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---
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title: "Dyalog APL Problem Solving Competition 2020"
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title: "Dyalog APL Problem Solving Competition 2020 — Phase I"
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subtitle: "Annotated Solutions"
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date: 2020-07-31
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toc: true
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---
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* Phase I
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#+begin_src default
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:Namespace Phase1
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#+end_src
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** 1. Let's Split!
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* 1. Let's Split!
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#+begin_quote
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Write a function that, given a right argument ~Y~ which is a scalar or
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@ -43,7 +41,7 @@ The second train ~(0>⊣)~ will return 1 if its left argument is
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positive. From this, we can use [[https://help.dyalog.com/18.0/index.htm#Language/Primitive%20Functions/Rotate.htm][Rotate]] (~⌽~) to correctly order the
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nested array, in the last train.
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** 2. Character Building
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* 2. Character Building
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#+begin_quote
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UTF-8 encodes Unicode characters using 1-4 integers for each
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@ -85,7 +83,7 @@ continuation character as 0, and all the others as 1. Next, we can use
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this binary array with [[https://help.dyalog.com/latest/#Language/Primitive%20Functions/Partitioned%20Enclose.htm][Partitioned Enclose]] (~⊂~) to return the correct
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output.
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** 3. Excel-lent Columns
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* 3. Excel-lent Columns
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#+begin_quote
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A Microsoft Excel spreadsheet numbers its rows counting up
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@ -106,7 +104,7 @@ the alphabet of every character. As a train, this can be done by
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letter from 1 to 26. The [[https://help.dyalog.com/latest/#Language/Primitive%20Functions/Decode.htm][Decode]] (~⊥~) function can then turn this
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base-26 number into the expected result.
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** 4. Take a Leap
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* 4. Take a Leap
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#+begin_quote
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Write a function that, given a right argument which is an integer
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@ -130,7 +128,7 @@ divisible by one of these numbers. If the count is 1 or 3, it is a
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leap year. Note that we use [[https://help.dyalog.com/latest/#Language/Primitive%20Operators/Commute.htm][Commute]] (~⍨~) to keep the dfn as a train,
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and to preserve the natural right-to-left reading of the algorithm.
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** 5. Stepping in the Proper Direction
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* 5. Stepping in the Proper Direction
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#+begin_quote
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Write a function that, given a right argument of 2 integers, returns a
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@ -151,7 +149,7 @@ decreasing if needed, so we multiply it by the opposite of the sign of
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~-/⍵~. Finally, we just have to start the sequence at the first
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element of ~⍵~.
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** 6. Please Move to the Front
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* 6. Please Move to the Front
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#+begin_quote
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Write a function that, given a right argument which is an integer
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@ -166,7 +164,7 @@ while all other elements keep their order.
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the left argument. Using this index to sort in the usual way with
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[[https://help.dyalog.com/latest/#Language/Primitive%20Functions/Grade%20Up%20Monadic.htm][Grade Up]] will return the expected result.
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** 7. See You in a Bit
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* 7. See You in a Bit
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#+begin_quote
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A common technique for encoding a set of on/off states is to use a
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@ -198,7 +196,7 @@ case). That is what the function ~f~ does.
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Next, we just need to check that all elements of ~f⍺~ are also in
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~f⍵~.
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** 8. Zigzag Numbers
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* 8. Zigzag Numbers
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#+begin_quote
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A zigzag number is an integer in which the difference in magnitude of
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@ -217,7 +215,7 @@ First, we decompose a number into an array of digits, using
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compute the difference between each pair of digits, take the sign, and
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ensure that the signs are indeed alternating.
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** 9. Rise and Fall
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* 9. Rise and Fall
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#+begin_quote
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Write a function that, given a right argument which is an integer
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@ -256,7 +254,7 @@ Next, ~(⍳∘≢≡⍋)~ on each of the two vectors will test if they are
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non-decreasing (i.e. if the ranks of all the elements correspond to a
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simple range from 1 to the size of the vector).
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** 10. Stacking It Up
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* 10. Stacking It Up
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#+begin_quote
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Write a function that takes as its right argument a vector of simple
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@ -317,238 +315,3 @@ final result.
