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jstevh@msn.com
science forum Guru
Joined: 21 Jan 2006
Posts: 951
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 Posted: Wed Jun 28, 2006 11:49 pm Post subject:
SF: Tandem factorization back, new approach
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Sorry but this is how the development process works with me.
I went down one path before, found out it didn't work like I thought it
did, so I called crap crap and started feeling sorry for myself. Then
I thought of another approach, so here we go, again.
The idea here is to use a tandem factorization of two numbers where
S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y))
and
T = (k_1*sqrt(x) - k_2*sqrt(y))*(k_3*sqrt(x) - k_4*sqrt(y))
which means that
2*sqrt(x)*k_1 = f_1 + g_1
2*sqrt(y)*k_2 = f_1 - g_1
2*sqrt(x)*k_3 = f_2 + g_2
2*sqrt(y)*k_4 = f_2 - g_2
and the initial approach I put forward was with a target composite T to
factor you pick some surrogate S, and I had various ideas for how you
then find all the other variables--when for that idea to work, you need
to already know the factorization of S and T.
So that was when I was ready to toss all of this and declared it crap.
However, why pick S?
Using the first two equations I have
S - T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy)
and
S+T = 2*k_1*k_3*x + 2*k_2*k_4*y
and focusing on that second equation if I let
k_1*k_3 = A
k_2*k_4 = B
and pick squares for x and y, then I determine S, if T is the target
factorization:
S = 2*k_1*k_3*x + 2*k_2*k_4*y - T
so
S = 2*A*x + 2*B*y - T
and you can use those equations to solve out two of the k's, and
substitute out S, to relate the remaining two k's in an equation that I
hope will give the approach that works.
| Quote: | From that equation the idea is you figure out how to pick A and B such
that everything is an integer, and that gives you S, with S, you have |
integer k's and an incidental factorization of T, if it works.
I'll have to work through the equations and see if it is another
crapshoot.
James Harris |
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Justin
science forum Guru Wannabe
Joined: 28 Apr 2005
Posts: 204
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| Posted: Thu Jun 29, 2006 12:58 am Post subject:
Re: SF: Tandem factorization back, new approach
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In sci.math jstevh@msn.com wrote:
: Sorry but this is how the development process works with me.
It's funny, I was about to post a response to your failure note suggesting
that I'd give it about a week until you were back in your usual way
blathering out nonsense.
: I'll have to work through the equations and see if it is another
: crapshoot.
Why bother? You've never worked through any of your other ideas.
Justin |
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UpChunky
science forum beginner
Joined: 03 Jun 2005
Posts: 36
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| Posted: Thu Jun 29, 2006 1:18 am Post subject:
Re: Tandem factorization back, new approach
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<jstevh@msn.com> wrote in message
news:1151538596.762044.301010@i40g2000cwc.googlegroups.com...
| Quote: | Sorry but this is how the development process works with me.
I went down one path before, found out it didn't work like I thought it
did, so I called crap crap and started feeling sorry for myself. Then
I thought of another approach, so here we go, again.
|
Forget about it. It is crap. Total crap, the foundation of it is indeed
"crap".
Try an entirely new approach, like squares, or cube roots. |
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fishfry
science forum Guru Wannabe
Joined: 29 Apr 2005
Posts: 299
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| Posted: Thu Jun 29, 2006 2:22 am Post subject:
Re: SF: Tandem factorization back, new approach
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In article <1151538596.762044.301010@i40g2000cwc.googlegroups.com>,
jstevh@msn.com wrote:
| Quote: | Sorry but this is how the development process works with me.
I went down one path before, found out it didn't work like I thought it
did, so I called crap crap and started feeling sorry for myself. Then
I thought of another approach, so here we go, again.
|
You left out the step where you insult everyone and threaten to bring
down the hammer. |
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Salami Man
science forum beginner
Joined: 05 Jun 2006
Posts: 38
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| Posted: Thu Jun 29, 2006 2:48 am Post subject:
Re: SF: Tandem factorization back, new approach
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"fishfry" <BLOCKSPAMfishfry@your-mailbox.com> wrote in message
news:BLOCKSPAMfishfry-
| Quote: | You left out the step where you insult everyone and threaten to bring
down the hammer.
|
.... and the stock market crash, plague of locusts, end of the world, etc. |
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Ed Weir (ComCast)
science forum beginner
Joined: 04 Feb 2006
Posts: 37
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| Posted: Thu Jun 29, 2006 3:03 am Post subject:
Re: SF: Tandem factorization back, new approach
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"Salami Man" <SalamiMan@anon.com> wrote in message
news:gcHog.192$I%1.67@fe07.lga...
| "fishfry" <BLOCKSPAMfishfry@your-mailbox.com> wrote in message
| news:BLOCKSPAMfishfry-
| > You left out the step where you insult everyone and threaten to bring
| > down the hammer.
|
| ... and the stock market crash, plague of locusts, end of the world, etc.
Oh well, might as well...
http://www.foxnews.com/story/0,2933,201355,00.html
before someone else does... what the hell |
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UpChunky
science forum beginner
Joined: 03 Jun 2005
Posts: 36
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| Posted: Thu Jun 29, 2006 3:11 am Post subject:
Re: SF: Tandem factorization back, new approach
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"Ed Weir (ComCast)" <Anon@Maus.duh> wrote in message
news:pNKdnexCHZZi3z7ZnZ2dnUVZ_sSdnZ2d@comcast.com...
| Quote: | "Salami Man" <SalamiMan@anon.com> wrote in message
news:gcHog.192$I%1.67@fe07.lga...
| "fishfry" <BLOCKSPAMfishfry@your-mailbox.com> wrote in message
| news:BLOCKSPAMfishfry-
| > You left out the step where you insult everyone and threaten to bring
| > down the hammer.
|
| ... and the stock market crash, plague of locusts, end of the world,
etc.
