Better display of problems and solutions

css/default.css

@@ -42,6 +42,14 @@ article .header {
 /*     background: #eee; */
 /* } */
 
+blockquote {
+    display: block;
+    border-left: 2px solid #808080;
+    padding-left: 1rem;
+    margin-top: 1rem;
+    margin-bottom: 1rem;
+}
+
 .definition, .proposition, .theorem {
     display: block;
     border-left: 2px solid #808080;

posts/dyalog-apl-competition-2020.org

@@ -23,9 +23,7 @@ integer so that its absolute value is less or equal to ~≢Y~, splits
   and the first vector contains the remaining elements.
 #+end_quote
 
-#+begin_src default
+*Solution:* ~(0>⊣)⌽((⊂↑),(⊂↓))~
-  split←(0>⊣)⌽((⊂↑),(⊂↓))
-#+end_src
 
 There are three nested trains here. The first one, ~((⊂↑),(⊂↓))~, uses
 the two functions [[https://help.dyalog.com/18.0/index.htm#Language/Primitive%20Functions/Take.htm][Take]] (~↑~) and [[https://help.dyalog.com/18.0/index.htm#Language/Primitive%20Functions/Drop.htm][Drop]] (~↓~) to build a nested array
@@ -73,9 +71,7 @@ character, like the result of ~'UTF-8'∘⎕UCS¨'UTF-8'∘⎕UCS~ but does not
 use any system functions (names beginning with ~⎕~)
 #+end_quote
 
-#+begin_src default
+*Solution:* ~{(~⍵∊127+⍳64)⊂⍵}~
-  characters←{(~⍵∊127+⍳64)⊂⍵}
-#+end_src
 
 First, we build a binary array from the string, encoding each
 continuation character as 0, and all the others as 1. Next, we can use
@@ -95,9 +91,7 @@ scalar or non-empty vector representing a valid character Excel column
 identifier between A and XFD, returns the corresponding column number
 #+end_quote
 
-#+begin_src default
+*Solution:* ~26⊥⎕A∘⍳~
-  columns←26⊥⎕A∘⍳
-#+end_src
 
 We use the alphabet ~⎕A~ and [[https://help.dyalog.com/latest/#Language/Primitive%20Functions/Index%20Of.htm][Index Of]] (~⍳~) to compute the index in
 the alphabet of every character. As a train, this can be done by
@@ -116,9 +110,7 @@ indicates that the corresponding year is a leap year (0 otherwise).
 A leap year algorithm can be found [[https://en.wikipedia.org/wiki/Leap_year#Algorithm][here]].
 #+end_quote
 
-#+begin_src default
+*Solution:* ~1 3∊⍨(0+.=400 100 4∘.|⊢)~
-  leap←1 3∊⍨(0+.=400 100 4∘.|⊢)
-#+end_src
 
 According to the algorithm, a year is a leap year in two situations:
 - if it is divisible by 4, but not 100 (and therefore not 400),
@@ -139,9 +131,7 @@ vector of the integers from the first element of the right argument to
 the second, inclusively.
 #+end_quote
 
-#+begin_src default
+*Solution:* ~{(⊃⍵)+(-×-/⍵)×0,⍳|-/⍵}~
-  stepping←{(⊃⍵)+(-×-/⍵)×0,⍳|-/⍵}
-#+end_src
 
 First, we have to compute the range of the output, which is the
 absolute value of the difference between the two integers ~|-/⍵~. From
@@ -161,9 +151,7 @@ right argument so any elements equal to the left argument come first
 while all other elements keep their order.
 #+end_quote
 
-#+begin_src default
+*Solution:* ~{⍵[⍋⍺≠⍵]}~
-  movefront←{⍵[⍋⍺≠⍵]}
-#+end_src
 
 ~⍺≠⍵~ will return a binary vector marking as 0 all elements equal to
 the left argument. Using this index to sort in the usual way with
@@ -187,9 +175,7 @@ that you want to query, returns 1 if all of the codes in the left
 argument are found in the right argument (0 otherwise).
 #+end_quote
 
-#+begin_src default
+*Solution:* ~{f←⍸∘⌽(2∘⊥⍣¯1)⋄∧/(f⍺)∊f⍵}~
-  bits←{f←⍸∘⌽(2∘⊥⍣¯1)⋄∧/(f⍺)∊f⍵}
-#+end_src
 
 The difficult part is to find the set of states for an integer. We
 need a function that will return ~1 8 128~ (or an equivalent
@@ -215,9 +201,7 @@ Write a function that takes a single integer greater than or equal to
 integer is a zigzag number, 0 otherwise.
 #+end_quote
 
-#+begin_src default
+*Solution:* ~∧/2=∘|2-/∘×2-/(10∘⊥⍣¯1)~
-  zigzag←∧/2=∘|2-/∘×2-/(10∘⊥⍣¯1)
-#+end_src
 
 TODO
 
@@ -233,9 +217,7 @@ conform to the following pattern (0 otherwise):
 - After the apex, any remaining values decrease or remain the same
 #+end_quote
 
-#+begin_src default
+*Solution:* ~{∧/(⍳∘≢≡⍋)¨(⊂((⊢⍳⌈/)↑⊢),⍵),⊂⌽((⊢⍳⌈/)↓⊢),⍵}~
-  risefall←{∧/(⍳∘≢≡⍋)¨(⊂((⊢⍳⌈/)↑⊢),⍵),⊂⌽((⊢⍳⌈/)↓⊢),⍵}
-#+end_src
 
 TODO
 
@@ -250,9 +232,7 @@ displays identically to what ~{⎕←⍵}¨~ displays when applied to the
 right argument.
 #+end_quote
 
-#+begin_src default
+*Solution:* ~{↑⊃,/↓¨⍕¨⍵}~
-  stacking←{↑⊃,/↓¨⍕¨⍵}
-#+end_src
 
 TODO