Solve the 15 Puzzle (the tile-sliding puzzle)

Question

The 15 Puzzle is a famous puzzle involving sliding 15 tiles around on a 4x4 grid. Starting from a random configuration, the goal is to arrange the tiles in the correct order. Here is an example of a solved 15 Puzzle:

01 02 03 04
05 06 07 08
09 10 11 12
13 14 15

Each move on the puzzle is of the form Up/Down/Left/Right. The move "Down" consists of sliding the tile that is above the empty spot downward. The move "Right" consists of sliding a tile to the right, into the empty spot. Here is how the board looks after the moves Down and Right.

01 02 03 04
05 06 07 08
09 10    11
13 14 15 12

---

The goal of this challenge is to write a program that can output the series of moves needed to solve the 15 Puzzle. The winner is the program who solves the five test cases (below) in the fewest total moves. The generated solution does not need to be a perfect solution, it merely has to be better than the competitors. For each individual test case, the program should not take more than ten seconds on a reasonable machine.

Your program must be able to solve any puzzle that is solvable, I'm just using these five test cases as the scoring.

Your program will receive the unsolved 15 Puzzle as input in the format of a 2D array. The 2D array can be formatted according to the language used, or changed if the language does not have 2D arrays. The first element of the first sub-array will be the number in the upper left, and the last element of the first sub-array will be the number in the upper right. A 0 will be the empty space.

As output, your program should print a list of moves in the order that they need to be performed. Each step should be numbered in order to increase the usability of the results.

EDIT: Based on comments, I will allow output to be in either the form of Down/Up/etc or in the form of the coordinates of the piece to move. As this is not code golf, the most important part is to solve the puzzle.

Some other general rules involve no using an outside source, etc.

---

Test Case 1

([5,1,7,3],[9,2,11,4],[13,6,15,8],[0,10,14,12])

Example Output:

1: Down
2: Down
3: Down
4: Left
....

Test Case 2

([2,5,13,12],[1,0,3,15],[9,7,14,6],[10,11,8,4])

Test Case 3

([5,2,4,8],[10,0,3,14],[13,6,11,12],[1,15,9,7])

Test Case 4

([11,4,12,2],[5,10,3,15],[14,1,6,7],[0,9,8,13])

Test Case 5

([5,8,7,11],[1,6,12,2],[9,0,13,10],[14,3,4,15])

Answer 1

PyPy, 195 moves, ~12 seconds computation

Computes optimal solutions using IDA* with a 'walking distance' heuristic augmented with linear conflicts. Here are the optimal solutions:

 5  1  7  3
 9  2 11  4
13  6 15  8
 0 10 14 12
Down, Down, Down, Left, Up, Up, Up, Left, Down, Down, Down, Left, Up, Up, Up

 2  5 13 12
 1  0  3 15
 9  7 14  6
10 11  8  4
Left, Down, Right, Up, Up, Left, Down, Down, Right, Up, Left, Left, Down, Right, Right, Right, Up, Up, Left, Left, Down, Left, Up, Up, Right, Down, Down, Left, Up, Up, Right, Right, Right, Down, Left, Up, Right, Down, Down, Left, Left, Down, Left, Up, Up, Right, Up, Left

 5  2  4  8
10  0  3 14
13  6 11 12
 1 15  9  7
Left, Up, Up, Right, Right, Down, Left, Up, Left, Left, Down, Down, Right, Right, Up, Left, Left, Down, Down, Right, Right, Up, Right, Up, Left, Left, Up, Right, Down, Down, Right, Down, Left, Left, Up, Up, Left, Up

11  4 12  2
 5 10  3 15
14  1  6  7
 0  9  8 13
Down, Left, Down, Right, Up, Left, Left, Left, Down, Down, Right, Right, Right, Up, Left, Left, Left, Down, Right, Right, Up, Left, Up, Up, Left, Down, Down, Right, Down, Right, Up, Up, Right, Up, Left, Left, Left, Down, Right, Right, Right, Up, Left, Down, Left, Down, Left, Up, Up

 5  8  7 11
 1  6 12  2
 9  0 13 10
14  3  4 15
Up, Right, Down, Left, Left, Down, Left, Up, Right, Up, Right, Down, Down, Right, Up, Up, Left, Left, Left, Down, Down, Down, Right, Right, Up, Right, Down, Left, Up, Left, Up, Left, Down, Right, Down, Left, Up, Right, Down, Right, Up, Up, Left, Left, Up

And the code:

import random


class IDAStar:
    def __init__(self, h, neighbours):
        """ Iterative-deepening A* search.

        h(n) is the heuristic that gives the cost between node n and the goal node. It must be admissable, meaning that h(n) MUST NEVER OVERSTIMATE the true cost. Underestimating is fine.

