\b;Exercise This exercise is very similar to the previous one. This time the bot should find its way alone from the start to the goal; you will have to execute the program only once. \b;Remark The labyrinth is not exactly the same, but this should be of no importance, as the program adapts to what it "sees". \image tlaby1 10 10; \b;General principle Use an infinite \c;\l;while\u cbot\while;\n; loop in order to execute the previous program several times: \s;\c;while ( true ) \s;{ \s; \n;If there is nothing in front, move forward\c; \s; \n;If there is nothing on your left hand, turn left\c; \s; \n;If there is nothing on your right hand, turn right\c; \s;} \n; Inside this \c;while\n; loop, replace the \c;return\n; instructions by \c;\l;continue\u cbot\continue;\n; instructions. \c;return\n; would quit the program, which is not what we want here. \c;continue\n; will just resume the execution at the beginning of the \c;\l;while\u cbot\while;\n; loop: \s;\c;if ( front == null ) \s;{ \s; move(5); \s; continue; \s;} \n; \b;Remember Here is again the program of the previous exercise : \c; \s;object front, left, right; \s; \s;front = radar(Barrier, 0, 45, 0, 5); \s;left = radar(Barrier, 90, 45, 0, 5); \s;right = radar(Barrier, -90, 45, 0, 5); \s; \s;if ( front == null ) \s;{ \s; move(5); \s; return; \s;} \s;if ( left == null ) \s;{ \s; turn(90); \s; move(5); \s; return; \s;} \s;if ( right == null ) \s;{ \s; turn(-90); \s; move(5); \s; return; \s;} \n; \b;Help If you need some help, just click on the hyperlinks of the instructions \c;\l;radar\u cbot\radar;\n;, \c;\l;if\u cbot\if;\n;, \c;\l;move\u cbot\move;\n; or \c;\l;turn\u cbot\turn;\n;. \t;See also \l;Programming\u cbot;, \l;types\u cbot\type; and \l;categories\u cbot\category;.