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:EndNamespace
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#+end_src
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* Phase II
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#+begin_src default
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:Namespace Contest2020
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:Namespace Problems
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(⎕IO ⎕ML ⎕WX)←1 1 3
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#+end_src
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** Problem 1 -- Take a Dive
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#+begin_src default
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∇ score←dd DiveScore scores
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:If 7=≢scores
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scores←scores[¯2↓2↓⍋scores]
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:ElseIf 5=≢scores
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scores←scores[¯1↓1↓⍋scores]
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:Else
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scores←scores
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:EndIf
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score←2(⍎⍕)dd×+/scores
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∇
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#+end_src
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** Problem 2 -- Another Step in the Proper Direction
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#+begin_src default
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∇ steps←{p}Steps fromTo;segments;width
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width←|-/fromTo
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:If 0=⎕NC'p' ⍝ No left argument: same as Problem 5 of Phase I
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segments←0,⍳width
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:ElseIf p<0 ⍝ -⌊p is the number of equally-sized steps to take
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segments←(-⌊p){0,⍵×⍺÷⍨⍳⍺}width
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:ElseIf p>0 ⍝ p is the step size
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segments←p{⍵⌊⍺×0,⍳⌈⍵÷⍺}width
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:ElseIf p=0 ⍝ As if we took zero step
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segments←0
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:EndIf
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⍝ Take into account the start point and the direction.
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steps←fromTo{(⊃⍺)+(-×-/⍺)×⍵}segments
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∇
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#+end_src
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** Problem 3 -- Past Tasks Blast
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#+begin_src default
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∇ urls←PastTasks url;r;paths
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r←HttpCommand.Get url
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paths←('[a-zA-Z0-9_/]+\.pdf'⎕S'&')r.Data
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urls←('https://www.dyalog.com/'∘,)¨paths
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∇
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#+end_src
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** Problem 4 -- Bioinformatics
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#+begin_src default
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⍝ Test if a DNA string is a reverse palindrome.
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isrevp←{⍵≡⌽'TAGC'['ATCG'⍳⍵]}
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⍝ Generate all subarrays (position, length) pairs, for
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⍝ 4 ≤ length ≤ 12.
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subarrays←{⊃,/(⍳⍵),¨¨3↓¨⍳¨12⌊1+⍵-⍳⍵}
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∇ r←revp dna;positions
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positions←subarrays⍴dna
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⍝ Filter subarrays which are reverse palindromes.
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r←↑({isrevp dna[¯1+⍵[1]+⍳⍵[2]]}¨positions)/positions
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∇
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#+end_src
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#+begin_src default
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sset←{((1E6|2∘×)⍣⍵)1}
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#+end_src
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** Problem 5 -- Future and Present Value
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#+begin_src default
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⍝ First solution: ((1+⊢)⊥⊣) computes the total return
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⍝ for a vector of amounts ⍺ and a vector of rates
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⍝ ⍵. It is applied to every prefix subarray of amounts
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⍝ and rates to get all intermediate values. However,
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⍝ this has quadratic complexity.
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⍝ rr←(,\⊣)((1+⊢)⊥⊣)¨(,\⊢)
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⍝ Second solution: We want to be able to use the
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⍝ recurrence relation (recur) and scan through the
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⍝ vectors of amounts and rates, accumulating the total
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⍝ value at every time step. However, APL evaluation is
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⍝ right-associative, so a simple Scan
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⍝ (recur\amounts,¨values) would not give the correct
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⍝ result, since recur is not associative and we need
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⍝ to evaluate it left-to-right. (In any case, in this
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⍝ case, Scan would have quadratic complexity, so would
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⍝ not bring any benefit over the previous solution.)
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⍝ What we need is something akin to Haskell's scanl
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⍝ function, which would evaluate left to right in O(n)
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⍝ time. This is what we do here, accumulating values
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⍝ from left to right. (This is inspired from
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⍝ dfns.ascan, although heavily simplified.)
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rr←{recur←{⍵[1]+⍺×1+⍵[2]} ⋄ 1↓⌽⊃{(⊂(⊃⍵)recur⍺),⍵}/⌽⍺,¨⍵}
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#+end_src
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#+begin_src default
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⍝ Simply apply the formula for cashflow calculations.
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pv←{+/⍺÷×\1+⍵}
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#+end_src
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** Problem 6 -- Merge
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#+begin_src default
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∇ val←ns getval var
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:If ''≡var ⍝ literal '@'
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val←'@'
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:ElseIf (⊂var)∊ns.⎕NL ¯2
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val←⍕ns⍎var
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:Else
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val←'???'
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:EndIf
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∇
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#+end_src
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#+begin_src default
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∇ text←templateFile Merge jsonFile;template;ns
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template←⊃⎕NGET templateFile 1
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ns←⎕JSON⊃⎕NGET jsonFile
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⍝ We use a simple regex search and replace on the
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⍝ template.
