diff options
Diffstat (limited to 'prolog/problems')
-rw-r--r-- | prolog/problems/lists_advanced/len_2/en.py | 6 | ||||
-rw-r--r-- | prolog/problems/lists_advanced/len_2/sl.py | 4 | ||||
-rw-r--r-- | prolog/problems/lists_advanced/sum_2/en.py | 87 | ||||
-rw-r--r-- | prolog/problems/lists_advanced/sum_2/sl.py | 6 |
4 files changed, 93 insertions, 10 deletions
diff --git a/prolog/problems/lists_advanced/len_2/en.py b/prolog/problems/lists_advanced/len_2/en.py index 33b1675..6518eec 100644 --- a/prolog/problems/lists_advanced/len_2/en.py +++ b/prolog/problems/lists_advanced/len_2/en.py @@ -15,14 +15,14 @@ plan = ['''\ is of length <code>LenT + 1</code>.</p> ''', '''\ <p>If I take away the head, and the recursion solves this smaller problem (tail), and if I add 1 to the -result returned by the recursion, then I got the length of the whole list.</p> +result returned by the recursion, then I get the length of the whole list.</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>conc/3</code> more general (e.g. able to work with output arguments becoming inputs).</p> +you can make the predicate <code>len/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> ''', @@ -78,7 +78,7 @@ second operand and only then attempts the unification of both operands.</p> 'forcing_result_onto_recursion': ''' <p>Don't force the result onto recursion, don't tell it what it should return. Just assume it will do its job. If this assumption is correct, then the rule will work for a larger case.</p> -<p>Is your recursive call of the form <code>len(T, LenT + 1)</code>? This forces the recursive call to +<p>Is your recursive call of the form <code>len(Tail, LenTail + 1)</code>? This forces the recursive call to return the length of <em>the whole</em> list, not just the tail! This will not work. It is your job to increase by one the result returned by the recursion. In short, add one outside the recursive call.</p> ''', diff --git a/prolog/problems/lists_advanced/len_2/sl.py b/prolog/problems/lists_advanced/len_2/sl.py index 49b9eea..28fe25e 100644 --- a/prolog/problems/lists_advanced/len_2/sl.py +++ b/prolog/problems/lists_advanced/len_2/sl.py @@ -75,8 +75,8 @@ svoji levi strani.</p> 'forcing_result_onto_recursion': ''' <p>Ne vsiljuj rekurziji kaj naj vrne, prepusti se ji. To je tisti del, ko narediš predpostavko, če je ta izpolnjena, potem bo tvoje pravilo delovalo za večji primer.</p> -<p>Je tvoj rekurzivni klic oblike <code>len(T, LenT + 1)</code>? S tem vsiljuješ rekurziji -da mora <emph>vrniti</emph> dolžino celega seznama in ne samo repa. To ni v redu, za ena moraš ti povečati +<p>Je tvoj rekurzivni klic oblike <code>len(Tail, LenTail + 1)</code>? S tem vsiljuješ rekurziji +da mora vrniti dolžino <em>celega</em> seznama in ne samo repa. To ni v redu, za ena moraš ti povečati rezultat, ki ti ga rekurzija vrne. Skratka, prištevanje naredi izven rekurzivnega klica.</p> ''', diff --git a/prolog/problems/lists_advanced/sum_2/en.py b/prolog/problems/lists_advanced/sum_2/en.py index 767a6f4..3abc348 100644 --- a/prolog/problems/lists_advanced/sum_2/en.py +++ b/prolog/problems/lists_advanced/sum_2/en.py @@ -2,10 +2,93 @@ name = 'sum/2' slug = 'find the sum of all elements in list' description = '''\ -<p><code>sum(L, Sum)</code>: <code>Sum</code> is the sum of all elements in the list <code>L</code>.</p> +<p><code>sum(L, Sum)</code>: <code>Sum</code> is the sum of all elements in list <code>L</code>.</p> <pre> ?- sum([1,2,3], Sum). Sum = 6. </pre>''' -hint = {} +plan = ['''\ +<p>The sum of an empty list is really not all that large. And if it's not empty, then we add it up, +element by element, recursively.