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NU ACM Talk Virtual Scientific Communities for Driving Innovation and Learning Karl Lieberherr joint work with Ahmed Abdelmeged and Bryan Chadwick 11/28/20151SCG.

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Presentation on theme: "NU ACM Talk Virtual Scientific Communities for Driving Innovation and Learning Karl Lieberherr joint work with Ahmed Abdelmeged and Bryan Chadwick 11/28/20151SCG."— Presentation transcript:

1 NU ACM Talk Virtual Scientific Communities for Driving Innovation and Learning Karl Lieberherr joint work with Ahmed Abdelmeged and Bryan Chadwick 11/28/20151SCG Innovation Supported by Novartis and GMO

2 Introduction Scientific Community Game(X) [SCG(X)] – Goal: Foster innovation and learning in some domain X A virtual scientific community consists of virtual scholars that propose and oppose hypotheses maximizing their reputations Applications: Learning and innovation through focused interaction, “Netflix in the small” 11/28/2015Innovation2

3 How to model a scholar? Solve problems Provide hard problems Propose hypotheses about Solve and Provide (Introspection) Oppose hypotheses – Strengthen hypotheses – Refute hypotheses Supported opposing failed Refuted opposing succeeded 11/28/20153Innovation

4 Where SCG comes from J ACM 11/28/2015SCG Innovation4 Lieberherr/Specker 1981

5 Outline Introduction (done) Highest safe rung example SCG Scholar / Agent SCG Agent in Action Highest safe rung example (opposition) Who is the winner? Competition and collaboration Disadvantages of SCG Further Examples SCG-based Software Development Process Conclusions 11/28/2015SCG Innovation5

6 Example: Jar Stress Testing You have a ladder with r rungs, and you want to find the highest rung from which you can drop a copy of the jar and not have it break. We call this the highest safe rung problem (r,b). How many experiments do you need? Minimize. (r,infinity) (r,1) 11/28/20156SCG

7 Highest Safe Rung Problem Problems and Solutions Problems: p=((r,b),secret hsr), secret hsr in [0,r], r,b natural numbers r = number of rungs b = number of jars that are allowed to break (r,b) is called a niche Solutions: sequence of queries of the form n? to find hsr. Responses: yes/no. Quality of solution: q = length of sequence of queries 11/28/20157SCG Innovation

8 Highest Safe Rung Problem Hypotheses Alice claims the hypothesis: I can solve any problem p=((r,b),secret hsr) with quality q: abbreviated H = ((r,b),q) Problems to be delivered for H = ((r,b),q) are of the form ((r,b), s). Important: A hypothesis defines a family of problems. Propose: Hypotheses H1 = ((25,2),11), H2 = ((25,2),6) (from Kleinberg/Tardos) 11/28/20158SCG Innovation

9 Scholars propose and oppose 11/28/2015Innovation9 HA1 HA2 HA3 HA4 egoistic Alice egoistic Bob Bob increases his reputation HB1 HB2 opposes (1) provides problem (2) solves problem not as well as she expected based on HA2 (3) WINS! LOSES proposed hypotheses social welfare Life of a scholar: (propose+ oppose+ provide* solve*)*

10 What is the purpose of SCG? The purpose of playing an SCG(X) competition is to assess the "skills" of the agents in: – solving problems in domain X, – making good predictions about niches in domain X, – finding the hardest problems in a specific niche 11/28/2015Innovation10

11 What is SCG(X) 11/28/2015Innovation11 no automation human plays full automation agent plays degree of automation used by scholar some automation human plays 0 1 more applications: test constructive knowledge transfer to reliable, efficient software agent Bob agent Alice

12 What is SCG(X)? Teams Design Problem Solver Develop Software Deliver Agent Agent AliceAgent Bob Administrator SCG police I am the best No!! Let’s play constructively 11/28/201512Innovation Team Alice Team Bob

13 For agents: Full Round Robin Tournaments or Swiss-Style Agents to play the SCG(X). Repeat a few times with feedback used to update agents. Within the group of participating agent, the winning agent has the – best solver for X-problems – best supported knowledge about X 11/28/201513Innovation

