diff --git a/posts/dyalog-apl-competition-2020.org b/posts/dyalog-apl-competition-2020-phase-1.org
similarity index 67%
rename from posts/dyalog-apl-competition-2020.org
rename to posts/dyalog-apl-competition-2020-phase-1.org
index 150254b..ae11d37 100644
--- a/posts/dyalog-apl-competition-2020.org
+++ b/posts/dyalog-apl-competition-2020-phase-1.org
@@ -1,17 +1,15 @@
 ---
-title: "Dyalog APL Problem Solving Competition 2020"
+title: "Dyalog APL Problem Solving Competition 2020 — Phase I"
 subtitle: "Annotated Solutions"
 date: 2020-07-31
 toc: true
 ---
 
-* Phase I
-
 #+begin_src default
   :Namespace Phase1
 #+end_src
 
-** 1. Let's Split!
+* 1. Let's Split!
 
 #+begin_quote
 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
 nested array, in the last train.
 
-** 2. Character Building
+* 2. Character Building
 
 #+begin_quote
 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
 output.
 
-** 3. Excel-lent Columns
+* 3. Excel-lent Columns
 
 #+begin_quote
 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
 base-26 number into the expected result.
 
-** 4. Take a Leap
+* 4. Take a Leap
 
 #+begin_quote
 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,
 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
 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
 element of ~⍵~.
 
-** 6. Please Move to the Front
+* 6. Please Move to the Front
 
 #+begin_quote
 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
 [[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
 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
 ~f⍵~.
 
-** 8. Zigzag Numbers
+* 8. Zigzag Numbers
 
 #+begin_quote
 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
 ensure that the signs are indeed alternating.
 
-** 9. Rise and Fall
+* 9. Rise and Fall
 
 #+begin_quote
 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
 simple range from 1 to the size of the vector).
 
-** 10. Stacking It Up
+* 10. Stacking It Up
 
 #+begin_quote
 Write a function that takes as its right argument a vector of simple
@@ -317,238 +315,3 @@ final result.
   :EndNamespace
 #+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
diff --git a/posts/dyalog-apl-competition-2020-phase-2.org b/posts/dyalog-apl-competition-2020-phase-2.org
new file mode 100644
index 0000000..e0c8576
--- /dev/null
+++ b/posts/dyalog-apl-competition-2020-phase-2.org
@@ -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