Simulated annealing

From Rosetta Code
Simulated annealing is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Quoted from the Wikipedia page : Simulated annealing (SA) is a probabilistic technique for approximating the global optimum of a given function. Simulated annealing interprets slow cooling as a slow decrease in the probability of temporarily accepting worse solutions as it explores the solution space.

Pseudo code from Wikipedia

Notations :
  T : temperature. Decreases to 0.
  s : a system state
  E(s) : Energy at s. The function we want to minimize
  ∆E : variation of E, from state s to state s_next
  P(∆E , T) : Probability to move from s to s_next. 
  	if  ( ∆E < 0 ) P = 1
  	      else P = exp ( - ∆E / T) . Decreases as T →  0
  
Pseudo-code:
    Let s = s0  -- initial state
    For k = 0 through kmax (exclusive):
        T ← temperature(k , kmax)
        Pick a random neighbour state , s_next ← neighbour(s)
        ∆E ← E(s) - E(s_next) 
        If P(∆E , T) ≥ random(0, 1), move to the new state:
            s ← s_next
    Output: the final state s

Problem statement

We want to apply SA to the travelling salesman problem. There are 100 cities, numbered 0 to 99, located on a plane, at integer coordinates i,j : 0 <= i,j < 10 . The city at (i,j) has number 10*i + j. The cities are all connected : the graph is complete : you can go from one city to any other city in one step.

The salesman wants to start from city 0, visit all cities, each one time, and go back to city 0. The travel cost between two cities is the euclidian distance between there cities. The total travel cost is the total path length.

A path s is a sequence (0 a b ...z 0) where (a b ..z) is a permutation of the numbers (1 2 .. 99). The path length = E(s) is the sum d(0,a) + d(a,b) + ... + d(z,0) , where d(u,v) is the distance between two cities. Naturally, we want to minimize E(s).

Definition : The neighbours of a city are the closest cities at distance 1 horizontally/vertically, or √2 diagonally. A corner city (0,9,90,99) has 3 neighbours. A center city has 8 neighbours.

Distances between cities
d ( 0, 7) → 7
d ( 0, 99) → 12.7279
d ( 23, 78) → 7.0711
d ( 33, 44) → 1.4142 // sqrt(2)

Task

Apply SA to the travelling salesman problem, using the following set of parameters/functions :

  • kT = 1
  • temperature (k, kmax) = kT * (1 - k/kmax)
  • neighbour (s) : Pick a random city u > 0 . Pick a random neighbour city v > 0 of u , among u's 8 (max) neighbours on the grid. Swap u and v in s . This gives the new state s_next.
  • kmax = 1000_000
  • s0 = a random permutation


For k = 0 to kmax by step kmax/10 , display k, T, E(s). Display the final state s_final, and E(s_final).

You will see that the Energy may grow to a local optimum, before decreasing to a global optimum.

Illustrated example Temperature charts

Numerical example

kT = 1
E(s0) = 529.9158

k:  0         T:  1       Es:  529.9158
k:  100000    T:  0.9     Es:  201.1726
k:  200000    T:  0.8     Es:  178.1723
k:  300000    T:  0.7     Es:  154.7069
k:  400000    T:  0.6     Es:  158.1412 <== local optimum
k:  500000    T:  0.5     Es:  133.856
k:  600000    T:  0.4     Es:  129.5684
k:  700000    T:  0.3     Es:  112.6919
k:  800000    T:  0.2     Es:  105.799
k:  900000    T:  0.1     Es:  102.8284
k:  1000000   T:  0       Es:  102.2426

E(s_final) =    102.2426    
Path  s_final =   ( 0 10 11 21 31 20 30 40 50 60 70 80 90 91 81 71 73 83 84 74 64 54 55 65 75 76 66
 67 77 78 68 58 48 47 57 56 46 36 37 27 26 16 15 5 6 7 17 18 8 9 19 29 28 38 39 49 59 69 
79 89 99 98 88 87 97 96 86 85 95 94 93 92 82 72 62 61 51 41 42 52 63 53 43 32 22 12 13 
23 33 34 44 45 35 25 24 14 4 3 2 1 0)  

Extra credit

Tune the parameters kT, kmax, or use different temperature() and/or neighbour() functions to demonstrate a quicker convergence, or a better optimum.

EchoLisp[edit]

 
(lib 'math)
;; distances
(define (d ci cj)
(distance (% ci 10) (quotient ci 10) (% cj 10) (quotient cj 10)))
(define _dists
(build-vector 10000 (lambda (ij) (d (quotient ij 100) (% ij 100)))))
(define-syntax-rule (dist ci cj)
[_dists (+ ci (* 100 cj))])
 
;; E(s) = length(path)
(define (Es path)
(define lpath (vector->list path))
(for/sum ((ci lpath) (cj (rest lpath))) (dist ci cj)))
 
;; temperature() function
(define (T k kmax kT)
(* kT (- 1 (// k kmax))))
#|
;; alternative temperature()
;; must be decreasing with k increasing and → 0
(define (T k kmax kT)
(* kT (- 1 (sin (* PI/2 (// k kmax))))))
|#

