10 subtraction

7/21/2019 10 Subtraction

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Subtraction (lvk) 1

Negative Numbers and Subtraction

The adders we designed can add only non-negative numbers If we can reresent negative numbers! then subtraction is "#ust$

the ability to add two numbers (one of which may be negative)% &e'll look at three different ways of reresenting signed numbers% ow can we decide reresentation is better

The best one should result in the simlest and fastest oerations% This is #ust like choosing a data structure in rogramming%

&e're mostly concerned with two articular oerations* +egating a signed number! or converting , into -,% dding two signed numbers! or comuting , . y% So! we will comare the reresentation on how fast (and how easily)

these oerations can be done on them


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Subtraction (lvk) /

Signed magnitude representation

umans use a signed-magnitudesystem* we add .or -in front of amagnitude to indicate the sign%

&e could do this in binary as well! by adding an e,tra sign bitto thefront of our numbers% 0y convention*

sign bit reresents a ositive number% 1sign bit reresents a negative number%

2,amles*

111/ 3 141 (a 5-bit unsigned number)

111 3 .141 (a ositive number in 6-bit signed magnitude)

1 111 3 -141 (a negative number in 6-bit signed magnitude)

1/3 51 (a 5-bit unsigned number)1 3 .51 (a ositive number in 6-bit signed magnitude)

1 1 3 -51 (a negative number in 6-bit signed magnitude)


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Subtraction (lvk) 4

Signed magnitude operations

+egating a signed-magnitude number is trivial* #ust change the sign bitfrom to 1! or vice versa%

dding numbers is difficult! though% Signed magnitude is basically whateole use! so think about the grade-school aroach to addition% It'sbased on comaring the signs of the augend and addend*

If they have the same sign! add the magnitudes and kee that sign% If they have different signs! then subtract the smaller magnitude

from the larger one% The sign of the number with the largermagnitude is the sign of the result%

This method of subtraction would lead to a rather comle, circuit%

. 4 7 8. - 9 5 7 -/ 9 :

6 14 17

9 5 7- 4 7 8/ 9 :

because


4/30

Subtraction (lvk) 5

Ones complement representation

different aroach! one's comlement! negates numbers bycomlementing each bit of the number%

&e kee the sign bits* for ositive numbers! and 1 for negative% Thesign bit is comlemented along with the rest of the bits%

2,amles*


111 3 .141 (a ositive number in 6-bit one's comlement)1 1 3 -141 (a negative number in 6-bit one's comlement)

1/3 51 (a 5-bit unsigned number)

1 3 .51 (a ositive number in 6-bit one's comlement)

1 111 3 -51 (a negative number in 6-bit one's comlement)


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Subtraction (lvk) 6

Why is it called ones complement?

;omlementing a single bit is e


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Subtraction (lvk) 9

Ones complement addition

To add one's comlement numbers* =irst do unsigned addition on the numbers! includingthe sign bits%

Then take the carry out and add it to the sum% Two e,amles*

This is simler and more uniform than signed magnitude addition%

111 (.7). 111 . (-5)

1 1

1. 1

11 (.4)

11 (.4). 1 . (./)

11

11.

11 (.6)


7/30

Subtraction (lvk) 7

Twos complement

>ur final idea is two's comlement% To negate a number! comlementeach bit (#ust as for ones' comlement) and then add 1%

2,amles*


111 3 .141 (a ositive number in 6-bit two's comlement)

1 1 3 -141 (a negative number in 6-bit ones'comlement)

1 11 3 -141 (a negative number in 6-bit two's comlement)

1/3 51 (a 5-bit unsigned number)

1 3 .51 (a ositive number in 6-bit two's comlement)

1 111 3 -51

(a negative number in 6-bit ones'comlement)

1 11 3 -51 (a negative number in 6-bit two's comlement)


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Subtraction (lvk) :

Two other eften! eole talk about "taking the two's comlement$ of a number% This is aconfusing hrase! but it usually means to negate some number that's alreadyintwo's comlement format%

More about twos complement

1 - 1 1 1 (.141)

1 1 1 (-141)

1 - 1 (.51)

1 1 1 (-51)


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Subtraction (lvk) 8

Twos complement addition

+egating a two's comlement number takes a bit of work! but additionis much easier than with the other two systems%

To find . 0! you #ust have to* @o unsigned addition on and 0! including their sign bits% Ignore any carry out%

=or e,amle! to find 111 . 11! or (.7) . (-5)* =irst add 111 . 11 as unsigned numbers*

@iscard the carry out (1)% The answer is 11 (.4)%

1 1 1. 1 1

1 1 1


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Subtraction (lvk) 1

nother twos complement e!ample

To further convince you that this works! let's try adding two negativenumbersA111 . 111! or (-4) . (-/) in decimal%

dding the numbers gives 1111*

@roing the carry out (1) leaves us with the answer! 111 (-6)%

1 1 1. 1 1 1

1 1 1 1


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Subtraction (lvk) 11

Why does this wor"?

