Gaussian Elimination Learning

1. Proving that XOR satisfies the Exchange law, the binding law, is its own inverse.　　　　　　For example, 1^0 = 1 1^1 = 0 0^1 = 1 0^0 = 0 1^1^0 = 0 = 1^0^1 = 0. A^b^a=b is a number XOR or two times equivalent to invalid

2. Select two numbers from the N number to make XOR and maximum. Solution: We know that the difference or operation between two numbers is bitwise XOR, that is, they are each turned into a binary form, and then each time the number is found to be the largest. Like now, 0111,0000,0101,1010. We look for the maximum XOR result.   Can build a trie tree. Binary comparison, to be from high to low, to make XOR and maximum, then we enumerate the number I, to find the number of the largest difference between the number of the first, high priority. For example, what is the number of the maximum values that are found to be different or later than 0 1 1 1, we must start from the root, and then start looking for numbers that are different from the 0 1 1 1, and deal with them if they encounter the same position. 3. ln a tree with the right edge of the point, find a path to make XOR and Max. Solution: Optionally a root, h_i represents the XOR of the path from the root node to the node I and the path xor of X to Y and is represented as H_x xor h_y 4. Select several of the n numbers to make the XOR and the K, give the scheme or indicate that it is not OK. Solution: X_i is 0, indicating that the number of I is not selected, X_i is 1, indicating the number of the first I selected. Now consider the case of the P-bit of k: if the P-bit of k is 1, then the number of P-bits for 1 has an odd number of choices if the P-bit of k is 0, then the number of P-bits for 1 has an even number of selected to get the equation x_i1+x_i2+x_i3+...+x_is = Kp (+ is XOR ) 60 equations, the solution of the equation is equivalent to the solution of the original problem.

Gaussian Elimination Learning