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Welcome back.
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So in last few lectures, we have seen how
to use the recurrence relations to model various
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counting problems.
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Now, as we have told recurrence relations
are very essential part of mathematics or
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particularly in counting.
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So in such, recurrence relation is an equation
that recursively defines a sequence or multi-dimensional
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array of values where there are some initial
terms and the nth term is defined as a function
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of the preceding terms.
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Recurrence relation is extensively used for
combinatorics, analysis of algorithms, in
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computational biology, in theoretical economics
and many other subjects.
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So we have already seen some of the recurrence
relations.
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And we have seen how recurrence relations
can be used to model various problems, particularly
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combinatorics problems.
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But we have not seen how to solve recurrence
relations.
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In this video, we will be focusing on how
to solve recurrence relations.
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So here, some of the recurrence relations
that do appear in real life problems.
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So the first one says, T(1) equals to 1 and
T(n) equals to 2 plus T(n minus 1).
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The second one say, T(1) equals to 1, T(2)
equals to 2 and Tn is equals to T(n minus
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1) plus T(n minus 2).
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Or this one is what we got from the Tower
of Hanoi problem H(1) equals to 1, H(2) equals
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to 2, and H(n) equals to 2 times H(n minus
1) plus 1.
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This one is basically what we have the Fibonacci
series where F(1) equals 1, F(2) equals to
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1 and F(n) equals to F(n minus 1) plus F(n
minus 2).
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This is what is known as the Fibonacci series
is quite a famous series that appears again
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and again in real life.
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Then this one b(1) equals to 1 and b(n) equals
to b (n over 2) plus 1.
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Then we have M(1) equals to 1 and M(n) equals
to 2 times M(n over 2) plus n.
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So these two appears in various algorithms,
particularly, the binary search algorithm
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and the merge sort algorithm that are very
popular in the algorithm’s literature.
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And then we have applied some of complicated
one.
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C(1) equals to 1 and C(n plus 1) is equals
to submission of i equals to 0 to n.
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C(i) C( n minus i).
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Now for all of them, we have to now understand
how one can solve them.
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So what is the technique for solving any of
these recurrences?
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So, how to solve these recurrences?
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Now, the first technique that we are going
to look at is, the simple thing of guess the
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solution and prove using induction.
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In this video, we will see how this technique
is useful and then we will - In next video,
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we will see how one can guess the solution.
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So say here we have this example.
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T(1) equals to 1, T(n) equals to 2 plus T(n
minus 1).
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Now how do you solve this particular problem?
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Now first of all, if somehow magically you
can guess this number, then we are great.
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For example, if I tell you, that guess where
T(n) equals to (2n minus 1).
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Now if this is the guess that we made, then
we can try to prove the statement using induction.
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So the technique is first to guess and then
prove by induction.
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Now I have skipped a big jump of how to guess
this number.
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We will see that one – see the technique
of guessing in the next video as well as in
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the next whole week.
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Guessing the solution for the recurrence relation
is possibly the most challenging part of solving
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the recurrence relation.
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But in this video, we will be focusing on
how to solve the guess if we have the induction,
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if we have the guess right.
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If we guess the thing right how to prove it?
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So how do you prove by induction?
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Now if you remember, so we should have a base
case.
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In this case, base case is say n = 1 and we
have T1 is equals to 1, this is a something
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that is given and which is of course same
as T2 times 1 minus 1 right?
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So this value is correct for T1 right?
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And now, we have the induction hypothesis,
induction hypothesis which says that say for
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some n T(n) equals to 2 times n minus 1.
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And in the case, what is the inductive step?
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Inductive step is to prove the same statement
for T(n plus 1) which is 2 times (n plus 1)
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minus 1 which is 2n plus 1.
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How do you prove it?
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Now, of course by the thing that is given
to us T of n plus 1 equals to 2 times 2 plus
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T of n which is equals to 2 plus 2n minus
1 by the induction hypothesis which is equals
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to of course 2n plus 1 and that is what we
had to prove.
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Right?
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Hence, we are done.
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Hence, we have proved the inductive step that
means T(n) equals to 2n minus 1 which is for
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all n greater than or equal to 1.
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So this is the proof by induction for how
once we have the guess right.
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Correct?
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So let us go over the next one.
