Answer : $n$ First i will prove that regardless of the initial alignment of the cows Tim is able to fulfill his wish in at most $n$ moves. Induct on $n$. For $n=2 $ just take all initial alignments: $WWPP\to PPWW$,$WPWP \to WWPP\to PPWW$,$WPPW\to PWPW\to PPWW$, etc Suppose that it's true for $n=1,2,..,k-1$ and the alignment is $a_1a_2...a_{2n}$ If $a_1=P,a_2n=W$ then by induction hypothesis he can transform $a_2..a_{2n-1}$ to $\underset{n-1}{\underbrace{PP..P}}\underset{n-1}{\underbrace{WW..W}}$ using at most $n-1<n $ moves etc If $a_1=P,a_2n=P$ then let $k$ be the largest index such that $a_k=W$, since $a_1=P$ we have $k\geq n+1$ and so Tim can make the move $P... \underline{a_{2k-2n+1}..a_{k-1}W}\overline{\underset{ 2n-k}{\underbrace{PP..P}}}\to P..\underset{2n-k}{\underbrace{PP..P}} {a_{2k-2n+1}..a_{k-1}W}$ and then $n-1$ moves left and go to the previous case. If $a_1=W,a_2n=P$ then by induction hypothesis Tim can transform $a_2..a_{2n-1}$ to $\underset{n-1}{\underbrace{WW..W}}\underset{n-1}{\underbrace{PP..P}}$ using at most $n-1$ moves and now he has one more move for $\underset{n}{\underbrace{WW..W}}\underset{n}{\underbrace{PP..P}}\longrightarrow \underset{n}{\underbrace{PP..P}} \underset{n}{\underbrace{WW..W}}$ If $a_1=W,a_2n=W$ then use the same method as in case 2 to make the first cow purple ( 1 move) and the use the induction hypothesis. Now in order to finish the proof it's enough to show that there is an initial alignment that can't be transormed into $ \underset{n}{\underbrace{PP..P}} \underset{n}{\underbrace{WW..W}}$ using less than $n$ moves. I claim that $WPWP..WP$ works. First condider any alignment $S=a_1a_2..a_{2n}$ and define the number $m(S)=|a_2-a_1|+|a_3-a_2|+...+|a_{2n}-a_{2n-1}|$ where $a_i=1$ if $a_i$ represents a purple cow and $a_i=0$ otherwise. Observe that $m(WPWP..WP)=2n-1$ and $m(\underset{n}{\underbrace{PP..P}} \underset{n}{\underbrace{WW..W}})=1$ Now consider a sequence of moves $S_0\to S_1\to S_2\to....$ I claim that $m(S_{i+1})\geq m(S_i)-2$ Let $S_i=a_1a_2..c_0\underline{c_1c_2..c_t}\overline{b_1b_2..b_t}b_0..\to a_1a_2..c_0{b_1b_2..b_t}{c_1c_2..c_t}b_0...=S_{i+1}$ so only the differences $|c_1-c_0|,|b_1-c_t|,|b_0-b_t|$ are possible change in $m(S_i)$ to $m(S_i+1)$ and thus $m(S_i+1)\geq m(S_1)-3$. Suppose that $m(S_i+1)= m(S_1)-3$, then $|c_1-c_0|=1,|b_1-c_t|=1,|b_0-b_t|=1$ and $|c_0-b_1|=0,|c_1-b_t|=0,|b_0-c_t|=0$ so $c_1\neq c_0=b_1\neq c_t\Rightarrow c_1=c_t$ and $b_0 \neq b_t=c_1=c_t\Rightarrow b_0\neq c_t$ which is impossible since $|b_0-c_t|=0$ So $m(S_i+1)\geq m(S_1)-2$ and since $m(WPWP..WP)-m(PP..PWW..W)=2n-2$ we need at least $(2n-2)/2=n-1$ moves. Suppose that we can do it in $n-1$ moves, then in every move $m(S)$ must be decreased by exactly $2$. Initialy the first cow was White so since at the end is purple there is a move after which the first cow is always white, consider the first such move : ${Wa_2a_3..a_t}{Pb_2b_3..b_t}b_0...\to {Pb_2b_3..b_t}{Wa_2a_3..a_t}b_0...$ and we must have $m({Pb_2b_3..b_t}{Wa_2a_3..a_t}b_0...)=m({Wa_2a_3..a_t}{Pb_2b_3..b_t}b_0...)-2$ so $a_t=0,b_t\neq b_0$ and $b_t=0,a_t=b_0$ but then $a_t=b_0\neq b_t=0=a_t$ a condtradiction. So we need at least $n$ moves and the proof is complete.