Bivalue AIC Ring
A Bivalue AIC Ring — traditionally listed as Bidirectional Y-Cycle — is a closed AIC whose strong links all run through bivalue cells: each node is the pair of digits of one cell, joined to the next by weak inferences, with a final weak inference closing the loop. The candidate propositions and links define the deduction; a count of highlighted cells or a familiar drawing does not.
What is a Bivalue AIC Ring?
Each bivalue node is a strong link (XOR): its two digits cannot both be true and cannot both be false. A weak inference (NAND) from one node to the next is valid when the digit leaving the first cell conflicts with the digit entering the second — for example, two occurrences of the same digit in peer cells.
Walkthrough example
A walkthrough of one Bivalue AIC Ring: strong links through bivalue cells, closed by a weak inference verified independently of the path.
Step 1
Consider these cells.
Step 2
The highlighted candidates form the closed Bivalue AIC Ring (B3:5=4)-(G3:4=1)-(I2:1=2)-(I9:2=4)-(H7:4=7)-(A7:7=4)-(A1:4=5). Each equals sign is a strong link (XOR); each dash is a weak inference (NAND). The last link, A1=5 back to B3=5, is a weak inference that holds independently of the path — it closes the loop, and the endpoint theorem supplies its complementary strong inference, so every weak edge of the ring becomes an exact-one pair. The numbered implication trace walks the ring in one direction to display the alternating links; it illustrates the closed structure and is not an extra forcing assumption. 1. If A1 contains the value 5, then B3 cannot contain the value 5 (the value can occur only once in the block) 2. If B3 does not contain the value 5, then B3 must contain the value 4 (only remaining possible value in the cell) 3. If B3 contains the value 4, then G3 cannot contain the value 4 (the value can occur only once in the row) 4. If G3 does not contain the value 4, then G3 must contain the value 1 (only remaining possible value in the cell) 5. If G3 contains the value 1, then I2 cannot contain the value 1 (the value can occur only once in the block) 6. If I2 does not contain the value 1, then I2 must contain the value 2 (only remaining possible value in the cell) 7. If I2 contains the value 2, then I9 cannot contain the value 2 (the value can occur only once in the column) 8. If I9 does not contain the value 2, then I9 must contain the value 4 (only remaining possible value in the cell) 9. If I9 contains the value 4, then H7 cannot contain the value 4 (the value can occur only once in the block) 10. If H7 does not contain the value 4, then H7 must contain the value 7 (only remaining possible value in the cell) 11. If H7 contains the value 7, then A7 cannot contain the value 7 (the value can occur only once in the row) 12. If A7 does not contain the value 7, then A7 must contain the value 4 (only remaining possible value in the cell) 13. If A7 contains the value 4, then A1 cannot contain the value 4 (the value can occur only once in the column) 14. If A1 does not contain the value 4, then A1 must contain the value 5 (only remaining possible value in the cell)
Step 3
So value 4 can be removed: closure turns every weak edge of the ring into an exact-one pair, so exactly one of its two alternating assignments holds — and each red candidate conflicts with a true ring proposition in both.
A bivalue ring runs every strong link (XOR) through the two digits of one cell and closes through a weak inference (NAND) verified independently of the path. The endpoint theorem saturates the closing pair, so candidates that conflict with both sides of any weak edge can be removed. You can play sudoku online to practise with the current candidate state visible.
How to spot it
List the bivalue cells, choose for each node which digit exits and which enters, and verify every weak connection against the current candidate state — including the closing weak inference from the last node back to the first. Geometric closure alone proves nothing.
The logic
With the closing weak inference verified, the endpoint theorem supplies the missing strong inference between the closing pair, so every weak edge around the ring gains a complementary strong inference. Remove candidates that conflict with both sides of any weak edge; the ring itself never selects which alternating assignment is the solution.