 experchange on solving recurrences

 Daniel Bastos (03-21-17, 09:13 PM)
Can you suggest a book that concerns precisely on solving recurrences
given my background? You'll see my background as I try my best at this
one. (I know Concrete Mathematics by Knuth et al, but as much I liked
their presentation of the Josephus problem, I think that book is too
advanced for me.) Thank you!

(*) Problem

Solve

T(n) = 1 if n = 0
= 2 T(n - 1) + 1 if n > 0

Remark. The only way I know how to solve this is by guessing. Here's
how I do it.

Solution. First I look at say T(8).

T(8) = 2 T(7) + 1
= 2 (2 T(6) + 1) + 1
= 2 (2 (2 T(5) + 1) + 1) + 1
= 2 (2 (2 (2 T(4) + 1) + 1) + 1) + 1

So I see a pattern emerging. I ``went down 4 levels'' and I found this:

= 2 (2 (2^2 T(4) + (1*2 + 1)) + 1) + 1
= 2 (2^3 T(4) + (1*2*2 + 1*2) + 1) + 1
= 2^4 T(4) + 1*2*2*2 + 1*2*2 + 1*2 + 1
= 2^4 T(4) + 2^3 + 2^2 + 2^1 + 2^0

So I must now discover what is Sum(2^i for i=0..n-1). Programming
verification (see below) makes clear that we don't actually need to find
that because it gets easy to see --- numerically --- that we have a case
for 2^n - 1.

Somehow

2^n - 1 = 2^n + Sum(2^i for i=0..n-1).

Thank you.

(*) Programming verification

def xrange(beg, end):
# Python's range(x,y) goes up to y - 1. We usually want to be
# inclusive.
return range(beg, end + 1)

def T0(n):
if n == 0:
return 1
elif n == 1:
return 3
else:
return 2**n + sum([2**i for i in xrange(0, n - 1)])

def T0r(n):
if n == 0:
return 1
else:
return 2 * T0r(n - 1) + 1

>>> [(t0(i), t0r(i)) for i in range(8)]

[(1, 1), (3, 3), (7, 7), (15, 15), (31, 31), (63, 63), (127, 127), (255, 255)]
 Ed Prochak (03-22-17, 03:35 PM)
On Tuesday, March 21, 2017 at 3:13:37 PM UTC-4, Daniel Bastos wrote:
[..]
> else:
> return 2 * T0r(n - 1) + 1
> [(1, 1), (3, 3), (7, 7), (15, 15), (31, 31), (63, 63), (127, 127), (255, 255)]

I am not sure what your question is.

You did a nice analysis.
Both versions T0r() and T0() are recursive.
I note you unrolled the loop one step only in the T0r() method
> elif n == 1:
> return 3

So is there something you think is wrong here?

Ed
 David Brown (03-22-17, 03:58 PM)
On 21/03/17 20:13, Daniel Bastos wrote:
[..]
> = 2 (2^3 T(4) + (1*2*2 + 1*2) + 1) + 1
> = 2^4 T(4) + 1*2*2*2 + 1*2*2 + 1*2 + 1
> = 2^4 T(4) + 2^3 + 2^2 + 2^1 + 2^0

That's good so far.

At this point, it should be rather obvious that the solution involves
2^n. Let's look at some actual values:

n t(n) 2^n
-------------------
0 1 1
1 3 2
2 7 4
3 15 8
4 31 16

It is rather obvious now that t(n) = 2^(n + 1) - 1.

Obvious is not a proof, of course, but it's a lot easier to prove a
result if you know what you want to prove!

Prove the assertion by induction. It is easy to see that the result
works for 0.

