6.2 Box-Based Puzzles & Vertical Stacks

6.2 Box-Based Puzzles & Vertical Stacks

Vertical box-stack puzzles are a fundamental component of the Reasoning Ability section in both the Preliminary and Main phases of the SBI Clerk Examination. While floor puzzles provide a rigid architectural skeleton with numbered storeys firmly established at the outset, vertical box puzzles present a floating relative order. A collection of distinct boxes (typically 7 to 9 boxes in Prelims, expanding to 10 to 12 boxes or stacks of unknown total height in Mains) are stacked one above another.

Candidates must determine the absolute vertical coordinates—traditionally numbered from Position 1 at the bottom of the stack up to Position $N$ at the top—by interpreting relative distance clues, symmetrical boundary conditions, and rigid adjacency pairs. A single miscalculated gap will derail the entire 5-question puzzle cluster, making mathematical rigor essential.

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Deciphering Critical Box-Stack Terminology

Precision in box puzzle solving begins with converting linguistic clue phrases into exact algebraic equations.

1. The Gap-to-Index Conversion Formula

When a clue specifies an intervening gap between two boxes: $\text{"Exactly } k \text{ boxes are kept between Box P and Box Q"}$ This means there are $k$ physical boxes separating $P$ and $Q$. In terms of stack index numbers: $|\text{Index}(P) - \text{Index}(Q)| = k + 1$

[!WARNING] The Common Off-by-One Trap: If three boxes are kept between Box $P$ and Box $Q$, and Box $Q$ is at Position 2, the position of Box $P$ is NOT $2 + 3 = 5$. The correct position is $2 + (3 + 1) = \mathbf{6}$. Positions 3, 4, and 5 contain the three intervening boxes. Forgetting to add 1 to the gap size is the most common clerical exam error.

2. Immediate Adjacency vs. Open Precedence

• Immediate Adjacency: "Box R is kept immediately above Box S" forms an unbreakable, rigid 2-box block: $\text{Index}(R) = \text{Index}(S) + 1 \quad \implies \quad \begin{bmatrix} R \\ S \end{bmatrix}$
• Open Precedence: "Box R is kept above Box S" indicates general superiority without fixed distance: $\text{Index}(R) > \text{Index}(S)$ Any number of boxes (from zero intervening boxes up to $N-2$ boxes) may lie between them.

3. The Symmetrical Inversion Invariant

Clues establishing symmetrical balance across stack boundaries appear in nearly every SBI Clerk box set: $\text{"As many boxes are kept above Box T as below Box U"}$ In a stack containing $N$ total boxes numbered 1 (bottom) to $N$ (top):

• Number of boxes above $T = N - \text{Index}(T)$
• Number of boxes below $U = \text{Index}(U) - 1$ Equating the two quantities yields the Symmetry Invariant Formula: $N - \text{Index}(T) = \text{Index}(U) - 1 \implies \mathbf{\text{Index}(T) + \text{Index}(U) = N + 1}$

This mathematical rule allows candidates to immediately place Box $U$ the instant Box $T$ is identified, bypassing manual counting.

4. Unknown Total Boxes in a Stack

In SBI Clerk Mains, puzzles often withhold the total count $N$, giving clues like:

• "Box X is 4th from the top."
• "Box Y is 5th from the bottom."
• "Two boxes are kept between Box X and Box Y."

To resolve $N$, analyze whether the segments are disjoint or overlapping:

• Disjoint Arrangement ($X$ is above $Y$): $N = (\text{Rank from Top}) + (\text{Gap}) + (\text{Rank from Bottom}) = 4 + 2 + 5 = \mathbf{11 \text{ boxes}}$
• Overlapping Arrangement ($Y$ is above $X$): $N = (\text{Rank from Top}) + (\text{Rank from Bottom}) - (\text{Gap}) - 2 = 4 + 5 - 2 - 2 = \mathbf{5 \text{ boxes}}$ If other clues state "at least 8 boxes are in the stack," the overlapping possibility is ruled out instantly, fixing $N = 11$.

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The Bounding Anchor Method

When attacking an 8-box puzzle under time constraints, follow this structured execution pipeline:

1. Locate the Widest Rigid Span (The Bounding Anchor): Scan the clue list for the largest gap. A statement like "Three boxes between Box P and Box K, where K is at an odd position" binds 5 consecutive slots ($P, -, -, -, K$).
2. Determine Finite Placements: In an 8-slot stack, a 5-slot span has very few valid configurations. If $K$ is on an odd position, $K$ can only be at Position 1 (forcing $P$ to 5) or Position 3 (forcing $P$ to 7). Position 5 would place $P$ at $5 + 4 = 9$, exceeding the 8-box ceiling.
3. Spin Off Parallel Cases: Set up Case 1 ($K=1, P=5$) and Case 2 ($K=3, P=7$).
4. Attach Compound Rigid Blocks: Connect secondary clues that bind directly to $P$ or $K$ (e.g., "Box M is immediately above Box P").
5. Apply Boundary Inversions: Use symmetrical conditions ("as many above M as below O") to fix outer coordinates.
6. Execute Case Termination: When an incoming clue requires a position that is already occupied by another box, immediately terminate that branch.

