Wednesday, February 20, 2008

Filmation math

In 1984, European teenagers were thunderstruck by the appearance of the ZX Spectrum hit Knight Lore from a firm called Ultimate Play the Game. This adventure game featured a rendering technique like we hadn't ever dreamed of before.

In a world of 48 KB, 2D platform games, Knight Lore represented a quantum leap forward with its strikingly realistic 3D world based on isometric perspective. Ultimate called this rendering technique Filmation. I used to wonder how they could implement Filmation in such a limited computer as the ZX Spectrum, but the math behind it turns out to be simpler than it might seem.

A given room consists of a background set (the stone walls in the example above) and a bunch of objects. We can assume that the objects are roughly convex: if an object is not convex, for instance if it has holes like the arched gateways, we can just decompose it into a number of convex elements.

Since isometric perspective does not introduce any scaling effect (distant objects are drawn the same size as closer ones), rendering a frame boils down to drawing the background set and then all the room objects, applying a simple projection rule:

(x,y,z) → (x0 + √3 x/2 − √3 y/2, y0x/2 − y/2 + z)

which is generally approximated by the following:

(x,y,z) → (x0 + xy, y0x/2 − y/2 + z)

so as to simplify computations as well as graphic design (the x and y axes are projected to easily pixelizable lines wih slope 1/2). But there is one detail left: if an object A appears behind B, we must draw A before we render B. So, room objects have to be ordered for drawing purposes according to the "is behind" relation. In which situations is A behind B? As objects are assumed to be convex and they do not overlap (i.e. AB = ∅), we can approximate them by prisms with opposite vertices (xA,yA,zA), (x'A,y'A,z'A) and (xB,yB,zB), (x'B,y'B,z'B), with xA<x'A, yA<y'A, zA<z'A, and similarly for B. These prisms are projected to hexagons as depicted in the figure:

A little calculation shows that the hexagons for A and B overlap if and only if:

  1. [xAy'A,x'AyA] and [xBy'B,x'ByB] overlap, and
  2. [xAz'A,x'AzA] and [xBz'B,x'BzB] overlap, and
  3. [−y'A+zA,−yA+z'A] and [−y'B+zB,−yB+z'B] overlap.

If the hexagons do not overlap, it is immaterial which object we draw first; if they do overlap, we can apply the following easily provable properties:

  1. x'A < xBA is behind B,
  2. x'B < xAB is behind A,
  3. y'A < yBA is behind B,
  4. y'B < yAB is behind A,
  5. z'A < zBA is behind B,
  6. z'B < zAB is behind A.

Since the prisms A and B do not overlap (in 3D), it can be seen that at least one of the clauses 1-6 can be applied, so we can use these properties to sort all the room objects and then proceed to drawing. Note that the relation "is behind" defined by the algorithm above is not total: if the projections of two blocks do not intersect, then neither block is behind the other; thus, in general blocks can be sorted in more than one acceptable manner.

Are we done then? It looks so, and actually a rendering algorithm can be implemented along these lines and seemingly works fine. But there is a fundamental flaw in our analysis. The relation "is behind" is not transitive:

In the figure, A is behind B, B is behind C and C is behind A! There is no way that these objects can be sorted so that they are rendered properly; if we wish to keep our basic algorithm, at least one of the objects should be split in two so as to restore transitivity, but knowing when and how to do this is not trivial. How did Ultimate programmers solve this problem? Well, the answer is that they did not solve it. Consider for instance the following snapshot, taken from Nigel Barford's Java Spectrum Emulator:

One table is on top of the other and slightly displaced with respect to it. If we position our hero so that he stands behind the bottom table and in front of the top one, something unexpected happens:

The top table is suddenly rendered as if it were behind the bottom table: if you examine this configuration you will see that we have indeed an antitransitive cycle composed of the two tables and the main character. As these situations are uncommon, the programmers of the Filmation engine decided to just ignore them. So, next time you play a Filmation game and observe one of these rendering quirks, you will know the math behind them.


  1. I am trying to apply your overlap formula to a project I am working on and am having an issue with understand some of the logic/vernacular used in your post.

    Could you explain a bit more about what you mean by prisms? Also by opposite vertices are you meaning something like [Ax, Ay, Az] and [Ax + Awidth, Ay + Alength, Az + Aheight]. Or are you referring to actual screen coordinates of certain vertices on the objects?

