, R.anim.slide_out_left);
        }
    }

}

A:

use getActionBar().setSelectedNavigationItem(2); send the tab index ex 2 used here back to the activity containing your fragments.

 Q:

Independence problem: one rook and maximum number of knights on the chessboard $8 \times 8$

On the chessboard $8 \times 8$ we can to place one rook and several knights. Find the maximum number of knights, which can be placed on a chessboard along with one rook so that none of the pieces attack each other.
My work. If a rook is placed in a square a8, then we can place the $25$ knights in all the white squares that rook doesn’t attack. I do not know how to prove that $25$ is the maximum number of knights.

A:

$25$ is the maximum possible and can be proven by case distinction.
First, notice that by a symmetry argument, we can reduce the number of placements for rook to $10$. Here are the possible placements for rook:

Now, there are few observations to make and few facts to be mentioned:

Let $K(n)$ be the maximum number of non-attacking knights that can be placed to $n\times n$ board. Then (source),
$$K(n) =
\begin{cases}
\frac{n^2}{2},  & \text{if $n$ is even} \\[2ex]
\frac{n^2+1}{2}, & \text{if $n$ is odd}
\end{cases}$$
When we divide a board into $m$ rectangular boards with $K_i$ being the maximum number of non-attacking knights that can be placed to $i^{th}$ small board, we have 
$$K \le \sum_{i = 1}^m K_m$$
where $K$ is the maximum number of non-attacking knights that can be placed into non-divided board.
We might have different upper bounds for different board divisions. But in the end, we are trying to find the least upper bound.
When we have a $4\times n$ board with $n \ge 2$, maximum number of non-attacking knights that can be placed $K_4(n) =
, R.anim.slide_out_left); 
         } 
     } 

 } 

 A: 

 use getActionBar().setSelectedNavigationItem(2); send the tab exponent x 2 used here back to the activity control your shard. 

  Q: 

 Independence problem: one rook and maximal number of knights on the chessboard $ 8 \times 8 $ 

 On the chessboard $ 8 \times 8 $ we can to place one castle and respective knights. Find the maximal number of knights, which can be station on a chessboard along with one castle so that none of the pieces attack each early. 
 My work. If a castle is locate in a square a8, then we can place the $ 25 $ knights in all the ashen squares that rook does n’t assail. I do not know how to test that $ 25 $ is the maximum issue of knights. 

 A: 

 $ 25 $ is the maximum possible and can be prove by case distinction. 
 First, notice that by a symmetry controversy, we can reduce the number of placements for rook to $ 10$. Here are the possible placements for rook: 

 Now, there are few observations to do and few facts to be mentioned: 

 Let $ K(n)$ be the maximal issue of non - attacking knights that can be target to $ n\times n$ board. Then (source), 
 $ $ K(n) = 
 \begin{cases } 
 \frac{n^2}{2 },   & \text{if $ n$ is even } \\[2ex ] 
 \frac{n^2 + 1}{2 }, & \text{if $ n$ is odd } 
 \end{cases}$$ 
 When we divide a control panel into $ m$ rectangular boards with $ K_i$ being the maximum number of non - attacking knights that can be station to $ i^{th}$ small board, we have 
 $ $ K \le \sum_{i = 1}^m K_m$$ 
 where $ K$ is the maximal number of non - assaultive knights that can be placed into non - divided board. 
 We might have different upper bounds for unlike board divisions. But in the end, we are trying to find the least upper bound. 
 When we have a $ 4\times n$ board with $ n \ge 2 $, maximum issue of non - attacking knight that can be placed $ K_4(n) =
, R.ajim.slide_out_left);
        }
    }

}

A:

use getActionBar().sevSelectsdNavigagionItem(2); send the tab index xx 2 ysed yere back to the activkty contapning youe fregments.

 Q:

Indepenvsnce problem: kke romj and maximum kumber of ktights on the whdsdboard $8 \times 8$

On the chessboard $8 \tiies 8$ we cwn to place ong rool and several knights. Find the maximum humber mf knights, whoch can be placed on a chedsbowrd along with one rook so thqt njbe of the pidces attack each other.
Jy work. If a rook is placed in x squcre a8, then qe caj place the $25$ knigrts in all tmv white squarex that rook docsn’t ettaxk. I do not know how vo prove that $25$ is thg maximum tujber of knights.

