¬…       Complexity note

o      model assumes source of random bits

o      we will assume primitives: biased coins, uniform sampling

o      in homework, see equivalent

Adelman's Theorem.

Consider RP (one sided error)

• Does randomness help?
• In practice YES
• in one theory model, no
• in another, yes!
• in another, maybe
• Size n problems (2ⁿ of them)
• matrix of advice rows by input columns
• some advice row witnesses half the problems.
• delete row and all its problems
• remaining matrix still RP (all remaining rows didn't have witness)
• halves number of inputs. repeat n times.

Result: on RP of size n, exists n witnesses that cover all problems.

• polytime algorithm: try n witnesses.

o      Nonuniformity: witnesses not known.

o      RP $\subseteq$P/poly

oblivious versus nonoblivious adversary and algorithms.

Review tree evaluation. Moving LOE through a (linear) recurrence.

• define. algo cost is number of leaves. n = 2^h
• NOR model
• deterministic model: must examine all leaves. time 2^h = 4^h/2 = n
• by induction: on any tree of height h, as questions are asked, can answer such that root is not determined until all leaves checked.
• Note: bad instance being constructed on the fly as algorithm runs.
• But, since algorithm deterministic, bad instance can be built in advance by simulating algorithm.
• nondeterministic/checking
• T(0) = 1
• winning position can guess move. W(h) = L(h - 1)
• losing must check both. L(h) = 2W(h - 1)
• follows W(h) = 2*W(h - 2) = 2^h/2 = n^1/2
• randomized-guess which leaf wins.
• explain W(T) = L(T₁) + ${\frac{1}{2}}$L(T₂)
• T(0) = 1
• W(T) is a random variable: time it takes to verify T is a win. Ditto L(T).
• W(h) = max over all height-h winning trees of E[W(T)]
• Suppose T₁ wins.

$$\le$$

$$\approx$$

• 
  

Lower Bound

• Game Theory
• Zero sum games. Scissors Paper Stone. Roberta, Charles.
• Payoff Matrix M. Entries are (large) strategies. chess.
• Optimal strategies
• row wants to maximize, column to minimze
• suppose Roberta picks i. Guarantees $\min_{j}^{}$M_ij.
• (Pessimistic) R-optimal strategy: choose i to $\max_{i}^{}$$\min_{j}^{}$M_ij.
• (Pessimistic) C-optimal strategy: choose j to $\min_{j}^{}$$\max_{i}^{}$M_ij.
• When C-optimal and R optimal strategies match, gives solution of game.
• if solution exists, knowing opponents strategy useless.
• Sometimes, no solution using these pure strategies
• Randomization:
• mixed strategy: distribution over pure ones
• R uses dist p, C uses dist q, expected payoff p^TMq
• Von Neumann:

$$\max_{p}^{}$$$$\min_{q}^{}$$p^TMq = $$\min_{q}^{}$$$$\max_{p}^{}$$p^TMq

that is, always exists solution in mixed strategies.

• Once p fixed, exists optimal pure q, and vice versa
• Yao's minimax method:
• Column strategies algorithms, row strategies inputs
• payoff is expected running time
• randomized algorithm is mixed strategy
• optimum algorithm is optimum randomized strategy
• worst case input is corresponding optimum pure strategy
• Thus:
• worst case expected runtime of optimum rand. algorithm
• is payoff of game
• instead, consider randomized inputs
• payoff of game via optimum pure strategy
• which is detemrinistic algorithm!
• Worst case expected runtime of randomized algorithm for any input equals best case running time of a deterministic algorithm for worst distribution of inputs.
• Thus, for lower bound on runtime, show an input distribution with no good deterministic algorithm

¬…       Game tree evaluation.

o      input distribution: each node 1 with probability p = ${\frac{1}{2}}$(3 - $\sqrt{5}$).

o      every node is 1 with probability p

o      lemma: any deterministic alg showld finish evaluating one child of a node before doing other: depth first pruning algorithm

o      Such algorithm has probability p of finding 1 on first child, so

W(h) = W(h - 1) + (1 - p)W(h - 1) = (2 - p)^h = n^0.694

Game tree evaluation lower bound.

• Recall Yao's minimax principle.
• lemma: any deterministic alg showld finish evaluating one child of a node before doing other: depth first pruning algorithm. proof by induction.
• input distribution: each leaf 1 with probability p = ${\frac{1}{2}}$(3 - $\sqrt{5}$).
• every node is 1 with probability p
• let T(h) be expected number of leaves evaluated from height h.
• with probablity p, eval one child. else eval 2.
• So

T(h) = pT(h - 1) + 2(1 - p)T(h - 1) = (2 - p)^h = n^0.694

Stable marriage.

• complete preference lists
• stable if no two unmarried people prefer each other.
• med school
• always exists.
• proposal algorithm:
• rank men
• lowest unmarried man proposes in order of preference
• woman accepts if unmarried or prefers new proposal to current mate.
• Time Analysis:
• womean's state only improves with time
• only n improvements per woman
• while unattached man, proposals continue
• must eventually all be attached
• Stability Analysis
• suppose X-y are dissatisfied with pairing X-x, Y-y.
• X proposed to y first
• y prefers current Y to X.

¬…       Average case analysis

o      nonstandard for our course

o      random preference lists

o      how many proposals?

o      principle of deferred decisions

¬ß       random choices all made in advance

¬ß       random choices made algorithm needs them.

o      used while discussing autopartition, quicksort

o      Proposal algorithm:

¬ß       each proposal is random among unchosen women

¬ß       still hard

¬ß       Each proposal among all women

¬ß       stochastic domination

¬ß       done when all women get a proposal.

¬ß       at each step 1/n chance women gts proposal

o      Coupon collection.

¬ß       expected bound

¬ß       upper bound