Stanford编程方法学公开课作业 1 ----关于Karel 中点寻找问题的讨论 （Midpoint finding）
问题描述：Karel需要寻找到第一行的中点位置，并在那里放下一个beeper，具体描述请参考英文原文
Problem 3
As an exercise in solving algorithmic problems, program Karel to place a single beeper at the center of 1st Street. For example, if Karel starts in the world
it should end with Karel standing on a beeper in the following position:
Note that the final configuration of the world should have only a single beeper at the midpoint of 1st Street. Along the way, Karel is allowed to place additional beepers wherever it wants to, but must pick them all up again before it finishes.
In solving this problem, you may count on the following facts about the world:
- Karel starts at 1st Avenue and 1st Street, facing east, with an infinite number of beepers in its bag.
- The initial state of the world includes no interior walls or beepers.
- The world need not be square, but you may assume that it is at least as tall as it is wide.
Your program, moreover, can assume the following simplifications:
- If the width of the world is odd, Karel must put the beeper in the center square. If the width is even, Karel may drop the beeper on either of the two center squares.
- It does not matter which direction Karel is facing at the end of the run.
There are many different algorithms you can use to solve this problem. The interesting part of this assignment is to come up with a strategy that works.
解决思路：（可以提出以下四种解题思路）
1. 先将第一行填满，然后左右各取出一个，并在第二行放下一个，当第一行beeper取完的时候，第二行的beeper数目即为第一行长度的一半
代码如下：
/* * File: MidpointFindingKarel.java * ------------------------------- * THe MidpointFindingKarel class will leave a beeper on the corner * closest to the center of 1st Street * (or either of the two central corners if 1st Street has an even * number of corners). The world may be of any size, but you are allowed to * assume that it is at least as tall as it is wide. */import stanford.karel.*;public class MidpointFindingKarel extends SuperKarel { public void run() { fillFirstRow(); //midpoint finding part while (beepersPresent()) { collectLeftEnd(); secondRowAddOne(); collectRightEnd(); //step forward one more to make sure there is no more beeper if (frontIsClear()) { //check whether all beepers have been collected move(); //if width is odd number Karel will finally arrive (1,1) } //if width is even number, Karel will finally arrive mid-point } //go to the midpoint of the first row, put beeper then collect beepers //on the 2nd row goToMidpoint(); putBeeper(); collectSecondRow(); goToOrigin(); moveToBeeper(); } /* * fillFirstRow method is defined to fill the 1st row with beepers * Karel will put only one beeper at each corner * precondition: facing east, at (1,1) point * postcondition: facing east, (1,1) point */ private void fillFirstRow() { goToOrigin(); //go to (1,1) facing east while (frontIsClear()) { putBeeper(); move(); } putBeeper(); //Off-By-One-Bug //make sure go back to (1,1) point facing east turnAround(); moveToWall(); turnAround(); } /* * Collect the most left beeper on the 1st row, if there is any * Precondition: any case * Postcondition: facing east, stop at the position where Karel * just collected one beeper, 1st row */ private void collectLeftEnd() { goToOrigin(); moveToBeeper(); if (beepersPresent()) { pickBeeper(); } } /*Similar as collectLeftEnd, while in this method, Karel will start from the most right end, and try to collect the beeper at the most right end */ private void collectRightEnd() { goToOrigin(); moveToWall(); //Reach the right end of the 1st row turnAround(); //To face west moveToBeeper(); if (beepersPresent()) { pickBeeper(); } } /* * go to 2nd row, add one more beeper * precondition: anywhere at 1st row facing east * postcondition: on 2nd row, at the most end where no more beeper exist */ private void secondRowAddOne() { goToOrigin(); bigTurnLeft(); turnRight(); while( beepersPresent()) { //arrive at the end of existing beepers move(); } putBeeper(); } /* * Make sure Karel will go to the midpoint of the first row * Precondition: any case * Postcondition: arrive at the mid point of the 1st row, facing south */ private void goToMidpoint() { goToOrigin(); moveToWall(); //reach right end bigTurnLeft(); turnLeft(); moveToBeeper(); bigTurnLeft(); } private void collectSecondRow() { goToOrigin(); bigTurnLeft(); //facing north turnRight(); //facing east while (beepersPresent() || frontIsBlocked()) { pickBeeper(); move(); } } /*Move Karel back to origin of map (1,1), make it facing east * Precondition: can be any case * Postcondition: standing on (1,1) facing east */ private void goToOrigin() { while (notFacingSouth()) { turnLeft(); } moveToWall(); while (notFacingWest()) { turnLeft(); } moveToWall(); while (notFacingEast()) { turnLeft(); //Karel will face east } } /* * move Karel to first beeper on its way * precondition: facing east/west * postcondition: share same orientation as in precondition, standing on * the corner where first beeper appears on its way */ private void moveToBeeper() { while ( noBeepersPresent() && frontIsClear() ) { move(); } } /* * go to upper/lower row above without changing column number, * make a big turn left * precondition: any case * postcondition: same column, arrive at upper/lower row * which is one above/under the row in precondition, * direction will turn left */ private void bigTurnLeft() { turnLeft(); move(); } // Similar as in bigTurnLeft, but in this case make a big //turn right private void bigTurnRight() { turnRight(); move(); } private void moveToWall() { //move Karel to wall while (frontIsClear()) { move(); } } }// end of public class
