1. Clojure and the Mandelbrot set on LEGO Mindstorms EV3

    The LeJOS operating system for LEGO Mindstorms EV3 is, as I said before, a small Linux with an embedded JVM.

    Running Clojure:

    To run Clojure on the EV3 I had to download the Clojure runtime, unzip it and throw away everything but the actual jar-file clojure-1.8.0.jar.

    Then the jar-file needs to be transfered to the EV3 (via scp for example) and can be executed via ssh with jrun -jar clojure-1.8.0.jar (I already have an alias in my ~/.ssh/config for the EV3). After about two minutes we have a Clojure repl that we can use to answer the important questions of our time.

    Takes about two minutes to start the repl

    Actually coding something:

    The first thing we have to notice is, that the leJOS menu stays visible when we opened the clojure repl. This is a problem insofar, that when we use the Clojure code to interact with (for example) the display, both the leJOS menu and our application want to use the display and their interactions will overlap in unpredictable ways (for me both images were "fighting", but the system might as well crash).

    So I created a shell script that kills the existing java process and opens it again once our command ran.

    #!/bin/sh
killall java
jrun # ... we will look at that later
~/lejos/bin/startmenu &

    Now we want to actually write Clojure code where we calculate members of the Mandelbrot set. The Mandelbrot set is the set of all numbers \(c\) for which \(z_n\) with \(z_0 = 0\) and \(z_{n+1} = z_{n}^2 + c\) does not diverge (when \(n\) goes to infinity). This is calculated in the complex plane. The numbers for which this doesn't diverge are usually drawn in black while the diverging numbers remain white.

    I looked for a Java-based library for the complex numbers and found one with Apache. This library was insofar underwhelming, that taking a complex number to the power of an arbitrary Integer doesn't work as one expects. The library always uses the following equivalence: \(y^x = exp(x \times log(y))\) which is fine in general but doesn't work if \(y\) is zero, which is the base case but also no problem with positive integer powers. Took me an hour because that is not at all well documented (for the integer-case). So the first thing is, to write our own pow function: (defn pow [y, x] (if (.equals y Complex/ZERO) y (.pow y x)))

    Now we can define both \(z_n\) and whether any \(z_n\) diverges:

    (defn zn [n, c] ( if (== n 0) Complex/ZERO (.add c (pow (zn (- n 1) c) 2.0 ) ) ))
(defn diverges [x] (or (.isNaN x) (.isInfinite x) (< 2.0 (.abs x))))

    And the idea is, to set the whole display to black and evaluate for every black pixel \(z_1\) and then for every remaining black pixel \(z_2\). Once a pixel was set to white, we no longer need to evaluate it, because we already know it diverges.

    So here's the full code for mb.clj:

    (import [org.apache.commons.math3.complex Complex])
(import [lejos.hardware.lcd LCD])
(import [lejos.utility Delay])

(defn pow [y, x] (if (.equals y Complex/ZERO) y (.pow y x) ))
(defn zn [n, c] ( if (== n 0) Complex/ZERO (.add c (pow (zn (- n 1) c) 2.0 ) ) ))
(defn diverges [x] (or (.isNaN x) (.isInfinite x) (< 2.0 (.abs x))))

(defn scale_x [x] ( - (/ (* 2.5 x) LCD/SCREEN_WIDTH ) 2.0 ))
(defn scale_y [y] ( - (/ (* 2.0 y) LCD/SCREEN_HEIGHT) 1.0 ))

(doseq [x (range  LCD/SCREEN_WIDTH )
        y (range  LCD/SCREEN_HEIGHT)] (LCD/setPixel x y 1))

(doseq [rep (range 1 1000)]
  (doseq [x (range  LCD/SCREEN_WIDTH )
          y (range  LCD/SCREEN_HEIGHT)]
          (
              if (== 1 (LCD/getPixel x y))
              (if (diverges (zn rep (Complex. (scale_x x) (scale_y y)))) (LCD/setPixel x y 0))
          )
  )
)

    We don't need a delay at the end because this will take long enough.

    But now we have to think about running this. Basically we need to include every jar we use into the java classpath and then run the clojure main with our script as the parameter. For org.apache.commons.math3 we need the commons-math3-3.6.1.jar from Apache and for the lejos namespace we need the ev3classes.jar from leJOS (which is not included on the system because it is usually compiled into the finished jar).

    Once this is all done we can basically run our application with jrun -cp "../ev3classes.jar:./commons-math3-3.6.1.jar:../clojure-1.8.0.jar" clojure.main mb.clj.

    After a few hours, it will look like this

    I am pretty sure it is possible to compile the whole jar with the correct main using eclipse/ant and whatnot. But that's the first successful experiment in this direction. Here's a timelapse of the program in action.

