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Dimensional MCC right this algorithm is based
on Bitonic sorting techniques. I hope you
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have not yet forgotten there what if I can
be sort so the algorithm is a by Biotonic
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sorting technique and that has been implemented
on two-dimensional MCC does idea now in the
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body being sort what it does Game one in the
bad news much what do we do then AI is compared
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with A and + size.
The smaller one you keep it one side and glad
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the one goes to the another side and then
both the secrets in
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00:01:25,880 --> 00:01:32,340
the Bitonic sequence if you can say and again
you excessively so now here we assume the
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00:01:32,340 --> 00:01:47,249
image is having n constant n process that
is n constant process and each processor contains
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00:01:47,249 --> 00:02:32,900
one element this process has at least 3 registers
Rr, Rs and Rt so these are 3 registers 3 memory
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location and cluster needs this is
storage registered.
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00:02:46,290 --> 00:03:03,250
Where this is a septum and this is all routing
in so in order to transfer one day tap from
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one processor to another processor the data
must be in this register now each processor
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is capable or performing.
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Three types of operation one is routing operations
between Rr change on the same processor comparing
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00:04:14,870 --> 00:04:55,010
it compare exchange between two registers
of the same processor compare exchange means
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that content of order and content of Rs will
be compared and if it is the winner will be
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kept in Rs loser will be kept in are opposed
in that case of increasing order minimum is
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the winner and that case of decreasing order
maximum will be the peanuts and then another
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operation is only exchange only exchange operations
between 2 registers right so these are 3 operations
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the processor can perform the time needs.
I have already told about Rr right okay let
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us put tr ,tc and te so time need for one
routine takes here times tc is the time needs
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for compare exchange and te is the only exchange
this is the time it is up to this definition
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is clear in case you need a clear time means
say a presser pi has data Ti in here in Rr
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register and b is in Rs register so if you
perform exchange operations what will happen
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bt he will be here a will be here B is in
Rr register and A is in Rs register and do
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this you have is it okay.
Now before we start about this algorithm how
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we do it let us first revisit the Bitonic
sorting technique.
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What you need a 1 a 2 a 3 a 4 a 5 a6 a7 these
are the eight elements you have and every
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time you are arranging in such a way that
it becomes a Bitonic difference for the next
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00:07:56,700 --> 00:08:08,630
step and also it is known to you that 2 sequence
with 2 elements is always a by Biotonic sequence
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so this is 1 Biotonic 3 sequence there are
4 Biotonic sequence so you have to arrange
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in such a the next time you get a Biotonic
sequence so you have to have some idea that
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how to create it so if I sequence.
Similarly is that case so now you get the
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two Biotonic sequence this is one Biotonic
sequence and this is another one so you arrange
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it in the second phase and then finally you
get right this is the study now in the case
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of two dimensional Merge suppose we have the
16 element this elements are here and each
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processor is connected with these four neighbors
right at least at most so in the it consists
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of you name one is that it consists of three
steps there it consists of four parallel phase
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first phase.
You will be meeting making a Biotonic sequence
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of size you have Biotonic sequence of size
2 if you consider this is one size 2 1 1 1
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1 1 11 like that 8 Biotonic sequence you have
and how did you know that there is a class
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going on through telephone or message guessing
after 40 minutes you cannot guess just okay
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those were quite as put three of you join
they let me tell you today we are discussing
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on sorting and the model is to put a 2 dimensional
mesh here we assume that there are n cross
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n processors each processor contains one element.
And each process has at least three registers
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or three memory locations one is RR another
one is RS and RT are these routing register
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R this is the sport is resistor and Rt temporary
storage register the result final result will
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be in Rs register now each process is capable
of doing these three types of operations routing
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operation between to order Rr processor Rr
processor J must be neighbor and time needs
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for that is Tr compare exchange operations
between the two register of a process and
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time order T C.
And similarly exchange operation takes order
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te time this is an example what I then we
told discuss about the strategy follow in
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the Biotonic sorting technique for eight elements
finally we are discussing about the how we
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are going to use it on two-dimensional mesh
where you have 4 cross 4 mesh and 16 elements
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now this if you absorb the 16 elements you
can think that they form the 8 Biotonic sequences
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each of size 2 now you have to go to the next
level which will give you 4Biotonic sequences
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of size right so you have to arrange in such
edge that one is in increasing order and another
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one is decreasing order.
