The mem ers of the team are um ered from to When the game egins no further comm unications et een team mem ers are er mitted The game host creates oard with little do ors um ered on their fron ts from to Behind the do ors he also writes the um ers f ID: 23489 Download Pdf

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The mem ers of the team are um ered from to When the game egins no further comm unications et een team mem ers are er mitted The game host creates oard with little do ors um ered on their fron ts from to Behind the do ors he also writes the um ers f

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Random erm utations What are the prop erties of randomly generated erm utations of ob jects? Suc erm utations are sp ecial case of random graphs, whic ha applications in man elds including co ding theory and puzzles lik this one: Example team of 100 con testan ts ust ho ose strategy for the follo wing game. The mem ers of the team are um ered from to When the game egins, no further comm unications et een team mem ers are er- mitted. The game host creates oard with little do ors, um ered, on their fron ts, from to Behind the do ors, he also writes the um ers from to in

random erm utation. (So ehind do or um er 1, migh nd the hidden um er is 42; all erm utations are equally probable.) Eac hidden um er is written either in red ink or in blue ink, hosen indep enden tly and randomly Eac con testan is allo ed to op en up to do ors, lo oking at the um er hidden ehind eac h. (He ma ho ose his sequence of do ors in an manner, for example, in that dep ends on what he sees ehind the do ors he has op ened, or on his wn supply of random um ers.) Then all the con testan ts ust guess the colour of their hidden um er. or example, con testan um er 42 ust guess the

colour of the hidden um er 42, wherev er it is lo cated. The team wins big prize if al the con testan ts guess their hidden um ers colours correctly Find strategy that giv es the team substan tial probabilit of winning. The erm utation denes graph of ertices and directed edges. Ev ery ertex has one directed edge lea ving it, and one directed edge arriving at it. This graph is called cycle gr aph ecause it consists of one or more directed cycles. 1.1 One-cycles and -cycles Call the erm utation If the erm utation maps particular ertex to itself, i.e. then e’ll describ that little part of

the graph as one-cycle. Ex.1 Sho that the probabilit that particular ertex is in one-cycle is If the graph con tains only one long cycle, linking all no des, e’ll sa that the graph is an -cycle. Ex.2 Sho that the probabilit that the graph con tains one big -cycle is (n um er of cycles 1) Solution: Start from ertex and follo the erm utation. The probabilit that there is one long -cycle is the probabilit that: is not and is not and is not and is not and is not This probabilit is 1)

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1.2 -cycles e’v found that the probabilit that giv en ertex is in 1-cycle is and the probabilit

that it is in an -cycle is What ab out other lengths of cycles? Ex.3 What is the probabilit that giv en ertex is in an -cycle? Solution: require that: is not and is not and is not and is So is in an -cycle That’s nice simple result: the probabilit that is in an -cycle is indep enden of Ex.4 What is the exp ected length of the cycle in whic ertex nds itself Solution: The ertex is in cycle of length dra wn from with equal probabilit y; the erage of these um ers is 1) 2. There is probabilit of that ’s cycle has length et een and 2. 1.3 Pr obability of lar ge cycle or this section, let

cycle length greater than 2. e’ll call suc cycle large cycle. kno that the probabilit that there is cycle of the maxim um length is and that the probabilit that particular ertex is in an -cycle is What is the probabilit that there exists an -cycle? (Remem er that the -cycle do esn’t necessarily con tain ertex if is smaller than .) Call the set of ertices in the -cycle or an giv en erm utation, there can at most one suc set, since decided momen ago that as constrained to bigger than 2. No w, the sum rule, the probabilit that there exists one -cycle can decomp osed summing er all ossible subsets

of size (an -cycle exists con tains an -cycle (1) (2) The um er of terms in the sum, the um er of subsets, is So (an -cycle exists (3) Ex.5 What is the probabilit that there is cycle with length larger than 2? Solution: (large cycle 2+1 ln 69 (4) So the probabilit that the graph con tains no large cycles (where ‘large means ‘ha ving length bigger than 2 is 0.31. Ho do es this relate to the puzzle? Hop efully ou can complete that step no w. In case ou don’t an to the solution to the puzzle et, let’s discuss random erm utations little more.