=or n-bit numbers! the negation of 0 in two's comlement is /n- 0 (thisis one of the alternative ways of negating a two's-comlement number)%

- 0 3 . (-0)3 . (/n- 0)3 ( - 0) . /n

If 0! then ( - 0) is a ositive number! and /nreresents a carry

out of 1% @iscarding this carry out is e


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Subtraction (lvk) 1/

#omparing the signed number systems

ere are all the 5-bit numbersin the different systems%

Bositive numbers are the same inall three reresentations%

Signed magnitude and one'scomlement have twoways ofreresenting % This makesthings more comlicated%

Two's comlement hasasymmetric rangesC there is onemore negative number thanositive number% ere! you canreresent -:but not .:%

owever! two's comlement isreferred because it has onlyone ! and its addition algorithmis the simlest%

@ecimal S%D% 1's com% /'s com%7 0111 0111 0111

6 0110 0110 01105 0101 0101 0101

4 0100 0100 0100

3 0011 0011 0011

2 0010 0010 0010

1 0001 0001 0001

0 0000 0000 0000-0 1000 1111

-1 1001 1110 1111

-2 1010 1101 1110

-3 1011 1100 1101

-4 1100 1011 1100

-5 1101 1010 1011

-6 1110 1001 1010

-7 1111 1000 1001

-8 1000


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ow many negative and ositive numbers can be reresented in each ofthe different systems on the revious age

In general! with n-bit numbers including the sign! the ranges are*

Ensigned

Signed

Dagnitude

>ne's

comlement

Two's

comlement

Smallest -(/n-1

-1) -(/n-1

-1) -/n-1

Fargest /n-1 .(/n-1-1) .(/n-1-1) .(/n-1-1)

$anges o% the signed number systems

Ensigned

Signed

Dagnitude

>ne's

comlement

Two's

comlement

Smallest () 1111 (-7) 1 (-7) 1 (-:)

Fargest 1111 (16) 111 (.7) 111 (.7) 111 (.7)


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#onverting signed numbers to decimal

;onvert 1111 to decimal! assuming this is a number in*

(a) signed magnitude format

(b) ones' comlement

(c) two's comlement


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&!ample solution

;onvert 1111 to decimal! assuming this is a number in*

Since the sign bit is 1! this is a negative number% The easiest way tofind the magnitude is to convert it to a ositive number%

(a) signed magnitude format

+egating the original number! 1111! gives 111! which is ./1 indecimal% So 1111 must reresent -/1%

(b) ones' comlement

+egating 1111 in ones' comlement yields 11 3 .11! so theoriginal number must have been -11%

(c) two's comlement

+egating 1111 in two's comlement gives 111 3 111! whichmeans 1111 3 -111%

The most imortant oint here is that a binary number has differentmeanings deending on which reresentation is assumed%


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's #omplement

&hat is the negative of 111

* 111

0* 1111 ;* 111 @* 1111

Gune 4th! /8 19 Subtraction


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's #omplement

&hat is the negative of 1

* 111111 0* ;* 11111

@* 1

Gune 4th! /8 17 Subtraction


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Subtraction (lvk) 1:

(nsigned numbers over%low

;arry-out can be used to detect overflow The largest number that we can reresent with 5-bits using unsigned

numbers is 16 Suose that we are adding 5-bit numbers* 8 (11) and 1 (11)%

The value 18 cannot be reresented with 5-bits &hen oerating with unsigned numbers! a carry-out of 1 can be used to

indicate overflow

1 1 (8). 1 1 (1)

1 1 1 (18)


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Signed over%low

&ith two's comlement and a 5-bit adder! for e,amle! the largestreresentable decimal number is .7! and the smallest is -:%

&hat if you try to comute 5 . 6! or (-5) . (-6)

&e cannot #ust include the carry out to roduce a five-digit result! asfor unsigned addition% If we did! (-5) . (-6) would result in ./4H lso! unlike the case with unsigned numbers! the carry out cannotbe

used to detect overflow%

In the e,amle on the left! the carry out is but there isoverflow%

;onversely! there are situations where the carry out is 1 but thereis no overflow%

1 (.5). 1 1 (.6)

1 1 (-7)

1 1 (-5). 1 1 1 (-6)

1 1 1 1 (.7)


