blog/posts/dyalog-apl-competition-2020-phase-2.org

• Problem 1 – Take a Dive
• Problem 2 – Another Step in the Proper Direction
• Problem 3 – Past Tasks Blast
• Problem 4 – Bioinformatics
• Problem 5 – Future and Present Value
• Problem 6 – Merge
• Problem 7 – UPC
• Problem 8 – Balancing the Scales
• Problem 9 – Upwardly Mobile

— title: "Dyalog APL Problem Solving Competition 2020 — Phase II" subtitle: "Annotated Solutions" date: 2020-07-31 toc: true —

  :Namespace Contest2020

	  :Namespace Problems
		  (⎕IO ⎕ML ⎕WX)←1 1 3

Problem 1 – Take a Dive

  ∇ score←dd DiveScore scores
    :If 7=≢scores
	    scores←scores[¯2↓2↓⍋scores]
    :ElseIf 5=≢scores
	    scores←scores[¯1↓1↓⍋scores]
    :Else
	    scores←scores
    :EndIf
    score←2(⍎⍕)dd×+/scores
  ∇

Problem 2 – Another Step in the Proper Direction

  ∇ steps←{p}Steps fromTo;segments;width
    width←|-/fromTo
    :If 0=⎕NC'p' ⍝ No left argument: same as Problem 5 of Phase I
	    segments←0,⍳width
    :ElseIf p<0 ⍝ -⌊p is the number of equally-sized steps to take
	    segments←(-⌊p){0,⍵×⍺÷⍨⍳⍺}width
    :ElseIf p>0 ⍝ p is the step size
	    segments←p{⍵⌊⍺×0,⍳⌈⍵÷⍺}width
    :ElseIf p=0 ⍝ As if we took zero step
	    segments←0
    :EndIf
    ⍝ Take into account the start point and the direction.
    steps←fromTo{(⊃⍺)+(-×-/⍺)×⍵}segments
  ∇

Problem 3 – Past Tasks Blast

  ∇ urls←PastTasks url;r;paths
    r←HttpCommand.Get url
    paths←('[a-zA-Z0-9_/]+\.pdf'⎕S'&')r.Data
    urls←('https://www.dyalog.com/'∘,)¨paths
  ∇

Problem 4 – Bioinformatics

  ⍝ Test if a DNA string is a reverse palindrome.
  isrevp←{⍵≡⌽'TAGC'['ATCG'⍳⍵]}

  ⍝ Generate all subarrays (position, length) pairs, for
  ⍝ 4 ≤ length ≤ 12.
  subarrays←{⊃,/(⍳⍵),¨¨3↓¨⍳¨12⌊1+⍵-⍳⍵}

  ∇ r←revp dna;positions
    positions←subarrays⍴dna
    ⍝ Filter subarrays which are reverse palindromes.
    r←↑({isrevp dna[¯1+⍵[1]+⍳⍵[2]]}¨positions)/positions
  ∇

  sset←{((1E6|2∘×)⍣⍵)1}

Problem 5 – Future and Present Value

  ⍝ First solution: ((1+⊢)⊥⊣) computes the total return
  ⍝ for a vector of amounts ⍺ and a vector of rates
  ⍝ ⍵. It is applied to every prefix subarray of amounts
  ⍝ and rates to get all intermediate values. However,
  ⍝ this has quadratic complexity.
  ⍝ rr←(,\⊣)((1+⊢)⊥⊣)¨(,\⊢)

  ⍝ Second solution: We want to be able to use the
  ⍝ recurrence relation (recur) and scan through the
  ⍝ vectors of amounts and rates, accumulating the total
  ⍝ value at every time step. However, APL evaluation is
  ⍝ right-associative, so a simple Scan
  ⍝ (recur\amounts,¨values) would not give the correct
  ⍝ result, since recur is not associative and we need
  ⍝ to evaluate it left-to-right. (In any case, in this
  ⍝ case, Scan would have quadratic complexity, so would
  ⍝ not bring any benefit over the previous solution.)
  ⍝ What we need is something akin to Haskell's scanl
  ⍝ function, which would evaluate left to right in O(n)
  ⍝ time. This is what we do here, accumulating values
  ⍝ from left to right. (This is inspired from
  ⍝ dfns.ascan, although heavily simplified.)
  rr←{recur←{⍵[1]+⍺×1+⍵[2]} ⋄ 1↓⌽⊃{(⊂(⊃⍵)recur⍺),⍵}/⌽⍺,¨⍵}