 
;; ∆E = Es_new - Es_old > 0
;; probability to move if ∆E > 0, → 0 when T → 0 (frozen state)
(define (P ∆E k kmax kT)
(exp (// (-E ) (T k kmax kT))))
 
;; ∆E from path ( .. a u b .. c v d ..) to (.. a v b ... c u d ..)
;; ∆E before swapping (u,v)
;; Quicker than Es(s_next) - Es(s)
 
(define (dE s u v)
;;old
(define a (dist [s (1- u)] [s u]))
(define b (dist [s (1+ u)] [s u]))
(define c (dist [s (1- v)] [s v]))
(define d (dist [s (1+ v)] [s v]))
;; new
(define na (dist [s (1- u)] [s v]))
(define nb (dist [s (1+ u)] [s v]))
(define nc (dist [s (1- v)] [s u]))
(define nd (dist [s (1+ v)] [s u]))
 
(cond
((= v (1+ u)) (- (+ na nd) (+ a d)))
((= u (1+ v)) (- (+ nc nb) (+ c b)))
(else (- (+ na nb nc nd) (+ a b c d)))))
 
;; all 8 neighbours
(define dirs #(1 -1 10 -10 9 11 -11 -9))
 
(define (sa kmax (kT 10))
(define s (list->vector (cons 0 (append (shuffle (range 1 100)) 0))))
(printf "E(s0) %d" (Es s)) ;; random starter
(define Emin (Es s)) ;; E0
 
(for ((k kmax))
(when (zero? (% k (/ kmax 10)))
(printf "k: %10d T: %8.4d Es: %8.4d" k (T k kmax kT) (Es s))
)
 
(define u (1+ (random 99))) ;; city index 1 99
(define cv (+ [s u] [dirs (random 8)])) ;; city number
#:continue (or (> cv 99) (<= cv 0))
#:continue (> (dist [s u] cv) 5) ;; check true neighbour (eg 0 9)
(define v (vector-index cv s 1)) ;; city index
 
(definee (dE s u v))
(when (or
(<e 0) ;; always move if negative
(>= (P ∆e k kmax kT) (random)))
(vector-swap! s u v)
(+= Emin ∆e))
 
;; (assert (= (round Emin) (round (Es s))))
) ;; for
 
(printf "k: %10d T: %8.4d Es: %8.4d" kmax (T (1- kmax) kmax kT) (Es s))
(s-plot s 0)
(printf "E(s_final) %d" Emin)
(writeln 'Path s))
 
Output:
(sa 1000000 1)

E(s0) 501.0909

k:  0         T:  1       Es:  501.0909
k:  100000    T:  0.9     Es:  167.3632
k:  200000    T:  0.8     Es:  160.7791
k:  300000    T:  0.7     Es:  166.8746
k:  400000    T:  0.6     Es:  142.579
k:  500000    T:  0.5     Es:  131.0657
k:  600000    T:  0.4     Es:  116.9214
k:  700000    T:  0.3     Es:  110.8569
k:  800000    T:  0.2     Es:  103.3137
k:  900000    T:  0.1     Es:  102.4853
k:  1000000   T:  0       Es:  102.4853

E(s_final)     102.4853    
Path     #( 0 10 20 30 40 50 60 70 71 61 62 53 63 64 54 44 45 55 65
 74 84 83 73 72 82 81 80 90 91 92 93 94 95 85 75 76 86 96 97 98 99
 88 89 79 69 59 49 48 47 57 58 68 78 87 77 67 66 56 46 36 35 25 24
 34 33 32 43 42 52 51 41 31 21 11 12 22 23 13 14 15 16 17 26 27 37 38
 39 29 28 18 19 9 8 7 6 5 4 3 2 1 0)

J[edit]

Implementation:

dist=: +/&.:*:@:-"1/~10 10#:i.100
 
satsp=:4 :0
kT=. 1
pathcost=. [: +/ 2 {&y@<\ 0 , ] , 0:
neighbors=. 0 (0}"1) y e. 1 2{/:~~.,y
s=. (?~#y)-.0
d=. pathcost s
step=. x%10
for_k. i.x+1 do.
T=. kT*1-k%x
u=. ({~ [email protected]#)s
v=. ({~ [email protected]#)I.u{neighbors
sk=. (<s i.u,v) C. s
dk=. pathcost sk
dE=. dk-d
if. (^-dE%T) >?0 do.
s=.sk
d=.dk
end.
if. 0=step|k do.
echo k,T,d
end.
end.
0,s,0
)

Notes:

E(s_final) gets displayed on the kmax progress line.

We do not do anything special for negative deltaE because the exponential will be greater than 1 for that case and that will always be greater than our random number from the range 0..1.

Also, while we leave connection distances (and, thus, number of cities) as a parameter, some other aspects of this problem made more sense when included in the implementation:

We leave city 0 out of our data structure, since it can't appear in the middle of our path. But we bring it back in when computing path distance.

Neighbors are any city which have one of the two closest non-zero distances from the current city (and specifically excluding city 0, since that is anchored as our start and end city).

