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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Best to see it with an example.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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bobbym wrote:

You do not have to stop there. When 31 does not work you try 30 which does. Algorithm is very fast.

Hm, okay, that does work. And it is very fast! Constant speed!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Hmmm, it is fast but its worst case might be quite slow!

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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bobbym wrote:

Best to see it with an example.

Okay

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'Humanity is still kept intact. It remains within.' -Alokananda

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**anonimnystefy****Real Member**- From: The Foundation
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I do not think so.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,304

Here is one where it works fine:

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Have you found an example in which it does not work quickly?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Heck no, why should I hurt my own feelings?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
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Well, in any case, I think the speed will be sufficient for any number.

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**bobbym****Administrator**- From: Bumpkinland
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Obviously the algorithm must get there the first time or by continually decrementing find the largest square that would work. But he might now have a problem with the subproblem of 3 squares.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
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Oh, yeah, you need to decrement there as well. Can you find an example when the decrement will not work immediately?

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**bobbym****Administrator**- From: Bumpkinland
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You mean where you would have to decrement the second one? No, not yet.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
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Actually, we need an example where both the first and the second number are decremented.

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**bobbym****Administrator**- From: Bumpkinland
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Why would we need an example like that?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Think about what would happen if only the second number needs decrementing. We could just keep the algorithm running until we get to the second number. We subtract it, the next largest square is the number we started with, all is well. But if both the first and the second numbers need decrementing then running the algorithm to the second number will still need the first number decremented (and it is now the second number).

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**bobbym****Administrator**- From: Bumpkinland
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I do not think that is true, at least if I am understanding it. The first two are an ordered pair. They are in >= order.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
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Not necessarily. If we need to, we can find the second number first!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Before you get the first one?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
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Yes. Imagine that the quartuplet for n is (m,k-1,...,...) and that k^2 is the largest square below n-m^2. Then we won't find anything until we run the algorithm for k-1. Then the next largest square below it is n^2 and we get exactly what we needed.

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**bobbym****Administrator**- From: Bumpkinland
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The algorithm will run sequentially. It is possible to get an expression for the second square by running it twice symbolically.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,507

What?

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**bobbym****Administrator**- From: Bumpkinland
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The way you can run it with numbers, you can run it with letters. Say n = a; Then the second square according to the recurrence is:

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Ok, so?

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**bobbym****Administrator**- From: Bumpkinland
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I am not sure but there it is. We were talking getting the second largest square without getting the first weren't we?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,507

No. I was talking about getting the second number from the answer before the first.

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