Right one possible way could be that you put
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it increasing order you put decreasing going
to so you got to Biotonic size for this side
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this side is there and so on but in that process
you may find little difficulties what we do
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that we arrange this is an increasing order
this is decreasing order and this forms a
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Bitonic sequence so in the first phase
you get basically you arrange the data in
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such a way that you get four Biotonic sequences
each of size four in the second phase .
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Now you to arrange the data again in such
that you get two by two Biotonic sequence
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in sub size 4 right each of size 8 so you
arrange the data in increasing order and this
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becomes decreasing order so you observe that
this gives a Biotonic sequence of the size
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how you will be arranging that parties would
be discussed later second phase, no third
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phase.
You arranged the date I know you have the
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a Biotonic sequence here another Biotonic
sequence here right so you arranged the data
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in such a that you get to Biotonic sequence
upside a right basically you will be getting
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this one and then this one so you get Biotonic
sequence of stage 60 this is increasing order
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you put the decreasing order so combined if
I C from this point of view that will become
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a battle in sequence in the third phase.
So in the fourth phase you have a Biotonic
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sequence so you can easily get so basically
at the end of the foot page you get the sorted
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sequence at the end of the third phase you
get the sorted sequence so your strategy says
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that you should be such that at the end of
every phase you get some number of Biotonic
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sequences this is first one and second one
it should collapse in such a way that after
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long in time you will be able to get the sorted
sequence right.
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So what do you observe first thing is that
you do the margin with respect to the row
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right that is horizontal you margin then you
get you get Biotonic sequence consisting of
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two 1/2 which you need to model vertically
right ,here you get this one and this sub
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blocks so you need to model horizontally,
right now here again you have one sorted sequence
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and other sorted sequence this forms a Biotonic
sequence and you have to merge it vertically
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and so on so once you do the horizontal one
next time you need to do the vertical one
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horizontal vertical like that to get the ultimately
after login intelligence.
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You will find the sorted sequence is it okay
the strategies here right so first let us
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discuss about so basically.
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We need to know one is horizontal merge another
one is vertical merge right if I know this
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one then we do one after another one to get
the sequence now vertical much of a botanic
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sequence it consists of two algorithm one
is known as Row merge another one id column
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merge okay so first let us discuss about the
row merge let us assume that there are K processors
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K process p 0 P 1 P 2 P K p0 p1 p2 pk - 1.
In the form of row Pi contains a0 ,a1 ak-1
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is Biotonic sequence right so what do you
have basically you have the process like that
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and the data is like this now what I am going
to do in their Biotonic sequence case Pi is
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to be computed with Pi+ky2 agreed so basically
this data has to be moved here this data has
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to be moved here and then compare right there
will be one accepted element another one is
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rejected element.
The rejected element you send back to this
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side again right, so what we will get this
is the Biotonic sequence this will become
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another Biotonic Sequence .Agreed terribly
again you call row merge this one row merge
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and so on we will be getting the ultimately
sorted sequence so if it is the case so I
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can write row Merge k and you assume the data
initially stored Rr register assume the data
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is in Rr register contains ai assume the data
so once you want to bring this data from here
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to here.
This state has to be safe first agree or not
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know because the data is also not Rr register
and routing has to be done through Rr own
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so you move this data first to Rs area so
each processor P I is 0 1 2 py 2 - 1 moves
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ai from Rr to Rs then next one is each processor
Pi I is K / 2 to K-1 instance the data once
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I write the census data it must be in Rr register
to P I - K/ 2 next is each processors P I
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is 0 1 K / 2 - 1 what it does he compare right
performs compare exchange operation between
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Rr and Rs.
Now one side performs Rr comparison operation
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between Rr,Rs the accepted element is in RS
rejected element is in Rr that is to always
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accepted element will be in Rs area and rejected
element will be Rr area now if it is the case
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now this rejected element has to be sent back
to the other part of the show so what I do
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is process CI is 0 1 K / 2 - 1 since data
2 p I + 1 K /2 so the rejected element has
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come here and that is in Rr area now RS contains
e first K / 2 process mean way.
