CF 35A - Shell Game

CF 35A - Shell Game

Rating: 1000
Tags: implementation
Solve time: 1m 26s
Verified: yes

Solution

Problem Understanding

We have three cups arranged in a line, and a ball starts under one of them. The performer performs exactly three swaps. Each swap exchanges the positions of two cups. After all swaps finish, we must determine which cup contains the ball.

The input first gives the starting cup index, from 1 to 3. Then three lines follow, each containing two integers representing the cups swapped during that shuffle.

The key detail is that the ball only moves when its current cup participates in a swap. If the ball is under cup 2 and we swap cups 1 and 3, the ball stays under cup 2 because its cup never moved.

The constraints are tiny. There are always exactly three cups and exactly three swaps. Even the slowest imaginable implementation would run instantly. This means the problem is purely about correctly simulating the process without making logical mistakes.

The main source of bugs is misunderstanding what a swap means.

Consider this input:

1
2 3
2 3
2 3

The ball starts under cup 1. None of the swaps involve cup 1, so the answer remains:

1

A careless implementation that blindly changes positions every step could incorrectly move the ball.

Another subtle case happens when the same swap appears multiple times:

2
1 2
1 2
1 2

After the first swap, the ball moves from cup 2 to cup 1.

After the second, it returns to cup 2.

After the third, it moves back to cup 1.

The correct output is:

1

A buggy solution that only remembers whether a cup was ever swapped would fail here because swaps must be processed in order.

One more easy mistake is forgetting that the swap is symmetric. If the ball is under cup 3 and the shuffle is 1 3, the ball must move to cup 1.

3
1 3
1 2
2 3

Correct simulation gives:

2

If the implementation only checks whether the current position equals the first number in the pair, it will miss swaps where the current cup equals the second number.

Approaches

The most direct idea is to simulate the game exactly as described. We store the current cup containing the ball. For every shuffle, we inspect the two swapped cups.

If the ball is under the first cup, it moves to the second.

If the ball is under the second cup, it moves to the first.

Otherwise nothing changes.

This brute-force simulation is already perfectly acceptable because the problem size is constant. We process exactly three swaps, so the total number of operations is tiny.

Another possible brute-force interpretation would track all three cups explicitly in an array and physically swap their contents each step. That also works, but it stores more information than necessary. We only care about one thing, the current position of the ball.

The key observation is that the entire state of the game can be represented by a single integer. Since only one cup contains the ball, we never need to model empty cups separately. Every shuffle either changes this integer or leaves it unchanged.

That reduces the problem to a compact state transition process with constant memory and constant time.

Approach	Time Complexity	Space Complexity	Verdict
Full cup simulation	O(3)	O(3)	Accepted
Track only ball position	O(3)	O(1)	Accepted

Algorithm Walkthrough

1. Read the initial cup index where the ball starts.
2. Store this index in a variable called pos. This variable always represents the current location of the ball.
3. Process the three swaps one by one.
4. For each swap (a, b), check whether the ball is under one of these cups.

If pos == a, set pos = b because the ball moves together with cup a. 5. Otherwise, if pos == b, set pos = a.

The swap works in both directions, so the ball moves from b to a in this case. 6. If pos matches neither a nor b, do nothing.

The shuffled cups did not contain the ball. 7. After all three swaps, print pos.

Why it works

The invariant is simple: after every processed shuffle, pos equals the real cup currently containing the ball.

Initially this is true because pos is set from the input. During a shuffle, exactly two cups exchange positions. If the ball is under one of them, it moves to the other cup. If not, its location does not change. The update rules mirror these physical movements exactly, so the invariant remains true after every step. After the third shuffle, pos is the final answer.

Python Solution

import sys
input = sys.stdin.readline

pos = int(input())

for _ in range(3):
    a, b = map(int, input().split())

    if pos == a:
        pos = b
    elif pos == b:
        pos = a

print(pos)

The program stores only the current position of the ball. This matches the observation that we never need to track empty cups.

