Tôi đã bắt đầu làm việc với vấn đề này và muốn đóng góp với mã của tôi cho đến nay. Như Gareth đã nêu, vấn đề này có thể so sánh với câu đố 8 ô và do đó, mã được dựa trên giải pháp tuyệt vời của Keith Randall và do đó trong Python. Giải pháp này có thể giải quyết tất cả 5 trường hợp thử nghiệm với tổng số ít hơn 400 lần di chuyển và các câu đố khác nữa. Nó chứa một giải pháp tối ưu và vũ phu. Mã bây giờ là một chút cồng kềnh. Đầu ra được viết tắt như "llururd .." Hy vọng nó hữu ích.
http://www.penschuck.org/j Joomla / tmp / 15Tile.txt (giải thích)
http://www.penschuck.org/j Joomla / tmp / tile15.txt (mã python)
# Author: Heiko Penschuck
# www.penschuck.org
# (C) 2012
# import os;os.chdir('work')
# os.getcwd()
# def execfile(file, globals=globals(), locals=locals()):
# with open(file, "r") as fh: exec(fh.read()+"\n", globals, locals)
#
#
# execfile("tile15.py");
#
## run these
# solve_brute();
# solve();
# some boards to play with
board2=(15,14,7,3,13,10,2,9,11,12,4,6,5,0,1,8);
# best: 76(52)
# 72(56)
# 68(51) uurddlurrulldrrdllluuruldrddlururulddruurdllldrurddlurdruuldrdluurdd
board3=(13, 8, 9, 4, 15, 11, 5, 3, 14, 6, 12, 7, 1, 10, 2, 0)
# best: 106(77)
#best: 90(64) ullldruuldrrdrlluurulldrrdldluruulddrulurrdrddlluuurdldrrulddrulldrurullldrdluurrrddllurdr
board4=(4, 8, 12, 1, 13, 7, 3, 11, 9, 15, 6, 14, 5, 2, 10, 0) ;# best 100(74)
board5=(15,2,3,4,5,6,7,8,9,10,11,12,13,1,14,0); # best 44(32)
board6=( 1, 2, 3, 4, 6, 11, 0, 12, 8, 14, 9, 13, 5, 10, 7, 15);
# testcases
board7=(5,1,7,3,9,2,11,4,13,6,15,8,0,10,14,12); # 15 (7)
board8=(2,5,13,12,1,0,3,15,9,7,14,6,10,11,8,4); # 124 (94)
board9=(5,2,4,8,10,0,3,14,13,6,11,12,1,15,9,7) ; # 72 (56)
board10=(11,4,12,2,5,10,3,15,14,1,6,7,0,9,8,13) ;# 71 (57)
board11=(5,8,7,11,1,6,12,2,9,0,13,10,14,3,4,15) ;# 99 (73)
board12=(1,2,3,4,5,6,7,8,9,10,11,12,13,0,14,15); #pretty simple board
board13=(4, 10, 5, 12, 11, 7, 15, 2, 13, 1, 14, 8, 6, 3, 9, 0)
board=board3 ; # used by solve()
bboard=list(board) ;# used by solve_brute()
# init
clean=(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,0)
i=0;
solution={};
invsolution={};
E={board:0}
# derived from Keith Randall 8-tile solution
# a: a board, d: offset to move from i: index in board
def Y(a,d,i):
b=list(a); # b is now an indexable board
b[i],b[i+d]=b[i+d],0; # make a move (up down left right)
b=tuple(b); # now back to searchable
if b not in E:E[b]=a;# store new board in E
def Calc():
ii=0;
# memory error when x is 21
for x in ' '*14:
if ii>10:
print(ii);
ii+=1
for a in E.copy():
# for all boards, make possible moves (up,left,right,down) and store the new boards
i=list(a).index(0)
if i>3:Y(a,-4,i)
if i%4:Y(a,-1,i)
if i%4 <3:Y(a,1,i)
if i<12:Y(a,4,i)
def weigh(a,goal):
factor=[26,8,4,6, 8,8,4,4, 4,4,1,1, 3,2,1,0]
weight=0;
for element in a:
i=list(a).index(element);
ix,iy=divmod(i,4); # ist
if element == 0:
# special for gap
weight=weight+ix;
#weight+=(ix+iy)
continue;
i=list(a).index(element);
ix,iy=divmod(i,4); # ist
j=list(goal).index(element);
sx,sy=divmod(j,4); # soll
#k=list(a).index(0); # gap
#kx,ky=divmod(k,4)
# try solving from topleft to bottom right (because clean board has gap at bottomright)
tmp= abs(sx-ix)*abs(sx-ix)*factor[j]+ abs(sy-iy)*abs(sy-iy)*factor[j]
