Question

Easy problem and known algorithm:

I have a big array with 100 members. First X members are 0, and the rest are 1. Find X.

I am solving it by a binary search: Check member 50, if it is 0 – check member 75, etc, until I find adjacent 0 and 1.

I am looking for an optimized algorithm for the same problem in 2-dimensions:

I have 2-dimensional array 100*100. Those members that are on rows 0-X AND on columns 0-Y are 0, and the rest are 1. How to find Y and X?

Answer 1

Simple solution: go first in X-direction and then in Y-direction.

Check (0,50); If it is 0, check (0,75); until You find adjacent 0 and 1. Then go to Y direction from there.

Second solution:

Check member (50,50). If it is 1, check (25,25), until You find 0. Continue, until You find adjacent (X,X) and (X+1,X+1) that are 0 and 1. Then test (X,X+1) and (X+1,X). Neither or one of them will be 1. If neither, You are finished. If only one, say for example (X+1,X), then You know that the box’s size is between (X+1,X) and (100,X). Use binary search to find box’s height.

EDIT: As Chris pointed out, it seems that the simple approach is faster.

Second solution (modified):

Check member (50,50). If it is 1, check (25,25), until You find 0. Continue, until You find adjacent (X,X) and (X+1,X+1) that are 0 and 1. Then test (X,X+1). If it is 1, then do binary search on line (X,X+1)…(X,100). Else do binary search on line (X,X)…(100,X).

Even then I am probably beating a dead horse here. If it will be faster, then by neglible amount. This is just for theoretical fun. 🙂

EDIT 2 As Fezvez and Chris put it, binary search divides the search space in two most efficiently; My approach divides the area to 1/4 and 3/4 pieces. Fezvez pointed out that this could be remedied by calculating the dividing factor beforehand (but that would be extra calculation). In modified version of my algorithm I choose the direction where to go (X or Y direction), which effectively also divides the search space in two, and then conduct binary search. To conclude, this shows that this approach will always be a bit slower. (and more complicated to implement.)

Thank You, Igor Oks, for interesting question. 🙂