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Fittoniya [83]
3 years ago
10

A computer is performing a binary search on the sorted list of 7 numbers below.

Computers and Technology
2 answers:
suter [353]3 years ago
6 0

Answer:

The answer is "Option D".

Explanation:

Binary search is also an efficient mechanism for only a sorted item list. It separates the element into one half of the checklist repeatedly, till you have narrowed it down to the only single place.

In this search, the number of comparisons is much more efficient than in the series to use the binary search approach, the items also should be shown, that's why the last choice is correct.

zhenek [66]3 years ago
6 0

Answer:

B

Explanation:

Binary search will start in the middle of the list, find which side the value is at, then look at the middle of those values.  This is incredibly efficient for larger values (Although it needs to be sorted) and practical in many instances.  Let's say that the desired number is 1, the binary search would look at "50" first, then proceed to compare with "5" and finally on it's third iteration, pick 1.

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4 0
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Here are some instructions in English. Translate each of them into the Simple Machine language.
Evgen [1.6K]

Answer:

a. LOAD register 3 with the hex value 56

2356

b. ROTATE register 5 three bits to the right.

A503

c. JUMP to the instruction at location F3 if the contents of register 7 are equal to the contents of register 0

B7F3

d. AND the contents of register A with the contents of register 5 and leave the result in register 0

80A5

Explanation:

To translate the English instructions to Machine language we have to consider Appendix C of A Simple Machine Language sheet.

a. LOAD register 3 with the hex value 56

The OP code for LOAD is:

2

Operand is:

RXY

Description is:

LOAD the register R with the bit pattern XY.

In a.

3 represents R

56 represents XY

Now following this we can join OP and Operand together to translate a.

2356

b. ROTATE register 5 three bits to the right

The OP code for ROTATE is:

A

Operand is:

R0X

Description is:

ROTATE the bit pattern in register R one bit to the right X times. Each time place the  bit that started at the low-order end at the high-order end

In b.

5 represents R

three represents X

Now following this we can join OP and Operand together to translate b.

A503

c. JUMP to the instruction at location F3 if the contents of register 7 are equal to the contents of register 0.

The OP code for JUMP is:

B

Operand is:

RXY

Description is:

JUMP to the instruction located in the memory cell at address XY if the bit pattern in  register R is equal to the bit pattern in register number 0.

In c.

7 represents R register

F3 represents XY

Now following this we can join OP and Operand together to translate c.

B7F3

d. AND the contents of register A with the contents of register 5 and leave the result in register 0.

The OP code for AND is:

8

Operand is:

RST

Description is:

AND the bit patterns in registers S and T and place the result in register R

In d.

A represents register S

5 represents register T

0 represents register R

Now following this we can join OP and Operand together to translate d.

80A5

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Harlamova29_29 [7]
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Answer:

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Explanation:

I think so this is what you had asked!!!!!!

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