From 25688997a271f6019ff74fcd22f075a337d1c869 Mon Sep 17 00:00:00 2001 From: Dimitri Lozeve Date: Sun, 2 Aug 2020 17:20:58 +0200 Subject: [PATCH] Dyalog APL competition phase 2: Add explanations for problems 1-4 --- _site/atom.xml | 272 +++++++++--------- .../dyalog-apl-competition-2020-phase-2.html | 272 +++++++++--------- _site/rss.xml | 272 +++++++++--------- posts/dyalog-apl-competition-2020-phase-2.org | 84 +++++- 4 files changed, 514 insertions(+), 386 deletions(-) diff --git a/_site/atom.xml b/_site/atom.xml index 49be1c1..30bc584 100644 --- a/_site/atom.xml +++ b/_site/atom.xml @@ -54,6 +54,8 @@ :EndIf score←2(⍎⍕)dd×+/scores ∇ +

This is a very straightforward implementation of the algorithm describe in the problem description. I decided to switch explicitly on the size of the input vector because I feel it is more natural. For the cases with 5 or 7 judges, we use Drop (↓) to remove the lowest and highest scores.

+

At the end, we sum up the scores with +/ and multiply them by dd. The last operation, 2(⍎⍕), is a train using Format (Dyadic) to round to 2 decimal places, and Execute to get actual numbers and not strings.

Problem 2 – Another Step in the Proper Direction

∇ steps←{p}Steps fromTo;segments;width
   width←|-/fromTo
@@ -69,147 +71,161 @@
   ⍝ Take into account the start point and the direction.
   steps←fromTo{(⊃⍺)+(-×-/⍺)×⍵}segments
 ∇
+

This is an extension to Problem 5 of Phase I. In each case, we compute the “segments”, i.e., the steps starting from 0. In a last step, common to all cases, we add the correct starting point and correct the direction if need be.

+

To compute equally-sized steps, we first divide the segment \([0, 1]\) in p equal segments with (⍳p)÷p. This subdivision can then be multiplied by the width to obtain the required segments.

+

When p is the step size, we just divide the width by the step size (rounded to the next largest integer) to get the required number of segments. If the last segment is too large, we “crop” it to the width with Minimum (⌊).

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
 ∇
+

I decided to use HttpCommand for this task, since it is simply one ]load HttpCommand away and should be platform-independent.

+

Parsing XML is not something I consider “fun” in the best of cases, and I feel like APL is not the best language to do this kind of thing. Given how simple the task is, I just decided to find the relevant bits with a regular expression using Replace and Search (⎕S).

+

After finding all the strings vaguely resembling a PDF file name (only alphanumeric characters and underscores, with a .pdf extension), I just concatenate them to the base URL of the Dyalog domain.

Problem 4 – Bioinformatics

+

The first task can be solved by decomposing it into several functions.

⍝ 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}
+isrevp←{⍵≡⌽'TAGC'['ATCG'⍳⍵]} +

First, we compute the complement of a DNA string (using simple indexing) and test if its Reverse (⌽) is equal to the original string.

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

We first compute all the possible lengths for each starting point. For instance, the last element cannot have any (position, length) pair associated to it, because there is no three element following it. So we crop the possible lengths to \([3, 12]\). For instance for an array of size 10:

+
        {3↓¨⍳¨12⌊1+⍵-⍳⍵}10
+┌──────────────┬───────────┬─────────┬───────┬─────┬───┬─┬┬┬┐
+│4 5 6 7 8 9 10│4 5 6 7 8 9│4 5 6 7 8│4 5 6 7│4 5 6│4 5│4││││
+└──────────────┴───────────┴─────────┴───────┴─────┴───┴─┴┴┴┘
+

Then, we just add the corresponding starting position to each length (1 for the first block, 2 for the second, and so on). Finally, we flatten everything.

