homework hints

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Homework Hints

Algorithms

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Converting Positive Numbers

Base 2 to base 16 Base 16 to base 2 Base 16 to base 10 Base 2 to base 10 Base 10 to base 2 Base 10 to base 16

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Converting positive numbers: base 2 to base 16 Put the digits into groups of 4 starting at

the right If the last group has last than 4 digits,

extend it with leading 0s Convert each group of 4 according to

the following translation 0000 -> 0, 0001 -> 1, 0010 -> 2, 0011 ->3 0100 -> 4, 0101 -> 5, 0110 -> 6, 0111 -> 7 1000 -> 8, 1001 -> 9, 1010 -> A, 1011 -> B 1100 -> C, 1101 -> D, 1110 -> E, 1111 -> F

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Example:0110010010010012 to base 16

011 0010 0100 10013 2 4 93249

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Converting positive numbers: base 16 to base 2 Convert each digit into a group of 4

according to the following translation 0 -> 0000, 1 -> 0001, 2 -> 0010, 3 -> 0011 4 -> 0100, 5 -> 0101, 6 -> 0110, 7 -> 0111 8 -> 1000, 9 -> 1001, A -> 1010, B -> 1011 C -> 1100, D -> 1101, E -> 1110, F -> 1111

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Example:8EF16 to base 2

8 = 1000E = 1110F = 1111

100011101111

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Converting positive numbers: base 16 to base 10 Expand the number into its component values:

face-value x position value Position value

rightmost digit, position value = 160

Moving right to left, the exponent in each position increases by 1 (161, 162, 163, … )

Face value 0 - 9 are their face values A - F are 10 – 15 respectively

Convert each exponent to its decimal equivalent Multiply each face-value pair together Add the values together

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Example:CAB16 to base 10

C x 162 + A x 161 + B x 160 = 12 x 256 + 10 x 16 + 11 x 1 = 3072 + 160 + 11 = 3243

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Converting positive numbers: base 2 to base 10 Expand the number into its component values:

face-value x position value Position value

rightmost digit, position value = 20

Moving right to left, the exponent in each position increases by 1 (21, 22, 23, … )

Face value 0 = 0, 1 = 1

Eliminate all the 0 terms Simplify each 1x2n to 2n

Convert each exponent to its decimal equivalent

Add the values together

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Example:1100101010112 to base 10

1x211 + 1x210 + 0x29 + 0x28 + 1x27 + 0x26 + 1x25 + 0x24 + 1x23 + 0x22 + 1x21 + 1x20 =

1x211 + 1x210 + 1x27 + 1x25 + 1x23 + 1x21 + 1x20 =

211 + 210 + 27 + 25 + 23 + 21 + 20 = 2048 + 1024 + 128 + 32 + 8 + 2 + 1

= 3243

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Converting positive numbers: base 10 to base 2

1. Divide the number by 22. Use the remainder as the next

number (beginning at the right)3. If the result is 0 then stop4. Use the result as the next

number and go to step 1

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327 / 2 = 163 r 1 answer 1163 / 2 = 81 r 1 answer 1181 /2 = 40 r 1 answer 11140 / 2 = 20 r 0 answer 011120 / 2 = 10 r 0 answer 0011110 / 2 = 5 r 0 answer 0001115 / 2 = 2 r 1 answer 10001112 / 2 = 1 r 0 answer 010001111 / 2 = 0 r 1 answer 101000111

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Converting positive numbers: base 10 to base 16

1. Divide the number by 162. Use the remainder as the next number

(beginning at the right)1. If the remainder is less than 10, use the

number directly, else convert it to a letter using 10->A, 11->B, 12->C, 13->D, 14->E, 15->F

3. If the result is 0 then stop4. Use the result as the next number and

go to step 1

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327 / 16 = 20 r 7 answer 716

20 / 16 = 1 r 4 answer 4716

1 / 16 = 0 r 1 answer 14716

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Two’s Complement Convert a positive base 10 number

to binary two’s complement Convert a negative base 10 number

to binary two’s complement Convert a positive base 10 number

to hexadecimal two’s complement Convert a negative base 10 number

to hexadecimal two’s complement

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Convert a positive base 10 number to binary two’s complement