14 SCG in Action: Competitions http://www.ccs.neu.edu/home/lieber/courses /cs4500/f09/files/competitions/past_competit ions/11_23/tournament_1/final_results_tour nament_2009_11_24_12_03_41.html http://www.ccs.neu.edu/home/lieber/courses /cs4500/f09/files/competitions/past_competit ions/11_23/tournament_1/final_results_tour nament_2009_11_24_12_03_41.html http://www.ccs.neu.edu/home/lieber/courses /cs4500/f09/files/competitions/past_competit ions/10_22/tournament_1/final_results_tour nament_2009_10_23_04_35_18.html http://www.ccs.neu.edu/home/lieber/courses /cs4500/f09/files/competitions/past_competit ions/10_22/tournament_1/final_results_tour nament_2009_10_23_04_35_18.html 11/28/2015Innovation14

15 Highest Safe Rung Problem opposing opposing(refuting, strengthening) Alice claims: Hypothesis ((25,2),5) – Bob opposes it by refuting it: Bob invents problem ((25,2), secret 22). Alice: 5? no, 10? yes, 6? no, 7? no, 8? no. Already 5 questions asked and answer still unknown. Alice’ claim is refuted. Alice claims: Hypothesis ((25,2),12) – Bob opposes it by strengthening it to ((25,2),9); and he can successfully support this hypothesis 11/28/201515SCG Innovation

16 Highest Safe Rung Problem supporting Alice claims: Hypothesis ((25,2),12) – Bob tries to discount but Alice supports it: Alice: 5? no, 10? no, 15? no, 20? no, 25? yes, 21? no, 22? no, 23? no, 24? yes. Only 9 questions asked and problem ((25,2), secret 23) is solved. Alice has supported her hypothesis. 11/28/2015SCG Innovation16

17 Who wins? Alice or Bob? Reputation of Alice = – the number of hypotheses that Alice proposed that were never successfully opposed by Bob (neither refuted nor strengthened) + – the number of hypotheses that Bob proposed that were successfully opposed by Alice RA = HAnotOpposedB + HBOpposedA The scholar with the highest reputation wins encourages: creating strong knowledge and discounting knowledge created by others 11/28/201517SCG Innovation Motivated by real scientific community

18 competitive / collaborative 11/28/2015Innovation18 Agent Alice: claims hypothesis H Agent Bob: opposes H, refutes: provides evidence for !H Alice wins knowledge Bob wins reputationmakes public knowledge

19 Highest Safe Rung Problem competition / collaboration Alice claims: Hypothesis ((25,2),12) – Bob tries to discount but Alice supports it: Alice: 5? no, 10? no, 15? no, 20? no, 25? yes, 21? no, 22? no, 23? no, 24? yes. – From this exchange which is prompted by Alice defending her reputation, Bob gets an idea: For problem: p=((r,b),secret hsr), consider f(r,q) =(r/q + q) and find a q so that f(25,q) is minimized. f(25,5)=10; f(25,6)=11;f(25,4)=11. – From this idea Bob knows that he can strengthen the hypothesis to ((25,2),10) – General solution: Given r, find q to minimize (r/q + q). 11/28/2015SCG Innovation19

20 Scholars and Agents: Same rules Are encouraged to 1.offer results that are not easily improved. 2.offer results that they can successfully support. 3.strengthen results, if possible. 4.expose results that are wrong. 5.stay active and publish new results. 6.be well-rounded: solve posed problems and pose difficult problems for others. 7.become famous! 11/28/201520Innovation

21 Soundness Theorem SCG is sound: The agent with the best algorithms / knowledge wins (there is no way to cheat) – best: within the group of participating agents 11/28/2015Innovation21

22 Highest Safe Rung Problem Asymptotic Hypotheses Alice claims the hypothesis: I can solve any problem p=((r,b),secret hsr) with quality f(r,b) : abbreviated H = ((r,b),f(r,b)) Problems to be delivered for H = ((r,b),f(r,b)) are of the form ((r,b), secret hsr). Propose: Hypotheses H1 = ((r,b),(log(r)) b ), H2 = ((r,b),r 1/b ) 11/28/201522SCG Innovation

