Lambdia

Computational complexity

3 articles
P versus np problemNp completeNp complexityP complexityR complexityPolynomial hierarchyExponential hierarchyPolynomial timeExponential timeCooks theoremTime hierarchy theoremSpace hierarchy theoremBoolean satisfiability problemTrue quantified boolean formulaCounting problem complexityInteractive proof systemNatural proofTime complexitySpace complexityP vs npNp completenessAssignment problemCnf satCollision problemConstraint satisfactionExact coverStable matching problemSubgraph isomorphism problemSubset sum problemTraveling salesman problemParameterized approximation algorithmPadding argumentAanderaa karp rosenberg conjectureAnalysis of algorithmsApproximation algorithmAsymptotic computational complexityAveraging argumentBest worst and average caseBoolean circuitCircuit complexityClaw finding problemCobhams thesisCommunication complexityComputation treeComputational complexity of mathematical operationsComputational complexity of matrix multiplicationConfiguration graphComputational resourceComputing the permanentDecision tree modelExistential theory of the realsGap hamming problemGeneric case complexityGeometric complexity theoryGraph isomorphism problemHamiltonian complexityHardness of approximationHartmanis stearns conjectureInformation based complexityLog rank conjectureParameterized complexityPebble gamePseudo polynomial transformationQuasi polynomial growthSmoothed analysisStrong np completenessSwitching lemmaTractable problemUnique games conjectureWeak np completenessYaos principleCircuit computer scienceBig o notation

Twelve Marbles, Three Weighings, and Twenty-Seven Ways to Land

Twenty-four possible answers against 3^3 = 27 outcome sequences leaves just enough room, and four against four splits those answers into exactly 8, 8 and 8. Six against six always tips, so it wastes the level outcome and leaves twelve answers for nine remaining sequences, which makes halving impossible rather than merely slow. The counting argument bounds outcome sequences rather than strategies, so it rules out every adaptive continuation at once, and the explicit schedule closes the positive half.

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One Gold Bar, Two Cuts, Seven Evenings: 1, 2 and 4

Seven pieces cost six cuts and the schedule pays correctly, so the six-cut answer breaks one constraint and nothing else. Because the worker can hand pieces back, the contract is on his holding rather than on the transfer, and the ledger turns out to be a three-bit counter. Brute force finds 1-2-4 is the only three-piece solution.

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23 personnes suffisent pour un anniversaire commun, et 253 paires l'expliquent

La réponse réflexe tourne autour de 180, la moitié du calendrier. Le vrai seuil est 23, parce qu'une coïncidence demande une paire et que 23 personnes en portent 253. Le même raisonnement met la question « quelqu'un partage-t-il MON anniversaire » à 253 personnes, onze fois plus de monde, et le seuil en racine carrée derrière les deux explique qu'un identifiant aléatoire de 64 bits se répète après cinq milliards de tirages et non dix-huit trillions.

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