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Step-by-Step Worked 8-Box Stack Puzzle

Problem Statement

Eight boxes—J, K, L, M, N, O, P, and Q—are stacked one above another in a vertical rack, but not necessarily in the same order. The bottommost position is numbered 1, and the topmost position is numbered 8.

Clues:

1. Exactly three boxes are kept between Box $P$ and Box $K$, where Box $P$ is kept above Box $K$.
2. Box $K$ is kept on an odd-numbered position.
3. Box $M$ is kept immediately above Box $P$.
4. As many boxes are kept above Box $M$ as below Box $O$.
5. Box $J$ is kept immediately above Box $O$.
6. Box $N$ is kept immediately above Box $Q$.
7. Box $Q$ is kept immediately above Box $K$.
8. Box $L$ occupies the remaining position in the stack.

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Step-by-Step Deduction and Case Resolution

Step 1: Anchor the Widest Span (Clues 1 and 2)

• Clue 1 states: 3 boxes between $P$ and $K$, with $P$ above $K$. Mathematically: $\text{Index}(P) - \text{Index}(K) = 4$.
• Clue 2 states: $K$ is on an odd-numbered position (${1, 3, 5, 7}$).
  • If $K = 1$: $\text{Index}(P) = 1 + 4 = 5$.
  • If $K = 3$: $\text{Index}(P) = 3 + 4 = 7$.
  • If $K = 5$: $\text{Index}(P) = 5 + 4 = 9$ (Exceeds stack limit of 8; invalid).
• This produces two viable starting cases:
  • Case 1: $K = \text{Position } 1$, $P = \text{Position } 5$.
  • Case 2: $K = \text{Position } 3$, $P = \text{Position } 7$.

Step 2: Integrate Clue 3 (Immediate Adjacency)

• Clue 3 states: Box $M$ is kept immediately above Box $P$ ($\text{Index}(M) = \text{Index}(P) + 1$):
  • In Case 1: $P = 5 \implies M = \text{Position } 6$.
  • In Case 2: $P = 7 \implies M = \text{Position } 8$.

Step 3: Apply Symmetrical Inversion (Clue 4)

• Clue 4 states: As many boxes are kept above Box $M$ as below Box $O$ ($\text{Index}(M) + \text{Index}(O) = 8 + 1 = 9$):
  • In Case 1: $M = 6$. Boxes above $M = 8 - 6 = 2$. Therefore, boxes below $O$ must be 2, placing $O = \text{Position } 3$ (Check: $6 + 3 = 9$).
  • In Case 2: $M = 8$. Boxes above $M = 8 - 8 = 0$. Therefore, boxes below $O$ must be 0, placing $O = \text{Position } 1$ (Check: $8 + 1 = 9$).

Step 4: Attach Clue 5 (Immediate Adjacency)

• Clue 5 states: Box $J$ is kept immediately above Box $O$ ($\text{Index}(J) = \text{Index}(O) + 1$):
  • In Case 1: $O = 3 \implies J = \text{Position } 4$.
  • In Case 2: $O = 1 \implies J = \text{Position } 2$.

Step 5: Test Rigid Block Clues 6 and 7 (Case Elimination)

• Clue 7 states: Box $Q$ is kept immediately above Box $K$ ($\text{Index}(Q) = \text{Index}(K) + 1$).
• Clue 6 states: Box $N$ is kept immediately above Box $Q$ ($\text{Index}(N) = \text{Index}(Q) + 1$).
• Combining Clues 6 and 7 yields a rigid 3-box vertical sequence: $[N / Q / K]$ where: $\text{Index}(Q) = \text{Index}(K) + 1 \quad \text{and} \quad \text{Index}(N) = \text{Index}(K) + 2$

Now test this rigid sequence against both cases:

• Testing Case 1:

  • In Case 1, $K = \text{Position } 1$.
  • Thus, $Q$ must be at Position 2, and $N$ must be at Position 3.
  • However, look at Case 1 coordinates from Step 3: Position 3 is already occupied by Box $O$!
  • A cell collision occurs: Box $N$ and Box $O$ cannot both occupy Position 3.
  • Case 1 is definitively contradictory and eliminated.

• Testing Case 2:

  • In Case 2, $K = \text{Position } 3$.
  • Thus, $Q$ must be at $\text{Position } 3 + 1 = \mathbf{4}$.
  • $N$ must be at $\text{Position } 3 + 2 = \mathbf{5}$.
  • Let us verify if Positions 4 and 5 are vacant in Case 2:
    • Position 8: Box $M$
    • Position 7: Box $P$
    • Position 5: Vacant $\implies$ Box $N$ placed
    • Position 4: Vacant $\implies$ Box $Q$ placed
    • Position 3: Box $K$
    • Position 2: Box $J$
    • Position 1: Box $O$
  • Positions 4 and 5 accommodate Box $Q$ and Box $N$ perfectly without conflict!

Step 6: Finalize Remaining Box (Clue 8)

• In Case 2, the only unoccupied coordinate is Position 6.
• Clue 8 states that Box $L$ occupies the remaining position.
• Therefore, Box $L = \text{Position } 6$.

Final Verified Box Stack Order