Algorithm to generate Diagonal Latin Square matrix - algorithm

Question

I need for given N create N*N matrix which does not have repetitions in rows, cells, minor and major diagonals and values are 1, 2 , 3, ...., N.
For N = 4 one of matrices is the following:
1 2 3 4
3 4 1 2
4 3 2 1
2 1 4 3

Answer 1

Problem overview
The math structure you described is Diagonal Latin Square. Constructing them is the more mathematical problem than the algorithmic or programmatic.
To correctly understand what it is and how to create you should read following articles:
Latin squares definition
Magic squares definition
Diagonal Latin square construction <-- p.2 is answer to your question with proof and with other interesting properties
Short answer
One of the possible ways to construct Diagonal Latin Square:
Let N is the power of required matrix L.
If there are exist numbers A and B from range [0; N-1] which satisfy properties:
A relativly prime to N
B relatively prime to N
(A + B) relatively prime to N
(A - B) relatively prime to N
Then you can create required matrix with the following rule:
L[i][j] = (A * i + B * j) mod N

Answer 2

It would be nice to do this mathematically, but I'll propose the simplest algorithm that I can think of - brute force.
At a high level
we can represent a matrix as an array of arrays
for a given N, construct S a set of arrays, which contains every combination of [1..N]. There will be N! of these.
using an recursive & iterative selection process (e.g. a search tree), search through all orders of these arrays until one of the 'uniqueness' rules is broken
For example, in your N = 4 problem, I'd construct
S = [
[1,2,3,4], [1,2,4,3]
[1,3,2,4], [1,3,4,2]
[1,4,2,3], [1,4,3,2]
[2,1,3,4], [2,1,4,3]
[2,3,1,4], [2,3,4,1]
[2,4,1,3], [2,4,3,1]
[3,1,2,4], [3,1,4,2]
// etc
]
R = new int[4][4]
Then the algorithm is something like
If R is 'full', you're done
Evaluate does the next row from S fit into R,
if yes, insert it into R, reset the iterator on S, and go to 1.
if no, increment the iterator on S
If there are more rows to check in S, go to 2.
Else you've iterated across S and none of the rows fit, so remove the most recent row added to R and go to 1. In other words, explore another branch.
To improve the efficiency of this algorithm, implement a better data structure. Rather than a flat array of all combinations, use a prefix tree / Trie of some sort to both reduce the storage size of the 'options' and reduce the search area within each iteration.

Answer 3

Here's a method which is fast for N <= 9 : (python)
import random
def generate(n):
a = [[0] * n for _ in range(n)]
def rec(i, j):
if i == n - 1 and j == n:
return True
if j == n:
return rec(i + 1, 0)
candidate = set(range(1, n + 1))
for k in range(i):
candidate.discard(a[k][j])
for k in range(j):
candidate.discard(a[i][k])
if i == j:
for k in range(i):
candidate.discard(a[k][k])
if i + j == n - 1:
for k in range(i):
candidate.discard(a[k][n - 1 - k])
candidate_list = list(candidate)
random.shuffle(candidate_list)
for e in candidate_list:
a[i][j] = e
if rec(i, j + 1):
return True
a[i][j] = 0
return False
rec(0, 0)
return a
for row in generate(9):
print(row)
Output:
[8, 5, 4, 7, 1, 6, 2, 9, 3]
[2, 7, 5, 8, 4, 1, 3, 6, 9]
[9, 1, 2, 3, 6, 4, 8, 7, 5]
[3, 9, 7, 6, 2, 5, 1, 4, 8]
[5, 8, 3, 1, 9, 7, 6, 2, 4]
[4, 6, 9, 2, 8, 3, 5, 1, 7]
[6, 3, 1, 5, 7, 9, 4, 8, 2]
[1, 4, 8, 9, 3, 2, 7, 5, 6]
[7, 2, 6, 4, 5, 8, 9, 3, 1]