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

For each possible (position, length) pair, we get the corresponding DNA substring with dna[¯1+⍵[1]+⍳⍵[2]] (adding ¯1 is necessary because ⎕IO←1). We test if this substring is a reverse palindrome using isrevp above. Replicate (/) then selects only the (position, length) pairs for which the substring is a reverse palindrome.

+

The second task is just about counting the number of subsets modulo 1,000,000. So we just need to compute \(2^n \mod 1000000\) for any positive integer \(n\leq1000\).

+
sset←{((1E6|2∘×)⍣⍵)1}
+

Since we cannot just compute \(2^n\) directly and take the remainder, we use modular arithmetic to stay mod 1,000,000 during the whole computation. The dfn (1E6|2∘×) doubles its argument mod 1,000,000. So we just apply this function \(n\) times using the Power operator (⍣), with an initial value of 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+⍵}
+
⍝ 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
-∇
+
∇ 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
-∇
+
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
-∇
+
∇ 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
-∇
-
    :EndNamespace
-:EndNamespace
+
∇ 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
+∇
+
    :EndNamespace
+:EndNamespace
]]> diff --git a/_site/posts/dyalog-apl-competition-2020-phase-2.html b/_site/posts/dyalog-apl-competition-2020-phase-2.html index d25811b..fbe43aa 100644 --- a/_site/posts/dyalog-apl-competition-2020-phase-2.html +++ b/_site/posts/dyalog-apl-competition-2020-phase-2.html @@ -88,6 +88,8 @@ :EndIf score←2(⍎⍕)dd×+/scores ∇ +

This is a very straightforward implementation of the algorithm describe in the problem description. I decided to switch explicitly on the size of the input vector because I feel it is more natural. For the cases with 5 or 7 judges, we use Drop (↓) to remove the lowest and highest scores.

+

At the end, we sum up the scores with +/ and multiply them by dd. The last operation, 2(⍎⍕), is a train using Format (Dyadic) to round to 2 decimal places, and Execute to get actual numbers and not strings.

Problem 2 – Another Step in the Proper Direction

∇ steps←{p}Steps fromTo;segments;width
   width←|-/fromTo
@@ -103,147 +105,161 @@
   ⍝ Take into account the start point and the direction.
   steps←fromTo{(⊃⍺)+(-×-/⍺)×⍵}segments
 ∇
+

This is an extension to Problem 5 of Phase I. In each case, we compute the “segments”, i.e., the steps starting from 0. In a last step, common to all cases, we add the correct starting point and correct the direction if need be.

+

To compute equally-sized steps, we first divide the segment \([0, 1]\) in p equal segments with (⍳p)÷p. This subdivision can then be multiplied by the width to obtain the required segments.

+

When p is the step size, we just divide the width by the step size (rounded to the next largest integer) to get the required number of segments. If the last segment is too large, we “crop” it to the width with Minimum (⌊).

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
 ∇
+

I decided to use HttpCommand for this task, since it is simply one ]load HttpCommand away and should be platform-independent.

+

Parsing XML is not something I consider “fun” in the best of cases, and I feel like APL is not the best language to do this kind of thing. Given how simple the task is, I just decided to find the relevant bits with a regular expression using Replace and Search (⎕S).

+

After finding all the strings vaguely resembling a PDF file name (only alphanumeric characters and underscores, with a .pdf extension), I just concatenate them to the base URL of the Dyalog domain.

Problem 4 – Bioinformatics

+

The first task can be solved by decomposing it into several functions.

⍝ 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}
+isrevp←{⍵≡⌽'TAGC'['ATCG'⍳⍵]} +

First, we compute the complement of a DNA string (using simple indexing) and test if its Reverse (⌽) is equal to the original string.

