26 April 2013 Lecture 6: Problem Solving 1

Problem Solving

Jorge CruzDI/FCT/UNL

April 2013

Lecture 6: Problem Solving 2

Problem Solving

Variable Elimination

Dependency Reduction

Modelling Techniques

System Scaling

Constraint Solving Systems

Interval Libraries

Some Languages and Tools

Some Common Benchmarks

Some Problems

26 April 2013

Final Project

Lecture 6: Problem Solving 3

Modelling Techniques

Frequently a single problem may be modelled by several equivalent CCSPs

The behaviour of a constraint solver may change drastically even with equivalent CCSPs

To choose the best model for a particular problem is important to understand the underlying constraint propagation algorithms

Some modelling techniques are commonly adopted for improving the accuracy and efficiency of the continuous constraint solvers

26 April 2013

Lecture 6: Problem Solving 4

Modelling Techniques

A fundamental problem of interval arithmetic is the dependency problem (see lecture 2).

Dependency Reduction

Dependency Problem. In the interval arithmetic evaluation of an interval expression, each occurrence of the same variable is treated as a different variable. The dependency between the different occurrences of a variable in an expression is lost.

Some expressions may be rewritten into equivalents that minimize the dependency problem

Factorize as much as possible polynomial expressions:Instead of using constraint x2y2+xy2+xy=0 use constraint xy(y(x+1)+1)=0

Use better interval extensions (mean value form, Taylor form,…):Instead of using constraint xx2=0 use constraint 0.25(x0.5)2=0

Examples:

26 April 2013

Lecture 6: Problem Solving 5

Modelling Techniques

Precision and efficiency may be improved if the number of variables is reduced

Variable Elimination

Sometimes a set of constraints may be rewritten into an equivalent set with less variables

Example:

Continuous constraint solvers rely on the efficiency of branch and prune algorithms for enforcing consistency on the CCSP variables

Instead of using the constraint system:

x1 + x2 + x3 = 1(x1 + x1x2 + x2x3) x4= c1

(x2 + x1x3) x4= c2

x3x4 = c3

Consider x4 = c3/x3 and use the constraint system:x1 + x2 + x3 = 1(x1 + x1x2 + x2x3)c3 = c1x3

(x2 + x1x3)c3 = c2x326 April 2013

Lecture 6: Problem Solving 6

Modelling Techniques

A consequence of numerical errors is the amplification of the variable domains and poor pruning results

System Scaling

Two major sources of numerical errors are:operations with large numbers (lower density of F-Numbers at this ranges)operands with different magnitudes

Scaling the system and making some variable substitutions may avoid such situations as much as possible

Continuous constraint solvers rely on interval techniques for dealing with numerical errors.

Consider x = 1020y and use the constraint: y2+3y+2=0

Example:

Instead of using constraint: 10-20x2+3x+2×1020=0

26 April 2013

Lecture 6: Problem Solving 7

Some Languages and Tools

PROFIL/BIAS (Programmer's Runtime Optimized Fast Interval Library) http://www.ti3.tu-harburg.de/Software/PROFILEnglisch.html

Interval Libraries

GAOL: NOT Just Another Interval Library

http://sourceforge.net/projects/gaol

smath (routines for interval arithmetic and constraint narrowing)

http://interval.sourceforge.net/interval/prolog/clip/clip/smath/README.html

CFILIB (Fast Interval LIBrary)

http://www.math.uni-wuppertal.de/org/WRST/software/filib.html

FILIB++ (Fast Interval LIBrary in C++)

http://www.math.uni-wuppertal.de/org/WRST/software/filib.html

C++

26 April 2013

Lecture 6: Problem Solving 8

Some Languages and Tools

ILOG Solver (C++)http://www.ilog.com/products/solver

Constraint Solving Systems

Elisa (C++)

http://sourceforge.net/projects/elisa

RealPaver (C++)

http://www.sciences.univ-nantes.fr/info/perso/permanents/granvil/realpaver

AcademicCLIP (Prolog)

http://interval.sourceforge.net/interval/prolog/clip/README.html

Commercial

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Lecture 6: Problem Solving 9

Some Problems

Many continuous problems may be modelled and solved as CCSPs.

Some Common Benchmarks

Several benchmarks are commonly used to address the quality of the constraint solvers.

The following problems illustrate some common benchmarks in interval constraints.

