;; My recursion tells me that n planes can divide space into ;; 2, 4, 8, 15, 26, and finally 42 regions. ;; But I'd like some sort of empirical confirmation of this result. ;; Forms in 3-space can be represented as ordered sets of three numbers (def a-form [1 2 3]) ;; As can vectors (def a-point [2 3 5]) ;; And we evaluate the form on the vector (defn contract [form vector] (reduce + (map * form vector))) ;; (or if you prefer, take the dot product, or contract the tensors) (contract a-form a-point) ;-> 23 ;; Any plane can be defined in terms of a 1-form and a number. (def a-plane {:form [1 2 3] :val 4}) ;; Now if we have a plane or a vector, we can evaluate the form on the ;; vector, and compare the result with the value. This tells us which ;; side of the plane the vector is on. (defn side [plane point] (- (contract (plane :form) point) (plane :val))) (side a-plane [2 3 5]) ;-> 19 (this point is on the positive side) (side a-plane [2 3 -5]) ;-> -11 (on the negative side) (side a-plane [2 3 -4/3]) ;-> 0N (in the plane itself) ;; Ok, now we need a way of taking vectors and forms at random. ;; The cauchy distribution is easy to sample from and has nice fat tails (defn cauchy[] (Math/tan (* Math/PI (- (rand) 0.5)))) (repeatedly 20 cauchy) ;-> (-0.43989542100474244 -0.6517139433588255 1.58518947555566 0.001268073580101198 3.6164981498788262 0.44928717656825584 0.3365831420089349 0.4646894211443393 0.8802485518044282 1.8146747880005754 0.1608864471756546 -0.23538854021056904 8.836583912257565 3.8174659241864703 0.5387819323291936 -0.18830386363467239 -1.0430272980416788 0.3310448308016341 -0.10735190850334911 0.3426157380908667) (defn make-point [] (repeatedly 3 cauchy)) (defn make-plane [] {:form (repeatedly 3 cauchy) :val (cauchy)}) (make-point) ;-> (33.032354006369815 -29.428219536044043 -37.796430533111334) (make-plane) ;-> {:form (-45.36910184399889 -1.6741101969009575 9.952054197916382), :val 0.9505471630252558} (def points (repeatedly #(make-point))) (def planes (repeatedly #(make-plane))) ;; And we'll need a function to tell us the sign of a number (defn sign[x] (if (< x 0) '- '+)) (map sign [ -1 -2 -3 0 -0.5 1.3]) ;-> (- - - + - +) ;; Now if we take a set of planes and a point, (defn sig [point planes] (for [p planes] (sign (side p point)))) ;; We can check which side of each plane the point is on (sig (first points) (take 3 planes)) ;-> (+ - +) ;; Every different region gives a different signature. ;; The more planes, the more signatures. (count (frequencies (take 10 (map #(sig % (take 1 planes)) points)))) ;-> 2 (count (frequencies (take 10 (map #(sig % (take 2 planes)) points)))) ;-> 4 (count (frequencies (take 10 (map #(sig % (take 3 planes)) points)))) ;-> 6 (count (frequencies (take 10 (map #(sig % (take 4 planes)) points)))) ;-> 7 (count (frequencies (take 10 (map #(sig % (take 5 planes)) points)))) ;-> 7 (count (frequencies (take 10 (map #(sig % (take 6 planes)) points)))) ;-> 7 ;; But the more planes we have, the smaller the smallest regions are ;; and thus the chance of a point falling in every one goes down. ;; The more points we take, the more likely we are to get one in every region (count (frequencies (take 100 (map #(sig % (take 1 planes)) points)))) ;-> 2 (count (frequencies (take 100 (map #(sig % (take 2 planes)) points)))) ;-> 4 (count (frequencies (take 100 (map #(sig % (take 3 planes)) points)))) ;-> 7 (count (frequencies (take 100 (map #(sig % (take 4 planes)) points)))) ;-> 11 (count (frequencies (take 100 (map #(sig % (take 5 planes)) points)))) ;-> 18 (count (frequencies (take 100 (map #(sig % (take 6 planes)) points)))) ;-> 21 (count (frequencies (take 1000 (map #(sig % (take 1 planes)) points)))) ;-> 2 (count (frequencies (take 1000 (map #(sig % (take 2 planes)) points)))) ;-> 4 (count (frequencies (take 1000 (map #(sig % (take 3 planes)) points)))) ;-> 8 (count (frequencies (take 1000 (map #(sig % (take 4 planes)) points)))) ;-> 15 (count (frequencies (take 1000 (map #(sig % (take 5 planes)) points)))) ;-> 26 (count (frequencies (take 1000 (map #(sig % (take 6 planes)) points)))) ;-> 38 (count (frequencies (take 10000 (map #(sig % (take 1 planes)) points)))) ; 2 (count (frequencies (take 10000 (map #(sig % (take 2 planes)) points)))) ; 4 (count (frequencies (take 10000 (map #(sig % (take 3 planes)) points)))) ; 8 (count (frequencies (take 10000 (map #(sig % (take 4 planes)) points)))) ; 15 (count (frequencies (take 10000 (map #(sig % (take 5 planes)) points)))) ; 26 (count (frequencies (take 10000 (map #(sig % (take 6 planes)) points)))) ; 40 (count (frequencies (take 100000 (map #(sig % (take 1 planes)) points)))) ; 2 (count (frequencies (take 100000 (map #(sig % (take 2 planes)) points)))) ; 4 (count (frequencies (take 100000 (map #(sig % (take 3 planes)) points)))) ; 8 (count (frequencies (take 100000 (map #(sig % (take 4 planes)) points)))) ; 15 (count (frequencies (take 100000 (map #(sig % (take 5 planes)) points)))) ; 26 (count (frequencies (take 100000 (map #(sig % (take 6 planes)) points)))) ; 41 (count (frequencies (take 1000000 (map #(sig % (take 1 planes)) points)))) ; 2 (count (frequencies (take 1000000 (map #(sig % (take 2 planes)) points)))) ; 4 (count (frequencies (take 1000000 (map #(sig % (take 3 planes)) points)))) ; 8 (count (frequencies (take 1000000 (map #(sig % (take 4 planes)) points)))) ; 15 (count (frequencies (take 1000000 (map #(sig % (take 5 planes)) points)))) ; 26 (count (frequencies (take 1000000 (map #(sig % (take 6 planes)) points)))) ; 42 ;; I'm painfully conscious of having stopped the experiment at the ;; exact point where I got the answer I expected. But my poor little ;; computer is not going to be up to running this for 10000000 points. ;; But this can't just be coincidence, surely?