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

We first compute all the possible lengths for each starting point. For instance, the last element cannot have any (position, length) pair associated to it, because there is no three element following it. So we crop the possible lengths to \([3, 12]\). For instance for an array of size 10:

+
        {3↓¨⍳¨12⌊1+⍵-⍳⍵}10
+┌──────────────┬───────────┬─────────┬───────┬─────┬───┬─┬┬┬┐
+│4 5 6 7 8 9 10│4 5 6 7 8 9│4 5 6 7 8│4 5 6 7│4 5 6│4 5│4││││
+└──────────────┴───────────┴─────────┴───────┴─────┴───┴─┴┴┴┘
+

Then, we just add the corresponding starting position to each length (1 for the first block, 2 for the second, and so on). Finally, we flatten everything.

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

For each possible (position, length) pair, we get the corresponding DNA substring with dna[¯1+⍵[1]+⍳⍵[2]] (adding ¯1 is necessary because ⎕IO←1). We test if this substring is a reverse palindrome using isrevp above. Replicate (/) then selects only the (position, length) pairs for which the substring is a reverse palindrome.

+

The second task is just about counting the number of subsets modulo 1,000,000. So we just need to compute \(2^n \mod 1000000\) for any positive integer \(n\leq1000\).

+
sset←{((1E6|2∘×)⍣⍵)1}
+

Since we cannot just compute \(2^n\) directly and take the remainder, we use modular arithmetic to stay mod 1,000,000 during the whole computation. The dfn (1E6|2∘×) doubles its argument mod 1,000,000. So we just apply this function \(n\) times using the Power operator (⍣), with an initial value of 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+⍵}
+
⍝ 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
-∇
+
∇ 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
-∇
+
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
-∇
+
∇ 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
-∇
-
    :EndNamespace
-:EndNamespace
+
∇ 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
+∇
+
    :EndNamespace
+:EndNamespace
diff --git a/_site/rss.xml b/_site/rss.xml index dcc0f47..2fed2f1 100644 --- a/_site/rss.xml +++ b/_site/rss.xml @@ -50,6 +50,8 @@ :EndIf score←2(⍎⍕)dd×+/scores ∇ +

This is a very straightforward implementation of the algorithm describe in the problem description. I decided to switch explicitly on the size of the input vector because I feel it is more natural. For the cases with 5 or 7 judges, we use Drop (↓) to remove the lowest and highest scores.

+

At the end, we sum up the scores with +/ and multiply them by dd. The last operation, 2(⍎⍕), is a train using Format (Dyadic) to round to 2 decimal places, and Execute to get actual numbers and not strings.

Problem 2 – Another Step in the Proper Direction

∇ steps←{p}Steps fromTo;segments;width
   width←|-/fromTo
@@ -65,147 +67,161 @@
   ⍝ Take into account the start point and the direction.
   steps←fromTo{(⊃⍺)+(-×-/⍺)×⍵}segments
 ∇
+

This is an extension to Problem 5 of Phase I. In each case, we compute the “segments”, i.e., the steps starting from 0. In a last step, common to all cases, we add the correct starting point and correct the direction if need be.

+

To compute equally-sized steps, we first divide the segment \([0, 1]\) in p equal segments with (⍳p)÷p. This subdivision can then be multiplied by the width to obtain the required segments.

+

When p is the step size, we just divide the width by the step size (rounded to the next largest integer) to get the required number of segments. If the last segment is too large, we “crop” it to the width with Minimum (⌊).

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
 ∇
+

I decided to use HttpCommand for this task, since it is simply one ]load HttpCommand away and should be platform-independent.

+

Parsing XML is not something I consider “fun” in the best of cases, and I feel like APL is not the best language to do this kind of thing. Given how simple the task is, I just decided to find the relevant bits with a regular expression using Replace and Search (⎕S).

+

After finding all the strings vaguely resembling a PDF file name (only alphanumeric characters and underscores, with a .pdf extension), I just concatenate them to the base URL of the Dyalog domain.

Problem 4 – Bioinformatics

+

The first task can be solved by decomposing it into several functions.