#tmp += ((sx!=ix )& (sy!=iy)) *(4-sx)*(4-sy)*4
weight+=tmp
#(10-sx-sy-sy)
# 8*abs(sx-ix) + (16-j)*(sx!=ix)
#print('%2d %2d_%2d (%2d_%2d)=> %d'%(element,i,j,(sx-ix),(sy-iy),weight))
return weight
# read numbers seperated by a whitespace
def readboard():
global E,D,board,clean,i
reset()
g=[]
for x in' '*4:g+=map(int,input().split())
board=tuple(g)
# read 'a' till 'o'
def readasciiboard():
global E,D,board,clean,i
trans={"0":0,"a":1,"b":2,"c":3,"d":4,"e":5,"f":6,"g":7,"h":8,"i":9,"j":10,"k":11,"l":12,"m":13,"n":14,"o":15}
reset()
g=[]
vec=tuple(input().split());
for x in vec: g.append(trans[x])
board=tuple(g)
def printasciiboard(a):
trans={"0":0,"a":1,"b":2,"c":3,"d":4,"e":5,"f":6,"g":7,"h":8,"i":9,"j":10,"k":11,"l":12,"m":13,"n":14,"o":15}
itrans={}
for x in trans: itrans[trans[x]]=x
g=[]
for x in a: g.append(itrans[x])
for i in(0,4,8,12): print('%s %s %s %s'%tuple(g[i:i+4]))
# find the board with the smallest weight
def minimum():
global minn,E,clean
minn=1111111;# start with a huge number
qq=board
for q in E:
if weigh(q,clean) < minn:
minn=weigh(q,clean)
qq=q
return qq
# run this and printsolution()
# (you might have to reverse the order of the printed solution)
def solve():
global start,board,E,clean,minn,solution
start=board;
solution={};
E={ board:0 }
for x in range(0,11):
Calc(); # walks all possible moves starting from board to a depth of 10~20 moves
if clean in E:
print('Solution found')
q=clean;
tmp=[];
while q:
tmp.append(q)
q=E[q]
for x in reversed(tmp):
solution[len(solution)]=x;
printsolution();
return
q=minimum(); # calculates the "weight" for all Calc()-ed boards and returns the minimum
#print("Len %3d"%len(E))
print("weight %d"%minn)
# stitch solution
newboard=q;
tmp=[];
while q:
tmp.append(q)
q=E[q]
for x in reversed(tmp):
solution[len(solution)]=x;
board=newboard;
E={board:0}; #reset the Calc()-ed boards
print("No Solution")
# collects and prints the moves of the solution
# from clean board to given board
# (you have to reverse the order)
def printsolution():
global invsolution,solution,moves,clean,start
moves=""
g=start; # start from board to clean
y=g
#invsolution[clean]=0;
for x in solution:
# uncomment this if you want to see each board of the solution
#print(g);
g=solution[x];
#sys.stdout.write(transition(y,g))
if (transition(g,y)=="E"): continue
moves+=transition(g,y)
# or as squares
#print('%10s %d %s'%("step",len(moves),transition(g,y)));
#print(" %s -- %s "%(y,g))
#for i in(0,4,8,12): print('%2d %2d %2d %2d'%g[i:i+4])
y=g
llen=len(moves)
print(" moves%3d "%llen)
print(moves)
# processing moves. funny, but occysionally ud,du,lr or rl appears due to the stitching
while 'lr' in moves:
a,b,c=moves.partition('lr')
moves=a+c
llen-=2
while 'rl' in moves:
a,b,c=moves.partition('rl')
moves=a+c
llen-=2
while 'ud' in moves:
a,b,c=moves.partition('ud')
moves=a+c
llen-=2
while 'du' in moves:
a,b,c=moves.partition('du')
moves=a+c
llen-=2
# processing moves. concatenating lll to 3l
while 'lll' in moves:
a,b,c=moves.partition('lll')
moves=a+' 3l '+c
llen-=2
while 'rrr' in moves:
a,b,c=moves.partition('rrr')
moves=a+' 3r '+c
llen-=2
while 'uuu' in moves:
a,b,c=moves.partition('uuu')
moves=a+' 3u '+c
llen-=2
while 'ddd' in moves:
a,b,c=moves.partition('ddd')
