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IGCSE Computer Science 2023-25 - (1) Data Representation - Number Systems 1.1(c) 2's Complement — Transcript

by Mr Bulmer's Learning Zone · 1,879 words · 333 segments · language en · Watch on YouTube

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0:00okay welcome back everybody and we're
0:02going to be looking at the um the third
0:04and final part of 1.1 number systems
0:08following the new computer science igcse
0:11syllabus
0:12um in this we're going to be looking at
0:15how to add
0:16two positive eight bit numbers
0:18we're going to be looking at overflow
0:20when we and and how this is caused when
0:22we do add to positive 8-bit binary
0:25numbers we're going to look at logical
0:27shifts
0:28on positive
0:31binary numbers and how we can multiply
0:34binary numbers and divide binary numbers
0:36using this method
0:38we're then going to finish off with um
0:40something called two's complement to
0:42complement notation and this is new
0:44um to the to the new syllabus but this
0:47this allows us to represent both
0:49positive and negative
0:51binary numbers
0:53once we've done that we should be moving
0:54on sticking with the um data
0:56representation but we'll be moving on to
0:58text sound and images
1:02okay
1:03so
1:04adding binary numbers together
1:06well in the normal system in an idenary
1:09system would be
1:10adding we'll be looking at the units
1:12column first of all and if we have one
1:15plus two obviously that equals three
1:18it's only when we carry over and we
1:19maybe add nine plus four and we get four
1:22lots of units and one lot of ten that we
1:25can sort of draw on that and see
1:27similarities
1:28in the
1:29in the danery system
1:31so there are some rules
1:34and here we go we've got four rules
1:36um if we have zero plus zero obviously
1:39that's going to be a zero it's the same
1:40in any number system
1:42um one plus zero of course is still one
1:46but when we get to the next two one plus
1:48one would equal two
1:50which in binary would be one lot of two
1:52and no lots of ones
1:54and then if we have one plus one plus
1:56one obviously this is three but three is
1:59represented as um one lot of two and one
2:02lot of one
2:04so if we follow this method
2:07we can look at this sum we've we've got
2:08here this um
2:10eight bits
2:11um
2:13from two registers
2:15and register one and i've put register
2:16two underneath it as a simple sort of
2:18addition sum
2:20we can go step by step through this
2:22together
2:24so we have one plus zero obviously
2:26equals one
2:28now we get to the third rule
2:30one plus one
2:31is
2:32two
2:33but that would be
2:35no units no ones
2:38so we carry over the two
2:40so one plus one again we're going to do
2:42the same thing zero
2:44carry the one
2:45here we've only got one so we pop that
2:48into the um into into the answer
2:50one plus one again
2:52zero carry the one
2:54one plus one again
2:56zero carry the one and now we've got the
2:58final rule one plus one plus one
3:01obviously equals three so that's one lot
3:04of one
3:05and one lot of two that we carry over
3:07and we can put this into the final
3:09column to get the answer
3:11one one zero zero one zero zero one
3:16hopefully look at that you could you can
3:18remember how you might change this first
3:20of all into a idenary number how we
3:23could convert that and also if we would
3:26divide it in half
3:27i'm splitting into two nibbles how we
3:29might change this into an hexadecimal
3:33now if we were to change the first
3:36number
3:37the the first term the first digit
3:39in that first byte of information and
3:41change it to a one
3:43obviously when we add all these numbers
3:45together again we'd end up with
3:47something
3:48with a ninth bit something that's called
3:51an overflow
3:52um this obviously would lead to errors
3:55in the sense that we can't have
3:57nine digits nine numbers represented in
4:00an eight bit number system so we'd end
4:02up with problems there and the system
4:04might
4:05might crash or it might ignore
4:08that extra digit so you wouldn't get the
4:09correct answer
4:12i would then move on
4:14so logical binary shifts 1.1.5
4:20okay we're going to use this technique
4:23to shift numbers um left and right
4:26backwards and forwards in a register
4:30um
4:31obviously with binary numbers things
4:33need to be added together taken away
4:35removed
4:36but in this case we're going to be
4:38multiplying
4:39or dividing um bytes in this case of
4:43information
4:46so
4:48if we were to shift
4:50the numbers in this system i'll shift
4:53these ones and zeros across to the left
4:56we would be multiplying you'll see that
4:58in a moment and if we shift them to the
5:00right going down the radius so to speak
5:03then we will be dividing by two what
5:06does that mean
5:07we have a an eight bit register here
5:10below and it contains the daenery number
5:1221 you can see that it's one lot of 16
5:16one lot of four and one lot of one added
5:18together obviously 21.
5:21now if we shift these digits to the left
5:26if we move them
5:28one space
5:30okay
5:31um we're going to get one lot of 32
5:33one out of eight and one left two now
5:35that's obviously 42
5:38which would be twice
5:41the value of 21. so 21 times 2
5:44equals 42.
5:46when we're shifting we've got spaces we
5:48automatically put in a 0 into the end
5:51column now you mind some of you might
5:54see this and think
5:55ah this could cause problems and it does
5:58cause problems
