I have 1000 bottles of wine and exactly one bottle contains a lethal poison. The poison has no detectable effect on the wine and causes death exactly 24 hours after ingestion. I have 10 test subjects and only 24 hours to identify the poisoned bottle.

Design a testing strategy that identifies the poisoned bottle using the minimum number of test subjects possible. Explain the encoding scheme and why it works.

Working time: 2m 28s

Binary Encoding Solution

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Strategy: Number each of the 1000 bottles from 0 to 999 in decimal, then express each number as a 10-bit binary code. Assign one test subject to each bit position (subject i corresponds to the ith bit).

Encoding Scheme: For each bottle with binary representation b₉b₈...b₀, have subject i taste wine from that bottle if and only if the i-th bit equals 1. This means:

• Subject 0 tastes from all bottles where the least significant bit is 1 (bottles 1, 3, 5, …, 999)
• Subject 1 tastes from all bottles where the second bit is 1 (bottles 2, 3, 6, 7, …)
• …
• Subject 9 tastes from all bottles where the most significant bit is 1 (bottles 512 to 999)

Each bottle is sampled by a unique combination of subjects based on its binary representation. The poisoned bottle is encoded as which subset of subjects consume it.

Why It Works: After 24 hours, observe which subjects died. If subject i dies, the poisoned bottle has a 1 in position i. The poisoned bottle’s number equals the sum of powers of two for each dead subject. For example, if subjects 0, 3, and 8 die, the poisoned bottle is 2⁰ + 2³ + 2⁸ = 1 + 8 + 256 = 265 (binary 0001000001).

Minimum Subjects: Exactly 10 test subjects are necessary and sufficient. With N subjects, there are 2^N possible outcome patterns. Since log₂(1000) ≈ 9.97, we need at least 10 bits to uniquely identify among 1000 bottles. Nine subjects would yield only 512 distinguishable outcomes—insufficient for our task. This binary encoding approach achieves identification with minimal resources by maximizing the information extracted per test.

Working time: 25s

Strategy – binary (base‑2) encoding

The goal is to obtain the unique identity of the single poisoned bottle from the pattern of dead subjects.
If each test subject can give only one piece of information (alive / dead), then with k subjects we can distinguish at most \(2^{k}\) different outcomes because each outcome is a binary string of length k (1 = subject died, 0 = subject alive).

So at least ten subjects are required; nine would not be enough to label all possible bottles.

Encoding scheme

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Number the bottles from 0 to 999.
Write each number in binary using exactly ten bits (bits \(b_{9}\) … \(b_{1}\), where \(b_{1}\) is the least‑significant bit).

Now, for each bit position \(i\) ( \(i=1\) is the right‑most bit) we let subject i drink only from those bottles whose binary representation has a 1 in that column.

• Example: subject 3 drinks from all bottles with \(b_{3}=1\).
  Bottles 3, 7, 11, …, 999 will be among her intake.

After the poison is ingested:

• A subject dies iff the poisonous bottle’s binary digit in that position is 1.
• If a subject is alive, the corresponding bit is 0.

Thus the pattern of dead subjects after 24 hours spells out the ten‑bit code of the poisoned bottle, which we decode directly to obtain its index (0–999).

Why it works

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Each subject supplies one bit. The union of all ten bits uniquely identifies any number from 0 to 1023. Because we have only one poisonous bottle, the observed binary string is exactly the code for that single bottle, and there is no ambiguity.

No other scheme can do better with a single‑bit per subject test because the information‑theoretic limit is \(\log_{2}(1000) \approx 9.97\) bits; rounding up gives ten bits (ten subjects). Hence this strategy uses the minimum possible number of test subjects—exactly ten.

Practical notes

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• All bottles are mixed well so that each subject’s drink contains a uniform concentration of wine from its assigned set.
• After 24 hours, observe which subjects are dead and record their numbers (1‑10).
• Convert the binary string (1 for dead, 0 for alive) to decimal; that gives the poisoned bottle number.

Thus, with ten test subjects and a binary encoding scheme, we identify the lethal bottle in exactly 24 hours.

Working time: 1m 6s

This puzzle requires applying binary encoding logic to maximize the information gained from each test subject. The minimum number of subjects required is 10.

Encoding Scheme and Explanation

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The core concept relies on representing every bottle’s identity with a unique, fixed-length binary code (a sequence of 1s and 0s). Since we have 10 test subjects, we can use each subject as one “bit” in our binary system. With \(K\) bits, we can differentiate up to \(2^K\) possibilities. To encode 1000 bottles (\(N=1000\)), we need at least \(\log_2(1000)\) bits. Since \(2^{9} = 512\) (not enough) and \(2^{10} = 1024\) (enough), exactly 10 bits are necessary and sufficient.