Oh well, might as well...
http://www.foxnews.com/story/0,2933,201355,00.html
before someone else does... what the hell
|
That means if JSH gets it right this time, then the world won't end ? |
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jstevh@msn.com
science forum Guru
Joined: 21 Jan 2006
Posts: 951
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| Posted: Fri Jun 30, 2006 12:14 am Post subject:
Re: SF: Tandem factorization back, new approach
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jstevh@msn.com wrote:
| Quote: | Sorry but this is how the development process works with me.
I went down one path before, found out it didn't work like I thought it
did, so I called crap crap and started feeling sorry for myself. Then
I thought of another approach, so here we go, again.
The idea here is to use a tandem factorization of two numbers where
S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y))
and
T = (k_1*sqrt(x) - k_2*sqrt(y))*(k_3*sqrt(x) - k_4*sqrt(y))
which means that
2*sqrt(x)*k_1 = f_1 + g_1
2*sqrt(y)*k_2 = f_1 - g_1
2*sqrt(x)*k_3 = f_2 + g_2
2*sqrt(y)*k_4 = f_2 - g_2
and the initial approach I put forward was with a target composite T to
factor you pick some surrogate S, and I had various ideas for how you
then find all the other variables--when for that idea to work, you need
to already know the factorization of S and T.
So that was when I was ready to toss all of this and declared it crap.
However, why pick S?
Using the first two equations I have
S - T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy)
and
S+T = 2*k_1*k_3*x + 2*k_2*k_4*y
and focusing on that second equation if I let
k_1*k_3 = A
k_2*k_4 = B
and pick squares for x and y, then I determine S, if T is the target
factorization:
S = 2*k_1*k_3*x + 2*k_2*k_4*y - T
so
S = 2*A*x + 2*B*y - T
and you can use those equations to solve out two of the k's, and
substitute out S, to relate the remaining two k's in an equation that I
hope will give the approach that works.
|
Ok, so making that substitution into the first equation I have
2A*x + 2*B*y - 2T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy)
and now I can solve out two of the k's, and divide by 2 to get
A*x + B*y - T =(k_2*(A/k_1) + k_1*(B/k_2))*sqrt(xy)
where there were a couple of possible ways to substitute out and I just
picked one.
Now multiplying both sides by k_1*k_2 gives
(A*x + B*y - T)*k_1*k_2 = (A*k_2^2 + B*k_1^2)*sqrt(xy)
so I have
A*sqrt(xy)*k_2^2 - (A*x + B*y - T)*k_1*k_2 + B*sqrt(xy)*k_1^2 = 0
and completing the square with respect to k_2, I have
A*sqrt(xy)*k_2^2 - (A*x + B*y - T)*k_1*k_2 + (A*x + B*y -
T)^2*k_1^2/(4A*sqrt(xy)) + B*sqrt(xy)*k_1^2 = (A*x + B*y -
T)^2*k_1^2/(4A*sqrt(xy))
multiplying both sides by 4*A*sqrt(xy), and simplifying a bit gives
(A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 + 4*A* B*(xy)*k_1^2 = (A*x +
B*y - T)^2*k_1^2
which is
(A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y)^2 -
2T(A*x+B*y) + T^2)*k_1^2
which yuck, shows that I need to know the factorization of T, to know
how to pick A and B, so that
((A*x - B*y)^2 - 2T(A*x+B*y) + T^2)
and you need the factorization of B and T, or A and T to go further so
it's another crap idea.
Went in a big freaking circle, again.
James Harris |
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jstevh@msn.com
science forum Guru
Joined: 21 Jan 2006
Posts: 951
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| Posted: Fri Jun 30, 2006 12:49 am Post subject:
Re: SF: Tandem factorization back, new approach
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Necessity is the mother of invention...
jstevh@msn.com wrote:
| Quote: | jstevh@msn.com wrote:
Sorry but this is how the development process works with me.
I went down one path before, found out it didn't work like I thought it
did, so I called crap crap and started feeling sorry for myself. Then
I thought of another approach, so here we go, again.
The idea here is to use a tandem factorization of two numbers where
S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y))
and
T = (k_1*sqrt(x) - k_2*sqrt(y))*(k_3*sqrt(x) - k_4*sqrt(y))
which means that
2*sqrt(x)*k_1 = f_1 + g_1
2*sqrt(y)*k_2 = f_1 - g_1
2*sqrt(x)*k_3 = f_2 + g_2
2*sqrt(y)*k_4 = f_2 - g_2
and the initial approach I put forward was with a target composite T to
factor you pick some surrogate S, and I had various ideas for how you
then find all the other variables--when for that idea to work, you need
to already know the factorization of S and T.
So that was when I was ready to toss all of this and declared it crap.
However, why pick S?
Using the first two equations I have
S - T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy)
and
S+T = 2*k_1*k_3*x + 2*k_2*k_4*y
and focusing on that second equation if I let
k_1*k_3 = A
k_2*k_4 = B
and pick squares for x and y, then I determine S, if T is the target
factorization:
S = 2*k_1*k_3*x + 2*k_2*k_4*y - T
so
S = 2*A*x + 2*B*y - T
and you can use those equations to solve out two of the k's, and
substitute out S, to relate the remaining two k's in an equation that I
hope will give the approach that works.
Ok, so making that substitution into the first equation I have
2A*x + 2*B*y - 2T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy)
and now I can solve out two of the k's, and divide by 2 to get
A*x + B*y - T =(k_2*(A/k_1) + k_1*(B/k_2))*sqrt(xy)
where there were a couple of possible ways to substitute out and I just
picked one.