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text←↑('@[a-zA-Z]*@'⎕R{ns getval ¯1↓1↓⍵.Match})template
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∇
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#+end_src
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** Problem 7 -- UPC
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#+begin_src default
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CheckDigit←{10|-⍵+.×11⍴3 1}
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#+end_src
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#+begin_src default
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⍝ Left and right representations of digits. Decoding
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⍝ the binary representation from decimal is more
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⍝ compact than writing everything explicitly.
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lrepr←⍉(7⍴2)⊤13 25 19 61 35 49 47 59 55 11
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rrepr←~¨lrepr
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#+end_src
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#+begin_src default
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∇ bits←WriteUPC digits;left;right
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:If (11=≢digits)∧∧/digits∊0,⍳9
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left←,lrepr[1+6↑digits;]
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right←,rrepr[1+6↓digits,CheckDigit digits;]
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bits←1 0 1,left,0 1 0 1 0,right,1 0 1
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:Else
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bits←¯1
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:EndIf
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∇
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#+end_src
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#+begin_src default
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∇ digits←ReadUPC bits
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:If 95≠⍴bits ⍝ incorrect number of bits
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digits←¯1
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:Else
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⍝ Test if the barcode was scanned right-to-left.
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:If 0=2|+/bits[3+⍳7]
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bits←⌽bits
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:EndIf
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digits←({¯1+lrepr⍳⍵}¨(7/⍳6)⊆42↑3↓bits),{¯1+rrepr⍳⍵}¨(7/⍳6)⊆¯42↑¯3↓bits
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:If ~∧/digits∊0,⍳9 ⍝ incorrect parity
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digits←¯1
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:ElseIf (⊃⌽digits)≠CheckDigit ¯1↓digits ⍝ incorrect check digit
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digits←¯1
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:EndIf
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:EndIf
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∇
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#+end_src
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** Problem 8 -- Balancing the Scales
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#+begin_src default
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∇ parts←Balance nums;subsets;partitions
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⍝ This is a brute force solution, running in
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⍝ exponential time. We generate all the possible
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⍝ partitions, filter out those which are not
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⍝ balanced, and return the first matching one. There
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⍝ are more advanced approach running in
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⍝ pseudo-polynomial time (based on dynamic
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⍝ programming, see the "Partition problem" Wikipedia
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⍝ page), but they are not warranted here, as the
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⍝ input size remains fairly small.
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⍝ Generate all partitions of a vector of a given
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⍝ size, as binary mask vectors.
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subsets←{1↓2⊥⍣¯1⍳2*⍵}
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⍝ Keep only the subsets whose sum is exactly
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⍝ (+/nums)÷2.
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partitions←nums{((2÷⍨+/⍺)=⍺+.×⍵)/⍵}subsets⍴nums
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:If 0=≢,partitions
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⍝ If no partition satisfy the above
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⍝ criterion, we return ⍬.
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parts←⍬
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:Else
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⍝ Otherwise, we return the first possible
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⍝ partition.
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parts←nums{((⊂,(⊂~))⊃↓⍉⍵)/¨2⍴⊂⍺}partitions
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:EndIf
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∇
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#+end_src
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** Problem 9 -- Upwardly Mobile
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#+begin_src default
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∇ weights←Weights filename;mobile;branches;mat
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⍝ Put your code and comments below here
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⍝ Parse the mobile input file.
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mobile←↑⊃⎕NGET filename 1
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branches←⍸mobile∊'┌┴┐'
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⍝ TODO: Build the matrix of coefficients mat.
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⍝ Solve the system of equations (arbitrarily setting
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⍝ the first variable at 1 because the system is
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⍝ overdetermined), then multiply the coefficients by
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⍝ their least common multiple to get the smallest
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⍝ integer weights.
