• 沒有找到結果。

# Direct Methods for Solving Linear Systems

N/A
N/A
Protected

Share "Direct Methods for Solving Linear Systems"

Copied!
100
0
0

(1)

## Hung-Yuan Fan (范洪源)

Department of Mathematics, National Taiwan Normal University, Taiwan

(2)

. . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . .

. .

. . . . .

## (線性系統; 線性聯立⽅程組)

(3)

### Objective

To solve a system of n linear equations with n unknowns:

## E

1: a11x1+ a12x2+· · · + a1nxn= b1

## E

2: a21x1+ a22x2+· · · + a2nxn= b2

...

## E

n: an1x1+ an2x2+· · · + annxn= bn.

(1)

Definition

The vectorx = [x1, x2,· · · , xn]T∈ Rn is called a solution to the linear system (1).

(4)

. . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . .

. .

. . . . .

### Matrix-Vector Forms (矩陣-向量形式)

The linear system (1) can be rewritten as the followg matrix-vector form:

Ax = b,

where coefficient matrix A∈ Rn×n and right-hand side vector b∈ Rn are defined by

A =





a11 a12 · · · a1n

a21 a22 · · · a2n

... ... ... an1 an2 · · · ann



, b =



 b1 b2 ... bn



.

## Assumption: The matrix A is nonsingular throughout the

context.

(5)

### The Linear Solvers (線性系統的數值⽅法)

1 Direct Methods: (直接法)

Used for solving small- or medium-sized linear systems with full and dense coefficient matrices. (適⽤於求解中⼩型線性系 統)

Floating-point operation count (簡稱 flop)≈ O(n3).

### Gaussian Elimination (⾼斯消去法，簡稱 GE) is an efficient

and stable algorithm for solving this type of linear systems.

2 Iterative Methods: (迭代法)

Used for solving large and sparse linear systems with problem size n≥ 104. (適⽤於求解⼤型稀疏線性系統)

flop≈ O(n) per iteration if A is a sparse matrix.

Jacobi, Gauss-Seidel, SOR and CG-based methods,. . ..

(6)

. . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . .

. .

. . . . .

### Elementary Row Operations (基本列運算)

Three Row Operations

In order to simplify linear system (1), we use

1 (λEi)→ (Ei): eq. Ei is replaced by λ· Ei for any λ̸= 0.

2 (Ei+ λEj)→ (Ei): eq. Ej is multiplied by any λ∈ R and added to eq. Ei.

3 (Ei)↔ (Ej): exchange eqs. Ei and Ej for i̸= j.

(7)

### Gaussian Elimination (GE)

The Procedure of GE

Given a linear system Ax = b with A∈ Rn×n and b∈ Rn.

1 Form the augmented matrix ˜A(1)= [A| b] ∈ Rn×(n+1).

2 Applying row operations continuously, we obtain a finite sequence of augmented matrices, i.e.,

A˜(1) → ˜A(2)→ · · · → ˜A(n), where ˜A(n) is upper triangular. (上三⾓矩陣)

3 Use backward substitution (向後代入) to obtain xn, xn−1,· · · , x2, x1.

(8)

. . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . .

. .

. . . . .

(1)

### to ˜ A

(2)

From the augmented matrix ˜A(1)= [A| b] = [a(1)ij ], we have

A˜(1) =







a(1)11 a(1)12 · · · a(1)1,n+1

a(1)21 a(1)22 · · · a(1)2,n+1 ... ... ... a(1)n1 a(1)n2 · · · a(1)n,n+1





 .

(9)

(1)

(2)

### (Conti’d)

If a(1)11 ̸= 0, do(

Ei a(1)i1

a(1)11E1)

→ (Ei) for i = 2, 3, . . . , n⇒

A˜(2) =







a(1)11 a(1)12 · · · a(1)1,n+1 0 a(2)22 · · · a(2)2,n+1

... ... ... 0 a(2)n2 · · · a(2)n,n+1





 .

(10)

. . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . .

. .

. . . . .

(2)

### to ˜ A

(3)

Next, if a(2)22 ̸= 0, do(

Ei a(2)i2

a(2)22 E2)

→ (Ei) for i = 3, 4, . . . , n

A˜(3) =











a(1)11 a(1)12 a(1)13 · · · a(1)1,n+1 0 a(2)22 a(2)23 · · · a(2)2,n+1 ... 0 a(3)33 · · · a(3)3,n+1 ... ... ... ... 0 0 a(3)n3 · · · a(3)n,n+1









 .