Now multiplying both sides by k_1*k_2 gives
(A*x + B*y - T)*k_1*k_2 = (A*k_2^2 + B*k_1^2)*sqrt(xy)
so I have
A*sqrt(xy)*k_2^2 - (A*x + B*y - T)*k_1*k_2 + B*sqrt(xy)*k_1^2 = 0
and completing the square with respect to k_2, I have
A*sqrt(xy)*k_2^2 - (A*x + B*y - T)*k_1*k_2 + (A*x + B*y -
T)^2*k_1^2/(4A*sqrt(xy)) + B*sqrt(xy)*k_1^2 = (A*x + B*y -
T)^2*k_1^2/(4A*sqrt(xy))
multiplying both sides by 4*A*sqrt(xy), and simplifying a bit gives
(A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 + 4*A* B*(xy)*k_1^2 = (A*x +
B*y - T)^2*k_1^2
which is
(A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y)^2 -
2T(A*x+B*y) + T^2)*k_1^2
which yuck, shows that I need to know the factorization of T, to know
how to pick A and B, so that
((A*x - B*y)^2 - 2T(A*x+B*y) + T^2)
and you need the factorization of B and T, or A and T to go further so
it's another crap idea.
Went in a big freaking circle, again.
|
Maybe, maybe not. After all, I have
S = 2*A*x + 2*B*y - T
so what if I just decide that S is the target?
Then I need to solve out further with
(A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y)^2 -
2T(A*x+B*y) + T^2)*k_1^2
as that is
(A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y - T)^2 -
4T*B*y)*k_1^2
and if I just pick B and T, as surrogates, and assume I know the
factorizations
a_1 * a_2 = B*y, and b_1*b_2 = T, then I have
A*x = B*y + T + a_1*b_1 + a_2*b_2
and I can now substitute to find that S is given by
S = 4*B*y + T + 2*a_1*b_1 + 2*a_2*b_2
and making my substitutions that is
S = 4*y*a_1 * a_2 + b_1*b_2 + 2*a_1*b_1 + 2*a_2*b_2
and let's collect with respect to the a's, so I have
S = 4*y*a_1 * a_2 + 2*a_1*b_1 + b_1*b_2 + 2*a_2*b_2
and simplify to get
S - b_1*b_2 - 2*a_2*b_2 = (4*y* a_2 + 2*b_1)*a_1
so I have finally that
a_1 = (S - b_1*b_2 - 2*a_2*b_2)/(4*y* a_2 + 2*b_1)
and it looks like I can just pick b_1 and b_2, and then I just need to
look for integer values for a_2 such that a_1 is an integer.
One last thing to do then, with
4*y* a_2 + 2*b_1 = h_1
then
4*y* a_2 + 2*b_1 = 0 mod h_1
S - b_1*b_2 - 2*a_2*b_2 = 0 mod h_1
so multiplying the first by -2*b_2 and the second by 4*y, I have
-8*y*b_2* a_2 - 4*b_1*b_2 = 0 mod h_1
4*y*S - 4*y* b_1*b_2 - 8*y*a_2*b_2 = 0 mod h_1
and subtracting the top one from the bottom one I get
4*y*S - 4*y* b_1*b_2 + 4*b_1*b_2= 0 mod h_1
which is a surpise to me, as it looks like if I go ahead and use
b_1*b_2 = T, I just have
4*y*S + 4*T*(y-1)= 0 mod h_1
whci is the kind of result I don't believe in (thinking I made an
algebra mistake somewhere) as it says that if y = 1, then you need the
factorization of S, but otherwise, you can factor this other thing.
Going back over it to find someplace where I left off y...
Didn't see it on a quick run back, but I guess I'll find something
later. In any event, will post in the meantime!
Of course, if by some chance I did the algebra right, then you have
this odd result that you need to pick a square for y other than 1, like
y=4, and then the damn thing will work!
But that seems too arbitrary, so I must have made a mistake somewhere?
James Harris |
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jstevh@msn.com
science forum Guru
Joined: 21 Jan 2006
Posts: 951
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| Posted: Fri Jun 30, 2006 1:27 am Post subject:
Re: SF: Tandem factorization back, new approach
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jstevh@msn.com wrote:
| Quote: | Necessity is the mother of invention...
jstevh@msn.com wrote:
jstevh@msn.com wrote:
Sorry but this is how the development process works with me.
I went down one path before, found out it didn't work like I thought it
did, so I called crap crap and started feeling sorry for myself. Then
I thought of another approach, so here we go, again.
The idea here is to use a tandem factorization of two numbers where
S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y))
and
T = (k_1*sqrt(x) - k_2*sqrt(y))*(k_3*sqrt(x) - k_4*sqrt(y))
which means that
2*sqrt(x)*k_1 = f_1 + g_1
2*sqrt(y)*k_2 = f_1 - g_1
2*sqrt(x)*k_3 = f_2 + g_2
2*sqrt(y)*k_4 = f_2 - g_2
and the initial approach I put forward was with a target composite T to
factor you pick some surrogate S, and I had various ideas for how you
then find all the other variables--when for that idea to work, you need
to already know the factorization of S and T.
So that was when I was ready to toss all of this and declared it crap.
However, why pick S?
Using the first two equations I have
S - T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy)
and
S+T = 2*k_1*k_3*x + 2*k_2*k_4*y
and focusing on that second equation if I let
k_1*k_3 = A
k_2*k_4 = B
and pick squares for x and y, then I determine S, if T is the target
factorization:
S = 2*k_1*k_3*x + 2*k_2*k_4*y - T
so
S = 2*A*x + 2*B*y - T
and you can use those equations to solve out two of the k's, and
substitute out S, to relate the remaining two k's in an equation that I
hope will give the approach that works.
Ok, so making that substitution into the first equation I have
2A*x + 2*B*y - 2T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy)
and now I can solve out two of the k's, and divide by 2 to get
A*x + B*y - T =(k_2*(A/k_1) + k_1*(B/k_2))*sqrt(xy)
where there were a couple of possible ways to substitute out and I just
picked one.