We will number the bottles from 001 to 1000. Each bottle’s number is converted into its unique 10-digit binary sequence (\(b_{10}, b_9, ..., b_1\)).

The testing protocol assigns each of the 10 subjects (let’s call them \(S_1\) through \(S_{10}\)) to a specific bit position:

1. Subject Assignment: Subject \(S_j\) will be given samples from all bottles whose corresponding \(j\)-th bit in their binary code is a ‘1’.
2. The Test Round: All 10 subjects drink their designated samples simultaneously. Since the poison only kills after exactly 24 hours, we wait the full period to observe results.

Why It Works

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The system works because each subject’s outcome provides one unique piece of binary information (a ‘0’ or a ‘1’) about the poisoned bottle:

• If Subject \(S_j\) survives: This means that every single bottle they drank from had a ‘0’ in the \(j\)-th position. Therefore, we know for certain that the poisoned bottle has a ‘0’ in this bit position.
• If Subject \(S_j\) dies: This means at least one bottle they consumed contained the poison. Since only one bottle is poisoned, the poisoned bottle must have a ‘1’ in the \(j\)-th position.

By recording the death/survival status of all 10 subjects (a sequence of ten outcomes), we create a unique 10-digit binary signature. This signature perfectly matches the binary code of the single poisoned bottle number, identifying it with certainty.

Working time: 43m 37s

Chemistry Problem: Radical Reaction Analysis

When two radicals, \(\text{A}\cdot\) and \(\text{B}\cdot\), approach each other to form a covalent bond (\(\text{A}-\text{B}\)), the unpaired electrons in their respective \(p\) or \(d\) orbitals pair up. Each radical has one electron with a specific spin orientation (e.g., spin-up, \(\uparrow\)). For a stable bond to form, these two electrons must occupy the same spatial molecular orbital. According to the Pauli Exclusion Principle, they cannot have the same quantum numbers; therefore, they must adopt opposite spin states (\(\uparrow\) and \(\downarrow\)) while sharing the bonding region.

The relationship between the initial spin state and bond formation is determined by the multiplicity of the radical pair. If the system starts in a singlet state (\(S=0\), antiparallel spins), it can form an ordinary covalent product directly because the total spin remains \(S=0\). However, if the radicals are in a triplet state (\(S=1\), parallel spins), they cannot immediately form a standard singlet product. The transition from a triplet to a singlet state is forbidden under the Born-Oppenheimer approximation because electronic states of different multiplicities are orthogonal and do not mix under the non-relativistic Hamiltonian.

A triplet pair cannot directly form an ordinary closed-shell singlet product without a change in spin state due to the conservation of angular momentum. Without interaction, the system remains trapped in the triplet manifold. One mechanism facilitating this required spin evolution is Intersystem Crossing (ISC) driven by Spin-Orbit Coupling (SOC). SOC is a relativistic effect where an electron’s intrinsic spin interacts with its orbital motion around the nucleus. This interaction mixes singlet and triplet wavefunctions, allowing for a non-radiative transition from a higher-energy triplet state to a lower-energy singlet state.

Essay: Technological Progress – Capability vs. Dependence

Working time: 24m 20s

To identify one poisoned bottle among 1,000 using only 10 test subjects in a single round of testing, you should use an information theory approach based on binary encoding.

The Strategy

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You can identify the poisoned bottle with 100% certainty using exactly 10 test subjects.

The Encoding Scheme

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The core of this strategy is to treat each test subject as a “bit” in a binary number. Because \(2^{10} = 1,024\), ten bits are sufficient to create a unique identification code for every bottle from 1 to 1,000.

1. Assignment: Assign each of the 1,000 bottles a unique ID number from 1 to 1,000.
2. Binary Conversion: Convert each ID number into its equivalent 10-digit binary representation (e.g., Bottle #1 is 0000000001, Bottle #512 is 1000000000, and Bottle #1,000 is 1111101000).
3. Preparation: Each of the 10 subjects will drink a “mixture” containing samples from different bottles. The contents of their mixture are determined by the bit they represent:
   • Subject 1 drinks a mixture containing all bottles that have a 1 in the first binary position (bit \(2^0\)).
   • Subject 2 drinks a mixture containing all bottles that have a 1 in the second binary position (bit \(2^1\)).
   • …and so on, until Subject 10, who drinks from all bottles with a 1 in the tenth position (bit \(2^9\)).

Why It Works

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Because there is exactly one poisoned bottle in the entire collection, the resulting pattern of deaths will act as a “map” to the bottle’s unique ID.