⍝ 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}
+isrevp←{⍵≡⌽'TAGC'['ATCG'⍳⍵]} +

First, we compute the complement of a DNA string (using simple indexing) and test if its Reverse (⌽) is equal to the original string.

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

We first compute all the possible lengths for each starting point. For instance, the last element cannot have any (position, length) pair associated to it, because there is no three element following it. So we crop the possible lengths to \([3, 12]\). For instance for an array of size 10:

+
        {3↓¨⍳¨12⌊1+⍵-⍳⍵}10
+┌──────────────┬───────────┬─────────┬───────┬─────┬───┬─┬┬┬┐
+│4 5 6 7 8 9 10│4 5 6 7 8 9│4 5 6 7 8│4 5 6 7│4 5 6│4 5│4││││
+└──────────────┴───────────┴─────────┴───────┴─────┴───┴─┴┴┴┘
+

Then, we just add the corresponding starting position to each length (1 for the first block, 2 for the second, and so on). Finally, we flatten everything.

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

For each possible (position, length) pair, we get the corresponding DNA substring with dna[¯1+⍵[1]+⍳⍵[2]] (adding ¯1 is necessary because ⎕IO←1). We test if this substring is a reverse palindrome using isrevp above. Replicate (/) then selects only the (position, length) pairs for which the substring is a reverse palindrome.

+

The second task is just about counting the number of subsets modulo 1,000,000. So we just need to compute \(2^n \mod 1000000\) for any positive integer \(n\leq1000\).

