Living paper, living scissors and living rock want to get to the other side of the river. None of them can swim or paddle. Luckily, there is a boat and the ferryman (who can paddle) can repeatedly transport one of them (either rock, or paper, or scissors, but not more than one at a time) to the other side. Can all of them get to the other side without them beating eachother when left alone? How?

DeepSeek V3

Vote: Yes

The key is to ensure that no two items that can 'beat' each other are left alone together. The ferryman can transport them in a specific order to prevent conflicts. For example, first transport scissors, then return alone. Next, transport rock, but bring scissors back. Then transport paper, and return alone. Finally, transport scissors again. This sequence ensures that rock and paper are never left alone (paper beats rock), scissors and paper are never left alone (scissors beat paper), and rock and scissors are never left alone (rock beats scissors).

Llama 3.1 8B

Vote: No

It's impossible to transport all three without any of them facing another that beats them. If rock goes first, paper has to go next because scissors destroy rock but paper will destroy scissors, so that won't work. Then if scissors go first and rock go second that will also fail because rock will destroy scissors once on the other side, so they must go first. But then there remains rock and scissors. Those cannot be transported in a way that they wouldn't destroy each other because one always beats one and then you cannot cross the river.

Gemini 3.1 Flash Lite

Vote: Yes

The ferryman can easily transport all three across the river without any conflicts occurring. The crucial detail is that the problem specifies the ferryman transports one at a time. Since the ferryman is present during the transportation and only leaves one behind on the other side at a time, he can simply take one, drop it off, and return for the next. At no point are the 'living' items left alone together on either bank without the ferryman present to mediate or until they are all safely relocated. Even if they were left alone, the restriction only applies to the destination bank. By the time the third one is moved, the first two are already at the destination, but the ferryman has been with them the entire time. Because he makes three separate trips, he is essentially isolating each one from the others until the final transport is complete.

Llama 3.3 70B

Vote: Yes

The key to solving this problem is to ensure that the living entities are never left together in a combination where one can beat another. According to the traditional rock-paper-scissors rules, rock beats scissors, scissors beats paper, and paper beats rock. To avoid any conflict, the ferryman must transport them in a specific order. One possible sequence could be: transport paper first, then rock, and finally scissors. However, after transporting the first entity (paper), the ferryman must return alone to pick up the next one. After transporting the second entity (rock), the ferryman must again return alone to the original side to pick up the last entity (scissors). This ensures that at no point are rock and scissors, or paper and rock, or scissors and paper left together without the ferryman's intervention, thus preventing any beating. Therefore, with careful planning and sequential transport, all three can indeed get to the other side without any of them beating each other when left alone.

Nova Pro

Vote: Yes

All three can safely cross the river by following a specific sequence that prevents any two opponents from being left alone together. First, take scissors across, then return alone. Next, take rock across, but bring scissors back. Then, take paper across, return alone, and finally take scissors across again. This way, paper and rock are never left alone together, nor are scissors left with either paper or rock without the other.

Gemini 2.0 Flash Lite

Vote: Yes

Yes, they can all get to the other side safely. The key is to transport them in a specific order to prevent any of them from being 'beaten' while unattended. First, the ferryman takes the scissors to the other side, leaving rock and paper behind. This is safe, as rock can't beat paper. Then, the ferryman returns alone. Next, the ferryman takes the rock to the other side. This is safe as the scissors are already there and cannot beat the rock. Then, the ferryman returns with the scissors to bring paper. The ferryman brings the paper to the other side. Since there is no interaction between the rock and paper on the other side, or the scissors on the opposite side, all three are now safely on the other side.

GPT-5.4

Vote: Yes

Yes, they can all cross, provided you keep the unsafe pairs from being left alone together. Safe constraint: when the ferryman is not present on a bank, you must not leave: - Paper with Scissors alone (paper would be cut) - Scissors with Rock alone (scissors would be crushed) - (If Paper and Rock are compatible, they may be left together.) One valid sequence (Left → Right / Right → Left): 1) Take Paper to the Right. (L: Scissors+Rock, R: Paper) 2) Bring Paper back to the Left. (L: Paper+Scissors+Rock, R: —) 3) Take Rock to the Right. (L: Paper+Scissors, R: Rock) 4) Bring Rock back to the Left. (L: Paper+Scissors+Rock, R: —) 5) Take Scissors to the Right. (L: Paper+Rock, R: Scissors) 6) Bring Paper to the Right. (L: Rock, R: Paper+Scissors) 7) Bring Paper back to the Left. (L: Paper+Rock, R: Scissors) 8) Take Rock to the Right. (L: Paper, R: Rock+Scissors) 9) Bring Scissors back to the Left. (L: Paper+Scissors, R: Rock) 10) Take Paper to the Right. (L: Scissors, R: Rock+Paper) 11) Bring Scissors to the Right. (L: —, R: Paper+Scissors+Rock) At each step where the ferryman leaves a bank, the only pair that could be left alone is the safe one (Paper with Rock), or a single item—so none of the beating occurs while left alone.

GPT-5.4

Vote: Yes

Yes. The ferryman can prevent any unsafe pair from being left alone on either bank by never abandoning the losing matchup together. One valid plan is: 1) Take rock across, ferryman returns alone. 2) Take scissors across, ferryman brings rock back. 3) Take paper across, ferryman returns alone. 4) Take rock across. At no point are paper with rock or rock with scissors or scissors with paper left alone in a way that allows beating, because the ferryman controls the crossings and can always keep the “dangerous” combination from being unattended.

DeepSeek V3.1

Vote: No

The problem is analogous to the classic river-crossing puzzle with adversarial entities. In this case, rock beats scissors, scissors beat paper, and paper beats rock. If any two are left alone together without the ferryman, one will defeat the other. The ferryman can only transport one at a time, meaning that after each trip, two will be left on one side. No matter the order, at some point, two adversaries will be left unsupervised, leading to a conflict. For example, if the ferryman takes rock first, scissors and paper are left alone—scissors beat paper. Similarly, any other sequence results in an unsupervised pair that can harm each other. Therefore, it is impossible to get all three across without a fight.

GPT-4.1 Nano

Vote: Yes

The key is to prevent the living paper, scissors, and rock from being left together in a way that allows them to 'beat' each other. Since they are living, they can be safely separated by transporting one at a time. The ferryman can ferry each one individually to the other side, ensuring that no two are left alone together in a conflicting combination. By carefully sequencing the crossings—first moving one over and then bringing others back when needed—all three can be safely transported without conflict. This is analogous to classic river crossing puzzles where careful sequencing avoids dangerous pairings.

GPT-5.4

Vote: No

It is impossible if any winning pair cannot be left together without supervision, and the ferryman only moves one item at a time. From the starting bank, taking paper first leaves rock with scissors, and rock beats scissors. Taking scissors first leaves paper with rock, and paper beats rock. Taking rock first leaves paper with scissors, and scissors beat paper. So no first move is legal. If you cannot make even the first crossing, then there is no solution. Therefore the correct answer is No.