        neighbours(n) is an iterable giving a pair (cost, node, descr) for each node neighbouring n
        IN ASCENDING ORDER OF COST. descr is not used in the computation but can be used to
        efficiently store information about the path edges (e.g. up/left/right/down for grids).
        """

        self.h = h
        self.neighbours = neighbours
        self.FOUND = object()


    def solve(self, root, is_goal, max_cost=None):
        """ Returns the shortest path between the root and a given goal, as well as the total cost.
        If the cost exceeds a given max_cost, the function returns None. If you do not give a
        maximum cost the solver will never return for unsolvable instances."""

        self.is_goal = is_goal
        self.path = [root]
        self.is_in_path = {root}
        self.path_descrs = []
        self.nodes_evaluated = 0

        bound = self.h(root)

        while True:
            t = self._search(0, bound)
            if t is self.FOUND: return self.path, self.path_descrs, bound, self.nodes_evaluated
            if t is None: return None
            bound = t

    def _search(self, g, bound):
        self.nodes_evaluated += 1

        node = self.path[-1]
        f = g + self.h(node)
        if f > bound: return f
        if self.is_goal(node): return self.FOUND

        m = None # Lower bound on cost.
        for cost, n, descr in self.neighbours(node):
            if n in self.is_in_path: continue

            self.path.append(n)
            self.is_in_path.add(n)
            self.path_descrs.append(descr)
            t = self._search(g + cost, bound)

            if t == self.FOUND: return self.FOUND
            if m is None or (t is not None and t < m): m = t

            self.path.pop()
            self.path_descrs.pop()
            self.is_in_path.remove(n)

        return m


def slide_solved_state(n):
    return tuple(i % (n*n) for i in range(1, n*n+1))

def slide_randomize(p, neighbours):
    for _ in range(len(p) ** 2):
        _, p, _ = random.choice(list(neighbours(p)))
    return p

def slide_neighbours(n):
    movelist = []
    for gap in range(n*n):
        x, y = gap % n, gap // n
        moves = []
        if x > 0: moves.append(-1)    # Move the gap left.
        if x < n-1: moves.append(+1)  # Move the gap right.
        if y > 0: moves.append(-n)    # Move the gap up.
        if y < n-1: moves.append(+n)  # Move the gap down.
        movelist.append(moves)

    def neighbours(p):
        gap = p.index(0)
        l = list(p)

        for m in movelist[gap]:
            l[gap] = l[gap + m]
            l[gap + m] = 0
            yield (1, tuple(l), (l[gap], m))
            l[gap + m] = l[gap]
            l[gap] = 0

    return neighbours

def slide_print(p):
    n = int(round(len(p) ** 0.5))
    l = len(str(n*n))
    for i in range(0, len(p), n):
        print(" ".join("{:>{}}".format(x, l) for x in p[i:i+n]))

def encode_cfg(cfg, n):
    r = 0
    b = n.bit_length()
    for i in range(len(cfg)):
        r |= cfg[i] << (b*i)
    return r


def gen_wd_table(n):
    goal = [[0] * i + [n] + [0] * (n - 1 - i) for i in range(n)]
    goal[-1][-1] = n - 1
    goal = tuple(sum(goal, []))

    table = {}
    to_visit = [(goal, 0, n-1)]
    while to_visit:
        cfg, cost, e = to_visit.pop(0)
        enccfg = encode_cfg(cfg, n)
        if enccfg in table: continue
        table[enccfg] = cost

        for d in [-1, 1]:
            if 0 <= e + d < n:
                for c in range(n):
                    if cfg[n*(e+d) + c] > 0:
                        ncfg = list(cfg)
                        ncfg[n*(e+d) + c] -= 1
                        ncfg[n*e + c] += 1
                        to_visit.append((tuple(ncfg), cost + 1, e+d))

    return table

def slide_wd(n, goal):
    wd = gen_wd_table(n)
    goals = {i : goal.index(i) for i in goal}
    b = n.bit_length()

    def h(p):
        ht = 0 # Walking distance between rows.
        vt = 0 # Walking distance between columns.
        d = 0
        for i, c in enumerate(p):
            if c == 0: continue
            g = goals[c]
            xi, yi = i % n, i // n
            xg, yg = g % n, g // n
            ht += 1 << (b*(n*yi+yg))
            vt += 1 << (b*(n*xi+xg))

            if yg == yi:
                for k in range(i + 1, i - i%n + n): # Until end of row.
                    if p[k] and goals[p[k]] // n == yi and goals[p[k]] < g:
                        d += 2

            if xg == xi:
                for k in range(i + n, n * n, n): # Until end of column.
                    if p[k] and goals[p[k]] % n == xi and goals[p[k]] < g:
                        d += 2