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I contains all the accept elements so each
processor so that you can start again P I
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I is 0 1 K / 2 - 1 performs exchange operations
between Rr,Rs so now all the data R in Rr
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registers so you can call RM K /2 on p0 P
KY 2 - 1 and RM k / 2 on p k / 2 p K - 1 simultaneously
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yes this one
compare exchange is doing that though RS contains
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all accepted element Rr all the rejected now
this rejected element has to come here so
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now once the you have sent the rejected element
here now these are all contains all junk things.
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I have to get back this oddest thing here
so that recursive call I can do because in
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the initial thing that I decided that all
the data are in RR register right initially
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see the data Re,Rr so if they gained Rr to
call this one I must have the data must be
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Rr registers so to do that that time performing
this operation now you know to compute the
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time complexity.
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For this T r row merge K compare, row merge
K exchange so now the thing is that you have
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to know where I have mention about the boundary
condition because this once it is a recursive
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I have to tell about the boundary conditions
right so what should be the boundary condition
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if whenever k is more than one you do it so
if K >1 if it is nothing so if K is 1 how
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many routine 0 otherwise or if you get the
you are also supporting that .
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You have now you have to move this data from
here to here how much time you need how many
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routines you need came by to routing surely
again you pump on the data so another K / 2
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so you have K + time of size k/2 agree now
so routing part is over so come over is the
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rotate this one and this one now can you tell
me compare exchange how many compare exchange
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how one this all this is a simultaneously+Tc
row k/2 agreed so compare exchange is also
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over the exchange operations it is in parallel
while.
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I am sending these data from here to here
this comes here to here this comes here to
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here but how are you going to bring this element
here this will come here ,here ,here, here
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while it is here this element is here this
element is here so it is see it is a pipeline
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type thing so you need k/ this distance to
cover for him this distance to be covered
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for him this is to go which is k/2o k=1 2+p
rn k/2 otherwise.
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Now first let us find out the solution of
this one from this suppose K = 2 I should
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I or I - 1 I do not know I or I -1100% right
I do not know I fine then log k then it will
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writing log k to please say I or I - 1 otherwise
it is log k agreed well this is 2 times constant
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times so it is masters theorem what about
this one here okay oh okay then why what for
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I am doing all this thing with n log n is
shorting time and situation margin times days
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here okay l log n.
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Let us see Tr k RM = k + PR RM k / 2 which
is k +K / 2 + TR and K / 4 +s K /2 + K / 4
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when K this is K then I get 1 K means 2i so
when this is one what I will write here K
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/ no very sure when I am writing 4 here I
am writing 21 when I am writing 2i it will
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00:37:47,960 --> 00:38:27,349
be 2 i-1 right and this is 0 this is 0 so
you get 1 + 1/2 + 4 2 times K - 2 K - 2 so
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this solution is okay.
So this is the number of routines you need
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now if I tell you that you can you prove it
that it gives the sorted sequence for Biotonic
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sequence right your input is in the row merge
input I told that the row contains a Biotonic
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sequence after row merge call I get a sorted
sequence now obviously this algorithm you
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have to prove that it gives you the sort sequence
any idea not on me because this is the is
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compared to a TI + N and then I am making
me to row this row and this row and then he
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gave me recursively calling they say.
I am not changing any or I am not making any
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new strategy or new things have introduced
you into that right now if it is the case
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so row merge gives you the sorted sequence
now the second thing is that column merge
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in the column merge.
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Let us put the names K M and C M and J is
the number of elements in that column right
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and same way I can do it instead of assuming
that it is in the form of row merge you mean
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let it is in the form of column right so column
merge also will give you the sorted sequence
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from if the elements form a Biotonic sequence
in that column so the time complexity for
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this I can write Tr CM j = 2j - 2t c CM j
it is log j tg cmj is 2 times now if you know
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this row merge column might you ready for
suppose.
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You have and this form say Biotonic sequence
this is an increasing one is a decreasing
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order combining this with forms a power to
do so right and let us issue right this is
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j/2 okay, so first you what you have to do
this element has to be compared with this
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element so this position of this is the Q
/ 2 x - 1 and these elements positions if
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I expand it in one row sequence this position
is a J / 2 -1 K +1 right and this is your
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a 1 so this a 1 has to be compared with these
elements agreed or not similarly this element
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00:43:17,529 --> 00:43:38,640
has to be compared different and so on.