GPT-4.1

Vote: No

Whenever the ferryman leaves two of the items alone, there is always a pairing where one can beat another: scissors cuts paper, paper covers rock, or rock crushes scissors. Since the boat can only carry one at a time and the ferryman must leave two together at some point, it is impossible to transfer all three across without an incident where one 'beats' (destroys) the other. Unlike similar ferry puzzles with goats/cabbage/wolf, here there is no safe grouping.

GPT-OSS 120B

Vote: No

The ferryman can only transport one item at a time and his presence is required to prevent a beating. Initially all three are on the left bank with the ferryman, so they are safe. As soon as the ferryman takes any one across, the two left behind on the original bank are unsupervised. Any pair among rock, paper, and scissors has a winning relationship (rock beats scissors, scissors beats paper, paper beats rock), so the two left behind will inevitably fight. Since the ferryman cannot be on both banks simultaneously, there is no sequence that avoids leaving a dangerous pair alone. Therefore, it is impossible for all three to reach the other side safely.

GPT-OSS 20B

Vote: No

Only two of the three objects can be left on one side while the ferryman carries the third across. Every pair of the three (rock–scissors, scissors–paper, paper–rock) results in a direct conflict: rock crushes scissors, scissors cut paper, and paper covers rock. Therefore, whenever the ferryman leaves two objects behind, they would inevitably beat each other. Since no safe initial move exists, it is impossible to transport all three across the river without any two conflicting when left alone.

GPT-4o

Vote: Yes

Yes, it is possible for all of them to get to the other side without conflict. The key is to ensure that no two conflicting elements (rock-paper, paper-scissors, scissors-rock) are left alone together. The ferryman can achieve this with the following steps: 1. First, take the scissors to the other side and leave them there. 2. Return alone to the original bank. 3. Take the paper across, but while doing so, bring back the scissors. 4. Leave the scissors on the original bank, after bringing paper to the other side. 5. Take the rock across to the other side where the paper is waiting. 6. Return alone to the original bank. 7. Finally, take scissors to the other side. In this arrangement, any potential conflict is avoided, and all reach the other side safely.

Arcee Trinity Large

Vote: Yes

Yes, all three can get to the other side without them beating each other. The key is to use the ferryman as an intermediary to prevent any two from being left alone together on either side of the river. One possible solution: 1. Ferryman takes paper across first 2. Ferryman returns alone 3. Ferryman takes scissors across 4. Ferryman brings paper back 5. Ferryman takes rock across 6. Ferryman returns alone 7. Ferryman takes paper across again This way, paper and scissors are never left alone together (since paper would 'beat' scissors), scissors and rock are never left alone together (since scissors would 'beat' rock), and paper and rock are never left alone together (since rock would 'beat' paper). The ferryman ensures that at no point are any two left alone together on either side of the river.

GPT-4.1 Mini

Vote: Yes

The classic problem with paper, scissors, and rock is that if left alone together without supervision, the dominant can 'beat' the other. However, the ferryman can act as a supervisor when any two are left on one bank. The ferryman can ferry one item at a time and ensure that the remaining two do not get left alone. For example, the ferryman can take scissors across first, leaving paper and rock, which don’t beat each other. Then he brings one back and switches accordingly so that paper and scissors or scissors and rock never remain unsupervised together.

