5x5 bimagic search technique
by Lee Morgenstern, January 2014.
A B C D E
F
G H I J
K L M N P
Q
R S T U
V W X Y Z
for MaxNo = 12 to 1000 [and hopefully greater]
for PartialSum
= (MaxNo + 66)/3 to 2(MaxNo + 1) [see Note 1]
List all 4-entry series adding to PartialSum using 1 to MaxNo.
Sort the
series by their sum of squares.
for each group of three disjoint series
with the same
sum and sum of squares
that contain both 1 and MaxNo [see
Note 2]
assign them to A,G,T,Z and
E,I,R,V and C,H,S,X such that
E + I + R + V = A + G + T + Z = C
+ H + S + X
E² + I² + R² + V² = A² + G² + T²
+ Z² = C² + H² + S² + X²
for each permutation of the 12 entries that satisfies the above,
Avoid permutations that can be reached by a rotation, reflection,
etc.
Delay permutations until they are needed. [see
Note 3]
3 choices for C,H,S,X line
3 choices
for (A,Z),(G,T) combo AGTZ AGZT AZTG
6
choices for (E,V),(I,R) combo
6 choices for (C,X),(H,S)
combo
Solve for L and N using Formula 1.
L
+ N = A + Z + E + V + C + X - G - T - I - R
L² + N² = A²
+ Z² + E² + V² + C² + X² - G² - T² - I² - R² Reject if not rational.
Solve for K and P using Formula 1.
K + P = A + Z + G
+ T - L - N
K² + P² = A² + Z² + G² + T² - L² - N² Reject
if not rational.
for each permutation involving swapping
E,V; I,R;
K,P; H,S [16 permutations]
Solve for J and Q using Formula 2.
J
- Q = A + K + V - G - H - I
J² - Q² = A² + K² + V² - G²
- H² - I² Reject if not integer.
Solve for U and M using Formula 2.
U - M = A + G + Z - Q - R - S
U²
- M² = A² + G² + Z² - Q² - R² - S² Reject if not integer.
Compute F = G + M + T + Z - K - V - Q
Verify F² = G² + M² + T² + Z² -
K² - V² - Q² Reject if not equal.
for each permutation involving swapping
L,N; C,X
[4 permutations]
Solve for W and D using Formula 2.
W - D = A + C + E
- G - L - R
W² - D² = A² + C² + E² - G² - L² - R² Reject
if not integer.
Compute Y = A + G + Z + M - D - I - N
Verify Y² = A² + G² + Z² + M² -
D² - I² - N² Reject if not equal.
Compute B = G + T + Z + M - C - D - E
Verify B² = G² + T² + Z² + M² -
C² - D² - E² Reject if not equal.
Output solution.
Given a and b, find x and y such that
x + y = a
x²
+ y² = b
x = (a + c)/2
y = (a - c)/2
where
c² = 2b - a²
Reject if c is not rational.
Note that if a and b are integer and c is
rational, then x and y will be integers.
Note also that the values of x and
y can be swapped.
Formula 2
Given a and b, find x and y such that
x - y = a
x²
- y² = b
This always has the rational solution
x = (b/a + a)/2
y
= (b/a - a)/2
but may not be integer.
The highest PartialSum value is 2(MaxNo + 1).
If a bimagic square existed
with a higher partial sum, you could subtract all of its entries from (MaxNo
+ 1) and get another bimagic square with a partial sum smaller than 2(MaxNo
+ 1) for the initial three series, which would have been found by an earlier
search.
Note 2.
1 and MaxNo are required to be in the initial 12 entries.
If they weren't
and a bimagic square was found, you could translate this solution to a range
using a smaller value of MaxNo, which would have been found by an earlier search.
Note 3.
The computation of L,N and K,P results in the same values if E and V are swapped, for example. If these extra swaps are done and the computation of L,N or K,P is rejected, you would be rejecting the same thing multiple times.
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