Sunday, March 24, 2013


SPOJ Problem Set (Classical) 224. Vonny and her dominos

This is exactly the same problem from 2006 TopCoder Collegiate Challenge, problem DominoesFinding. I am not sure whether this problem can be solved by bipartite matching algorithm or dynamic programming, probably both will run out of time limit. But it can be solved by straight forward backtracking with a little bit pruning. The backtracking idea is pretty simple, just keep track of which tiles are used (each tile must be used exactly once), and try filling the grid in row major fashion. So there is no point discussing the solution, it is better to discuss why backtracking can be used here. I am not going to write these on my own words as it has already been written, so I will repost the analysis from TopCoder


by soul-net

Backtracking. Yes, that's it. Knowing that a problem is in fact solvable with a backtracking approach is most times a matter of intuition gained with experience. Anyway, in this and some other cases, there can be found more formal estimators that the idea is in fact THE idea.

I'll describe a possible backtracking approach, possibly the easiest to implement, but there are other possibilities. The idea is based on the fact that all squares must be used. For example, if we take the upper-left square of the board, we can see that we must connect it with one of its two neighbors. With this in mind, we can iterate over all squares and, each time we find an unused one, we know that we must match it with one of its two (or one) remaining neighboors -- or both, if we iterate in a column-row or a row-column fashion; when we find an unused square, we know that everybody in its upper-left rectangle is already used.

As we do this, we go marking each used piece and only continue trying if the new piece made by each new matching is "new". In this way, if we finally get all squares to be used, we know also that all pieces are used (because we managed to get no repeats) and then, we add 1 to the counter.

To be sure this approach works perfectly in time, you can conduct a little experiment and run the algorithm over an empty board without the "new piece" pruning. This will show you that there are less than 1.3 million ways to divide the board (1,292,697 actually), so it is perfectly feasible to try every one of them. Of course, the pruning of the "new piece" will reduce the running time dramatically in most cases.

There is also a good theoretical estimator that the approach will work in time, to convince ourselves before programming anything (many programmers think this is a must). There is a total of 56 squares in the board, our algorithm does nothing for half of them (when it finds them already used) and tries 2 or less cases for the other half (the ones it finds unused). This means the total number of leaves in the search tree will be bounded by 256/2 which is roughly 256 millions. This is pretty big, but considering it is a wide margin upper bound, it can be pretty well used as a "proof" that time limits won't bother.

View original analysis page from TopCoder.


  1. Hi Zobayer thanks a lot for this great blog...I am Big fan of those who are great programmer..I read your blog daily this helps me a lot to understand..

    1. Thank you man! The honor and pleasure is mine :D