English translation for sum/2 added.

4 files changed, 93 insertions, 10 deletions

diff --git a/prolog/problems/lists_advanced/len_2/en.py b/prolog/problems/lists_advanced/len_2/en.py
index 33b1675..6518eec 100644
--- a/prolog/problems/lists_advanced/len_2/en.py
+++ b/prolog/problems/lists_advanced/len_2/en.py
@@ -15,14 +15,14 @@ plan = ['''\
 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>
+result returned by the recursion, then I get 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>
+you can make the predicate <code>len/2</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>
 ''',
@@ -78,7 +78,7 @@ second operand and only then attempts the unification of both operands.</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
+<p>Is your recursive call of the form <code>len(Tail, LenTail + 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>
 ''',
diff --git a/prolog/problems/lists_advanced/len_2/sl.py b/prolog/problems/lists_advanced/len_2/sl.py
index 49b9eea..28fe25e 100644
--- a/prolog/problems/lists_advanced/len_2/sl.py
+++ b/prolog/problems/lists_advanced/len_2/sl.py
@@ -75,8 +75,8 @@ svoji levi strani.</p>
     'forcing_result_onto_recursion': '''
 <p>Ne vsiljuj rekurziji kaj naj vrne, prepusti se ji. To je tisti del, ko narediš predpostavko,
 če je ta izpolnjena, potem bo tvoje pravilo delovalo za večji primer.</p>
-<p>Je tvoj rekurzivni klic oblike <code>len(T, LenT + 1)</code>? S tem vsiljuješ rekurziji
-da mora <emph>vrniti</emph> dolžino celega seznama in ne samo repa. To ni v redu, za ena moraš ti povečati
+<p>Je tvoj rekurzivni klic oblike <code>len(Tail, LenTail + 1)</code>? S tem vsiljuješ rekurziji
+da mora vrniti dolžino <em>celega</em> seznama in ne samo repa. To ni v redu, za ena moraš ti povečati
 rezultat, ki ti ga rekurzija vrne. Skratka, prištevanje naredi izven rekurzivnega klica.</p>
 ''',
 
diff --git a/prolog/problems/lists_advanced/sum_2/en.py b/prolog/problems/lists_advanced/sum_2/en.py
index 767a6f4..3abc348 100644
--- a/prolog/problems/lists_advanced/sum_2/en.py
+++ b/prolog/problems/lists_advanced/sum_2/en.py
@@ -2,10 +2,93 @@ name = 'sum/2'
 slug = 'find the sum of all elements in list'
 
 description = '''\
-<p><code>sum(L, Sum)</code>: <code>Sum</code> is the sum of all elements in the list <code>L</code>.</p>
+<p><code>sum(L, Sum)</code>: <code>Sum</code> is the sum of all elements in list <code>L</code>.</p>
 <pre>
 ?- sum([1,2,3], Sum).
   Sum = 6.
 </pre>'''
 
-hint = {}
+plan = ['''\
+<p>The sum of an empty list is really not all that large. And if it's not empty, then we add it up,
+element by element, recursively.</p>
+''', '''\
+<p>If the sum of the elements in the tail (list without a head) equals <code>SumT</code>, then the sum
+of the elements in the whole list equals <code>SumT + H</code>.</p>
+''', '''\
+<p>If I take away the head, and the recursion solves this smaller problem (tail), and if I add the value
+of the head to the result returned by the recursion, then I get the sum 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>sum/2</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? What's the sum of the elements in an empty list?</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 sum of the elements in 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>
+''',
+
+    '+1_instead_of_+H': '''\
+<p>Did you really add one instead of the element's value? Copy/paste operation from the previous exercise? ;)</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>sum(Tail, SumTail + H)</code>? This forces the recursive call to
+return the sum of <em>the whole</em> list, not just the tail! This will not work. It is your job to
+increase (by value of the head) the result returned by the recursion. In short, perform the addition
+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/sum_2/sl.py b/prolog/problems/lists_advanced/sum_2/sl.py
index 539271b..d3bddcf 100644
--- a/prolog/problems/lists_advanced/sum_2/sl.py
+++ b/prolog/problems/lists_advanced/sum_2/sl.py
@@ -22,7 +22,7 @@ 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>sum/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>
 ''',
@@ -76,8 +76,8 @@ svoji levi strani.</p>
     'forcing_result_onto_recursion': '''
 <p>Ne vsiljuj rekurziji kaj naj vrne, prepusti se ji. To je tisti del, ko narediš predpostavko,
 če je ta izpolnjena, potem bo tvoje pravilo delovalo za večji primer.</p>
-<p>Je tvoj rekurzivni klic oblike <code>len(T, LenT + H)</code>? S tem vsiljuješ rekurziji
-da mora <emph>vrniti</emph> vsoto celega seznama in ne samo repa. To ni v redu, za vrednost glave moraš ti povečati
+<p>Je tvoj rekurzivni klic oblike <code>sum(Tail, SumTail + H)</code>? S tem vsiljuješ rekurziji
+da mora vrniti vsoto <em>celega</em> seznama in ne samo repa. To ni v redu, za vrednost glave moraš ti povečati
 rezultat, ki ti ga rekurzija vrne. Skratka, prištevanje naredi izven rekurzivnega klica.</p>
 ''',