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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,646

Best to see it with an example.

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,600

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
- Posts: 88,646

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.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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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'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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I do not think so.

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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: 88,646

Here is one where it works fine:

**In mathematics, you don't understand things. You just get used to them.**

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**anonimnystefy****Real Member**- From: The Foundation
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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
- Posts: 88,646

Heck no, why should I hurt my own feelings?

**In mathematics, you don't understand things. You just get used to them.**

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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
- Registered: 2009-04-12
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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.

**In mathematics, you don't understand things. You just get used to them.**

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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
- Registered: 2009-04-12
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You mean where you would have to decrement the second one? No, not yet.

**In mathematics, you don't understand things. You just get used to them.**

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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
- Registered: 2009-04-12
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Why would we need an example like that?

**In mathematics, you don't understand things. You just get used to them.**

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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
- Registered: 2009-04-12
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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.

**In mathematics, you don't understand things. You just get used to them.**

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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?

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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.

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**anonimnystefy****Real Member**- From: The Foundation
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What?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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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:

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**anonimnystefy****Real Member**- From: The Foundation
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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?

**In mathematics, you don't understand things. You just get used to them.**

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**anonimnystefy****Real Member**- From: The Foundation
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No. I was talking about getting the second number from the answer before the first.

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