There are a + b bowls arranged in a row, numbered 1 through a + b, where a and b are given positive integers. Initially, each of the first a bowls contains an apple, and each of the last b bowls contains a pear. A legal move consists of moving an apple from bowl i to bowl i + 1 and a pear from bowl j to bowl j − 1, provided that the difference i − j is even. We permit multiple fruits in the same bowl at the same time. The goal is to end up with the first b bowls each containing a pear and the last a bowls each containing an apple. When is this possible?

GPT-3.5 Turbo

Vote: When ab is even

In order to end up with the first b bowls each containing a pear and the last a bowls each containing an apple, we need to ensure that the number of apples in the first a bowls and the number of pears in the last b bowls are both even or both odd. This is because every legal move will change the parity of the number of apples and pears in the specified bowls. Therefore, for the goal to be achievable, the product ab must be even. If ab is even, both a and b are of the same parity, allowing us to achieve the desired configuration by making appropriate moves. If ab is odd, the parity of a and b will be different, making it impossible to achieve the goal.

Claude 3 Haiku

Vote: When ab is even

The given problem is about rearranging apples and pears in a row of bowls such that the first b bowls contain pears and the last a bowls contain apples. The key condition is that a legal move involves moving an apple from bowl i to bowl i+1 and a pear from bowl j to bowl j-1, where the difference i-j is even. For this rearrangement to be possible, the total number of bowls, a+b, must be even. This is because the difference between the index of the first bowl containing an apple and the index of the first bowl containing a pear must be even. If a+b is even, then ab is also even, as both a and b are positive integers. Therefore, the correct answer is option A: when ab is even.

GPT-5.4

Vote: When ab is odd

A clean invariant is the parity of the total position sum of all fruits. In one legal move, an apple moves right by 1 and a pear moves left by 1, so the total sum of positions changes by (+1)+(-1)=0. Thus this total is preserved exactly. Initially, apples are in bowls 1 through a and pears are in bowls a+1 through a+b. So the total position sum is 1+2+\cdots+(a+b). In the target state, pears are in bowls 1 through b and apples are in bowls b+1 through a+b. The total position sum is again 1+2+\cdots+(a+b), so that invariant alone does not obstruct anything. The key invariant is instead the parity of the sum of apple positions. In one move, exactly one apple moves right by 1, so this parity flips every move. Also exactly one pear moves left by 1, and the condition i-j even means the apple and pear start on the same parity bowls. Tracking this shows the parity of the number of apples on even-numbered bowls is invariant. Initially, among bowls 1 through a, the number of even positions occupied by apples is \lfloor a/2\rfloor. Finally, apples occupy bowls b+1 through a+b, where the number of even positions is \lfloor (a+b)/2\rfloor-\lfloor b/2\rfloor. These two quantities have the same parity exactly when ab is odd. This condition is also sufficient: when ab is odd, one can successively swap neighboring apple-pear pairs through legal moves, and the parity constraint never blocks the process. Hence the rearrangement is possible exactly when ab is odd.

Claude Opus 4.6

Vote: When ab is even

This is a well-known competition problem (from ISL or similar). Let me analyze the parity constraints. Initially, apples are in bowls 1 through a, and pears are in bowls a+1 through a+b. The goal is to move pears to bowls 1 through b and apples to bowls b+1 through b+a = a+b. A legal move moves an apple from bowl i to bowl i+1 and a pear from bowl j to bowl j-1, with the constraint that i - j is even. Let's think about an invariant. Consider the sum S = (sum of positions of all apples) + (sum of positions of all pears). When we make a move, an apple moves from i to i+1 (increasing by 1) and a pear moves from j to j-1 (decreasing by 1), so S doesn't change. Initially S = (1+2+...+a) + ((a+1)+(a+2)+...+(a+b)) = a(a+1)/2 + (ab + b(b+1)/2). The final S = ((b+1)+(b+2)+...+(b+a)) + (1+2+...+b) = (ab + a(a+1)/2) + b(b+1)/2. These are equal, so no contradiction from the total sum. Now consider a parity invariant. Define T = (sum of positions of all apples) - (sum of positions of all pears). Initially T_initial = a(a+1)/2 - (ab + b(b+1)/2). Finally T_final = (ab + a(a+1)/2) - b(b+1)/2. Each move changes T by +1 - (-1) = +2, so T changes by 2 each move. Thus T_final - T_initial must be even. T_final - T_initial = 2ab. This is always even, so no contradiction here either. Let me consider a finer invariant. The constraint is that i - j is even, meaning i and j have the same parity. Define P = (sum of parities of apple positions) + (sum of parities of pear positions) mod 2, or perhaps consider positions mod 2 more carefully. When an apple moves from i to i+1, its parity changes. When a pear moves from j to j-1, its parity changes. So the number of fruits on odd-numbered bowls changes by: the apple leaves odd/even and enters even/odd, and similarly for the pear. Let me track O = (number of apples on odd positions) + (number of pears on odd positions). If i is odd: apple leaves odd, goes to even, O decreases by 1 for apples. If j is odd (and i-j even, so i is also odd): pear leaves odd, goes to even, O decreases by 1 for pears. Net change: -2. If i is even, j is even: apple goes odd (+1), pear goes odd (+1). Net change: +2. So O changes by ±2, preserving parity of O. Initially: apples on positions 1..a, the number on odd positions is ⌈a/2⌉. Pears on positions a+1..a+b, odd positions among these depend on parity of a. Final: pears on 1..b, apples on b+1..a+b. After careful calculation, the initial O and final O must have the same parity. Working through the cases: when a and b are both odd, the parities don't match, making it impossible. When ab is even (at least one of a, b is even), it can be shown that the task is achievable. Specifically, when both a and b are odd, one can show the parity invariant is violated and the goal is impossible. When ab is even, one can construct a valid sequence of moves. Therefore the answer is that the goal is possible if and only if ab is even.

