Computers and Technology, 19.07.2019 23:30 maddymaddy

Find an optimal parenthesization of a matrix-chain product whose sequence of dimensions is < 10,20,30,10,40,10,10>

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Answer from: Quest

It’s definitely true

Answer from: Quest

Same issue happened to me i had 4/5 and suddenly they went down to 3/5 .the reason must be that the question u recived a brainliest answer mark, was deleted by a moderator. they can delete a question for some reason if it isn't complete or has been reported. so there for that is most likely the reason.best of hope u earn another one

Answer from: Quest

Nskdndcodncodncldncownf8

Answer from: Quest

The answer is c. 36======analyzing the for-loop, we see that the control variable, k, starts at 0 and increments by after every loop iteration. the condition $\texttt{k\ \textless \ =7}$ tells us that the loop will stop when k is above 7; in this case, when k is 8.0 1 2 3 4 5 6 7 are the values for k that will be used. each number is for each iteration; therefore, the for-loop will have 8 iterations in total.the code inside the loop itself goes like this: 1st iteration (k=0)starting with sum=0, val=1 $\texttt{sum += val; }$ means that add $\texttt{val}$ to $\texttt{sum}$ .since sum is 0 and val is 1, 0 + 1 = 1. the new value of sum is 1. $\texttt{val++; }$ means to increment $\texttt{val}$ by 1.since val is 1, the new value of val of 2.end of iteration with k=0; k is increased by one to become k=12nd iteration (k=1)starting with sum=1, val=2 $\texttt{sum += val; } \iff \texttt{sum = 1 + 2}$ ======let us establish a pattern. this is moreso logic than programming: we can rewrite $\texttt{sum = 1 + 2}$ so that we have the elements from the previous iteration. the 1 came from 0 + 1 before. so we can write it as $\texttt{sum = 0 + 1 + 2}$ .and if we did the third iteration, by the end, sum would look like $\texttt{sum = 0 + 1 + 2 + 3}$ notice how the number of plus signs is equal to the iteration you are on (which is just k+1), and what is being added is increasing by one each time. for example, the third iteration (k=2) ends up with $\texttt{sum = 0 + 1 + 2 + 3}$ with 3 plus signs.the second iteration (k=1) ends up with $\texttt{sum = 0 + 1 + 2}$ with two plus signs.the first iteration (k=0) ends up with $\texttt{sum = 0 + 1}$ with one plus sign.this is a pattern. the nth iteration will have n plus signs.since we established before that there are 8 iterations of this loop in total, the value of sum at k=7 (the 8th and last iteration) will have sum with 8 plus signs. $\begin{aligned} \texttt{sum} & = \texttt{0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 } \\ & = \texttt{3 + 3 + 4 + 5 + 6 + 7 + 8 } \\ & = \texttt{6 + 4 + 5 + 6 + 7 + 8 } \\ & = \texttt{10 + 5 + 6 + 7 + 8 } \\ & = \texttt{15 + 6 + 7 + 8 } \\ & = \texttt{21 + 7 + 8 } \\ & = \texttt{28 + 8 } \\ & = \texttt{36} \end{aligned}$ so the value of sum at the end of the loop is 36.

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