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## Tuesday, May 20, 2014

### Planes In Space : Random Sampling

### Fizz Buzz : An Interview Question

;; Fizz Buzz ;; My sources ;; http://blog.codinghorror.com/why-cant-programmers-program/ ;; inform me that: ;; The majority of computer science graduates can't solve this problem: ;; Write a program that prints the numbers from 1 to 100. But for ;; multiples of three print "Fizz" instead of the number and for the ;; multiples of five print "Buzz". For numbers which are multiples of ;; both three and five print "FizzBuzz". ;; And Brother Downing of this parish, who actually hires people to ;; program in Java and Clojure learns me that he does indeed use this ;; to screen job applicants, and that most of them can't do it. ;; It is hard to read a thing like that without thinking: 'hang on, is that harder than it looks?' ;; So I did it, just to check: ;; I decided to use pull it out your ass driven development, where ;; you just pull the answer out of your ass. ;; First bit, print out the numbers from 1 to 100 (range 100) ;-> (0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 ...) ;; Bugger (range 1 101) ;-> (1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 ...) ;; paranoid now (last (range 1 101)) ;-> 100 ;; Although I guess really (print (range 1 101)) is what I want ;; here. In PIOYADD, we defer this important user interface question for later. ;; Next, print 'Fizz' instead of all the multiples of three (map (= 0 #(% quot 3)) (range 1 101)) ;; ClassCastException java.lang.Boolean cannot be cast to clojure.lang.IFn clojure.core/map/fn--4207 (core.clj:2485) ;; bugger (map #(= 0 (quot % 3)) (range 1 101)) ;-> (true true false false false false false false false false false false false false false false false false false false false false false false false false false ...) ;; ok (map #(if (= 0 (quot % 3)) "Fizz" %) (range 1 101)) ;-> ("Fizz" "Fizz" 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 ...) ;; oh for fuck's sake (map #(if (= 0 (mod % 3)) "Fizz" %) (range 1 101)) ;-> (1 2 "Fizz" 4 5 "Fizz" 7 8 "Fizz" 10 11 "Fizz" 13 14 "Fizz" 16 17 "Fizz" 19 20 "Fizz" 22 23 "Fizz" 25 26 "Fizz" ...) ;; payday (map #(case (= 0 (mod % 3)) "Fizz" (= 0 (mod % 5)) "Buzz" %) (range 1 101)) ;; IllegalArgumentException No matching clause: false user/eval1234/fn--1235 (NO_SOURCE_FILE:1) ;; ok, I always screw that up (map #(cond (= 0 (mod % 3)) "Fizz" (= 0 (mod % 5)) "Buzz" %) (range 1 101)) ;; CompilerException java.lang.IllegalArgumentException: cond requires an even number of forms, compiling:(NO_SOURCE_PATH:1:7) ;; twice usually (map #(cond (= 0 (mod % 3)) "Fizz" (= 0 (mod % 5)) "Buzz" :else %) (range 1 101)) ;-> (1 2 "Fizz" 4 "Buzz" "Fizz" 7 8 "Fizz" "Buzz" 11 "Fizz" 13 14 "Fizz" 16 17 "Fizz" 19 "Buzz" "Fizz" 22 23 "Fizz" "Buzz" 26 "Fizz" ...) ;; Bwahhahhhahhh! No, Mr Bond, I expect you to die. ;; And now, for the tricky bit of the problem, we unsheathe the ;; superweapon, copy-and-paste-driven development (map #(cond (or (= 0 (mod % 3)) (= 0 (mod % 3))) "FizzBuzz" (= 0 (mod % 5)) "Buzz" :else %) (range 1 101)) ;-> (1 2 "FizzBuzz" 4 "Buzz" "FizzBuzz" 7 8 "FizzBuzz" "Buzz" 11 "FizzBuzz" 13 14 "FizzBuzz" 16 17 "FizzBuzz" 19 "Buzz" "FizzBuzz" 22 23 "FizzBuzz" "Buzz" 26 "FizzBuzz" ...) ;; hmmm (map #(cond (and (= 0 (mod % 3)) (= 0 (mod % 3))) "FizzBuzz" (= 0 (mod % 5)) "Buzz" :else %) (range 1 101)) ;-> (1 2 "FizzBuzz" 4 "Buzz" "FizzBuzz" 7 8 "FizzBuzz" "Buzz" 