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Subtraction (lvk) /

)etecting signed over%low

The easiest way to detect signed overflow is to look at all the sign bits%

>verflow occurs only in the two situations above* If you add two ositivenumbers and get a negativeresult% If you add two negativenumbers and get a ositiveresult%(The only e,cetion is when you add two negativenumbers in Signed

Dagnitude% In Signed Dagnitude! you #ust add the magnitudes and

kee the same sign% In this case there is overflow if there is acarry% Try to add -5 . -6 using Signed Dagnitude)%

>verflow cannot occur if you add a ositive number to a negativenumber% @o you see why

1 (.5). 1 1 (.6)

1 1 (-7)

1 1 (-5). 1 1 1 (-6)

1 1 1 1 (.7)


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Subtraction (lvk) /1

Sign e!tension

In everyday life! decimal numbers are assumed to have an infinitenumber of s in front of them% This hels in "lining u$ numbers%

To subtract /41 and 4! for instance! you can imagine*/41

- 4

//:

?ou need to be careful in e,tending signed binary numbers! because theleftmost bit is the signand not art of the magnitude% If you #ust add s in front! you might accidentally change a negative

number into a ositive oneH

=or e,amle! going from 5-bit to :-bit numbers*

11 (.6) should become 11 (.6)%

0ut 11 (-5) should become 1111 11 (-5)% The roer way to e,tend a signed binary number is to relicate the

sign bit! so the sign is reserved%


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Subtraction (lvk) //

Signed Numbers are Sign*&!tended to +n%inity,

Demorie this idea% It will save you every time when you are doing binary arithmetic by

hand% 111 in twos comlement is actually J111 111 in twos comlement is actually J1111111

111 . 11 3 1111% This is overflow% 0ut can you see it Fook again*

J1111111 . J111111 3 J1111111 The sign of the result (1) is different than bit 4 ()% &e cannot reresent the result in 5 bits

The sign of the result is e,tended to infinity ;heck this every time in your answers and you will always see the

recision of the result (number of significant bits) very clearly% Sign-e,tend inuts to infinity @o the math on an infinite number of bits (concetually) Truncate the result to the desired outut tye


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)oes over%low occur in my program?

ublic class test K

ublic static void main(StringMN args) K

int i 3 Cwhile (i O3 ) K

i..C

P

System%out%rintln(Qi 3 Q . i)C

i--C


i..C


PP

>utut*

i 3 -/1575:495: /41

i 3 /1575:4957 /41-1

i 3 -/1575:495:



24/30


Ma"ing a subtraction circuit

&e could build a subtraction circuit directly! similar to the way wemade unsigned adders%

owever! by using two's comlement we can convert any subtractionroblem into an addition roblem% lgebraically!

- 0 3 . (-0)

So to subtract0 from ! we can instead addthe negation of 0 to %

This way we can re-use the unsigned adder hardware from last time%


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twos complement subtraction circuit

To find - 0 with an adder! we'll need to* ;omlement each bit of 0%

Set the adder's carry in to 1% The net result is . 0' . 1! where 0' . 1 is the two's comlement negation of 0%

Remember that 4! 04 and S4 here are actually sign bits%


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Small di%%erences

The only differences between the adder and subtractor circuits are* The subtractor has to negate 04 0/ 01 0%

The subtractor sets the initial carry in to 1! instead of %

It's not too hard to make one circuit that does bothaddition and subtraction%


27/30


n adder*subtractor circuit

>R gates let us selectively comlement the 0 inut%

3 1 3 '

&hen Sub 3 ! the >R gates outut 04 0/ 01 0 and the carry in is %The adder outut will be . 0 . ! or #ust . 0%

&hen Sub 3 1! the >R gates outut 04' 0/' 01' 0' and the carry in is1% Thus! the adder outut will be a two's comlement subtraction! - 0%


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Subtraction (lvk) /:

Subtraction summary

good reresentation for negative numbers makes subtractionhardware much easier to design%

Two's comlement is used most often (although signed magnitudeshows u sometimes! such as in floating-oint systems! which we'lldiscuss later)%

Esing two's comlement! we can build a subtractor with minorchanges to the adder from last time%

&e can also make a single circuit which can both add and subtract% >verflow is still a roblem! but signed overflow is very different fromthe unsigned overflow we mentioned last time%

Sign e,tension is needed to roerly "lengthen$ negative numbers%

fter the midterm we'll use most of the ideas we've seen so far to buildan FE an imortant art of a rocessor%


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;onvert (-//)1to /'s comlement 9-bitbinary*

* 111

0* 111 ;* 111 @* 1111

Gune 4th! /8 /8 Subtraction


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In : bit twos comlement binary*1111113

* 11111

0* 1111 ;* 1111 @* 1111

Gune 4th /8 4 Subtraction

10 subtraction

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