PowerPoint Presentationbest.eng.buffalo.edu/Research/Seminar slides/pre_K4_Zack.pdf · •QR...
Transcript of PowerPoint Presentationbest.eng.buffalo.edu/Research/Seminar slides/pre_K4_Zack.pdf · •QR...
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Ax=bZack
10/4/2013
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Iteration method
Ax=b 𝑣1, 𝑣2… 𝑥𝑘 = 𝑉𝑘𝑦𝑘
Given (𝐴, 𝑏) standard orthonormal base 𝑣1, 𝑣2… 𝑥𝑘 = 𝑉𝑘𝑦𝑘
For symmetric A, the method to generate 𝑉𝑘 is called Lanczos Method
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Lanczos
• Goal of Lanczos:
Find an orthonormal base in 𝑠𝑝𝑎𝑛(𝑏, 𝐴𝑏 …𝐴𝑘−1𝑏)
Note: this orthonormal base is unique. At k+1 iteration step, we want to generate 𝑣𝑘+1, we have:
𝛽𝑘+1𝑣𝑘+1 = 𝛾0𝑏 + 𝛾1𝐴𝑏 + 𝛾2𝐴2𝑏…𝛾𝑘−1𝐴
𝑘−1𝑏 + 𝐴𝑘𝑏𝑣𝑖𝑇𝑣𝑘+1 = 0 ∀𝑖 ≠ 𝑘
𝑣𝑘𝑇𝑣𝑘+1 = 1
So 𝑣𝑘+1 is unique.
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Lanczos: Algorithm
• Given 𝑉𝑘 = 𝑣1, 𝑣2…𝑣𝑘𝛽𝑘+1𝑣𝑘+1= 𝛾0𝑏 + 𝛾1𝐴𝑏 + 𝛾2𝐴
2𝑏 …𝛾𝑘−1𝐴𝑘−1𝑏 + 𝛾𝑘𝐴
𝑘𝑏
= 𝛾′1𝑣1 + 𝛾′2𝑣2…𝛾′𝑘𝑣𝑘 + 𝛾𝑘𝐴𝑘𝑏
= 𝑧1𝑣1 + 𝑧2𝑣2…𝑧𝑘𝑣𝑘 + 𝐴𝑣𝑘= 𝐴𝑣𝑘 + 𝑉𝑘𝑧Where z is a vector, 𝛽 is a parameter to uniformization vector v
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Lanczos: Algorithm
• 𝛽𝑘+1𝑣𝑘+1 = 𝐴𝑣𝑘 + 𝑉𝑘𝑧
• 𝑣𝑘𝑇𝑣𝑘+1 = 0 𝑣𝑘
𝑇𝐴𝑣𝑘 + 𝑣𝑘𝑇𝑉𝑘𝑧 = 0
𝑒𝑘𝑇𝑧 = −𝑣𝑘
𝑇𝐴𝑣𝑘
• 𝑣𝑖𝑇𝑣𝑘+1 = 0 𝑖 < 𝑘𝑣𝑖
𝑇𝐴𝑣𝑘 + 𝑣𝑖𝑇𝑉𝑘𝑧 = 0
𝑒𝑖𝑇𝑧 = 𝑣𝑖
𝑇𝐴𝑣𝑘
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Lanczos: Algorithm
• Notice that we have:𝛽𝑖+1𝑣𝑖+1 = 𝐴𝑣𝑖 + 𝑉𝑖𝑧
𝑣𝑘𝑇𝛽𝑖+1𝑣𝑖+1 = 𝑣𝑘
𝑇𝐴𝑣𝑖 + 𝑣𝑘𝑇𝑉𝑖𝑧
