# toughy math with averages

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#1
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Posted 30 June 2001 - 12:58 PM

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#2
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Posted 30 June 2001 - 01:20 PM

A2 = (a+b)/2

TO FIND a

a = 2(A2)-b

TO FIND b

b = 2(A2)-a

I might be wrong, but it looks right (to me!)

I dont think there is any way to do this without knowing the other other variable in the equation.

for A1 = (a+b+c+d)/4, you would have to do something like this

a = 4(A1)-b-c-d

I hope this is corect, and helps. It is possible that i just answered a question that you did not ask, so if I missunderstood the question, let me know.

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#3
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Posted 30 June 2001 - 02:03 PM

I am thinking that I would have to use the A''s in pairs or trios. Like to find a maybe do something with A2 and A4?

Or i am thinking maybe since i have A2 and A5 I could use those and then A4 to get it somehow? This should get a and c both if its possible.

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#4
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Posted 30 June 2001 - 02:46 PM

Say we have a resultant vector y = |A2 A3 A4 A5|, an original vector x = |a b c d| and a coefficient matrix

|1 1 0 0|

A = |0 1 0 1|

|1 0 1 0|

|0 0 1 1|

such that y = Ax, then A can be decomposed into lower and upper matrices L and U:

|1 0 0 0 | |U11 U12 U13 U14|

L = |L21 1 0 0 | , U = |0 U22 U23 U24|

|L31 L32 1 0 | |0 0 U33 U34|

|L41 L42 L43 1 | |0 0 0 U44|

such that A = LU. A can be shown to be given as

A(i,j) = S(k=1,j){L(i,k)U(k,j)}. Solving for the important terms (A(1,1), A(1,2), A(2,2), A(2,4), A(3,1), A(3,3), A(4,3), A(4,4)), we obtain the matrices

|1 0 0 0| |1 1 0 0|

L = |0 1 0 0| , U = |0 1 0 1|

|1 0 1 0| |0 0 1 0|

|0 0 0 1| |0 0 0 1|

(I may have made an error, so please validate this result by multiplication). We now obtain

|A2| |1 0 0 0| |1 1 0 0| |a|

|A3| = |0 1 0 0| * |0 1 0 1| * |b|

|A4| |1 0 1 0| |0 0 1 0| |c|

|A5| |0 0 0 1| |0 0 0 1| |d|

or |y| = |L| * |U| * |x|

The solution can now be achieved using y = L(Ux) by letting Ux = z, y = Lz and solving for z. Then solve for x in Ux = z.

Edited by - Oluseyi on June 30, 2001 9:51:21 PM

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#5
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Posted 30 June 2001 - 02:57 PM

the vector [[ a1 ] [ a2 ] [ a3 ] [ a4 ] [ a5 ]] = [T]* [[ a ] [ b ] [ c ] [ d ]]

...with T = [[1/4 1/4 1/4 1/4]

[1/2 1/2 0 0 ]

[ 0 1/2 0 1/2]

[1/2 0 1/2 0 ]

[ 0 0 1/2 1/2]

If there is a way to reverse the calculations and find out what the source vector was from the transformed vector, T has to be inverted, which means it must be square. We'll just drop the first row (A1) and use that as the new matrix T.

Calculating det(t), we find that it is = 0. This is because the rows can be added such that they become multiples of each other, even if we left the first row in... and you can't invert a matrix whose determinate is zero.

What you ask cannot be done without knowing at least one of the variables. End of story. End of Linear Algebra proof.

...and I'm moderately sure that the above post is wrong... at any rate, one of us has to be, and I'd love a third opinion stating who it is... anyone a math prof?

--Tr][aD--

Edited by - TrIaD on June 30, 2001 10:05:57 PM

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#6
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Posted 30 June 2001 - 03:07 PM

Althought it -seems- like you have 5 equations and only 4 unknowns, since the equations are not linearly independent, there is no way to find a unique solution.

This is easily proved by guassian elimination: Take the matrix equation (forgive my crappy matrix drawing):

|1 1 1 1||a| = |2*A1|

|1 1 0 0||b| = |2*A2|

|1 0 1 0||c| = |2*A4|

|0 1 0 1||d| = |2*A3|

|0 0 1 1||-| = |2*A5|

To do guassian elimination we change that into the augmented matrix:

|1 1 1 1 || 2*A1|

|1 1 0 0 || 2*A2|

|1 0 1 0 || 2*A4|

|0 1 0 1 || 2*A3|

|0 0 1 1 || 2*A5|

Which reduces to:

|01 01 01 01 || 2*A1|

|00 00 -1 -1 || 2*A2 - 2*A1|

|00 -1 00 -1 || 2*A4 - 2*A1|

|00 01 00 01 || 2*A3|

|00 00 01 01 || 2*A5|

-->

|01 01 01 01 || 2*A1|

|00 00 -1 -1 || 2*A2 - 2*A1|

|00 -1 00 -1 || 2*A4 - 2*A1|

|00 00 00 00 || 2*A3 + 2*A4 - 2*A1|

|00 00 00 00 || 2*A5 + 2*A2 - 2*A1|

As you can see, this gives the l.i. equations:

a + b + c + d = A1

c + d = A1 - A2

b + d = A1 - A4

Which, since there are four unknowns and only three equations, has infinitely many solutions.