+
sset←{((1E6|2∘×)⍣⍵)1}
+

Since we cannot just compute \(2^n\) directly and take the remainder, we use modular arithmetic to stay mod 1,000,000 during the whole computation. The dfn (1E6|2∘×) doubles its argument mod 1,000,000. So we just apply this function \(n\) times using the Power operator (⍣), with an initial value of 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+⍵}
+
⍝ 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
-∇
+
∇ 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
-∇
+
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
-∇
+
∇ 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
-∇
-
    :EndNamespace
-:EndNamespace
+
∇ 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
+∇
+
    :EndNamespace
+:EndNamespace
]]> diff --git a/posts/dyalog-apl-competition-2020-phase-2.org b/posts/dyalog-apl-competition-2020-phase-2.org index 4860f89..bdacfb8 100644 --- a/posts/dyalog-apl-competition-2020-phase-2.org +++ b/posts/dyalog-apl-competition-2020-phase-2.org @@ -46,6 +46,17 @@ explanations for every problem below. ∇ #+end_src +This is a very straightforward implementation of the algorithm +describe in the problem description. I decided to switch explicitly on +the size of the input vector because I feel it is more natural. For +the cases with 5 or 7 judges, we use Drop (~↓~) to remove the lowest +and highest scores. + +At the end, we sum up the scores with ~+/~ and multiply them by +~dd~. The last operation, ~2(⍎⍕)~, is a train using [[https://help.dyalog.com/18.0/index.htm#Language/Primitive%20Functions/Format%20Dyadic.htm][Format (Dyadic)]] to +round to 2 decimal places, and [[https://help.dyalog.com/18.0/index.htm#Language/Primitive%20Functions/Execute.htm][Execute]] to get actual numbers and not +strings. + * Problem 2 -- Another Step in the Proper Direction #+begin_src default @@ -65,6 +76,20 @@ explanations for every problem below. ∇ #+end_src +This is an extension to [[./dyalog-apl-competition-2020-phase-1.html#stepping-in-the-proper-direction][Problem 5 of Phase I]]. In each case, we compute +the "segments", i.e., the steps starting from 0. In a last step, +common to all cases, we add the correct starting point and correct the +direction if need be. + +To compute equally-sized steps, we first divide the segment $[0, 1]$ +in ~p~ equal segments with ~(⍳p)÷p~. This subdivision can then be +multiplied by the width to obtain the required segments. + +When ~p~ is the step size, we just divide the width by the step size +(rounded to the next largest integer) to get the required number of +segments. If the last segment is too large, we "crop" it to the width +with Minimum (~⌊~). + * Problem 3 -- Past Tasks Blast #+begin_src default @@ -75,16 +100,55 @@ explanations for every problem below. ∇ #+end_src +I decided to use ~HttpCommand~ for this task, since it is simply one +~]load HttpCommand~ away and should be platform-independent. + +Parsing XML is not something I consider "fun" in the best of cases, +and I feel like APL is not the best language to do this kind of +thing. Given how simple the task is, I just decided to find the +relevant bits with a regular expression using [[https://help.dyalog.com/18.0/index.htm#Language/System%20Functions/r.htm][Replace and Search]] +(~⎕S~). + +After finding all the strings vaguely resembling a PDF file name (only +alphanumeric characters and underscores, with a =.pdf= extension), I +just concatenate them to the base URL of the Dyalog domain. + * Problem 4 -- Bioinformatics +The first task can be solved by decomposing it into several functions. + #+begin_src default ⍝ Test if a DNA string is a reverse palindrome. isrevp←{⍵≡⌽'TAGC'['ATCG'⍳⍵]} +#+end_src - ⍝ Generate all subarrays (position, length) pairs, for - ⍝ 4 ≤ length ≤ 12. +First, we compute the complement of a DNA string (using simple +indexing) and test if its Reverse (~⌽~) is equal to the original +string. + +#+begin_src default + ⍝ Generate all subarrays (position, length) pairs, for 4 ≤ length ≤ 12. subarrays←{⊃,/(⍳⍵),¨¨3↓¨⍳¨12⌊1+⍵-⍳⍵} +#+end_src +We first compute all the possible lengths for each starting point. For +instance, the last element cannot have any (position, length) pair +associated to it, because there is no three element following it. So +we crop the possible lengths to $[3, 12]$. For instance for an array +of size 10: + +#+begin_src default + {3↓¨⍳¨12⌊1+⍵-⍳⍵}10 +┌──────────────┬───────────┬─────────┬───────┬─────┬───┬─┬┬┬┐ +│4 5 6 7 8 9 10│4 5 6 7 8 9│4 5 6 7 8│4 5 6 7│4 5 6│4 5│4││││ +└──────────────┴───────────┴─────────┴───────┴─────┴───┴─┴┴┴┘ +#+end_src + +Then, we just add the corresponding starting position to each length +(1 for the first block, 2 for the second, and so on). Finally, we +flatten everything. + +#+begin_src default ∇ r←revp dna;positions positions←subarrays⍴dna ⍝ Filter subarrays which are reverse palindromes. @@ -92,10 +156,26 @@ explanations for every problem below. ∇ #+end_src +For each possible (position, length) pair, we get the corresponding +DNA substring with ~dna[¯1+⍵[1]+⍳⍵[2]]~ (adding ~¯1~ is necessary +because ~⎕IO←1~). We test if this substring is a reverse palindrome +using ~isrevp~ above. [[https://help.dyalog.com/18.0/index.htm#Language/Primitive%20Functions/Replicate.htm][Replicate]] (~/~) then selects only the (position, +length) pairs for which the substring is a reverse palindrome. + +The second task is just about counting the number of subsets modulo +1,000,000. So we just need to compute $2^n \mod 1000000$ for any +positive integer $n\leq1000$. + #+begin_src default sset←{((1E6|2∘×)⍣⍵)1} #+end_src +Since we cannot just compute $2^n$ directly and take the remainder, we +use modular arithmetic to stay mod 1,000,000 during the whole +computation. The dfn ~(1E6|2∘×)~ doubles its argument mod +1,000,000. So we just apply this function $n$ times using the [[https://help.dyalog.com/18.0/index.htm#Language/Primitive%20Operators/Power%20Operator.htm][Power]] +operator (~⍣~), with an initial value of 1. + * Problem 5 -- Future and Present Value #+begin_src default