if 𝑖 = 𝑘 − 1
𝛽𝑘 = 𝑣𝑘𝑇𝐴𝑣𝑖
else 0 = 𝑣𝑘𝑇𝐴𝑣𝑖
• 𝑒𝑖𝑇𝑧 = 𝑣𝑖
𝑇𝐴𝑣𝑘 = 𝑣𝑘𝑇𝐴𝑣𝑖 =
−𝛽𝑘 (𝑖 = 𝑘 − 1)0 (𝑖 < 𝑘 − 1)
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Lanczos: Algorithm
• 𝛽𝑘+1𝑣𝑘+1 = 𝐴𝑣𝑘 − 𝑣𝑘𝑇𝐴𝑣𝑘 𝑣𝑘 − 𝛽𝑘𝑣𝑘−1
define 𝛼𝑘 = 𝑣𝑘𝑇𝐴𝑣𝑘
𝐴𝑣𝑘 = 𝛽𝑘+1𝑣𝑘+1 + 𝛼𝑘𝑣𝑘 + 𝛽𝑘𝑣𝑘−1 𝐴𝑉𝑘 = 𝑉𝑘+1𝐻𝑘
where, 𝐻𝑘 =
𝛼1 𝛽2𝛽2 𝛼2 𝛽3
⋱ ⋱ ⋱𝛽𝑘 𝛼𝑘
𝛽𝑘+1
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Lanczos: Algorithm
• We can also define a tridiagonal matrix 𝑇𝑘 :
𝑇𝑘 =
𝛼1 𝛽2𝛽2 𝛼2 𝛽3
⋱ ⋱ ⋱𝛽𝑘 𝛼𝑘
;
𝐴𝑉𝑘 = 𝑉𝑘𝑇𝑘 + 𝛽𝑘+1𝑣𝑘+1𝑒𝑘𝑇
Note: theoretically, iteration will stop when 𝑘 = 𝑙 < 𝑛, but in fact, 𝑙 is always far more bigger than 𝑛, we need to set a reasonable stop rule.
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• Define 𝑥𝑘 = 𝑉𝑘𝑦𝑘 is approximation of x.
• residual vector
𝑟𝑘 ≡ 𝑏 − 𝐴𝑥𝑘= 𝛽1𝑣1 − 𝐴𝑉𝑘𝑦𝑘= 𝑉𝑘+1 𝛽1𝑒1 −𝐻𝑘𝑦𝑘= 𝑉𝑘+1𝑡𝑘+1
where 𝑡𝑘+1 = 𝛽1𝑒1 −𝐻𝑘𝑦𝑘
Lanczos: Algorithm
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Method Definition of optimal 𝑥𝑘
CG min 𝑟𝑘 𝐴−12
MINRES min 𝑟𝑘 22
SYMMLQ min 𝑥 − 𝑥𝑘 22
Choose “Optimal” 𝑥𝑘
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• minimize function 𝑟𝑘 𝐴−12
• 𝑟𝑘 𝐴−12 = 𝑟𝑘
𝑇𝐴−1𝑟𝑘= 𝑏 − 𝐴𝑥𝑘
𝑇𝐴−1 𝑏 − 𝐴𝑥𝑘= 𝑏 − 𝐴𝑉𝑘𝑦𝑘
𝑇𝐴−1 𝑏 − 𝐴𝑉𝑘𝑦𝑘
• 𝑟𝑘 𝐴−12 has a stationary value at 𝑦𝑘 if
𝑉𝑘𝑇𝐴𝑉𝑘𝑦𝑘 = 𝑉𝑘
𝑇𝑏
𝑉𝑘𝑇(𝑉𝑘𝑇𝑘 +𝛽𝑘+1𝑣𝑘+1𝑒𝑘
𝑇)𝑦𝑘 = 𝑉𝑘𝑇𝛽1𝑣1
𝑇𝑘𝑦𝑘 = 𝛽1𝑒1
CG
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• 𝑇𝑘𝑦𝑘 = 𝛽1𝑒1
• Use LDLT decomposition, 𝑇𝑘 = 𝐿𝑘𝐷𝑘𝐿𝑘𝑇 . 𝐿𝑘 is a
bidiagonal lower matrix, 𝐷𝑘 is a diagonal matrix.
