Question 4

Step 1 · Fill Grid 1 using Binairo Rules

Let black dashes be $0$ and red vertical lines be $1$.

• Cells $(1,5)$ and $(1,6)$ are $0, 0 \implies$ cell $(1,4) = 1$.

• Cells $(1,3)$ and $(1,4)$ are $1, 1 \implies$ cell $(1,2) = 0$.

• Balancing Row 1 (needs three $1$s and three $0$s) $\implies$ cell $(1,1) = 1$. $\boxed{\text{R1: } 1\ 0\ 1\ 1\ 0\ 0}$

• Column 3 has $1$ at $(1,3)$, $(3,3)$, and $(5,3)$ (already three $1$s) $\implies (2,3) = (4,3) = (6,3) = 0$. $\boxed{\text{C3: } 1\ 0\ 1\ 0\ 1\ 0}$

• Column 2 has $0$ at $(1,2)$ and $(2,2) \implies (3,2) = 1$.

• Row 3 now has three $1$s at $(3,2), (3,3), (3,5) \implies (3,1) = (3,4) = (3,6) = 0$. $\boxed{\text{R3: } 0\ 1\ 1\ 0\ 1\ 0}$

• In Row 2, $(2,2)$ and $(2,3)$ are $0, 0 \implies (2,1) = 1$ and $(2,4) = 1$. $\boxed{\text{R2: } 1\ 0\ 0\ 1\ \_\ \_}$

• In Column 1, $(3,1)$ and $(4,1)$ are $0, 0 \implies (5,1) = 1$. Column 1 now has three $1$s $\implies (6,1) = 0$. $\boxed{\text{C1: } 1\ 1\ 0\ 0\ 1\ 0}$

• In Row 6, $(6,1)$ and $(6,3)$ are $0, 0 \implies (6,2) = 1$. $\boxed{\text{R6: } 0\ 1\ 0\ 1\ \_\ \_}$

• In Row 4, $(4,1)$ and $(4,3)$ are $0, 0 \implies (4,2) = 1$.

• Column 2 now has three $1$s $\implies (5,2) = 0$. $\boxed{\text{C2: } 0\ 0\ 1\ 1\ 0\ 1}$

• Column 4 has three $1$s $\implies (4,4) = 0$. $\boxed{\text{C4: } 1\ 1\ 0\ 0\ 0\ 1}$

• In Row 4, $(4,3)$ and $(4,4)$ are $0, 0 \implies (4,5) = 1$. Balancing Row 4 gives $(4,6) = 1$. $\boxed{\text{R4: } 0\ 1\ 0\ 0\ 1\ 1}$

• In Column 5, $(3,5)$ and $(4,5)$ are $1, 1 \implies (2,5) = 0$ and $(5,5) = 0$. Balancing Column 5 gives $(6,5) = 1$. $\boxed{\text{C5: } 0\ 0\ 1\ 1\ 0\ 1}$

• Balancing remaining rows:

  • Row 2: $(2,6) = 1 \implies \boxed{\text{R2: } 1\ 0\ 0\ 1\ 0\ 1}$
  • Row 5: $(5,6) = 1 \implies \boxed{\text{R5: } 1\ 0\ 1\ 0\ 0\ 1}$
  • Row 6: $(6,6) = 0 \implies \boxed{\text{R6: } 0\ 1\ 0\ 1\ 1\ 0}$

Step 1 · Fill Grid 2 using Binairo Rules

Let black dashes be $0$ and red vertical lines be $1$.

• Column 1 has $(1,1) = 0$ and $(2,1) = 0 \implies (3,1) = 1$. $\boxed{\text{C1: } 0\ 0\ 1\ \_\ \_\ \_}$

• Row 3 already contains three $1$s at $(3,1), (3,3), (3,5) \implies (3,2) = (3,4) = (3,6) = 0$. $\boxed{\text{R3: } 1\ 0\ 1\ 0\ 1\ 0}$

• Column 3 has three $1$s at $(3,3), (4,3), (6,3) \implies (1,3) = (2,3) = (5,3) = 0$. $\boxed{\text{C3: } 0\ 0\ 1\ 1\ 0\ 1}$

• In Row 1, $(1,1)$ and $(1,3)$ are $0, 0 \implies (1,2) = 1$. $\boxed{\text{R1: } 0\ 1\ 0\ \_\ \_\ \_}$

• In Row 2, $(2,1)$ and $(2,3)$ are $0, 0 \implies (2,2) = 1$. $\boxed{\text{R2: } 0\ 1\ 0\ \_\ \_\ \_}$

• In Row 4, $(4,3)$ and $(4,5)$ are $1, 1 \implies (4,4) = 0$. $\boxed{\text{R4: } \_\ \_\ 1\ 0\ 1\ \_}$

• In Row 5, $(5,3)$ and $(5,6)$ are $0, 0 \implies (5,4) = 1$. $\boxed{\text{R5: } \_\ \_\ 0\ 1\ \_\ 0}$

