Prolog: add introduction for the sets group and make it visible

3 files changed, 152 insertions, 0 deletions

diff --git a/prolog/problems/sets/count_3/common.py b/prolog/problems/sets/count_3/common.py
new file mode 100644
index 0000000..e770eaa
--- /dev/null
+++ b/prolog/problems/sets/count_3/common.py
@@ -0,0 +1,96 @@
+from operator import itemgetter
+import socket
+
+import prolog.engine
+import prolog.util
+from server.hints import Hint, HintPopup
+
+id = 120
+number = 10
+visible = True
+facts = None
+
+solution = '''\
+count(_, [], 0).
+count(X, [X|T], N) :-
+  count(X, T, NT),
+  N is NT + 1.
+count(X, [Y|T], NT) :-
+  X \== Y,
+  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'}]),
+    ('count(r, [a, r, b, c, r], X)',
+        [{'X': '2'}]),
+    ('count(l, [l, l, l, 1, 2, 3], X)',
+        [{'X': '3'}]),
+    ('count(z, [a, b, c, z, z], X)',
+        [{'X': '2'}]),
+]
+
+def test(code, aux_code):
+    n_correct = 0
+    engine_id = None
+    try:
+        engine_id, output = prolog.engine.create(code=code+aux_code, timeout=1.0)
+        if engine_id is not None and 'error' not in map(itemgetter(0), output):
+            # Engine successfully created, and no syntax error in program.
+            for query, answers in test_cases:
+                if prolog.engine.check_answers(engine_id, query=query, answers=answers, timeout=1.0):
+                    n_correct += 1
+    except socket.timeout:
+        pass
+    finally:
+        if engine_id:
+            prolog.engine.destroy(engine_id)
+
+    hints = [{'id': 'test_results', 'args': {'passed': n_correct, 'total': len(test_cases)}}]
+    return n_correct, len(test_cases), hints
+
+def hint(code, aux_code):
+    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/sets/count_3/en.py b/prolog/problems/sets/count_3/en.py
new file mode 100644
index 0000000..481fd1c
--- /dev/null
+++ b/prolog/problems/sets/count_3/en.py
@@ -0,0 +1,11 @@
+name = 'count/3'
+slug = 'find the number of occurrences of an element in list'
+
+description = '''\
+<p><code>count(X, L, N)</code>: <code>N</code> is the number of times the element <code>X</code> appears in the list <code>L</code>.</p>
+<pre>
+?- count(1, [1,2,1,3,1], N).
+  N = 3.
+</pre>'''
+
+hint = {}
diff --git a/prolog/problems/sets/count_3/sl.py b/prolog/problems/sets/count_3/sl.py
new file mode 100644
index 0000000..1e3f5d2
--- /dev/null
+++ b/prolog/problems/sets/count_3/sl.py
@@ -0,0 +1,45 @@
+name = 'count/3'
+slug = 'Preštej kolikokrat se element pojavi v seznamu'
+
+description = '''\
+<p><code>count(X, L, N)</code>: <code>N</code> je število kolikokrat se element <code>X</code> pojavi v seznamu <code>L</code>.</p>
+<pre>
+?- count(1, [1,2,1,3,1], N).
+  N = 3.
+</pre>'''
+
+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>
+''',
+}