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259
posts/dyalog-apl-competition-2020.org → posts/dyalog-apl-competition-2020-phase-1.org
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@ -1,17 +1,15 @@
--- ---
title: "Dyalog APL Problem Solving Competition 2020" title: "Dyalog APL Problem Solving Competition 2020 — Phase I"
subtitle: "Annotated Solutions" subtitle: "Annotated Solutions"
date: 2020-07-31 date: 2020-07-31
toc: true toc: true
--- ---
* Phase I
#+begin_src default #+begin_src default
:Namespace Phase1 :Namespace Phase1
#+end_src #+end_src
** 1. Let's Split! * 1. Let's Split!
#+begin_quote #+begin_quote
Write a function that, given a right argument ~Y~ which is a scalar or Write a function that, given a right argument ~Y~ which is a scalar or
@ -43,7 +41,7 @@ The second train ~(0>⊣)~ will return 1 if its left argument is
positive. From this, we can use [[https://help.dyalog.com/18.0/index.htm#Language/Primitive%20Functions/Rotate.htm][Rotate]] (~⌽~) to correctly order the positive. From this, we can use [[https://help.dyalog.com/18.0/index.htm#Language/Primitive%20Functions/Rotate.htm][Rotate]] (~⌽~) to correctly order the
nested array, in the last train. nested array, in the last train.
** 2. Character Building * 2. Character Building
#+begin_quote #+begin_quote
UTF-8 encodes Unicode characters using 1-4 integers for each UTF-8 encodes Unicode characters using 1-4 integers for each
@ -85,7 +83,7 @@ continuation character as 0, and all the others as 1. Next, we can use
this binary array with [[https://help.dyalog.com/latest/#Language/Primitive%20Functions/Partitioned%20Enclose.htm][Partitioned Enclose]] (~⊂~) to return the correct this binary array with [[https://help.dyalog.com/latest/#Language/Primitive%20Functions/Partitioned%20Enclose.htm][Partitioned Enclose]] (~⊂~) to return the correct
output. output.
** 3. Excel-lent Columns * 3. Excel-lent Columns
#+begin_quote #+begin_quote
A Microsoft Excel spreadsheet numbers its rows counting up A Microsoft Excel spreadsheet numbers its rows counting up
@ -106,7 +104,7 @@ the alphabet of every character. As a train, this can be done by
letter from 1 to 26. The [[https://help.dyalog.com/latest/#Language/Primitive%20Functions/Decode.htm][Decode]] (~⊥~) function can then turn this letter from 1 to 26. The [[https://help.dyalog.com/latest/#Language/Primitive%20Functions/Decode.htm][Decode]] (~⊥~) function can then turn this
base-26 number into the expected result. base-26 number into the expected result.
** 4. Take a Leap * 4. Take a Leap
#+begin_quote #+begin_quote
Write a function that, given a right argument which is an integer Write a function that, given a right argument which is an integer
@ -130,7 +128,7 @@ divisible by one of these numbers. If the count is 1 or 3, it is a
leap year. Note that we use [[https://help.dyalog.com/latest/#Language/Primitive%20Operators/Commute.htm][Commute]] (~⍨~) to keep the dfn as a train, leap year. Note that we use [[https://help.dyalog.com/latest/#Language/Primitive%20Operators/Commute.htm][Commute]] (~⍨~) to keep the dfn as a train,
and to preserve the natural right-to-left reading of the algorithm. and to preserve the natural right-to-left reading of the algorithm.
** 5. Stepping in the Proper Direction * 5. Stepping in the Proper Direction
#+begin_quote #+begin_quote
Write a function that, given a right argument of 2 integers, returns a Write a function that, given a right argument of 2 integers, returns a
@ -151,7 +149,7 @@ decreasing if needed, so we multiply it by the opposite of the sign of
~-/⍵~. Finally, we just have to start the sequence at the first ~-/⍵~. Finally, we just have to start the sequence at the first
element of ~⍵~. element of ~⍵~.