Convert the number to base 2 as described on slide 11

Fill in all leading spaces with 0’s

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Example: 52 to binary two’s complement

52/2 = 26 r 1 126/2 = 13 r 0 0113/2 = 6 r 1 101 6/2 = 3 r 0 0101 3/2 = 1 r 1 10101 1/2 = 0 r 1 110101Extend to 16-bit field: 0000000000110101

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Convert a negative base 10 number to binary two’s complement

Convert the positive value of number to binary as described on slide 11

Fill in all leading spaces with 0’s Flip all digits: 0 -> 1, 1 -> 0 Add binary 1 to the number,

remembering the following rules: 0+0=0, 0+1=1, 1+0=1, 1+1=10

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Example: -52 to binary two’s complement52/2 = 26 r 1 126/2 = 13 r 0 0113/2 = 6 r 1 101 6/2 = 3 r 0 0101 3/2 = 1 r 1 10101 1/2 = 0 r 1 110101

extend to 16 bits 0000000000110101

Flip the bits and add 1 1111111111001010+ 1 1111111111001011

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Convert a positive base 10 number to hexadecimal two’s complement Convert the number to base 2 as described on

slide 11 Fill in all leading spaces with 0’s Put the digits into groups of 4 starting at the

right Convert each group of 4 according to the

following translation 0000 -> 0, 0001 -> 1, 0010 -> 2, 0011 ->3 0100 -> 4, 0101 -> 5, 0110 -> 6, 0111 -> 7 1000 -> 8, 1001 -> 9, 1010 -> A, 1011 -> B 1100 -> C, 1101 -> D, 1110 -> E, 1111 -> F

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Example: 52 to hexadecimal two’s complement52/2 = 26 r 1 126/2 = 13 r 0 0113/2 = 6 r 1 101 6/2 = 3 r 0 0101 3/2 = 1 r 1 10101 1/2 = 0 r 1 1101010000000000110101

0000 0000 0011 0101003516

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Convert a negative base 10 number to binary two’s complement Convert the positive value of number to binary as

described on slide 11 Fill in all leading spaces with 0’s Flip all digits: 0 -> 1, 1 -> 0 Add binary 1 to the number, remembering the following

rules: 0+0=0, 0+1=1, 1+0=1, 1+1=10

Put the digits into groups of 4 starting at the right Convert each group of 4 according to the following

translation 0000 -> 0, 0001 -> 1, 0010 -> 2, 0011 ->3 0100 -> 4, 0101 -> 5, 0110 -> 6, 0111 -> 7 1000 -> 8, 1001 -> 9, 1010 -> A, 1011 -> B 1100 -> C, 1101 -> D, 1110 -> E, 1111 -> F

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Example: -52 to binary two’s complement52/2 = 26 r 1 126/2 = 13 r 0 0113/2 = 6 r 1 101 6/2 = 3 r 0 0101 3/2 = 1 r 1 10101 1/2 = 0 r 1 110101

0000000000110101

1111111111001010+ 1 1111111111001011

1111 1111 1100 1011FFCB16

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Two’s Complement Operations

Binary addition Binary subtraction Binary comparison Hexadecimal addition Hexadecimal subtraction Hexadecimal comparison

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Binary addition

Add the two numbers If the carry into the leftmost digit is

not the same as the carry out, define it as an overflow

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Example

00000000101110102

+00000000101010112

00000001011001012

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Binary subtraction Convert the second number into it’s

negative by Flipping each bit (0->1, 1->0) Add binary 1 to the number, remembering

the following rules: 0+0=0, 0+1=1, 1+0=1, 1+1=10

Add the two numbers If the carry into the leftmost digit is not

the same as the carry out, define it as an overflow and stop

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Example 00000000101110102

- 00000000101010112

negate 00000000101010112 -> 1111111101010100+ 1 1111111101010101

00000000101110102

+11111111010101012

00000000000011112

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Binary comparison

Subtract the two numbers as described in slide 27

Comparison If result is positive (first bit 0), first

number is greater If result is negative (first bit 0 and not

all others are 0), first number is smaller If result is zero, numbers are equal

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Example

00000000101110102

- 00000000101010112

00000000000011112

First number is greater than the

second

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Hexadecimal addition Convert both numbers into binary as described

in slide 5 Add the two numbers If the carry into the leftmost digit is not the

same as the carry out, define it as an overflow and stop

Put the digits into groups of 4 Convert each group of 4 according to the

following translation 0000 -> 0, 0001 -> 1, 0010 -> 2, 0011 ->3 0100 -> 4, 0101 -> 5, 0110 -> 6, 0111 -> 7 1000 -> 8, 1001 -> 9, 1010 -> A, 1011 -> B 1100 -> C, 1101 -> D, 1110 -> E, 1111 -> F