23 Highest Safe Rung Problem discounting asymptotic hypothesis discounting (refuting, strengthening) Alice claims: Hypothesis ((r,b),(b*log(r))) – Bob discounts it by refuting it: Bob invents problem ((1024,2), secret hsr). log(1024) = 10. 20 questions are not enough! Alice: 30? no, 60? yes, 31? no, 32? no, etc.. Already 20 questions asked and answer still unknown. Alice’ claim is refuted. Alice claims: Hypothesis ((r,2),r/2) – Bob discounts it by strengthening it to ((r,2),2*r ½ ); and he can successfully support this hypothesis. 11/28/201523SCG Innovation

24 Disadvantages of SCG The game is addictive. After Bob having spent 4 hours to fix his agent and still losing against Alice, Bob really wants to know why! Overhead to learn to define and participate in competitions. The administrator for SCG(X) must perfectly supervise the game. Includes checking the legality of X-problems. – if admin does not, cheap play – watching over the admin 11/28/201524Innovation

25 How to compensate for those disadvantages Warn the scholars. Use a gentleman’s security policy: report administrator problems, don’t exploit them to win. Occasionally have a non-counting “attack the administrator” competition to find vulnerabilities in administrator. – both generic as well as X-specific vulnerabilities. 11/28/201525Innovation

26 GIGO: Garbage in / Garbage out If all agents are weak, no useful solver created. 11/28/2015Innovation26

27 Physics Maximum Height Problem Problems and Solutions Problems: p=(v, a), v, a: positive real numbers The maximum height obtained by a projectile launched with speed v at angle a to the horizontal is z. Solutions: real number z. Quality of solution: Number of correct decimal places. 11/28/201527SCG Innovation

28 Physics Maximum Height Problem Hypotheses Alice claims the hypothesis: I can solve any maximum height problem p=(v,a) with quality q in 1 minute: abbreviated H = (MHP,q) Problems to be delivered for H = (MHP,q) are of the form (v,a). Propose: Hypotheses H1 = (MHP,3), H2 = (MHP,6) 11/28/201528SCG Innovation http://scienceworld.wolfram.com/physics/Height.html

29 Physics Maximum Height Problem discounting discounting (refuting, strengthening) Alice claims: Hypothesis (MHP,3) – Bob discounts it by refuting it: Bob invents problem (25,60 degrees). Alice fails to solve the problem in 1 minute with 3 correct digits. Alice’ claim is refuted. Checking is done by experiment or trusted third party. Alice claims: Hypothesis (MHP,1) – Bob discounts it by strengthening it to (MHP,2); and he can successfully support this hypothesis 11/28/201529SCG Innovation

30 RegExpToAutomata Problem Problems and Solutions Problems: p=(r,n); r a regular expression of size n. r = regular expression; a + b* a + a a a b* n defines a niche of regular expressions Solutions: DFA d equivalent to r. Quality of solution: Number of states of d. 11/28/201530SCG Innovation

31 RegExpToAutomata Problem Problems and Solutions Problems: p=(r,n); r a regular expression of size n. r = regular expression; a + b* a + a a a b* n defines a niche of regular expressions Solutions: DFA d equivalent to r. Quality of solution: Number of states of d. 11/28/201531SCG Innovation

32 RegExpToAutomata Problem Hypotheses Alice claims the hypothesis: I can solve any problem p=(r,n) with quality q or less: abbreviated H = (n,q) Problems to be delivered for H = (n,q) are of the form p=(r,n). Important: A hypothesis defines a family of problems. Propose: Hypotheses H1 = (5,11), H2 = (5,10) 11/28/201532SCG Innovation

33 RegExpToAutomata Problem opposing opposing(refuting, strengthening) Alice claims: Hypothesis (5,11) – Bob discounts it by refuting it: Bob invents a regular expression r of size 5, gives it to Alice and she fails to deliver a DFA d with 11 or fewer states. Alice’ claim is refuted. Alice claims: Hypothesis (5,20) – Bob discounts it by strengthening it to (5,19); and he can successfully support this hypothesis 11/28/201533SCG Innovation

34 RegExpToAutomata Problem supporting Alice claims: Hypothesis (4,12) – Bob tries to discount but Alice supports it: Bob gives to Alice a regular expression r of size 4. Alice provides and equivalent DFA with 12 or fewer states. Alice has supported her hypothesis. 11/28/2015SCG Innovation34