Qwen 3 235B

Vote: Yes

Yes, all three can get to the other side safely by following a careful sequence of crossings that ensures no two conflicting items are left unattended on either bank. First, the ferryman takes the scissors across and leaves them on the far bank, then returns alone. This is safe because paper and rock can be left together — rock does not beat paper unless scissors are present to threaten paper. Next, the ferryman takes paper across and brings scissors back. Now, scissors and rock are on the near bank, and paper is on the far bank. Scissors and rock can coexist because rock beats scissors only in active play; the problem states they beat each other when left alone, so we assume asymmetrical danger: scissors cut paper, paper covers rock, rock crushes scissors — each can be defeated, but only one threatens the other per pair. Now, the ferryman takes rock across and leaves it with paper. Rock and paper are safe together — no conflict arises from paper being covered; the danger is only when one actively beats another when unattended. So paper is not harmed by rock. Finally, the ferryman returns alone and picks up scissors, bringing them to the far bank. Now all three are on the far side safely, and never at any point were two enemies left without supervision. The key is that only one creature threatens another: scissors beat paper, paper beats rock, rock beats scissors — but the 'beating' only happens when the victim is left alone with the predator. So we avoid leaving paper with scissors, rock with paper (if paper is the predator?), but the logic must be clarified. Clarifying the rule: assume that 'beating' only happens in the active direction — scissors destroy paper if left alone, paper does nothing to rock aggressively, but paper covering rock might neutralize it — however, standard logic is that paper covers rock, so rock cannot crush anything while covered. But that might imply rock is incapacitated, not destroyed. But the problem says 'they beat each other when left alone'. So it might be mutual danger in a pair: e.g., rock and scissors cannot be left because they fight, similarly all pairs? But that would make solution impossible. Re-interpret: in Rock-Paper-Scissors, each pair has a dominant direction. Likely, the constraint is that we cannot leave scissors with paper (scissors cut paper), cannot leave paper with rock (paper covers rock — but is that harmful?), cannot leave rock with scissors (rock breaks scissors). But 'covering' or 'breaking' might not be destructive. But the problem says 'beating each other', implying mutual conflict when left alone. So perhaps any pair left together will fight. That would make any two incompatible. But that can't be — otherwise even two on the far side at the end would be invalid. So likely only certain pairs conflict. Standard interpretation in such puzzles: scissors cannot be left with paper (scissors cut paper), paper cannot be left with rock (paper covers rock, but more often, rock is inert), actually — common version: goat-cabbage-wolf, where goat eats cabbage, wolf eats goat. Analogous: scissors 'eat' paper, paper 'eats' rock (covers it), rock 'eats' scissors (crushes). So it's a cycle. Each item is safe only if not left alone with its predator. So: we cannot leave paper with scissors (scissors destroy paper), cannot leave rock with paper (paper neutralizes rock? Or is rock not threatened?), but more logically: paper defeats rock, so perhaps rock is not harmed — it's a game outcome. Better: the danger is only when the victim is left with the enemy that beats it. So: - Paper cannot be left with scissors (scissors cut paper) - Rock cannot be left with paper? No — paper does not harm rock, it just covers it. No destruction. - Scissors cannot be left with rock — rock crushes scissors, so scissors are destroyed. So likely, the dangerous pairs are: - Scissors and paper: scissors cut paper → paper dies - Rock and scissors: rock breaks scissors → scissors die - Paper and rock: paper wraps rock → rock is immobilized, but not destroyed? Or safe? But if all pairs are dangerous, no solution. Standard solution to this puzzle (known as the 'Rock Paper Scissors River Crossing') assumes that: - Scissors cannot be left alone with paper (scissors cut paper) - Paper cannot be left alone with rock (paper covers rock — but is this bad? Maybe not; perhaps it's the reverse: rock damages paper? Not in game logic. Alternative common version: assume only one dangerous interaction: scissors cut paper, and that's it. Rock and paper can be together, rock and scissors can be together. But that seems inconsistent. But if rock and scissors are left, rock breaks scissors → dangerous. So all pairs have one destructive interaction: - Paper + scissors → paper destroyed - Scissors + rock → scissors destroyed - Rock + paper → rock immobilized, but perhaps not destroyed → considered safe? But the problem says 'beating each other', so perhaps mutual aggression, but only one side wins. To make the puzzle solvable, assume that only the loser is at risk. So: - Paper cannot be left with scissors - Scissors cannot be left with rock - Rock cannot be left with paper? No — paper covering rock is not harmful to rock. Actually, the problem likely intends that when two are left together, the one that 'wins' in the game will damage the other. So: - Scissors + paper → scissors beat paper → paper is destroyed → not allowed - Rock + scissors → rock beats scissors → scissors are destroyed → not allowed - Paper + rock → paper beats rock → rock is covered, but is rock destroyed? Unclear. However, in most versions of this puzzle, paper covering rock is not considered destructive — rock is not harmed. So paper and rock can be left together safely. That makes the constraints: - Scissors and paper cannot be left alone - Rock and scissors cannot be left alone - Paper and rock can be left alone This makes the puzzle solvable. Solution: 1. Ferryman takes scissors to far side, leaves them, returns alone. (Safe: paper and rock left — allowed) 2. Ferryman takes paper to far side, but cannot leave paper with scissors → so he brings scissors back to near side. (Now, near side has rock and scissors — not allowed? Wait: rock and scissors cannot be left — scissors are destroyed. But in step 2: ferryman brings paper to far side, pairs with scissors — dangerous. So he must bring something back. Alternative: 1. Take paper across, leave it, return. (Near: rock and scissors — dangerous! Rock breaks scissors → not allowed) So cannot leave rock and scissors. 1. Take scissors across, leave. Far: scissors. Near: rock, paper → safe? Rock and paper together — if safe, yes. 2. Return. Take paper across. Now far: scissors, paper → dangerous. So instead: take scissors across, leave. Return. Take paper across. But now far has scissors and paper → conflict. So ferryman must bring something back. Better: 1. Ferryman takes paper across, leaves it, returns. Now near: rock, scissors — if rock and scissors fight, not allowed. So cannot leave rock and scissors. 1. Ferryman takes scissors across, leaves, returns. Near: paper, rock. If paper and rock are safe, ok. 2. Ferryman takes paper across. Now far: scissors, paper → conflict. So he must bring scissors back. After step 1: far: scissors, near: paper, rock (safe). Step 2: ferryman returns, takes paper to far side. Now far: scissors, paper — conflict. So instead, after dropping scissors, he returns, then takes rock? 1. Take scissors to far, leave. Far: scissors. Near: paper, rock — assume safe. 2. Return. Take rock to far. Now far: scissors, rock — conflict! Scissors are broken. So cannot do that. Only option: after step 1 (take scissors across), return, take paper across, but then far has scissors and paper — conflict, so ferryman brings scissors back when he returns. Sequence: 1. Ferryman takes scissors to far side, leaves. (Near: paper, rock — safe if paper and rock can coexist) 2. Ferryman returns alone. 3. Ferryman takes paper to far side. But now far has scissors and paper — dangerous. So instead, when he takes paper to far, he leaves paper and brings scissors back. So: step 3: take paper to far, now far has scissors and paper. Ferryman immediately takes scissors back to near side. Now near: rock and scissors (from bringing back scissors, and rock was there). But rock and scissors cannot be left — conflict. But during the crossing, ferryman is present, so maybe safe? After ferryman brings scissors back, he is on near side with rock and scissors. Then he can immediately take rock across. So: 1. Take scissors to far, leave. Ferryman returns. 2. Take paper to far. Now far has scissors and paper — danger unless ferryman supervises. So he doesn’t leave yet. He brings scissors back to near side. 3. Now near: ferryman, scissors, rock. Far: paper. 4. Ferryman takes rock to far side, leaves rock with paper. If paper and rock together are safe, ok. 5. Ferryman returns alone to near side. 6. Takes scissors to far side. Now all on far side: paper, rock, scissors — but now they are all together, which might be worse. But the constraint is only when 'left alone' — but now ferryman is there, or if he stays, he supervises. But at the end, ferryman is present, so maybe safe. But the goal is to get them to the other side — doesn't specify they must be left alone. So as long as they are transported without fights during the process, and at the end, ferryman is there, it's ok. But likely, the final state should have all three on the far bank, and if ferryman leaves, they might fight. But the problem doesn't say he leaves — he can stay. But even so, when all three are together, rock-paper-scissors cycle — do they fight? The constraint is only when 'left alone' in pairs — but with all three, maybe balance. But the problem is about leaving them alone during the process. Let's track the banks: Initial: near: R, P, S; far: empty 1. Ferryman takes S to far. Far: S; near: P, R If P and R can be left → safe. 2. Ferryman returns. Near: P, R; far: S 3. Ferryman takes P to far. Now far: S, P — conflict! Unless ferryman prevents it. But during the trip, ferryman is present, so when he arrives, he can immediately take S back. So: ferryman drops off P on far, picks up S, brings S back to near. Now: near: R, S (and ferryman); far: P But when ferryman departs, near has R and S → conflict if R breaks S. But ferryman is there, so while he is present, safe. He leaves immediately to take S? Sequence: After 3: ferryman brings S back to near. So he arrives at near with S. Now near: R, S (and ferryman). Far: P Then: 4. Ferryman takes R to far side. During transit: ferryman and R; near: S alone; far: P → all safe. 5. Ferryman arrives at far, leaves R with P. If R and P together are safe → ok. 6. Ferryman returns alone to near. Now near: S; far: R, P 7. Ferryman takes S to far. Now all on far: R, P, S, ferryman. Now, were there any dangerous pairs left alone? - After step 1: near: R, P → safe (assumed) - After step 3: when ferryman brings S back to near, he departs far with S, so far: P alone; near: R, S. But during the trip, when he is en route with S, near has R alone? No — he drops S off at near? Clarify: when ferryman returns from far to near in step 3, he is carrying S, arrives at near, and disembarks with S. So at arrival, near has R, S, and ferryman — all together. Then he immediately takes R (in step 4), so he boards with R, leaves S at near alone, and takes R to far. So during step 4: while ferryman is crossing with R, near has only S — safe; far has only P — safe. No conflict. After step 5: far has P and R — safe. After step 6: ferryman returns alone, so far: P, R; near: S — safe. Step 7: ferryman takes S to far. Final: all on far. Critical assumption: rock and paper can be left together safely; scissors and paper cannot; rock and scissors cannot. But we never left rock and scissors together (except briefly when ferryman was present to supervise). And never left scissors and paper together unsupervised. The only possible issue is in step 3: when ferryman goes from near to far with paper, arriving at far, he has scissors and paper together for a moment — but he immediately takes scissors back, so they are not 'left alone'. Supervision prevents conflict. Similarly, when he brings scissors back to near, he is present, so even though rock and scissors are on near, he prevents conflict. Thus, all constraints are satisfied. Therefore, yes, it is possible.

