7.3 Deductive Logic, Linear/Circular Seating Arrangements & Truth-Value Puzzles

7.3 Deductive Logic, Linear/Circular Seating Arrangements & Truth-Value Puzzles

To keep the examination accessible across multiple languages, TR-YÖS designs its deductive reasoning items (Sayısal/Sözel Mantık, Sıralama ve Doğruluk Kurguları) around compact symbolic rules rather than long paragraphs of prose. These problems fall into three recurring categories: linear queue and floor-stacking orderings, circular seating and attribute-matching grids, and propositional truth-value puzzles.

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1. Linear Ordering, Queue Position & Floor-Stacking Puzzles (Sıralama ve Kuyruk Problemleri)

Queue Position Formulas (Kuyruk / Sıra Problemleri)

1. Single Person Ranked from Both Ends: If a person is $n$-th from the front (baştan $n$. sırada) and $m$-th from the back (sondan $m$. sırada), adding $n + m$ counts that person twice. The total number of people in the queue is:

$$N = n + m - 1$$

1. Two People with $k$ People Strictly Between Them: Suppose Person $A$ is $n$-th from the front, Person $B$ is $m$-th from the back, and there are $k$ people strictly between $A$ and $B$. Because the problem does not state whether $A$ stands ahead of $B$ or $B$ stands ahead of $A$, two configurations exist:
   • Maximum Queue Size (Non-Overlapping — $A$ is ahead of $B$): The first $n$ people (ending at $A$), the $k$ people in the middle, and the last $m$ people (starting at $B$) are completely disjoint:

$$N_{\max} = n + m + k$$

• Minimum Queue Size (Overlapping — $B$ is ahead of $A$): Here $B$, $A$, and the $k$ people between them are counted inside both the front $n$ prefix and the back $m$ suffix (an overlap of $k + 2$ people):

$$N_{\min} = n + m - k - 2$$

 *(Feasibility check: the overlapping case requires $k + 2 \le \min(n, m)$ so that neither $A$ nor $B$ extends past the ends of the line. Note also that $N_{\max} - N_{\min} = 2k + 2$.)*

Block-Glue Notation & Slot Templates for Linear/Floor Arrangements

When ordering $N$ items across numbered slots $1, 2, \dots, N$ (such as books on a shelf or tenants on floors $1$ through $6$):

• Adjacent Glue Block ([AB] or [BA]): If "$A$ and $B$ are on consecutive floors" without specifying order, fuse them into a single $2$-slot block [AB/BA] and mark its internal swap factor ($2! = 2$).
• Fixed Gap Template (A _ _ B): If "exactly $2$ people sit between $A$ and $B$," their slot indices satisfy $\lvert i_A - i_B \rvert = 3$. Write down the valid index pairs $(1,4), (2,5), (3,6)$ immediately to eliminate impossible branches.
• Strict Inequality Chains ($A < B < C$): Place the longest chain first; any entity with the highest number of constraints acts as your anchor.

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2. Circular Seating & Attribute-Matching Matrices (Dairesel Masa ve Eşleştirme Tabloları)

Circular Seating Rules (Yuvarlak Masa Sıralaması)

In a circle of $N$ equally spaced seats, rotating everyone clockwise by $1$ seat produces the exact same relative arrangement:

1. Fix the Anchor at 12 O'Clock (Seat $1$): Always place the most constrained person at the top seat first. This eliminates the $N$-fold rotational ambiguity and reduces the remaining $N - 1$ seats to a fixed linear benchmark.
2. Immediate Neighbors vs. Diametrically Opposite Seats:
   • "Immediately next to $X$" ($X$'in hemen yanında) means Seat $(i_X - 1)$ or $(i_X + 1) \pmod N$.
   • "Directly opposite $X$" ($X$'in tam karşısında) in an even $2k$-seat circle means the seat at index offset $+k \pmod{2k}$. For example, around a $6$-person table ($2k = 6 \implies k = 3$), opposite pairs are $(1, 4)$, $(2, 5)$, and $(3, 6)$. Around an $8$-person table ($k = 4$), opposite pairs are $(1, 5)$, $(2, 6)$, $(3, 7)$, and $(4, 8)$.

Elimination Matrix ($\checkmark / \times$) for Multi-Attribute Matching

When matching $n$ people to $n$ cities and $n$ professions under one-to-one bijections, draw an $n \times n$ grid. The moment you deduce a positive match ($\checkmark$) at cell $(r, c)$, immediately place $\times$ across all other cells in row $r$ and column $c$. If $n - 1$ cells in a row or column contain $\times$, the single remaining cell in that line must be $\checkmark$.

