Magic Shapes - Solved

Method for Triangles

First, label the nodes sequentially starting from any vertex:

Labelled Magic Triangle

The solution involves the following steps:

1. Generate all permutations of numbers 1 to 6 as a, b, c, d, e, f.
2. Filter permutations to satisfy the magic shape condition: a + b + c = c + d + e = e + f + a.
3. Apply the final condition: a + b + c = 10.

Using Haskell

Generate all permutations of the numbers 1 to 6:

import Data.List

permutations [1..6]

This yields 6! = 720 permutations.

Filter for sides with equal sums:

[
  [(a,b,c), (c,d,e), (e,f,a)] | [a,b,c,d,e,f] <- permutations [1..6],
    a+b+c == c+d+e &&
    c+d+e == e+f+a
]

This produces all solutions where the sides have equal sums:

  [(3,2,5),(5,4,1),(1,6,3)]
  [(2,4,3),(3,5,1),(1,6,2)]
  [(1,5,3),(3,4,2),(2,6,1)]
  [(1,4,5),(5,2,3),(3,6,1)]
  [(5,3,4),(4,2,6),(6,1,5)]
  [(6,2,4),(4,3,5),(5,1,6)]
  [(4,5,2),(2,3,6),(6,1,4)]
  [(6,3,2),(2,5,4),(4,1,6)]
  [(2,3,6),(6,1,4),(4,5,2)]
  [(4,1,6),(6,3,2),(2,5,4)]
  [(3,4,2),(2,6,1),(1,5,3)]
  [(1,6,2),(2,4,3),(3,5,1)]
  [(5,4,1),(1,6,3),(3,2,5)]
  [(4,3,5),(5,1,6),(6,2,4)]
  [(6,1,5),(5,3,4),(4,2,6)]
  [(3,6,1),(1,4,5),(5,2,3)]
  [(5,2,3),(3,6,1),(1,4,5)]
  [(2,6,1),(1,5,3),(3,4,2)]
  [(3,5,1),(1,6,2),(2,4,3)]
  [(1,6,3),(3,2,5),(5,4,1)]
  [(5,1,6),(6,2,4),(4,3,5)]
  [(2,5,4),(4,1,6),(6,3,2)]
  [(6,1,4),(4,5,2),(2,3,6)]
  [(4,2,6),(6,1,5),(5,3,4)]

Filter for sides summing specifically to 10:

[
  [(a,b,c), (c,d,e), (e,f,a)] | [a,b,c,d,e,f] <- permutations [1..6],
    a+b+c == 10 &&
    c+d+e == 10 &&
    e+f+a == 10
]

This yields the following list of solutions:

[(3,2,5),(5,4,1),(1,6,3)]
[(1,4,5),(5,2,3),(3,6,1)]
[(5,4,1),(1,6,3),(3,2,5)]
[(3,6,1),(1,4,5),(5,2,3)]
[(5,2,3),(3,6,1),(1,4,5)]
[(1,6,3),(3,2,5),(5,4,1)]

These solutions are not unique, as the list includes rotations and reflections. To isolate the unique solution, impose an ordering constraint on the vertices:

[
  [(a,b,c), (c,d,e), (e,f,a)] | [a,b,c,d,e,f] <- permutations [1..6],
    a+b+c == 10 &&
    c+d+e == 10 &&
    e+f+a == 10 &&
    a > c &&
    c > e
]

This yields the unique solution:

[(5,2,3),(3,6,1),(1,4,5)]

Refined Haskell implementation:

[
  [(a,b,c), (c,d,e), (e,f,a)] | [a,b,c,d,e,f] <- permutations [1..6],
    all (==10) [a+b+c, c+d+e, e+f+a] &&
    a > c && c > e
]

Solved Magic Triangle

Verify the answer on the CSIRO page.

Using Python

Python supports list comprehensions comparable to Haskell’s. The permutations function from the itertools module generates the candidates:

import itertools

[
  [(a,b,c),(c,d,e),(e,f,a)]
    for a,b,c,d,e,f in list(itertools.permutations(range(1,7)))
      if a+b+c == 10 and c+d+e == 10 and e+f+a == 10
        and a > c and c > e
]

This yields the same result:

>>> [[(5, 2, 3), (3, 6, 1), (1, 4, 5)]]

Method for Pentagons

To find solutions where all sides sum to an equal value, first label the vertices:

Labelled Magic Pentagon

Similar to the magic triangle approach, we search all permutations for those where side sums are equal. The search is limited to the first result for each total sum.

import Data.List
import Data.Maybe

solve :: Int -> Maybe [(Int,Int,Int)]
solve n
    | null ns   = Nothing
    | otherwise = Just (head ns)
    where ns = [ [(a,b,c), (c,d,e), (e,f,g), (g,h,i), (i,j,a)]
                  | [a,b,c,d,e,f,g,h,i,j] <- permutations [1..10],
                    all (==n) [a+b+c, c+d+e, e+f+g, g+h+i, i+j+a] ]

Solutions are calculated for side sums from 10 to 27:

-- solve for sums 10 to 27
solutions = map solve [10..27]

The range 10–27 is generous. While theoretical bounds for a side sum with inputs 1-10 are 6 and 27, the pentagon’s constraints narrow this significantly (likely 14-19).

Show the solutions:

-- show solutions
zip [10..27] solutions

The solutions are:

10: Nothing
11: Nothing
12: Nothing
13: Nothing
14: Just [(3,6,5),(5,7,2),(2,8,4),(4,9,1),(1,10,3)]
15: Nothing
16: Just [(5,2,9),(9,4,3),(3,6,7),(7,8,1),(1,10,5)]
17: Just [(6,9,2),(2,7,8),(8,5,4),(4,3,10),(10,1,6)]
18: Nothing
19: Just [(8,5,6),(6,4,9),(9,3,7),(7,2,10),(10,1,8)]
20: Nothing
21: Nothing
22: Nothing
23: Nothing
24: Nothing
25: Nothing
26: Nothing
27: Nothing

Dr Stephen Weissenhofer from the School of Computing, Data and Mathematical Sciences at Western Sydney University kindly provided these problems.