</p> +''', '''\ +<p>If the sum of the elements in the tail (list without a head) equals <code>SumT</code>, then the sum +of the elements in the whole list equals <code>SumT + H</code>.</p> +''', '''\ +<p>If I take away the head, and the recursion solves this smaller problem (tail), and if I add the value +of the head to the result returned by the recursion, then I get the sum of the whole list.</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>sum/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? What's the sum of the elements in an empty list?</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> +''', + + 'arbitrary_base_case': '''\ +<p>What's the sum of the elements in an empty list? Give a number!</p> +''', + + 'args_not_instantiated': '''\ +<p>The error that prolog reported means that when it encountered an arithmetic operation, not all the +values were known. Unfortunately, the ordering of goals is important when dealing with arithmetics.</p> +<p>Perhaps you can try moving the arithmetic operation more towards the end of the predicate?</p> +''', + + '=_instead_of_is': '''\ +<p>Did you use the operator <code>=</code> instead of <code>is</code>? Operator <code>=</code> is used for +unification and tries to leave both its operands with as little modifications as possible while still making +them equal. Operator <code>is</code>, on the other hand, performs actual arithmetic evaluation of its +second operand and only then attempts the unification of both operands.</p> +''', + + '+1_instead_of_+H': '''\ +<p>Did you really add one instead of the element's value? Copy/paste operation from the previous exercise? ;)</p> +''', + + 'forcing_result_onto_recursion': ''' +<p>Don't force the result onto recursion, don't tell it what it should return. Just assume it will do its job. +If this assumption is correct, then the rule will work for a larger case.</p> +<p>Is your recursive call of the form <code>sum(Tail, SumTail + H)</code>? This forces the recursive call to +return the sum of <em>the whole</em> list, not just the tail! This will not work. It is your job to +increase (by value of the head) the result returned by the recursion. In short, perform the addition +outside the recursive call.</p> +''', + + 'same_var_on_both_sides_of_is': '''\ +<p>Does one of your goals look similar to <code>N is N + 1</code>? Let's assume <code>N</code> is equal to 3. +With this goal you just stated that 3 must be equal to 4 (3+1). Prolog is a logical language and will +gladly say "false" to such a statement! Just use a new variable. The garbage collector will take care of +those not needed anymore automatically.</p> +''', +} diff --git a/prolog/problems/lists_advanced/sum_2/sl.py b/prolog/problems/lists_advanced/sum_2/sl.py index 539271b..d3bddcf 100644 --- a/prolog/problems/lists_advanced/sum_2/sl.py +++ b/prolog/problems/lists_advanced/sum_2/sl.py @@ -22,7 +22,7 @@ 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>sum/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> ''', @@ -76,8 +76,8 @@ svoji levi strani.</p> 'forcing_result_onto_recursion': ''' <p>Ne vsiljuj rekurziji kaj naj vrne, prepusti se ji. To je tisti del, ko narediš predpostavko, če je ta izpolnjena, potem bo tvoje pravilo delovalo za večji primer.</p> -<p>Je tvoj rekurzivni klic oblike <code>len(T, LenT + H)</code>? S tem vsiljuješ rekurziji -da mora <emph>vrniti</emph> vsoto celega seznama in ne samo repa. To ni v redu, za vrednost glave moraš ti povečati +<p>Je tvoj rekurzivni klic oblike <code>sum(Tail, SumTail + H)</code>? S tem vsiljuješ rekurziji +da mora vrniti vsoto <em>celega</em> seznama in ne samo repa. To ni v redu, za vrednost glave moraš ti povečati rezultat, ki ti ga rekurzija vrne. Skratka, prištevanje naredi izven rekurzivnega klica.</p> ''', |