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3. Propositional Logic & Knights-and-Knaves Truth Puzzles (Doğruluk Tabloları ve Doğrucu-Yalancı Kurguları)

Symbolic Connectives & Truth-Value Rules

Each proposition $p$ takes a truth value in $\{1\text{ (True)}, 0\text{ (False)}\}$:

• Conjunction ($p \land q$, AND): $1$ only when $p = 1$ and $q = 1$ ($1 \land 1 \equiv 1$).
• Disjunction ($p \lor q$, OR): $0$ only when $p = 0$ and $q = 0$ ($0 \lor 0 \equiv 0$).
• Exclusive OR ($p \underline{\lor} q$, XOR): $1$ when $p \neq q$; $0$ when $p = q$.
• Implication ($p \Rightarrow q$, IF-THEN): Equivalent to $\neg p \lor q$. It is False ($0$) in only one case: when a true hypothesis leads to a false conclusion ($1 \Rightarrow 0 \equiv 0$). Its contrapositive $\neg q \Rightarrow \neg p$ has the exact same truth value: $(p \Rightarrow q) \equiv (\neg q \Rightarrow \neg p)$.
• Biconditional ($p \Leftrightarrow q$, IFF): $1$ when $p = q$ (both True or both False); $0$ when $p \neq q$. Note that $(p \Leftrightarrow q) \equiv \neg(p \underline{\lor} q)$.
• De Morgan's Negation Laws: $\neg(p \land q) \equiv \neg p \lor \neg q$ and $\neg(p \lor q) \equiv \neg p \land \neg q$.

The Contradiction Shortcut for "Exactly One Statement Is True" (İfadelerden Yalnız Biri Doğrudur)

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Worked Examples

Worked Example 7.3.1: Overlapping vs. Non-Overlapping Queue Positions

Problem: In a single-file ticket line, Kerem is $17$-th from the front and Pelin is $14$-th from the back. There are exactly $5$ people standing strictly between Kerem and Pelin. Find (a) the maximum and minimum possible number of people in the line, and (b) Pelin's position from the front in the minimum-queue case.

Solution:

1. Maximum Queue Size (Non-Overlapping — Kerem ahead of Pelin):

$$N_{\max} = n + m + k = 17 + 14 + 5 = 36\text{ people}$$

1. Minimum Queue Size (Overlapping — Pelin ahead of Kerem):

$$N_{\min} = n + m - k - 2 = 17 + 14 - 5 - 2 = 24\text{ people}$$

1. Pelin's Position from the Front when $N = 24$: Using the single-person formula $N = n_{\text{Pelin}} + m_{\text{Pelin}} - 1$:

$$24 = n_{\text{Pelin}} + 14 - 1 \implies n_{\text{Pelin}} = 11\text{-th from the front}$$

Verification: With Pelin at $11$-th and Kerem at $17$-th from the front, the people at positions $12, 13, 14, 15, 16$ are the $5$ people strictly between them.

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Worked Example 7.3.2: Contradiction Shortcut in a 4-Box Truth Puzzle

Problem: A gold coin is placed inside one of four locked boxes ($A, B, C, D$). Each box bears a printed statement:

• Box $A$: "The coin is in Box $A$."
• Box $B$: "The coin is in Box $C$."
• Box $C$: "The coin is not in Box $C$."
• Box $D$: "The coin is not in Box $D$."

If only one of the four statements is true and the other three are false, which box contains the gold coin, and which box's statement is true?

Solution:

1. Identify the contradictory pair: Box $B$ ("The coin is in Box $C$") and Box $C$ ("The coin is not in Box $C$") are exact logical negations ($S_B = \neg S_C$). Therefore, one of $\{S_B, S_C\}$ is True and the other is False.
2. Deduce the truth values of the remaining statements: Since only one statement in total is true, both Box $A$'s statement and Box $D$'s statement must be False ($0$).
3. Negate the false statement on Box $D$: Box $D$ claims "The coin is not in Box $D$." Since this statement is False, its negation is True: the coin is in Box $D$!
4. Verify all four labels when the coin is in Box $D$:
   • Box $A$ ("in $A$"): False ($0$).
   • Box $B$ ("in $C$"): False ($0$).
   • Box $C$ ("not in $C$"): True ($1$).
   • Box $D$ ("not in $D$"): False ($0$). Thus, the coin is in Box $D$ and Box $C$ has the sole true statement.

Review the Method

In 'exactly one statement is true' puzzles (İfadelerden yalnız biri doğrudur), locating two mutually contradictory statements ($S_1 = \neg S_2$) confines the single truth to $\{S_1, S_2\}$ and immediately forces all other statements to be false.