  2. Working and Working on LEGO Mindstorms EV3

    Presentation Space at a Networking Event for Educators

    Since October I am working at the Leipzig University of Applied Science as a researcher and project manager in a robotics project. The project is called "Roberta" and is aimed at young people, specifically girls and young women. With consumer-grade robotics sets, like LEGO Mindstorms, we try to get them excited about robotics, computer science and STEM in general. What we're doing is working with educators to get this project to schools.

    I am doing project management but I am also doing research and testing the limits of what we can do with those robots. Not from an engineering (or software engineering) perspective. I am not throwing C-code at the machines.

    "programming"

    The EV3 Bricks (the brain of the LEGO Mindstorms EV3) are running a small Linux. Using the original proprietary LEGO Mindstorms programming software is a bit lacking. You use the graphical interface to create a flow-graph that describes the actions the robot should take. Sadly, this language doesn't support variable scoping (meaning that every variable is global) and only support non-recursive functions. Indirect recursion is prohibited as well (no tricks!). That's no fun.

    Alternatively there are multiple (two) alternative Linux-based operating systems. I have yet to try the other one, but the first one is basically a Linux with an embedded JavaVM. This is usually used to, well, program Java. I've tried that and it's boring because Java is boring. But having the JVM is, even with 300Mhz and 64MB RAM, pretty powerful.

    I've tried to run Jython on the brick but there wasn't enough memory to do it. Using a USB Stick formatted as swap I was able to get it to run, but it took almost an hour to even start the repl. Running any actual code took prohibitively long.

    16GB of additional very slow main memory ought to be enough for everybody

    But Clojure also runs on the JVM and this worked quite nicely. The next article will be about me, running Clojure on a LEGO Mindstorms Ev3.

  4. Solving Tatham's Puzzles - Signpost (backtracking)

    I've started writing solvers for games of Simon Tatham's Portable Puzzle Collection. Those Problems can usually be reduced to some NP-complete problem. This is quite a fun exercise.

    Unsolved Signpost Puzzle

    Unsolved Signpost Puzzle

    We will be continuing with the puzzle "Signpost", which is a variant of Hamiltonian paths in a directed graph where, for some fields or nodes, the position within the path is given.

    The goal of the puzzle is to connect the Node "1" to some node, which is then node "2" in the direction the arrow points to. All numbers need to be connected to their successor until all nodes have a single unique number from "1" to (in this example) "25".

    We will be solving this problem in two different ways, both with the help of the programming language Python. But first of all we'll need to convert a puzzle into a machine readable format. Luckily, Signpost has an easy and accessible format. The problem instance from the image is the instance 5x5:1cceefcfggeeccghcac3e12hch10ah25a. It starts off with the size of the problem and then a list of letters that show the directions of the arrow where a is north or 12 o'clock and all other letters go clockwise with h being north-west. A letter can be prefixed by a number which is the hint for that node. That node's number is fixed. The cells or nodes are defined line by line from the top left to the bottom right.

    Problems don't need to start with 1 or end with the last number. The start and the end of the path can be anywhere but puzzles generated with the ends at opposite corner look nicer.

    First we define what our data looks like. We implement a simple form of directed graph that we define as a set of nodes and their successors. From that we derive following definition:

    class Node(object):
    hint = None
    following = frozenset()

    def __init__(self, direction):
        self.direction = direction

    We don't want to use any of that code in a library or something something so we don't do anything fancy here and we use the simplest class definition that comes to mind.

    Railroad diagram of regex

    ^(?P<size_x>\d+)x(?P<size_y>\d+)$

    Railroad diagram of regex

    (?P<defn>\d*[a-h])

    Now we go on to parsing the problem. The parsing function should get the problem as a string and return a 3-tuple containing width and height of the problem as well as a mapping of positions within the grid to ``Node`` instances.

    Onto actually parsing. We can split at the colon and then regular expressions to the rescue. The size definition has the form ^(?P<size_x>\d+)x(?P<size_y>\d+)$ while a single node definition has the form (?P<defn>\d*[a-h]). As we can see, the digits are optional but we need to greedily accept them. All non-overlapping occurences in order will be our cell definitions.

    In code we, again, do nothing too too fancy:

    def parse(puzzle_str):
    directions = {
        'a': (0,-1), 'b': (1,-1), 'c': (1,0), 'd': (1,1),
        'e': (0,1),  'f': (-1,1), 'g': (-1,0),'h': (-1,-1)
    }

    size, _, definition = puzzle_str.partition(':')
    r_size = re.compile(r"^(?P<size_x>\d+)x(?P<size_y>\d+)$")
    r_defn = re.compile(r"(?P<defn>\d*[a-h])")
    size_t = tuple(map(int, r_size.match(size).groups()))
    w, h = size_t
    nodes = {}
    for n, m in enumerate(r_defn.finditer(definition)):
        pos = (n % w, n // w)
        nodes[pos] = Node(directions[m.group(0)[-1:]])
        hint = m.group(0)[:-1]
        if hint:
            nodes[pos].hint = int(hint)
    return w, h, nodes

