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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Looking at your transformation, is there any rule to determine how it is to be done?

**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
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Didn't you read the definition of the transformation in #2?

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**anonimnystefy****Real Member**- From: The Foundation
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Do you understand the transformation now?

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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
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Not yet. I will need more time to look at it.

**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****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 88,723

Hi;

I have not even verified the truth that your transformation has the same D number.

**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.**

**Online**

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,603

Do you have doubt in me?

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
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Not in you but mathematics is a prove it sort of endeavor. Everyone makes mistakes.

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

It's easy for to show that making a loop disappear does nothing to the drawability of a graph.

It is also easy to prove the same thing for deleting any cycle.

And it is easy to prove the general statement by showing that it works for any three general nodes.

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**bobbym****Administrator**- From: Bumpkinland
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First, I like to convince myself of an ideas validity. This can be done empirically.

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

Than do that.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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One thing at a time. Is it not very, very late over there?

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

Yes, yes it is.

Now convince yourself that what I conjectured is true.

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

For your example:

Drawability of the left is 5 and of the right is 6. Drawability property does not hold under the transformation.

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

My mistake.First, I didn't draw it correctly, because there should remain only one of those on the right picture, and that's the lefter one.

Second, how did you manage to get the drawability when not all nodes are given?

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**bobbym****Administrator**- From: Bumpkinland
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I have a technique that seems to be working.

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

i edited #2.

and what technique?

on the pic i posted that shows the definition of the transformation (that's the second picture) those edges that have an open end have nodes on that side but on those nodes that are not shown you might have more nodes with more edges.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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I use a pen and paper!

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

so what?there's still no way to calculate the drawability of the given graphs.

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**bobbym****Administrator**- From: Bumpkinland
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I disagree. 5 lines do the first one and 6 do the second one.

I am having connection problems so I am just going to shut down and get some rest see you later.

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**anonimnystefy****Real Member**- From: The Foundation
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hi bobbym

it's 5 on both.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Which both? The old ones or the new ones?

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

The drawability is 5 on both sides of the second pic I posted.

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

Agreed, the new picture has 5 on both sides. Now how do we know that always holds?

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

Like I said, we could check that out holds for three arbitrary nodes on a graph with k, l and m edges sticking out of those three nodes.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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How about more nodes than 3?

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