(11)

(k−1)

(k)

### , k≥ 2

For each k≥ 2, suppose augmented matrix ˜A(k−1) has the form

A˜(k−1)=

















a(1)11 a(1)12 · · · a(1)1,k−2 a(1)1,k−1 a(1)1k · · · a(1)1,n+1 0 a(2)22 · · · a(2)2,k−2 a(2)2,k−1 a(2)2k · · · a(2)2,n+1

... 0 ... ... ... ...

... ... . .. a(k−2)k−2,k−2 a(k−2)k−2,k−1 a(k−2)k−2,k · · · a(k−2)k−2,n+1 ... .

.. 0 a(kk−1,k−1−1) a(kk−1,k−1) · · · a(kk−1,n+1−1) ... .

.. .

.. a(kk,k−1−1) a(kk,k−1) · · · a(kk,n+1−1)

... ... ... ... ... ...

0 0 · · · 0 a(kn,k−1)−1 a(kn,k−1) · · · a(kn,n+1−1)

















(12)

. . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . .

. .

. . . . .

(k−1)

(k)

### , k≥ 2 (Conti’d)

If a(kk−1,k−1−1) ̸= 0, do (

Ei a(ki,k−1−1)

a(kk−1,k−1−1) Ek−1)

→ (Ei) for k≤ i ≤ n ⇒

A˜(k) =

















a(1)11 a(1)12 · · · a(1)1,k−2 a(1)1,k−1 a(1)1k · · · a(1)1,n+1 0 a(2)22 · · · a(2)2,k−2 a(2)2,k−1 a(2)2k · · · a(2)2,n+1

... 0 ... ... ... ...

... .

.. . .. a(kk−2,k−2−2) a(kk−2,k−1−2) a(kk−2,k−2) · · · a(kk−2,n+1−2) ... .

.. 0 a(k−1)k−1,k−1 a(k−1)k−1,k · · · a(k−1)k−1,n+1 ... .

.. .

.. 0 a(k)k,k · · · a(k)k,n+1

... ... ... ... ... ...

0 0 · · · 0 0 a(k)n,k · · · a(k)n,n+1

















(13)

### At Final Stage of GE

When k = n, the GE will produce an augmented matrix in upper

A˜(n)







## a

11 a12 · · · a1,n−1 a1n | a1,n+1 0

## a

22 · · · a2,n−1 a2n | a2,n+1

... 0 . .. ... ... | ... ... ... . .. an−1,n−1 an−1,n | an−1,n+1

0 0 · · · 0

nn | an,n+1







.

## Note: Entries a

ii̸= 0 are called pivot elements (軸元) for 1≤ i ≤ n.

(14)

. . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . .

. .

. . . . .

### Backward Substitution (向後代入)

From the special form of ˜A(n), we obtain the solution x to linear system (1) as follows:

Firstly, compute xn= an,n+1/ann.

Then compute xn−1, xn−2,· · · , x1 successively by

xi =

ai,n+1n

j=i+1

aijxj

aii , i = n− 1, n − 2, . . . , 1.

## Note: This process is called the backward substitution of GE.

(15)

Backward Substitution with n = 3

Applying Backward Substitution for the 3× 3 linear system a11x1+ a12x2+ a13x3= b1

a22x2+ a23x3 = b2 a33x3 = b3,

the unique solution to above linear system is computed via x3= b3/a33,

x2= b2− a23x3 a22 , x1= b3− a12x2− a13x3

a11 .

(16)

. . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . .

. .

. . . . .

### Will GE break down? (1/2)

Example 2, p. 363 (GE 無法執⾏的例⼦) Use GE to find the solution of the linear system

E1 : x1− x2+ 2x3− x4=−8, E2 : 2x1− 2x2+ 3x3− 3x4 =−20, E3 : x1+ x2+ x3 =−2,

E4 : x1− x2+ 4x3+ 3x4 = 4.

## Sol: Form the augmented matrix ˜

A(1) as

A˜(1) =



1 −1 2 −1 | −8

2 −2 3 −3 | −20

1 1 1 0 | −2

1 −1 4 3 | 4



 .