        d += wd[ht] + wd[vt]

        return d
    return h




if __name__ == "__main__":
    solved_state = slide_solved_state(4)
    neighbours = slide_neighbours(4)
    is_goal = lambda p: p == solved_state

    tests = [
        (5,1,7,3,9,2,11,4,13,6,15,8,0,10,14,12),
        (2,5,13,12,1,0,3,15,9,7,14,6,10,11,8,4),
        (5,2,4,8,10,0,3,14,13,6,11,12,1,15,9,7),
        (11,4,12,2,5,10,3,15,14,1,6,7,0,9,8,13),
        (5,8,7,11,1,6,12,2,9,0,13,10,14,3,4,15),
    ]

    slide_solver = IDAStar(slide_wd(4, solved_state), neighbours)

    for p in tests:
        path, moves, cost, num_eval = slide_solver.solve(p, is_goal, 80)
        slide_print(p)
        print(", ".join({-1: "Left", 1: "Right", -4: "Up", 4: "Down"}[move[1]] for move in moves))
        print(cost, num_eval)

Answer 2

JavaScript (ES6) total steps 329 for all 5 test cases in ~1min

Edit Same strategy, different targets, better solution. Slower ...

This is more or less how I solve it by hand: using intermediate targets After each target the relative tiles are not moved again Each intermediate target is reached using a parametric BSF function. The 2 params are the loop condition L (repeat while true) and the select condition S (select what tile can be moved). The steps:

1. Place 1 top/left
2. Place 2
3. Place 5
4. Place 3,4 - top row ok
5. Place 9,13 - left column ok
6. All the rest

Side note I don't check the position of tiles 14 and 15. Unsolvable puzzles like [11,4,12,2,,15,10,3,5,,14,1,6,7,,0,9,8,13] will have 14 and 15 swapped.

F=b=>(
  s=[],
  [[_=>b[0]!=1, (o,p)=>b[o+p]]
  ,[_=>b[1]!=2, (o,p)=>(p=b[o+p])>1&&p]
  ,[_=>b[5]!=5, (o,p)=>(p=b[o+p])>2&&p]
  ,[_=>b[2]!=3|b[3]!=4, (o,p)=>(p=b[o+p])>2&&p!=5&&p]
  ,[_=>b[10]!=9|b[15]!=13, (o,p)=>(p=b[o+p])>5&&p]
  ,[_=>b[6]!=6|b[7]!=7|b[8]!=8|b[11]!=10|b[12]!=11|b[13]!=12|b[18]!=0, (o,p)=>(p=b[o+p])>5&&p!=9&&p!=13&&p]
  ].forEach(([L,S])=>{
    for(v={},v[b]=1,t=0,m=[];L();)
    {
      b.forEach((x,p)=>
        x=='0'&&[-1,5,1,-5].forEach((o,d)=>
          (x=S(o,p))&&(c=b.slice(0),c[p]=x,c[o+p]=0,v[k=''+c]?0:v[k]=m.push([c,s.concat(d)]))
        )
      );[b,s]=m[t++]
    }
  }),
  ,s.map((d,i)=>i+': '+'RULD'[d]).join('\n') // multi line output
  // ,s.map(d=>'RULD'[d]).join(' ') // single line output (easier to test)
)

Open snippet to test or play (Firefox only)

F=b=>(
  s=[],
  [[_=>b[0]!=1, (o,p)=>b[o+p]]
  ,[_=>b[1]!=2, (o,p)=>(p=b[o+p])>1&&p]
  ,[_=>b[5]!=5, (o,p)=>(p=b[o+p])>2&&p]
  ,[_=>b[2]!=3|b[3]!=4, (o,p)=>(p=b[o+p])>2&&p!=5&&p]
  ,[_=>b[10]!=9|b[15]!=13, (o,p)=>(p=b[o+p])>5&&p]
  ,[_=>b[6]!=6|b[7]!=7|b[8]!=8|b[11]!=10|b[12]!=11|b[13]!=12|b[18]!=0, (o,p)=>(p=b[o+p])>5&&p!=9&&p!=13&&p]
  ].forEach(([L,S])=>{
    for(v={},v[b]=1,t=0,m=[];L();)
    {
      b.forEach((x,p)=>
        x=='0'&&[-1,5,1,-5].forEach((o,d)=>
          (x=S(o,p))&&(c=b.slice(0),c[p]=x,c[o+p]=0,v[k=''+c]?0:v[k]=m.push([c,s.concat(d)]))
        )
      );[b,s]=m[t++]
    }
  }),
  //,s.map((d,i)=>i+': '+'RULD'[d]).join('\n') // multi line output
  //,s.map(d=>'RULD'[d]).join(' ') // single line output (easier to test)
  s
)
B.value=PZ.value,show()