For each benchmark we would like to know where are the solutions (if any) satisfying the constraints within the specified ranges

26 April 2013

Lecture 6: Problem Solving 10

Some Problems

2x1=2 2x2=1 2x3=2 2x4=1 2x5=x6

x5+2x6=2 (x1-x7)2+x8

2-x72=(x8-x2)2

(x1-x7)2+x82-(x3-x7)2=(x4-x8)2

x12+x2

2-2x1x7+2x2x8=0 x1

2-x33-x4

2-2x7(x1-x3)+2x4x8=0

Appolonius

Ranges:

Constraints:

x1, x2, x3, x4, x5, x6, x7, x8[-108,108]

26 April 2013

Lecture 6: Problem Solving 11

Some Problems

x2 + y2 + z2 = 36x + y = zxy + z2 = 1

Bronstein

Ranges:

Constraints:

x, y, z[-108,108]

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Lecture 6: Problem Solving 12

Some Problems

Brown

Ranges:

Constraints:

n = 5

x1, x2, x3, x4, x5[-108,108]

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Lecture 6: Problem Solving 13

Some Problems

R = 10 R5 = 0.193 R6 = 0.002597/40 R7 = 0.003448/40 R8 = 0.00001799/40 R9 = 0.0002155/40 R10 = 0.00003846/40 3x5=x1(x2+1) x2(2x1+x3

2+R8+2R10x2+R7x3+R9x4)+x1=Rx5

x3(2x2x3+2R5x3+R6+R7x2)=8x5

x4(R9x2+2x4)=4Rx5

x2(x1+R10x2+x32+R8+R7x3+R9x4)+x1+x3(R5x3+R6)+x4

2=1

Chem

Ranges:

Constraints:

x1, x2, x3, x4, x5[0,108]

26 April 2013

Lecture 6: Problem Solving 14

Some Problems

x12 = x2

x42 = x3

2.177e7x2 + 0.55x1 x4 + 0.45x1 = x4 + 1.697e7x2x41.585×1014x2x4+4.126×107x1x3+2.284×107x3 x4+48.4x4 =

8.282×106x1 x4+1.918×107x3+27.73

Chemk

Ranges:

Constraints:

x1, x2, x3, x4[0,1]

26 April 2013

Lecture 6: Problem Solving 15

Some Problems

y2z2 + y2 + z2 + 13 = 24yzx2z2 + x2 + z2 + 13 = 24xzx2y2 + x2 + y2 + 13 = 24xy

Cyclo

Ranges:

Constraints:

x, y, z[0,105]

26 April 2013

Lecture 6: Problem Solving 16

Some Problems

x2 + 2x6 + x9 +2x10 = 10-5

x3 + x8 = 3×10-5

x1 + x3 +2x5 + 2x8 + x9 + x10 = 5×10-5

x4 + 2x7 = 10-5

0.5140437×10-7x5 = x12

0.1006932×10-6x6 = 2x22

0.7816278×10-15x7 = x42

0.1496236×10-6x8 = x1x3

0.6194411×10-7x9 = x1x2

0.2089296×10-14x10 = x1x22

Combustion

Ranges:

Constraints:

x1, x2, x3, x4, x5, x6, x7, x8 , x9 , x10 [-108,108]

26 April 2013

Lecture 6: Problem Solving 17

Some Problems

x1 +5x22 = x2

3 + 2x2 + 13x1 + x2

3 + x22

= 14x2 + 29

Freudenstein

Ranges:

Constraints:

x1, x2[-108,108]

26 April 2013

Lecture 6: Problem Solving 18

Some Problems

2x22 + 4x1x2 + 4x1

3 = 42x1 + 142x1

2 + 4x1x2 + 4x23 = 26x2 + 22

Himmelblau

Ranges:

Constraints:

x1, x2[-108,108]

26 April 2013

Lecture 6: Problem Solving 19

Some Problems

x1 = 0.25428722 + 0.183247757 x4 x3 x9 x2 = 0.37842197 + 0.16275449 x1 x10 x6 x3 = 0.27162577 + 0.16955071 x1 x2 x10 x4 = 0.19807914 + 0.15585316 x7 x1 x6 x5 = 0.44166728 + 0.19950920 x7 x6 x3 x6 = 0.14654113 + 0.18922793 x8 x5 x10 x7 = 0.42937161 + 0.21180486 x2 x5 x8 x8 = 0.07056438 + 0.17081208 x1 x7 x6 x9 = 0.34504906 + 0.19612740 x10 x6 x8 x10 = 0.42651102 + 0.21466544 x4 x8 x1