    And we also need to fill in the successor Nodes. This is no problem at all.

    w, h, nodes = parse(sys.argv[1])
from itertools import product
for x, y in product(range(w), range(h)):
    this_node = nodes[(x,y)]
    dx, dy = this_node.direction
    x, y = x + dx, y + dy
    while x >= 0 and x < w and y >= 0 and y < h:
        this_node.following = this_node.following | frozenset([nodes[x,y]])
        x, y = x + dx, y + dy

    And now on to actually solving this thing. The idea is, that we slowly build a list of Nodes that is our solution. The first element is the Node with the hint "1" and we iterate through all nodes that follow that Node and do not have a hint that is other than "2", and so on. And of course, the Node that we add to the solution is not allowed to have already been part of the solution and if there is a node with that hint, it should be this exact one. If we find such a conflict in our solution, we reject that solution and return to an earlier partial solution.

    Instead of using recursion, we build a queue of partial solutions until we get a complete solution. This is functionally identical to restarting a solver function with partial solutions of increasing size. In that case, the call stack manages the backtracking. But we'll do it with a queue this time. We build the queue so that every element in the list is either

    1. a partial solution (beginning from 1) without conflicts
    2. a partial solution (beginning from 1) where the last element is in some conflict
    3. a complete solution

    For case 1, we remove the partial solution that has the length n and add all partial solutions of length n+1 by iterating through all successor Nodes that have not yet been part of the partial solution. For case 2, we reject that partial solution. For case 3, we immediately return that solution and are finished.

    def solve_dhc(nodes):
    # get the Node that has hint 1
    first = next(iter(n for n in nodes.values() if n.hint == 1))
    # get the set of all hints, so it is easier to check for conflicts
    hints = frozenset(n.hint for n in nodes.values() if n.hint is not None)
    from collections import deque
    q = deque([[first]]) # initializing queue
    while q:
        curr = q.pop()
        idx = len(curr)
        if (idx in hints or curr[-1].hint is not None) and curr[-1].hint != idx:
            continue # case 2
        if idx == len(nodes):
            return curr # case 3
        for _next in curr[-1].following: # case 1
            if _next not in curr:
                q.append(curr + [_next])

    This algorithm terminates because we always remove one element from the queue of some length n and possibly add a large but finite amount of elements of length n+1 and the algorithm terminates when we have an element in the queue of some possibly large but finite length.

    Sidenote: if you use popleft instead of pop you get a breadth-first search instead the depth-first search, that is implemented here. Add to the right and pop from the right for depth-first and add to the right and pop from the left for breadth-first. Since every proper solution has the exact same length/depth (that we know and never go further anyways), searching depth-first is strictly better than breadth-first.

    And given a solution, we still have to print it. That's the easy part.

    dhc = solve_dhc(nodes)

fill = len(str(w*h))
for l in range(h):
    for c in range(w):
        node = nodes[(c,l)]
        print(str(dhc.index(node) + 1).rjust(fill), end=" ")
    print("")
    Solved Puzzle

    Solved Puzzle

    And there's the solution for our puzzle.

    $ python3 signpost_dhc.py 5x5:1cceefcfggeeccghcac3e12hch10ah25a
 1 20  9  2 21
23 14 13 22 24
15  5  7  6  8
18 19 11  3 12
16 17 10  4 25

  5. Solving Tatham's Puzzles - Pattern

    I've started writing solvers for games of Simon Tatham's Portable Puzzle Collection. Those Problems can usually be reduced to some NP-complete problem. This is quite a fun exercise.

    We will be starting with the puzzle "Pattern", also known as Nonograms that we reduce to SAT.

    Unsolved Nonogram

    Unsolved Nonogram

    First to explain the puzzle: The cells of the puzzle can be filled with black or white squares. The numbers on the sides of the rows and colums denote the number of consecutive black squares. Every run of black squares has to be seperated from another run by white squares.

    First we want to define a type for our problem.

    type PSize = (Int, Int)
type Run = [[Int]]
type Runs = (Run, Run)
data Problem = Problem PSize Runs deriving Show

    The puzzle on the side is defined in the Tatham format by the string 10x10:2/3.2/1.1.3/2.5/2.2.4/1.1.4/1.3/1.1.2/4/4.1/5.2/2.3.3/1.2/6/3/1/4/6/7/4.1. It has a size information and then the list of runs. Another format we want to support is the Haskell Runs format which would be ([[2],[3,2],[1,1,3],[2,5],[2,2,4],[1,1,4],[1,3],[1,1,2],[4],[4,1]],[[5,2],[2,3,3],[1,2],[6],[3],[1],[4],[6],[7],[4,1]]). This format seperates the runs for rows and columns and ommits the size information.