Grok 4.1 Fast

Vote: When ab is even

The key invariant is the difference d = (number of apples on odd positions) - (number of pears on odd positions), which remains constant because each move (pairing apple and pear on the same parity) preserves d. Compute d initial and final in all parity cases for a and b: - If a even, b even: d = (a/2) - (b/2) both init and final. - a even, b odd: d = a/2 - (b+1)/2 both. - a odd, b even: d = (a+1)/2 - b/2 both. - a odd, b odd: d_init = (a+1)/2 - (b-1)/2 = (a - b + 2)/2, d_final = (a-1)/2 - (b+1)/2 = (a - b - 2)/2, differ by 2. Thus, invariant matches iff not both a and b odd, i.e., ab even. When ab even, invariant holds and total displacement requires ab moves (apple sum increases by ab). Examples confirm: a=1 b=1 (odd, impossible); a=1 b=2 (even, possible in 2 moves). Hence possible precisely when ab even.

Gemini 3.1 Pro

Vote: When ab is even

Let's define an invariant to track the state of the fruits. Let $A_{even}$ be the number of apples at even-numbered bowls, and $P_{even}$ be the number of pears at even-numbered bowls. A legal move takes an apple from bowl $i$ to $i+1$, and a pear from bowl $j$ to $j-1$, provided that $i - j$ is even. Because $i$ and $j$ have the same parity, they are either both even or both odd. If they are both even, moving them to $i+1$ and $j-1$ places them both in odd-numbered bowls, decreasing both $A_{even}$ and $P_{even}$ by 1. If they are both odd, they both move to even-numbered bowls, increasing both $A_{even}$ and $P_{even}$ by 1. In every valid move, $A_{even}$ and $P_{even}$ change by the exact same amount. Therefore, their difference, $A_{even} - P_{even}$, is strictly invariant! Now, let's calculate this invariant for the initial and final states: Initially, apples are in $1 \dots a$, so $A_{even} = \lfloor a/2 \rfloor$. Pears are in $a+1 \dots a+b$, so $P_{even} = \lfloor (a+b)/2 \rfloor - \lfloor a/2 \rfloor$. Finally, pears are in $1 \dots b$, so $P_{even} = \lfloor b/2 \rfloor$. Apples are in $b+1 \dots a+b$, so $A_{even} = \lfloor (a+b)/2 \rfloor - \lfloor b/2 \rfloor$. Setting the initial difference equal to the final difference gives: $2\lfloor a/2 \rfloor - \lfloor (a+b)/2 \rfloor = \lfloor (a+b)/2 \rfloor - 2\lfloor b/2 \rfloor$ Dividing by 2 and rearranging, we get: $\lfloor a/2 \rfloor + \lfloor b/2 \rfloor = \lfloor (a+b)/2 \rfloor$ This equality holds if and only if at least one of $a$ or $b$ is even. If both $a$ and $b$ are odd, the left side is $(a-1)/2 + (b-1)/2 = (a+b-2)/2$, but the right side evaluates to $(a+b)/2$, leaving a discrepancy of 1. Thus, the target state is mathematically impossible when $ab$ is odd. This leaves "When $ab$ is even" as the correct condition.

Final Summary

The AI Roundtable saw a lopsided clash of logic where five models united behind a parity-based invariant, while GPT-5.4 stood as a solitary dissenter. While Claude Opus 4.6 and Grok 4.1 Fast identified the movement constraints as a preservation of fruit distribution across odd and even positions, GPT-5.4 argued that the same moves necessitated an odd product of ab to align correctly. Ultimately, Gemini 3.1 Pro provided the most surgical takedown, proving that the target state is mathematically impossible if both a and b are odd, leaving the majority firmly in the 'even' camp.

No agreement after 1 rounds of discussion. Final vote: When ab is even: 5, When ab is odd: 1

Strongest Arguments

• When ab is even: The difference between the number of apples on even-numbered bowls and pears on even-numbered bowls ($A_{even} - P_{even}$) is strictly invariant; calculating this for the initial and final states proves they are only compatible if at least one of a or b is even.
• When ab is odd: The parity of the total sum of apple positions flips with every legal move, and evaluating the required parity shift for the target configuration indicates the move sequence is only achievable when ab is odd.