    This is a great article. Thank you so much for shedding some practical math on topic.

  2. This comment has been removed by the author.

  3. I think I understand by what you mean prisms now (those being the actual psuedo-3D objects we're drawing).

    What I don't understand what you mean by overlap:

    1. [xA−y'A,x'A−yA] and [xB−y'B,x'B−yB] overlap, and
    2. [xA−z'A,x'A−zA] and [xB−z'B,x'B−zB] overlap, and
    3. [−y'A+zA,−yA+z'A] and [−y'B+zB,−yB+z'B] overlap.

    I have expressed these as two dimensional points (6 total) but I am not sure how you compare them to determine if an overlap exists.

    Please advise.


  4. What I don't understand what you mean by overlap:

    1. [xA−y'A,x'A−yA] and [xB−y'B,x'B−yB] overlap, and
    2. [xA−z'A,x'A−zA] and [xB−z'B,x'B−zB] overlap, and
    3. [−y'A+zA,−yA+z'A] and [−y'B+zB,−yB+z'B] overlap.

    What I mean by "[a,b] and [c,d] overlap" is:

    (acb) or (cad)

    Hope this helps.

  5. That makes alot more sense to me. I am doing this in actionScript3 so the explanation was hard to put into context of AS3.

    Thanks again for the great posting. I have now gotten my isometric rendering engine doing depth sorting correctly.

  6. Glad to be helpful. Just in case you haven't found it yet, please take a look at a JavaScript implementation of these ideas here.

  7. Hi. Wanted to say thanks to your post I have gotten past a major hurdle in developing my isometric library built in actionscript 3. You can check it out here -

  8. Hi jwopitz, I'm glad the stuff's been helpful! When will you provide a demo app for your library?

  9. Tho there isn't a live demo of the lib in use, there are some tutorials posted as well as a users' group.

  10. Hi, great post. I've developed an Actionscript library that uses more or less the Filmation approach, although this implementation is O(n) so if the original Filmation algorithm was O(n^3) then maybe this is an improvement? Check it out here:

    Your thoughts?

  11. Hi Brian,

    Very nice demo! Can you describe your O(n) algorithm?

    Best regards,

  12. I think the sorting error that you show in the snapshot is not because of the case you mention.

    Is because the Filmation algorithm is not transitive for the set { all cases except intersection and “ABC overlap”}. It´s because:

    “If the hexagons do not overlap, it is immaterial which object we draw first”

    This is right, but the algorithm assuming that because “it is immaterial which object we draw first”, it´s not necessary to sort it, and this is wrong:

    Let´s name A the table at the back, B the character and C the table at the front, and X the array in which we are going to sort the objects by depth. Because you don´t know in which order you are going to compare the objects, it´s possible to end in this situation:

    X is empty

    Add C to X

    Is C behind B? Because they don´t overlap, it doesn´t matter, so add B to X.

    Now you have the X=(C, B)

    Is B behind A? Because they don´t overlap, it doesn´t matter, so add A to X.

    Now you have the X=(C, B, A)

    So you display the objects in that order and because A is displayed after C, so the table at the back appears over the table at the front.

    To solve this problem you need a transitive algorithm for the set { all cases except intersection and “ABC overlap”}

    I think this can easily achieved by making some extra comparisons if the objects don´t intersect.

  13. Hi Satan,

    I think your analysis is not correct: A, B and C all overlap each other, so the algorithm does not assign them arbitrary rendering orders as your post suggests. Please take a look again at the figure with three prisms and the text above:

    A is behind B, B is behind C and C is behind A.

    There is no way that one can order A, B and C so that if X is behind Y then X is ordered after Y. It's just impossible, not a defect of the algorithm we happen to be using.

  14. Hi,

    first of all thanks for a great explanation of filmation math! I've managed to implement it in ActionScript 3. However I'm experiencing one bug which I can't fix.

    For some reason detecting if items overlaps fails in one case. See screenshot below:

    If I move the block in the top little bit close to the bigger one then it works:

    Did you ever met this problem?

    Best Regards,

  15. Hi Bartosz,

    I don't know why you're getting the glitch you show, since there are no transitive cycles in your example. I think a correct implementation of the ordering algorithm should not fail here. Can you post or provide a link to the relevant portion of your code?

  16. Bartosz,

    Additionally, have a look at the function depth_sort in this article. Comparing with your code might give you some clue.