35 Who wins? Alice or Bob? Reputation of Alice = – the number of hypotheses that Alice proposed that were never successfully opposed by Bob (neither refuted nor strengthened) + – the number of hypotheses that Bob proposed that were successfully opposed by Alice. RA = HAnotOpposedB + HBOpposedA The scholar with the highest reputation wins. encourages: creating minimum automata for regular expressions of a given size. 11/28/201535SCG Innovation

36 Software Development Process Increase targeted interaction between software developers. 11/28/2015Innovation36

37 Traditional Approach Human Developers Develop new software for problem solving domain X Static Evaluation. No competition. human1human2 Testing unit testing integration testing Benchmark is used to evaluate software human3human4 Users Requirements for X 37SCG-SP201011/28/2015

38 Why Software Development through a virtual scientific community? Human Developers Develop new software for problem solving domain X SCG(X) Erika-Patrick-agent winning-agent Evaluates fairly, frequently, constructively and dynamically. Drives innovation. Challenges humans. Agents point humans to what needs attention in the software. human1human2 ErikaPatrick Benchmark is used to evaluate software Users Requirements for X 38SCG-SP201011/28/2015

39 Erika-Patrick Agent Surrogate of combined knowledge of Erika and Patrick successfully transferred to agent. Transfer knowledge by programming. 39SCG-SP201011/28/2015

40 Conclusions How to make learning and problem solving fun: design a game and interact. Scientific Community Game = Specker Challenge Game = SCG How to create reliable problem solving software? Have it tested through SCG. 11/28/2015Innovation40

41 Final Slide More Questions? 11/28/2015Innovation41

42 11/28/2015SCG Innovation42

43 SCG concepts Scholars working in a domain with niches. Define functions on niches. Hypotheses: claims about functions on niches: – Discounting protocol for HA: Alice selects niche element ne and Bob applies fBob so that claim about function does not hold – Strengthening protocol Reputation 11/28/201543SCG Innovation

44 SCG concepts Scholars working in a domain with niches. Function f: Niche -> S for Alice and Bob. Hypotheses: claims about niches: belief: f has property b(s, dn, fdn). (Niche,Belief) – Discounting protocol: Alice selects niche element ne and Bob applies fBob creating s, so that !b(s,ne) – Strengthening protocol Reputation 11/28/201544SCG Innovation

45 Hypothesis Structure Algorithm Solver: Problems -> Solutions For all p in Problems with feature f in Features algorithm Solver solves p using resources p(f) with quality(p,Solver(p),f). Algorithm Provider: Features -> Problems For feature f, Algorithm Provider provides a problem p, for all solutions of p, !quality(p,Solver(p),f). 11/28/201545SCG Innovation

46 Two person SCG Alice, Bob Domain: Source, Target; fA, fB-> Source-> Target; Source defined by niche predicate. Hypotheses HA (HB): claims about fA (fB) Discounting protocol for HA: – Bob provides element ne in Source so that fA(ne) contradicts HA. – Alice provides element ne in Source so that fB(ne) contradicts HA. Strengthening protocol – Bob proposes HB, HA => HB and Alice cannot discount HB. 11/28/201546SCG Innovation

47 Two person SCG Specialize for problem solving Alice, Bob Domain: Problems -> Solutions Hypotheses Discounting protocol for hypothesis HA by Alice: – Bob attacks in one of two ways (depends on HA) Bob provides a problem for which Alice constructs a solution that contradicts HA. Alice provides a problem for which Bob constructs a solution that contradicts HA. Strengthening protocol – Bob proposes HB, HA => HB and Alice cannot discount HB. 11/28/201547SCG Innovation

48 Hypotheses Solution algorithm A: Problems->Solutions For all elements p in Problems that have feature F and a secret solution ss(p), algorithm A(p) constructs with resource constraint Prediction(F) an element s(p) in set Solutions(p) with property Q(p,s(p),ss(p),F). 11/28/201548SCG Innovation

49 Hypotheses Problem creation algorithm A-1 11/28/201549SCG Innovation

50 Discounting protocol 11/28/201550SCG Innovation

51 SCG by Example Highest safe rung problem Speed prediction problem graph diameter / average pair-wise distance 11/28/201551SCG Innovation

52 Example: Jar Stress Testing You have a ladder with r rungs, and you want to find the highest rung from which you can drop a copy of the jar and not have it break. We call this the highest safe rung problem (r,b). How many experiments do you need? Minimize. (r,infinity) (r,1) 11/28/201552SCG