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Num er of cycles, and their exp ected

lengths e’v found the exp ected length of the cycle con taining one particular ertex, and the proba- bilit that there are no large cycles. What do exp ect the numb er of -cycles to e? And what do exp ect the total numb er of cycles to e? What’s the probabilit distribution of that total um er? already kno the probabilit that the um er of cycles is 1. 2.1 Pr obability tha the number of cycles is It’s 2.2 Pr obability tha the number of cycles is Ex.6 Sho that the probabilit that all ertices are in one-cycles is (n um er of cycles [Hin t: it’s the probabilit that the erm utation is the iden tit

mapping.] 2.3 Pr obability tha the number of cycles is kno the answ ers for and Ex.7 What happ ens in et een? can’t see an easy to answ er this question. Ho ev er, can establish what the mean um er of cycles is. 2.4 Number of -cycles Ex.8 What is the exp ected um er of -cycles? Solution: The probabilit that is on an -cycle is If ask all ertices to sa ‘eek if they are on an -cycle, then the exp ected um er of ‘eeks will hear is 1. The eeks will come along in clumps of size eac clump corresp onding to single -cycle. So the exp ected um er of -cycles is (W sa this expression efore, when ask ed

for the pr ob ability that there is an -cycle, for greater than 2.) 2.5 Number of cycles Ex.9 What is the exp ected um er of cycles? Solution: The exp ected um er is exactly =1 =1 ln 2.6 vera ge cycle length already computed the erage length of the cycle con taining giv en ertex, and found it as 1) 2. But can dene another erage. (See the busstop parado and the rolling sixes examples in textb ok for further examples of these sorts of erage.) Ex.10 What is the erage length of cycle, selecting uniformly randomly from all cycles? Solution: The total length of edges is shared et een cycles.

So, eraging uniformly er all cycles in the ensem ble, the erage length is ln

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Solution to puzzle Here are some strategies the team migh adopt. Strategy Ev ery one op ens 50 random do ors. If ou don’t nd our wn hidden um er, ou guess its colour at random. The probabilit that giv en erson nds their um er is 2; the probabilit that they sa the righ colour is th us 4. The do ors op ened are indep enden t, so the probabilit of team success is (3 4) 100 10 13 Strategy Ev ery one op ens do ors 1{50. If ou don’t nd our wn hidden um er, ou guess its colour at

random. The probabilit that giv en erson nds their um er is 2; the probabilit that they sa the righ colour is 4. Ho ev er, is it at all lik ely that the team will all guess correctly? Exactly half of them are guaran teed to learn their wn um er’s colour the 50 of them whose um ers lie ehind do ors 1{50. The other 50 will certainly guess. The probabilit of team success is (1 2) 50 10 16 That’s orse than the random strategy strategy 1! It ould nice if there ere hance that all the eople ould learn their wn um er’s colour. Strategy The th erson op ens do ors 1){( 50), mo dulo If ou don’t

nd our wn hidden um er, ou guess its colour at random. What’s the probabilit of team success? ha en’t ork ed it out. exp ect it’s similar to strategy 1. Strategy Ho ab out using the erm utation to determine whic do ors are op ened? The th erson op ens do or rev ealing ); he then op ens the do or um er rev ealing ); and so on. No w, if ertex is on ‘short cycle (of length equal to or less) then he will, at the th step, op en do or that rev eals his um er Recall from section 1.2 that there is hance of that he is on cycle of length et een and 2. So there is hance that he will nd

his wn um er. If his ertex is on ‘long cycle, then he will, sadly nev er reac his wn um er; and neither will an of the other eople on the same long cycle. This is the secret of success of strategy 4: the shared source of randomness, causes failures of the participan ts to ositiv ely correlated. And if they are more lik ely to sim ultaneously fail; they ust also more lik ely to sim ultaneously succeed! The team succeeds if the erm utation con tains no ‘long cycle. This ccurs with probabilit 0.31. This is an addition to Information The ory, Infer enc e, and arning lgorithms (Cam bridge Univ.

Press, 2003), whic is ailable online from http://www.aims.ac.za/ acka y/

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