Assume that it works for i up to n-1, where n > 0. Then

t(n) = 2 * t(n - 1) + 1
= 2 * (2^(n - 1 + 1) - 1) + 1 // By induction assumption
= 2 * 2 ^ n - 2 + 1
= 2 ^ (n + 1) - 1

QED.
[..]
 Daniel Bastos (03-24-17, 12:06 AM)
Ed Prochak <edprochak> writes:

> On Tuesday, March 21, 2017 at 3:13:37 PM UTC-4, Daniel Bastos wrote:

Something must be wrong here. Otherwise -1 = Sum(2^i for i=0..n-1).
And I don't know where I went wrong, but there's something wrong here.

> I am not sure what your question is.

My question was the first sentence. I showed my struggle to resolve
problems like this in order to help someone suggest me a book. I lack
strategies to resolve recurrences. My only strategy is to unfold the
recursion and see a pattern. That's all I can do. I'm getting the
feeling that it all boils down to solving a lot of sums, products and
things like that. So it's the kind of work that involves studying
discrete mathematics, calculus and whatnot.

> You did a nice analysis.

Thank you!

> Both versions T0r() and T0() are recursive.

Both? T0() is not recursive. It doesn't mention T0 in its body.

> I note you unrolled the loop one step only in the T0r() method
>> elif n == 1:
>> return 3

That was an accident. In one of my attempts I used (n - 2), so I had to
worry about not going negative. Thanks for pointing it out.
 Daniel Bastos (03-24-17, 12:11 AM)
David Brown <david.brown> writes:

> On 21/03/17 20:13, Daniel Bastos wrote:
> That's good so far.
> At this point, it should be rather obvious that the solution involves
> 2^n.

Yes.

> Let's look at some actual values:
> n t(n) 2^n
> -------------------
> 0 1 1
> 1 3 2
> 2 7 4
> 3 15 8
> 4 31 16
> It is rather obvious now that t(n) = 2^(n + 1) - 1.

Okay. :-)

> Obvious is not a proof, of course, but it's a lot easier to prove a
> result if you know what you want to prove!
> Prove the assertion by induction. It is easy to see that the result
> works for 0.
> Assume that it works for i up to n-1, where n > 0. Then
> t(n) = 2 * t(n - 1) + 1
> = 2 * (2^(n - 1 + 1) - 1) + 1 // By induction assumption
> = 2 * 2 ^ n - 2 + 1
> = 2 ^ (n + 1) - 1
> QED.

I like constructions. Guessing and proving requires insight. I don't
have a whole lot of that. Thanks for your attention.
 Daniel Bastos (03-24-17, 12:16 AM)
Daniel Bastos <dbastos> writes:

> Ed Prochak <edprochak> writes:
> Something must be wrong here. Otherwise -1 = Sum(2^i for i=0..n-1).
> And I don't know where I went wrong, but there's something wrong here.

After seeing David Brown's message: I must have got my counters wrong
since I thought I verified the result numerically. It's not 2^n - 1.
It's 2^(n+1) - 1. The equation is therefore

2^(n + 1) - 1 = 2^n + Sum(2^i for i=0..n-1).

Thanks.
 David Brown (03-24-17, 10:48 AM)
On 23/03/17 23:11, Daniel Bastos wrote:
> David Brown <david.brown> writes:
> Yes.
> Okay. :-)
> I like constructions. Guessing and proving requires insight. I don't
> have a whole lot of that. Thanks for your attention.

It may look like guessing, but it is a matter of knowing the type of
difference equation - a difference equation that involves multiplying by
k at each step is going to involve k^n somewhere, just as one that
involves adding k at each step is going to involve k*n somewhere. So
you can start with that term, with some flexibility on the multiplier,
and pull that bit out. Then you can work on the parts that are left,
and see what extras are needed to make everything correct. You don't
have to start with a lucky guess of the correct answer.
 David Brown (03-24-17, 10:53 AM)
On 23/03/17 23:16, Daniel Bastos wrote:
> Daniel Bastos <dbastos> writes:
> After seeing David Brown's message: I must have got my counters wrong
> since I thought I verified the result numerically. It's not 2^n - 1.
> It's 2^(n+1) - 1. The equation is therefore
> 2^(n + 1) - 1 = 2^n + Sum(2^i for i=0..n-1).
> Thanks.