Now multiplying both sides by k_1*k_2 gives
(A*x + B*y - T)*k_1*k_2 = (A*k_2^2 + B*k_1^2)*sqrt(xy)
so I have
A*sqrt(xy)*k_2^2 - (A*x + B*y - T)*k_1*k_2 + B*sqrt(xy)*k_1^2 = 0
and completing the square with respect to k_2, I have
A*sqrt(xy)*k_2^2 - (A*x + B*y - T)*k_1*k_2 + (A*x + B*y -
T)^2*k_1^2/(4A*sqrt(xy)) + B*sqrt(xy)*k_1^2 = (A*x + B*y -
T)^2*k_1^2/(4A*sqrt(xy))
multiplying both sides by 4*A*sqrt(xy), and simplifying a bit gives
(A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 + 4*A* B*(xy)*k_1^2 = (A*x +
B*y - T)^2*k_1^2
which is
(A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y)^2 -
2T(A*x+B*y) + T^2)*k_1^2
which yuck, shows that I need to know the factorization of T, to know
how to pick A and B, so that
((A*x - B*y)^2 - 2T(A*x+B*y) + T^2)
and you need the factorization of B and T, or A and T to go further so
it's another crap idea.
Went in a big freaking circle, again.
Maybe, maybe not. After all, I have
S = 2*A*x + 2*B*y - T
so what if I just decide that S is the target?
Then I need to solve out further with
(A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y)^2 -
2T(A*x+B*y) + T^2)*k_1^2
as that is
(A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y - T)^2 -
4T*B*y)*k_1^2
and if I just pick B and T, as surrogates, and assume I know the
factorizations
a_1 * a_2 = B*y, and b_1*b_2 = T, then I have
A*x = B*y + T + a_1*b_1 + a_2*b_2
and I can now substitute to find that S is given by
S = 4*B*y + T + 2*a_1*b_1 + 2*a_2*b_2
and making my substitutions that is
S = 4*y*a_1 * a_2 + b_1*b_2 + 2*a_1*b_1 + 2*a_2*b_2
|
That was dumb. That's why y is in at the end, as I put it in here when
it was substituted out.
Corrected it goes as follows.
and let's collect with respect to the a's, so I have
S = 4*a_1 * a_2 + 2*a_1*b_1 + b_1*b_2 + 2*a_2*b_2
and simplify to get
S - b_1*b_2 - 2*a_2*b_2 = (4* a_2 + 2*b_1)*a_1
so I have finally that
a_1 = (S - b_1*b_2 - 2*a_2*b_2)/(4a_2 + 2*b_1)
and it looks like I can just pick b_1 and b_2, and then I just need to
look for integer values for a_2 such that a_1 is an integer.
One last thing to do then, with
4a_2 + 2*b_1 = h_1
then
4a_2 + 2*b_1 = 0 mod h_1
S - b_1*b_2 - 2*a_2*b_2 = 0 mod h_1
so multiplying the first by -2*b_2 and the second by 4, I have
-8b_2* a_2 - 4*b_1*b_2 = 0 mod h_1
4S - 4b_1*b_2 - 8*y*a_2*b_2 = 0 mod h_1
and subtracting the top one from the bottom one I get
4S = 0 mod h_1
so not surprisingly I need the factorization of S to go further so it's
another big freaking circle.
James Harris |
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Canaan Banana
science forum beginner
Joined: 30 Jun 2006
Posts: 11
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| Posted: Fri Jun 30, 2006 4:03 am Post subject:
Re: SF: Tandem factorization back, new approach
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jstevh@msn.com wrote:
| Quote: | Necessity is the mother of invention...
jstevh@msn.com wrote:
jstevh@msn.com wrote:
Sorry but this is how the development process works with me.
I went down one path before, found out it didn't work like I thought it
did, so I called crap crap and started feeling sorry for myself. Then
I thought of another approach, so here we go, again.
The idea here is to use a tandem factorization of two numbers where
S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y))
and
T = (k_1*sqrt(x) - k_2*sqrt(y))*(k_3*sqrt(x) - k_4*sqrt(y))
which means that
2*sqrt(x)*k_1 = f_1 + g_1
2*sqrt(y)*k_2 = f_1 - g_1
2*sqrt(x)*k_3 = f_2 + g_2
2*sqrt(y)*k_4 = f_2 - g_2
and the initial approach I put forward was with a target composite T to
factor you pick some surrogate S, and I had various ideas for how you
then find all the other variables--when for that idea to work, you need
to already know the factorization of S and T.
So that was when I was ready to toss all of this and declared it crap.
However, why pick S?
Using the first two equations I have
S - T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy)
and
S+T = 2*k_1*k_3*x + 2*k_2*k_4*y
and focusing on that second equation if I let
k_1*k_3 = A
k_2*k_4 = B
and pick squares for x and y, then I determine S, if T is the target
factorization:
S = 2*k_1*k_3*x + 2*k_2*k_4*y - T
so
S = 2*A*x + 2*B*y - T
and you can use those equations to solve out two of the k's, and
substitute out S, to relate the remaining two k's in an equation that I
hope will give the approach that works.
Ok, so making that substitution into the first equation I have
2A*x + 2*B*y - 2T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy)
and now I can solve out two of the k's, and divide by 2 to get
A*x + B*y - T =(k_2*(A/k_1) + k_1*(B/k_2))*sqrt(xy)
where there were a couple of possible ways to substitute out and I just
picked one.
Now multiplying both sides by k_1*k_2 gives
(A*x + B*y - T)*k_1*k_2 = (A*k_2^2 + B*k_1^2)*sqrt(xy)
so I have
A*sqrt(xy)*k_2^2 - (A*x + B*y - T)*k_1*k_2 + B*sqrt(xy)*k_1^2 = 0
and completing the square with respect to k_2, I have
A*sqrt(xy)*k_2^2 - (A*x + B*y - T)*k_1*k_2 + (A*x + B*y -
T)^2*k_1^2/(4A*sqrt(xy)) + B*sqrt(xy)*k_1^2 = (A*x + B*y -
T)^2*k_1^2/(4A*sqrt(xy))
multiplying both sides by 4*A*sqrt(xy), and simplifying a bit gives
(A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 + 4*A* B*(xy)*k_1^2 = (A*x +
B*y - T)^2*k_1^2
which is
(A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y)^2 -
2T(A*x+B*y) + T^2)*k_1^2
which yuck, shows that I need to know the factorization of T, to know
how to pick A and B, so that
((A*x - B*y)^2 - 2T(A*x+B*y) + T^2)
and you need the factorization of B and T, or A and T to go further so
it's another crap idea.