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weights←((1∘,)×(∧/÷))mat[;1]⌹1↓[2]mat
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∇
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#+end_src
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#+begin_src default
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:EndNamespace
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:EndNamespace
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#+end_src
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* General Remarks
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238
posts/dyalog-apl-competition-2020-phase-2.org
Normal file
238
posts/dyalog-apl-competition-2020-phase-2.org
Normal file
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@ -0,0 +1,238 @@
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---
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title: "Dyalog APL Problem Solving Competition 2020 — Phase II"
|
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subtitle: "Annotated Solutions"
|
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date: 2020-07-31
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toc: true
|
||||
---
|
||||
|
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#+begin_src default
|
||||
:Namespace Contest2020
|
||||
|
||||
:Namespace Problems
|
||||
(⎕IO ⎕ML ⎕WX)←1 1 3
|
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#+end_src
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* Problem 1 -- Take a Dive
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#+begin_src default
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∇ score←dd DiveScore scores
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:If 7=≢scores
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scores←scores[¯2↓2↓⍋scores]
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:ElseIf 5=≢scores
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scores←scores[¯1↓1↓⍋scores]
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:Else
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scores←scores
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:EndIf
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score←2(⍎⍕)dd×+/scores
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∇
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#+end_src
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* Problem 2 -- Another Step in the Proper Direction
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#+begin_src default
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||||
∇ steps←{p}Steps fromTo;segments;width
|
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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
|
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segments←(-⌊p){0,⍵×⍺÷⍨⍳⍺}width
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:ElseIf p>0 ⍝ p is the step size
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||||
segments←p{⍵⌊⍺×0,⍳⌈⍵÷⍺}width
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||||
:ElseIf p=0 ⍝ As if we took zero step
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segments←0
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||||
:EndIf
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⍝ Take into account the start point and the direction.
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steps←fromTo{(⊃⍺)+(-×-/⍺)×⍵}segments
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∇
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||||
#+end_src
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* Problem 3 -- Past Tasks Blast
|
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|
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#+begin_src default
|
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∇ urls←PastTasks url;r;paths
|
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r←HttpCommand.Get url
|
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paths←('[a-zA-Z0-9_/]+\.pdf'⎕S'&')r.Data
|
||||
urls←('https://www.dyalog.com/'∘,)¨paths
|
||||
∇
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||||
#+end_src
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* Problem 4 -- Bioinformatics
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|
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#+begin_src default
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||||
⍝ 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+⍵-⍳⍵}
|
||||
|
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∇ r←revp dna;positions
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positions←subarrays⍴dna
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⍝ Filter subarrays which are reverse palindromes.
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r←↑({isrevp dna[¯1+⍵[1]+⍳⍵[2]]}¨positions)/positions
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∇
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||||
#+end_src
|
||||
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||||
#+begin_src default
|
||||
sset←{((1E6|2∘×)⍣⍵)1}
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||||
#+end_src
|
||||
|
||||
* Problem 5 -- Future and Present Value
|
||||
|
||||
#+begin_src default
|
||||
⍝ 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⍺),⍵}/⌽⍺,¨⍵}
|
||||
#+end_src
|
||||
|
||||
#+begin_src default
|
||||
⍝ Simply apply the formula for cashflow calculations.
|
||||
pv←{+/⍺÷×\1+⍵}
|
||||
#+end_src
|
||||
|
||||
* Problem 6 -- Merge
|
||||
|
||||
#+begin_src default
|
||||
∇ val←ns getval var
|
||||
:If ''≡var ⍝ literal '@'
|
||||
val←'@'
|
||||
:ElseIf (⊂var)∊ns.⎕NL ¯2
|
||||
val←⍕ns⍎var
|
||||
:Else
|
||||
val←'???'
|
||||
:EndIf
|
||||
∇
|
||||
#+end_src
|
||||
|
||||
#+begin_src default
|
||||
∇ 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
|
||||
∇
|
||||
#+end_src
|
||||
|
||||
* Problem 7 -- UPC
|
||||
|
||||
#+begin_src default
|
||||
CheckDigit←{10|-⍵+.×11⍴3 1}
|
||||
#+end_src
|
||||
|
||||
#+begin_src default
|
||||
⍝ 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
|
||||
#+end_src
|
||||
|
||||
#+begin_src default
|
||||
∇ 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
|
||||
∇
|
||||
#+end_src
|
||||
|
||||
#+begin_src default
|
||||
∇ 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
|
||||
∇
|
||||
#+end_src
|
||||
|
||||
* Problem 8 -- Balancing the Scales
|
||||
|
||||
#+begin_src default
|
||||
∇ 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
|
||||
∇
|
||||
#+end_src
|
||||
|
||||
* Problem 9 -- Upwardly Mobile
|
||||
|
||||
#+begin_src default
|
||||
∇ 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
|
||||
∇
|
||||
#+end_src
|
||||
|
||||
#+begin_src default
|
||||
:EndNamespace
|
||||
:EndNamespace
|
||||
#+end_src
|
||||
Loading…
Add table
Add a link
Reference in a new issue