(17)

### Will GE break down? (2/2)

Since a(1)11 = 1, do (Ei− a(1)i1 E1)→ Ei for i = 2, 3, 4, we have

A˜(2) =



1 −1 2 −1 | −8

0

## 0

−1 −1 | −4

0 2 −1 1 | 6

0 0 2 4 | 12



 .

Because a(2)22 = 0, GE will break down here and STOP!

How to fix it? Partial Pivoting Strategy!

(18)

. . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . .

. .

. . . . .

### Partial Pivoting (1/2)

If we do (E2)↔ (E3), then ˜A(2) becomes

A˜(3) =



1 −1 2 −1 | −8

0

## 2

−1 1 | 6

0 0 −1 −1 | −4

0 0 2 4 | 12



 .

Applying (E4+ 2E3)→ (E4) =

A˜(4) =



1 −1 2 −1 | −8

0

## 2

−1 1 | 6

0 0 −1 −1 | −4

0 0 0 2 | 4



 .

(19)

### Partial Pivoting (2/2)

Use backward substitution x4= 4

2 = 2

x3= [−4 − (−1)x4]

−1 = 2

x2= [6− (−1)x3− x4]

2 = 3

x1= [−8 − (−1)x2− 2x3− (−1)x4]

1 =−7.

It seems that GE with partial pivoting works well in this case!

(20)

. . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . .

. .

. . . . .

### Pseudocode of Gaussian Elimination

To solve the n× n linear system (1).

Algorithm 6.1: GE with Backward Substitution INPUT dimension n; augmented matrix A = [aij]∈ Rn×(n+1). OUTPUT solution x1, x2,· · · , xn.

Step 1 For i = 1, . . . , n− 1 do Steps 2–4 Step 2 Find smallest i≤ p ≤ n s.t. api̸= 0.

If not, OUTPUT(‘No unique solution exists.’); STOP.

Step 3 If p̸= i, perform (Ep)↔ (Ei).

Step 4 For j = i + 1, . . . , n do Steps 5–6 Step 5 Set mji= aji/aii.

Step 6 Perform (Ej− mjiEi)→ (Ej).

Step 7 If ann = 0, OUTPUT(‘No unique solution exists.’); STOP.

Step 8 Set xn= an,n+1/ann. (Start backward substitution.) Step 9 For i = n− 1, . . . , 1 set xi= [ai,n+1n

j=i+1aijxj]/aii. Step 10 OUTPUT(x1, x2,· · · , xn); STOP.

(21)

## Multiplications/Divisions:

n−1

i=1

[(n− i) + (n − i)(n − i + 1)] =

n−1

i=1

(n− i)(n − i + 2)

=

n−1

i=1

(n2− 2ni + i2+ 2n− 2i) =

3+ 3n2− 5n

## 6

.

n−1

i=1

(n− i)(n − i + 1) =

n−1

i=1

(n2− 2ni + i2+ n− i) =

3− n

.

(22)

. . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . .

. .

. . . . .

## Multiplications/Divisions:

1 +

n−1

i=1

[(n− i) + 1] = 1 +

n−1

i=1

(n− i) + (n − 1) =

2+ n

## 2

.

n−1

i=1

[(n− i − 1) + 1] =

n−1

i=1

(n− i) =

2− n

.

(23)

2n3+ 3n2− 5n

6 +n2+ n 2 =

3

+ n2

.

n3− n

3 +n2− n 2 =

3

+

2

## 6

. Total flop of GE ≈O(23n3) (as n→ ∞).

(24)

. . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . .

. .

. . . . .

## Pivoting Strategies

(25)

### Occurrence of Small Pivot Elements (1/2)

Example 1, p. 372

Apply GE to solve the linear system

E1 :

## 0.003000x

1+ 59.14x2 = 59.17 E2 : 5.291x1− 6.130x2 = 46.78,

using 4-digit rounding arithmetic and the exact solution is

## x

1 = 10.00 and x2 = 1.000.

## Sol: Note that m

21= fl(0.0030005.291 ) = fl(1763.6¯6) = 1764. Then do (E2− m21E1)→ (E2)

fl(−6.130 + fl(−1764 · 59.14))x2= fl(46.78 + fl(−1764 · 59.17)) or−104300x2=−104400 and hence x2 ≈ 1.001.

