A rigorous proof of an algorithm for finding the minimum number of interchanges is strictly proof, not taken for granted

Problem Description:

There is an array of 1~n, but this sequence has been scrambled to order, if we just "exchange any two elements", then to implement the elements from the 1~n order, "the minimum number of exchanges?" ”

Solution process:

First of all, we can write a simple case to try, such as arranging: 4 3 1 2, the number of exchanges = 3; we can test in multiple groups, the edge of testing, the real implementation needs to meet: 4 should be 2, 2 should be 3, 3 should be 1, 1 should be to 4, just a loop.

So, consider:

Lemma: Is there a minimum number of exchanges equal to 1 of elements in order to satisfy such a circular arrangement?

Mathematical induction can be considered.

First k=1,2,3 When we can know is established;

Assuming that when n <=k, the minimum number of exchanges for this single loop arrangement is n-1;

Consider the situation of n=k+1: At this time, we arbitrarily exchange the k+1 of the two elements, we will find that the single cycle split into two single loops (such as above, 4 and 1 after the exchange, the sequence becomes 1 3 4 2 then left <1> and <3 4 2> two sub-sequences, <1> exchange The number of times =0 <3 4 2> interchange = 2, then the total Exchange is 3 times, if we exchange 1 and 2 (this exchange does not have any element in the bit), the remainder is the sequence 4 3 2 1, is two single-cycle sequence <4 1> and <3 2>, the total Exchange or 1+ (1 + 1) = 3 = 4-1), then there are two different approaches:

Method 1: Perform recursive calls to two single-loop sequences, the minimum number of interchanges is 1 + (number of sequence 1-1) + (number of sequence 2-1) = total number of elements-1

Method 2: We merge the separate two single-cycle sequence, that is, the elements in sequence 1 and sequence 2 in the element Exchange, after the exchange of a new composition sequence, the exchange of the original sequence is separated and merged, in vain to Exchange 2 times, the remainder of the sequence is k+1 a single cycle, this exchange is obviously not the optimal mode of exchange

In this view, the exchange of Method 1 is the way to the least exchange.

Description when N=k + 1 o'clock, we can be better in accordance with the Mode 1 exchange, at this time can guarantee the number of exchanges =k=n-1 times

At this point we prove that: single cycle sequence, the minimum number of exchanges is n-1 times.

scenarios where multiple single-loop sequences are considered

Because of the different two single-cycle sequence merge exchange called a single-cycle sequence, the above method 2 indicates that the invalid exchange is actually

Therefore, the exchange of multiple single-cycle sequences is the sum of the number of times each single-cycle sequence is exchanged.

For example 4 3 2 1 7 6 5

is two single-cycle sequence <4 3 2 1> <7 6 5> The sum of their respective exchanges = 3 + 2 = 5 = 7-The number of single-cycle sequences

It can be seen that, for an n-length sequence, the optimal number of exchanges is the number of =n-n decomposed into a single cycle.

Understand this process, the code is naturally a cloud ...

A rigorous proof of an algorithm for finding the minimum number of interchanges is strictly proof, not taken for granted