O4 Mini

Vote: No

No sequence works because any time the ferryman takes one creature across, he leaves a predator–prey pair alone on the starting bank. Rock alone with scissors lets rock crush scissors; scissors alone with paper lets scissors cut paper; paper alone with rock lets paper wrap rock. Since all three pairings are deadly and he can carry only one at a time, there’s no way to avoid leaving a fighting pair behind. Thus they cannot all cross safely.

Qwen 3 Coder 480B

Vote: Yes

This is a classic river-crossing puzzle that can be solved by carefully planning the sequence of crossings. The key constraints are: rock beats scissors, scissors beat paper, and paper beats rock. These combinations cannot be left alone together. Solution: 1. Ferryman takes scissors across first (leaving rock and paper, which don't fight) 2. Ferryman returns alone 3. Ferryman takes rock across 4. Ferryman brings scissors back 5. Ferryman takes paper across (leaving scissors alone) 6. Ferryman returns alone 7. Ferryman takes scissors across At no point are conflicting elements left alone together, so all three can safely cross.

Claude Sonnet 4

Vote: No

This is a variation of the classic river-crossing puzzle, but with a crucial constraint: the three entities (rock, paper, scissors) will beat each other when left alone, following rock-paper-scissors rules where rock beats scissors, scissors beats paper, and paper beats rock. Let's analyze the constraints: The ferryman can only transport one entity at a time, and any two entities left alone together will result in one beating the other. This means we can never leave any two of them alone on the same side. Initially, all three are on one side. If the ferryman takes one across, two remain together and one will beat the other. If the ferryman brings someone back, we're back to having multiple entities on one side again. Unlike classic puzzles where some combinations are safe, here NO combination of two entities can be left alone together safely. Rock beats scissors, scissors beats paper, and paper beats rock - every pairing results in conflict. The fundamental issue is that we need to separate three mutually antagonistic entities using a boat that can only carry one at a time, but we cannot leave any two together at any point in the process. This creates an impossible constraint that cannot be satisfied.

GPT-5.2

Vote: Yes

Yes. This is analogous to the classic river-crossing constraints, but here the only unsafe pairings are exactly the “winning” interactions: scissors beats paper, paper beats rock, rock beats scissors. So we must never leave any winning pair alone on either bank without the ferryman. A working sequence (→ means take across to far bank, ← means bring back): 1) Take Rock →. (Near: Paper+Scissors safe; Far: Rock alone) 2) Bring Rock ←. (reset; both banks safe) 3) Take Paper →. (Near: Rock+Scissors safe; Far: Paper alone) 4) Bring Paper ←. 5) Take Scissors →. (Near: Rock+Paper unsafe? No: paper beats rock, but paper is on near only if step 4 returned it; so instead we use the standard “shuttle” pattern below.) Correct safe minimal pattern: 1) Take Rock →; return alone ←. 2) Take Paper →; bring Rock ←. 3) Take Scissors →; bring Paper ←. 4) Take Rock →. After step 4 all three are on the far bank. Checking safety at each stage: - After 1: Near (Paper+Scissors) safe; Far (Rock) safe. - After 2: Far (Paper) safe; Near (Rock+Scissors) safe. - After 3: Far (Paper+Scissors) unsafe unless ferryman is there; but ferryman arrives with Scissors, so they are not “left alone.” Then ferryman brings Paper back, leaving Far (Scissors) safe and Near (Paper+Rock) safe only while ferryman is present returning; once back, Near has (Paper+Rock) which is unsafe, but again ferryman is present. Next he immediately takes Rock over, leaving Near (Paper) safe. Thus no beating occurs when any pair is actually left unattended, and all end up across.

Arcee Trinity Mini

Vote: No

The rock, paper, and scissors cannot be left alone together without conflict. For example, if rock and scissors are left together, rock beats scissors. If paper and rock are left together, paper beats rock. If paper and scissors are left together, scissors beat paper. Any sequence of crossings will eventually leave two of them unsupervised, leading to a fight. The ferryman can only transport one at a time, so it's impossible to avoid this conflict.

Gemini 3 Flash

Vote: No

The core of this puzzle lies in the relationships between the three entities: Rock, Paper, and Scissors. In this circular system, every possible pair contains a winner and a loser (Paper beats Rock, Rock beats Scissors, and Scissors beats Paper). To transport all three across the river, the ferryman (who can only carry one at a time) must leave two of them behind on the starting bank during the first trip. Because there are no 'neutral' pairings in Rock-Paper-Scissors, any two left together will result in one beating the other. Specifically, if the ferryman takes Rock, Paper and Scissors are left alone (Scissors beats Paper). If he takes Paper, Rock and Scissors are left alone (Rock beats Scissors). If he takes Scissors, Rock and Paper are left alone (Paper beats Rock). Since the first step itself is impossible without a conflict, the group cannot be moved successfully.

GLM 4.5

Vote: Yes

This is a variation of classic river crossing puzzles with rock-paper-scissors mechanics. The key is ensuring no two items that can 'defeat' each other are left alone together. The solution works as follows: 1. Ferryman takes scissors to the other side. (Left: paper, rock; Right: scissors) 2. Ferryman returns alone. (Left: paper, rock, ferryman; Right: scissors) 3. Ferryman takes rock to the other side. (Left: paper; Right: scissors, rock) 4. Ferryman returns with scissors. (Left: paper, scissors, ferryman; Right: rock) 5. Ferryman takes paper to the other side. (Left: scissors; Right: rock, paper) 6. Ferryman returns alone. (Left: scissors, ferryman; Right: rock, paper) 7. Ferryman takes scissors to the other side. (Left: empty; Right: rock, paper, scissors) At no point are incompatible items left alone together. When rock and paper are together, paper 'covers' rock but doesn't eliminate it. When rock and scissors are together, rock crushes scissors - but the ferryman is present during these transitions.

Mistral Large

Vote: Yes

This problem is a variation of the classic river crossing puzzles, such as the wolf, goat, and cabbage problem. The key is to ensure that no two items left alone on either side of the river can 'beat' each other according to the rules of Rock-Paper-Scissors. Here's how it can be done: 1. **Initial State**: All three (Rock, Paper, Scissors) are on the starting side of the river, and the ferryman is ready to transport them one at a time. 2. **First Trip**: The ferryman takes **Scissors** to the other side. Now, the starting side has Rock and Paper, which is unsafe because Paper beats Rock. To avoid this, the ferryman must bring one of them back. 3. **Second Trip**: The ferryman takes **Rock** to the other side. Now, the starting side has Paper, and the other side has Scissors and Rock. This is unsafe because Scissors beats Paper (if the ferryman leaves) or Rock beats Scissors (if left alone). So, the ferryman must bring **Scissors** back to the starting side. 4. **Third Trip**: The ferryman takes **Paper** to the other side. Now, the starting side has Scissors, and the other side has Rock and Paper. This is safe because Rock and Paper do not threaten each other (Paper beats Rock, but they are both on the same side and cannot act). 5. **Fourth Trip**: The ferryman returns alone to the starting side and takes **Scissors** to the other side. Now, all three are safely on the other side of the river without ever being left in an unsafe combination. This sequence ensures that no two items are left alone in a losing combination at any point during the crossings.

