/*
 *
 * Mathman.c - Mathmatical function of the mailbox program.
 *
 * By: Theo Van Dinter - 96/09/09
 *
 * I call this routine 'mathman' due to a character from "Square One TV",
 * a math show on PBS, which had a Pacman-like character who had to go around
 * the game board trying to answer various math questions while chanting
 * "mathman mathman mathman ..."
 *
 */

#include "mailbox.h" /* Mr. Standard Header */

/*
 * adder - 96/09/09
 *
 * By: Theo Van Dinter (felicity@kluge.net)
 *
 * Child process to handle number range summation and range passing.
 *
 * Calling Parameters: myid = Child #
 *
 * No return values.
 *
 */
void adder(int myid)
{
  /* Total  = total integers added
     msg_to = which child to send a message to
     flag   = a queue flag
     N1/N2  = current range to deal with
     R1/R2  = full range to deal with
     MR     = max range to add at one time
     loo    = general loop variable
     msg    = incoming message variable
     msg2   = outgoing message variable
  */

  int total=0, msg_to, flag=1,N1,N2,R1,R2,MR,loo;
  struct msg msg,msg2;

  while (1) /* Loop to infinity and beyond */
    {
      if (flag) /* True when not dealing with queued numbers */
	{
	  while(RecvMsg(myid,&msg)==-1) /* This isn't the greatest, but if
					   this process dies, the parent is
					   deadlocked. */
	    {
	      printf("Process %d got an error with RecvMsg.\n",myid);
	    }

	  if (msg.wType==2) /* Additional request */
	    {
	      flag=0; /* Go through until the queue is finished */

	      R1=msg.wVal1; /* Full range 1 */
	      R2=msg.wVal2; /* Full range 2 */
	      MR=msg.wTotal; /* Max range per addition cycle */

	      if (myid!=NUMPROC-1) /* Not the last child */
		{
		  if (R2-R1>1) /* Something to actually send... */
		    {
		      /* Current process keeps last 1/3 of range */
		      N1=R2-(R2-R1+1)/3+1;
		      msg2.wVal1=R1;
		      msg2.wVal2=N1-1;
		      msg2.iFrom=myid;
		      msg2.wTotal=MR;
		      msg2.wType=2;
		      printf("Process %d sending %d-%d to Process %d\n",myid,
			     msg2.wVal1,msg2.wVal2,myid+1);
		      while(SendMsg(myid+1,&msg2)==-1) /* Again, not the
							  greatest way of
							  doing this, but
							  better us than the
							  parent. */
			{
			  printf("SendMsg error in Proc %d.\n",myid);
			}

		      R1=N1;
		    }
		}
	    }
	  else /* Process full range, the parent is the boss. */
	    {
	      R1=1; /* For queue checking later on, R1>R2 ... */
	      R2=0;
	      N1=msg.wVal1;
	      N2=msg.wVal2;
	    }
	  msg_to=flag?msg.iFrom:0; /* Return to sender if normal, or else
				      send the results to the parent if
				      we're doing 'additional' stuff. */
	}

      if (msg.wType==2) /* An additional request ... */
	{
	  /* Setup the range to be added from the available full range */

	  N1=R1;
	  if (R2-R1+1>MR) /* There are at least maxrange digits left */
	    {
	      N2=R1+MR-1;
	      R1+=MR;
	    }
	  else /* There are less than (or equal to) maxrange digits left */
	    {
	      N2=R2;
	      R1=R2+1;
	    }
	}	  

      if (msg.wType==-1) /* Terminate */
	{
	  msg2.iFrom=myid;
	  msg2.wType=-1;
	  msg2.wTotal=total;

	  while(SendMsg(msg_to,&msg2)==-1)
	    printf("SendMsg failed for Proc %d.\n",myid);

	  exit(0);
	}

      else if (msg.wType==0||msg.wType==2) /* Request */
	{
	  total+=N2-N1+1; /* Total integers summed is increased */
	  msg2.iFrom=myid;
	  msg2.wType=1;
	  msg2.wTotal=addrange(N1,N2); /* total=sum of the range */
	  msg2.wVal1=N1;
	  msg2.wVal2=N2;
	  sleep(myid); /* sleep the correct amount */

	  while(SendMsg(msg_to,&msg2)==-1) /* return the result */
	    printf("SendMsg failed for Proc %d.\n",myid);

	}
      else if (msg.wType==1) /* A Result, but I shouldn't get one of these.. */
	{
	  printf("Child %d (%d) received a result message from process ID ",
		 myid,getpid);
	  printf("%d.\n",msg.iFrom);
	}

      if (msg.wType==2) /* Add. Request, check to reset flag. */
	{
	  if (R1>R2) /* If the full range is finished, set flag=1 so that */
	    flag=1;  /* another message can be received. */
	}

    } /* While(1) */
} /* adder(int) */

/*
 * addrange - 96/09/09
 *
 * By: Theo Van Dinter (felicity@kluge.net)
 *
 * Returns the sum of the range from N1-N2 inclusive.
 *
 * Calling Parameters: N1 = First number in range
 *                     N2 = Last number in range
 *
 * Return Values: Sum of the range inclusively.
 *
 * Math:
 *
 * If the sum from 0-N = n(n+1)/2, then from N1-N2 = n2(n2+1)/2 - (n1-1)(n1)/2
 * so, 2*sum = n2(n2+1) - (n1-1)(n1) = n2^2+n2 - n1^2-n1
 * and finally, sum = (n2*n2 + n2 + n1 - n1*n1)/2
 *
 */
int addrange(int N1,int N2)
{
  return((N2*N2+N2+N1-N1*N1)/2);
}