From 878daf66ab9c6257c0ef7e677e6c444813efca85 Mon Sep 17 00:00:00 2001
From: Aleksander Sadikov <aleksander.sadikov@fri.uni-lj.si>
Date: Thu, 1 Sep 2016 16:19:09 +0200
Subject: English translation for shiftleft/2 added.

---
 prolog/problems/lists_advanced/shiftleft_2/en.py | 58 +++++++++++++++++++++++-
 prolog/problems/lists_advanced/shiftleft_2/sl.py |  2 +-
 2 files changed, 58 insertions(+), 2 deletions(-)

(limited to 'prolog/problems/lists_advanced/shiftleft_2')

diff --git a/prolog/problems/lists_advanced/shiftleft_2/en.py b/prolog/problems/lists_advanced/shiftleft_2/en.py
index 6adbc12..9905cac 100644
--- a/prolog/problems/lists_advanced/shiftleft_2/en.py
+++ b/prolog/problems/lists_advanced/shiftleft_2/en.py
@@ -8,4 +8,60 @@ description = '''\
   X = [2,3,4,5,1].
 </pre>'''
 
-hint = {}
+plan = ['''\
+<p>I take the first element of a list, let's call it <code>H</code>, and add it <em>at the end</em> of
+the list's remainder (let's call the remainder <code>T</code>). As simple as that! You probably still
+remember how we took the last element of the list? Adding an element is the same operation as taking it,
+just from the opposite view ;)</p>
+''', '''\
+<p>A list of length one is represented as a pattern <code>[X]</code>. This might come in handy, as well
+as the predicate <code>conc/3</code>.</p>
+''', '''\
+<p>If the given list <code>L</code> is composed of head <code>H</code> and tail <code>T</code>, and if
+we add <code>H</code> at <em>the end</em> of <code>T</code>, then we get list <code>L</code>
+shifted left.</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>shiftleft/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>
+''',
+
+    '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>
+''',
+
+    'conc_arg_not_list': '''\
+<p>All three arguments of predicate <code>conc/3</code> are <em>lists</em>.
+Are you sure you used it properly?</p>
+''',
+
+    'arbitrary_result': '''\
+<p>Did you connect (use) all the variables? It seems as if you're returning an arbitrary result
+(a variable without an assigned value). It's usually not a good idea to ignore the warning about
+"singleton variables".</p>
+''',
+
+    'tail_must_be_list': '''\
+<p>The list's tail is always another <em>list</em> and never an element!</p>
+''',
+}
diff --git a/prolog/problems/lists_advanced/shiftleft_2/sl.py b/prolog/problems/lists_advanced/shiftleft_2/sl.py
index aa83672..aa116ee 100644
--- a/prolog/problems/lists_advanced/shiftleft_2/sl.py
+++ b/prolog/problems/lists_advanced/shiftleft_2/sl.py
@@ -24,7 +24,7 @@ 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>shiftleft/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>
 ''',
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