(26)

. . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . .

. .

. . . . .

### Occurrence of Small Pivot Elements (2/2)

Substituting x2 = 1.001 into E1, we obtain x1 ≈ fl(59.17− (59.14)(1.001)

0.003000

)

=−10.00, which gives a totally wrong answer for x1! Why?

(27)

### Partial Pivoting (部分選軸元)

For each i = 1, 2, . . . , n− 1, find smallest integer p ≥ i s.t.

|a(i)pi| = max

i≤j≤n|a(i)ji |.

Then perform the row operation

(Ei)↔ (Ep) if p̸= i.

This strategy is also called maximal column pivoting.

(28)

. . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . .

. .

. . . . .

### Improvement of Accuracy for Example 1

Example 2, p. 373 (改進例題 1 的計算精度)

Apply GE with partial pivoting to solve the linear system E1 :

## 0.003000x

1+ 59.14x2 = 59.17 E2 : 5.291x1− 6.130x2 = 46.78, using 4-digit rounding arithmetic.

## Sol: Since

|a11| < |a21|, we first perform (E1)↔ (E2):

E1 : 5.291x1− 6.130x2 = 46.78 E2 : 0.003000x1+ 59.14x2= 59.17.

Then m21= fl(0.0030005.291 ) = 0.0005670. Do

(E2− m21E1)→ (E2)⇒ 59.14x2≈ 59.14 and hence x2= 1.000.

Moreover, x1 = 10.00 is correct now!

(29)

To solve the n× n linear system (1).

Algorithm 6.2: GE with Partial Pivoting

INPUT dimension n; augmented matrix A = [aij]∈ Rn×(n+1). OUTPUT solution x1, x2,· · · , xn.

Step 1 For i = 1, . . . , n− 1 do Steps 2–5

Step 2 Find smallest i≤ p ≤ n s.t.|api| = maxi≤j≤n|aji|.

Step 3 Ifapi= 0, OUTPUT(‘No unique solution exists.’); STOP.

Step 4 If p̸= i, perform (Ep)↔ (Ei).

Step 5 For j = i + 1, . . . , n do Steps 6–7 Step 6 Set mji= aji/aii.

Step 7 Perform (Ej− mjiEi)→ (Ej).

Step 8 If ann = 0, OUTPUT(‘No unique solution exists.’); STOP.

Step 9 Set xn= an,n+1/ann. (Start backward substitution.) Step 10 For i = n− 1, . . . , 1 set xi= [ai,n+1n

j=i+1aijxj]/aii. Step 11 OUTPUT(x1, x2,· · · , xn); STOP.

(30)

. . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . .

. .

. . . . .

### Occurrence of Coefficients of Large Magnitude

Example 1’

Apply GE with partial pivoting to solve the linear system E1 :

## 30.00x

1+ 591400x2 = 591700 E2 : 5.291x1− 6.130x2 = 46.78,

using 4-digit rounding arithmetic. Exact solution is x1 = 10.00 and x2= 1.000.

## Sol: Note that

m21= fl(5.291

30.00) = 0.1764.

Perform (E2− m21E1)→ (E2)⇒ −104300x2 ≈ −104400. Hence,

## x

2 ≈ 1.001 and x1 ≈ −10.00! Why?

(31)

### Scaled Partial Pivoting

At the start of GE, compute n scale factors once as follows.

si = max

1≤j≤n|aij|, i = 1, 2, . . . , n.

For each i = 1, 2, . . . , n− 1, find smallest integer p ≥ i s.t.

|a(i)pi|

sp = max

i≤j≤n

|a(i)ji | sj . If i̸= p, perform (Ei)↔ (Ep).

This strategy is also called scaled-column pivoting.

(32)

. . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . .

. .

. . . . .

Example 2’

Apply GE with scaled partial pivoting to solve the linear system E1 :

## 30.00x

1+ 591400x2 = 591700

E2 : 5.291x1− 6.130x2 = 46.78, using 4-digit rounding arithmetic.

## Sol: First compute scale factors s

1= 591400 and ss2 = 6.130. For i = 1, we see that

|a11|

s1 = fl( 30.00

591400) = 0.5073× 10−4< |a21|

s2 = fl(5.291

6.130) = 0.8631.

