Efficient algorithm for existence of permutation with differences sequence?

Question

This question is motivated by this post, Can you identify the sum of two permutations in polynomial time? , and my interest in computational properties of permutations.

A differences sequence $a_1, a_2, \ldots a_n$ of a permutation $\pi$ of numbers $1, 2, \ldots n+1$ is formed by finding the difference between every two adjacent numbers in the permutation $\pi$. In other words, $a_i= |\pi(i+1)-\pi(i)|$ for $1 \le i \le n$

For example, sequence $1, 1, 3$ is the differences sequence of permutation $2 3 4 1$. While, sequences $2, 2, 3$ and $ 3, 1, 2$ are not the differences sequence of any permutation of numbers $1, 2, 3, 4$.

Is there an efficient algorithm to determine whether a given sequence is the differences sequence for some permutation $\pi$, or is it NP-hard?

EDIT: We get computationally equivalent problem if we formulate the problem using circular permutations.

EDIT2: Cross posted on MathOverflow, How hard is reconstructing a permutation from its differences sequence?

EDIT3 Awarded the bounty to the proof sketch and I would accept the answer after getting the complete formal proof.

EDIT 4: Marzio's nice $NP$-completeness proof has been published in the Electronic Journal of Combinatorics.

Answer 1

This is a sketch of a possible reduction to prove that it is NP-hard:

1) $a_i$ subsequences made of 1s (e.g. $...11111...$) (I call them 1SEQ) force a subsequence of increasing or decreasing numbers in the permutation;

2) if a value of $2$ is put in a long 1SEQ, it forces a hole (a missing number) and doesn't change the direction of the 1SEQ. For example: $1112112111$ forces two holes:

 a_i seq.:     1 1 1  2  1 1  2   1  1  1  forces
 permutation: 1 2 3 4 _ 6 7 8 _ 10 11 12 13 (or its decreasing equivalent)
 (from 4 you cannot go back to 2,
 from 8 you cannot go back to 6)

The holes must be filled in the rest of the permutation.

3) using a large enough 1SEQ, followed by a 1SEQ with some holes, followed by another large 1SEQ you can build a forced line;

4) putting together many forced lines you can build a permutation grid graph in which the nodes correspond to the missing numbers in the underlying forced permutation.

For example the sequence 1111111112111111111112111111111, forces the following 5x7 permutation grid graph:

29 30 31 32 33 34 35
22 23 24    26 27 28
15 16 17 18 19 20 21
 8  9 10    12 13 14   
 1  2  3  4  5  6  7

(or its symmetric version). Note that if the grid has size $w \times w$ and $a,b$ are two missing numbers in the same vertical column then $|a-b|=kw$.

5) the Hamiltonian cycle on grid graphs problem is NP-complete; so given a grid graph $G$ (with holes) you can build the equivalent permutation grid graph;

6) from the last number of the permutation you can "jump" to a number corresponding to a hole (a node in $G$), and with a fixed sequence of moves you can simulate the traversal of $G$; this requires a rather complex gadget - the "selection gadget" - that must be created in another part of the permutation grid graph;

7) you can fill all the holes (i.e. complete the permutation) if and only if the original grid graph has an Hamiltonian cycle

EDIT: July, 27 2013

I tried to formally prove the NP completeness of the problem: I introduced a new problem (Crazy Frog problem) which is NPC. The Permutation Reconstruction from Differences problem is equivalent to the "1-D Crazy Frog problem without blocked cells" (which is also NPC).

For the details of the reduction see my question/answer on cstheory "Two Hamiltonian path variants" or download a draft of the proof "When a frog meets a permutation" :)) (I'm still checking/completing it)