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# (To avoid the trivial step [which is tricky] we ensure that we call recursively, there are at least two squares.
#  This means at least two green intervals on the previous stage.)
# Provided that this case is handled in the caller, the additional green lines appear when the length of the start/end rectangle
# is s+1; if it is above s+1, this is an impossible situation.

# In particular, every case is reduced to a “unique solution” one, or the “trivial” one.  The last one is equivalent to
# having 3 equidistant paralle lines with an an interval [AB] on the middle one (the preimage [AB] of the last red line), and
# opposite to each other rays XA' and YB' on the other two lines.  The fitting line must intersect all 3 of them. 
# Oone of ray may be the whole line).  It is easy to see that this is equivalent to the line intersecting intervals [XA], [AB] 
# and [BY].  If the quadrilateral XAYB is not convex, it may be decreased (so that X,A,B,Y are 3 vertices of a â–³, and a point
# on a side.  If it is convex, then intersecting [AB] is a corollary of other two.  ???

# Possibly unknown squares: the 2nd and 3rd row “share” a x-coordinate; assume that intersecting a red line in any of them is OK.
#                â–¡	This is equivalent to having the red line of double length, which is equivalent to
#             â–¡â–¡â–¡		the green line of double length at this position (assuming it is not at edge).
#         â–¡â–¡â–¡â–¡		Hence this allows the induction step as well.
#     â–¡â–¡â–¡â–¡â–¡		(Encode ignoring the extra square at second line, with second line marked as: $extended->{1}=1.)
#   â–¡â–¡

# Returns empty or a,b,db of the line y=ax+b-db which rasterizes to the rectangles of widths @$CC (connected UR ↔ LL corners);
# The other two arguments as as above.  LL corner of the leftmost of bottom squares is at 0,0.