SudokuSolves/TheHub

OP = Pallando

A solve path for the SudokuPuzzle?: The Hub

Substituting in some variables gives:

r3c1 = a
r6c8 = a
r1c3 = b
r8c6 = b
r1c6 = c
r8c3 = c
r6c1 = d
r3c8 = d

Using mod8(0)=0 mod8(1)=1 ... mod8(7)=7 mod8(8)=0 mod8(9)=1 etc.

r2c2 = 1 + mod8( J-1 ) = J
r2c7 = 1 + mod8( J+1 )
r7c7 = 1 + mod8( J+3 )
r7c2 = 1 + mod8( J+5 )

r3c4 = 1 + mod8( L-1 ) = L
r4c6 = 1 + mod8( L+1 )
r6c5 = 1 + mod8( L+3 )
r5c3 = 1 + mod8( L+5 )

r4c3 = 1 + mod8( K-1 ) = K
r3c5 = 1 + mod8( K+1 )
r5c6 = 1 + mod8( K+3 )
r6c4 = 1 + mod8( K+5 )

https://f-puzzles.com/?id=yj2ahvq9

Setting up the little killer sums as equations we get:

9 = a + b + J
11 = c + d + 1 + mod8( J+1 )
13 = a + b + 1 + mod8( J+3 )
15 = c + d + 1 + mod8( J+5 )

27 = a + b + e + f + 1 + mod8( L+5 ) + 1 + mod8( K+5 )
23 = c + d + g + h + 1 + mod8( L+3 ) + 1 + mod8( K+3 )
27 = a + b + e + f + 1 + mod8( L+1 ) + 1 + mod8( K+1 )
31 = c + d + g + h + 1 + L + K

That's 8 equations and 11 unknowns...
But actually we have four pairs of variables that always and only appear together.
a+b, c+d, e+f, g+h
So we can substitute in P=a+b, Q=c+d, R=e+f, S=g+h
giving

[Eq1.] 9 = P + J
[Eq2.] 10 = Q + mod8( J+1 )
[Eq3.] 12 = P + mod8( J+3 )
[Eq4.] 14 = Q + mod8( J+5 )

[Eq5.] 25 = P + R + mod8( L+5 ) + mod8( K+5 )
[Eq6.] 21 = Q + S + mod8( L+3 ) + mod8( K+3 )
[Eq7.] 25 = P + R + mod8( L+1 ) + mod8( K+1 )
[Eq8.] 31 = Q + S + L + K

So what can we deduce?

By default: P,Q,R,S all have a min value of 3 (1+2) and a max value of 15 (7+8)

However we can narrow that down...

Looking at [Eq1.] and [Eq3.]
1+mod8(J+3) is 4 larger than J so J can't be 5,6,7 or 8

Looking at [Eq2.] and [Eq4.]
1+mod8(J+5) is 4 larger than 1+mod8(J+1) so J can't be 3,4,5 or 6

Which, together, tells us that J is either 1 or 2.

Therefore:
P is 7 or 8 (making one of a/b 1-3 and the other 4-7)
Q is 7 or 8 (making one of c/d 1-3 and the other 4-7)
However it also tells us that P = Q (ie both are 7 or both are 8)

But what about K and L?

Well, if L and K were both 1, then we'd have:
Q + S + 2 = 31, where Q is 7 or 8, making S=22 or S+23

But at most S can be 7+8 = 15, which gives us a minimum on K+L...
(7 or 8) + 15 + (K+L) = 31
K+L = 8 or 9

In the other direction, the least S can be is 3, which gives us a maximum on K+L...
(7 or 8) + 3 + (K+L) = 31
of 20 or 21. Not helpful. How about using one of the other diagonals?
The lowest valued long diagonal is
[Eq6.] 21 = Q + S + mod8( L+3 ) + mod8( K+3 )
substitute in S=3 gives us
(7 or 8) + 3 + ( mod8( L+3 ) + mod8( K+3 ) = 21
mod8( L+3 ) + mod8( K+3 ) = 10 or 11

Similar calculations on all 4 long diagonals gives us:

8 or 9 <= K+L <= 20 or 21
4 or 5 <= 2 + (mod(K+1) + mod(L+1) <= 16 or 17
0 or 1 <= 2 + (mod(K+3) + mod(L+3) <= 12 or 13
4 or 5 <= 2 + (mod(K+5) + mod(L+5) <= 16 or 17

As we rotate by 90 degrees, the total of K+L will go up or down as follows,
depending on whether 0, 1 or 2 of them increase above 8:

0: total -> total + 4
1: total -> total - 4
2: total -> total -12

from this we can deduce that

: from K,L -> K+2,L+2 one of them ticks over
: from K+2,L+2 -> K+4,L+4 one of them ticks over
: from K+4,L+4 -> K+6,L+6 neither of them ticks over
: from K+6,L+6 -> K,L neither of them ticks over

So either K = 7/8 and L = 5/6
or K = 5/6 and L = 7/8

Either way,

  K + L = 12,13 or 14
  2 + (mod(K+1) + mod(L+1) = 8,9 or 10
  2 + (mod(K+3) + mod(L+3) = 4,5 or 6
  2 + (mod(K+5) + mod(L+5) = 8,9 or 10

Which lets us narrow down R and S:

27 = (7 or 8) + R + (8,9 or 10)
23 = (7 or 8) + S + (4,5 or 6)

R = 9,10,11,12
S = 9,10,11,12

But, looking at the hub, not all combinations of L and K are possible. If they differ by exactly 2, then a 90 degree turn will cause conflict. Therefore, whether they are 5+8 or 6+7, their total is precisely 13.

R = 10,11
S = 10,11

K=7,L=6 --> https://f-puzzles.com/?id=yewaksaw
K=6,L=7 --> https://f-puzzles.com/?id=yjtw2vje
K=8,L=5 --> https://f-puzzles.com/?id=ydlah5vh
L=5,L=8 --> https://f-puzzles.com/?id=yh8nxhgy

Added a cage to disambiguate, then test again for uniquness:

SOLUTION 1 = https://f-puzzles.com/?id=ygboowrc (K=5,L=8)

good, it's unique. However, it takes bifurcation to find.
So I should give a direct clue to the inner circle instead of the cage.
A given digit?

Version with the cage: https://f-puzzles.com/?id=ydrx5q47

Confirmed - it is possible to solve, as long as you do some maths :-)

https://f-puzzles.com/?id=yj5txs4t

https://f-puzzles.com/?id=yhnn4k6l