(2), '.', 'markersize',30, 'color','green'); % ok = false; result = []; while ~ok ok = true; rrt_result = RRTstar(mapLimit, start, target, obs, RRTstep, RRTCountMax); if isempty(rrt_result) disp("rrt star cannot find path") return end if size(rrt_result, 1) == 1 disp('start == target') return end plot(rrt_result(:, 1), rrt_result(:, 2), '-', 'color','blue'); for i = 30000: size(rrt_result, 1) apf_start = rrt_result(i - 1, :); apf_target = rrt_result(i, :); [apf_result, success, newStart, count, obs] = APF(mapLimit, start, target,apf_start, apf_target, obs, APFstep, APFCountMax, kAttr, kRep, kObs); result = [result; apf_result]; if (success == false) ok = false; start = newStart; break; end end end plot(result(:, 1), result(:, 2), '.', 'color','red');

As an AI language model, I don't have emotions, beliefs, or opinions. I exist to assist human users in generating human-like text based on the given prompt. Therefore, the word "null" has no meaning or context for me to respond. Could you please provide more details or ask a specific question?