function show(s='') {
  var b = eval(B.value)
  var t = b.map((c,i)=>'<td>'+c+'</td>').join()
  .replace(/,,/g,'</tr><tr>').replace(/,/g,'')
  G.innerHTML='<tr>'+t+'</tr>'
  S.value=s
}
function solve() {
  show('... solving ...')
  var b = eval(B.value)
  setTimeout(_=>{
    var s = F(b), zp = b.indexOf(0), sp = 0
    S.value=s.map(d=>'RULD'[d]).join(' ');
    (A=_=>{
      d=[-1,5,1,-5][m=s[sp++]]
      b[zp]=b[zp+d],zp+=d,b[zp]=0
      var t = b.map((c,i)=>'<td>'+c+'</td>').join()
      .replace(/,,/g,'</tr><tr>').replace(/,/g,'')
      G.innerHTML='<tr>'+t+'</tr>'
      if (sp<s.length)
        setTimeout(A, 300);  
    })()
  },100)
}

#D { position: relative }
#D input { position: absolute; width: 300px; top:2px; left:2px; border:0 none }
#D select { position: relative; width: 400px; height:1.8em; top:0; left:0; }
textarea{ width: 600px; height: 40px }
td{ width: 1.5em; text-align:right }

Puzzles (select or edit)
<div id=D>
<select id=PZ onchange="B.value=PZ.value,show()">
<option>[5,1,7,3,,9,2,11,4,,13,6,15,8,,0,10,14,12]</option>
<option>[2,5,13,12,,1,0,3,15,,9,7,14,6,,10,11,8,4]</option>
<option>[5,2,4,8,,10,0,3,14,,13,6,11,12,,1,15,9,7]</option>
<option>[11,4,12,2,,5,10,3,15,,14,1,6,7,,0,9,8,13]</option>
<option>[5,8,7,11,,1,6,12,2,,9,0,13,10,,14,3,4,15]</option>
</select>
<button onclick="solve()">Solve</button>
<input id=B>
</div>
<textarea id=S></textarea>
<table id=G></table>

Test suite In Firefox/FireBug console

T=~new Date
;[[5,1,7,3,,9,2,11,4,,13,6,15,8,,0,10,14,12]
,[2,5,13,12,,1,0,3,15,,9,7,14,6,,10,11,8,4]
,[5,2,4,8,,10,0,3,14,,13,6,11,12,,1,15,9,7]
,[11,4,12,2,,5,10,3,15,,14,1,6,7,,0,9,8,13]
,[5,8,7,11,,1,6,12,2,,9,0,13,10,,14,3,4,15]]
.forEach(t=>console.log(t+'',F(t)))
console.log('Time ms ',T-=~new Date)

Output

"5,1,7,3,,9,2,11,4,,13,6,15,8,,0,10,14,12" "D D D L U L D L U R R U U L D D L U U"
"2,5,13,12,,1,0,3,15,,9,7,14,6,,10,11,8,4" "D R U L U L L U R D L D R D L U R U L D R D L U R U L U R R R D L L U R D R U L L D L D R U U L D R U R D L U L D D R R U L U L D R U L"
"5,2,4,8,,10,0,3,14,,13,6,11,12,,1,15,9,7" "R U U L D D R U L D D R U U L L D D R U L D L U U R R D L U R R D L L U L D D R U U L D D R U U U R R D L L U R R D L L L U R D D L U R D R U U L L D R D L U U"
"11,4,12,2,,5,10,3,15,,14,1,6,7,,0,9,8,13" "D L D R U L D D R U L L D L U R R D L U R U R D L U R U L L D R D L L D R U U L D R D L U R U U L D R R U L D R R U L L D L D R U U L D R R D L L U U R D R U L L"
"5,8,7,11,,1,6,12,2,,9,0,13,10,,14,3,4,15" "D D R U L L L D R U R D L U U R R D L U L U R D D L U U L D D D R U U L D D R U U U R D R U L D D L U U R D R U L D L L D R U L U R D L D R R U L L U R D D L U U"
"Time ms " 62234

Answer 3

I started working on this problem and wanted to contribute with my code so far. As stated by Gareth, the problem is comparable to the 8-tile puzzle and so the code is based on the magnificient solution of Keith Randall and thus in Python. This solution can solve all 5 test cases with a total sum of less than 400 moves, and other puzzles, too. It contains an optimized and a brute force solution. The code is a bit bloated by now. Output is abbrevated like "llururd.." Hope its helpful. http://www.penschuck.org/joomla/tmp/15Tile.txt (explanation) http://www.penschuck.org/joomla/tmp/tile15.txt (python code)

# 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