When the 24 hours pass, you examine which subjects died:

• If a subject dies, it indicates that their corresponding bit in the binary sequence is a 1.
• If a subject survives, it means their corresponding bit is a 0.

Example: If only Subject 10 dies, you know the poisoned bottle is \(1000000000\) (Bottle #512). If Subjects 10 and 9 both die, you know the poisoned bottle is \(1100000000\) (Bottle #768). By looking at the specific combination of deaths, you can read the binary code directly to identify exactly which bottle contains the poison.

Working time: 11m 18s

To identify the poisoned bottle among 1000 bottles using the minimum number of test subjects, you need 10 subjects.

The Encoding Scheme: Binary Representation

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The most efficient way to solve this is through binary encoding. Each of the 1000 bottles will be assigned a unique binary number from \(0\) to \(999\). Since \(2^{10} = 1024\), a 10-bit binary number is sufficient to give every bottle a unique “ID” (e.g., Bottle 1 is 0000000001, Bottle 2 is 0000000010, Bottle 500 is 0111110100, and so on).

The Testing Procedure

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Each of the 10 test subjects corresponds to one specific bit position in the 10-bit binary ID (from the 1st bit to the 10th bit).

1. Distribution: For every bottle, check its 10-bit binary ID. If a bit at a specific position is “1”, the corresponding test subject drinks a drop from that bottle.
2. Execution: Each subject will drink from a different subset of bottles. For example, Subject 1 (representing the first bit) will drink from every bottle whose ID ends in 1. Subject 2 (representing the second bit) will drink from every bottle whose second-to-last bit is 1.
3. Observation: After 24 hours, you observe which subjects have died.

How it Works

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Each subject provides a binary output: Dead (1) or Alive (0).

• If Subject 1 dies, the poisoned bottle must have a 1 in the first bit.
• If Subject 1 survives, the poisoned bottle must have a 0 in the first bit.

By looking at the results of all 10 subjects, you can construct a 10-bit binary number. For example, if only subjects 3, 5, and 8 die, the poisoned bottle is the one with the binary ID 0010101000.

Why 10 is the Minimum

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This is a problem of information theory. You have 1000 possible outcomes (bottles). Each test subject provides 1 bit of information (Alive/Dead). To distinguish \(N\) possibilities, you need \(\log_2(N)\) bits of information.

Since you cannot have a fraction of a test subject, you must round up to the nearest whole number, which is 10. With 9 subjects, you could only distinguish up to \(2^9 = 512\) bottles.

Working time: 34s

This problem is a classic application of information theory, specifically requiring an encoding scheme to map 1000 possibilities onto the outcomes of 10 tests.

The testing strategy requires exactly 10 test subjects. Since \(2^{10} = 1024\), we have enough distinct outcomes (binary results) to uniquely identify any bottle from the 1000 available options.

Encoding Scheme and Explanation

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The encoding scheme is a base-2 representation where each of the 1000 bottles is assigned a unique 10-digit binary number, ranging from 0000000000 (Bottle 0) to 1111101001 (Bottle 1000). We can map the bottle numbers from 0 to 999.

Why it works: Each test subject is assigned a specific bit position in this binary code. For instance, we assign Test Subject \(J\) to test the state of the \(J\)-th bit of the bottle’s binary address.

Let the 10 subjects be indexed \(T_1\) through \(T_{10}\). We can map these directly to the 10 positions in the binary representation:

• \(T_1\) tests the \(2^0\) (least significant) bit.
• \(T_{10}\) tests the \(2^9\) (most significant) bit.

The Process: For each of the 1000 bottles, we determine its 10-bit binary code. We then instruct the test subjects to test only the bottles whose corresponding bits are ‘1’. If a subject is assigned to test all bottles where their corresponding bit is set, they will identify exactly one bottle (or more, depending on the assignment) that contains the poison.

For the system to guarantee identification of exactly one poisoned bottle using the minimum number of tests, we must ensure that the poisoning event results in a unique outcome across the 10 subjects. The most direct interpretation is:

1. Assignment: Assign each bottle \(B_i\) an index \(i \in \{0, 1, \dots, 999\}\).
2. Test Protocol: Each test subject \(T_j\) is assigned to test the bottles whose index \(i\) has the \(j\)-th bit set in its binary representation of \(i\).

Since there are exactly 10 subjects, and \(\log_2(1000) \approx 9.965\), we use 10 bits. This method allows us to uniquely map the bottle index (the poisoned one) to a specific combination of test results from the 10 subjects, effectively using the 10 tests to probe the 1024 possible states and isolate the single poisoned item among the 1000.