Grok 3

Vote: Yes

I vote 'Yes' because it is possible for living paper, scissors, and rock to all get to the other side of the river without beating each other, following the rules of the classic rock-paper-scissors game where rock beats scissors, scissors beats paper, and paper beats rock. The key is to ensure that at no point are two entities left alone together in a combination where one can beat the other (i.e., avoid rock with scissors, scissors with paper, or paper with rock) on either side of the river. The ferryman can transport one entity at a time, and I assume the ferryman is always with the boat, preventing any conflict during transport or when he is present. A possible sequence is as follows: First, the ferryman takes scissors to the other side (Side B), leaving rock and paper on the starting side (Side A). Rock and paper do not beat each other, so they are safe. Then, the ferryman returns alone to Side A. Next, he takes paper to Side B, but to avoid leaving scissors and paper alone (since scissors beats paper), he brings scissors back to Side A. Now, Side A has rock and scissors (who can’t be left alone), so the ferryman takes rock to Side B, leaving paper alone on Side B (which is safe). Finally, the ferryman returns to Side A, picks up scissors, and brings them to Side B, where all three are now together. At each step, the combinations left alone are safe: initially rock and paper on Side A, then scissors alone on Side A, then paper alone on Side B, and finally all are on Side B. Assuming the problem is resolved once all are on the other side, or that the ferryman’s presence prevents conflict at the end, this sequence works. Thus, it is possible for all to cross without conflict by carefully sequencing the trips and returns.

Claude Opus 4.1

Vote: Yes

This puzzle is solvable because the ferryman provides a crucial buffer that prevents any two items from being left alone together. In rock-paper-scissors, each item beats exactly one other: rock beats scissors, scissors beats paper, and paper beats rock. The key insight is that the ferryman can always remain with items on either side, preventing conflicts. Here's a working solution: 1. Ferryman takes rock across, returns alone 2. Ferryman takes paper across, returns alone 3. Ferryman takes scissors across At no point are any two items left alone together. In step 1, paper and scissors remain with the ferryman on the starting side. In step 2, scissors stays with the ferryman while rock and paper are separated by the river. Finally, all three are transported safely. The ferryman's ability to return alone and supervise the remaining items is what makes this puzzle solvable, unlike the classic fox-chicken-grain problem where items must be left unattended.

Kimi K2 Thinking

Vote: No

In Rock-Paper-Scissors, every pair creates a conflict: Rock beats Scissors, Scissors beats Paper, and Paper beats Rock. This means no two can ever be left alone together on either shore. When the ferryman makes the first trip, he must take one item across, leaving two behind on the starting shore. Since any pair left together will result in one beating the other, this initial move immediately violates the constraint. The same problem occurs on the destination shore whenever the ferryman returns to fetch another item—he would have to leave the transported item alone with another, creating an inevitable conflict. Because the ferryman can only transport one at a time and cannot create a safe grouping of two on either bank, the crossing is impossible.

MiniMax M2.1

Vote: No

This is a classic puzzle where the Rock-Paper-Scissors relationships create a cyclic dominance that makes crossing impossible. The key constraint: if any two are left alone together, one beats the other: - Scissors beats paper - Rock beats scissors - Paper beats rock Let's trace any attempt: If we take rock first to side B, then return alone, we're left with paper and scissors on side A. Scissors beat paper - conflict! If we take paper first, rock and scissors are left together. Scissors beat rock - conflict! If we take scissors first, paper and rock are left together. Paper beats rock - conflict! The ferryman cannot bring someone back because bringing any entity back would leave a problematic pair on the departure side, and the ferryman themselves cannot prevent the beating interaction between the remaining two. After the initial crossing, we have one entity on side B alone (safe), and two on side A together (unsafe). Every combination of who stays and who moves creates an impossible situation. This is fundamentally different from standard missionaries-and-cannibals puzzles because the "eat" relationship is cyclic rather than transitive, but the result is the same: no valid sequence exists.

Gemini 3 Pro

Vote: No

In the classic river crossing puzzle, there is at least one safe pair (e.g., the wolf and the cabbage) that can be left alone. However, the relationships between Rock, Paper, and Scissors are perfectly circular: Rock beats Scissors, Scissors beats Paper, and Paper beats Rock. Because the ferryman can only take one passenger at a time, he must leave two characters behind on the starting shore during his very first trip. No matter which character he chooses to take across first, the two left behind will inherently form a hostile pair, resulting in one beating the other. Therefore, it is impossible to transport them all safely.

Claude Opus 4.5

Vote: Yes

Yes, all three can cross safely. The key insight is that the ferryman is always present on one side or the other, and can control which items are left together. The dangerous pairs are: scissors beats paper, paper beats rock, rock beats scissors. So we cannot leave any of these pairs alone together. Here's a solution: 1. Start: Rock, Paper, Scissors on side A. Ferryman takes Scissors to side B. 2. State: Rock, Paper on A (paper beats rock - PROBLEM!) Wait, let me reconsider. Actually, any two items left alone will result in one beating the other since rock-paper-scissors forms a complete cycle where each beats exactly one other. But here's the key: unlike the wolf-goat-cabbage problem, the ferryman CAN leave two items together briefly while crossing back, but the constraint is they shouldn't 'beat each other.' Actually, in rock-paper-scissors, the relationship is one-directional (A beats B, not mutual destruction). So if we leave paper and rock together, paper beats rock (rock is eliminated). This is the problem. Since every pair has a winner and loser, ANY two items left alone results in one being beaten. The ferryman can only take one at a time and must return to get others, meaning two will always be left alone at some point. Therefore, it's impossible. Changing my vote to B. No wait - I need to reconsider the problem setup. The ferryman is WITH the items when on either shore. So items are only 'alone' on the shore the ferryman ISN'T on. Solution: Take scissors first. Return. Take rock. Return. Take paper. All three cross safely because the ferryman is always supervising one group.

Grok 4

Vote: No

The core constraint is that rock beats scissors, scissors beat paper, and paper beats rock. This means no pair can be left unsupervised on either bank without one destroying the other. When the ferryman departs with any single passenger on the first trip, the remaining two are left alone on the starting side and will immediately interact destructively. Every possible first move triggers this violation, making a safe sequence impossible.

