name = 'len/2' slug = 'find the length of a list' description = '''\

len(L, Len): Len is the length of list L.

?- len([1,2,3], Len).
  Len = 3.
''' plan = ['''\

A list is not very long if it's empty, and if it's not empty it must have a head and a tail.

''', '''\

If the tail (list without a head) is of length LenT, then the whole list is of length LenT + 1.

''', '''\

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 get the length of the whole list.

'''] 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 len/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? Which list is the shortest list in the world?

''', '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?

''', 'arbitrary_base_case': '''\

What's the length of an empty list? Give a number!

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

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.

Perhaps you can try moving the arithmetic operation more towards the end of the predicate?

''', '=_instead_of_is': '''\

Did you use the operator = instead of is? Operator = is used for unification and tries to leave both its operands with as little modifications as possible while still making them equal. Operator is, on the other hand, performs actual arithmetic evaluation of its second operand and only then attempts the unification of both operands.

''', '+H_instead_of_+1': '''\

Did you really add the element's value instead of its length (one)? ;)

''', 'forcing_result_onto_recursion': '''

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.

Is your recursive call of the form len(Tail, LenTail + 1)? This forces the recursive call to return the length of the whole 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.

''', 'same_var_on_both_sides_of_is': '''\

Does one of your goals look similar to N is N + 1? Let's assume N 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.

''', }