• In Row 6, $(6,3)$ and $(6,6)$ are $1, 1 \implies (6,5) = 0$. $\boxed{\text{R6: } \_\ \_\ 1\ \_\ 0\ 1}$

• Column 5 has $(3,5) = 1, (4,5) = 1 \implies (5,5) = 0$. $\boxed{\text{C5: } \_\ \_\ 1\ 1\ 0\ 0}$

• Row 5 has three $0$s $\implies (5,2) = 1$ and $(5,1) = 1$. $\boxed{\text{R5: } 1\ 1\ 0\ 1\ 0\ 0}$

• Column 1 has $(3,1) = 1, (5,1) = 1 \implies (4,1) = 0$, giving $(6,1) = 1$. $\boxed{\text{C1: } 0\ 0\ 1\ 0\ 1\ 1}$

• Column 2 has $(2,2) = 1, (5,2) = 1 \implies (4,2) = 0$, giving $(6,2) = 0$. $\boxed{\text{C2: } 1\ 1\ 0\ 0\ 1\ 0}$

• Balancing remaining cells:

  • Row 4: $(4,6) = 1 \implies \boxed{\text{R4: } 0\ 0\ 1\ 0\ 1\ 1}$
  • Column 4: $(2,4) = 1, (6,4) = 0 \implies \boxed{\text{C4: } 1\ 1\ 0\ 0\ 1\ 0}$
  • Row 1: $(1,5) = 0, (1,6) = 1 \implies \boxed{\text{R1: } 0\ 1\ 0\ 1\ 0\ 1}$
  • Column 6: $(2,6) = 0 \implies \boxed{\text{C6: } 1\ 0\ 0\ 1\ 0\ 1}$
  • Row 2: $(2,5) = 1 \implies \boxed{\text{R2: } 0\ 1\ 0\ 1\ 1\ 0}$

Step 1 · Fill Grid 3 using Binairo Rules

Let black dashes be $0$ and red vertical lines be $1$.

• In Column 6, $(2,6)$ and $(3,6)$ are $0, 0 \implies (1,6) = 1$. $\boxed{\text{C6: } \_\ \_\ \_\ \_\ \_\ 1\ 0\ 0}$

• In Row 1, $(1,5) = 0 \implies (1,4) = 1$. $\boxed{\text{R1: } \_\ \_\ \_\ 1\ 0\ 1}$

• In Row 3, $(3,2) = 0 \implies (3,1) = 1$. $\boxed{\text{R3: } 1\ 0\ \_\ \_\ \_\ 0}$

• Column 1 has $(3,1) = 1, (6,1) = 1 \implies (1,1) = 0, (2,1) = 0$. $\boxed{\text{C1: } 0\ 0\ 1\ \_\ \_\ 1}$

• In Row 2, $(2,6) = 0 \implies (2,5) = 1$. $\boxed{\text{R2: } 0\ \_\ \_\ \_\ 1\ 0}$

• In Column 5, $(2,5) = 1 \implies (3,5) = 0$. $\boxed{\text{C5: } 0\ 1\ 0\ \_\ \_\ \_}$

• In Row 3, $(3,2) = 0, (3,5) = 0 \implies (3,3) = 1$. Balancing Row 3 gives $(3,4) = 1$. $\boxed{\text{R3: } 1\ 0\ 1\ 1\ 0\ 0}$

• Column 3 already contains three $1$s $\implies (2,3) = (4,3) = (6,3) = 0$. $\boxed{\text{C3: } 1\ 0\ 1\ 0\ 1\ 0}$

• In Column 4, to avoid violating Row 2 rules, $(2,4) = 1$. Column 4 now has three $1$s $\implies (4,4) = (5,4) = (6,4) = 0$. $\boxed{\text{C4: } 1\ 1\ 1\ 0\ 0\ 0}$

• Row 2 has three $0$s $\implies (2,2) = 1$. $\boxed{\text{R2: } 0\ 1\ 0\ 1\ 1\ 0}$

• Column 2 has $(3,2) = 0 \implies (4,2) = 1$. Balancing gives $(6,2) = 0$. $\boxed{\text{C2: } 0\ 1\ 0\ 1\ 1\ 0}$

• Balancing remaining rows:

  • Row 1: $(1,3) = 1 \implies \boxed{\text{R1: } 0\ 0\ 1\ 1\ 0\ 1}$
  • Row 4: $(4,5) = 1, (4,6) = 1 \implies \boxed{\text{R4: } 0\ 1\ 0\ 0\ 1\ 1}$
  • Row 5: $(5,5) = 0, (5,6) = 0 \implies \boxed{\text{R5: } 1\ 1\ 1\ 0\ 0\ 0}$
  • Row 6: $(6,5) = 1, (6,6) = 1 \implies \boxed{\text{R6: } 1\ 0\ 0\ 0\ 1\ 1}$