2. 对于方法1中的思路进行进一步优化，其实左侧取一个右侧再取一个完全可以兼并成为从左侧出发一次取两个，同时放一个到第二行，这将大量节省Karel的运动过程，优化后代码如下：
/* * File: MidpointFindingKarel.java * ------------------------------- * THe MidpointFindingKarel class will leave a beeper on the corner * closest to the center of 1st Street * (or either of the two central corners if 1st Street has an even * number of corners). The world may be of any size, but you are allowed to * assume that it is at least as tall as it is wide. */import stanford.karel.*;public class MidpointFindingKarel extends SuperKarel { public void run() { fillFirstRow(); //midpoint finding part while (beepersPresent()) { secondRowAddOne(); collectLeftEnd(); //step forward one more to make sure there is no more beeper if (frontIsClear()) { //check whether all beepers have been collected move(); //if width is odd number Karel will finally arrive (1,1) } //if width is even number, Karel will finally arrive mid-point } //go to the midpoint of the first row, put beeper then collect beepers //on the 2nd row goToMidpoint(); putBeeper(); collectSecondRow(); goToOrigin(); moveToBeeper(); } /* * fillFirstRow method is defined to fill the 1st row with beepers * Karel will put only one beeper at each corner * precondition: facing east, at (1,1) point * postcondition: facing east, at the most right end of 1st row */ private void fillFirstRow() { goToOrigin(); //go to (1,1) facing east while (frontIsClear()) { putBeeper(); move(); } putBeeper(); //Off-By-One-Bug } /* * Collect the most left beeper on the 1st row, if there is any * Precondition: any case * Postcondition: facing east, stop at the position where Karel * just collected one beeper, 1st row */ private void collectLeftEnd() { goToOrigin(); moveToBeeper(); if ( beepersPresent() ) { pickBeeper(); } if ( frontIsClear() ) { move(); } if ( beepersPresent() ) { pickBeeper(); } } /* * go to 2nd row, add one more beeper * precondition: anywhere at 1st row facing east * postcondition: on 2nd row, at the most end where no more beeper exist */ private void secondRowAddOne() { goToOrigin(); bigTurnLeft(); turnRight(); while( beepersPresent()) { //arrive at the end of existing beepers move(); } putBeeper(); } /* * Make sure Karel will go to the midpoint of the first row * Precondition: any case * Postcondition: arrive at the mid point of the 1st row, facing south */ private void goToMidpoint() { goToOrigin(); moveToWall(); //reach right end bigTurnLeft(); turnLeft(); moveToBeeper(); bigTurnLeft(); } private void collectSecondRow() { goToOrigin(); bigTurnLeft(); //facing north turnRight(); //facing east while (beepersPresent() || frontIsBlocked()) { pickBeeper(); move(); } } /*Move Karel back to origin of map (1,1), make it facing east * Precondition: can be any case * Postcondition: standing on (1,1) facing east */ private void goToOrigin() { while (notFacingSouth()) { turnLeft(); } moveToWall(); while (notFacingWest()) { turnLeft(); } moveToWall(); while (notFacingEast()) { turnLeft(); //Karel will face east } } /* * move Karel to first beeper on its way * precondition: facing east/west * postcondition: share same orientation as in precondition, standing on * the corner where first beeper appears on its way */ private void moveToBeeper() { while ( noBeepersPresent() && frontIsClear() ) { move(); } } /* * go to upper/lower row above without changing column number, * make a big turn left * precondition: any case * postcondition: same column, arrive at upper/lower row * which is one above/under the row in precondition, * direction will turn left */ private void bigTurnLeft() { turnLeft(); move(); } private void moveToWall() { //move Karel to wall while (frontIsClear()) { move(); } } }// end of public class
3. 对于方法2中的思路进行进一步优化，其实karel完全可以将用于中点位置计数的beeper直接放在第一行，这样Karel的运动过程将会得到进一步的简化，换言之，Karel将从（1,1）出发，一次取两个beeper，然后折回（1,1），放下一个beeper，之后再去取两个beeper，折回在（1,2）放下一个beeper，以此类推，直至开始时放置的beeper全部取完。具体代码如下：