moves=a+' 3d '+c
llen-=2
while 'll' in moves:
a,b,c=moves.partition('ll')
moves=a+' 2l '+c
llen-=1
while 'rr' in moves:
a,b,c=moves.partition('rr')
moves=a+' 2r '+c
llen-=1
while 'uu' in moves:
a,b,c=moves.partition('uu')
moves=a+' 2u '+c
llen-=1
while 'dd' in moves:
a,b,c=moves.partition('dd')
moves=a+' 2d '+c
llen-=1
print(" processed:%3d "%llen)
print(moves)
return
def transition(a,b):
# calculate the move (ie up,down,left,right)
# between 2 boards (distance of 1 move and a weight of 1 only)
i=list(a).index(0);
j=list(b).index(0);
if (j==i+1): return "l"
if (j==i-1): return "r"
if (j==i-4): return "d"
if (j==i+4): return "u"
#print("transition not possible")
return "E"
###################################################
# below this line are functions for the brute force solution only
# added for comparision
#
# its using a global variable bboard and works destructively on it
def solve_brute():
global bboard,board;
bboard=list(board); # working copy
move(1,0);move(2,1);
move(3,14); # <== additional move, move 3 out of way
move(4,2);move(3,6);
gap_down();gap_down();gap_right();gap_right();gap_up();gap_up();gap_up();gap_left();gap_down();
#first line solved
print("first line");printbboard();
move(5,4);move(6,5);move(7,14);move(8,6);move(7,10);
gap_down();gap_down();gap_right();gap_right();gap_up();gap_up();gap_left();gap_down();
#second line solved (upper half)
print("2nd line");printbboard();
move(9,15);move(13,8);move(9,9)
gap_down();gap_left();gap_left();gap_up();gap_right();
print("left border");printbboard();
#left border solved
move(10,15);move(14,9);move(10,10);
gap_down();movegap(1+3*4);gap_up();gap_right();
print("left half");printbboard();
#left half solved
#rotating last 4 tiles 5 times
for x in ' '*5:
gap_right();gap_down(); # gap is now on 15
if (bboard==[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,0]):
print("solution found");printbboard();
return;
gap_left();gap_up();
print("No solution found");
printbboard();
return
def printbboard():
global bboard
for i in(0,4,8,12): print('%2d %2d %2d %2d'%tuple(bboard[i:i+4]))
def gap_up():
global bboard
i=bboard.index(0);
if (i<4):
print("Err up()")
return
bboard[i],bboard[i-4] = bboard[i-4] , 0 ;
def gap_down():
global bboard
i=bboard.index(0);
if (i>11):
print("Err down()")
return
bboard[i],bboard[i+4] = bboard[i+4] , 0 ;
def gap_left():
global bboard
i=bboard.index(0);
if (i%4<1):
print("Err left()")
return
bboard[i],bboard[i-1]= bboard[i-1] , 0 ;
def gap_right():
global bboard
i=bboard.index(0);
if (i%4>2):
print("Err right()")
return
bboard[i],bboard[i+1] = bboard[i+1] , 0 ;
def movegap(d):
global bboard;
# d: destination location (0-15)
k=bboard.index(0);
ky,kx=divmod(k,4);
dy,dx=divmod(d,4);
# moving the gap
while (ky>dy):
gap_up();ky-=1;
while (ky<dy):
gap_down();ky+=1;
while (kx>dx):
gap_left();kx-=1;
while (kx<dx):
gap_right();kx+=1;
def move(s,d):
global bboard
i=bboard.index(s);
iy,ix=divmod(i,4);
dy,dx=divmod(d,4);
#moving a number
while (ix<dx):
move1right(s);
print("1right ");
ix+=1;
while (ix>dx):
move1left(s);
ix-=1;
print("1left ");
while(iy<dy):
move1down(s);
print("1down ");
iy+=1;
while(iy>dy):
move1up(s);
print("1up");
iy-=1;
def move1up(s):
global bboard
i=bboard.index(s);
iy,ix=divmod(i,4);