• 𝐿𝑘𝐷𝑘𝐿𝑘𝑇𝑦𝑘 = 𝛽1𝑒1
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• Define 𝑧𝑘 = 𝐿𝑘𝑇𝑦𝑘 , 𝐿𝑘𝐷𝑘𝑧𝑘 = 𝛽1𝑒1
• Define 𝑊𝑘𝑇 = 𝐿𝑘
−1𝑉𝑘𝑇
• 𝑥𝑘 = 𝑉𝑘𝑦𝑘 = 𝑊𝑘𝐿𝑘𝑇𝑦𝑘 = 𝑊𝑘𝑧𝑘
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1 1 1
0 1
T Tk k k
k T Tk k k
L W VL
w v
1
T T T
k k k kw w v
1 1
1 1 1 1 1
2
1 1
0 0
0 101
k k T TTk k k k k k kk k k k
k k
k k k k k
L DL L D L dT L D L
dd d d
1
2
1
k k k
k k k k
d
d d
1 1 1 11 1
1 11
=0 01
k k k kk k
k k kk k k k kk k
L D L Dz zL D z e
d d d
1 1 0k k k k kd d
1
1 1 1 1
k
k k k k k k k k k k k k
k
zx W z W w W z w x w
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𝐴,𝑏, 𝑣𝑘−2, 𝑣𝑘−1 𝑣𝑘 ……… (Lanczos)
𝑑𝑘−1, 𝛽𝑘λ𝑘𝑑𝑘−1, λ𝑘 , 𝛼𝑘𝑑𝑘 ………(LDL decomposition)
λ𝑘 , 𝑤𝑘−1, 𝑣𝑘𝑤𝑘 ……… (𝑊𝑘𝑇 = 𝐿𝑘
−1𝑉𝑘𝑇)
𝑑𝑘−1, 𝑑𝑘 , λ𝑘 , ζ 𝑘−1ζ
𝑘……… (𝑧𝑘 = 𝐿𝑘
𝑇𝑦𝑘)
𝑥𝑘−1, ζ 𝑘, 𝑤𝑘𝑥𝑘 ……… (𝑥𝑘 = 𝑊𝑘𝑧𝑘 )
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Problem in CG method
• A should be positive definite• If A is indefinite, in practice, CG would still give out
answer but no longer be relied upon numerically.
• It used LDL decomposition • if A is an indefinite symmetric matrix, LDLT
decomposition can be tried, often success, but it does not always exist.
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Method Definition of optimal 𝑥𝑘
Subproblem Factorization Estimate of x
CG min 𝑟𝑘 𝐴−12 𝑇𝑘𝑦𝑘 = 𝛽1𝑒1 𝑇𝑘 = 𝐿𝑘𝐷𝑘𝐿𝑘
𝑇 𝑥𝑘𝐶 = 𝑉𝑘𝑦𝑘
MINRES min 𝑟𝑘 22 min 𝛽1𝑒1 −𝐻𝑘𝑦𝑘 𝑄𝑘𝐻𝑘 =
𝑅𝑘
0𝑥𝑘𝑀 = 𝑉𝑘𝑦𝑘
SYMMLQ min 𝑥 − 𝑥𝑘 22 min 𝑦𝑘 s.t.
