2 files changed, 84 insertions, 3 deletions
diff --git a/prolog/problems/lists_advanced/len_2/en.py b/prolog/problems/lists_advanced/len_2/en.py
index 9202c2d..33b1675 100644
--- a/prolog/problems/lists_advanced/len_2/en.py
+++ b/prolog/problems/lists_advanced/len_2/en.py
@@ -2,10 +2,91 @@ name = 'len/2'
 slug = 'find the length of a list'
 
 description = '''\
-<p><code>len(L, Len)</code>: <code>Len</code> is the length of the list <code>L</code>.</p>
+<p><code>len(L, Len)</code>: <code>Len</code> is the length of list <code>L</code>.</p>
 <pre>
 ?- len([1,2,3], Len).
   Len = 3.
 </pre>'''
 
-hint = {}
+plan = ['''\
+<p>A list is not very long if it's empty, and if it's not empty it must have a head and a tail.</p>
+''', '''\
+<p>If the tail (list without a head) is of length <code>LenT</code>, then the whole list
+is of length <code>LenT + 1</code>.</p>
+''', '''\
+<p>If I take away the head, and the recursion solves this smaller problem (tail), and if I add 1 to the
+result returned by the recursion, then I got the length of the whole list.</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>conc/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>Did you think of a base case? Which list is the shortest list in the world?</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 <code>N</code> is equal to <code>N + 1</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>
+''',
+
+    'arbitrary_base_case': '''\
+<p>What's the length of an empty list? Give a number!</p>
+''',
+
+    'args_not_instantiated': '''\
+<p>The error that prolog reported means that when it encountered an arithmetic operation, not all the
+values were known. Unfortunately, the ordering of goals is important when dealing with arithmetics.</p>
+<p>Perhaps you can try moving the arithmetic operation more towards the end of the predicate?</p>
+''',
+
+    '=_instead_of_is': '''\
+<p>Did you use the operator <code>=</code> instead of <code>is</code>? Operator <code>=</code> is used for
+unification and tries to leave both its operands with as little modifications as possible while still making
+them equal. Operator <code>is</code>, on the other hand, performs actual arithmetic evaluation of its
+second operand and only then attempts the unification of both operands.</p>
+''',
+
+    '+H_instead_of_+1': '''\
+<p>Did you really add the element's value instead of its length (one)? ;)</p>
+''',
+
+    'forcing_result_onto_recursion': '''
+<p>Don't force the result onto recursion, don't tell it what it should return. Just assume it will do its job.
+If this assumption is correct, then the rule will work for a larger case.</p>
+<p>Is your recursive call of the form <code>len(T, LenT + 1)</code>? This forces the recursive call to
+return the length of <em>the whole</em> list, not just the tail! This will not work. It is your job to
+increase by one the result returned by the recursion. In short, add one outside the recursive call.</p>
+''',
+
+    'same_var_on_both_sides_of_is': '''\
+<p>Does one of your goals look similar to <code>N is N + 1</code>? Let's assume <code>N</code> is equal to 3.
+With this goal you just stated that 3 must be equal to 4 (3+1). Prolog is a logical language and will
+gladly say "false" to such a statement! Just use a new variable. The garbage collector will take care of
+those not needed anymore automatically.</p>
+''',
+}
diff --git a/prolog/problems/lists_advanced/len_2/sl.py b/prolog/problems/lists_advanced/len_2/sl.py
index c95796b..49b9eea 100644
--- a/prolog/problems/lists_advanced/len_2/sl.py
+++ b/prolog/problems/lists_advanced/len_2/sl.py
@@ -21,7 +21,7 @@ hint = {
     'eq_instead_of_equ': '''\
 <p>Operator <code>==</code> je strožji od operatorja <code>=</code> v smislu, da je za slednjega dovolj,
 da elementa lahko naredi enaka (unifikacija). Morda z uporabo <code>=</code> narediš predikat
-<code>memb/2</code> delujoč tudi v kakšni drugi smeri.</p>
+<code>len/2</code> delujoč tudi v kakšni drugi smeri.</p>
 <p>Seveda pa lahko nalogo rešiš brez obeh omenjenih operatorjev, spomni se, da lahko unifikacijo narediš
 implicitno že kar v argumentih predikata (glavi stavka).</p>
 ''',