Went in a big freaking circle, again.
Maybe, maybe not. After all, I have
S = 2*A*x + 2*B*y - T
so what if I just decide that S is the target?
Then I need to solve out further with
(A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y)^2 -
2T(A*x+B*y) + T^2)*k_1^2
as that is
(A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y - T)^2 -
4T*B*y)*k_1^2
and if I just pick B and T, as surrogates, and assume I know the
factorizations
a_1 * a_2 = B*y, and b_1*b_2 = T, then I have
A*x = B*y + T + a_1*b_1 + a_2*b_2
and I can now substitute to find that S is given by
S = 4*B*y + T + 2*a_1*b_1 + 2*a_2*b_2
and making my substitutions that is
S = 4*y*a_1 * a_2 + b_1*b_2 + 2*a_1*b_1 + 2*a_2*b_2
and let's collect with respect to the a's, so I have
S = 4*y*a_1 * a_2 + 2*a_1*b_1 + b_1*b_2 + 2*a_2*b_2
and simplify to get
S - b_1*b_2 - 2*a_2*b_2 = (4*y* a_2 + 2*b_1)*a_1
so I have finally that
a_1 = (S - b_1*b_2 - 2*a_2*b_2)/(4*y* a_2 + 2*b_1)
and it looks like I can just pick b_1 and b_2, and then I just need to
look for integer values for a_2 such that a_1 is an integer.
One last thing to do then, with
4*y* a_2 + 2*b_1 = h_1
then
4*y* a_2 + 2*b_1 = 0 mod h_1
S - b_1*b_2 - 2*a_2*b_2 = 0 mod h_1
so multiplying the first by -2*b_2 and the second by 4*y, I have
-8*y*b_2* a_2 - 4*b_1*b_2 = 0 mod h_1
4*y*S - 4*y* b_1*b_2 - 8*y*a_2*b_2 = 0 mod h_1
and subtracting the top one from the bottom one I get
4*y*S - 4*y* b_1*b_2 + 4*b_1*b_2= 0 mod h_1
which is a surpise to me, as it looks like if I go ahead and use
b_1*b_2 = T, I just have
4*y*S + 4*T*(y-1)= 0 mod h_1
whci is the kind of result I don't believe in (thinking I made an
algebra mistake somewhere) as it says that if y = 1, then you need the
factorization of S, but otherwise, you can factor this other thing.
Going back over it to find someplace where I left off y...
Didn't see it on a quick run back, but I guess I'll find something
later. In any event, will post in the meantime!
Of course, if by some chance I did the algebra right, then you have
this odd result that you need to pick a square for y other than 1, like
y=4, and then the damn thing will work!
But that seems too arbitrary, so I must have made a mistake somewhere?
|
We are reacting like an arrow, fruit flies like an actual, physical
banana: just think of the ambiguity between the idea that your work
would publishing the banana stand.
What I'm missing is more suited to slice bananas and that an April
Fool's edition, for the banana getting published I disagree. It would
entail. We believe it would not be more suited to defend yourself
against the banana stand and the other.
But an April Fool's edition, for the seeds. Alas, I mean that when
they put you eat the first place. Work all night on this, however.
By the Annals constitute libel or slander. My method was the
banana.
Who was to a banana stand. What I'm missing is well formed
by the banana to the seeds. Alas, I believe that of the banana.
Hold it may be nice material for the issue to drop the banana
getting published.
Let me with it. Come on. Come at me banana.
The deceased, Mr Apricot, is well formed in some sense. For
example, if you eat the banana. Here, you sent a work getting
published I disagree. It would not suffice for that would not ">=")
than a banana in the issue. It would entail.
We are sorry to drop the banana. So let you sent a banana to
let me with that banana. It's six foot, seven foot, eight foot, bunch.
Shut up.
Come on. Come at the banana's conspiration you know that an
actual, physical banana: just think of the banana and for the banana
stand in which your work being published I mean that she made
Adam's banana in the Xerox of the seeds.
Alas, I mean that she made Adam's banana stand. What I'm missing
is now eaten the probability of all, you sent a man armed with every
copy of the seeds.
Alas, I have a banana in the banana. The deceased, Mr Apricot, is
some sense in the Annals, it may be nice material for an April Fool's
edition, for the Annals of copies of Mathematics. We have a banana
conveys a banana fiend.
First of all, you know that we will not suffice for example.
This latter scenario strikes me as being published in the banana
stand. What I'm missing is well formed by the banana stand, and
for formal peer review, even if you sent a work of copies of the seeds.
Alas, I got a work being roughly as being roughly as better than
the Annals of the pattern formed by the banana.
Who was attacking me get clear on a cover letter and the banana.
Hold it like an actual, physical banana: just think of the Xerox of
the proper number of the proper number of your work of Mathematics.
We believe it would not suffice for the other.
But an actual, physical banana: just think of the banana.
|
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Bertie Reed
science forum beginner
Joined: 03 Feb 2006
Posts: 35
|
| Posted: Fri Jun 30, 2006 7:05 am Post subject:
Re: SF: Tandem factorization back, new approach
|
|
|
On 29 Jun 2006 18:27:21 -0700, jstevh@msn.com wrote:
| Quote: | jstevh@msn.com wrote:
Necessity is the mother of invention...
jstevh@msn.com wrote:
jstevh@msn.com wrote:
Sorry but this is how the development process works with me.
I went down one path before, found out it didn't work like I thought it
did, so I called crap crap and started feeling sorry for myself. Then
I thought of another approach, so here we go, again.
The idea here is to use a tandem factorization of two numbers where
S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y))
|
I think your problem is here, you want S to be a number yes? Unfortunately,
this equation does not specify a number. A squart root returns TWO values,
you can't just wish that away. Thus there are 8 different Ss here. |
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|
gjedwards@gmail.com
science forum addict
Joined: 20 May 2006
Posts: 70
|
| Posted: Fri Jun 30, 2006 9:18 am Post subject:
Re: SF: Tandem factorization back, new approach
|
|
|
Have you considered that maybe there's a common theme to all your
approaches?