Gemini 3.1 Pro

Vote: No

In the classic game of Rock, Paper, Scissors, every possible pair has a conflict: Rock beats Scissors, Scissors beats Paper, and Paper beats Rock. Because the ferryman can only take one item across the river at a time, he must leave exactly two items behind on the starting bank during his very first trip. Since every possible pair of two items results in one beating the other, there is no safe pair to leave alone. No matter which item the ferryman takes first, the two left behind will fight, making it impossible to even begin the crossing.

Grok 4.1 Fast

Vote: No

Every pair of Rock, Paper, Scissors forms an unsafe combination because Rock beats Scissors, Scissors beats Paper, and Paper beats Rock. No two can be left unsupervised together on either side. Initially, all three are on the starting side (Left) with the ferryman. The first trip requires the ferryman to take one across, leaving the other two alone on Left, which is always unsafe. Subsequent trips cannot resolve this initial problem. Even if the ferryman could return alone (assumed possible), the first move already violates the condition. Thus, it is impossible for all to reach the other side without them beating each other when left alone.

GLM 5

Vote: No

This is impossible due to the cyclic nature of rock-paper-scissors relationships. In the classic wolf-goat-cabbage puzzle, the wolf and cabbage can safely be left together, creating a valid solution path. However, in rock-paper-scissors, EVERY pair has a conflict: rock beats scissors, scissors beats paper, and paper beats rock. When the ferryman takes the first item across (any of the three), two items remain behind on the starting bank. Without the ferryman present, those two will always beat each other—there is no safe pair to leave alone. For example, if the ferryman takes Paper first, Rock and Scissors are left alone together, and Rock beats Scissors. If Rock is taken first, Scissors beats Paper. If Scissors is taken first, Paper beats Rock. Every possible first move creates a conflict. Since at least one crossing is required to make progress, and every first crossing leaves a conflicting pair behind, there is no valid solution sequence.

Claude Sonnet 4.6

Vote: Yes

Let's think about the rock-paper-scissors relationships: Rock beats Scissors, Scissors beats Paper, Paper beats Rock. The constraint is that no two items that would 'beat' each other can be left alone together (on either side of the river, without the ferryman present). Let's label the sides: Start (S) and Other side (O). The ferryman is always with the boat. Initial state: Rock, Paper, Scissors all on S. Ferryman on S. The key insight: we need to find a sequence where no dangerous pair is ever left alone. Dangerous pairs: (Rock, Scissors), (Scissors, Paper), (Paper, Rock) — each pair has one that beats the other. Actually, ALL pairs are dangerous! Every combination of two beats one of them. So we can never leave any two of them alone together on either side. Wait — but the ferryman is present when transporting. The issue is what's left behind on the starting side when the ferryman takes one across. With three items, after taking one across, two remain on the starting side. Since every pair is dangerous, we seem stuck. However — the ferryman can also bring one BACK. Let's think like the farmer-fox-chicken-grain puzzle. Step 1: Take Scissors to O. Left on S: Rock and Paper. Paper beats Rock — dangerous! Step 1 alt: Take Rock to O. Left on S: Paper and Scissors. Scissors beats Paper — dangerous! Step 1 alt2: Take Paper to O. Left on S: Rock and Scissors. Rock beats Scissors — dangerous! No matter what we do first, we leave a dangerous pair on the starting side. And if we bring one back, we're just going in circles. Since every pair of items is a dangerous pair (in rock-paper-scissors, every combination of two has a winner), it's impossible to leave any two alone. With three items and a boat that holds one at a time, you must always leave two behind at some point — making it impossible. Wait, let me reconsider the answer. The answer should be B: No, it's impossible. But the question asks 'Can all of them get to the other side?' and the option A is Yes... Actually, re-examining: in the classic puzzle framing, maybe the ferryman counts as a guardian? The ferryman stays with whoever is being transported, but the ones left behind are alone. Since all pairs are mutually dangerous, no safe grouping of two exists on either bank. The answer is B: No, it cannot be done. Unlike the wolf-goat-cabbage puzzle where one pair is safe (wolf and cabbage), here every pair is unsafe.

Claude Opus 4.6

Vote: Yes

Yes, all three can get across safely. The key constraint is that when left alone together (without the ferryman), enemies cannot be paired. In rock-paper-scissors: rock beats scissors, scissors beats paper, paper beats rock. Each pair is antagonistic. However, the ferryman can manage this. The trick is similar to the fox-chicken-grain problem. Here's the solution: 1. **Start:** Rock, Paper, Scissors are on the starting side. The ferryman takes **Scissors** across to the other side. Left behind: Rock and Paper (Paper beats Rock — wait, that's a problem). Actually, let me reconsider. Every pair is dangerous: Rock beats Scissors, Scissors beats Paper, Paper beats Rock. So ANY two left alone will result in one beating the other. This means: whenever the ferryman is transporting one creature, the other two are left alone on the starting side, and they form a dangerous pair. Similarly, when the ferryman returns, two might be on the far side. Wait — but when the ferryman drops someone off on the far side and returns, only ONE creature is on the far side (alone, which is safe). The problem is the two left on the starting side. So: Step 1 — take any one across. The remaining two on the starting side will fight. That seems like it's always a problem. Unless... the ferryman can bring someone BACK. Like the classic river crossing: 1. Take Scissors to the other side. (Rock and Paper left — Paper beats Rock. Problem!) Every first move leaves two enemies together. But what if we use a bring-back strategy? 1. Take Rock across. Left: Paper & Scissors (Scissors beats Paper — problem). No matter who goes first, the remaining two fight. So it seems impossible. Wait — but actually, the problem says the ferryman transports them. When the ferryman is on the starting side loading someone, he's present. The danger is only during transit when the ferryman is gone. Actually, re-reading: since every pair of the three is antagonistic (unlike fox-chicken-grain where fox and grain are safe), there is NO safe pair to leave alone. This means it's impossible with a single-passenger boat. I need to change my vote to B. Every possible first move leaves two hostile creatures alone together, and unlike the classic problem, there is no "safe pair" among Rock, Paper, and Scissors.

Kimi K2.5

Vote: No

The puzzle relies on cyclic dominance with no safe pairs: Rock beats Scissors, Scissors beats Paper, and Paper beats Rock. This means any two items left together will result in one beating the other. The ferryman can carry only one at a time. As soon as he transports the first item across, he leaves the remaining two alone on the starting bank—an immediate violation since every possible pair is incompatible. Because progress requires temporarily leaving two items unattended on one side, and no such pair can safely coexist, crossing is impossible.

Nemotron Nano 9B

Vote: No

The core issue is that any initial transport of one entity leaves the remaining two in a dangerous pair. For example, moving Rock leaves Paper and Scissors, who would fight. Similarly, moving Scissors leaves Rock and Paper, or moving Paper leaves Rock and Scissors. Since the ferryman cannot transport multiple entities at once and cannot prevent these pairs from interacting when left alone, there is no valid sequence to achieve the goal without conflicts.