/* * File: MidpointFindingKarel.java * ------------------------------- * THe MidpointFindingKarel class will leave a beeper on the corner * closest to the center of 1st Street * (or either of the two central corners if 1st Street has an even * number of corners). The world may be of any size, but you are allowed to * assume that it is at least as tall as it is wide. */import stanford.karel.*;public class MidpointFindingKarel extends SuperKarel { public void run() { fillFirstRow(); goToOrigin(); //midpoint finding part while (beepersPresent()) { collectLeftEnd(); firstRowAddOne(); move(); //go to blank place without any beeper moveToBeeper(); } //go to the midpoint of the first row, put beeper then collect beepers //on the 2nd row goToMidpoint(); } /* * fillFirstRow method is defined to fill the 1st row with beepers * Karel will put only one beeper at each corner * precondition: facing east, at (1,1) point * postcondition: facing east, at the most right end of 1st row */ private void fillFirstRow() { goToOrigin(); //go to (1,1) facing east while (frontIsClear()) { putBeeper(); move(); } putBeeper(); //Off-By-One-Bug } /* * Collect the most left beeper on the 1st row, if there is any * Precondition: any case * Postcondition: facing east, stop at the position where Karel * just collected one beeper, 1st row */ private void collectLeftEnd() { if ( beepersPresent() ) { pickBeeper(); } if ( frontIsClear() ) { move(); } if ( beepersPresent() ) { pickBeeper(); } } /* * go to 2nd row, add one more beeper * precondition: anywhere at 1st row facing east * postcondition: on 2nd row, at the most end where no more beeper exist */ private void firstRowAddOne() { goToOrigin(); while( beepersPresent()) { //arrive at the end of existing beepers move(); } putBeeper(); } /* * Make sure Karel will go to the midpoint of the first row * Precondition: any case * Postcondition: arrive at the mid point of the 1st row, facing west */ private void goToMidpoint() { goToOrigin(); while ( beepersPresent() ) { pickBeeper(); move(); } turnAround(); //OBOB, Karel will move one more step move(); //at the end of this part putBeeper(); } /*Move Karel back to origin of map (1,1), make it facing east * Precondition: can be any case * Postcondition: standing on (1,1) facing east */ private void goToOrigin() { while (notFacingSouth()) { turnLeft(); } moveToWall(); while (notFacingWest()) { turnLeft(); } moveToWall(); while (notFacingEast()) { turnLeft(); //Karel will face east } } /* * move Karel to first beeper on its way * precondition: facing east/west * postcondition: share same orientation as in precondition, standing on * the corner where first beeper appears on its way */ private void moveToBeeper() { while ( noBeepersPresent() && frontIsClear() ) { move(); } } /* * go to upper/lower row above without changing column number, * make a big turn left * precondition: any case * postcondition: same column, arrive at upper/lower row * which is one above/under the row in precondition, * direction will turn left */ private void bigTurnLeft() { turnLeft(); move(); } private void moveToWall() { //move Karel to wall while (frontIsClear()) { move(); } } }// end of public class
4. 对于方案1还有另一种优化的思路，那就是摆脱计数器的限制，左右各取出一个，一直这样取下去，直至取完，那么Karel此时必然自然位于中点位置从而无需计数，其运动过程将更为简单快捷。编程的要点在于要保证Karel始终执行往复运动直至beeper被全部取尽，代码如下：
/* * File: MidpointFindingKarel.java * ------------------------------- * THe MidpointFindingKarel class will leave a beeper on the corner * closest to the center of 1st Street * (or either of the two central corners if 1st Street has an even * number of corners). The world may be of any size, but you are allowed to * assume that it is at least as tall as it is wide. */import stanford.karel.*;public class MidpointFindingKarel extends SuperKarel { public void run() { fillFirstRow(); goToOrigin(); //midpoint finding part while (beepersPresent()) { pickBeeper(); move(); moveToEnd(); } putBeeper(); } /* * fillFirstRow method is defined to fill the 1st row with beepers * Karel will put only one beeper at each corner * precondition: facing east, at (1,1) point * postcondition: facing east, at the end of 1st row */ private void fillFirstRow() { goToOrigin(); //go to (1,1) facing east while (frontIsClear()) { putBeeper(); move(); } putBeeper(); //Off-By-One-Bug } /* * Karel will move to the last beeper of the beeper chain * Precondition: standing at one end facing that chain * Postcondition: standing at another end facing that chain * which means the orientation is changed oppositely */ private void moveToEnd() { while ( beepersPresent() && frontIsClear() ) { move(); } if ( frontIsBlocked() && beepersPresent() ) { //for the case arrive at the wall turnAround(); } else { turnAround(); move(); } } /*Move Karel back to origin of map (1,1), make it facing east * Precondition: can be any case * Postcondition: standing on (1,1) facing east */ private void goToOrigin() { while (notFacingSouth()) { turnLeft(); } moveToWall(); while (notFacingWest()) { turnLeft(); } moveToWall(); while (notFacingEast()) { turnLeft(); //Karel will face east } } private void moveToWall() { //move Karel to wall while (frontIsClear()) { move(); } } }// end of public class
讨论：
从思想性上，方案4远优于其他方案，而且实现过程简洁漂亮；但就扩展简易性而言不如方案2、3，方案2，3稍作修改就可以实现在1/3，1/4... 处放置一个beeper，而方案4的扩展性相对而言不是特别明显（以1/3为例，需要左侧取一个，右侧取两个）