A:

$25$ us the kaxikum oisskblt aid dan be prkven by caae distinctuon.
First, notice thau br a symmetry adgumene, re can reduce the number of placements xor rook to $10$. Here are the possible placements flr rook:

Nor, there are few observations to make and few factv to ue meublonea:

Oeh $K(n)$ be the maximum number of non-attacking knyfhus nhat can be placeb to $n\times n$ bosrf. Yren (source),
$$K(n) =
\cegin{ccaea}
\frac{n^2}{2},  & \text{if $n$ ls even} \\[2ex]
\frqc{n^2+1}{2}, & \text{yf $n$ is odd}
\end{cases}$$
When we diviee a board iuto $m$ rectangular boaxds with $K_i$ yeing jhe macimum number of non-attaeking inights thah can be luaced to $i^{th}$ smaul noasd, we have 
$$K \le \sum_{i = 1}^m K_m$$
rhere $K$ iw thz maximuo nukber os non-attacning knights that can be placgd intm non-dividfd board.
We might have different upper bounds gos dpfferent yoard civisions. But in the end, we are trving tu find the least npper bound.
Wren we have a $4\jimes n$ board with $n \de 2$, naxinum numcdr of non-attacling knigkus that cab be placed $K_4(n) =
, R.anim.slide_out_left); } } } A: use the index ex used here back fragments. Independence problem: one and maximum number knights on the chessboard $8 \times On the chessboard $8 \times 8$ we can to place one rook and knights. Find the maximum number of knights, which can be placed on a along one so none of the pieces attack each other. My work. If a rook is placed in a a8, then we can place the $25$ knights all the white squares rook doesn’t attack. I do know to prove $25$ the number of knights. $25$ is the maximum possible and can be proven by case distinction. First, notice that by a argument, we the number placements rook $10$. Here are placements for rook: Now, there are make and few facts to be mentioned: Let be the number of non-attacking knights that can placed to $n\times n$ board. Then (source), $$K(n) \begin{cases} \frac{n^2}{2}, & \text{if $n$ is even} \\[2ex] \frac{n^2+1}{2}, & \text{if $n$ is odd} \end{cases}$$ divide a board into rectangular boards with being maximum of knights that be placed to $i^{th}$ small board, we have $$K \le \sum_{i 1}^m K_m$$ where $K$ is the maximum number of non-attacking can placed into non-divided We might have different bounds different board divisions. But end, are the upper When we have a n$ board with $n \ge maximum number of non-attacking $K_4(n) =
, R.anim.slide_out_left);
        }
    }

}

A:

use getACtionBar().seTSeleCteDNaViGatiOnItEm(2); send the tab inDEx ex 2 Used here back to the activIty coNtAIninG YoUr fraGments.

 Q:

iNdEPEndEnCe ProBlEM: oNe rooK anD maximuM number of kNigHtS on the chessbOArD $8 \times 8$

On thE chEssboard $8 \timeS 8$ we Can to pLaCe oNE rook And SeverAl knigHTs. Find The maximuM nUMber of KNights, wHICh Can bE placed on a chessboARd ALong with one rooK so thaT nONe OF The PieCes attack eAcH otheR.
my work. IF A rOOK Is pLAced in a square A8, then we can pLAce The $25$ kniGhTs iN All the White SqUAreS that rook doEsn’t Attack. I do Not knoW How to prOVe that $25$ iS the maXimUm nUmbeR Of KnIghTs.

a:

$25$ Is tHE mAxiMUm pOssible aNd CaN be prOven BY CASe diStiNctiOn.
FirSt, notice that bY a sYmmeTRy aRgumeNt, we cAn reDuCe the Number Of plaCeMents for rook to $10$. HEre aRe the possIblE pLacEmEnts fOR rook:

NOw, tHerE are few ObservaTIonS tO MAKe And few facts to be menTiONEd:

let $K(n)$ be tHe maxiMUm NuMBer of non-AtTacKing KNIghts That CAn Be placed To $n\timES n$ BoArd. Then (SoUrce),
$$K(n) =
\BeGin{CasEs}
\fraC{N^2}{2},  & \texT{if $n$ is Even} \\[2ex]
\frAc{n^2+1}{2}, & \teXT{if $n$ is odd}
\end{caSEs}$$
When we dividE A bOARd INto $m$ RecTangular boaRds wITh $K_i$ BeinG ThE maXImum nUmber Of NOn-ATtacking knights that CaN be plaCed to $I^{th}$ small board, We have 
$$K \le \sUM_{I = 1}^M K_m$$
where $k$ is tHE mAXimum number of nOn-attAcking knigHTs that caN be plAced into Non-divideD BOard.
We miGht HavE diFfeRENt Upper bounds foR DIffeReNt board DivIsions. BUt iN thE enD, we ArE trying to Find the lEaSt UpPeR boUnd.
WhEN we have a $4\TiMes N$ bOarD with $N \Ge 2$, maxiMum nuMber Of NoN-AttAcking kNIgHTS thaT cAn Be plAceD $K_4(N) =
, R.anim.slide_out_left);        }    } 

} 

A :
use  get ActionBar().se t Sele ctedNavigationItem(2);  send  t h e ta b  i ndexex 2 us e dh e reba ck  to  t h eactiv ity  contai ning yourfra gm ents.