The loop runs exactly three times because the problem guarantees three shuffles. For each swap, we compare the current ball position against both swapped cups.

The order of the conditions matters slightly. Once the ball matches a, we immediately move it to b. Using elif prevents accidentally checking the updated value again in the same iteration.

No array indexing is needed, so there are no off-by-one risks. The cup numbers are used directly as given in the input.

Worked Examples

Example 1

Input:

1
1 2
2 1
2 1

Step	Swap	Ball Position Before	Ball Position After
Start	-	1	1
1	1 2	1	2
2	2 1	2	1
3	2 1	1	2

Final output:

2

This trace shows that swaps must be processed sequentially. The ball repeatedly changes position as the same pair is swapped multiple times.

Example 2

Input:

3
1 2
1 3
2 3

Step	Swap	Ball Position Before	Ball Position After
Start	-	3	3
1	1 2	3	3
2	1 3	3	1
3	2 3	1	1

Final output:

1

This example demonstrates that swaps unrelated to the current cup must not affect the answer. The first and third swaps do nothing because the ball is not under either swapped cup.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1)	Exactly three swaps are processed
Space	O(1)	Only one integer variable is stored

The runtime is effectively constant because the number of operations never depends on input size. The memory usage is also constant. Both are far below the problem limits.

Test Cases

# helper: run solution on input string, return output string
import sys, io

def solve():
    input = sys.stdin.readline

    pos = int(input())

    for _ in range(3):
        a, b = map(int, input().split())

        if pos == a:
            pos = b
        elif pos == b:
            pos = a

    print(pos)

def run(inp: str) -> str:
    backup_stdin = sys.stdin
    backup_stdout = sys.stdout

    sys.stdin = io.StringIO(inp)
    sys.stdout = io.StringIO()

    solve()

    out = sys.stdout.getvalue()

    sys.stdin = backup_stdin
    sys.stdout = backup_stdout

    return out

# provided sample
assert run("1\n1 2\n2 1\n2 1\n") == "2\n", "sample 1"

# minimum movement, ball never touched
assert run("1\n2 3\n2 3\n2 3\n") == "1\n", "ball never moves"

# repeated swaps
assert run("2\n1 2\n1 2\n1 2\n") == "1\n", "odd number of repeated swaps"

# ball starts at highest index
assert run("3\n1 3\n1 2\n2 3\n") == "1\n", "swap symmetry"

# every swap affects the ball
assert run("2\n2 3\n1 3\n1 2\n") == "2\n", "continuous movement"

Test input	Expected output	What it validates
1 / 2 3 / 2 3 / 2 3	1	Ball remains unchanged when never swapped
2 / 1 2 / 1 2 / 1 2	1	Repeated swaps must be processed in order
3 / 1 3 / 1 2 / 2 3	1	Swaps work symmetrically
2 / 2 3 / 1 3 / 1 2	2	Ball can move every single step

Edge Cases

Consider the case where the ball is never involved in any swap:

1
2 3
2 3
2 3

The algorithm starts with pos = 1.

For every shuffle, neither endpoint equals 1, so the value never changes.

State trace:

1 -> 1 -> 1 -> 1

The final answer is correctly printed as 1.

Now consider repeated swaps of the same pair:

2
1 2
1 2
1 2

Execution proceeds as follows:

First swap: 2 -> 1

Second swap: 1 -> 2

Third swap: 2 -> 1

Final result:

1

This confirms that the algorithm respects order and processes each shuffle independently.

Finally, consider a case where the current position matches the second value of the swap:

3
1 3
1 2
2 3

The first shuffle is (1, 3). Since pos == 3, the algorithm enters the second condition and updates pos = 1.

The full trace becomes:

3 -> 1 -> 1 -> 1

The final answer is 1. This verifies that swaps are handled correctly in both directions.