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3 years ago

In a bubble sort for a list of length n >= 2, in each iteration, the first step is to compare_______.

Computers and Technology

1 answer:

Hunter-Best [27]3 years ago

7 0

Answer:

the consecutive two elements and swap them based on the value which is larger if we go for ascending order

Explanation:

bubble sort compares 2 consecutive elements of the list and swap them. in one iteration one value will be placed correctly depending on the sorting order either ascending or descending

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3 years ago

Question No. 5:

Mkey [24]

Answer:

The answers are:

$a) (1001111)_2 = (79 )_{10}\\b) (11000001)_2 = (C1 )_{16}\\c) (E16)_{16} = ( 3606)_{10}\\d) (56)_{10} = (38 )_{16}\\e) (63)_{10}= (111111 )_2$

Explanation:

a. (1001111) 2 = ( ) 10

In order to convert a base 2 number to base 10 number system the place values are used.

the procedure is as follows:

$(1001111)_2\\=(1*2^6)+(0*2^5)+(0*2^4)+(1*2^3)+(1*2^2)+(1*2^1)+(1*2^0)\\=(1*64)+(0*32)+(0*16)+(1*8)+(1*4)+(1*2)+(1*1)\\=64+0+0+8+4+2+1\\=79$

b) (11000001) 2 = ( ) 16

In order to convert a base 2 number into base 16, group of 4-bits are made starting from right to left

The groups from the given number are:

1100 0001

Then the groups are changed in to decimal/hexa

So,

$(1100)_2 = (1*2^3)+(1*2^2)+(0*2^1)+(0*2^0)\\=(1*8)+(1*4)+(0*2)+(0*1)\\=8+4+0+0=12=C\\\\0001=1$

Writing in the same order in which groups were:

$(C1)_{16}$

c) (E16) 16 = ( ) 10

$(E16)_{16}\\=(E*16^2)+(1*16^1)+(6*16^0)\\=(E*256)+(1*16)+(6*1)\\=3584+16+6\\=3606$

d) (56) 10 = ( ) 16

Dividing by 16 and noting remainders

16 56

3 - 8

So,

The number in base 16 is 38

e) (63) 10 = ( ) 2

Dividing by 2

2 63

2 31 - 1

2 15 - 1

2 7 - 1

2 3 - 1

1 - 1

So the number after converting in base 2 is:

111111

Hence,

The answers are:

$a) (1001111)_2 = (79 )_{10}\\b) (11000001)_2 = (C1 )_{16}\\c) (E16)_{16} = ( 3606)_{10}\\d) (56)_{10} = (38 )_{16}\\e) (63)_{10}= (111111 )_2$

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