11 "FizzBuzz" 13 14 "FizzBuzz" 16 17 "FizzBuzz" 19 "Buzz" "FizzBuzz" 22 23 "FizzBuzz" "Buzz" 26 "FizzBuzz" ...) ;; still wrong (map #(cond (and (= 0 (mod % 3)) (= 0 (mod % 5))) "FizzBuzz" (= 0 (mod % 5)) "Buzz" :else %) (range 1 101)) ;-> (1 2 3 4 "Buzz" 6 7 8 9 "Buzz" 11 12 13 14 "FizzBuzz" 16 17 18 19 "Buzz" 21 22 23 24 "Buzz" 26 27 ...) ;; aargh, where did the fizzes go? (map #(cond (and (= 0 (mod % 3)) (= 0 (mod % 5))) "FizzBuzz" (= 0 (mod % 3)) "Fizz" (= 0 (mod % 5)) "Buzz" :else %) (range 1 101)) ;-> (1 2 "Fizz" 4 "Buzz" "Fizz" 7 8 "Fizz" "Buzz" 11 "Fizz" 13 14 "FizzBuzz" 16 17 "Fizz" 19 "Buzz" "Fizz" 22 23 "Fizz" "Buzz" 26 "Fizz" ...) ;; Wahey! That looks done. Three minutes. ;; And now, pretending you're not a filthy hacker driven development: (map #(cond (and (= 0 (mod % 3)) (= 0 (mod % 5))) "FizzBuzz" (= 0 (mod % 3)) "Fizz" (= 0 (mod % 5)) "Buzz" :else %) (range 1 101)) ;-> (1 2 "Fizz" 4 "Buzz" "Fizz" 7 8 "Fizz" "Buzz" 11 "Fizz" 13 14 "FizzBuzz" 16 17 "Fizz" 19 "Buzz" "Fizz" 22 23 "Fizz" "Buzz" 26 "Fizz" ...) (defn divides? [n m] (= 0 (mod n m))) (divides? 3 15) ;-> false ;; sigh (defn divides? [n m] (= 0 (mod m n))) (divides? 3 15) ;-> true (divides? 15 3) ;-> false ;; rah! ;; and so, behold: beauty is truth, and truth, beauty (map #(cond (divides? 15 %) "FizzBuzz" (divides? 3 %) "Fizz" (divides? 5 %) "Buzz" :else %) (range 1 101)) ;-> (1 2 "Fizz" 4 "Buzz" "Fizz" 7 8 "Fizz" "Buzz" 11 "Fizz" 13 14 "FizzBuzz" 16 17 "Fizz" 19 "Buzz" "Fizz" 22 23 "Fizz" "Buzz" 26 "Fizz" ...) ;; finally, smugness driven development: (def fizzbuzz (map #(cond (divides? 15 %) "FizzBuzz" (divides? 3 %) "Fizz" (divides? 5 %) "Buzz" :else %) (map inc (range)))) ;; There are those who would call this 'premature abstraction', but they deserve not the names of men. (print (take 100 fizzbuzz)) ;; (1 2 Fizz 4 Buzz Fizz 7 8 Fizz Buzz 11 Fizz 13 14 FizzBuzz 16 17 Fizz 19 Buzz Fizz 22 23 Fizz Buzz 26 Fizz ...) ;; And I suppose a regression test would be nice, if I'm trying to ;; give some sort of professional impression: (= (take 27 fizzbuzz) (list 1 2 "Fizz" 4 "Buzz" "Fizz" 7 8 "Fizz" "Buzz" 11 "Fizz" 13 14 "FizzBuzz" 16 17 "Fizz" 19 "Buzz" "Fizz" 22 23 "Fizz" "Buzz" 26 "Fizz")) ;; And some paranoid checks, if I'm going to put this travesty on my blog: (count (filter #(= "Fizz" %) (take 1000 fizzbuzz))) ; -> 267 (count (filter #(= "FizzBuzz" %) (take 1000 fizzbuzz))) ;-> 66 (count (filter #(= "Buzz" %) (take 1000 fizzbuzz))) ;-> 134 (* (+ 134 66) 5) ;-> 1000 (* (+ 267 66) 3) ;-> 999 ;; bah, that's close enough for government work. I declare myself ;; done. Three minutes of flail and two minutes of tidying up and ;; checking it works. ;; So my question to the wider community is: Does my three minutes of ;; flailing trying to remember the semantics of my favourite language ;; (which I can perfectly imagine looking dreadful at an interview) ;; count as a fail or a pass? ;; In a previously unknown assembly language, but given a library to ;; print numbers on the screen, I can imagine taking half an hour to ;; get this working. Without the library, it's a research project. ;; And obviously, if I tried to do it in Haskell, I'd have to spend ;; three weeks remembering what a monad was in order to print the ;; damned thing out, but at least the type system would magically ;; guarantee the correctness of the final program. ;; In other words, is fizzbuzz really actually quite hard, or is ;; everyone out there a complete idiot?