    For the second format we can use Haskell's own read function. For Tatham's format we have to define a parser. I've added type information for most functions which is often needed for code involving read.

    import Text.ParserCombinators.ReadP
number' = many1 (choice (map char ['0'..'9']))

r_size :: ReadP PSize
r_size = do l <- number'
            char 'x'
            r <- number'
            return (read l, read r)

r_run :: ReadP [Int]
r_run = do { li <- sepBy number' (char '.'); return $ map read li }

r_runs :: ReadP [[Int]]
r_runs = sepBy r_run (char '/')

r_problem :: ReadP Problem
r_problem = do size <- r_size; char ':'
               runs <- r_runs
               skipSpaces; eof
               return $ Problem size (take (fst size) runs, drop (fst size) runs)

    What do we now do with such a problem? First we want to check it for consistency. If we got size information it should fit the runs and the numbers in the runs (and the spaces between them) shouldn't add up to more than the size.

    import qualified Prelude as P (all, (&&), not)
consistent :: Problem -> Bool
consistent (Problem size (xin, yin)) = consistent' size xin P.&& consistent' (swap size) yin
consistent' (w, h) xs = (w == length xs) P.&& P.all (\x -> (sum x) + (length x - 1) <= h) xs
    Calling our solver

    Calling our solver

    But now on to solving the problem. We will reduce this problem to SAT using the Ersatz library. The encoding of a problem, I guess, is obvious. We say that every cell within the Nonogram is a Variable Bit and a bit being True has the semantics of that field being colored black while False would mean white.

    We want to build constraints of single rows or columns and then solve the conjunction of all those row/column-constraints for satisfiability. Therefore we define a function of the type [Int] -> [Bit] -> Bit that takes a list of Integers (the numbers of one column or row) and the list of remaining Bits that do not yet are constrained.

    First we begin with the end of the recursion. If no number is left or the remaining number is zero (for empty columns or rows), we want all remaining variables to be False. (we take for to be defined as flip map)

    build_runs :: [Int] -> [Bit] -> Bit
build_runs nums vars = case nums of
  [] -> all not vars
  [0] -> all not vars
  x:xs -> let max_d = length vars - (length xs + sum xs) - x
              run   = [not] ++ replicate x id ++ [not]
              v     = [false] ++ vars ++ [false]
          in any id $ for [0..max_d] $ \ d ->
             (all not $ take d vars)
             && (all id $ zipWith id run $ take (x + 2) $ drop d v)
             && (build_runs xs $ drop (x + d + 1) vars)
    Solved Nonogram

    Solved Nonogram

    The recursion is something to think about a bit. We're given a number of consecutive black squares x and all the variables in the column/row vars. The x Trues can be put anywhere from 0 to max_d where max_d is all the rest of the variables take the minimum necessary space of the remaining number of numbers xs, while all variables before that number of black squares and one after need to be white. For every possible position the rest of the variables need to fulfill the same condition with the rest of the numbers.

    Now we have to combine all the constraints from a given problem.

    nono :: (MonadState s m, HasSAT s) => Problem -> m (Map (Int,Int) Bit)
nono (Problem (sizex, sizey) (xin, yin)) = do
  let indices = [(x, y) | x <- [1..sizex], y <- [1..sizey] ]
  vars' <- replicateM (length indices) exists
  vars <- return $ Map.fromList $ zip indices vars'
  let getx x = for [1..sizey] $ \y -> (vars ! (x,y))
      gety y = for [1..sizex] $ \x -> (vars ! (x,y))
      xs = zipWith build_runs xin (map getx [1..])
      ys = zipWith build_runs yin (map gety [1..])
  assert (all id xs && all id ys)
  return vars

    Internally we let Ersatz identify variables by their position in the grid Map (Int,Int) Bit. Now we can solve our problem easily using minisat with Ersatz's native minisat binding. Given a solution, we just format it correctly and give it to the user.

    main = do
  input <- getContents
  let parsed = parse_tat input
      problem = case parsed of
        Just p -> p
        Nothing -> let (xin, yin) = read input :: ([[Int]],[[Int]])
                   in Problem (length xin, length yin) (xin, yin)

  putStrLn $ "Size: " ++ (show $ problem_size problem)
  unless (consistent problem) $ do {putStrLn "Problem inconsitent"; exitFailure}

  (Satisfied, Just solution) <- solveWith minisat (nono problem)
  let sizey = snd $ problem_size problem
      lines = for [1..sizey] $ \i -> filter (\ ((x, y), b) -> y == i) $ toList solution
      conv  = map $ \((x,y), b)-> if b then 'X' else ' '
      join = foldr (:) ""
  forM_ (map (join . conv) lines) print

    And we're done.

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