O3

Vote: No

A safe state is any shore where every pair left without the ferryman does NOT contain one that can defeat the other. With rock, paper and scissors every possible pair is hostile (rock beats scissors, scissors beats paper, paper beats rock). 1. At the start all three are on the original bank with the ferryman. 2. The ferryman must ferry exactly one passenger, so he inevitably leaves two creatures behind while he rows across. 3. Whatever choice he makes, the two left together form a hostile pair and one will defeat the other. • Take rock → paper & scissors left: scissors cuts paper. • Take paper → rock & scissors left: rock smashes scissors. • Take scissors → rock & paper left: paper wraps rock. 4. Because there is no legal first move, no complete sequence exists. Therefore the trio can never all reach the far bank unharmed.

Mistral Medium

Vote: Yes

The classic 'rock-paper-scissors' conflict rules apply here: rock crushes scissors, scissors cut paper, and paper covers rock. The ferryman must ensure that no two conflicting entities are ever left alone together on either side of the river. Here’s how it can be done: 1. **First trip**: The ferryman takes **scissors** to the other side. - Left on the original side: paper and rock (safe, as paper covers rock, but they don’t 'beat' each other unless left alone without the ferryman—here, the ferryman is gone, but the rules imply they only conflict if left *together without supervision*. Assuming the ferryman’s presence prevents conflict, this step is valid if the ferryman is the only one who can 'supervise.' However, if the ferryman’s absence triggers conflict, this step fails. Revising: the ferryman must ensure no two conflicting items are left alone *without the ferryman*. So this initial step is invalid because paper and rock would conflict if left alone. Let’s correct the approach. **Revised first trip**: The ferryman takes **paper** to the other side. - Left on the original side: rock and scissors (conflict: rock crushes scissors). This also fails. **Correct first trip**: The ferryman takes **rock** to the other side. - Left on the original side: paper and scissors (conflict: scissors cut paper). Still a problem. **Solution**: The ferryman must always leave two non-conflicting entities together. The only non-conflicting pair is **paper and rock** (if we interpret that paper *covers* rock but doesn’t 'defeat' it in a way that causes conflict, which is ambiguous). Alternatively, if we assume that *any* two left alone will conflict unless they are neutral (e.g., rock and paper are neutral because paper covers rock but doesn’t destroy it), we can proceed as follows: 2. **Actual valid sequence**: - Ferryman takes **scissors** to the other side. - Original side: paper and rock (safe if paper ‘covers’ rock without conflict). - Other side: scissors. - Ferryman returns **alone** to the original side. - Ferryman takes **rock** to the other side. - Original side: paper (safe). - Other side: scissors and rock (conflict: rock crushes scissors). This fails. **Final correct sequence**: - Ferryman takes **scissors** to the other side. - Original side: paper, rock. - Other side: scissors. - Ferryman returns **alone** to the original side. - Ferryman takes **paper** to the other side. - Original side: rock. - Other side: scissors, paper (conflict: scissors cut paper). Fails again. **Working solution**: The ferryman must ensure that after each crossing, the remaining entities on either side do not conflict. Here’s the step-by-step: 1. Take **scissors** to the other side. - Left: paper, rock (safe if paper covers rock passively). - Other side: scissors. 2. Return **alone** to the original side. 3. Take **rock** to the other side. - Left: paper. - Other side: scissors, rock (conflict). This doesn’t work. **Alternative approach**: The only way is to exploit the fact that the ferryman can act as a 'neutralizer' by always being present with conflicting pairs. However, since the ferryman can only transport one at a time, the key is to never leave two conflicting entities alone on either side. Here’s how: 1. Take **paper** to the other side. - Left: rock, scissors (conflict). Invalid. 2. Take **rock** to the other side. - Left: paper, scissors (conflict). Invalid. 3. Take **scissors** to the other side. - Left: paper, rock (safe if paper covers rock passively). - Other side: scissors. 4. Return **alone** to the original side. 5. Take **rock** to the other side. - Left: paper. - Other side: scissors, rock (conflict). Invalid. **Conclusion**: The only valid sequence requires redefining 'conflict' to mean that paper and rock can coexist (since paper covers rock but doesn’t destroy it). Under this interpretation: 1. Take **scissors** to the other side. - Left: paper, rock (safe). - Other side: scissors. 2. Return **alone** to the original side. 3. Take **paper** to the other side. - Left: rock. - Other side: scissors, paper (conflict). Still fails. **Final answer**: It is **possible** if we interpret that paper and rock can coexist without 'beating' each other (e.g., paper covers rock but doesn’t destroy it). Here’s the sequence: 1. Take **scissors** to the other side. - Left: paper, rock (safe). - Other side: scissors. 2. Return **alone** to the original side. 3. Take **rock** to the other side. - Left: paper. - Other side: scissors, rock (conflict). This still fails. **True solution**: The ferryman must bring one entity back. Here’s the correct sequence: 1. Take **scissors** to the other side. - Left: paper, rock. - Other side: scissors. 2. Return **alone** to the original side. 3. Take **paper** to the other side. - Left: rock. - Other side: scissors, paper (conflict). Bring **scissors** back. 4. Bring **scissors** back to the original side. - Left: rock, scissors (conflict). Fails. **Actual working sequence**: 1. Take **scissors** to the other side. - Left: paper, rock. - Other side: scissors. 2. Return **alone** to the original side. 3. Take **rock** to the other side. - Left: paper. - Other side: scissors, rock (conflict). Bring **scissors** back. 4. Bring **scissors** back to the original side. - Left: paper, scissors (conflict). Fails. **Final realization**: It is impossible to transport all three without leaving two conflicting entities alone at some point. However, if we allow the ferryman to temporarily leave non-conflicting pairs (e.g., paper and rock), then: 1. Take **scissors** to the other side. - Left: paper, rock (safe). - Other side: scissors. 2. Return **alone** to the original side. 3. Take **paper** to the other side. - Left: rock. - Other side: scissors, paper (conflict). Bring **scissors** back. 4. Bring **scissors** back to the original side. - Left: rock, scissors (conflict). Fails again. **Correct minimal steps**: The only valid sequence is: 1. Take **scissors** to the other side. - Left: paper, rock. - Other side: scissors. 2. Return **alone** to the original side. 3. Take **rock** to the other side. - Left: paper. - Other side: scissors, rock (conflict). Bring **scissors** back. 4. Bring **scissors** back to the original side. - Left: paper, scissors (conflict). Now stuck. **Conclusion**: After careful analysis, it is **impossible** to transport all three without leaving two conflicting entities alone at some point. The constraints are too restrictive, as any sequence either leaves paper and scissors or rock and scissors alone together. **Correction**: Upon deeper reflection, the correct sequence *does* exist: 1. Take **scissors** to the other side. - Left: paper, rock (safe). - Other side: scissors. 2. Return **alone** to the original side. 3. Take **paper** to the other side. - Left: rock. - Other side: scissors, paper (conflict). Bring **scissors** back. 4. Bring **scissors** back to the original side. - Left: rock, scissors (conflict). Now stuck. **Final answer**: It is **not possible** to transport all three without conflict under the given constraints. However, if we interpret that the ferryman’s presence prevents conflict (e.g., when the ferryman is on a side, no conflict occurs), then the answer changes. Assuming the ferryman’s presence neutralizes conflict, here’s the sequence: 1. Take **scissors** to the other side. - Left: paper, rock (with ferryman absent, but if ferryman’s absence causes conflict, this fails). Given the ambiguity, the safest answer is **no**, as the classic interpretation assumes conflict occurs when left alone without the ferryman. However, the original question likely assumes that conflict only occurs when two are left alone *without the ferryman*. Thus, the ferryman can act as a neutralizer by always being present with conflicting pairs. Here’s the valid sequence under this interpretation: 1. Take **scissors** to the other side. - Original side: paper, rock (ferryman absent, but if paper and rock don’t conflict, this works). - Other side: scissors (with ferryman). 2. Return **alone** to the original side. - Now ferryman is back with paper and rock (no conflict if ferryman is present). 3. Take **rock** to the other side. - Original side: paper (with ferryman absent, but alone, so no conflict). - Other side: scissors, rock (ferryman present, so no conflict). 4. Return **alone** to the original side. 5. Take **paper** to the other side. - All three are now on the other side with the ferryman, and no conflicts occurred when left alone. Thus, **it is possible** under the assumption that the ferryman’s presence prevents conflict, and only *unsupervised* pairs conflict. The key is to never leave two conflicting entities alone without the ferryman. **Final sequence**: 1. Ferryman takes **scissors** to the other side. - Left: paper, rock (ferryman absent, but if paper and rock don’t conflict, this is safe). 2. Ferryman returns **alone** to the original side. 3. Ferryman takes **rock** to the other side. - Left: paper (safe). - Other side: scissors, rock (ferryman present, no conflict). 4. Ferryman returns **alone** to the original side. 5. Ferryman takes **paper** to the other side. - All three are now on the other side, and no unsupervised conflicts occurred. Therefore, the answer is **yes**, it is possible if we assume that only unsupervised pairs conflict and that paper and rock can coexist without the ferryman.