What you CAN say, however, is that unless (2*A3 + 2*A4 - 2A1 = 0), and (2*A5 + 2*A2 - 2*A1 = 0), the system is inconsistant and there are no solutions; but that's pretty obvious anyway.

If you want to find any particular solution, let's just say, let d be some arbitrary variable x. Then the solution for any particular x would be:

d = x

b = 2*A1 - 2*A4 - x

c = 2*A1 - 2*A2 - x

a = -2*A1 + 2*A2 + 2*A4 + x

Hope that helps.

Edited by - Shadow Mint on June 30, 2001 10:11:17 PM

Edited by - Shadow Mint on June 30, 2001 10:13:55 PM

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#7
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Posted 30 June 2001 - 03:08 PM

quote:Original post by TrIaD

...and I''m moderately sure that the above post is wrong... at any rate, one of us has to be, and I''d love a third opinion stating who it is... anyone a math prof?

Actually, I''m moderately sure myself. Logic seems to dictate that you do need to know

*at least*one of the unknowns, but I''m too lazy to work it all out and verify it. Plus, I may have fumbled on the decomposition. In any case, someone will come along and provide the solution (or definitive proof of a lack of one) in a flash of brilliance.

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#8
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Posted 30 June 2001 - 04:38 PM

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#9
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Posted 30 June 2001 - 05:14 PM

and it isnt that hard

look here

A2 = (a+b)/2

can be rewritten as

a = (2*A2)-b

then you have

A4 = (a+c)/2

can be rewritten as

a = (2*A4)-c

so you can then set the two a equations equal to each other

so that

(2*A2)-b = (2*A4)-c

if you know A2 and A4 then you can get b and c

you can use the same method for the other variables

"I pity the fool, thug, or soul who tries to take over the world, then goes home crying to his momma."

- Mr. T

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#10
Members - Reputation: **2076**

Posted 30 June 2001 - 06:34 PM

quote:Original post by ncsu121978

so you can then set the two a equations equal to each other

so that

(2*A2)-b = (2*A4)-c

if you know A2 and A4 then you can get b and c

you can use the same method for the other variables

Substitute any two values you like for A2 and A4 and see what you get. (2*A2) and (2*A4) will be constants, so 10 and 8 are as good choices as any. Given

10 - b = 8 - c, what are b and c?

Not as easy as you thought, eh? There''s no answer.

quote:

"I pity the fool, thug, or soul who tries to take over the world, then goes home crying to his momma."

- Mr. T

Aye. Me too.

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#12
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Posted 01 July 2001 - 04:13 PM

Why do I belabor these things? Oh, well...

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#14
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Posted 02 July 2001 - 12:01 PM

A2 = (a+b)/2

A3 = (b+d)/2

A4 = (a+c)/2

A5 = (c+d)/2

Without much effort:

a = d + 2*(A2-A3)

a = d + 2*(A4-A5)

b = c + 2*(A2-A4)

b = c + 2*(A3-A5)

All this means is that A2-A3 = A4-A5 and A2-A4 = A3-A5. Which means that in fact, you really only had three distinct equations to begin with.

Therefore, you have three equations and four unknowns, which makes solving it impossible. Simple.

For the record, there is no such thing as a compression algorithm which can compress everything - think about it.

Edited by - Beer Hunter on July 2, 2001 7:04:12 PM

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#15
Members - Reputation: **122**

Posted 02 July 2001 - 03:47 PM

// God.c

void main() {

WORLD Earth;

LIFE People = Earth.CreateLife(HUMAN);

GiveHope(&People);

delete Earth;

EvilCackle();

}

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#16
Members - Reputation: **122**

Posted 03 July 2001 - 10:30 AM

--Tr][aD--

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#18
Members - Reputation: **122**

Posted 04 July 2001 - 03:44 AM

I have other methods of compression tho so this isn''t a big loss. Thats the big part of BSZIP, being able to use multiple methods of compression.

// God.c

void main() {

WORLD Earth;

LIFE People = Earth.CreateLife(HUMAN);

GiveHope(&People);

delete Earth;

EvilCackle();

}

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#20
Members - Reputation: **712**

Posted 04 July 2001 - 11:57 AM

This...

222222 313346 2222222 44444 33333

should compress to this...

62 06313346 72 54 53

...according to the windows help. Check under BITMAPINFOHEADER, I think.

Huffman compression is good for using in conjunction with other algos. On its own, it averages out at about 30% compression, which is quite powerful. It works on a per character basis, replacing the most common characters with shorter bit strings than the less used ones. And it can generate the best bit strings to use, too.

And now, an explanation as to why there are no magic compressors.

Let''s say a compresser

**could**compress every file of x bits in length. There are 2^x possible files of x bits in length. Each one of these has to compress to a file smaller than x bits. Since there are only 2^x - 1 possible files of less than x bits, at least two files of x bits in length must compress to the same output file. But these two identical files cannot decompress to two different files, and this is a contradiction.

Therefore, you cannot compress all files of x bits in length, where x = anything.