?? problems.txt~
字號:
1 1 12 21
1 1 1 1
0
0
0
0
0
1
1 3
30
30 -20 20 0
0
0 0.00001
0
0 0 100
-40000 -40000 -1
--------------
Put the problem parameter set(s) BEFORE this line. See examples below
If you want to run several problems, precise how many
(very first number after the strategy set) >
WARNING. A multiojective problem has to be ran alone
( some pb with the run.txt and run_cleaned.txt files)
DATA FORMAT
------
<strategy set> Just for test purpose. Will be hard coded, later.
4 integers = 4 strategies,
strategy 1 is for "bad" particles
strategy 2 for "normal" ones
strategy 3 for "good" ones
strategy 4 for "excellent" ones
For most of problems, the set (1 1 12 21) seems quite good
You may also try (1 1 12 25) (1 1 12 12)
or simply (23 23 23 23) or even (25 25 25 25) for some particular cases
For hybrid problems, you may use (21 21 21 21) (independent gaussian distributions)
(9 9 9 9) is also not bad, but needs the hard coded phi parameter
------
<nonunif set>
4 floats = 4 nonunif coefficients, one for each strategy.
nonunif coeffs are for distributions of possibilities.
1 => uniform distribution
If a nonunif coeff is # 1, for the corresponding strategy,
the distribution is not uniform
(see rand_in_hypersphere()).
if nonunif<0 => Gaussian distribution. Standard deviation = -nonunif
nonunif = (1 1 1 1) is often OK
------
<confin_interv> If <0, no interval (boundary) confinement
------
<circular_hood> If >0, it is the size of a circular neighbourhood. Then tribes and i-groups are not
taken into account Mainly useful for comparison.
For example, by setting it at 3, using always strategy 9, and a constant swarm size (see below)
you obtain the classical "constricted PSO".
If <-1, it means "choose a random i-group whose size is -circular_hood"
If =-1, "choose a random i-group whose size is (2/N)*ln(eps)/ln(1-1/N)". N=swarm size
------
<no_best> 0, normal best neighbour
1, the "best neighbour" is in fact chosen at random
2, use pseudo gradient to define the "best" informer
3
------
<adapt> If 0, complete adaptive method
If >0, it is in fact the constant swarm size used for the non adaptive method
------
<linkreorg> Use (1) or not (0) information link reorganization
------
<nb_pb_max> Number of different problems to solve
------
------
<nb_f> number of objective functions to simultaneously optimize
(needed for multiobjective optimization)
if <0, use lexicographical order.
------
<function_code> (see functions.txt or, to be sure, myfunction.c)
if <0, it means that fuzzification has to be used. The real function code is then abs(function_code)
------
<dimension >
(use <n> times <xmin xmax granularity (0 if continuous)>)*
* you can have several lines like this one, for you can have different granularities
along different dimensions
Be SURE that the sum of the all n values is equal to the total dimension
If granularity <0, the problem is seen as a special binary one.
In that case, be sure xmin>-0.5 and xmax<1.5
------
<constrain> For special constraint cases. Usually 0. See constrain.c
1 for special list of discrete values on the file discrete.txt
------
<target> Warning. For a multiobjective problem
"target" is only for the first function
------
<precision>
------
< all components of the solution must be different (1), or not necessary (0)>
------
<print_level>
------
< save > 1 => save swarm on swarm.txt
>1 => save also some data on trace.txt
In any case, see run.txt for result
-1 => save energies on energy.txt
-2 => save the number of evaluations and the best result on file trace_run.txt
(in order to plot the convergence curve)
------
<number of runs>
------
<Max_Eval> Max number of clock ticks if >0. If <0, the absolute value is the max number of evaluations
----
<Max_Eval_2> Second "max number" if you want a whole sequence
<Max_Eval_delta> Step from Max_Eval to Max_Eval_delta
----
if function_code==68 (binary multimodal)
<number_of_peaks>
============ Examples you can copy/paste at the beginning of this file
DO NOT copy/paste the name nor the comment (first text line(s) )
1) DD-sum (sum=200 with 10 different integers in [1,100])
Best parameters for classical PSO (strategy 9): hood 3, size 9
1 1
10
10 1 100 1
0
200 0.1
1
0 0 20
-10000 -10000 -1
2) DD-product
3) Parabola (Sphere)
1 3
30
30 -20 20 0
0
0 0.00001
0
0 0 100
-40000 -40000 -1
4) Rosenbrock
1 4
30
30 -10 10 0
0
0 0.00001
0
0 0 100
-40000 -40000 -1
5) Alpine
1 5
10
10 -10 10 0
0
0 0.00001
0
0 0 100
-15000 -15000 -1
6) Griewank nD, n>2 (min 0 en (100, ... 100))
1 6
30
30 -300 300 0
0
0 0.00001
0
0 0 100
-40000 -40000 -1
7) Rastrigin
1 7
30
30 -5.12 5.12 0
0
0 0.0000
0
0 0 50
-20000 -20000 -1
8) Fifty_fifty (non combinatorial)
with fuzzification
1 -8
20
20 1 100 1
0
0 0.01
1
0 0 10
-1000 -1000 -1
9) Ackley
1 9
30
30 -30 30 0
0
0 0.00001
0
0 0 100
-40000 -40000 -1
10) Foxholes (with fuzzification)
1 -10
2
2 -50 50 0
0
0 0.998004
0
0 0 10
-40000 -40000 -1
11) Apple trees
1 11
3
3 0 1 0
0
1 0.1
0
0 0 1
-200 -200 -1
12) El Farol problem. 100 Irishmen => best solution = 30 at the pub (function value = 15)
after about 50 evaluations
Note that if the problem is considered as not combinatorial,
it is more difficult to find a solution
1 12
100
100 0 1 1
0
15 0.000001
0
1 0 1
-40000 -40000 -1
13) Fermat (find, for example (3,4,5), (24,32,40), (84,13,85) etc.)