5:59so as a bit shift an empty position
6:03is replaced with a zero and obviously
6:06there are limits in an 8-bit number
6:07system as to how many zeros you can put
6:09in before it completely zeros out
6:13so for example if i was to take this
6:15number here
6:16um one lot of 64 one of 32 and one love
6:1916 and i was to move this
6:23five places to the left
6:28then we would end up with um a register
6:30full of zeros okay but we've multiplied
6:33this number
6:34um and obviously when we multiply this
6:36number the equivalent of
6:38the value 1 1 2 or 112
6:41we're not when we times it by 2 and 2
6:44and 2 and 2 and 2 again
6:47we're not going to get
6:48purely zeros
6:51okay
6:52so here's some more examples
6:54um
6:56what we're going to do first of all
6:57write 24 into the 8-bit register well
7:00that's obviously um one lot of 16 and
7:03one lot of eight
7:05okay so that's that's straightforward we
7:07know that
7:08now we're going to shift to a logical
7:10shift going three places to the left
7:13so we're going up the register
7:16so
7:17shifting them one two three
7:20as you can see there is some
7:22times two
7:23times two times two
7:26which would equal
7:27192. so we've times it to um 24 times 2
7:32to the 3 which represents the three
7:34shifts
7:35okay 192.
7:37now if we go back to the existing
7:39um results the existing register which
7:41is 24 and we go
7:44two places to the right
7:47you can see there we've gone one two
7:49we're basically dividing so 24 divided
7:52by two to the two two shifts would equal
7:55six
7:56basically 24
7:58divided by 2 is 12 and divided by 2
8:01again would be 6.
8:04okay so that's how we multiply
8:06and divide using logical binary shifts
8:13okay and for the final bit we're going
8:14to be looking at two's complement
8:17on binary numbers how to represent
8:19negative integers
8:21um when we're using two's complement
8:24so in this section again like the others
8:25we're going to assume
8:27um the eight bit registers are being
8:29used for the
8:30in terms of the questions
8:33the only major difference
8:35that we're gonna we're gonna use in
8:37terms of these registers is we're going
8:39to change the heading of one of the
8:41numbers that being
8:43minus one to eight one two eight becomes
8:46minus one to eight
8:48of course that means that we no longer
8:49have
8:51255
8:53different character representations
8:56different sequences of ones and zeroes
8:58because we've got the minus one to eight
9:01this refers to
9:03everything
9:04which has a one under
9:06minus one to eight
9:08is a negative number
9:10everything with a zero under the minus
9:12one to eight
9:13is a positive number so we basically
9:15split
9:17two five five
9:18into two into two halves
9:21the positives being 0 up to 1 127
9:25and the negatives being minus 128 all
9:29the way up to minus 1.
9:33and here we can see that minus 128 would
9:36be one lot of minus 128 and nothing in
9:40the other
9:41columns
9:42whereas plus 127
9:44the sort of biggest positive number we
9:46can have would be no loss of 128
9:50and all the rest of the um the columns
9:53being filled in with ones
9:57it is important to realize when applying
9:59two's complement to a binary number that
10:02the leftmost bit always determines the
10:04sign of the binary number as i've just
10:07said a one value in the leftmost bit
10:09indicates a negative number and a zero
10:11value in the left most bit indicates a
10:14positive number i can't emphasize this
10:16enough i'm going to try and show you a
10:18number here
10:19um a representation of 51.
10:22and of course if we look at this you
10:24look and see okay we've got one lot of
10:2532 one lot of 16
10:28a two and a one
10:30it's exactly the same as how you would
10:32normally do it there are no
10:34negative 128s there are no positive 128s
10:37in the same way there's no 64s
10:40um it is exactly the same
10:43so if we want to represent the number
10:45minus 49
10:48we're going to put a 1 under the minus
10:49128 this shows that it's a minus number
10:52and we're going to start with that
10:54number minus 128. we're then going to
10:56add to it
10:58positive 64.
11:00positive 64 plus minus 128 would give us
11:04minus 64.
11:06if we then add positive 8
11:08to minus 64
11:10we would get minus 56
11:12at positive 4 we would get minus 52
11:15add positive 2 would get minus 50 and
11:18finally add positive 1
11:20and we would get minus 49 and this is
11:24how we rep how we represent
11:26the minus numbers we're adding the
11:29positive numbers
11:30the ones in greens you can see here to
11:32the minus 128.
11:37well that's it that's how you would do
11:41the two's complement representation of
11:42negative numbers
11:45so
11:46have a practice
11:47good luck with that and remember if
11:49there's a one there
11:51it's a minus number and if there's a
11:53zero it is a positive number but the
11:56question will always say
11:58this
11:59is a representation of a two's
12:01complement binary number
12:04that's it for now
12:05thank you very much indeed um that as i
12:08said at the beginning is the end
12:11of this little bit 1.1
12:15so thank you very much for watching
12:17next time the next video will be 1.2
12:20text sound and images
12:22if you haven't already please subscribe
12:24i will be doing these as quickly as
12:26possible i'm trying for
12:28one video a week but
12:30um bear with me
12:32so um thank you very much indeed and i
12:34will see you next time
12:36bye for now

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