So, perform (E2)↔ (E1)⇒ we obtain the correct solution x1 = 10.00 and x2 = 1.000!

(33)

To solve the n× n linear system (1).

Algorithm 6.3: GE with Scaled Partial Pivoting INPUT dimension n; augmented matrix A = [aij]∈ Rn×(n+1). OUTPUT solution x1, x2,· · · , xn.

Step 1 For i = 1, . . . , n set si = max1≤j≤n|aij|;

Ifsi= 0, OUTPUT(‘No unique solution exists.’); STOP.

Step 2 For i = 1, . . . , n− 1 do Steps 3–6

Step 3 Find smallest i≤ p ≤ n s.t. |aspip|= maxi≤j≤n|asji|

j .

Step 4 Ifapi= 0, OUTPUT(‘No unique solution exists.’); STOP.

Step 5 If p̸= i, perform (Ep)↔ (Ei).

Step 6 For j = i + 1, . . . , n do Steps 7–8 Step 7 Set mji= aji/aii.

Step 8 Perform (Ej− mjiEi)→ (Ej).

Step 9 If ann = 0, OUTPUT(‘No unique solution exists.’); STOP.

Step 10 Set xn= an,n+1/ann. (Start backward substitution.) Step 11 For i = n− 1, . . . , 1 set xi= [ai,n+1n

j=i+1aijxj]/aii.

(34)

. . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . .

. .

. . . . .

### Complete Pivoting

For each k = 1, 2, . . . , n− 1, find integers k ≤ p, q ≤ n s.t.

|a(k)pq| = max

k≤i,j≤n|a(k)ij |.

If p̸= i or q ̸= i, row and/or column interchanges are performed to bring a(k)pq to the pivot position a(k)kk.

This strategy is also called the maximal pivoting at the kth step.

(35)

## (矩陣分解)

(36)

. . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . .

. .

. . . . .

Motivation

For solving a linear system Ax = b, it requires O(13n3) arithmetic operations to determine x∈ Rn.

If the right-hand vector b∈ Rn is changed to another vector ˜b (and coeff. matrix A is unchanged), how can we solve this linear system efficiently using some matrix factorization of A generated from GE?

In fact, if A has been factored into the triangular form A = LU,

where L is lower triangular and U is upper triangular, then the operation counts can be reduced toO(2n2)!

(37)

### Comparison of Arithmetic Calculations

The relative rate of reduction of the operation counts O(2n2) compared with O(13n3) becomes larger and larger for n = 10, 102 and 103, respectively. The results are shown in the following table.

(38)

. . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . .

. .

. . . . .

### The 1st Step of GE

Let A(1)= A∈ Rn×n and b(1) = b∈ Rn for a linear system.

If a(1)1,1 ̸= 0, do(

Ei a(1)i,1

a(1)1,1E1)

→ (Ei) for i = 2, 3, . . . , n⇒

A(2) =







a(1)1,1 a(1)1,2 · · · a(1)1,n 0 a(2)2,2 · · · a(2)2,n ... ... ... 0 a(2)n,2 · · · a(2)n,n







, b(2) =





 b(2)1 b(2)2 ... b(2)n





 .

The corresponding multipliers are given by

mi,1= a(1)i,1

a(1)11 , i = 2, 3, . . . , n.

(39)

### The 1st Step of GE (Conti’d)

This is equivalent to

A(2)= M(1)A(1) and b(2)= M(1)b(1), where the first Gaussian transformation matrix M(1)∈ Rn×n is defined by

M(1)=









1 0 · · · 0

−m2,1 1 0 · · · 0

−m3,1 0 . .. ... ...

... ... . .. ... 0

−mn,1 0 · · · 0 1







 .

(40)

. . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . .

. .

. . . . .

### At the kth Step of GE, 2≤ k ≤ n − 1

If a(k)k,k ̸= 0, do( Ei a

(k) i,k

a(k)k,kEk)

→ (Ei) for k + 1≤ i ≤ n ⇒

A(k+1) =









a(1)1,1 · · · a(1)1,k a(1)1,k · · · a(1)1,n 0 . ..

.. .

.. .

.. . ..

. a(k)k,k a(k)k,k+1 · · · a(k)k,n ..

