From 464f5fd3eebfc30fa487eebb29b24727b98276f1 Mon Sep 17 00:00:00 2001
From: Aleksander Sadikov <aleksander.sadikov@fri.uni-lj.si>
Date: Mon, 29 Aug 2016 00:45:09 +0200
Subject: English translation for insert/3 added.

---
 prolog/problems/lists/insert_3/en.py | 83 +++++++++++++++++++++++++++++++++++-
 prolog/problems/lists/insert_3/sl.py |  2 +-
 2 files changed, 83 insertions(+), 2 deletions(-)

(limited to 'prolog')

diff --git a/prolog/problems/lists/insert_3/en.py b/prolog/problems/lists/insert_3/en.py
index 5553445..0de098b 100644
--- a/prolog/problems/lists/insert_3/en.py
+++ b/prolog/problems/lists/insert_3/en.py
@@ -10,4 +10,85 @@ description = '''\
   L = [2,3,1].
 </pre>'''
 
-hint = {}
+plan = ['''
+<p><img src="[%@resource plan.svg%]" /></p>
+<p>Where in the list can we insert a new element <code>X</code>? Remember that a list has two parts: head and tail.
+That means there are two possibilies. That's right, only two -- but in the tail we can again insert either as its
+new head or into the tail of the tail. And so on. Recursion to the rescue!</p>
+''', '''\
+<p>What is the simplest option? Insert at the beginning!</p>
+''', '''\
+<p>How do I insert into the list's tail? Just divide the list into its head and tail, recursively (as
+the problem is one element smaller now) insert into the tail, and at the end don't forget about the
+head previously taken away.</p>
+''', '''\
+<p>Recursive step: if we assume <code>NewTail</code> is the tail with already inserted element <code>X</code>,
+then <code>[H|NewTail]</code> is the whole list with the element <code>X</code> inserted.</p>
+''']
+
+hint = {
+    'eq_instead_of_equ': '''\
+<p>The operator <code>==</code> is "stricter" than operator <code>=</code> in the sense that
+for the latter it is enough to be able to make the two operands equal (unification). Perhaps by using <code>=</code>
+you can make the predicate <code>insert/3</code> more general (e.g. able to work with output arguments becoming inputs).
+This might come in handy later on!</p>
+<p>Of course, you can also solve the exercise without explicit use of either of these two operators, just
+remember that unification is implicitly performed with the predicate's arguments (head of clause).</p>
+''',
+
+    'eq_instead_of_equ_markup': '''\
+<p>Perhaps the operator for unification (=) would be better?</p>
+''',
+
+    'base_case': '''\
+<p><img src="[%@resource base_case.svg%]" /></p>
+<p>Did you think of a base case? In which place in the list is it the easiest to insert a new element?</p>
+''',
+
+    'recursive_case': '''\
+<p>The base case is ok. However, what about the general recursive case?</p>
+''',
+
+    'predicate_always_false': '''\
+<p>It seems your predicate is <emph>always</emph> "false". Did you give it the correct name,
+or is it perhaps misspelled?</p>
+<p>If the name is correct, check whether something else is misspelled, perhaps there is a full stop instead of
+a comma or vice versa, or maybe you typed a variable name in lowercase?</p>
+<p>It is, of course, also possible that your conditions are too restrictive, or even impossible to satisfy
+(as would be, for example, the condition that an empty list <code>[]</code> is equal to a list with
+exactly three elements <code>[A,B,C]</code>,
+or something similarly impossible).</p>
+''',
+
+    'timeout': '''\
+<p>Is there an infinite recursion at work here? How will it ever stop?</p>
+<p>Or perhaps is there a missing, faulty, or simply incompatible (with the general recursive case) base case?</p>
+''',
+
+    'ins_results_in_empty_list': '''\
+<p>How can the result of the insertion be an empty list?</p>
+<p>If that is your base case, rethink it! What is the result of the insertion?</p>
+''',
+
+    'ins_results_in_arbitrary_result': '''\
+<p>How can the result of the insertion be an arbitrary list or an unassigned variable?</p>
+<p>If that is your base case, rethink it! What is the result of the insertion?</p>
+''',
+
+    'lost_heads': '''\
+<p><img src="[%@resource lost_heads.svg%]" /></p>
+<p>The element has been successfully inserted, but all the others before it are lost, right?
+Did you forget to put the head back at the front of the list after returning from recursion?</p>
+<p>Try asking prolog the following query and check <em>all</em> the solutions:</p>
+<p><code>?- insert(q, [a,b,c,d], L).</code></p>
+''',
+
+    'leading_heads_all_x': '''\
+<p><img src="[%@resource leading_heads_all_x.svg%]" /></p>
+<p>Did you forget (copy/paste?) and used <code>[X|T]</code> instead of the more general <code>[H|T]</code>
+in the recursive case?</p>
+<p>Of the following two queries one works and the other doesn't.</p>
+<p><code>?- insert(d, [d,d,d,d,e,f,g], L).</code></p>
+<p><code>?- insert(d, [a,b,c,d,e,f,g], L).</code></p>
+''',
+}
diff --git a/prolog/problems/lists/insert_3/sl.py b/prolog/problems/lists/insert_3/sl.py
index 17efb5b..7f62f7e 100644
--- a/prolog/problems/lists/insert_3/sl.py
+++ b/prolog/problems/lists/insert_3/sl.py
@@ -74,7 +74,7 @@ da je <code>X</code> hkrati starš in sestra od <code>Y</code> ali kaj podobno z
 <p><img src="[%@resource lost_heads.svg%]" /></p>
 <p>Element je vstavljen, ampak vsi pred njim so se pa izgubili, kajne?
 Si pozabil dati glavo nazaj na začetek seznama, ko se vračaš iz rekurzije?</p>
-<p>Poskusi postaviti naslednje vprašanje prologu in preglej <emph>vse</emph> rešitve:</p>
+<p>Poskusi postaviti naslednje vprašanje prologu in preglej <em>vse</em> rešitve:</p>
 <p><code>?- insert(q, [a,b,c,d], L).</code></p>
 ''',
 
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