... I searched some patterns to build a reduction from a NPC problem, but didn't find a way to represent a "flow" with a "fork" ...
So (after some work) this is a polynomial algorithm ...
ALGORITHM
The starting list can be viewed as an array of $N*2$ consecutive "holes". For each initial pair $(a_j,b_j)$, put the "element" $b_j$ at hole number $a_j$. Each pair can be viewed as a directed edge from position $a_j$ to position $b_j$. A move consists in picking an element $b_j$ at position $a_j$ and moving it to its destination position $b_j$ (the destination hole becomes an unmovable peg). We delete the edge, and proceed to choose the next move which will start from one of the two nearest reachable elements $b_k$ from position $b_j$ (only holes between $b_j$ and $b_k$ are allowed). We must find a sequence of $N$ consecutive moves.
When you make a move you fix a peg at position $b_j$ and the array is splitted in two partitions $L$ (left) and $R$ (right) and the only way to go from $L$ to $R$ (or from $R$ to $L$) is using an edge that jump across the peg. Set
- $edgesLR$ = number of edges from left to right (do not count the final edge)
- $edgesRL$ = number of edges from right to left (do not count the final edge)
- $flow$ = $edgesLR - edgesRL$
Cases:
A) if $| flow | > 1$ then one of the two partitions will become unreachable, stop
Now suppose that $end > b_j$, i.e. $end \in R$
B) if $flow = 1$ then there is an extra edge from left to right, you must go left (pick the nearest element of $L$), otherwise you will never reach $end$
C) if $flow = -1$ then there is an extra edge from right to left and whatever node you pick you will never reach $end$, stop
D) if $flow = 0$ you must go right (pick the nearest element of $R$), otherwise you will neve reach $end$
If $end < b_j$ ($end \in L$), B,C,D are inverted.
NOTE: when moving left or right, you must consider $end$ as a peg. For example, if you must go right, but the nearest element on $R$ is $end$ then the move is impossible (and you must proceed with another pair $(start,end)$)
Apply the same resoning at every move.
COMPLEXITY
The flows over each hole can be precalculated in O(N) and reused at every scan.
The loops are:
for start = 1 to N
for end = 1 to N
for move = 1 to N
make a move (fix a peg and update flows)
check if another move can be done using flow
No choices are made during the computation, so the complexity of the algorithm is $O(N^3)$
CODE
This is a working Java implementation of the algorithm:
public class StrangeSort {
static int PEG = 0xffffff, HOLE = 0x0;
static int M = 0, N = 0, choices = 0, aux = 0, end;
static int problem[][], moves[], edgeflow[], field[];
boolean is_hole(int x) { return x == HOLE; }
boolean is_peg(int x) { return x == PEG; }
boolean is_ele(int x) { return ! is_peg(x) && ! is_hole(x); };
int []cp(int src[]) { // copy an array
int res[] = new int[src.length];
System.arraycopy(src, 0, res, 0, res.length);
return res;
}
/* find the first element on the left (dir=-1) right (dir=1) */
int find(int pos, int dir, int nm) {
pos += dir;
while (pos >= 1 && pos <= M ) {
int x = field[pos];
if ( is_peg(x) || (pos == end && nm < N-1) ) return 0;
if ( is_ele(x) ) return pos;
pos += dir;
}
return 0;
}
void build_edges() {
edgeflow = new int[M+1];
for (int i = 1; i<=M; i++) {
int start = i;
int b = field[start];
if (! is_ele(b)) continue;
if (i == end) continue;
int dir = (b > start)? 1 : -1;
start += dir;
while (start != b) { edgeflow[start] += dir; start += dir; }
}
}
boolean rec_solve(int start, int nm) {
boolean f;
int j;
int b = field[start];
moves[nm++] = b;
if (nm == N) return true;
//System.out.println("Processing: " + start + "->" + field[start]);
field[start] = HOLE;
field[b] = PEG;
int dir = (b > start)? 1 : -1;
int i = start + dir;
while (i != b) { edgeflow[i] -= dir; i += dir; } // clear edge
int flow = edgeflow[b];
if (Math.abs(flow) > 2) return false;
if (end > b) {
switch (flow) {
case 1 :
j = find(b,-1,nm);
if (j <= 0) return false;
return rec_solve(j,nm);
case -1 :
return false;
case 0 :
j = find(b,1,nm);
if (j <= 0) return false;
return rec_solve(j,nm);
}
} else {
switch (flow) {
case -1 :
j = find(b,1,nm);
if (j <= 0) return false;
return rec_solve(j,nm);
case 1 :
return false;
case 0 :
j = find(b,-1,nm);
if (j <= 0) return false;
return rec_solve(j,nm);
}
}
return false;
}
boolean solve(int demo[][]) {
N = demo.length;
for (int i = 0; i < N; i++)
M = Math.max(M, Math.max(demo[i][0], demo[i][1]));
moves = new int[N];
edgeflow = new int[M+1];
field = new int[M+1];
problem = demo;
for (int i = 0; i < problem.length; i++) {
int a = problem[i][0];
int b = problem[i][1];
if ( a < 1 || b < 1 || a > M || b > M || ! is_hole(field[a]) || ! is_hole(field[b])) {
System.out.println("Bad input pair (" + a + "," + b + ")");
return false;
}
field[a] = b;
}
for (int i = 1; i <= M; i++) {
end = i;
build_edges();
if (!is_ele(field[i])) continue;
for (int j = 1; j <= M; j++) {
if (!is_ele(field[j])) continue;
if (i==j) continue;
int tmp_edgeflow[] = cp(edgeflow);
int tmp_field[] = cp(field);
choices = 0;
//System.out.println("START: " + j + " " + " END: " + i);
if (rec_solve(j, 0)) {
return true;
}
edgeflow = tmp_edgeflow;
field = tmp_field;
}
}
return false;
}
void init(int demo[][]) {
}
public static void main(String args[]) {
/**** THE INPUT ********/
int demo[][] = {{4,2},{5,7},{6,3},{10,12},{11,1},{13,8},{14,9}};
/***********************/
String r = "";
StrangeSort sorter = new StrangeSort();
if (sorter.solve(demo)) {
for (int i = 0; i < N; i++) { // print it in clear text
int b = moves[i];
for (int j = 0; j < demo.length; j++)
if (demo[j][1] == b)
r += ((i>0)? " -> " : "") + "(" + demo[j][0] + "," + demo[j][1] + ")";
}
r = "SOLUTION: "+r;
}
else
r = "NO SOLUTIONS";
System.out.println(r);
}
}
