Solution for "crackless wall" problem

Question

Hi all,

Here's the problem statement : Consider the problem of building a wall out of 2x1 and 3x1 bricks (horizontal×vertical dimensions) such that, for extra strength, the gaps between horizontally-adjacent bricks never line up in consecutive layers, i.e. never form a "running crack".

There are eight ways of forming a crack-free 9x3 wall, written W(9,3) = 8.

Calculate W(32,10). < Generalize it to W(x,y) >

http://www.careercup.com/question?id=67814&amp;form=comments

The above link gives a few solutions , but 'm unable to underastand the logic behind them.. I'm trying to code this in perl and have done so far :

input : W(x,y)
find all possible i's and j's such that x == 3(i) + 2(j);
for each pair (i,j) ,
find n = (i+j)C(j)            # C:combinations

Adding all these n's should give the count of all possible combinations . But i have no idea on how to find the real combinations for one row and how to proceed further.. Pl help..

Answer 1

Based on the claim that W(9,3)=8, I'm inferring that a "running crack" means any continuous vertical crack of height two or more. Before addressing the two-dimensional problem as posed, I want to discuss an analogous one-dimensional problem and its solution. I hope this will make it more clear how the two-dimensional problem is thought of as one-dimensional and eventually solved.

Suppose you want to count the number of lists of length, say, 40, whose symbols come from a reasonably small set of, say, the five symbols {a,b,c,d,e}. Certainly there are 5^40 such lists. If we add an additional constraint that no letter can appear twice in a row, the mathematical solution is still easy: There are 5*4^39 lists without repeated characters. If, however, we instead wish to outlaw consonant combinations such as bc, cb, bd, etc., then things are more difficult. Of course we would like to count the number of ways to choose the first character, the second, etc., and multiply, but the number of ways to choose the second character depends on the choice of the first, and so on. This new problem is difficult enough to illustrate the right technique. (though not difficult enough to make it completely resistant to mathematical methods!)

To solve the problem of lists of length 40 without consonant combinations (let's call this f(40)), we might imagine using recursion. Can you calculate f(40) in terms of f(39)? No, because some of the lists of length 39 end with consonants and some end with vowels, and we don't know how many of each type we have. So instead of computing, for each length n<=40, f(n), we compute, for each n and for each character k, f(n,k), the number of lists of length n ending with k. Although f(40) cannot be calculated from f(39) alone, f(40,a) can be calculated in terms of f(30,a), f(39,b), etc.

The strategy described above can be used to solve your two-dimensional problem. Instead of characters, you have entire horizontal brick-rows of length 32 (or x). Instead of 40, you have 10 (or y). Instead of a no-consonant-combinations constraint, you have the no-adjacent-cracks constraint.

You specifically ask how to enumerate all the brick-rows of a given length, and you're right that this is necessary, at least for this approach. First, decide how a row will be represented. Clearly it suffices to specify the locations of the 3-bricks, and since each has a well-defined center, it seems natural to give a list of locations of the centers of the 3-bricks. For example, with a wall length of 15, the sequence (1,8,11) would describe a row like this: (ooo|oo|oo|ooo|ooo|oo). This list must satisfy some natural constraints:

1. The initial and final positions cannot be the centers of a 3-brick. Above, 0 and 14 are invalid entries.
2. Consecutive differences between numbers in the sequence must be odd, and at least three.
3. The position of the first entry must be odd.
4. The difference between the last entry and the length of the list must also be odd.

There are various ways to compute and store all such lists, but the conceptually easiest is a recursion on the length of the wall, ignoring condition 4 until you're done. Generate a table of all lists for walls of length 2, 3, and 4 manually, then for each n, deduce a table of all lists describing walls of length n from the previous values. Impose condition 4 when you're finished, because it doesn't play nice with recursion.

You'll also need a way, given any brick-row S, to quickly describe all brick-rows S' which can legally lie beneath it. For simplicity, let's assume the length of the wall is 32. A little thought should convince you that

1. S' must satisfy the same constraints as S, above.
2. 1 is in S' if and only if 1 is not in S.
3. 30 is in S' if and only if 30 is not in S.
4. For each entry q in S, S' must have a corresponding entry q+1 or q-1, and conversely every element of S' must be q-1 or q+1 for some element q in S.

For example, the list (1,8,11) can legally be placed on top of (7,10,30), (7,12,30), or (9,12,30), but not (9,10,30) since this doesn't satisfy the "at least three" condition. Based on this description, it's not hard to write a loop which calculates the possible successors of a given row.

Now we put everything together:

First, for fixed x, make a table of all legal rows of length x. Next, write a function W(y,S), which is to calculate (recursively) the number of walls of width x, height y, and top row S. For y=1, W(y,S)=1. Otherwise, W(y,S) is the sum over all S' which can be related to S as above, of the values W(y-1,S').

This solution is efficient enough to solve the problem W(32,10), but would fail for large x. For example, W(100,10) would almost certainly be infeasible to calculate as I've described. If x were large but y were small, we would break all sensible brick-laying conventions and consider the wall as being built up from left-to-right instead of bottom-to-top. This would require a description of a valid column of the wall. For example, a column description could be a list whose length is the height of the wall and whose entries are among five symbols, representing "first square of a 2x1 brick", "second square of a 2x1 brick", "first square of a 3x1 brick", etc. Of course there would be constraints on each column description and constraints describing the relationship between consecutive columns, but the same approach as above would work this way as well, and would be more appropriate for long, short walls.