You can try as many variations on these themes as you like until the
end of time and you'll still be stuck in your circles of nonsense.
Imagine how much math you could have learnt in the last 5 years. Are
you planning to waste the next 5 too? Why not actually go to college
and learn something?
jstevh@msn.com wrote:
| Quote: | jstevh@msn.com wrote:
Necessity is the mother of invention...
jstevh@msn.com wrote:
jstevh@msn.com wrote:
Sorry but this is how the development process works with me.
I went down one path before, found out it didn't work like I thought it
did, so I called crap crap and started feeling sorry for myself. Then
I thought of another approach, so here we go, again.
The idea here is to use a tandem factorization of two numbers where
S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y))
and
T = (k_1*sqrt(x) - k_2*sqrt(y))*(k_3*sqrt(x) - k_4*sqrt(y))
which means that
2*sqrt(x)*k_1 = f_1 + g_1
2*sqrt(y)*k_2 = f_1 - g_1
2*sqrt(x)*k_3 = f_2 + g_2
2*sqrt(y)*k_4 = f_2 - g_2
and the initial approach I put forward was with a target composite T to
factor you pick some surrogate S, and I had various ideas for how you
then find all the other variables--when for that idea to work, you need
to already know the factorization of S and T.
So that was when I was ready to toss all of this and declared it crap.
However, why pick S?
Using the first two equations I have
S - T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy)
and
S+T = 2*k_1*k_3*x + 2*k_2*k_4*y
and focusing on that second equation if I let
k_1*k_3 = A
k_2*k_4 = B
and pick squares for x and y, then I determine S, if T is the target
factorization:
S = 2*k_1*k_3*x + 2*k_2*k_4*y - T
so
S = 2*A*x + 2*B*y - T
and you can use those equations to solve out two of the k's, and
substitute out S, to relate the remaining two k's in an equation that I
hope will give the approach that works.
Ok, so making that substitution into the first equation I have
2A*x + 2*B*y - 2T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy)
and now I can solve out two of the k's, and divide by 2 to get
A*x + B*y - T =(k_2*(A/k_1) + k_1*(B/k_2))*sqrt(xy)
where there were a couple of possible ways to substitute out and I just
picked one.
Now multiplying both sides by k_1*k_2 gives
(A*x + B*y - T)*k_1*k_2 = (A*k_2^2 + B*k_1^2)*sqrt(xy)
so I have
A*sqrt(xy)*k_2^2 - (A*x + B*y - T)*k_1*k_2 + B*sqrt(xy)*k_1^2 = 0
and completing the square with respect to k_2, I have
A*sqrt(xy)*k_2^2 - (A*x + B*y - T)*k_1*k_2 + (A*x + B*y -
T)^2*k_1^2/(4A*sqrt(xy)) + B*sqrt(xy)*k_1^2 = (A*x + B*y -
T)^2*k_1^2/(4A*sqrt(xy))
multiplying both sides by 4*A*sqrt(xy), and simplifying a bit gives
(A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 + 4*A* B*(xy)*k_1^2 = (A*x +
B*y - T)^2*k_1^2
which is
(A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y)^2 -
2T(A*x+B*y) + T^2)*k_1^2
which yuck, shows that I need to know the factorization of T, to know
how to pick A and B, so that
((A*x - B*y)^2 - 2T(A*x+B*y) + T^2)
and you need the factorization of B and T, or A and T to go further so
it's another crap idea.
Went in a big freaking circle, again.
Maybe, maybe not. After all, I have
S = 2*A*x + 2*B*y - T
so what if I just decide that S is the target?
Then I need to solve out further with
(A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y)^2 -
2T(A*x+B*y) + T^2)*k_1^2
as that is
(A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y - T)^2 -
4T*B*y)*k_1^2
and if I just pick B and T, as surrogates, and assume I know the
factorizations
a_1 * a_2 = B*y, and b_1*b_2 = T, then I have
A*x = B*y + T + a_1*b_1 + a_2*b_2
and I can now substitute to find that S is given by
S = 4*B*y + T + 2*a_1*b_1 + 2*a_2*b_2
and making my substitutions that is
S = 4*y*a_1 * a_2 + b_1*b_2 + 2*a_1*b_1 + 2*a_2*b_2
That was dumb. That's why y is in at the end, as I put it in here when
it was substituted out.
Corrected it goes as follows.
and let's collect with respect to the a's, so I have
S = 4*a_1 * a_2 + 2*a_1*b_1 + b_1*b_2 + 2*a_2*b_2
and simplify to get
S - b_1*b_2 - 2*a_2*b_2 = (4* a_2 + 2*b_1)*a_1
so I have finally that
a_1 = (S - b_1*b_2 - 2*a_2*b_2)/(4a_2 + 2*b_1)
and it looks like I can just pick b_1 and b_2, and then I just need to
look for integer values for a_2 such that a_1 is an integer.
One last thing to do then, with
4a_2 + 2*b_1 = h_1
then
4a_2 + 2*b_1 = 0 mod h_1
S - b_1*b_2 - 2*a_2*b_2 = 0 mod h_1
so multiplying the first by -2*b_2 and the second by 4, I have
-8b_2* a_2 - 4*b_1*b_2 = 0 mod h_1
4S - 4b_1*b_2 - 8*y*a_2*b_2 = 0 mod h_1
and subtracting the top one from the bottom one I get
4S = 0 mod h_1
so not surprisingly I need the factorization of S to go further so it's
another big freaking circle.
James Harris |
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Znod Grimpo
science forum beginner
Joined: 30 Jun 2006
Posts: 16
|
| Posted: Fri Jun 30, 2006 4:58 pm Post subject:
Re: SF: Tandem factorization back, new approach
|
|
|
"Canaan Banana" <bananacanaan@aol.com> wrote in message
news:1151640198.041986.47730@d56g2000cwd.googlegroups.com...
| Quote: | jstevh@msn.com wrote:
Necessity is the mother of invention...
jstevh@msn.com wrote:
jstevh@msn.com wrote:
Sorry but this is how the development process works with me.