## Monday, May 12, 2014

### Planes in Space : Checking by Hand

;; Ok, so I reckon: (defn regions [n m] (cond (= n 1) (list m (inc m)) (= m 0) (concat (repeat n 0) '(1)) :else (map + (regions n (dec m)) (concat '(0) (regions (dec n) (dec m))) (concat (regions (dec n) (dec m)) '(0))))) ;; In fact, I reckon that I can prove it. ;; It's quite easy to prove something that isn't true. Usually when ;; you run across a counterexample, that shows you where the ;; unsuspected false step in your reasoning was. ;; So I'd like to verify the formula on lots of examples. ;; But how? ;; These I can count in my head. (regions 3 0) ;-> (0 0 0 1) ; just space, no planes (regions 3 1) ;-> (0 0 1 2) ; a plane cuts space in half (regions 3 2) ;-> (0 1 4 4) ; two planes intersect in one line, cutting space into four (regions 3 3) ;-> (1 6 12 8) ; 1 point where three planes meet, ; six coordinate half-axes, ; three coordinate planes divided into four quadrants each, ; and eight octants ;; I am sort of confident that four planes make fifteen regions and intersect in four points. (regions 3 4) ;-> (4 18 28 15) ;; But every attempt to count the 18 lines and 28 regions I make ends up relying on ;; the sort of arguments I made to make the recursion in the first place. ;; And at this point my intuition breaks. (regions 3 5) ;-> (10 40 55 26) ;; I mean, five planes, any three intersect in a point, ;; 10 ways to choose 3 planes from five, so 10 points, ;; but after that I'm dead. ;; And this? Six choose 3 is 20, I can see that.... (regions 3 6) ;-> (20 75 96 42) ;; And I defy anyone to even picture (regions 5 8) ;-> (56 350 896 1176 792 219) ;; 8 choose 5 is (/ (* 8 7 6) (* 1 2 3)) = 56, which is fair enough. ;; 8 choose 4 is (/ (* 8 7 6 5) (* 1 2 3 4)) is 70 ;; If we take any 4 hyperplanes from our 8, they'll define a line. ;; The four remaining hyperplanes then divide each line like: (regions 1 4) ;-> (4 5) ;; So if each of those lines is sliced into 5 pieces then that's our 350. ;; But I'm just using the same recursive argument again. ;; So I don't even know if that's evidence or not. ;; One nice thing about it, the formula has an alternating sum property, ;; like the Euler index. (defn signature [lst] (reduce + (map * (apply concat (repeat '(+1 -1))) (reverse lst)))) (signature (regions 5 8)) ;-> 1 (def regions (memoize regions)) (regions 7 23) ;-> (245157 1817046 5787628 10271800 10973116 7057688 2531288 390656) (signature (regions 7 23)) ;-> 1 (regions 23 20) ;-> (0 0 0 1 40 760 9120 77520 496128 2480640 9922560 32248320 85995520 189190144 343982080 515973120 635043840 635043840 508035072 317521920 149422080 49807360 10485760 1048576) (signature (regions 23 20)) ;-> 1 ;; But annoyingly, you can just read that property straight from the recursion!

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