# Other articles

1. Published: Mon 27 August 2018

I love watching 8 Out Of 10 Cats Does Countdown. It's a british panel show where a few comedians solve word puzzles and mathematics puzzles. It's based on the game show Countdown that's been running since 1982.

And I also like the host Jimmy Carr very much.

## The Puzzles

The show consists of three different puzzles:

In the Letters round a team (or single contestant) picks 9 consonants and vowels from their respective piles. After the team chooses from one of both piles, the letter is revealed and put up on the board. They then choose again until 9 letters are on the board. After all letters are on the board, the teams have 30 seconds to form a word out of as many of the letters as possible. The team with the longest word scores that many points.

Letters Round

In the Conundrum a list of letters that already spell out a funny group of words is given by the host and the contestants try to find a 9 letter anagram. The first team to buzzer in the correct answer gains 10 points. This is the final round of the game.

Conundrum

In the Numbers round a team (or single contestant) picks 6 numbers from two piles. A random number is picked and the contestants have to create a mathematical formula out of the picked numbers that equal the target number. The specific rules follow.

Numbers Round

## Numbers Round

According to Wikipedia there are 6 columns of 4 numbers (4 rows). One column contains the numbers 25, 50, 75, and 100 and the other 5 columns contain the numbers 1 to 10 twice each. That's not exactly the same as on the show, as there appear to be 7 columns of 4 numbers but not every stack has all 4 number cards.

Contestants choose a total of 6 numbers from the "small" and "large" columns. The usual pick would be 2 to 3 "large" numbers and the rest small ones.

Then a random number between 100 und 999 is generated. The contestants try to create a mathematical formula that has the random number as the result. In the formulare only the following operations are allowed:

• Multiplication
• Subtraction (that does not result in a negative number)
• Division (that results in an integer)

The contestants do not have to use all the numbers to get the target number.

Finished Numbers Round

Since I am really not good at this, I wrote a solver for the numbers game in Haskell.

## Solver

How do we go about this problem? First of all, we have to decide on a data structure. Let's start with a traditional term structure where a Term is a binary operation and two terms, or an integer value. An Operation is the function mapping two integers to a third, and a string to pretty-print the operation. The mapping is not total and might fail.

data Term = Op Operation Term Term | Val Integer
data Operation = Operation {fun :: (Integer -> Integer -> Maybe Integer), name :: String}

instance Show Term where
show (Val x) = show x
show (Op op l r) = "(" ++ (show l) ++ (name op) ++ show r ++ ")"


Now we have to define our operations and what they should do. Since we're using Maybe, and it's a monad, we can use monadic notation where it suits us:

operations = [  Operation (\x y -> return (x + y)) "+"
,  Operation (\x y -> return (x * y)) "*"
,  Operation (\x y -> guard (x > y) >> return (x - y)) "-"
,  Operation (\x y -> guard (x mod y == 0) >> return (x div y)) "/"
]
eval :: Term -> Maybe Integer
eval (Val v) = Just v
eval (Op o l r) = do l' <- eval l
r' <- eval r
(fun o) l' r'


As we can see, the preconditions to the subtraction and division are encoded using the guard function. If the function returns true, the whole operation returns Nothing, otherwise it returns Just the value.

Since Haskell uses lazy evaluation, we can define a list of all possible terms given a list of integers and filter for the ones that evaluate to the correct number.

As we create a Term we have to be careful to not use any of the numbers again. So we will split our numbers in two sets. One that can be used in the left Term of an Operation and the rest can be used in the right Term. We will create all terms using our definition. A Term is either the Value of one of the numbers for that subterm, or an operation with some numbers reserved for the left term and some numbers reserved for the right term.

terms :: [Integer] -> [Term]
terms nums =  do  n <- nums -- choose a number
[Val n] -- one solution
++
do  op <- operations -- choose an operation
(l, r) <- subset_split nums -- choose a split
guard $l /= [] guard$ r /= []
ls <- (terms l) -- choose a term for the left side
guard (eval ls /= Nothing)
rs <- (terms r) -- choose a term for the right side
guard (eval rs /= Nothing)
[Op op ls rs] -- one solution


For performance reasons we already evaluate the generated subterms in order to check whether the operations are valid. There are a lot of terms that can not be evaluated and we want to throw them out as early as possible (or not even generate them).

We haven't defined subset_split yet. The idea is, to just split the list of numbers in two groups in every possible way:

subsets :: [Integer] -> [[Integer]]
subsets []  = [[]]
subsets (x:xs) = subsets xs ++ map (x:) (subsets xs)

subset_split nums = do l <- subsets nums
return (l, nums \\ l)


Now all that remains is to generate all terms given a set of numbers, evaluate them, filter out the ones that don't have the correct value, and return the first one:

solve :: [Integer] -> Integer -> Maybe Term
solve nums target = let possible_solutions = map (\x -> (x, eval x)) (terms nums)
solutions = filter (\x -> (snd x) == (Just target)) possible_solutions
in listToMaybe $map fst solutions  And in our main we only read the list of numbers and the target as arguments and print the solution: main = do args <- getArgs let nums = (read (args !! 0))::[Integer] target = (read (args !! 1))::Integer print$ solve nums target


And there we go:

$./countdown "[100, 75, 2, 10, 3, 8]" 746 Just (8+(3+((75*(100-2))/10)))  Our solution is $$8+3+{{75*(100-2)} \over 10}$$ while the solution from the show was $$10*75 - {8 \over 2}$$. As you can see, we do not always get the easiest solution but we usually have a solution quite fast. This one took about a tenth of a second. The 30 seconds given to the contestant are usually quite enough. To make sure we had the "nicest" solution, we probably would need to generate all terms and sort them by size. The smallest one should be the nicest one. 2. Published: Wed 18 July 2018 # Prolog Programs, Order of Evaluation, and Functional Equivalents While teaching Prolog this semester I had a few thoughts about Prolog programs and their equivalent programs in a functional programming language like Haskell. ## The Peano Naturals The Peano numbers are a unary number system for positive natural numbers where the zero is represented by a Z (or 0, or any atom - it doesn't matter) and every successor of a natural number N is the function S applied to N. We would therefore represent the natural number 3 with the term S(S(S(Z))). To look at it differently: in the Peano axioms, the naturals are defined as the set that contains zero and a closure over the successor function. In haskell we would define the type of the naturals like this: data N = Z | S N  In Prolog we don't need a type definition for the naturals, as terms are generated without types. Now we want a predicate that checks whether our input is a natural number as we've defined it. The idea is that the term is a natural, iff either the term is a Z or it has an S as its root and its parameter is a natural. In Prolog we would write it intuitively like this: naturals(z). naturals(s(X)) :- naturals(X).  And indeed this checks whether our input is a natural number. The query naturals(s(s(z))). returns true while naturals(foo). does not. Writing down this program in Haskell is not very useful since the type system does not allow us to have anything else but a natural as the function parameter but let's pretend: naturals v = case v of S x -> naturals x Z -> True _ -> False  And of course, calling naturals$ S \$ S Z in haskell has the correct type and returns true. But this haskell program only corresponds to the case where we actually call the Prolog predicate with a bound parameter. We could call the prolog predicate with a free binding pattern like naturals(X). and get bindings for X for every solution, so every natural number. The corresponding Haskell programm would be:

naturals_f  = [ Z ] ++ map S naturals_f


This might need some explaining: The list corresponds to some notion of non-determinism. In Prolog the non-determinism is baked into the language due to backtracking from failed predicate evaluations and us asking for more answers. The Prolog program essentially says: Z is a solution. Everything with an S is a solution, if the "content" of the S is a solution as well. That's the same for our haskell programm. In our list of solutions we know that Z is in it and for every solution we have, adding an S leads to another one. And evaluating naturals_f starts to calculate the infinite list of natural numbers.

Now consider the following Prolog program where we switched the order of the clauses:

naturals(s(X)) :- naturals(X).
naturals(z).


This still works nicely for the bound case where we call naturals(s(s(z)). But if we now call naturals(X) we quickly get ERROR: Out of local stack. The natural intuition might be that the order of clauses does not matter but it does in this case. Writing down the equivalent Haskell programm we are able to see why:

naturals_f  = map S naturals_f ++ [ Z ]


We have the recursion before we have any solution. Well, the solution is easy: just write the simple case in the first line and the recursive case in the second. But let's consider another problem.

## Repeated Printing

Consider the problem of repeatedly printing a character:

repeat(0, _).
repeat(N, C) :- write(C), NN is N - 1, repeat(NN, C).


This looks very correct and is very wrong.

?- repeat(5, f).
fffff
true ;
ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff...


Now we see why it's nice to represent the naturals als Peano numbers. After hitting 0 there's a choice point left and asking for another solution leads to the negatives and printing our character in an infinite loop. So let's augment the second rule to prevent us from going into the negatives.

repeat(0, _).
repeat(N, C) :- N > 0, write(C), NN is N - 1, repeat(NN, C).
% query:
?- repeat(5, f).
fffff
true ;
false.


This looks better but we still have a choice point left over which might be not so nice. The problem is that the matched patterns are not disjoint and after the end of the recursion there's still a pattern left over that fits. We can't encode that N should be larger than 0 in the head of the rule. Remember that we can match naturals larger than 0 (or any other number) in the rule head using Peano numbers - that's why they are so cool.

But how do we fix our problem with the left-over choice point? There are two ways we can do it. We can either introduce a cut or reorder the rules.

% Variant 1a: The order doesn't matter but we introduce a cut
repeat(0, _) :- !.
repeat(N, C) :- N > 0, write(C), NN is N - 1, repeat(NN, C).

% Variant 1b: If we promise to be good and only call repeat with a positive numeric argument, we can even leave out the range check
repeat(0, _) :- !.
repeat(N, C) :- write(C), NN is N - 1, repeat(NN, C).

% Variant 2: The order does matter
repeat(N, C) :- N > 0, write(C), NN is N - 1, repeat(NN, C).
repeat(0, _).


I personally think that the variant without the cut is nicer. So for terminating rules, it might be better to have the "smaller" rule last as to not leave choice-points hanging around. Of course, this only works if the matched terms in the rules are proper subsets of each other so that we actually can order the rules in that fashion. In general, that might not be possible.

## Conclusion

• The order of rules can (in theory) not affect termination. But if our rules are already non-terminating for our query, the order and timeliness of results does matter. Then order does matter.
• Disjoint rule heads are important to not have left-over choice points.
• If rules are terminating but the rule heads are not disjoint, they still should be ordered carefully.
• Implicit conditions (N > 0) can lead to non-obvious non-termination if they are not made explicit.
• By encoding data properties in the term structure we can make some implicit conditions explicit and make rule-heads disjoint.
• Having numerical conditions in rule-heads would be really nice. Maybe some Prolog implementation in the future allows for that.

3. Published: Mon 05 December 2016

# 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


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.

Page 1 / 3 »