Sample run:

   1e6 satsp dist
0 1 538.409
100000 0.9 174.525
200000 0.8 165.541
300000 0.7 173.348
400000 0.6 168.188
500000 0.5 134.983
600000 0.4 121.585
700000 0.3 111.443
800000 0.2 101.657
900000 0.1 101.657
1e6 0 101.657
0 1 2 3 4 13 23 24 34 44 43 33 32 31 41 42 52 51 61 62 53 54 64 65 55 45 35 25 15 14 5 6 7 17 16 26 27 37 36 46 47 48 38 28 18 8 9 19 29 39 49 59 69 79 78 68 58 57 56 66 67 77 76 75 85 86 87 88 89 99 98 97 96 95 94 84 74 73 63 72 82 83 93 92 91 90 80 81 71 70 60 50 40 30 20 21 22 12 11 10 0

zkl[edit]

Translation of: EchoLisp
var [const] _dists=(0d10_000).pump(List,fcn(abcd){ // two points (a,b) & (c,d), calc distance 
ab,cd,a,b,c,d:=abcd/100, abcd%100, ab/10,ab%10, cd/10,cd%10;
(a-c).toFloat().hypot(b-d)
});
fcn dist(ci,cj){ _dists[cj*100 + ci] } // index into lookup table of floats
 
fcn Es(path) // E(s) = length(path): E(a,b,c)--> dist(a,b) + dist(b,c)
{ d:=Ref(0.0); path.reduce('wrap(a,b){ d.apply('+,dist(a,b)); b }); d.value }
 
// temperature() function
fcn T(k,kmax,kT){ (1.0 - k.toFloat()/kmax)*kT }
 
// deltaE = Es_new - Es_old > 0
// probability to move if deltaE > 0, -->0 when T --> 0 (frozen state)
fcn P(deltaE,k,kmax,kT){ (-deltaE/T(k,kmax,kT)).exp() } //-->Float
 
// deltaE from path ( .. a u b .. c v d ..) to (.. a v b ... c u d ..)
// deltaE before swapping (u,v)
fcn dE(s,u,v){ su,sv:=s[u],s[v]; //-->Float
// old
a,b,c,d:=dist(s[u-1],su), dist(s[u+1],su), dist(s[v-1],sv), dist(s[v+1],sv);
// new
na,nb,nc,nd:=dist(s[u-1],sv), dist(s[u+1],sv), dist(s[v-1],su), dist(s[v+1],su);
 
if (v==u+1) (na+nd) - (a+d);
else if(u==v+1) (nc+nb) - (c+b);
else (na+nb+nc+nd) - (a+b+c+d);
}
 
// all 8 neighbours
var [const] dirs=ROList(1, -1, 10, -10, 9, 11, -11, -9),
fmt="k:%10,d T: %8.4f Es: %8.4f".fmt; // since we use it twice
 
fcn sa(kmax,kT=10){
s:=List(0, [1..99].walk().shuffle().xplode(), 0); // random path from 0 to 0
println("E(s0) %f".fmt(Es(s))); // random starter
Emin:=Es(s); // E0
 
foreach k in (kmax){
if(0==k%(kmax/10)) println(fmt(k,T(k,kmax,kT),Es(s)));
u:=(1).random(100); // city index 1 99
cv:=s[u] + dirs[(0).random(8)]; // city number
if(not (0<cv<100)) continue; // bogus city
if(dist(s[u],cv)>5) continue; // check true neighbour (eg 0 9)
v:=s.index(cv,1); // city index
 
deltae:=dE(s,u,v);
if(deltae<0 or // always move if negative
P(deltae,k,kmax,kT)>=(0.0).random(1)){
s.swap(u,v);
Emin+=deltae;
}
// (assert (= (round Emin) (round (Es s))))
}//foreach
 
println(fmt(kmax,T(kmax-1,kmax,kT),Es(s)));
println("E(s_final) %f".fmt(Emin));
println("Path: ",s.toString(*));
}
sa(0d1_000_000,1);
Output:
E(s0) 540.897080
k:         0 T:   1.0000 Es: 540.8971
k:   100,000 T:   0.9000 Es: 181.5102
k:   200,000 T:   0.8000 Es: 167.1944
k:   300,000 T:   0.7000 Es: 159.0975
k:   400,000 T:   0.6000 Es: 170.2344
k:   500,000 T:   0.5000 Es: 130.9919
k:   600,000 T:   0.4000 Es: 115.3422
k:   700,000 T:   0.3000 Es: 113.9280
k:   800,000 T:   0.2000 Es: 106.7924
k:   900,000 T:   0.1000 Es: 103.7213
k: 1,000,000 T:   0.0000 Es: 103.7213
E(s_final) 103.721349
Path: L(0,10,11,21,20,30,40,50,60,70,80,81,71,72,73,63,52,62,61,51,41,31,32,22,12,13,14,15,25,16,17,18,28,27,26,36,35,45,34,24,23,33,42,43,44,54,53,64,74,84,83,82,90,91,92,93,94,95,85,86,96,97,87,88,98,99,89,79,69,68,78,77,67,66,76,75,65,55,56,46,37,38,48,47,57,58,59,49,39,29,19,9,8,7,6,5,4,3,2,1,0)