k=bboard.index(0);
ky,kx=divmod(k,4);
if (ky<iy):
# above: move 1 above, then leftorright, then 1 down
movegap(kx+4*(iy-1))
movegap(ix+4*(iy-1))
movegap(ix+4*iy)
return; # fin
if (ky==iy):
# if equal, then first try 1 down
# (not nescessary if gap is right of s)
if (kx<ix):
if (ky<=2):
movegap(kx+4*(iy+1))
movegap(ix+1+4*(iy+1)); # 1right 1down of s
movegap(ix+1+4*(iy-1)); # 1right 1up of s
movegap(ix+4*(iy-1));# right over s
gap_down(); # fin
return;
# bottom border, must go up first
movegap(kx+4*(iy-1));
movegap(ix+4*(iy-1));
gap_down();
return; # fin
else:
movegap(ix+1+4*iy); # move 1 right of s
gap_up()
gap_left()
gap_down();
return; # fin
movegap(ix+1+4*ky); # move 1 right of s
movegap(ix+1+4*(iy+1)); # move 1 right and 1 down of s
gap_up();
gap_up();
gap_left();
gap_down();
def move1left(s):
global bboard
i=bboard.index(s);
iy,ix=divmod(i,4);
k=bboard.index(0);
ky,kx=divmod(k,4);
if (ky<iy):
# if above gap move 1 over s
if (kx<ix):
movegap(kx+4*iy);
movegap(ix+4*iy);
return;# fin
if (kx==ix):
#gap over s
if (ix<3):
# try to move under s and then left
if (iy<3):
movegap(ix+1+4*ky)
movegap(ix+1+4*(iy+1))
movegap(ix-1+4*(iy+1))
movegap(ix-1+4*iy)
movegap(ix+4*iy)
return; #fin
# have to move left
movegap(kx-1+4*ky)
movegap(ix-1+4*iy)
movegap(ix+4*iy)
return;# fin
# move 1 right of s
if (iy==3):
# cant go under, have to go left over
movegap(kx+4*(iy-1))
movegap(ix-1+4*(iy-1))
movegap(ix-1+4*iy)
movegap(ix+4*iy);
return; #fin
movegap(ix+1+4*(iy-1))
gap_down();gap_down();gap_left();gap_left();gap_up();gap_right();
return; #fin
if (ky==iy):
if (kx<ix):
movegap(ix-1+4*iy)
gap_right();
return; # fin
if (ky<3):
gap_down();
ky+=1;
else:
#have to move up
movegap(ix+4*(iy-1))
movegap(ix-1+4*(iy-1))
movegap(ix-1+4*iy)
gap_right();
return; #fin
# gap below s
movegap(ix+4*(iy+1));
gap_left();gap_up();gap_right();
def move1right(s):
global bboard
i=bboard.index(s);
iy,ix=divmod(i,4);
k=bboard.index(0);
ky,kx=divmod(k,4);
if (ky<iy):
if (kx==ix):
movegap(kx+1+4*ky)
movegap(kx+1+4*iy)
movegap(ix+4*iy);
return; #fin
movegap(kx+4*iy)
if (kx>ix):
movegap(ix+4*iy);
return; #fin
movegap(kx+4*(iy+1))
movegap(ix+1+4*(iy+1))
movegap(ix+1+4*iy);
movegap(ix+4*iy);
return; #fin
if (ky==iy):
if (kx<ix):
if (ky>2):
# bottom row, left of s, have to move 1 up
gap_up()
# move 1 right 1 up of s
movegap(ix+1+4*(ky-1));
gap_down()
gap_left()
return; # fin
# first 1 down
movegap(kx+4*(ky+1))
# to the right of s
movegap(ix+1+4*(ky+1))
gap_up()
gap_left()
return; # fin
# already 1 right of s
movegap(ix+4*iy);
return; #fin
# move gap 1 right and 1 down of s
movegap(kx+4*(iy+1))
movegap(ix+1+4*(iy+1))
gap_up();
gap_left();
def move1down(s):
global bboard
i=bboard.index(s);
iy,ix=divmod(i,4);
k=bboard.index(0);
ky,kx=divmod(k,4);
if (ky<iy):
# gap is over s, move it below
if (kx==ix):
if (ix>2):
# right border, have to move 1 to the left
movegap(kx+4*(iy-1))
movegap(kx-1+4*(iy-1))
movegap(kx-1+4*(iy+1))
gap_up();
return; #fin
# move right of s
movegap(kx+4*(iy-1))
movegap(kx+1+4*(iy-1))
movegap(kx+1+4*(iy+1))
movegap(kx+4*(iy+1))
gap_up(); #fin
movegap(kx+4*(iy+1))
movegap(ix+4*(iy+1))
gap_up(); #fin
if (ky==iy):
gap_down();
ky+=1;
# gap is below s, move 1 under s
movegap(ix+4*(iy+1))
gap_up();
#fin