I went down one path before, found out it didn't work like I thought
it
did, so I called crap crap and started feeling sorry for myself.
Then
I thought of another approach, so here we go, again.
The idea here is to use a tandem factorization of two numbers where
S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y))
and
T = (k_1*sqrt(x) - k_2*sqrt(y))*(k_3*sqrt(x) - k_4*sqrt(y))
which means that
2*sqrt(x)*k_1 = f_1 + g_1
2*sqrt(y)*k_2 = f_1 - g_1
2*sqrt(x)*k_3 = f_2 + g_2
2*sqrt(y)*k_4 = f_2 - g_2
and the initial approach I put forward was with a target composite T
to
factor you pick some surrogate S, and I had various ideas for how you
then find all the other variables--when for that idea to work, you
need
to already know the factorization of S and T.
So that was when I was ready to toss all of this and declared it
crap.
However, why pick S?
Using the first two equations I have
S - T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy)
and
S+T = 2*k_1*k_3*x + 2*k_2*k_4*y
and focusing on that second equation if I let
k_1*k_3 = A
k_2*k_4 = B
and pick squares for x and y, then I determine S, if T is the target
factorization:
S = 2*k_1*k_3*x + 2*k_2*k_4*y - T
so
S = 2*A*x + 2*B*y - T
and you can use those equations to solve out two of the k's, and
substitute out S, to relate the remaining two k's in an equation that
I
hope will give the approach that works.
Ok, so making that substitution into the first equation I have
2A*x + 2*B*y - 2T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy)
and now I can solve out two of the k's, and divide by 2 to get
A*x + B*y - T =(k_2*(A/k_1) + k_1*(B/k_2))*sqrt(xy)
where there were a couple of possible ways to substitute out and I just
picked one.
Now multiplying both sides by k_1*k_2 gives
(A*x + B*y - T)*k_1*k_2 = (A*k_2^2 + B*k_1^2)*sqrt(xy)
so I have
A*sqrt(xy)*k_2^2 - (A*x + B*y - T)*k_1*k_2 + B*sqrt(xy)*k_1^2 = 0
and completing the square with respect to k_2, I have
A*sqrt(xy)*k_2^2 - (A*x + B*y - T)*k_1*k_2 + (A*x + B*y -
T)^2*k_1^2/(4A*sqrt(xy)) + B*sqrt(xy)*k_1^2 = (A*x + B*y -
T)^2*k_1^2/(4A*sqrt(xy))
multiplying both sides by 4*A*sqrt(xy), and simplifying a bit gives
(A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 + 4*A* B*(xy)*k_1^2 = (A*x +
B*y - T)^2*k_1^2
which is
(A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y)^2 -
2T(A*x+B*y) + T^2)*k_1^2
which yuck, shows that I need to know the factorization of T, to know
how to pick A and B, so that
((A*x - B*y)^2 - 2T(A*x+B*y) + T^2)
and you need the factorization of B and T, or A and T to go further so
it's another crap idea.
Went in a big freaking circle, again.
Maybe, maybe not. After all, I have
S = 2*A*x + 2*B*y - T
so what if I just decide that S is the target?
Then I need to solve out further with
(A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y)^2 -
2T(A*x+B*y) + T^2)*k_1^2
as that is
(A*sqrt(xy)*k_2 - (A*x + B*y - T)*k_1)^2 = ((A*x - B*y - T)^2 -
4T*B*y)*k_1^2
and if I just pick B and T, as surrogates, and assume I know the
factorizations
a_1 * a_2 = B*y, and b_1*b_2 = T, then I have
A*x = B*y + T + a_1*b_1 + a_2*b_2
and I can now substitute to find that S is given by
S = 4*B*y + T + 2*a_1*b_1 + 2*a_2*b_2
and making my substitutions that is
S = 4*y*a_1 * a_2 + b_1*b_2 + 2*a_1*b_1 + 2*a_2*b_2
and let's collect with respect to the a's, so I have
S = 4*y*a_1 * a_2 + 2*a_1*b_1 + b_1*b_2 + 2*a_2*b_2
and simplify to get
S - b_1*b_2 - 2*a_2*b_2 = (4*y* a_2 + 2*b_1)*a_1
so I have finally that
a_1 = (S - b_1*b_2 - 2*a_2*b_2)/(4*y* a_2 + 2*b_1)
and it looks like I can just pick b_1 and b_2, and then I just need to
look for integer values for a_2 such that a_1 is an integer.
One last thing to do then, with
4*y* a_2 + 2*b_1 = h_1
then
4*y* a_2 + 2*b_1 = 0 mod h_1
S - b_1*b_2 - 2*a_2*b_2 = 0 mod h_1
so multiplying the first by -2*b_2 and the second by 4*y, I have
-8*y*b_2* a_2 - 4*b_1*b_2 = 0 mod h_1
4*y*S - 4*y* b_1*b_2 - 8*y*a_2*b_2 = 0 mod h_1
and subtracting the top one from the bottom one I get
4*y*S - 4*y* b_1*b_2 + 4*b_1*b_2= 0 mod h_1
which is a surpise to me, as it looks like if I go ahead and use
b_1*b_2 = T, I just have
4*y*S + 4*T*(y-1)= 0 mod h_1
whci is the kind of result I don't believe in (thinking I made an
algebra mistake somewhere) as it says that if y = 1, then you need the
factorization of S, but otherwise, you can factor this other thing.
Going back over it to find someplace where I left off y...
Didn't see it on a quick run back, but I guess I'll find something
later. In any event, will post in the meantime!
Of course, if by some chance I did the algebra right, then you have
this odd result that you need to pick a square for y other than 1, like
y=4, and then the damn thing will work!
But that seems too arbitrary, so I must have made a mistake somewhere?
We are reacting like an arrow, fruit flies like an actual, physical
banana: just think of the ambiguity between the idea that your work
would publishing the banana stand.
