From c1cc16c4088cb68d205342450493957f17a71c9a Mon Sep 17 00:00:00 2001
From: Timotej Lazar <timotej.lazar@fri.uni-lj.si>
Date: Thu, 17 Mar 2016 15:41:44 +0100
Subject: Prolog: add generic recursion hints for lists_advanced problems

TODO modify texts in */sl.py where appropriate.
---
 prolog/problems/lists_advanced/count_3/common.py | 50 ++++++++++++++++++++++--
 prolog/problems/lists_advanced/count_3/en.py     |  2 -
 prolog/problems/lists_advanced/count_3/sl.py     | 38 ++++++++++++++++--
 3 files changed, 81 insertions(+), 9 deletions(-)

(limited to 'prolog/problems/lists_advanced/count_3')

diff --git a/prolog/problems/lists_advanced/count_3/common.py b/prolog/problems/lists_advanced/count_3/common.py
index 2953f06..2785d78 100644
--- a/prolog/problems/lists_advanced/count_3/common.py
+++ b/prolog/problems/lists_advanced/count_3/common.py
@@ -1,8 +1,9 @@
-# coding=utf-8
-
 from operator import itemgetter
+import socket
+
 import prolog.engine
-import server.problems
+import prolog.util
+from server.hints import Hint, HintPopup
 
 id = 120
 number = 27
@@ -19,6 +20,15 @@ count(X, [Y|T], NT) :-
   count(X, T, NT).
 '''
 
+hint_type = {
+    'eq_instead_of_equ_markup': HintPopup('eq_instead_of_equ_markup'),
+    'eq_instead_of_equ': Hint('eq_instead_of_equ'),
+    'predicate_always_false': Hint('predicate_always_false'),
+    'base_case': Hint('base_case'),
+    'recursive_case': Hint('recursive_case'),
+    'timeout': Hint('timeout'),
+}
+
 test_cases = [
     ('count(a, [], X)',
         [{'X': '0'}]),
@@ -50,5 +60,37 @@ def test(code, aux_code):
     return n_correct, len(test_cases), hints
 
 def hint(code, aux_code):
-    # TODO
+    tokens = prolog.util.tokenize(code)
+
+    try:
+        engine_id, output = prolog.engine.create(code=code+aux_code, timeout=1.0)
+
+        # strict equality testing instead of simple matching
+        # this is usually (but not necessarily) wrong
+        targets = [prolog.util.Token('EQ', '==')]
+        marks = [(t.pos, t.pos + len(t.val)) for t in tokens if t in targets]
+        if marks:
+            return [{'id': 'eq_instead_of_equ_markup', 'start': m[0], 'end': m[1]} for m in marks] + \
+                   [{'id': 'eq_instead_of_equ'}]
+
+        # missing/failed base case
+        if not prolog.engine.ask_truthTO(engine_id, 'count(42, [], 0)'):
+            return [{'id': 'base_case'}]
+
+        # target predicate seems to always be false
+        if not prolog.engine.ask_truthTO(engine_id, 'count(_, _, _)'):
+            return [{'id': 'predicate_always_false'}]
+
+        # base case works, the recursive doesn't (but it doesn't timeout)
+        # this may be left as the last, most generic hint
+        if not prolog.engine.ask_truth(engine_id, 'count(42, [2, 41, 42, -22, 42], 2)'):
+            return [{'id': 'recursive_case'}]
+
+    except socket.timeout as ex:
+        return [{'id': 'timeout'}]
+
+    finally:
+        if engine_id:
+            prolog.engine.destroy(engine_id)
+
     return []
diff --git a/prolog/problems/lists_advanced/count_3/en.py b/prolog/problems/lists_advanced/count_3/en.py
index dd13bc0..481fd1c 100644
--- a/prolog/problems/lists_advanced/count_3/en.py
+++ b/prolog/problems/lists_advanced/count_3/en.py
@@ -1,5 +1,3 @@
-# coding=utf-8
-
 name = 'count/3'
 slug = 'find the number of occurrences of an element in list'
 
diff --git a/prolog/problems/lists_advanced/count_3/sl.py b/prolog/problems/lists_advanced/count_3/sl.py
index 98cd6f8..1e3f5d2 100644
--- a/prolog/problems/lists_advanced/count_3/sl.py
+++ b/prolog/problems/lists_advanced/count_3/sl.py
@@ -1,5 +1,3 @@
-# coding=utf-8
-
 name = 'count/3'
 slug = 'Preštej kolikokrat se element pojavi v seznamu'
 
@@ -10,4 +8,38 @@ description = '''\
   N = 3.
 </pre>'''
 
-hint = {}
+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>
+<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>
+''',
+
+    'eq_instead_of_equ_markup': '''\
+<p>Morda bi bil bolj primeren operator za unifikacijo (=)?</p>
+''',
+
+    'base_case': '''\
+<p>Si pomislil na robni pogoj? Kaj je najbolj enostaven primer, ko je element v seznamu?
+Do katerega elementa najlažje prideš?</p>
+''',
+
+    'recursive_case': '''\
+<p>Robni primer deluje. Kaj pa rekurzivni, splošni, primer?</p>
+''',
+
+    'predicate_always_false': '''\
+<p>Vse kaže, da tvoj predikat vedno vrne "false". Si mu dal pravilno ime, si se morda pri imenu zatipkal?</p>
+<p>Če je ime pravilno, se morda splača preveriti tudi, če se nisi zatipkal kje drugje,
+je morda kakšna pika namesto vejice ali obratno, morda kakšna spremenljivka z malo začetnico?</p>
+<p>Možno je seveda tudi, da so tvoji pogoji prestrogi ali celo nemogoči (kot bi bila npr. zahteva,
+da je <code>X</code> hkrati starš in sestra od <code>Y</code> ali kaj podobno zlobnega).</p>
+''',
+
+    'timeout': '''\
+<p>Je morda na delu potencialno neskončna rekurzija? Kako se bo ustavila?</p>
+<p>Morda pa je kriv tudi manjkajoč, neustrezen ali preprosto nekompatibilen (s splošnim primerom) robni pogoj?</p>
+''',
+}
-- 
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