1. Tower of Hanoi problem
- Mathematical induction: ① the establishment of the minimum size; ② disposed on \ (n = [1, k ] \) is established when, for proof \ (n = k + 1 \ ) also holds when. So the question of any size are established.
It can be combined with naturally recursive model.
The actual problem -> recursive -> Mathematical Induction -> General Term Formula
- By handling some items will make recursive mathematical induction easier.
2. The straight line on the plane
Recursive expansion is a good way to seek general term formula.
For example direct analysis is not easy, can discuss the "loss" of the number.
3.JOJO problem
- For a recursive, to replace part of its constant unknown, it is desired to obtain the coefficients for each term formula unknowns.
At this time, each of the unknowns can take some special value substituted into the recursion formula.
A more common approach is to find some simple functions, so that it meets recursive, whereby listed equation, solving for the value of unknowns. Look for the link between the unknown factor in this way, which lists the equations for the unknown coefficients.
Noting the number of independent unknowns, we need to list how many sets of equations.
At the same time, the lifting of recursion in some of the established limits (hexadecimal, etc.), recursion can be extended to more general cases.
4. warm-up questions
The problem in the "similar", mathematical induction can only assume \ ([1, k] \ ) horse in the same color, if extended to \ ([2, k + 1 ] \) case, it is necessary proof \ (k + 1 \) No. horse previous color are the same.
Find the size of \ (K \) problem, the process as follows: ① The \ ([1, k-1 ] \) is moved to the \ (B \) column; ② the \ (K \) to an intermediate column; ③ the [1, k-1] is moved to the \ (A \) column; ④ the \ (K \) is moved to the \ (B \) column; ⑤ the \ ([1, k-1] \) is moved to the \ ( B \) column.
Set \ (T_n \) is the size of \ (n \) is the minimum number of steps required to solve the problem, thereby obtaining recursion:
\[T_1=2\]
\ [T_n = 6T_ {n-1} + 2, n \ geq 2 \]
设\(G_n=T_n+\frac{2}{5}\)，则\(G_1=\frac{12}{5},G_n=6(T_{n-1}+\frac{2}{5})=6G_{n-1}=\frac{2}{5}6^n\)。
即\(T_n=\frac{2}{5}(6^n-1)\)。
Proof: \ (T_n \) is always an integer, i.e. \ (. 5 | (. 6-n-^. 1) \) .
For \ (n = 1 \) situation, it is clearly established.
Suppose that for \ ([1, n-1 ] \) were established. \ (. 1. 6-n-^ =. 6 \. 1-n-times6 ^ {-1} = (6-1) \. 6 Times-n-^ {+}. 1. 6-n-^ {-1}. 1 \) , due to the \ (n-1 \) the establishment, also set here.
Obviously \ (A, B \) column SUCH placed correctly, just consider the middle column. Found in the process ③, \ (n-\) necessarily in the middle column, and \ ([1, n-1 ] \) from \ (B \) to \ (A \) , integrally bound through the middle column.
Set \ (T_n \) is the minimum number of steps in the original rule, \ (G_n \) is the \ (n-\) Uncertain beginning and end of a discharge position of the disk good minimum number of steps. For the whole story problem into any position to prove that \ (G_n \ Leq T_n \) .
For \ (n = 1 \) case, if the whole story in the same position a pillar, the \ (= 0 G_1 <T_l \) , or \ (. 1-G_1 = T \) , the conclusion holds.
Suppose that for \ ([1, n-1 ] \) were established. For the first \ (n-\) discs, the start position is set \ (X \) , the end position \ (Y \) , an intermediate position \ (Z \) . If \ (X = Y \) , then \ (= G_n. 1-n-G_ {} \ Leq. 1-n-T_ {} <T_n \) . Otherwise, perform the following steps: ① \ ([. 1,. 1-n-] \) is moved to the \ (Z \) , provided this operation takes \ (1,0-n-G_ {} \) ; ② the \ (n-\ ) to move to the \ (Y \) ; ③ the \ ([1, n-1 ] \) is moved to the position corresponding to the end, provided for this operation takes \ (1,1-n-G_ {} \) . At this time, \ (G_ G_n = {1,0} +-n-n-G_ {1,1}-+ 1'd \ Leq 2T_ {+}. 1. 1-n-T_n = \) .
It is proved by the mathematical induction.