So basically I can think that these are several
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columns you have and if I do the column Merge
what I will get if I do the column merge I
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get I get a Biotonic sequence in this zone
no yes or no so but the size is J / 2 again
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you do this column once again here column
merge here again then you get like this size
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again you do the column merge you get this
side finally you will get Biotonic sequence
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in each row really and it says finds the property
that these elements are large smaller than
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this element this element smaller.
Because that is the property of but now you
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do the core row Merge you do the row merge
and then you combine this thing you get disordered
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suppose finally we will be getting there tell
me again yes but within the row that form
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a vital discipline that is the that is not
solving the content of this elements is by
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2d sequence and this elements are smaller
than this elements these elements are smaller
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than this element and so on now you produce
the row much here right so vertical merge.
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If I write J ,K first what do you do you do
the column merge and then you do the row merge
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so you do the first column merge oh then you
do the row merge this is your vertical module
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we will consider one example to see how it
works now the time complexity for that is
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00:46:06,739 --> 00:46:39,890
T V M routing J ,K is your 2 times J +s K
- 4 compare vertical merge J , K is log JK
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exchange vertical merge check from our case
this 2 Times log JK just I am combining these
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00:46:51,529 --> 00:47:16,829
two results one example.
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00:47:16,829 --> 00:47:38,180
You have suppose 4 cross 8 4 8 what do
you have to give me 32 elements okay, let
193
00:47:38,180 --> 00:48:16,259
us do it here 1 2 3 4 5 6 7 8 9 10 11 12 13
14 15 16 17 18 19 20 21 22 23 24 25 26 27
194
00:48:16,259 --> 00:48:25,319
28 29 30 so you have suppose this is Biotonic
sequence here right this is even increasing
195
00:48:25,319 --> 00:48:33,540
order and this is decreasing order now if
I do the column words assuming that this is
196
00:48:33,540 --> 00:48:39,690
one column this is another column there are
not for long so what I will get you what I
197
00:48:39,690 --> 00:48:59,839
will get tell me if I do the column merge
here.
198
00:48:59,839 --> 00:49:41,359
I will be getting 1 7 10 15 19 24 these 24
27 and then 31 what about this 1 2 5 11 14
199
00:49:41,359 --> 00:50:29,880
20 23 28 30 then you get 4 6 12 4 6 12 13
17 when to 21 ,26 ,29,3,9 389 16 18 22,25
200
00:50:29,880 --> 00:50:37,970
42 so it through column implicating that and
you observed that every element has got his
201
00:50:37,970 --> 00:50:47,829
own row but we need the row it forms a Biotonic
sequence right and if you do the row merge
202
00:50:47,829 --> 00:50:58,609
you will be getting this is column merge while
trying at home you use the all the steps of
203
00:50:58,609 --> 00:51:01,970
column merge because within the column odd
you will be getting column merge J /2.
204
00:51:01,970 --> 00:51:37,819
Then column J/4 and so on immediately row
merge 1,2,3,4 ,5,6,7,8 right because you know
205
00:51:37,819 --> 00:51:48,710
from here to here it involves first this type
of thing row merge 4 and then row merge 2
206
00:51:48,710 --> 00:52:02,480
row merge to will be this one yes or no for
this case 1 2,4,3.
207
00:52:02,480 --> 00:52:11,979
What is in between row merge is to so data
go back so you will be getting one 1,2,4,3
208
00:52:11,979 --> 00:52:26,599
then row merge 5 then it getting 1,2,3,4,5
right for this case you will be getting 10.11,12,9
209
00:52:26,599 --> 00:52:34,920
fist is row merge this 2 ,so you will be getting
12,would be coming here 1 1would be coming
210
00:52:34,920 --> 00:52:49,560
here 9 will be coming here except retain reject
so then 9 will be here 12 will be here then
211
00:52:49,560 --> 00:53:06,619
row merge size 2 so compare 9 10,11,12 this
is Rm 4.
212
00:53:06,619 --> 00:53:24,819
And this sis Rm 2 which one this one of this
one the first one is indicates yes and secondary
213
00:53:24,819 --> 00:53:35,989
no adds tilted by this direction or this direction
that is yes depending upon south Indian north
214
00:53:35,989 --> 00:53:48,520
Indian but in both the case if head this and
level is known.