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One more example say, so this example two
says that T(1) equals to 1 and T(n) equals
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to n plus T(n minus 1).
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Again first of all, you have to guess it and
let us imagine that somebody just comes up
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and manages to guess it correctly and say
somebody comes and says that T(n) equals to
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n into n plus 1 by 2.
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Now once someone has guessed it, we have to
prove it, we have to ensure that the guess
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is right and to get that is true, we have
to again use induction.
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So like in the earlier case, we have to again
prove this one by induction and let us see
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how we prove it again.
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Say base case, n equals to 1, of course T1
equals to 1 which is 1 times 1 plus 1 by 2
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right?
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Which is what, so the thing is correct for
the case where n equals to 1.
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Now we have the induction hypothesis, what
is this says that T(n) equals to Tn sorry
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Tn equals to n into n plus 1 by 2.
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Now in inductive step, we have to prove that
T of n plus 1 equals to n plus 1 into n plus
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2 by 2.
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Now how do you prove it?
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Now T of n plus 1 is given as n plus T(n-1)
which is n plus n into n plus 1 by 2.
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This is by induction hypothesis which is taking
- sorry I made a mistake here.
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This is not n, this should be n plus 1 right?
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So, this is also so Tn equals to n plus Tn
minus 1, Tn plus 1, has to be n plus 1 plus
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T(n) and this is equals to n plus 1 plus the
given induction hypothesis which is n into
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n plus 1 by 2.
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So now I can take n plus 1 common in that
case, I get 1 plus n by 2, so this is 2 plus
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n by 2 which is of course n plus 1 into n
plus 2 by 2 which is what we had to prove.
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So we have T of n equals to n into n plus
1 by 2 for all n greater than equal to 1.
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Again the idea is simple if you can guess
the value correctly for T(n), then you can
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prove what T(n) is by induction.
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Right?
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Let us see one more example, what can be the
various guesses?
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So, this is we say Tower of Hanoi problem,
right, so H(1) equals to 1 and H(n) equals
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to 1 plus H(n minus 1).
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Again, we first have to guess it.
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Now, what is the guess here?
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Say the guess is H(n) equals to 2 power n
minus 1 and again we have to prove this one
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by induction.
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Note here that if you guess it wrong, we will
not be able to prove it by induction or we
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are not be able to prove it.
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So thus, only if you guess it right we will
be able to prove this statement.
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So there are people who actually come up with
these cases by some various intuitions of
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their brain but and there are some techniques
also which will help to come up with the correct
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guesses which we will study in next few lectures.
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But again for this particular problem, how
do we prove this statement?
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Again and again, we have to look at the base
case, so base case n equals to 1.
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So here, H(1) equals to 1 which is 2 power
1 minus 1 which is 1 which is right.
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So base is correct.
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So induction hypothesis say H(n) equals to
2 power n minus 1, inductive step, so
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we have to prove, so to prove, H of n plus
1 equals to 2 power n plus 1 minus 1.
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Now let us see, H(n) equals to sorry, H of
n plus 1 equals to by is given 2 times H(n)
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which is 1 plus 2 times 2 power n minus 1.
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This is again by induction hypothesis which
is 1 plus 2 times n plus 1 minus 2 which is
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2 times n plus 1 minus 1 and this is what
we had to prove.
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So H of n equals to 2 power n minus 1 for
all n greater than or equal to 1.
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Note that, this is not only a way to proving
the recurrence, this also if you go back to
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our previous video, this gives us a compact
form for the number of moves required for
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the Tower of Hanoi problem.
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So the Tower of Hanoi problem, therefore requires
2 power n minus 1 moves and we got it by first
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modeling it as a recurrence relation and then
solving the recurrence relation.
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Now how did you solve the recurrence relation,
we first guess the recurrence relation and
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then we prove that the guess is right.
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So this is how most of the counting problems
work, you first model it a recurrence relation
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and then you solve the recurrence relation.
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But this is all this is fine, if you can guess
the recurrence relations correctly.
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You first guess the recurrence relation and
then prove it using induction.
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The main question is how do you guess the
solution?
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And we will be doing this problem of how to
guessing the solution to the recurrence relation
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in the next video.
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We will see one of the techniques and in the
the next few videos we will see the other
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techniques.
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Thank you.