  17. Hi Joaquin,
    first, many thanks for writing this nice article! It really gave me some insight to isometric rendering. I'm currently working on an iso engine and thinking about the occlusion computations. It is not using the classic "diamond" map, but a rectangular map with arbitrary shaped convex boundary polygons for the objects.

    I assume that objects are shaped like "towers", i.e. they never overlap with their x and y coordinates in 3D space (unlike the tables / prisms in your example, which is the reason for the circle in the draw order).

    While Satan T., in his comment, was wrong about the point of occlusion between the example objects A, B and C, he mentions a real problem which makes the whole thing complicated:

    "If the hexagons do not overlap, it is immaterial which object we draw first."

    This is, one the one hand, true, because we can not in general decide over the order if two objects don't overlap.

    On the other hand, like S.T. mentioned, the case of "equality" of two objects breaks the transitivity of the relation: if there are three objects A, B and C, and A and B do not overlap as well as B and C do not overlap, but A occludes C, this causes a problem:

    A sort algorithm might find that
    A == B and B == C
    and therefore do nothing. But instead, because A occludes C, it should actually swap A and C!!

    I have been using the (C++) STL sort algorithm and encountered that problem.

    Now, with an own algorithm that always compares every object with every other object, I have a O(n^2) complexity or sth. like that, which is much too slow - it brought my application from ~300 frames (with the std::sort) down to ~40 frames x-D

    I know this is due to the more complex case of comparing two arbitrary polygons, and I wish I could somehow introduce the transitivity into my isBehind relation... but I fear it's not possible. What do you think?

    Best Regards, Max

  18. Hi Max,

    First of all, using a sorting algorithm can't work because the "is behind" relation is not a strict weak order (a partial ≤-type order): for instance, if A->B (-> meaning "is behind") and B->C it does not follow that A->C --the offset between A and C can be so large that they not overlap. So, using std::sort might seem to work but will produce glitches and spurious results. You should need (transitive cycles aside) to replace -> with its transitive closure ->*, defined as A->*B if there's a path A->X1->...->Xn->B. ->* is a strict weak order (transitive cycles aside.)

    In graph theory's terms, the problem is: we have a bunch of objects related by -> and need to

    1. detect and eliminate -> cycles
    2. do a topological sorting of the remaining graph.

    This can be done in O(|V|+|E|), where V and E are the vertices and efges of the graph. The problem is that we don't have E given as a set of edges: we can only determine if there's an edge between A and B (i.e. if A->B), which is not the same thing --constructing the E set is O(|V|^2).

    I'm sure this algorithm can be improved to reach lineal or near-linear time, but I haven't done the anlasysis. Let me think of it for a while.

  19. Hi Joaquin,
    thanks for your fast reply! The thingy with the transitive closure sounds worth checking!
    I already got the clue that it can be solved with a topological sort, so I spent the last hours on using boost for that. It turned out that, as one might expect, the performance leak is not the topological sort of the graph, but its actual creation of the graph, i.e. finding out which object is occluding which object.
    For now, this is done by checking each object against all others.
    I think investing the possibility to turn the relation into a strict weak ordering relation might help at this point, I'll check that!

  20. Pardon my ignorance, but I dont get what you mean by prism with opposite vertices, nor I know what vertices are you talking about with [xa, ya, za], [xa', ya', za']. Those are only 2 vertices..I dont get that representation.
    With makes the info following that hard to grasp.
    If you would put the xa, xa', etc. on the figure (showing the overlapping of A and B), where those would be?

  21. Hi Suminsky

    In the figure you refer to, [xa, ya, za] would be the vertex right at the back of the prism, that is, the only one that is not visible, while [xa', ya', za'] would be the vertex opposite that, that is, the one that lies at the intersection of the three light gray edges. You will note that these two vertices univocally determine the whole prism.

  22. So you're checking if 2 hexagons overlap using 3d coordinates of the 'rectangular cube' they represent?
    Jeez..How is that possible? To me the only way to do this would be making a full collision test between the 2 hexagons (with 2d coords).
    How did you get to this man?

  23. Precisely, the conditions for the two hexagons to overlap (or not) are laid out in the article, equations 1. to 3. As it happens, the 2D coordinates of the hexagons can be expressed as functions of the 3D coordinates of the associated primsms, hence hexagon overlapping can be checked using the latter.