** 6. Please Move to the Front * 6. Please Move to the Front
#+begin_quote #+begin_quote
Write a function that, given a right argument which is an integer Write a function that, given a right argument which is an integer
@ -166,7 +164,7 @@ while all other elements keep their order.
the left argument. Using this index to sort in the usual way with the left argument. Using this index to sort in the usual way with
[[https://help.dyalog.com/latest/#Language/Primitive%20Functions/Grade%20Up%20Monadic.htm][Grade Up]] will return the expected result. [[https://help.dyalog.com/latest/#Language/Primitive%20Functions/Grade%20Up%20Monadic.htm][Grade Up]] will return the expected result.
** 7. See You in a Bit * 7. See You in a Bit
#+begin_quote #+begin_quote
A common technique for encoding a set of on/off states is to use a A common technique for encoding a set of on/off states is to use a
@ -198,7 +196,7 @@ case). That is what the function ~f~ does.
Next, we just need to check that all elements of ~f⍺~ are also in Next, we just need to check that all elements of ~f⍺~ are also in
~f⍵~. ~f⍵~.
** 8. Zigzag Numbers * 8. Zigzag Numbers
#+begin_quote #+begin_quote
A zigzag number is an integer in which the difference in magnitude of A zigzag number is an integer in which the difference in magnitude of
@ -217,7 +215,7 @@ First, we decompose a number into an array of digits, using
compute the difference between each pair of digits, take the sign, and compute the difference between each pair of digits, take the sign, and
ensure that the signs are indeed alternating. ensure that the signs are indeed alternating.
** 9. Rise and Fall * 9. Rise and Fall
#+begin_quote #+begin_quote
Write a function that, given a right argument which is an integer Write a function that, given a right argument which is an integer
@ -256,7 +254,7 @@ Next, ~(⍳∘≢≡⍋)~ on each of the two vectors will test if they are
non-decreasing (i.e. if the ranks of all the elements correspond to a non-decreasing (i.e. if the ranks of all the elements correspond to a
simple range from 1 to the size of the vector). simple range from 1 to the size of the vector).
** 10. Stacking It Up * 10. Stacking It Up
#+begin_quote #+begin_quote
Write a function that takes as its right argument a vector of simple Write a function that takes as its right argument a vector of simple
@ -317,238 +315,3 @@ final result.
:EndNamespace :EndNamespace
#+end_src #+end_src
* Phase II
#+begin_src default
:Namespace Contest2020
:Namespace Problems
(⎕IO ⎕ML ⎕WX)←1 1 3
#+end_src
** Problem 1 -- Take a Dive
#+begin_src default
∇ 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
∇
#+end_src
** Problem 2 -- Another Step in the Proper Direction
#+begin_src default
∇ 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
∇
#+end_src
** Problem 3 -- Past Tasks Blast
#+begin_src default
∇ urls←PastTasks url;r;paths
r←HttpCommand.Get url
paths←('[a-zA-Z0-9_/]+\.pdf'⎕S'&')r.Data
urls←('https://www.dyalog.com/'∘,)¨paths
∇
#+end_src
** Problem 4 -- Bioinformatics
#+begin_src default
⍝ 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
∇
#+end_src
#+begin_src default
sset←{((1E6|2∘×)⍣⍵)1}
#+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
* General Remarks

238
posts/dyalog-apl-competition-2020-phase-2.org Normal file
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@ -0,0 +1,238 @@
---
title: "Dyalog APL Problem Solving Competition 2020 — Phase II"
subtitle: "Annotated Solutions"
date: 2020-07-31
toc: true
---
#+begin_src default
:Namespace Contest2020
:Namespace Problems
(⎕IO ⎕ML ⎕WX)←1 1 3
#+end_src
* Problem 1 -- Take a Dive
#+begin_src default
∇ 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
∇
#+end_src
* Problem 2 -- Another Step in the Proper Direction
#+begin_src default
∇ 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
∇
#+end_src
* Problem 3 -- Past Tasks Blast
#+begin_src default
∇ urls←PastTasks url;r;paths
r←HttpCommand.Get url
paths←('[a-zA-Z0-9_/]+\.pdf'⎕S'&')r.Data
urls←('https://www.dyalog.com/'∘,)¨paths
∇
#+end_src
* Problem 4 -- Bioinformatics
#+begin_src default
⍝ 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
∇
#+end_src
#+begin_src default
sset←{((1E6|2∘×)⍣⍵)1}
#+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