𝐻𝑘−1𝑇 𝑦𝑘 = 𝛽1𝑒1
𝐻𝑘−1𝑇 𝑄𝑘−1
𝑇
= 𝐿𝑘−1 0𝑥𝑘𝐿 = 𝑉𝑘𝑦𝑘
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• CG• A must be positive definite matrix
• MINRES• Any symmetric A, 𝑟𝑘
𝑀 decreases monotonically
• Risk of cancellation error when A is indefinite
• SYMMLQ• QR factor, Any symmetric A, except Ax=b must be
consistent, Very little cancellation error, method of choice for indefinite consistent system
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Unsymmetric and rectangular iteration• Set 𝛼, 𝛽 as normalized parameters of 𝑢, 𝑣 , 𝛽1𝑢1 =𝑏, 𝛼1𝑣1 = 𝐴𝑇𝑢1
follow this iteration:
generate 2 orthonormal base
𝑈𝑘 = 𝑢1 𝑢2 …𝑢𝑘 𝑉𝑘 = 𝑣1 𝑣2 …𝑣𝑘
1 1
1 1 1 1
k k k k k
T
k k k k k
u Av u
v A u v
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Unsymmetric and rectangular iteration• PROOF:
assume that we have orthonormal vectors
𝑢1 𝑢2 …𝑢𝑘 , 𝑣1 𝑣2…𝑣𝑘generate 𝑢𝑘+1 , 𝑣𝑘+1 as:
1 1
1 1 1 1
k k k k k
T
k k k k k
u Av u
v A u v
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Unsymmetric and rectangular iteration
1
T T T T
k i i k i k i iv v v A u v v
1 1
T T T
i k k i k i k ku u u Av u u
When i=k
1 1 1+ 0T T T T T T
k k k k k k k k k k k k k k k k ku u u Av u u v v v v u u
When i<k
1 1 1= + =0T T T T T T
i k k i k i k k k i i k i i i k ku u u Av u u v v v v u u
So 1 1k ku u u 1 1k kv v v , we can also prove that
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Unsymmetric and rectangular iteration• 𝐴𝑉𝑘 = 𝑈𝑘+1𝐵𝑘 = 𝑈𝑘𝐿𝑘 + 𝛽𝑘+1𝑢𝑘+1𝑒𝑘
𝑇
• 𝐴𝑇𝑈𝑘+1 = 𝑉𝑘𝐵𝑘𝑇 + 𝛼𝑘+1𝑣𝑘+1𝑒𝑘+1
𝑇 = 𝑉𝑘+1𝐿𝑘+1𝑇
• 𝐿𝑘 =
𝛼1𝛽2 𝛼2
⋱ ⋱𝛽𝑘 𝛼𝑘
𝐵𝑘 =
𝛼1𝛽2 𝛼2
⋱ ⋱𝛽𝑘 𝛼𝑘
𝛽𝑘+1
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Direct method: LUSOL
• Factor:
:
:
/j ij
T
i
T
T
l A A
u A
L L l
UU
u
A A lu
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Direct method: LUSOL
• Pivot strategies
1. Preserve sparsity
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Direct method: LUSOL
a Markowitz strategies is used to select potential pivot 𝐴𝑖𝑗 . Pivot should have a low Markowitz merit function
𝑀𝑖𝑗 ≡ 𝑟𝑖 − 1 𝑐𝑗 − 1
where 𝑟𝑖 and 𝑐𝑗 are number of nonzero element in row 𝑖 and column 𝑗.
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The sparsest columns and rows are searched in turn: columns of length 1,rows of length 1, then columns of length 2, rows of length 2, and so on).
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Direct method: LUSOL
2. Preserve stability.
check (i,j) given by Markowitz strategies with stability test:
Strategy Name Stability test
Threshold Partial Pivoting TPP 𝑙 ∞ ≤ 𝐿𝑡𝑜𝑙
Threshold Rook Pivoting TRP 𝑙 ∞𝑎𝑛𝑑 𝑢/𝐴𝑖𝑗 ∞≤ 𝐿𝑡𝑜𝑙
Threshold Complete Pivoting TCP 𝐴𝑚𝑎𝑥 ≤ 𝐿𝑡𝑜𝑙 ∗ 𝐴𝑖𝑗
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Name Stability test
TPP 𝑙 ∞ ≤ 𝐿𝑡𝑜𝑙
TRP 𝑙 ∞𝑎𝑛𝑑 𝑢/𝐴𝑖𝑗 ∞≤ 𝐿𝑡𝑜𝑙
TCP 𝐴𝑚𝑎𝑥 ≤ 𝐿𝑡𝑜𝑙 ∗ 𝐴𝑖𝑗
Example, Ltol=3.0
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Thank you