• If a drone, or a "sub-completed" primary counter (who already counted to Cp, and cannot increment his 1st stage counter anymore) has to turn the light off at the end of stage 1, he will have to turn the light on whenever he has a chance during following occurrences of stage 1. This may result in a Drone having to turn the light on twice or more - it's OK.
  
• If a previously completed secondary counter leaves the light on
  at the end of stage 1, he will have to turn it on whenever he has a chance during stage 1.
  
• If anyone but the primary counter has to turn the light off at the end of stage 2, he will have to turn the light on whenever he has a chance during following occurrences of stage 2

  N   M   Mean Number of Days
10   10    3537
11   12    3544
 9   10    3537
12   16    3574
14   16    3633
 7    9    3654

  N   M   Mean Number of Days
11   12    3535.55
10   10    3537.20
 9   10    3537.21

/**      BanderSnatch - Jul 24th, 2002, 2:19am
  *
  * My first thought on this is rather simple, though in reality it would
  * (most likely) take far too long to be practical.  Starting the day of
  * the meeting, each prisoner starts counting the days.   As each prisoner
  * goes into the room, they leave the light in the off position.  However,
  * if they have been in the room once before in the hundred days, they
  * turn the light on.  Once the light is turned on, it stays on for the
  * remainder of the initial hundred days as a signal to the others that
  * someone has been in the room more than once, and hence someone will not
  * have been in the room at the end of the hundred days.  On the hundredth
  * day, the prisoner turns the light off, starting another hundred day
  * cycle.  If, on the last day of any hundred day cycle, the light remains
  * in the off position, it can be said with certainty that every prisoner
  * has been in the room, since no prisoner has been there twice.
  *
  * Though this solution would eventually work, probability says that it
  * would not work in a timely manner, since it goes in 100 day cycles,
  * rather than having the possibility of the hundred days starting at any
  * arbitrary time.
  */

package com.monahansen.prison;

public class BanderSnatchPrisoner extends ThinkingPrisoner
{
      long currentStage = -1;

      public BanderSnatchPrisoner(int id, int totalNumberOfPrisoners)
      {
            super(id, totalNumberOfPrisoners);
            this.currentStage = -1;
      }

      public boolean drawConclusion(IBooleanSwitch lightBulb, long currentDay)
      {
            boolean conclusion = false;

            do
            {
                  // if the day is 1, 101, 201, etc. we're into the next cycle - reset light
                  if ((currentDay % this.totalNumberOfPrisoners) == 1)
                  {
                        lightBulb.setState(false);
                  }

                  // if the light bulb is on, just leave the room - no point continuing with this stage
                  if (lightBulb.getState() == true)
                  {
                        conclusion = false;
                        break;
                  }

                  // if it is the 100th day in the cycle, and the light is off, declare freedom
                  if ((currentDay % this.totalNumberOfPrisoners) == 0)
                  {
                        conclusion = true;
                        break;
                  }

                  // if the current stage is equal to my recorded stage, then switch on the light and leave.
                  if (((long)(currentDay / this.totalNumberOfPrisoners)) == this.currentStage)
                  {
                        lightBulb.setState(true);
                        break;
                  }

                  // record visit in this stage
                  this.currentStage = (long)(currentDay / this.totalNumberOfPrisoners);

            } while (false);


            return conclusion;
      }
}

/** Captain Caveman - Jul 24th, 2002, 2:46am
  * Ok, here's 2 i snagged quickly from /. that I like.
  *
  * 1:  Designate 1 person to turn the light off.  Everyone else, on your
  * first time in the room, IF THE LIGHT IS OFF, turn it on.  (Otherwise,
  * leave it on.)  After the designated person turns it off 99 times (or
  * 100, if he turns it off for himself...), he can demand their freedom.
  *
  * This works even if they can't see the bulb from their cells and is
  * rather elegant.
  */

package com.monahansen.prison;

public class CaptainCavemanPrisoner extends LeaderElectingPrisoner
{
      public CaptainCavemanPrisoner(int id, int totalNumberOfPrisoners)
      {
            super(id, totalNumberOfPrisoners);

            if (this.id == 0)
            {
                  this.leader = true;
            }
            else
            {
                  this.leader = false;
            }
      }

      public boolean drawConclusion(IBooleanSwitch lightBulb, long currentDay)
      {
            boolean conclusion = false;

            do
            {
                  // initialise light bulb to off state on day 1
                  if (currentDay == 1)
                  {
                        lightBulb.setState(false);
                  }

                  // if the current prisoner is the leader (id=0), and the light is on, collect token and switch off light
                  if (this.leader)
                  {
                        if (lightBulb.getState() == true)
                        {
                              this.tokenBalance++;
                              lightBulb.setState(false);
                        }

                        // if the leader has all the tokens, then declare it.
                        if (this.tokenBalance == this.totalNumberOfPrisoners)
                        {
                              conclusion = true;
                        }

                        break;
                  }

                  // if current prisoner is not the leader, and the light is off, deposit token and switch on the light
                  if (this.tokenBalance == 1)
                  {
                        if (lightBulb.getState() == false)
                        {
                              this.tokenBalance--;
                              lightBulb.setState(true);
                        }
                        break;
                  }
            } while (false);

            return conclusion;
      }
}

/** sjbrown - Jul 25th, 2002, 10:09pm
  *
  * I think I've found another solution.  (the only other two so far were
  * the "Leader" solution and the one where it's off on the 101st (or
  * 201st or 301st ...) day. )
  *
  * 100-day Months Pretend every month has 100 days.  Each prisoner is
  * assigned a day of the month.  You only switch on the light if you're
  * in the room on your day, otherwise you switch the light off.  If the
  * light is on when you go in the room, you know prisoner d has been in
  * the room.  When you've seen the light on on *every* day of the month,
  * you know every prisoner has been in the room.
  *
  * badabing.DOT
  */

package com.monahansen.prison;

public class SJBrownPrisoner extends RegisterKeepingPrisoner
{
      public SJBrownPrisoner(int id, int totalNumberOfPrisoners)
      {
            super(id, totalNumberOfPrisoners);

            register[id] = true;
      }

      public boolean drawConclusion(IBooleanSwitch lightBulb, long currentDay)
      {
            boolean conclusion = false;

            do
            {
                  // initialise light bulb to off state on day 1
                  if (currentDay == 1)
                  {
                        lightBulb.setState(false);
                  }

                  // if the light is on, mark yesterday's prisoner off in the register and reset the light to off
                  if (lightBulb.getState() == true)
                  {
                        this.register[(int)((currentDay - 1) % totalNumberOfPrisoners)] = true;
                        lightBulb.setState(false);
                  }

                  // if the register is full, then everyone has visited.
                  if (this.registerFull())
                  {
                        conclusion = true;
                        break;
                  }

                  // if today represents a prisoner that this prisoner can claim has visited, turn on the light
                  if (this.isKnownToHaveVisited((int)(currentDay % totalNumberOfPrisoners)))
                  {
                        lightBulb.setState(true);
                        break;
                  }

            } while (false);

            return conclusion;
      }

      public boolean isKnownToHaveVisited(int prisonerId)
      {
            boolean result = false;

            if (this.id == prisonerId)
            {
                  result = true;
            }

            return result;
      }
}