  ⍝ Simply apply the formula for cashflow calculations.
  pv←{+/⍺÷×\1+⍵}

Problem 6 – Merge

  ∇ val←ns getval var
    :If ''≡var ⍝ literal '@'
	    val←'@'
    :ElseIf (⊂var)∊ns.⎕NL ¯2
	    val←⍕ns⍎var
    :Else
	    val←'???'
    :EndIf
  ∇

  ∇ text←templateFile Merge jsonFile;template;ns
    template←⊃⎕NGET templateFile 1
    ns←⎕JSON⊃⎕NGET jsonFile
    ⍝ We use a simple regex search and replace on the
    ⍝ template.
    text←↑('@[a-zA-Z]*@'⎕R{ns getval ¯1↓1↓⍵.Match})template
  ∇

Problem 7 – UPC

  CheckDigit←{10|-⍵+.×11⍴3 1}

  ⍝ Left and right representations of digits. Decoding
  ⍝ the binary representation from decimal is more
  ⍝ compact than writing everything explicitly.
  lrepr←⍉(7⍴2)⊤13 25 19 61 35 49 47 59 55 11
  rrepr←~¨lrepr

  ∇ bits←WriteUPC digits;left;right
    :If (11=≢digits)∧∧/digits∊0,⍳9
	    left←,lrepr[1+6↑digits;]
	    right←,rrepr[1+6↓digits,CheckDigit digits;]
	    bits←1 0 1,left,0 1 0 1 0,right,1 0 1
    :Else
	    bits←¯1
    :EndIf
  ∇

  ∇ digits←ReadUPC bits
    :If 95≠⍴bits ⍝ incorrect number of bits
	    digits←¯1
    :Else
	    ⍝ Test if the barcode was scanned right-to-left.
	    :If 0=2|+/bits[3+⍳7]
		    bits←⌽bits
	    :EndIf
	    digits←({¯1+lrepr⍳⍵}¨(7/⍳6)⊆42↑3↓bits),{¯1+rrepr⍳⍵}¨(7/⍳6)⊆¯42↑¯3↓bits
	    :If ~∧/digits∊0,⍳9 ⍝ incorrect parity
		    digits←¯1
	    :ElseIf (⊃⌽digits)≠CheckDigit ¯1↓digits ⍝ incorrect check digit
		    digits←¯1
	    :EndIf
    :EndIf
  ∇

Problem 8 – Balancing the Scales

  ∇ parts←Balance nums;subsets;partitions
    ⍝ This is a brute force solution, running in
    ⍝ exponential time. We generate all the possible
    ⍝ partitions, filter out those which are not
    ⍝ balanced, and return the first matching one. There
    ⍝ are more advanced approach running in
    ⍝ pseudo-polynomial time (based on dynamic
    ⍝ programming, see the "Partition problem" Wikipedia
    ⍝ page), but they are not warranted here, as the
    ⍝ input size remains fairly small.

    ⍝ Generate all partitions of a vector of a given
    ⍝ size, as binary mask vectors.
    subsets←{1↓2⊥⍣¯1⍳2*⍵}
    ⍝ Keep only the subsets whose sum is exactly
    ⍝ (+/nums)÷2.
    partitions←nums{((2÷⍨+/⍺)=⍺+.×⍵)/⍵}subsets⍴nums
    :If 0=≢,partitions
	    ⍝ If no partition satisfy the above
	    ⍝ criterion, we return ⍬.
	    parts←⍬
    :Else
	    ⍝ Otherwise, we return the first possible
	    ⍝ partition.
	    parts←nums{((⊂,(⊂~))⊃↓⍉⍵)/¨2⍴⊂⍺}partitions
    :EndIf
  ∇

Problem 9 – Upwardly Mobile

  ∇ weights←Weights filename;mobile;branches;mat
    ⍝ Put your code and comments below here

    ⍝ Parse the mobile input file.
    mobile←↑⊃⎕NGET filename 1
    branches←⍸mobile∊'┌┴┐'
    ⍝ TODO: Build the matrix of coefficients mat.

    ⍝ Solve the system of equations (arbitrarily setting
    ⍝ the first variable at 1 because the system is
    ⍝ overdetermined), then multiply the coefficients by
    ⍝ their least common multiple to get the smallest
    ⍝ integer weights.
    weights←((1∘,)×(∧/÷))mat[;1]⌹1↓[2]mat
  ∇

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