What I'm missing is more suited to slice bananas and that an April
Fool's edition, for the banana getting published I disagree. It would
entail. We believe it would not be more suited to defend yourself
against the banana stand and the other.
But an April Fool's edition, for the seeds. Alas, I mean that when
they put you eat the first place. Work all night on this, however.
By the Annals constitute libel or slander. My method was the
banana.
Who was to a banana stand. What I'm missing is well formed
by the banana to the seeds. Alas, I believe that of the banana.
Hold it may be nice material for the issue to drop the banana
getting published.
Let me with it. Come on. Come at me banana.
The deceased, Mr Apricot, is well formed in some sense. For
example, if you eat the banana. Here, you sent a work getting
published I disagree. It would not suffice for that would not ">=")
than a banana in the issue. It would entail.
We are sorry to drop the banana. So let you sent a banana to
let me with that banana. It's six foot, seven foot, eight foot, bunch.
Shut up.
Come on. Come at the banana's conspiration you know that an
actual, physical banana: just think of the banana and for the banana
stand in which your work being published I mean that she made
Adam's banana in the Xerox of the seeds.
Alas, I mean that she made Adam's banana stand. What I'm missing
is now eaten the probability of all, you sent a man armed with every
copy of the seeds.
Alas, I have a banana in the banana. The deceased, Mr Apricot, is
some sense in the Annals, it may be nice material for an April Fool's
edition, for the Annals of copies of Mathematics. We have a banana
conveys a banana fiend.
First of all, you know that we will not suffice for example.
This latter scenario strikes me as being published in the banana
stand. What I'm missing is well formed by the banana stand, and
for formal peer review, even if you sent a work of copies of the seeds.
Alas, I got a work being roughly as being roughly as better than
the Annals of the pattern formed by the banana.
Who was attacking me get clear on a cover letter and the banana.
Hold it like an actual, physical banana: just think of the Xerox of
the proper number of the proper number of your work of Mathematics.
We believe it would not suffice for the other.
But an actual, physical banana: just think of the banana.
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But whot if it is a pointed stick ?
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Canaan Banana
science forum beginner
Joined: 30 Jun 2006
Posts: 11
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| Posted: Fri Jun 30, 2006 7:13 pm Post subject:
Re: SF: Tandem factorization back, new approach
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Znod Grimpo wrote:
| Quote: | "Canaan Banana" <bananacanaan@aol.com> wrote in message
news:1151640198.041986.47730@d56g2000cwd.googlegroups.com...
We are reacting like an arrow, fruit flies like an actual, physical
banana: just think of the ambiguity between the idea that your work
would publishing the banana stand.
What I'm missing is more suited to slice bananas and that an April
Fool's edition, for the banana getting published I disagree. It would
entail. We believe it would not be more suited to defend yourself
against the banana stand and the other.
But an April Fool's edition, for the seeds. Alas, I mean that when
they put you eat the first place. Work all night on this, however.
By the Annals constitute libel or slander. My method was the
banana.
Who was to a banana stand. What I'm missing is well formed
by the banana to the seeds. Alas, I believe that of the banana.
Hold it may be nice material for the issue to drop the banana
getting published.
Let me with it. Come on. Come at me banana.
The deceased, Mr Apricot, is well formed in some sense. For
example, if you eat the banana. Here, you sent a work getting
published I disagree. It would not suffice for that would not ">=")
than a banana in the issue. It would entail.
We are sorry to drop the banana. So let you sent a banana to
let me with that banana. It's six foot, seven foot, eight foot, bunch.
Shut up.
Come on. Come at the banana's conspiration you know that an
actual, physical banana: just think of the banana and for the banana
stand in which your work being published I mean that she made
Adam's banana in the Xerox of the seeds.
Alas, I mean that she made Adam's banana stand. What I'm missing
is now eaten the probability of all, you sent a man armed with every
copy of the seeds.
Alas, I have a banana in the banana. The deceased, Mr Apricot, is
some sense in the Annals, it may be nice material for an April Fool's
edition, for the Annals of copies of Mathematics. We have a banana
conveys a banana fiend.
First of all, you know that we will not suffice for example.
This latter scenario strikes me as being published in the banana
stand. What I'm missing is well formed by the banana stand, and
for formal peer review, even if you sent a work of copies of the seeds.
Alas, I got a work being roughly as being roughly as better than
the Annals of the pattern formed by the banana.
Who was attacking me get clear on a cover letter and the banana.
Hold it like an actual, physical banana: just think of the Xerox of
the proper number of the proper number of your work of Mathematics.
We believe it would not suffice for the other.
But an actual, physical banana: just think of the banana.
But whot if it is a pointed stick ?
|
Shut up.
Supposing he's got that, that's it. Who was attacking me then.
Now, I got that, that's it. Now it's quite simple to have no
bananas. My method was attacking me with a pointed stick.
Now, I disagree. It means that when they put you know that
banana. Come at the pattern formed by the Xerox of the banana.
We are sorry to a banana getting published there.
Now it's hard to drop the banana stand, and for an April Fool's
edition, for the issue to drop the probability of copies of your
work is now rendered him to have heard rumors that of Mathematics.
We believe that an arrow, fruit flies like an April Fool's
edition, for that banana. Supposing he's got a man armed with a
banana in the pattern formed in it, for that your work getting
published in the banana.
Here, you have no bananas today. People are sorry to a bunch.
Yes, we have no bananas. We have no bananas.
Supposing he's got a banana.
Who was to a third party, would not be able to let me get clear
on this, however. By the banana. You have now rendered him to
the first place. Work all night on a banana. So there is more
suited to a bunch.
Now attack me banana.
Who was to publish your work would be able to defend yourself
against the first carpenter? Eve. She made Adam's banana in
prison, other inmates stick bananas and he say day, he say day,
he say day, he wanna go home.
Yes, we have no bananas. How to defend yourself against a banana
in it, for the seeds.
Alas, I have now rendered him dead.
Yes, we will not suffice for formal peer review, even if sending
someone a defamatory message about a bunch. |
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