Changing minds

First, let $\pi$ be some $k$ -cycle on $[n] = \left \{ 1 \cdots n \right \}$ . Then write in Cauchy's two-line notation:

$\pi = \begin{pmatrix} 1 \;\;\; 2 \;\;\; \cdots \;\;\; k \;\;\; k+1 \;\;\; n \\ 2 \;\;\; 3 \;\;\; \cdots \;\;\; 1 \;\;\; k+1 \;\;\; n \end{pmatrix}$

Now let $\wp (a,b)$ be the operation that makes the transposition of $a$ and $b$ . For our purposes, $\pi$ is generated by distinct switches on $[n]$ . Introduce then two new elements $\left \{x, y\right \}$ and then

$\pi* = \begin{pmatrix} 1 \;\;\; 2 \;\;\; \cdots \;\;\; k \;\;\; k+1 \;\;\; \cdots \;\;\; n \;\;\; x \;\;\; y \\ 2 \;\;\; 3 \;\;\; \cdots \;\;\; 1 \;\;\; k+1 \;\;\; \cdots \;\;\; n \;\;\; x \;\;\; y \end{pmatrix}$

For any $i = 1, \cdots, k$ let $\varphi$ be the series of switches

$\varphi = (\wp(x,1)\wp(x,2)\cdots\wp(x,i))(\wp(y,i+1)\wp(y,i+2)\cdots\wp(y,k))(\wp(x,i+1)\wp(y,1))$

Note that each switch exchanges one element of $[n]$ with one of $\left \{x, y\right \}$ , so they are all distinct from the switches within $[n]$ that generated $\pi$ and also from $\wp(x,y)$ . Verification leads us to

$\pi*\varphi = \begin{pmatrix} 1\;\;\; 2 \;\;\; \cdots \;\;\; n \;\;\; x \;\;\; y \\ 1 \;\;\; 2 \;\;\; \cdots \;\;\; n \;\;\; y \;\;\; x \end{pmatrix}$

i.e., $\varphi$ inverts the $k$ -cycle and leaves $\left \{x, y\right \}$ inverted, without switching $\left \{x, y\right \}$ not even once. Then we can invert $\left \{x, y\right \}$ , as desired.

This proves that two new elements (or bodies) are enough to bring everyone back to normal. Proving that they are the least number is a matter of trying only one new body.

This is known as Keeler's Theorem .

let's take a simple case where we switch a single pair of bodies (X,Y) suppose we have already switched X and Y , now we can't do that switch again so we need another body

let A be that new body , we switch (X,A) {Step 1} and switch (A,Y) {Step 2} , so now we know that Y has its mind back

and we know that A now has X's mind but we can't switch (X,A) again so we need another body B

now we switch (A,B) {Step 3} , B now has X's mind , now we switch (B,X) {Step 4} and X has its mind back

we can use that same process no matter X , no matter Y , which mean A and B are sufficient to do all the switches so that each pair get their minds back.

The solution is quite simple. Suppose we take another person A and make him exchange his mind with 1 of the 20 people . Now person A will exchange his acquired mind with a second person whose mind he took from A. Further he now gives the acquired mind to another person whose mind he took earlier and this cycle goes on till the second last person gets his mind back.(The first person still has the mind of person A) So now we need another person B so that it could first exchange his mind with A and then give the acquired mind to the first person