English translation for del/3 added.

2 files changed, 92 insertions, 3 deletions

diff --git a/prolog/problems/lists/del_3/en.py b/prolog/problems/lists/del_3/en.py
index 99f87ca..d6e1e30 100644
--- a/prolog/problems/lists/del_3/en.py
+++ b/prolog/problems/lists/del_3/en.py
@@ -1,5 +1,5 @@
 name = 'del/3'
-slug = 'delete an element from list'
+slug = 'delete element from list'
 
 description = '''\
 <p><code>del(X, L1, L2)</code>: the list <code>L2</code> is obtained from <code>L1</code> by deleting element <code>X</code>.</p>
@@ -15,4 +15,93 @@ description = '''\
   X = 3, L = [1,2].
 </pre>'''
 
-hint = {}
+plan = ['''
+<p style="text-align: center;">
+    <img style="display: inline-block; width: 40%; border-right: 1px solid lightgray;" src="[%@resource plan_a.svg%]" />
+    <img style="display: inline-block; width: 40%;" src="[%@resource plan_b.svg%]" />
+</p>
+<p>This exercise is very similar to the exercise <code>memb/2</code>, except that this time we also delete
+the element we are looking for. Where in the list can element <code>X</code> be hiding?
+Remember that a list has two parts: head and tail. Therefore there are two possibilies!</p>
+''', '''\
+<p>What is the simplest option? Perhaps deleting the first element?</p>
+''', '''\
+<p>How do I delete from the list's tail? Just divide the list into its head and tail, recursively (as
+the problem is one element smaller now) delete from 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 element <code>X</code> already deleted,
+then <code>[H|NewTail]</code> is the whole list with element <code>X</code> deleted.</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>del/3</code> more general (e.g. able to work with output arguments becoming inputs).</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? Where in the list is it the easiest to delete an 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>
+''',
+
+    'del_from_empty_list_success': '''\
+<p>You can't delete an element from an empty list!</p>
+<p>If I delete an element from an empty list, this doesn't result in an empty list or anything else, but it
+simply shouldn't succeed. You don't even have to include a rule for this as prolog by default acts like that.
+You know how prolog loves to say "no"! :)</p>
+<p>If that is your base case, rethink it! Where in the list is it the easiest to delete an element?</p>
+''',
+
+    'lost_heads': '''\
+<p><img src="[%@resource lost_heads.svg%]" /></p>
+<p>The element has been successfully deleted, but all the others before it got deleted too, right?
+Did you forget to put the head back at the front of the list after returning from recursion?</p>
+<p>Try asking the following query:</p>
+<p><code>?- del(d, [a,b,c,d,e,f,g], 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>?- del(d, [d,d,d,d,e,f,g], L).</code></p>
+<p><code>?- del(d, [a,b,c,d,e,f,g], L).</code></p>
+''',
+
+    'final_hint': '''\
+<p>Interesting tidbit. Inserting and deleting an element into/from the list are opposite operations.
+Just by playing around with the arguments you can solve <code>del/3</code> using <code>insert/3</code>.</p>
+<p>Logically the following holds: deleting element <code>X</code> from list <code>BigList</code> resulting
+in list <code>SmallList</code> is the same as <em>inserting</em> element <code>X</code>
+into list <code>SmallList</code> to get list <code>BigList</code> as a result. ;)</p>
+''',
+}
diff --git a/prolog/problems/lists/del_3/sl.py b/prolog/problems/lists/del_3/sl.py
index feee5e4..17df017 100644
--- a/prolog/problems/lists/del_3/sl.py
+++ b/prolog/problems/lists/del_3/sl.py
@@ -97,7 +97,7 @@ Si pozabil dati glavo nazaj na začetek seznama, ko se vračaš iz rekurzije?</p
 <p>Zanimivost: operaciji vstavljanja in brisanja iz seznama sta si ravno nasprotni. Če se malce poigraš z
 argumenti, lahko <code>del/3</code> rešiš kar z <code>insert/3</code>.</p>
 <p>Logično velja naslednje: če zbrišem <code>X</code> iz seznama <code>BigList</code> in dobim kot rezultat
-seznam <code>SmallList</code> je isto kot če <emph>vstavim</emph> <code>X</code> v seznam <code>SmallList</code> in dobim
+seznam <code>SmallList</code> je isto kot če <em>vstavim</em> <code>X</code> v seznam <code>SmallList</code> in dobim
 kot rezultat seznam <code>BigList</code>. ;)</p>
 ''',
 }