  24. Hi I am making an isometric game but I am having some trouble following this tutorial. Can you explain where these points are on the diagram: "(xA,yA,zA), (x'A,y'A,z'A) and (xB,yB,zB), (x'B,y'B,z'B)" I don't understand what all those letters refer to. Or could you post the is_behind method? Sorry if this is a stupid question.

    1. Hi Vincent:

      I explained this very issue to Suminsky some posts ago:

      In the figure you refer to, [xa, ya, za] would be the vertex right at the back of the prism, that is, the only one that is not visible, while [xa', ya', za'] would be the vertex opposite that, that is, the one that lies at the intersection of the three light gray edges. You will note that these two vertices univocally determine the whole prism.

      As for is_behind, you might want to take a look at the code in this entry.

    2. First of all thank you for responding so fast! That explanation makes more sense but how do I get a prism from a 2d image? If I just have 2d images I know how to get X and Y but not Z. :(

    3. Oh also I saw that article but it did not show code for is_behind, It just showed how to sort objects using the is_behind method.

    4. How do I get a prism from a 2d image? If I just have 2d images I know how to get X and Y but not Z.

      If you're writing an isometric game your characters, objects etc. move in a 3D world with X and Y (tipically ground coordinates) and Z (typically height above the ground). You can approximate these blocks as 3D prisms described by the coordinates (xA,yA,zA), (x'A,y'A,z'A) and project to the 2D screen using the formulas explained at the article.

    5. Oh also I saw that article but it did not show code for is_behind [...]

      Look at the source code of the frame where the game is embedded (this URL).

  25. Thank you so much I think I can figure it out from here (although I dont know javascript so its a bit hard to read). I have been searching for the answer for so long. Im so glad I happened to find this article!

  26. Sorry one more question. I am using a map system where the point (0, 0) is at the left not the top and I cant figure out how to get it to work with that system. (imagine that the y line is between the x and the z in your picture). I have tried figuring it out myself but I am a bit confused on how you came up with these calculations: "
    A little calculation shows that the hexagons for A and B overlap if and only if:

    [xA−y'A,x'A−yA] and [xB−y'B,x'B−yB] overlap, and
    [xA−z'A,x'A−zA] and [xB−z'B,x'B−zB] overlap, and
    [−y'A+zA,−yA+z'A] and [−y'B+zB,−yB+z'B] overlap."

    1. Sorry but I don't get how your coordinates are laid out. Anyway, the formulas for checking if two hexagons overlap will work just the same (technically, they're invariant on axis naming and sign).

    2. The coordinate system works like this: And since the x and y values start at the left instead of the top, when I try to use your equations they aren't drawn in the right order.

  27. I see.

    xc = 0.5 + 0.5 xi + 0.5 yi
    yc = 0.25 - 0.25 xi + 0.25 yi + 0.5 zi

    1. Dont the equations in this method need to be altered as well?
      function block_projection_overlap(obj)
      function interval_overlap(a1,a2,b1,b2)
      return a1>=b1&&a1=a1&&b1<a2;

      return interval_overlap(
      obj.x- obj.y- obj.yy, obj.x+ obj.xx- obj.y)&&
      obj.x- obj.z- obj.zz, obj.x+ obj.xx- obj.z)&&
      - obj.y- obj.yy+ obj.z,- obj.y+ obj.z+ obj.zz);

    2. No, these equations stay the same regardless of the orientation of 3D axes.

  28. I think I found a bug in your "isBehind" heuristic or I missed out on something here.
    Consider the blocks as defined by:
    [xA,yA,zA] = [0,0,0] (furthest vertex from camera)
    [x'A,y'A,z'A] = [1,1,1] (closest vertex to camera)

    [xB,yB,zB] = [0.5,0,0] furthest vertex from camera of block B
    [x'B,y'B,z'B] = [1.5,1,1] closest vertex to camera of block B

    Now they clearly intersect, but none of your 6 cases does apply.
    How do I proceed then?

    1. As you correctly point out, the projections of these two blocks overlap yet none of the conditions 1...6 apply. The reason is that the blocks intersect in 3D (for instance, the point [0.75,0.75,0.75] is inside both A and B), and isBehind requires that this is not the case:

      As objects are assumed to be convex and they do not overlap (i.e. A∩B = ∅)...

    2. > [...] if they do overlap, we can apply the following easily provable properties [...] Since the prisms A and B do not overlap (in 3D)[...]