I1

Ranges:

Constraints:

x1, x2, x3, x4, x5, x6, x7, x8 , x9 , x10 [-2,2]

26 April 2013

Lecture 6: Problem Solving 20

Some Problems

6(x1 x2 - x3 x4) + 10 x1 = 1 6(x1 x4 + x2 x3) + 10 x3 = 4 x1

2 + x32 = 1

x22 + x4

2 = 1

Kincox

Ranges:

Constraints:

x1, x2, x3, x4 [-1,1]

26 April 2013

Lecture 6: Problem Solving 21

Some Problems

eg + 2dh = 09e + 4b = 04ch + 2ef + 3dg = 07c + 8f = 9a4df + 5cg + 6h + 3e = 05d + 6cf + 7g - 9b = 09d + 6a = 5b7a = 9c + 8

Nauheim

Ranges:

Constraints:

a, b, c, d, e, f, g, h [-108,108]

26 April 2013

Lecture 6: Problem Solving 22

Some Problems

x12 + x3

2 =1x2

2 + x42 =1

x5 x33 + x6 x4

3 = 5x5 x1

3 + x6 x23 = 4

x5 x1 x32 + x6 x4

2 x2 = 3x5 x1

2 x3 + x6 x22 x4 = 2

x2 x1

Neuro

Ranges:

Constraints:

x1, x2, x3, x4, x5, x6 [-10,10]

26 April 2013

Lecture 6: Problem Solving 23

Some Problems

x - lt3 - Lw =0y - Lt - lw3 =0L = 1 l = 2x = a2a = 3y = bb = rtw2 + t2 = 1

Piano

Ranges:

Constraints:

x, y, a, b, r, t, l, L[-105,105] w [0,105]

26 April 2013

Lecture 6: Problem Solving 24

Some Problems

x12 + x2

2 =1x3

2 + x42 =1

x52 + x6

2 =1x7

2 + x82 =1

0.004731x1x3 + x7 = 0.3578x2x3 + 0.1238x1 + 0.001637x2 + 0.9338x4 + 0.35710.2238x1x3 + 0.7623x2x3 + 0.2638x1 = 0.07745x2 + 0.6734x4 + 0.6022 + x7

x6 x8 + 0.3578x1 + 0.004731x2 = 00.2238x2 + 0.3461 = 0.7623x1

Robot Kinematics

Ranges:

Constraints:

x1, x2, x3, x4, x5, x6, x7, x8 [-1,1]

26 April 2013

Lecture 6: Problem Solving 25

Project

Use interval constraint techniques to solve systems of nonlinear equations that models Social Argumentation Networks [1].

The problem

These are systems with n equations and n unknowns of the form:

with 1in, xi[0,1], ai(0,1) and mji{0,1}

[1] J. Leite and J. Martins. Social abstract argumentation. In Proceedings of IJCAI'11, volume 3, 2011.

26 April 2013

n

jjjiii xmax

1

)1(

Lecture 6: Problem Solving 26

Project

Solve the following systems:a)

b)

c)

Task 1

26 April 2013

)1(8.0)1(2.0

)1)(1(86.0)1)(1(5.0

)1(5.0

45

54

543

312

21

xxxx

xxxxxx

xx

)1)(1(9.0)1)(1(9.0)1)(1(9.0

323

212

311

xxxxxxxxx

)1)(1(8.0)1)(1(7.0)1)(1(9.0

213

312

321

xxxxxxxxx

Lecture 6: Problem Solving 27

Project

Solve the systems represented in the files: a) example1b) example2c) example3d) example4e) example5f) example6g) example7h) example8

with the following format: the first line contains an integer representing the number of variables n; the next n lines represent the values of ai; the next n lines represent the matrix m.

Task 2

26 April 2013

Lecture 6: Problem Solving 28

Project

a) Implement na application to generate and solve a random system of n variables.

b) Solve random systems of 5, 10, 15, 20, 25, … variables (10 instances of each) and produce a table with statistical information of its performance (execution time, number of splits, …)

Task 3

26 April 2013