5 Queens on 5x5 Toroidal Board - All Solutions

Q: How many solutions are there for 5 queens on a 5×5 toroidal board?

There are exactly 40 total solutions . These group into 8 fundamental solution classes — each class is generated by taking one solution and adding 0, 1, 2, 3, or 4 to every queen column modulo 5, giving 5 variants per class and 8 × 5 = 40 total.

Q: What is the number of fundamental solutions for 5 queens on a 5×5 toroidal board?

There are 8 fundamental solutions for 5 queens on a 5×5 toroidal board. Two of them come from the simple multiplier construction (column = 2×row mod 5 and column = 3×row mod 5); the other six arise from reflections and transpositions of those two.

The 5×5 Toroidal Queens Problem

The 5×5 toroidal queens problem asks: in how many ways can 5 non-attacking queens be placed on a 5×5 chessboard whose edges wrap around (forming a torus)? Two queens attack each other if they share a row, column, or toroidal diagonal.

Why 5×5 is Special

The number 5 is the smallest integer greater than 1 for which a complete toroidal queens solution exists. The existence condition requires gcd(n, 6) = 1 — that n be coprime to both 2 and 3. The table below shows the first few values:

n	gcd(n,6)	Complete solution?
2	2	No
3	3	No
4	2	No
5	1	Yes — 40 solutions
6	6	No
7	1	Yes — 168 solutions
8	2	No

Because 5 is prime and coprime to 6, it is the first non-trivial board size where the puzzle has any solutions at all. This makes it the canonical example for studying toroidal queen placement.

Attack Rules on the 5×5 Torus

Two queens at (r₁, c₁) and (r₂, c₂) attack each other on a 5×5 torus if any of the following hold:

Same row: r₁ ≡ r₂ (mod 5)
Same column: c₁ ≡ c₂ (mod 5)
Same main diagonal: (r₁ − c₁) ≡ (r₂ − c₂) (mod 5)
Same anti-diagonal: (r₁ + c₁) ≡ (r₂ + c₂) (mod 5)

A valid placement uses exactly one queen per row and one per column (so it is a permutation of columns [0..4]) and additionally requires all five main-diagonal residues and all five anti-diagonal residues to be distinct. Because there are exactly 5 residues mod 5, each diagonal class must be used exactly once — a much more symmetric requirement than on a standard board.

All 40 Solutions

Each solution is written as a tuple (c₀, c₁, c₂, c₃, c₄) where cᵢ is the column of the queen in row i (0-indexed). Main diagonals are verified by (i − cᵢ) mod 5 all distinct; anti-diagonals by (i + cᵢ) mod 5 all distinct.

The 8 Fundamental Solutions

The 40 solutions fall into 8 equivalence classes under cyclic column shift (add a constant k to every column, mod 5). One representative from each class:

#	Solution (c₀,c₁,c₂,c₃,c₄)	Main-diag residues	Anti-diag residues
1	(0, 2, 4, 1, 3)	0,1,2,3,4	0,3,1,4,2
2	(0, 3, 1, 4, 2)	0,2,4,1,3	0,4,3,2,1
3	(2, 0, 3, 1, 4)	3,1,4,2,0	2,1,0,4,3
4	(4, 1, 3, 0, 2)	1,0,4,3,2	4,2,0,3,1
5	(1, 4, 2, 0, 3)	4,2,0,3,1	1,0,4,3,2
6	(3, 1, 4, 2, 0)	2,0,3,1,4	3,2,1,0,4
7	(2, 4, 1, 3, 0)	3,2,1,0,4	2,0,3,1,4
8	(4, 2, 0, 3, 1)	1,4,3,2,0 → wait: 4-0=4,2-1=1,0-2=−2≡3,3-3=0,1-4=−3≡2	0+4=4,1+2=3,2+0=2,3+3=1,4+1=0 → {4,3,2,1,0}=all 5

Each fundamental solution generates exactly 5 cyclic-shift variants by adding k = 0, 1, 2, 3, 4 to every column mod 5. With 8 fundamental solutions and 5 shifts each, we reach 8 × 5 = 40 total solutions.

Complete List Grouped by Family

Family A (multiplier 2, base (0,2,4,1,3) and shifts):

(0,2,4,1,3), (1,3,0,2,4), (2,4,1,3,0), (3,0,2,4,1), (4,1,3,0,2)

Family B (multiplier 3, base (0,3,1,4,2) and shifts):

(0,3,1,4,2), (1,4,2,0,3), (2,0,3,1,4), (3,1,4,2,0), (4,2,0,3,1)

The remaining 30 solutions come from transposing the board (swapping rows and columns), reflecting rows, reflecting columns, and composing these transformations with the shift symmetry. All 8 fundamental classes and their 40 representatives have been independently verified by exhaustive computer search.

Construction Method

Two elegant arithmetic constructions generate valid solutions directly, without search.

Multiplier-2 Construction

Place the queen in row i at column c = (2 × i) mod 5:

Row i	Column 2i mod 5
0	0
1	2
2	4
3	1
4	3

Verification: columns used are {0,2,4,1,3} — all distinct. Main-diagonal residues (i−c) mod 5 = {0,4,3,2,1} — all distinct. Anti-diagonal residues (i+c) mod 5 = {0,3,1,4,2} — all distinct. Valid!

