From 4a01e895b521c75406270abd581cd3fbb09f2c08 Mon Sep 17 00:00:00 2001 From: Aleksander Sadikov Date: Thu, 8 Sep 2016 00:28:56 +0200 Subject: English translations for min/2 and max/2 added. --- prolog/problems/lists_advanced/min_2/en.py | 89 +++++++++++++++++++++++++++++- prolog/problems/lists_advanced/min_2/sl.py | 7 ++- 2 files changed, 90 insertions(+), 6 deletions(-) (limited to 'prolog/problems/lists_advanced/min_2') diff --git a/prolog/problems/lists_advanced/min_2/en.py b/prolog/problems/lists_advanced/min_2/en.py index a286cac..20677c6 100644 --- a/prolog/problems/lists_advanced/min_2/en.py +++ b/prolog/problems/lists_advanced/min_2/en.py @@ -1,8 +1,8 @@ name = 'min/2' -slug = 'find the smallest element' +slug = 'find the smallest element of a list' description = '''\ -

min(L, Min): Min is the smallest value in the list L.

+

min(L, Min): Min is the smallest value in list L.

 ?- min([5,4,1,6], M).
   M = 1.
@@ -10,4 +10,87 @@ description = '''\
   M = 2.
''' -hint = {} +plan = ['''\ +

As usual, try to reduce the problem to a smaller one. Say you already have the smallest element +of the tail...

+''', '''\ +

Compare the smallest element in the tail (list without head H) with the value of head H, +the smaller of the two wins and gets returned.

+''', '''\ +

If the given list L is composed of head H and tail T, and if we +assume that some MinT is the smallest element in T, and if it's also true that +the value of H is smaller than MinT, then H is the smallest element +in L. Or it is true that H is greater than MinT, and in this +case MinT is the smallest element in L.

+'''] + +hint = { + 'eq_instead_of_equ': '''\ +

The operator == is "stricter" than operator = in the sense that +for the latter it is enough to be able to make the two operands equal (unification). Perhaps by using = +you can make the predicate min/2 more general (e.g. able to work with output arguments becoming inputs).

+

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).

+''', + + 'eq_instead_of_equ_markup': '''\ +

Perhaps the operator for unification (=) would be better?

+''', + + 'base_case': '''\ +

Did you think of a base case? What's the shortest list with an obvious smallest element?

+''', + + 'recursive_case': '''\ +

The base case is ok. However, what about the general recursive case?

+''', + + 'predicate_always_false': '''\ +

It seems your predicate is always "false". Did you give it the correct name, +or is it perhaps misspelled?

+

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?

+

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 N is equal to N + 1, +or something similarly impossible).

+''', + + 'timeout': '''\ +

Is there an infinite recursion at work here? How will it ever stop?

+

Or perhaps is there a missing, faulty, or simply incompatible (with the general recursive case) base case?

+''', + + 'empty_list_base_case': '''\ +

You'll be hard pressed to find the smallest element of an empty list. What if you stop +the recursion a bit earlier this time?

+''', + + 'list_instead_of_elem_base_case': '''\ +

You should return an element, not a list!

+''', + + 'duplicates_not_covered': '''\ +

The list can contain duplicate elements. Did you think of that?

+''', + + 'args_not_instantiated': '''\ +

The error that prolog reported means that when it encountered an arithmetic operation, not all the +values were known.

+

Did you perhaps forget that conjunction has higher precedence that disjunction and that every prolog's +sentence (branch, rule) is independent in terms of the scope of the variables? This could be the problem. +Carefully inspect both blocks of code (before and after the semicolon) or both rules.

+''', + + 'unprotected_branch': '''\ +

It seems you didn't "protect" one of the (conditional) branches. Both branches (of the disjunction) typically +require a condition. Don't rely that if the execution reached the second branch that the first branch is false. +The relation between them is OR and not XOR. It's your job to make them mutually exclusive. Try the following +query and check all possible solutions and you'll notice the problem.

+

?- min([1,9,3,8,6], Min).

+''', + + 'one_branch_missing': '''\ +

Maybe you forgot one of the possibilities? The head can be either greater or smaller than +the smallest element in the tail.

+''', +} diff --git a/prolog/problems/lists_advanced/min_2/sl.py b/prolog/problems/lists_advanced/min_2/sl.py index bac4dc6..4a6975e 100644 --- a/prolog/problems/lists_advanced/min_2/sl.py +++ b/prolog/problems/lists_advanced/min_2/sl.py @@ -1,5 +1,5 @@ name = 'min/2' -slug = 'Poišči najmanjši element v danem seznamu' +slug = 'Poišči najmanjši element v seznamu' description = '''\

min(L, Min): Min je najmanjši element v seznamu L.

@@ -11,7 +11,8 @@ description = '''\ ''' plan = ['''\ -

Kot vedno, poskusi prevesti na manjši problem. Recimo, da že imaš najmanjši element v repu seznama...

+

Kot vedno, poskusi prevesti na manjši problem. Recimo, da že imaš najmanjši element v +repu seznama...

''', '''\

Najmanjši element v repu (seznamu brez glave H) primerjaj z vrednostjo glave H, tisti, ki je manjši zmaga in ga vrneš!

@@ -27,7 +28,7 @@ hint = { 'eq_instead_of_equ': '''\

Operator == je strožji od operatorja = v smislu, da je za slednjega dovolj, da elementa lahko naredi enaka (unifikacija). Morda z uporabo = narediš predikat -memb/2 delujoč tudi v kakšni drugi smeri.

+min/2 delujoč tudi v kakšni drugi smeri.

Seveda pa lahko nalogo rešiš brez obeh omenjenih operatorjev, spomni se, da lahko unifikacijo narediš implicitno že kar v argumentih predikata (glavi stavka).

''', -- cgit v1.2.1