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I knew the rejection sampling trick from somewhere in my past but proving it rigorously on the spot was harder than I expected.
Use rejection sampling with a base-5 representation: generate enough random digits to cover a range that is a multiple of 7, then reject values outside that range. To minimize calls, generate two digits at a time (25 outcomes), accept 21 values (3 full cycles of 7), and reject the remaining 4. This yields an expected number of calls of 50/21 ≈ 2.38, which is close to the information-theoretic lower bound of log_5(7) ≈ 1.21.
Pro tip: Mention that you can reduce the rejection rate by combining leftover entropy from rejected outcomes with new calls, but for simplicity and near-optimality, the two-digit method is sufficient. Also, explicitly compute the expected calls and compare to the lower bound to show awareness of optimality.
Clarify that the given function returns uniform integers 1-5, and we need a uniform integer 1-7. The solution must be unbiased, always terminate, and be close to optimal in calls.
Generate random numbers in a larger range using multiple calls, then map to 1-7 by rejection. For efficiency, use two calls to get 25 equally likely outcomes (5x5 grid).
Assign 21 of the 25 outcomes to the numbers 1-7 (3 outcomes per number). Reject the remaining 4 outcomes and retry. This ensures each number has probability 3/25 per trial, and conditional on acceptance, probability 1/7.
Show that each accepted outcome is equally likely because the 21 accepted outcomes are symmetric. Termination is guaranteed because each trial has acceptance probability 21/25 > 0, so the number of trials is geometric and finite almost surely.
Each trial uses 2 calls. The expected number of trials is 1/(21/25) = 25/21. Thus expected calls = 2 * (25/21) = 50/21 ≈ 2.38. Compare to the lower bound log_5(7) ≈ 1.21 to show near-optimality.
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This is where it got genuinely interesting and also where I started to feel the pressure.
Start by explaining rejection sampling: generate a random number in [1, M^k] by calling the M-sided generator k times, and accept if it falls within the largest multiple of N. Derive the expected number of calls as k / p where p = floor(M^k / N) * N / M^k, and choose k = ceil(log_M N) to minimize calls. Then discuss the two regimes: when N >> M, k is large and acceptance probability is high, so expected calls are O(log_M N); when M >> N, k=1 and expected calls are O(M/N).
Pro tip: Mention that you can optimize by using the leftover randomness from rejected samples (e.g., recycling the remainder) to reduce calls, but note the trade-off in complexity. This shows you think about efficiency beyond the basic algorithm.
Restate the problem: we have a uniform generator for 1..M and need a uniform generator for 1..N. Note that M and N are positive integers, and we want to minimize the expected number of calls to the M-generator.
Explain that we can simulate a uniform generator for 1..M^k by calling the M-generator k times, treating the results as digits in base M. Then accept the result if it is ≤ the largest multiple of N within M^k, otherwise reject and retry.
Let k = ceil(log_M N). The acceptance probability p = floor(M^k / N) * N / M^k. The expected number of calls is k / p. Show that p > 1/2, so expected calls < 2k = O(log_M N). Also consider the case when M > N: then k=1 and expected calls = M / (floor(M/N)*N) ≈ M/(M - (M mod N)) which is O(M/N) when M is much larger than N.
When N >> M, k is large, but acceptance probability is high (close to 1), so expected calls are about log_M N. When M >> N, k=1, and acceptance probability is N/M, so expected calls are about M/N. Discuss how the algorithm adapts naturally.
Mention that you can reduce calls by using the leftover randomness from rejected samples (e.g., if you generate a number in [1, M^k] and reject, you can use the remainder modulo N to generate additional randomness). Also note that for very large N, you might use a different approach like generating multiple N-sided numbers from one M^k sample.
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Von Neumann's trick, flip twice, if you get HT output 0, if TH output 1, otherwise repeat.
Start by explaining the von Neumann extractor: flip the biased coin twice, use HT as heads and TH as tails, and discard HH/TT. Prove fairness by showing P(HT) = P(TH) = p(1-p). Then compute the expected number of flips as 1/[2p(1-p)] by modeling the number of trials until a non-discarded pair appears. For extension to K, use rejection sampling: generate a random integer in [0, 2^n - 1] by flipping n times (mapping heads to 1 and tails to 0), and accept if the value is less than K; otherwise, discard and repeat. Choose n = ceil(log2 K) and prove uniformity by symmetry of the binary representation.
Pro tip: When discussing expected flips, emphasize that the method is optimal in terms of entropy (each fair bit costs 1/H(p) flips on average) and mention that for K, you can optimize by using the largest multiple of K less than 2^n to reduce rejection probability.
Clarify that the coin is biased with unknown p, and we need a fair coin flip without knowing p. Also, we need to extend to uniform sampling from 1 to K using only this biased coin.
Flip the coin twice. If the result is HT, output heads; if TH, output tails; if HH or TT, discard and repeat. This yields a fair outcome because P(HT) = P(TH) = p(1-p).
Show that conditional on a non-discarded pair, the probability of HT is 1/2. The number of pairs until success follows a geometric distribution with success probability 2p(1-p), so expected flips = 2 / [2p(1-p)] = 1/[p(1-p)]? Wait, careful: each trial is two flips, so expected flips = 2 / (2p(1-p)) = 1/(p(1-p)). Actually, success probability per pair is 2p(1-p), so expected pairs = 1/(2p(1-p)), expected flips = 2 * that = 1/(p(1-p)). But note: if p=0.5, expected flips = 2, which matches. However, the correct expected flips for von Neumann is 1/(p(1-p))? Let's check: For p=0.5, 1/(0.25)=4, but actually von Neumann with fair coin: each pair has success prob 0.5, so expected pairs = 2, expected flips = 4. Yes, so expected flips = 1/(p(1-p))? Wait, 1/(0.25)=4, correct. So expected flips = 1/(p(1-p)). But earlier I wrote 1/[2p(1-p)] for expected pairs? Let's recalc: success probability per pair = 2p(1-p). Expected number of pairs = 1/(2p(1-p)). Each pair is 2 flips, so expected flips = 2/(2p(1-p)) = 1/(p(1-p)). Yes. So expected flips = 1/(p(1-p)). But note: if p=0.5, expected flips = 4, which is correct for von Neumann. So the formula is 1/(p(1-p)). However, some sources say expected flips = 1/(2p(1-p))? That would be for expected pairs? Actually, expected flips = 1/(p(1-p)). Let's verify: For p=0.5, expected flips = 4. Yes. So the expected number of flips is 1/(p(1-p)). But wait, the question asks: 'find the expected number of flips as a function of p.' So answer: E = 1/(p(1-p)). But note: if p=0 or 1, infinite. So that's correct.
Use rejection sampling: let n = ceil(log2 K). Generate a random integer X in [0, 2^n - 1] by flipping n times (heads=1, tails=0). If X < K, output X+1; else discard and repeat. This yields uniform distribution over 1..K because each n-bit string is equally likely (by symmetry of the fair bits generated from the biased coin).
The expected number of fair bits needed is n * (2^n / K) (since acceptance probability is K/2^n). Each fair bit costs 1/(p(1-p)) flips on average, so total expected flips = n * (2^n / K) * 1/(p(1-p)). Mention that you can reduce rejection by using the largest multiple of K less than 2^n, or by using arithmetic coding for optimal efficiency.
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