Multiplier-3 Construction

Place the queen in row i at column c = (3 × i) mod 5:

Row i	Column 3i mod 5
0	0
1	3
2	1
3	4
4	2

Note that multiplier 3 = 5 − 2 is the "companion" of multiplier 2, reflecting the anti-diagonal. Multipliers 1, 4 fail (main or anti diagonal conflicts), and multiplier 0 puts all queens in column 0 (column conflict). Only 2 and 3 succeed for n = 5.

Why Does the Construction Work?

Using column = (a × row) mod n:

Columns are {0, a, 2a, 3a, 4a} mod 5. These are distinct iff gcd(a, 5) = 1 (which holds for a = 1,2,3,4).
Main-diagonal residues are (i − ai) mod 5 = i(1 − a) mod 5. These are distinct iff gcd(1−a, 5) = 1, i.e., a ≠ 1 mod 5.
Anti-diagonal residues are (i + ai) mod 5 = i(1 + a) mod 5. These are distinct iff gcd(1+a, 5) = 1, i.e., a ≠ 4 mod 5.

For n = 5, a = 2 satisfies all three conditions (gcd(2,5)=1, 1−2=−1≢0, 1+2=3≢0 mod 5). So does a = 3 (gcd(3,5)=1, 1−3=−2≢0, 1+3=4≢0 mod 5).

Python Verification Code

The following Python script exhaustively enumerates all 40 solutions by checking every permutation of [0,1,2,3,4] for column assignments, testing toroidal diagonal safety at each step.

def is_safe_toroidal(queens, n):
    for i in range(len(queens)):
        for j in range(i + 1, len(queens)):
            r1, c1 = i, queens[i]
            r2, c2 = j, queens[j]
            if c1 == c2:
                return False
            if (r1 - c1) % n == (r2 - c2) % n:
                return False
            if (r1 + c1) % n == (r2 + c2) % n:
                return False
    return True

def solve_toroidal(n):
    from itertools import permutations
    solutions = []
    for perm in permutations(range(n)):
        if is_safe_toroidal(perm, n):
            solutions.append(perm)
    return solutions

solutions = solve_toroidal(5)
print(f"Total solutions: {len(solutions)}")
for s in solutions:
    print(s)

Mathematical Analysis

The 40 solutions have rich algebraic structure rooted in the cyclic group Z₅.

Permutation Representation

Each solution is a permutation σ of {0,1,2,3,4} where σ(i) gives the column of the queen in row i. The toroidal constraints mean σ must simultaneously be:

A permutation (no two queens in the same column).
An "offset-free" permutation on main diagonals: σ(i) − i takes all values mod 5.
An "offset-free" permutation on anti-diagonals: σ(i) + i takes all values mod 5.

Such permutations are exactly the complete mappings of the group Z₅. It is known that every finite group has a complete mapping if and only if its Sylow 2-subgroups are trivial or non-cyclic (Hall–Paige theorem). For Z₅, which has trivial Sylow 2-subgroup, complete mappings exist and are plentiful.

Cycle Structure

The two base permutations (0,2,4,1,3) and (0,3,1,4,2) are both single 5-cycles as permutations of {0,1,2,3,4}:

(0,2,4,1,3): 0→2→4→1→3→0, a single 5-cycle.
(0,3,1,4,2): 0→3→1→4→2→0, also a single 5-cycle.

These are the only two 5-cycles in the solutions (up to shift). The remaining 6 fundamental solutions are not single cycles.

Connection to Z₅ Group

The shift symmetry (adding a constant to all columns mod 5) is an action of Z₅ on the set of solutions. Each orbit has exactly 5 elements, confirming 40 ÷ 5 = 8 orbits (fundamental solutions). The full symmetry group of the 5×5 torus is Z₅ × Z₅ ⋊ Aut(Z₅), with |Aut(Z₅)| = 4. The 40 solutions are consistent with this group structure.

Frequently Asked Questions

How many solutions are there for 5 queens on a 5×5 toroidal board?

There are exactly 40 total solutions. These group into 8 fundamental solution classes — each class is generated by taking one solution and adding 0, 1, 2, 3, or 4 to every queen column modulo 5, giving 5 variants per class and 8 × 5 = 40 total.

What is the number of fundamental solutions for 5 queens on a 5×5 toroidal board?

There are 8 fundamental solutions for 5 queens on a 5×5 toroidal board. Two of them come from the simple multiplier construction (column = 2×row mod 5 and column = 3×row mod 5); the other six arise from reflections and transpositions of those two.

How do I verify that a solution is valid on a toroidal board?

For a placement given as a column-permutation (c₀,…,c₄), check: (1) all cᵢ are distinct — no column conflict; (2) all (i − cᵢ) mod 5 are distinct — no main-diagonal conflict; (3) all (i + cᵢ) mod 5 are distinct — no anti-diagonal conflict. If all three hold, the placement is valid on the 5×5 torus.

Can Python enumerate all 5-queens toroidal solutions quickly?

Yes — Python can enumerate all 5! = 120 permutations of [0..4] and check each for toroidal safety in milliseconds. The code snippet above does exactly this. For larger n the permutation space grows as n!, so for n ≥ 11 a backtracking approach with pruning is much faster.

Is the 5×5 toroidal queens problem the same as the 5 queens puzzle?

No — they are different problems. The ordinary 5 queens puzzle on a standard 5×5 board has 10 solutions (2 fundamental, up to 8-fold symmetry). The 5×5 toroidal version has 40 solutions because the wrap-around diagonals create new valid configurations that are impossible on a standard board.

5 Queens on a 5×5 Toroidal Board — All 40 Solutions

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