      I think it would be better here to use "intersect in 3D" rather than overlap. This was pretty confusing for me. And it would be great if you could mention what [x,y,z] and [x',y',z'] refers to. Took me a while to figure it out in the comments.

      Anyways thanks for this article, it helped a lot!

    3. I just want to add here that its important that your collision geometry and sprite prism consider penetration errors. Generally collision solving (specially if physics based) have some amount of penetration (floating errors, cant solve everything in one frame, etc.) So you can have sorting errors(maybe for a few frames) if you use the exact same size for collision and your sprite prism. Its better to leave some room (geometry a lil bit bigger than the sprite prism)

  29. I just stumbled across your blog entry tonight! Thanks for the insight, it'll come in very useful in the next few weeks, I'm sure, as I'm currently reverse-engineering Knight Lore to port it to a number of retro platforms. Whilst I'm commenting the original code, I'm also porting it to generic C code, preserving as much of the original Z80 code and data structure as possible.

    Check out my Retro Ports blog (which I assume you can access from my id) right here on

    I've worked my way through most of the code (I can render rooms and have objects moving around) and am now getting down to the nitty gritty, such as the display order sorting and the wiping algorithm. Just now I've found what I believe is a bug in the wiping code, and so started searching for reports of a graphics glitch in Knight Lore (which led me here). It's obviously not a serious bug, but if I'm not mistaken, it doesn't calculate the correct number of lines to wipe a moving object in all cases.

    I'd be interested to hear how much (if any) reverse-engineering you did on the original code and whether you'd be willing to share your notes, as the display order is one of the remaining areas I've left until the end! ;)

    1. I'm afraid to say I didn't reverse-engineere the code at all: the Z-order stuff is basically self-evident, and the fact that Knight Lore shows a transitivity glitch as predicted by theory confirms that Ultimate guys followed the approach I describe in my blog entry.

      BTW, your project looks extremely interesting. I'd be delighted to have a look at the Z-order section once you crack it (and maybe post an addendum to my blog). Are you making your code available somewhere?

  30. I'll be releasing everything at some point. My immediate goal is to finish commenting the Spectrum disassembly and the C port on the PC simultaneously. Then I want to port it to one or perhaps two retro platforms, and then I'll probably release all the source code with it. My final goal is to port it to the 6809-based Coco3.

    But I'd be happy to share the Z-order stuff with you once it's complete, since I'll be taking advantage of your work shortly. I'll let you know when I get it done.

  31. A note on that "bug" I found. Turns out it isn't, I somehow missed a JP despite checking it several times. I was thrown because of the somewhat inconsistent implementation of the same algorithm in the same routine on the two different axes - and a completely unnecessary JP.

  32. A quick note to let you know that I've just started on the Z-order algorithm tonight (everything else is complete), and will hopefully have nutted it out within a few days, depending on spare time.

  33. Hi Joaquin

    Great article thanks!

    I'm a little confused how you pick the opposing vertices. The rule is that x1 < x2, y1 < y2, z1 < z2, and I see in another post you say the two picked for this are the front and back vertices in the center of the hexagon.

    If you rotated the camera slightly, you would have that x1 == x2 for those points, and the rule doesnt fit... what do you do in this case? Can we simply pick any two opposing vertices that fit the x1 < x2, y1 < y2, z1 < z2 rule?

    It would be brilliant if you could provide a proof for equations 1-3 also!

    1. Given a prism there are two and only two vertices (x1,y1,z1) and (x2,y2,z2) such that x1<x2, y1<y2, z1<z2. This has nothing to do with the fact that these two vertices coincide in their 2D projection --in fact, this could fail to be the case if the projection is skewed, as you point out.

      As for a proof of eqs. 1-3, it can be derived from the following. simple observation: label the six edges of an hexagon (or, rather, the lines they extend to) as a, b, c, d, e, f, starting from the top vertix and going clockwise. Do similarly with the second hexagon to get A, B, C, D, E, F. Then, it should be obvious that the hexagons overlap if and only if the followin three conditions are met:

      1. line A lies between a and d, or else line a lies between A and D.
      2. line B lies between b and e, or else line b lies between B and E.
      3. line C lies between c and f, or else line c lines between C and F.

      And 1, 2, 3 directly imply the overlapping conditions presented in the article.

