Construct Permutation with longest arithmetic progression subsequence of predefined length

Question

A subsequence of sequence A is a sequence that is obtained from A by removing several (zero or more) elements from it. Eg: {1,3,5}, {2,4},{ } are subsequence of sequence {1,2,3,4,5}.

An arithmetic subsequence of sequence A is a subsequence of A , that is an arithmetic progression. Eg: {1,3,5}, {1,2,3} are arithmetic subsequence of sequence {1,2,3,4,5} with length 3.

how to construct any permutation of first N positive integers such that the length of the longest arithmetic subsequence of the permutation is equal to K.

For example if N=5 and K=3, Different permutation will be N! ie, 5!=120 ways to represent first N positive numbers.

Now our answer will be any one permutation which will be having longest arithmetic subsequence in it of length K.

For example if N=5 and K=3 one of the many permutation will be {4,1,3,2,5} because in this sequence {4,3,2} & {1,3,5} are two arithmetic subsequence with length 3 and no other arithmetic subsequence with length > K is present here.

We can not take {4,5,3,2,1} as our answer because it is having {4,3,2,1} as its longest arithmetic subsequence which is exceeded the given limit K=3 . So we not consider it as our answer.

We just need any permutation which satisfies our given conditions.

For eg. If N=5 and K=3 then following can be valid permutations for our answer : {4,1,3,2,5} {2,3,5,1,4} {1,2,5,4,3}... Etc...

One obvious approach is to generate all permutations and then to compute the Longest AP Sequence and check if is equal to given length and print that permutation but it would be O(n! * n^2) which will be too large. I even tried observing some pattern but couldn't come up with any observation.

Answer 1

One way to do it is to first arrange the numbers $\{k-1,k,\dots,n\}$ such that there is no three-term arithmetic subsequence, by adding $k-2$ to one of the solutions described in answers to this question. Reverse this, if needed, so that $k-1$ comes before $k$. Then add $1, 2, \dots, k-2$ to the beginning, in that order.

For example, when $n=7$ and $k=5$, we first find a permutation of $\{4,5,6,7\}$ that works, by shifting a permutation of $\{1,2,3,4\}$. If we started with $(2,4,1,3)$, we'd end up with $(5,7,4,6)$. Since $4$ does not come before $5$, we'd reverse this to get $(6,4,7,5)$. Now we add $1,2,3$ at the beginning to get $(1,2,3,6,4,7,5)$.

There is a $k$-term arithmetic subsequence formed by $1, 2, \dots, k-1,k$.

There is no $(k+1)$-term arithmetic subsequence. If there were, then delete however many of the numbers $\{1,2,\dots,k-2\}$ appear at the beginning of that subsequence, and we're still left with at least a $3$-term arithmetic subsequence in our permutation of $\{k-1,k,\dots,n\}$, which we know doesn't exist.