Programming Praxis

Over at PrologSite, Werner Hett gives a list of Ninety-Nine Prolog Problems “to give you the opportunity to practice your skills in logic programming.” Today’s exercise is number 1.28 on the list. Hett gives the problem two stars, meaning that he expects a skilled Prolog programmer to spend thirty to ninety minutes to solve it. Here is Hett’s statement of the problem:

1.28 (**) Sorting a list of lists according to length of sublists
a) We suppose that a list (InList) contains elements that are lists themselves. The objective is to sort the elements of InList according to their length. E.g. short lists first, longer lists later, or vice versa.

Example:
?- lsort([[a,b,c],[d,e],[f,g,h],[d,e],[i,j,k,l],[m,n],[o]],L).
L = [[o], [d, e], [d, e], [m, n], [a, b, c], [f, g, h], [i, j, k, l]]

b) Again, we suppose that a list (InList) contains elements that are lists themselves. But this time the objective is to sort the elements of InList according to their length frequency; i.e. in the default, where sorting is done ascendingly, lists with rare lengths are placed first, others with a more frequent length come later.

Example:
?- lfsort([[a,b,c],[d,e],[f,g,h],[d,e],[i,j,k,l],[m,n],[o]],L).
L = [[i, j, k, l], [o], [a, b, c], [f, g, h], [d, e], [d, e], [m, n]]

Note that in the above example, the first two lists in the result L have length 4 and 1, both lengths appear just once. The third and forth list have length 3; there are two list of this length. And finally, the last three lists have length 2. This is the most frequent length.

Your task is to solve Hett’s problem 1.28; you need not use Prolog. Be sure to follow Hett’s advice:

Your goal should be to find the most elegant solution of the given problems. Efficiency is important, but logical clarity is even more crucial.

When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

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