diff options
| -rw-r--r-- | prolog/problems/lists_advanced/palindrome_1/en.py | 3 | ||||
| -rw-r--r-- | prolog/problems/lists_advanced/sublist_2/en.py | 53 | ||||
| -rw-r--r-- | prolog/problems/lists_advanced/sublist_2/sl.py | 4 |
3 files changed, 55 insertions, 5 deletions
diff --git a/prolog/problems/lists_advanced/palindrome_1/en.py b/prolog/problems/lists_advanced/palindrome_1/en.py index 27103d3..8471ae2 100644 --- a/prolog/problems/lists_advanced/palindrome_1/en.py +++ b/prolog/problems/lists_advanced/palindrome_1/en.py @@ -13,7 +13,8 @@ description = '''\ </pre>''' plan = ['''\ -<p>A palindrome is a list (ok, a word) that reads the same from front or back. Like aibohphobia. ;)</p> +<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> diff --git a/prolog/problems/lists_advanced/sublist_2/en.py b/prolog/problems/lists_advanced/sublist_2/en.py index 67b50a4..33e59c5 100644 --- a/prolog/problems/lists_advanced/sublist_2/en.py +++ b/prolog/problems/lists_advanced/sublist_2/en.py @@ -2,7 +2,7 @@ name = 'sublist/2' slug = 'generate sublists of a list' description = '''\ -<p><code>sublist(L, SL)</code>: <code>SL</code> is a continuous sublist of the +<p><code>sublist(L, SL)</code>: <code>SL</code> is a continuous sublist of list <code>L</code>. Your program should return every possible sublist; each answer may be returned more than once.</p> <pre> @@ -16,4 +16,53 @@ answer may be returned more than once.</p> X = [3]. </pre>''' -hint = {} +plan = ['''\ +<p>First a reminder: we're looking for sublists, not subsets. The difference is that sublists contain a number of +<em>consecutive</em> elements of the original list.</p> +<p>Perhaps you should look for some pattern? And which predicate is ideal to search for patterns? +You already know that. ;)</p> +''', '''\ +<p>Of course, predicate <code>conc/3</code> can be used to search for patterns in lists. Perhaps this time +we don't even need explicite recursion? Is that really possible in prolog? ;)</p> +''', '''\ +<p>So, what could the pattern be? Well, a part of the original list! Imagine that the original list is a tube +that you want to shorten -- a little bit from the left, a little bit from the right -- what's left is a sublist! +You chop off some elements from the front of the original list, and then you chop off some from the end.</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>sublist/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> +''', + + 'base_case': '''\ +<p>Did you think of a base case?</p> +''', + + 'recursive_case': '''\ +<p>The base case is ok. However, what about the general recursive case?</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> +''', +} diff --git a/prolog/problems/lists_advanced/sublist_2/sl.py b/prolog/problems/lists_advanced/sublist_2/sl.py index 4884665..b75aa0c 100644 --- a/prolog/problems/lists_advanced/sublist_2/sl.py +++ b/prolog/problems/lists_advanced/sublist_2/sl.py @@ -33,7 +33,7 @@ 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>sublist/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> ''', @@ -56,7 +56,7 @@ Do katerega elementa najlažje prideš?</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> +da je <code>N</code> enako kot <code>N + 1</code> ali kaj podobno logično zlobnega).</p> ''', 'timeout': '''\ |