From 1b2c914d8552f4af1676f6b63203513d949c7817 Mon Sep 17 00:00:00 2001
From: Aleksander Sadikov <aleksander.sadikov@fri.uni-lj.si>
Date: Thu, 1 Sep 2016 18:36:34 +0200
Subject: English translation for shiftright/2 added.

---
 prolog/problems/lists_advanced/shiftright_2/en.py | 66 ++++++++++++++++++++++-
 prolog/problems/lists_advanced/shiftright_2/sl.py |  2 +-
 2 files changed, 66 insertions(+), 2 deletions(-)

diff --git a/prolog/problems/lists_advanced/shiftright_2/en.py b/prolog/problems/lists_advanced/shiftright_2/en.py
index 4e0fec3..21b404d 100644
--- a/prolog/problems/lists_advanced/shiftright_2/en.py
+++ b/prolog/problems/lists_advanced/shiftright_2/en.py
@@ -8,4 +8,68 @@ description = '''\
   X = [5,1,2,3,4].
 </pre>'''
 
-hint = {}
+plan = ['''\
+<p>I take the last element from the given list and add it back at the start of the list's remainder.
+You probably still remember how we took the last element of the list?  And adding an element at the start
+is quite simple, isn't it?</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 last element <code>E</code> and the remainder <code>L1</code>,
+and if we put <code>E</code> at <em>the start</em> of <code>L1</code>, then we get list <code>L</code>
+shifted right.</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_2nd_argument_not_1elem_list': '''\
+<p>Do you remember how the "pattern" representing a list with exactly one element looks like? Not like
+the second argument you gave to predicate <code>conc/3</code>. ;)</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>
+''',
+
+'shiftleftish_solution': '''\
+<p>The tail of the list is always another list and never just an element. How did you get the last
+element? This will not work...</p>''',
+
+'last_used': '''\
+<p>By using predicate <code>last/2</code> it will be difficult to solve this exercise as it leaves the last
+element in the original list. Rather try using predicate <code>conc/3</code> instead.</p>''',
+
+'final_hint': '''\
+<p>Predicates <code>shiftleft/2</code> and <code>shiftright/2</code> perform exactly the opposite function.
+If you simply swap the order of their arguments, you get the other predicate. In this way you could
+program <code>shiftright/2</code> just by calling <code>shiftleft/2</code>. You know that in prolog inputs
+and outputs can often be interchanged.</p>''',
+}
diff --git a/prolog/problems/lists_advanced/shiftright_2/sl.py b/prolog/problems/lists_advanced/shiftright_2/sl.py
index ce9b9fa..784f4ed 100644
--- a/prolog/problems/lists_advanced/shiftright_2/sl.py
+++ b/prolog/problems/lists_advanced/shiftright_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>shiftright/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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