1 13
3
3 2 100 1
0
0 0.01
0
1 0 4
-40000 -40000 -1
14) Knap_sack (10 different integer numbers in [1,100] whose sum is equal to 100
See also DD-sum
1 14
10
10 1 100 1
0
100 0.01
1
0 0 20
-500 -500 -1
15)
16)
17)
18) 2D Linear_system
1 18
3
3 -3 3 0.001
0
0 0.001
0
1 0 1
-40000 -40000 -1
19)
20)
21) Magic square (3x3) (different integer numbers of [1,100])
You can also try 16 (4x4), 25 is still possible ... if you are lucky
1 21
9
9 1 100 1
0
0 0.001
1
0 0 10
-10000 -10000 -1
22) Model tuning
Note: hard coded in myfunction.c. Currently, the target is in fact 1918.95
for (10.073 1.786)
1 22
2
1 1 100 0.00
1 1 2 0.00
0
0 0.001
0
0 0 10
-5000 -5000 -1
23) Master Mind (solution (1,2,3,4) hard coded in myfunction.c)
Just for fun: it is very bad. You should use my "ultimate" unbeatable MasterMind program ;-)
1 23
4
4 1 6 1
0
0 0.001
0
1 0 1
-1000 -1000 -1
24) Catalan's conjecture
1 24
4
1 2 30 1
1 2 10 1
1 2 30 1
1 2 10 1
0
0 0.001
0
0 0 10
-40000 -40000 -1
25) //
26)
27) //
28)
29) Cognitive harmony
Note: need to have the matrix in matrix.txt
1 29
6
6 0 1 0
0
24 0.001
0
0 0 100
-1000 -1000 -1
30) Non specific TSP (just for test). You should use PSO_for_TSP
You need the file tsp_br17 for this problem
You should find a solution after about 2000 evaluations
Note the granularity set to 1
and also the option all_different
1 30
17
17 0.0 16 1
0
39.0 0.9
1
1 0 5
-10000 -10000 -1
With tsp_4
1 30
4
4 0.0 3 1
0
4 0.9
1
1 0 10
-10 -10 -1
With tsp_6
1 30
6
6 0.0 5 1
0
6 0.9
1
1 0 10
-20 -20 -1
31) Sum of absolute values. You can use it with granularity = 1 (integer problem)
1 31
5
5 -100 100 1
0
0 0.1
0
0 0 1
-40000 -40000 -1
32) Non specific QAP
(use data file qap_scr12)
1 32
12
12 0 11 1
0
31410.0 0.9
1
0 0 10
-5000 -5000 0
33) Multiobjective Lis-Eiben 1
Each run gives a point whose (f1,f2) is near of the Pareto front
See result in run.txt. (f1,f2) are in the Error columns
2 33
2
2 0 1 0
0
0 0.001
0
0 0 200
-20 -20 -1
34) Multiobjective Schaffer
2 34
2
2 -5 10 0
0
0 0.001
0
0 0 200
-20 -20 -1
35) Multiobjective F1
2 35
2
2 0 1 0
0
0 0.001
0
0 0 200
-20 -20 -1
36) Multiobjective F2
2 35
2
2 0 1 0
0
0 0.001
0
0 0 200
-20 -20 -1
37) Multiobjective F3 (simple front)
38) Multiobjective F4 (convex, concave)
39) Multiobjective F5 (5 Pareto fronts)
2 39
2
2 0 1 0
0
0 0.001
0
0 0 300
-100 -100 -1
40) Multiobjective F6
41) Multiobjective Coello F3
42)
43)
44) Jeannet-Messine (cf ROADEF 2003 proceedings p 273)
min -112.5 in position (3,-7.5, 10)
The authors define 5 different methods, and
their best result is: 3271 evaluations
1 44
3
1 0 5 1
1 -15 25 0
1 3 10 0
0
-112.5 0.000001
0
0 0 10
-2000 -2000 -1
45) MINLP X. Yan
2 45
72
36 0 1 1
12 0 600 0
12 0 400 0
12 0 200 0
0
0 0.01
0
1 0 2
-100000 -100000 -1
46) Tripod (Louis Gacogne)
on [-100, 100]^2, min 0 at (0,-50)
Seems simple but it is in fact difficult