    2. Thanks a lot for the clarification! Im still trying to get my head around it a little; do equations 1,2,3 in the article work just the same for skewed projections? I dont quite understand whether the rules [e.g. (x-y',x'-y)] are general rules which you apply to the projected points of the prism, or if it includes projecting the prisms to hexagons with the approximation you give also?

      For example, if I have a camera that could be projecting from any angle, can I use the same rule by applying 1/2/3 to my 2 vertices after the projection is done?

    3. The equations depend on the particular coefficients for the projection formula we are using

      (x,y,z) → (x0 + x − y, y0 − x/2 − y/2 + z),

      that is, (1,-1,0) for x, (-1/2,-1/2,1) for y. With other cofficients th resulting equations would be different. For instance, if you consider the image were the two prisms labeled A and B are depicted and try to imagine how it would change as you rotate the 3D scene around the z axis, it's easy to see that at some point the projections would not overlap --hence the equations are not "universal" for any orthogonal projection, but specific to its particular coefficients.

    4. Okay yes it makes sense, thanks very much for the discussion!

    5. Hi Joaquín,

      Thanks for the post, very interesting.

      I believe that your projection formula distorts the image a little, since your x and y vectors have different norms, ain't that the case?

      I tried the little linear algebra that I know and ended up with these transformation matrices, using an orthonormal basis for the screen (x,y):
      Projection matrix P = 2/3 on the diagonal, -1/3 elsewhere;

      Change of basis matrix B = [-sqrt(2)/2 sqrt(2)/2 0; -sqrt(6)/6 -sqrt(6)/6 sqrt(6)/3; sqrt(3)/3 sqrt(3)/3 sqrt(3)/3].

      If I haven't got anything wrong, B*M should be the linear transformation that takes the 3D vector and gives back the normalized coordinates on the screen (you still need to add the origin point).

    6. Hi Rox,

      I must admit I'm not fully getting your formulas since you mention an M (in B*M) that is not defined before. For what it's worth, in the article the following transformation is given

      (x,y,z) → (x0 + √3 x/2 − √3 y/2, y0 − x/2 − y/2 + z)

      that is usually approximated (maybe this is what you are referring?) by

      (x,y,z) → (x0 + x − y, y0 − x/2 − y/2 + z)

      which is simpler to handle from a pixelization point of view.

    7. Hi Joaquín,

      Sorry for that, I changed the variable names while editing and changed it wrong :)

      I meant B*P. What I did was project into the plane with basis [1 -1 0]' and [1 0 -1]', then express the projection in terms of the basis [-1 1 0]',[-1 -1 2], [1 1 1]' (after normalizing it). If I made some mistake, it was probably on the choice of these vectors. Also I inverted the 3D axes x and y because I wanted a right handed system.

      When I started, I expected to get the same result as your first formula, but something way uglier came up (although I can see that is only a relative scaling x/y, the transformation matrix has 0's on the right places).

    8. I played a little more with the problem, and found that if I scale your transformation matrix by sqrt(2)/2 and further scale the screen-y vector by 3/2, I get exactly the transformation matrix I had calculated earlier.

      That's clearly caused by the normalization of the screen basis vectors, so that's where we diverge.

      Incidentally, something interesting came up: the projection matrix turned out to be unnecessary! When I changed the basis to the screen vectors, all that I needed to project was to zero the z component, since it was made orthogonal to the screen plane.

      Another thing to note if you or anybody else wants to follow the proof (attempt) is that you can invert an orthogonal matrix simply by transposing it. Forgetting that the screen basis was orthonormal made me waste a lot of time computing Gauss-Jordan by hand.

    9. I have to admit that I couldn't follow you --for one, B*P looks like a 3x3 matrix (you don't state the dimensions of P, but you say it has a diagonal, so I assume is 3x3) where a 3D->2D projection would use a 2x3 matrix.

      Anyway, I'm glad your own calculations solved the problem for you :-)

    10. You are right about that. P is 3x3, but after we change basis with B we can discard the component that is orthogonal to the plane and end up with a 2x3 modified B*P matrix (in fact you can choose B to be 2x3 already and things work out automagically).

      If you are interested, I'm considering writing down the derivation on a blog post (I only have it on paper).

    11. I'll be happy to read it when you publish that!

    12. Here it is :)

  34. So I guess John Ritman (Head Over Heels) outsmarted us all by stating (in God's voice) "If you have a tall sprite, like the main character, better divide it vertically in two or you will definitely have problems..."