En = Fn(B, C) + ???Fn(D)???

[NotLKH]

Useing the information as presented in the attached Image,
Can you finish the Identities for En such that in the 9 Row Equations
En Em, Either:

1) All the D Elements sum to Zero

or

2) You result in the Identity D₁ + D₃ + D₅ = D₂ + D₄ + D₆

???

{ the "Such That" info displays what the current E equations sum to, useing the "current" values of En. }

[NotLKH]

The following extra identities might prove helpful, but I tend to think they
might be a last resort:

D₁ = – B₁ – B₂ – B₃ – B₄ – B₅ – B₆ + C₁ + C₂ + C₃ + C₄ + C₅ + C₆ – D₃ – D₅
D₂ = – B₁ – B₂ – B₃ – B₄ – B₅ – B₆ + C₁ + C₂ + C₃ + C₄ + C₅ + C₆ – D₄ – D₆

And,I have some exact identities for E, but they are dependent on E₆, which I'm trying to remove:

E₁ = – B₃ – B₄ + C₁ + C₂ + C₅ + C₆ – D₃ + D₆ – E₆
E₂ = – B₁ – B₂ – B₄ + –2*B₅ – B₆ + C₁ + C₂ + 2*C₃ + C₄ + C₆ – D₄ – D₅ – D₆ + E₆
E₃ = – B₃ – B₆ + C₁ + C₂ + C₄ + C₅ – E₆
E₄ = – B₁ + B₃ – C₁ + C₃ + D₃ – D₆ + E₆
E₅ = B₁ + B₄ + B₅ + B₆ – C₂ – C₃ + D₄ + D₅ + D₆ – E₆

Finally, every identity in this specific post is Rotatable Mod 6, ie...
Given:

E₃ = – B₃ – B₆ + C₁ + C₂ + C₄ + C₅ – E₆
E₄ = – B₁ + B₃ – C₁ + C₃ + D₃ – D₆ + E₆

We can also produce the following valid Identities by incrimenting the Index values by 1, {Then remap 7 to 1}

E₄ = – B₄ – B₁ + C₂ + C₃ + C₅ + C₆ – E₁
E₅ = – B₂ + B₄ – C₂ + C₄ + D₄ – D₁ + E₁

[kedaman]

What's this for? What does the diagram illustrate? just curious

[siyan]

i somehow think that if Lou is asking a question about math, we're all doomed.

-C

[sql_lall]

Part 2) The identity is 'given', so i don't know why you have to 'reach' it, as you already know it.

Also, what do the different shapes mean (like what Does the "B1 + C1" in the oval mean??? or all the other equations in the shapes

[NotLKH]

Originally posted by sql_lall
Part 2) The identity is 'given', so i don't know why you have to 'reach' it, as you already know it.

Good Question.
What I'm trying to do {Actually, I think I've done it} is balance an
arrangement of numbers in a symmetrical fashion. At the time I posted this thread, I was trying to balance every row so that each one adds to A + Sum(C), but I was unbalanced in terms of D. So, If I were to add elements of D to E such that every row resulted in an equation of D that equaled 0, then I would have succeeded. Certainly, if in one cell I had a +D₁ and in the same row there was a different cell containing -D₁,
then the sum of the D₁'s would automatically zero out.

However, since we also have the identity SumD_odd = SumD_even then we could zero +D₁ out by haveing -D₂ -D₄ -D₆ +D₃ +D₅ in another cell of the same row.

So, its not a matter of proveing SumD_odd = SumD_even, its a matter of useing SumD_odd = SumD_even to offer a second way of zero'ing the D's out.

Originally posted by sql_lall
Also, what do the different shapes mean (like what Does the "B1 + C1" in the oval mean??? or all the other equations in the shapes

Originally posted by kedaman
What's this for? What does the diagram illustrate? just curious

AhHa! 2 More good questions!
B1 is a variable representing a number. So is C1, and so is Every B and C, so when you see a cell with B1 + C1 then it means add B1 with C1.

Other than that, At this moment,
I'm afraid I'm unable to easily answer these questions.


Give me a day or two, I might have a satisfactory answer.

-Lou

[NotLKH]

Originally posted by NotLKH

....Actually, I think I've done it.....

Yep. I just thouroughly confirmed it, and appending the D offsets for E that Zero the D's out of every row equation, I've acheived the following Identities of E based on B, C, and D:

E₁= –B₄ + C₂ + C₆ + K₁ * (D₆ – D₃) + K₂ * (D₁ – D₄)
E₂= –B₅ + C₃ + C₁ + K₁ * (D₂ – D₅) + K₃ * (D₁ – D₄)
E₃= –B₆ + C₄ + C₂ + K₂ * (D₂ – D₅) + K₃ * (D₃ – D₆)
E₄= –B₁ + C₅ + C₃ + K₁ * (D₃ – D₆) + K₂ * (D₄ – D₁)
E₅= –B₂ + C₆ + C₄ + K₁ * (D₅ – D₂) + K₃ * (D₄ – D₁)
E₆= –B₃ + C₁ + C₅ + K₂ * (D₅ – D₂) + K₃ * (D₆ – D₃)

Where K₁+K₂+K₃=1

So, if you check every equation, you will see that all the D's zero out, but you must remember:

D₁+ D₃+ D₅=D₂+ D₄+ D₆ =
– B₁ – B₂ – B₃ – B₄ – B₅ – B₆ + C₁ + C₂ + C₃ + C₄ + C₅ + C₆

Why go to all the Bother?
Stay Tuned...

[kedaman]

sometimes.. its more than obvious that the answer always is in the question

[sql_lall]

Glad to hear you succeeded (i'm not sure what at, but it's always a good feeling )

Just wondering what this was for (if anything). I mean, it looks cool, but i was wondering if it has a purpose.