to obtain a success rate > 50%
Strategies 1 1 12 21 => success rate=52%
Strategies 1 1 1 1 => 64%
1 46
2
2 -100 100 0
0
0 0.00001
0
0 0 100
-40000 -40000 -1
(Compare with strategies 1 1 1 1)
47) Tripod, with circular constraint, and multiobjective optimization
2 47
2
2 -100 100 0
0
0 0.001
0
0 0 10
-3000 -3000 -1
48) DeJong f4
1 48
30
30 -20 20 0
0
0 0.0000
0
0 0 50
-20000 -20000 -1
49) Pressure vessel
1 49
4
1 1.125 12.5 0.0625
1 0.625 12.5 0.0625
2 0.01 240 0
0
0 0.0
0
0 0 100
-51818 -51818 -1
50) Coil compressing spring
Note the constrain=1 (see constrain.c) for discrete values list
1 50
3
1 1.0 70.0 1
1 0.6 3.0 0.0
1 0.2 0.5 0.0
1
0 0.0
0
0 0 100
-12500 -12500 -1
51) Gear train
1 51
4
4 12 60 1
0
0 0.0
0
0 0 100
-19800 -19800 -1
52) Pressure vessel as multiobjective
4 52
4
1 1.125 12.5 0.0625
1 0.625 12.5 0.0625
2 0.01 240 0
0
0 0.0
0
0 0 100
-51818 -51818 -1
53) Coil compressing spring as multiobjective
Note the constrain=1 (see constrain.c) for discrete values list
6 53
3
1 1.0 70.0 1
1 0.6 3.0 0.0
1 0.2 0.5 0.0
1
0 0.0
0
0 0 100
-12500 -12500 -1
54)
55) Neural Network Training. XOR
1 55
9
9 -30 30 0
0
0 0.001
0
0 0 30
-40000 -40000 -1
56)Neural Network Training. 4 bit parity
1 56
25
25 -100 100 0
0
0 0.001
0
0 0 30
-40000 -40000 0
57) Neural Network Training. Three Color Cube
1 57
46
46 -100 100 0
0
0 0.001
0
0 0 30
-40000 -40000 -1
58) Neural Network Training. Diabetes in Pima Indians
1 58
64
64 -100 100 0
0
0 0.001
0
0 0 30
-40000 -40000 -1
59) Neural Network Training. Sin Times Sin
1 59
26
26 -100 100 0
0
0 0.001
0
0 0 30
-40000 -40000 -1
60) Neural Network Training. Time Servomechanism
1 60
28
28 -100 100 0
0
0 0.001
0
0 0 30
-40000 -40000 -1
61) Move Peaks
Warning. Some parameters are in fact hard coded (see movpeaks_mc.c)
Also, the interval of values must be the same for all dimensions
1 61
5
5 0 100 0
0
0 0.00001
0
0 0 50
-500000 -500000 -1
62) Goldberg?s order 3 deceptive problem
For such a problem, strategy set 1 12 12 21 may be better
Note the granularity set to -1 (special biary problem)
1 62
60
60 0 1.4999 -1
0
0 0.00001
0
0 0 10
-100000 -100000 -1
63) M?hlenbein's order 5,
15 binary string
seen as a 1D problem
1 12 21 21
1 1 1 1
0
0
0
0
0
1
1 63
1
1 0 32767 1
0
0 0.01
0
0 0 1
-40000 -40000 -1
64) Schaffer's f6
1 1 12 21
1 1 1 1
0
0
0
0
0
1
1 64
2
2 -100 100 0
0
0 0.00
0
0 0 1
-40000 -40000 -1
68) Binary. Multimodal (D=100, 20 peaks)
1 68
100
100 0 1 1
0
0 0.000
0
0 0 100
-1000 -20000 -1000
20
======================================= to test
use data file tsp_x50.1 (solution <= 5.89)
21 21 21 21
1 1 1 1
0
0
0 1
30
50
50 0.0 49.0 1
0
0 0.01
1
1 0 2
-1000000 -1000000 -1
=================== current tests
abs(sin(x) , abs(x)
9 9 9 9
1 1 1 1
3
0
20 2
99
1
1 1 10 0
0
0 0.00001
0
1 0 2
-1000 -1000 -1
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