Analysis of three-way set disjoint (Three-way set disjointness) algorithm

Concept Introduction

In fact, the principle is similar, whether it is a three-way or a multi-way. The three-way set does not intersect, meaning that there are 3 sets of numbers, in any set, without the same element. and satisfies both: for any x, there is no x $ \in $ a,x $ \in $ B, and x $ \in $ C.

Achieve a

def disjoint1(A, B, C):    """Return True if there is no element common to all three lists"""    forin A:        forin B:            forin C:                if==== c:                    returnFalse   # we found a common vale    returnTrue                    # if we reach this, sets are disjoint

Obviously, three-tier nesting, for a,b,c with a set length of n, the total time complexity is O ($ n^3 $).

Implementation two

def  disjoint2 (A, B, C):  "" "Return True If There is no element common to all three lists "" " for  a in  A: for  b in  B: if  a = =  B: # only check C if we found match from A and b  for  C in  C: if  a ==< /span> c: # (and thus a = = b = = C)  return false  # we found a common value  return  span class= "va" >true  # if we reach this, sets is disjoint  

To calculate the total time complexity, layers of analysis are required.

• First, the outer loop A is O (n) and the inner loop b is O ($ n^2 $).

• For if a== B, a total of O ($ n^2 $) Judgments will be executed.

• The remaining time depends on the number of times (A, b) is equal. Because of the cross-specificity of the set, A and B are equal in number of N. Because once a = = B is established, then a is not equal to the element except B in B.

• Because the number of a = = B is n, the time complexity of the most inner loop and its loop body is O ($ n^2 $).

Overall, the total time complexity is still O ($ n^2 $).

Summarize

With different algorithm time, the time complexity of program execution can be reduced, and the program is optimized to improve the efficiency.

Analysis of three-way set disjoint (Three-way set disjointness) algorithm