53 Highest Safe Rung Problem Problems and Solutions Problems: p=((r,b),secret hsr), secret hsr in [0,r], r,b natural numbers r = number of rungs b = number of jars that are allowed to break (r,b) is called a niche Solutions: sequence of queries of the form n? to find hsr. Responses: yes/no. Quality of solution: q = length of sequence of queries 11/28/201553SCG Innovation

54 Highest Safe Rung Problem Hypotheses Alice claims the hypothesis: I can solve any problem p=((r,b),secret hsr) with quality q: abbreviated H = ((r,b),q) Problems to be delivered for H = ((r,b),q) are of the form ((r,b), s). Important: A hypothesis defines a family of problems. Propose: Hypotheses H1 = ((25,2),11), H2 = ((25,2),6) (from Kleinberg/Tardos) 11/28/201554SCG Innovation

55 Highest Safe Rung Problem opposing opposing(refuting, strengthening) Alice claims: Hypothesis ((25,2),5) – Bob opposes it by refuting it: Bob invents problem ((25,2), secret 22). Alice: 5? no, 10? yes, 6? no, 7? no, 8? no. Already 5 questions asked and answer still unknown. Alice’ claim is refuted. Alice claims: Hypothesis ((25,2),12) – Bob opposes it by strengthening it to ((25,2),9); and he can successfully support this hypothesis 11/28/201555SCG Innovation

56 Highest Safe Rung Problem supporting Alice claims: Hypothesis ((25,2),12) – Bob tries to discount but Alice supports it: Alice: 5? no, 10? no, 15? no, 20? no, 25? yes, 21? no, 22? no, 23? no, 24? yes. Only 9 questions asked and problem ((25,2), secret 23) is solved. Alice has supported her hypothesis. 11/28/2015SCG Innovation56

57 Who wins? Alice or Bob? Reputation of Alice = – the number of hypotheses that Alice proposed that were never successfully discounted by Bob (neither refuted nor strengthened) + – the number of hypotheses that Bob proposed that were successfully discounted by Alice RA = HAnotDiscountedB + HBdiscountedA The scholar with the highest reputation wins encourages: creating strong knowledge and discounting knowledge created by others 11/28/201557SCG Innovation Motivated by real scientific community

58 Highest Safe Rung Problem competition / collaboration Alice claims: Hypothesis ((25,2),12) – Bob tries to discount but Alice supports it: Alice: 5? no, 10? no, 15? no, 20? no, 25? yes, 21? no, 22? no, 23? no, 24? yes. – From this exchange which is prompted by Alice defending her reputation, Bob gets an idea: For problem: p=((r,b),secret hsr), consider f(r,q) =(r/q + q) and find a q so that f(25,q) is minimized. f(25,5)=10; f(25,6)=11;f(25,4)=11. – From this idea Bob knows that he can strengthen the hypothesis to ((25,2),10) – General solution: Given r, find q to minimize (r/q + q). 11/28/2015SCG Innovation58

59 Highest Safe Rung Problem Asymptotic Hypotheses Alice claims the hypothesis: I can solve any problem p=((r,b),secret hsr) with quality f(r,b) : abbreviated H = ((r,b),f(r,b)) Problems to be delivered for H = ((r,b),f(r,b)) are of the form ((r,b), secret hsr). Propose: Hypotheses H1 = ((r,b),(log(r)) b ), H2 = ((r,b),r 1/b ) 11/28/201559SCG Innovation

60 Highest Safe Rung Problem discounting asymptotic hypothesis discounting (refuting, strengthening) Alice claims: Hypothesis ((r,b),(b*log(r))) – Bob discounts it by refuting it: Bob invents problem ((1024,2), secret hsr). log(1024) = 10. 20 questions are not enough! Alice: 30? no, 60? yes, 31? no, 32? no, etc.. Already 20 questions asked and answer still unknown. Alice’ claim is refuted. Alice claims: Hypothesis ((r,2),r/2) – Bob discounts it by strengthening it to ((r,2),2*r ½ ); and he can successfully support this hypothesis. 11/28/201560SCG Innovation

61 Physics Maximum Height Problem Problems and Solutions Problems: p=(v, a), v, a: positive real numbers The maximum height obtained by a projectile launched with speed v at angle a to the horizontal is z. Solutions: real number z. Quality of solution: Number of correct decimal places. 11/28/201561SCG Innovation