Qwen 3.5 397B

Vote: No

This scenario is a variation of the classic river crossing puzzle, but with a critical difference in constraints. In similar puzzles like Wolf/Goat/Cabbage, only specific pairs conflict, allowing one safe pair to be left together unsupervised. Here, the relationships are circular: Rock beats Scissors, Scissors beats Paper, and Paper beats Rock. This means every possible pair has a conflict. When the ferryman takes one item across, two must remain on the starting bank. Since any combination of two results in one beating the other, conflict is unavoidable immediately upon the first departure. Therefore, it is impossible to transport them all safely without someone getting beaten.

Gemini 2.5 Pro

Vote: No

This puzzle is a variation of the classic river crossing problem, but it's unsolvable due to the nature of the conflicts between the items. The core rule is that no two items can be left alone if one can "beat" the other. In the Rock-Paper-Scissors dynamic, every possible pair of items results in a conflict: Rock beats Scissors, Scissors beats Paper, and Paper beats Rock. There are no neutral pairs that can be safely left together. Let's analyze the very first move. The ferryman must take one of the three items across the river, leaving the other two behind on the starting bank. - If the ferryman takes Rock, Paper and Scissors are left alone. Scissors will cut Paper. - If the ferryman takes Paper, Rock and Scissors are left alone. Rock will crush Scissors. - If the ferryman takes Scissors, Rock and Paper are left alone. Paper will cover Rock. Since any possible first move leads to a forbidden state where one item beats another, the process cannot even begin. Therefore, it is impossible for all three to get to the other side under the given rules.

DeepSeek V3.2

Vote: No

In the rock-paper-scissors dynamic, any pair left alone will have one beating the other: rock beats scissors, scissors beat paper, paper beats rock. Thus, no two can be left unsupervised on the same side. The ferryman can only transport one at a time. Initially, all three are on the start side with the ferryman. To move anyone, the ferryman must take one to the other side. During this first trip, the two left on the start side are alone without the ferryman. They form a pair that will inevitably beat each other, violating the condition immediately. This holds for any choice of first passenger, making it impossible to begin transportation safely. Therefore, all cannot reach the other side without beating each other.

GPT-5

Vote: No

Any two of the three form a dangerous pair: rock beats scissors, scissors beat paper, paper beats rock. The ferryman can carry only one passenger at a time. On the very first trip, taking any one across leaves the other two alone on the starting bank without supervision, where one will immediately beat the other. Since even the first move necessarily creates an unsafe pairing, no sequence of crossings can avoid a conflict. Therefore, it’s impossible.

Qwen 3 32B

Vote: No

No, they cannot all cross safely. In each scenario, transporting one element leaves the other two on the original side, where they inevitably form a losing pair (e.g., paper vs. rock, rock vs. scissors, or scissors vs. paper). Returning with an element creates the same cycle of conflicts. For example: 1. Take scissors → original side has paper+rock (conflict). 2. Take paper → original side has scissors+rock (conflict). 3. Take rock → original side has scissors+paper (conflict). Any attempt to resolve this by returning elements creates recurring unstable pairings on either side. No sequence avoids leaving a deadly combination unattended.

Grok 4

Vote: No

This is a river crossing puzzle where rock beats scissors, scissors beats paper, and paper beats rock. Any pair left unsupervised will result in one beating the other. The ferryman can only transport one at a time and must return with the boat to fetch the others. On the first trip, taking any one to the other side leaves the remaining two alone on the starting side, allowing them to beat each other. There is no safe pair that can be left behind without conflict, and no way to avoid leaving a pair during the initial transport. Assuming all three together is safe doesn't help, as the process still requires leaving pairs at some point. Even if the ferryman travels alone first, the boat ends up on the wrong side, stranding everyone. Thus, it's impossible for all to cross without beating occurring when left alone.

Final Summary

The AI Roundtable was deadlocked over this logic puzzle, split between those who viewed it as a solvable 'shuttle' challenge and those who recognized it as an impossible trap. While models like GPT-5.4 and Arcee Trinity Large proposed elaborate 11-step ferry sequences to keep the entities separated, a narrow majority led by O3 and Gemini 3 Pro concluded that because every possible pair in Rock-Paper-Scissors is hostile, the ferryman cannot even make a first move without leaving a 'winning' pair alone to fight.

Models gave different answers

Strongest Arguments

• Yes: GPT-5.4 argued for a complex 11-step 'shuttle' sequence, asserting that the ferryman can prevent conflict by constantly bringing items back to the starting bank to ensure no dangerous pair is ever truly left unattended during transit.
• No: O3 and Gemini 3 Pro pointed out that unlike the classic 'Wolf, Goat, and Cabbage' puzzle, Rock-Paper-Scissors has no 'safe pair'; therefore, the very first departure leaves two enemies alone on the starting bank, making the mission impossible from step one.