Merge branch 'master' of 192.168.15.97:codeq-problems

2 files changed, 82 insertions, 6 deletions

diff --git a/prolog/problems/lists_advanced/palindrome_1/en.py b/prolog/problems/lists_advanced/palindrome_1/en.py
index bc6342b..8471ae2 100644
--- a/prolog/problems/lists_advanced/palindrome_1/en.py
+++ b/prolog/problems/lists_advanced/palindrome_1/en.py
@@ -8,6 +8,81 @@ description = '''\
   true.
 ?- palindrome([1,2,3]).
   false.
+?- palindrome([a,b,b,a]).
+  true.
 </pre>'''
 
-hint = {}
+plan = ['''\
+<p>A palindrome is a list (ok, a word) that reads the same from front or back. Like aibohphobia!
+Was it a car or a cat I saw? ;)</p>
+''', '''\
+<p>As always we want to reduce the problem into a smaller one. Let's chop off the first and the last element
+of a list, and, if equal, proceed recursively.</p>
+''', '''\
+<p>If head <code>H</code> of list <code>L</code> is equal to the list's last element, and if the remainder
+(middle part) is a palindrome, then list <code>L</code> is also a palindrome.</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).</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>
+''',
+
+    'base_case': '''\
+<p>Did you think of a base case? Which list represents the shortest possible palindrome?</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>
+''',
+
+    '[X,X]_instead_of_[]_base_case': '''\
+<p>Well, <code>[X,X]</code> is definitely a good base case. However, it doesn't cover one special case,
+and that is an empty list. Of course, this is a matter of definition (taste even?), but please correct it
+so that we all have solutions working in the same way.</p>
+''',
+
+    'one_base_case_missing': '''\
+<p>Do you take <em>two</em> elements out with each recursive call? How does this end? Odd, even? ;)</p>
+<p>Try the following two queries. One will work nicely, the other one won't. What's the difference?</p>
+<p><code>?- palindrome([a,b,b,a]).</code></p>
+<p><code>?- palindrome([l,e,v,e,l]).</code></p>
+''',
+
+    'arbitrary_base_case': '''\
+<p>Note that <code>_</code> is not the same as <code>[_]</code>. The first pattern represents an arbitrary
+variable (anything), the second a list with <em>a single</em> arbitrary element.</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.</p>''',
+
+    'final_hint_1': '''\
+<p>Interesting tidbit: do you know that you could have solved this exercise with just a single predicate call?
+What happens with the palindrome if you... hmm, reverse it? ;)</p>
+''',
+
+    'final_hint_2': '''\
+<p>You can make the solution even shorter! How can you get rid of the operator <code>=</code> (<code>==</code>)
+or rather make it implicit?</p>
+''',
+}
diff --git a/prolog/problems/lists_advanced/palindrome_1/sl.py b/prolog/problems/lists_advanced/palindrome_1/sl.py
index 30a6bab..1e4e568 100644
--- a/prolog/problems/lists_advanced/palindrome_1/sl.py
+++ b/prolog/problems/lists_advanced/palindrome_1/sl.py
@@ -8,6 +8,8 @@ description = '''\
   true.
 ?- palindrome([1,2,3]).
   false.
+?- palindrome([a,b,b,a]).
+  true.
 </pre>'''
 
 plan = ['''\
@@ -22,8 +24,7 @@ palindrom, potem je tudi celoten seznam <code>L</code> palindrom.</p>
 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>
+da elementa lahko naredi enaka (unifikacija).</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>
 ''',
@@ -50,8 +51,8 @@ da je <code>N</code> enako kot <code>N + 1</code> ali kaj podobno logično zlobn
 ''',
 
     '[X,X]_instead_of_[]_base_case': '''\
-<p>Vsekakor je [X,X] povsem dober robni pogoj, a ne pokrije posebnega primera, ko je vhod kar prazen seznam. To je
-seveda stvar definicije, a da bomo imeli vsi enake rešitve, prosim, popravi.</p>
+<p>Vsekakor je <code>[X,X]</code> povsem dober robni pogoj, a ne pokrije posebnega primera, ko je vhod kar
+prazen seznam. To je seveda stvar definicije, a da bomo imeli vsi enake rešitve, prosim, popravi.</p>
 ''',
 
     'one_base_case_missing': '''\
@@ -62,7 +63,7 @@ seveda stvar definicije, a da bomo imeli vsi enake rešitve, prosim, popravi.</p
 ''',
 
     'arbitrary_base_case': '''\
-<p>Pazi <code>_</code> ni enako kot <code>[_]</code>. Prvo predstavlja poljubno spremenljivko, drugo seznam
+<p>Pazi, <code>_</code> ni enako kot <code>[_]</code>. Prvo predstavlja poljubno spremenljivko, drugo seznam
 z <em>enim</em> poljubnim elementom.</p>
 ''',