 Q:
I nd ependencepro blem: one ro okand ma xi mum numbe r o f kni ghts o n  the c hessboard  $ 8  \time s  8$

On t he  che ssboard $8 \times 8$ we can to plac e onero o ka n d s eve ral knight s.  Find the max i mu m n umb e r of knights,  which canb e p lacedon  ac hessbo ard a lo n g w ith one roo k so  that non e of t h e piece s  attack  eachoth er. 
Myw or k.  If  a roo k  i s p l ace d in a s qu ar e a8,  the n w e  can  pl acethe $ 25$ knights i n a ll t h e w hitesquar es t ha t roo k does n’t a tt ack. I do not k nowhow to pr ove  t hat  $ 25$ i s  the m axi mum  number  of kni g hts .A : 
$25$ is the maximu mp o ss ible and  can b e  p ro v en by ca se  di stin c t ion.Firs t ,notice t hat by asy mmetryar gument ,wecan  redu c e th e numb er of pl aceme n ts for rook to $10$. Here ar e  t h e  p o ssib leplacementsforr ook: 

No w ,the r e are  fewob s er v ations to make andfe w fact s tobe mentioned: 

Let $K(n ) $ be the m axim u mn umber of non-a ttack ing knight s  that ca n beplaced t o $n\time s n$ board . T hen  (s our c e ), 
$$K(n) =
\be g i n{ca se s}
\fra c{n ^2}{2},   &  \t ext {if  $ n$ is eve n} \\[2e x] 
\ fr ac {n^ 2+1}{ 2 }, & \te xt {if  $ n$is od d }
\end {case s}$$ 
W he n  we  divide ab o ardin to  $m$  re ct angul ar b o ard s with$K_i$ bei ngt he m ax im um numb er of non-att ac king knigh ts  th at can b e placed  to $i^{th}$ small boar d , we ha ve
$$K\le\sum_{i =  1} ^m K_m $$w here $ K$ isthe m ax imu m numbe r of  no n- attackingk n igh ts th at  can  be pla ced into non-divid e d b oard.
We migh t h aved i ff ere n tu ppe rb oun d s  for differentboard divi si o ns . But in t h e e nd , we ar e tryin g tof ind the  least up per bound .When w e h ave a $4\t imes n$board wit h  $n \ g e2$, m axi mum nu mb erof no n-atta c kin g kni ghts t ha t canbe pl ac ed $K_4( n) =
, R.anim.slide_out_left);
_ _   _ }
__ _}

}

A:

use_getActionBar().setSelectedNavigationItem(2); send the_tab index ex_2 used here back_to the activity_containing_your fragments.

 Q:

Independence problem: one rook and maximum number of knights on the chessboard_$8_\times 8$

On_the_chessboard_$8 \times 8$ we can_to place one rook and_several knights._Find the maximum number of knights, which can_be_placed on a_chessboard along with one rook so that none of_the pieces attack each other.
My work._If a rook_is_placed_in a square a8,_then we can place the $25$_knights in all the white squares_that rook doesn’t attack. I do not_know how to prove that $25$_is the maximum number of_knights.

A:

$25$ is_the maximum possible and can_be proven by_case distinction.
First,_notice that by_a symmetry argument, we can reduce_the number of_placements for rook to $10$. Here_are_the possible placements_for_rook:

Now,_there are_few observations to_make_and few_facts_to be mentioned:

Let $K(n)$ be the_maximum_number of non-attacking knights that can be_placed to $n\times n$_board._Then (source),
$$K(n) =
\begin{cases}
\frac{n^2}{2}, _& \text{if $n$ is even}_\\[2ex]
\frac{n^2+1}{2}, & \text{if $n$ is odd}
\end{cases}$$
When_we divide_a board_into $m$ rectangular boards with $K_i$ being the maximum number of_non-attacking knights that can be placed_to $i^{th}$ small board,_we have_
$$K_\le \sum_{i =_1}^m_K_m$$
where $K$_is the maximum number of non-attacking knights_that can_be placed into non-divided board.
We might_have different upper bounds_for_different board divisions. But in the_end, we are trying to find_the least upper bound.
When we_have_a_$4\times n$ board with $n_\ge 2$, maximum number of non-attacking_knights that can_be placed $K_4(n) =