/*- PICKNODE.C --------------------------------------------------------------* To be able to divide the nodes down, this routine must decide which is the best Seg to use as a nodeline. It does this by selecting the line with least splits and has least difference of Segs on either side of it. Credit to Raphael Quinet and DEU, this routine is a copy of the nodeline picker used in DEU5beta. I am using this method because the method I originally used was not so good. (Some stuff was optimized, check out 'picknode.old' for original code - Alex.) *---------------------------------------------------------------------------*/ #include "prottype.h" struct Seg *PickNode(struct Seg *ts) { int num_splits,num_left,num_right; int min_splits,min_diff,val; struct Seg *part,*check,*best; int max,new,grade = 0,bestgrade = 32767,seg_count; // min_splits = 32767; // min_diff = 32767; best = ts; /* Set best to first (failsafe measure)*/ for(part=ts;part;part = part->next) /* Use each Seg as partition*/ { progress(); /* Something for the user to look at.*/ psx = vertices[part->start].x; /* Calculate this here, cos it doesn't*/ psy = vertices[part->start].y; /* change for all the interations of*/ pex = vertices[part->end].x; /* the inner loop!*/ pey = vertices[part->end].y; pdx = psx - pex; /* Partition line DX,DY*/ pdy = psy - pey; num_splits = 0; num_left = 0; num_right = 0; seg_count = 0; for(check=ts;check;check=check->next) /* Check partition against all Segs*/ { seg_count++; if(check == part) num_right++; /* If same as partition, inc right count*/ else { lsx = vertices[check->start].x; /* Calculate this here, cos it doesn't*/ lsy = vertices[check->start].y; /* change for all the interations of*/ lex = vertices[check->end].x; /* the inner loop!*/ ley = vertices[check->end].y; val = DoLinesIntersect(); /* get state of lines relation to each other*/ if((val&2 && val&64) || (val&4 && val&32)) { num_splits++; /* If line is split, inc splits*/ num_left++; /* and add one line into both*/ num_right++; /* sides*/ } else { if(val&1 && val&16) /* If line is totally in same*/ { /* direction*/ if(check->flip == part->flip) num_right++; else num_left++; /* add to side according to flip*/ } else /* So, now decide which side*/ { /* the line is on*/ if(val&34) num_left++; /* and inc the appropriate*/ if(val&68) num_right++; /* count*/ } } } // if(num_splits > min_splits) break; /* If more splits than last, reject*/ } if(num_right > 0 && num_left > 0) /* Make sure at least one Seg is*/ { /* on either side of the partition*/ max = max(num_right,num_left); new = (num_right + num_left) - seg_count; grade = max+new*8; if(grade < bestgrade) { bestgrade = grade; best = part; /* and remember which Seg*/ } } } return best; /* all finished, return best Seg*/ } /*---------------------------------------------------------------------------* Because this is used a horrendous amount of times in the inner loops, the coordinate of the lines are setup outside of the routine in global variables psx,psy,pex,pey = partition line coordinates lsx,lsy,lex,ley = checking line coordinates The routine returns 'val' which has 3 bits assigned to the the start and 3 to the end. These allow a decent evaluation of the lines state. bit 0,1,2 = checking lines starting point and bits 4,5,6 = end point these bits mean 0,4 = point is on the same line 1,5 = point is to the left of the line 2,6 = point is to the right of the line There are some failsafes in here, these mainly check for small errors in the side checker. ( I made some changes here to eliminate sqrt() functions - Alex) *---------------------------------------------------------------------------*/ int DoLinesIntersect() { short int x,y,val = 0; long dx2,dy2,dx3,dy3,a,b; dx2 = psx - lsx; /* Checking line -> partition*/ dy2 = psy - lsy; dx3 = psx - lex; dy3 = psy - ley; a = pdy*dx2 - pdx*dy2; b = pdy*dx3 - pdx*dy3; if((a<0 && b>0) || (a>0 && b<0)) /* Line is split, just check that*/ { ComputeIntersection(&x,&y); dx2 = lsx - x; /* Find distance from line start*/ dy2 = lsy - y; /* to split point*/ if(dx2 == 0 && dy2 == 0) a = 0; else if( dx2*dx2 + dy2*dy2 < 4) a = 0; /* If either ends of the split */ /* are smaller than 2 pixs then */ /* assume this starts on part line*/ dx3 = lex - x; /* Find distance from line end*/ dy3 = ley - y; /* to split point*/ if(dx3 == 0 && dy3 == 0) b = 0; else if ( dx3*dx3 + dy3*dy3 < 4) b = 0; } if(a == 0) val = val | 16; /* start is on middle*/ if(a < 0) val = val | 32; /* start is on left side*/ if(a > 0) val = val | 64; /* start is on right side*/ if(b == 0) val = val | 1; /* end is on middle*/ if(b < 0) val = val | 2; /* end is on left side*/ if(b > 0) val = val | 4; /* end is on right side*/ return val; } /*---------------------------------------------------------------------------* Calculate the point of intersection of two lines. ps?->pe? & ls?->le? returns int xcoord, int ycoord ( Profiling revealed that this was most time-consuming routine here so I rewrote it completly and new one is more than 2 times faster than previous (w/o coprocessor) - Alex) *---------------------------------------------------------------------------*/ void _fastcall ComputeIntersection(short int *outx,short int *outy) { /* Now, algorithm is based on solving the system of 2 linear equaltions of given lines - Alex*/ double X,Y, a1,b1,a2; long dx1,dy1,dx2,dy2; dx1 = (long)pex - psx; dy1 = (long)pey - psy; dx2 = (long)lex - lsx; dy2 = (long)ley - lsy; if( (!dx1 && !dy1) || (!dx2 && !dy2) ) ProgError("Invalid line encountered"); if( !dx1 && !dy2) { *outx = psx; *outy = lsy; return; } if( !dy1 && !dx2) { *outx = lsx; *outy = psy; return; } if( !dx1 || !dx2) { a1 = (double)((long)pex-psx)/((long)pey-psy); a2 = (double)((long)lex-lsx)/((long)ley-lsy); if( a1 == a2 ) { *outx = lsx; *outy = lsy; return; } b1 = psx - a1*psy; X = (lsx - a2*lsy - b1)/(a1-a2); Y = a1*X + b1; *outx = (int)(Y + ((Y<0)?-0.5:0.5)); *outy = (int)(X + ((X<0)?-0.5:0.5)); } else { a1 = (double)((long)pey-psy)/((long)pex-psx); a2 = (double)((long)ley-lsy)/((long)lex-lsx); if( a1 == a2 ) { *outx = lsx; *outy = lsy; return; } b1 = psy - a1*psx; X = (lsy - a2*lsx - b1)/(a1-a2); Y = a1*X + b1; *outx = (short int)(X + ((X<0)?-0.5:0.5)); *outy = (short int)(Y + ((Y<0)?-0.5:0.5)); } } /*--------------------------------------------------------------------------*/