. 0 a(k+1)k+1,k+1 · · · a(k+1)k+1,n ..

.

.. .

.. .

.. . 0 · · · 0 a(k+1)n,k+1 · · · a(k+1)n,n









, b(k+1)=



b(k+1)1 b(k+1)2

.. . b(k+1)n

 .

The corresponding multipliers are given by

mi,k= a(k)i,k

a(k)k,k, i = k + 1, k + 2, . . . , n.

(41)

### At the kth Step of GE, 2≤ k ≤ n − 1 (Conti’d)

This is equivalent to

A(k+1) = M(k)A(k) and b(k+1)= M(k)b(k), where the kth Gaussian transformation matrix M(k)∈ Rn×n is defined by

M(k)=











0 · · · · · · 0

0 . .. . .. ...

...

0 ...

... mk+1,k

. .. ...

... ... . .. 0

0 mn,k









 .

(42)

. . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . .

. .

. . . . .

### The Inverse Matrix of M

(k)

It is easily seen that the inverse matrix of M(k) is given by

L(k)= [M(k)]−1=

0 · · · · · · 0

0 . .. . .. ...

...

0 ...

... mk+1,k

. .. ...

... ... . .. 0

0 mn,k

### 1

(2)

for each k = 1, 2, . . . , n− 1.

Check . . .

M(k)L(k)= I or L(k)M(k)= I, where I denotes the n× n identity matrix.

The simulation results of the DWT method and the least-squares matrix inversion method in x and y direction are shown in Table 1, The computer simulation results of the

In words, this says that the values of f(x) can be made arbitrarily close to L (within a distance ε, where ε is any positive number) by requiring x to be sufficiently large

In this paper, we provide new decidability and undecidability results for classes of linear hybrid systems, and we show that some algorithms for the analysis of timed automata can

 lower triangular matrix: 下三角矩陣.  upper triangular matrix:

Real Schur and Hessenberg-triangular forms The doubly shifted QZ algorithm.. Above algorithm is locally

Now, nearly all of the current flows through wire S since it has a much lower resistance than the light bulb. The light bulb does not glow because the current flowing through it

Because both sets R m  and L h i ði ¼ 1; 2; :::; JÞÞ are second-order regular, similar to [19, Theorem 3.86], we state in the following theorem that there is no gap between

Courtesy: Ned Wright’s Cosmology Page Burles, Nolette &amp; Turner, 1999?. Total Mass Density

The superlinear convergence of Broyden’s method for Example 1 is demonstrated in the following table, and the computed solutions are less accurate than those computed by

Since the subsequent steps of Gaussian elimination mimic the first, except for being applied to submatrices of smaller size, it suffices to conclude that Gaussian elimination

Since the subsequent steps of Gaussian elimination mimic the first, except for being applied to submatrices of smaller size, it suffices to conclude that Gaussian elimination

For R-K methods, the relationship between the number of (function) evaluations per step and the order of LTE is shown in the following

Now we assume that the partial pivotings in Gaussian Elimination are already ar- ranged such that pivot element a (k) kk has the maximal absolute value... The growth factor measures

Midpoint break loops are useful for situations where the commands in the loop must be executed at least once, but where the decision to make an early termination is based on

Large data: if solving linear systems is needed, use iterative (e.g., CG) instead of direct methods Feature correlation: methods working on some variables at a time (e.g.,

 If SAT can be solved in deterministic polynomial time, then so can any NP problems  SAT ∈ NP-hard..  If A is an NP-hard problem and B can be reduced from A, then B is an

For a 4-connected plane triangulation G with at least four exterior vertices, the size of the grid can be reduced to (n/2 − 1) × (n/2) [13], [24], which is optimal in the sense

3 recommender systems were proposed in this study, the first is combining GPS and then according to the distance to recommend the appropriate house, the user preference is used

The results showed that (1) in the evolution process of GNN, two phenomena can be observed that “the training period performance is truly relevant to test period performance” and

According to the simulation results, the sliding distances and the contact forces are the almost same on the upper and lower layers probe for the models of single pitch, ten

The purpose of this research is to develop an approach that uses the triangular distribution with the Fractile Method to estimate the optimistic and pessimistic duration of

Using the rational method and the triangular unit hydrograph for calculation, conferring the amount of the collected rainwater from different roof areas that is stored in