Exam 3: Algorithms

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For each string of length $n$ of the letters $\mathbf { T }$ and $\mathbf { F }$ (representing true and false), evaluate the given compound proposition when the $i ^ { \text {th } }$ variable is assigned the truth value given by the $i ^ { \text {th } }$ letter in the string. If any of these values of $P$ is $\mathbf { T }$ , then $P$ is satisfiable; otherwise it is not. This will take at least $2 ^ { n }$ steps in the worst case (the case in which $P$ is not satisfiable), once for each such string. If $\mathrm { P } = \mathrm { NP }$ , then there is a polynomial worst-case time algorithm, so the brute force algorithm is not the most efficient.

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To multiply $A$ by $B$ , we will need 3.4 $5 = 60$ multiplications. The result is a $3 \times 5$ matrix. To multiply it by $C$ will require $3 \cdot 5 \cdot 6 = 90$ multiplications. This gives a total of $60 + 90 = 150$ steps. On the other hand, if we multiply $\boldsymbol { B }$ by $\boldsymbol { C }$ first, we use $4 \cdot 5 \cdot 6 = 120$ multiplications and then another $3 \cdot 4 \cdot 6 = 72$ to multiply $\boldsymbol { A }$ by the $4 \times 6$ matrix $B C$ . This method uses a total of $120 + 72 = 192$ steps. Therefore the first method is a little faster.

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(a) We first compare the first and second integers in the sequence $a _ { 1 } , a _ { 2 } , \ldots , a _ { n }$ , setting the value of the variable firstmax equal to the larger, and the value of the variable secondmax equal to the smaller. For each successive integer $a _ { i }$ in the sequence, $i = 3,4 , \ldots , n$ , we first compare it to firstmax. If $a _ { i } >$ firstmax, then we make the assignments secondmax $: =$ firstmax and firstmax $: = a _ { i } .$ Otherwise, we compare $a _ { i }$ to secondmax, and if $a _ { i } > \operatorname { secondmax }$ , then we make the assignment secondmax := $a _ { i }$ . At the end of this procedure the value of secondmax will be the second largest integer in the sequence.
(b) We do one comparison at the beginning of the algorithm to determine whether $a _ { 1 }$ or $a _ { 2 }$ is larger. Then for each successive term, for $i = 3,4 , \ldots , n$ , we carry out at most two comparisons. Hence the largest number of comparisons used is $2 ( n - 2 ) + 1 = 2 n - 3$ , ignoring bookkeeping. This is $O ( n )$ .