Algorithm (ii)--------Divide and conquer law

Conditions for the Division and Treatment of Law:

• The size of the problem can be easily solved by scaling it down.
• The problem can be broken down into the same problem of a larger scale than the following: Recursive thinking should be?
• The problem is separate from each other, that is, there is no public problem between the issues.
• Benefit? The solution to the problem can be combined into the solution of the problem.

Case---Quick line:

(1) process

• Divide (Partition)
– Rearrange the elements in order to get such a partition
• All elements in a position s front? is equal to A[s]
• After position s? All elements are equal to A[s]
Conquer: When a partition is determined, the position of the a[s] is determined, and then the two arrays are recursively divided
combine: Sort in place in-place sort, need to merge

(2) algorithm analysis

Best case: All divisions are divided into equal size two? array

Tbest (n) = 2Tbest (N/2) + O (n), (T (1) = 0)= O (NLOGN)

Worst case: Dividing two of the produced? The array contains n-1 elements and 0 elements (the array to be sorted is ordered!). ）

Tworst (n) = Tworst (n-1) + O (n), (T (1) = 0) = O (n2)

Average: Assuming that the probability that the division may appear at any location is 1/n, then the average "?" is the recursive relationship of time t (N):

(3) Select a good division benchmark, production, good division:

Find the Dividing datum (approximate median): "Median median"
1. Divide n elements into N/5 groups, 5 elements per group (except the last? Group may not be 5 elements); O (1)
2. The random sort algorithm sorts the 5 elements in each group and takes out the median of each group, together with N/5. O (N)
3. Recursion to find out the median number of N/5, as the Dividing Datum; T (? N/5?)
divide
– the input to the output array by the approximate median number. Partition; O (n)
Conquer
If k = s, return s//s is divided by position
If P < K < S, recursive selection a[p...s-1] k element
If S < K < R, recursive selection A[S+1...R] K (s-p+1)? element
Combine:?

Board Overlay:

In the 2KX2K chess board, just the same as the other lattice, said the lattice is a special, and said the board is a special board. In the board coverage problem, want to figure 4 different forms of L-shaped cards covering the given special chessboard all except special lattice, and any 2 L-type cards must not overlap.

(1) The problem has a solution----mathematical inductive method

– When N=1 (2x2 chessboard), the problem has a solution;
– Assuming that when n=k (2kx2k chessboard), the problem has a solution;
– Then when N=k+1 (2k+1x2k+1 chessboard), divide the chessboard into 4
2KX2K Chess board, special lattice is located in 4 of the chessboard, the remaining 3 of the Board, special lattice.
– How do you convert these 3 special-grid chess boards into special chessboard?
• ?? An L-type card covering these 3 boards of the meeting place, the original problem converted to 4 n=k when the problem, because n=k when there is a solution, so n=k+1 also have a solution.

• When k=0 (1x1 chessboard), that is, special lattice, the number of cards is 0;
• When K>0,
-Divide: Divides the 2kx2k checkerboard into 4 2k-1x2k-1? chessboard. Special lattice is located in 4 of the more?? In the chess board, the other 3 of the board? Special lattice.
-Before recursion, convert the original question to the same 4-scale one.
- ?? A l-shaped card covering these 3? The rendezvous of the chessboard
-Conquer: Recursion to make this division, straight checkerboard shrink to 1x1.
-Combine:?

Algorithm (ii)--------Divide and conquer law