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Example: BA16 + AB16

BA16 AB16

1011 1010 1010 10110000000010111010 0000000010101011

0000000010111010+0000000010101011 0000000101100101

0000 0001 0110 0101016516

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Alternative algorithm:hexadecimal addition table

0 1 2 3 4 5 6 7 8 9 A B C D E F

0 0 1 2 3 4 5 6 7 8 9 A B C D E F

1 1 2 3 4 5 6 7 8 9 A B C D E F 10

2 2 3 4 5 6 7 8 9 A B C D E F 10 11

3 3 4 5 6 7 8 9 A B C D E F 10 11 12

4 4 5 6 7 8 9 A B C D E F 10 11 12 13

5 5 6 7 8 9 A B C D E F 10 11 12 13 14

6 6 7 8 9 A B C D E F 10 11 12 13 14 15

7 7 8 9 A B C D E F 10 11 12 13 14 15 16

8 8 9 A B C D E F 10 11 12 13 14 15 16 17

9 9 A B C D E F 10 11 12 13 14 15 16 17 18

A A B C D E F 10 11 12 13 14 15 16 17 18 19

B B C D E F 10 11 12 13 14 15 16 17 18 19 1A

C C D E F 10 11 12 13 14 15 16 17 18 19 1A 1B

D D E F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C

E E F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D

F F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E

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Example: BA16 + AB16

(1) BA16

+AB16

------------

516

BA16

+AB16

------------

16516

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Hexadecimal subtraction Convert both numbers into binary as described

in slide 5 Convert the second number into it’s negative

by Flipping each bit (0->1, 1->0) Add binary 1 to the number, remembering the

following rules: 0+0=0, 0+1=1, 1+0=1, 1+1=10

Add the two numbers If the carry into the leftmost digit is not the

same as the carry out, define it as an overflow

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Example: BA16 - AB16 BA16 AB16

1011 1010 1010 10110000000010111010 0000000010101011

00000000101110102

- 00000000101010112

negate 00000000101010112 -> 1111111101010100+ 1 1111111101010101

00000000101110102

+11111111010101012

00000000000011112

0000 0000 0000 11112

000F16

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Alternative algorithm:hexadecimal subtraction table borrow not needed borrow needed

0 1 2 3 4 5 6 7 8 9 A B C D E F

0 0 1 2 3 4 5 6 7 8 9 A B C D E F

1 F 0 1 2 3 4 5 6 7 8 9 A B C D E

2 E F 0 1 2 3 4 5 6 7 8 9 A B C D

3 D E F 0 1 2 3 4 5 6 7 8 9 A B C

4 C D E F 0 1 2 3 4 5 6 7 8 9 A B

5 B C D E F 0 1 2 3 4 5 6 7 8 9 A

6 A B C D E F 0 1 2 3 4 5 6 7 8 9

7 9 A B C D E F 0 1 2 3 4 5 6 7 8

8 8 9 A B C D E F 0 1 2 3 4 5 6 7

9 7 8 9 A B C D E F 0 1 2 3 4 5 6

A 6 7 8 9 A B C D E F 0 1 2 3 4 5

B 5 6 7 8 9 A B C D E F 0 1 2 3 4

C 4 5 6 7 8 9 A B C D E F 0 1 2 3

D 3 4 5 6 7 8 9 A B C D E F 0 1 2

E 2 3 4 5 6 7 8 9 A B C D E F 0 1

F 1 2 3 4 5 6 7 8 9 A B C D E F 0

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Example: BA16 - AB16

A B 1A16

- A B16

------------

F16

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