From d259285459176fc281193adb12483a6e6d527cf5 Mon Sep 17 00:00:00 2001
From: Aleksander Sadikov <aleksander.sadikov@fri.uni-lj.si>
Date: Tue, 6 Sep 2016 16:48:08 +0200
Subject: English translation for sum/2 added.

---
 prolog/problems/lists_advanced/len_2/en.py | 6 +++---
 prolog/problems/lists_advanced/len_2/sl.py | 4 ++--
 2 files changed, 5 insertions(+), 5 deletions(-)

(limited to 'prolog/problems/lists_advanced/len_2')

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>
 ''',
 
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