I can't understand why you keep wanting to put this "Sum(...)" bit in.
It does not simplify anything - it leaves the result as a mess.

Sum(2^i for i = 0 .. n-1) = 2^n - 1

This is a very well known result, and is easily verified. If you don't
see it, write out a table of the first 4 or 5 values here. Then do the
same, but write the numbers in binary rather than decimal. The pattern
will leap out at you. And you can practice your proof by induction
skills by proving it.
 Daniel Bastos (07-10-19, 04:29 PM)
> On 21/03/17 20:13, Daniel Bastos wrote:
>> Can you suggest a book that concerns precisely on solving recurrences
>> given my background? You'll see my background as I try my best at this
>> one. (I know Concrete Mathematics by Knuth et al, but as much I liked
>> their presentation of the Josephus problem, I think that book is too
>> advanced for me.) Thank you!

There's section 5.3, ``Solving Recurrences'', in ``Discrete
Mathematics'' by James L. Heine, Jones and Bartlett Publishers, 1996,
page 279. It shows a straightforward technique for solving
recurrences such as the one discussed in this thread. If you have no
technique at all, solving these simpler types of recursion will
definitely mean something to you.

>> (*) Problem
>> Solve
>> T(n) = 1 if n = 0
>> = 2 T(n - 1) + 1 if n > 0

Rewriting it to:

a(0) = 1
a(n) = 2 a(n - 1) + 1

--8<---------------cut here---------------start------------->8---
Notation. Note that ``a'' is a function whose argument is the index
of the sequence. So a(n) is the nth element of the sequence.
--8<---------------cut here---------------end--------------->8---

An easy way to expose the construction of the technique is to consider
a particular case of the recurrence and generalize. So consider a(5):

a(5) = 2 a(4) + 1

Now write an equation for 2 a(4):

2 a(4) = 2(2a(3) + 1)
= 2^2 a(3) + 2

Now write an equation for 2^2 a(3):

2^2 a(3) = 2^2 (2a(2) + 1)
= 2^3 a(2) + 2^2

Now write an equation for 2^3 a(2):

2^3 a(2) = 2^3 (2a(1) + 1)
= 2^4 a(1) + 2^3

No write an equantion for 2^4 a(1):

2^4 a(1) = 2^4 (2a(0) + 1)
= 2^5 a(0) + 2^4

Since a(0) is the base case of the recurrence, we're done. Now add
all the left hand side of these equations and add all the right hand
sides. We get

a(5) + 2 a(4) + 2^2 a(3) = 2^3 a(2) + 2^4 a(1) = 2 a(4) + 1 +
2^2 a(3) + 2 + 2^3 a(2) + 2^2 + 2^4 a(1) + 2^3 + 2^5 a(0) + 2^4

Notice now that all terms of the left hand side cancel out, except
a(5). So we get

a(5) = 1 + 2 + 2^2 + 2^3 + 2^4
= 2^0 + 2^1 + 2^2 + 2^3 + 2^4
= Sum 2^i from i = 0 to i= 5 - 1

This sum on the right hand side is (2^{5 + 1} - 1) / (2 - 1). So

a(5) = 2^6 - 1 = 63.

The computation of this particular case makes it clear that in
general

a(n) = 2^(n + 1) - 1

is the desired answer. The method is called ``the cancellation
technique'' by James L. Hein. It works for any recurrence of the form

a(0) = P(0)
a(n) = Q(n) a(n - 1) + P(n),

where P(n) and Q(n) are constants.
 David Brown (07-10-19, 05:25 PM)
On 10/07/2019 16:29, Daniel Bastos wrote:
[..]
> general
> a(n) = 2^(n + 1) - 1
> is the desired answer.

No, it most certainly does not.

What you have done here is found a likely candidate for an answer. Now
you can use that and prove it works, by induction. Until then, you only
have a guess that looks good in some tests, but you have no confirmed