10 small balls with a balance to find the one with different weights.

Last week was an algorithm topic, the topic is: There are 10 small balls, the appearance of the same, of which 1 balls and other 9 ball weight is different, please use the balance to find the minimum number of times the weight of the ball.

Problem-solving ideas: In fact, this is a very typical use of the partition algorithm example, because do not know this particular ball is heavier than other balls or light, so we can not simply use the binary method to solve, so I choose 3 for the mold.

Problem Solving steps:

1. Divide 10 balls into 4 arrays, namely A[A1, A2, A3],b[b1, B2, B3], c[c1, C2, C3, D[d1]; Suppose that the weight of the ball is N, and any other ordinary ball is m;

2, first take A and B, a and C said, if A==b && a== C, then directly get the result n = d1; otherwise we can get two results, the first n in [a| b| C], the second one is n is heavier or lighter than M.

Description

If a > B && a > C, then N in A,n > m

If a > B && a = = C, then n in b,n < m

If a > B && a < C, this situation does not exist

if A = = B && A > C, then N in C, n < m

if A = = B && A = = C, this situation we have obtained the result

if A = = B && A < C, then n in C, n > M

If a < B && a < C, then N in a,n < m

If a < B && a = = C, then n in B,n > m

If a < B && a > C, this situation does not exist

3. We get n in [a| b| C] After taking the first two balls in a, B, C, it is known that n is heavier or lighter than m, so you can get n directly after the call. Suppose N in A

Description

If a1 > A2 && n > M, then n = A1

If a1 > A2 && N < m, then n = A2

If a1 = = A2, then n = A3

If A1 < A2 && n > M, then n = A2

If A1 < A2 && n < m, then n = A1

On the code:

<?PHP$data=Array(0, 0, 0, 0, 1, 0, 0, 0, 0, 0);classFilterball { Public $theSpecial;//Special Ball 1 = heavy,-1 = Light     Public $balanceTimes= 0; /** param integer & array $left left ball * param integer & array $right right ball * return left ball weight =-1 same weight = 0 right Ball weight =1*/     Public functionBalance$left,$right) {        $this->balancetimes++; if(Is_array($left)) {            $left=Array_sum($left); }        if(Is_array($right)) {            $right=Array_sum($right); }        if($left==$right) {            return0; } ElseIf($left>$right) {            return-1; } Else {            return1; }    }    functionExecute$data) {        $mod=intval(Count($data)/3);//Modulo //divide the array into 3-4 arrays of each $mod element        $i= 0; $j= 0; $newArray=Array();  while($i<Count($data)) {            if(isset($newArray[$j]) &&Count($newArray[$j]) >=$mod) {                $j++; }            $newArray[$j][] =$data[$i]; $i++; }        $return 0=$this->balance ($newArray[0],$newArray[1]); $return 1=$this->balance ($newArray[0],$newArray[2]); Switch((string)$return 0. (string)$return 1) {             Case' 11 ':$this->thespecial =-1; return0 +$this->comparethird ($newArray[0]);  Case' 10 ':$this->thespecial = 1; return3 +$this->comparethird ($newArray[1]);  Case' 01 ':$this->thespecial = 1; return6 +$this->comparethird ($newArray[2]);  Case' 00 ':return9;  Case' 0-1 ':$this->thespecial =-1; return6 +$this->comparethird ($newArray[2]);  Case'-10 ':$this->thespecial =-1; return3 +$this->comparethird ($newArray[1]);  Case'-1-1 ':$this->thespecial = 1; return0 +$this->comparethird ($newArray[0]); }    }    /** Through the first two pairs, know whether the special ball is heavy or light, and then in three balls in a pair, know three balls in the index of the special ball*/     Public functionComparethird ($childArray) {        $return=$this->balance ($childArray[0],$childArray[1]); Switch($return) {             Case1: Case-1:if($return==$this-thespecial) {                    return1; } Else {                    return0; }                 Break;  Case0:return2;  Break; }    }}Echo' <pre> ';Print_r($data);$filter=NewFilterball ();Echo sprintf(' Key value:%s,%s times ',$filter->execute ($data),$filter->balancetimes);

10 small balls with a balance to find the one with different weights.