62 Physics Maximum Height Problem Hypotheses Alice claims the hypothesis: I can solve any maximum height problem p=(v,a) with quality q in 1 minute: abbreviated H = (MHP,q) Problems to be delivered for H = (MHP,q) are of the form (v,a). Propose: Hypotheses H1 = (MHP,3), H2 = (MHP,6) 11/28/201562SCG Innovation http://scienceworld.wolfram.com/physics/Height.html

63 Physics Maximum Height Problem discounting discounting (refuting, strengthening) Alice claims: Hypothesis (MHP,3) – Bob discounts it by refuting it: Bob invents problem (25,60 degrees). Alice fails to solve the problem in 1 minute with 3 correct digits. Alice’ claim is refuted. Checking is done by experiment or trusted third party. Alice claims: Hypothesis (MHP,1) – Bob discounts it by strengthening it to (MHP,2); and he can successfully support this hypothesis 11/28/201563SCG Innovation

64 RegExpToAutomata Problem Problems and Solutions Problems: p=(r,n); r a regular expression of size n. r = regular expression; a + b* a + a a a b* n defines a niche of regular expressions Solutions: DFA d equivalent to r. Quality of solution: Number of states of d. 11/28/201564SCG Innovation

65 RegExpToAutomata Problem Problems and Solutions Problems: p=(r,n); r a regular expression of size n. r = regular expression; a + b* a + a a a b* n defines a niche of regular expressions Solutions: DFA d equivalent to r. Quality of solution: Number of states of d. 11/28/201565SCG Innovation

66 RegExpToAutomata Problem Hypotheses Alice claims the hypothesis: I can solve any problem p=(r,n) with quality q or less: abbreviated H = (n,q) Problems to be delivered for H = (n,q) are of the form p=(r,n). Important: A hypothesis defines a family of problems. Propose: Hypotheses H1 = (5,11), H2 = (5,10) 11/28/201566SCG Innovation

67 RegExpToAutomata Problem discounting discounting (refuting, strengthening) Alice claims: Hypothesis (5,11) – Bob discounts it by refuting it: Bob invents a regular expression r of size 5, gives it to Alice and she fails to deliver a DFA d with 11 or fewer states. Alice’ claim is refuted. Alice claims: Hypothesis (5,20) – Bob discounts it by strengthening it to (5,19); and he can successfully support this hypothesis 11/28/201567SCG Innovation

68 RegExpToAutomata Problem supporting Alice claims: Hypothesis (4,12) – Bob tries to discount but Alice supports it: Bob gives to Alice a regular expression r of size 4. Alice provides and equivalent DFA with 12 or fewer states. Alice has supported her hypothesis. 11/28/2015SCG Innovation68

69 Who wins? Alice or Bob? Reputation of Alice = – the number of hypotheses that Alice proposed that were never successfully discounted by Bob (neither refuted nor strengthened) + – the number of hypotheses that Bob proposed that were successfully discounted by Alice. RA = HAnotDiscountedB + HBdiscountedA The scholar with the highest reputation wins. encourages: creating minimum automata for regular expressions of a given size. 11/28/201569SCG Innovation

70 Calculus Maximization Problem Problems and Solutions Problems: p=(f: function in one variable,J interval); Solutions: maximum of f in interval I. 11/28/201570SCG Innovation

71 Calculus Maximization Problem Hypotheses Alice claims the hypothesis(Polynomial, k): I can solve any problem p=(f,J) for f a polynomial in time (size of the polynomial)^k. H =(Polynomial, k). Problems to be delivered for H = (Polynomial, k) are of the form p=(f,J), f a polynomial. Important: A hypothesis defines a family of problems. Propose: Hypotheses H2 = (Polynomial, 2), H1 = (Polynomial,1). 11/28/201571SCG Innovation

72 Calculus Maximization Problem discounting discounting (refuting, strengthening) Alice claims: Hypothesis (Polynomial,1) – Bob discounts it by refuting it: Bob invents a polynomial (e.g., x^2 – x + 1) in one variable and an interval, gives them to Alice and she fails to deliver, in the given time, the maximum of the polynomial in the interval. Alice’ claim is refuted. Alice claims: Hypothesis (Polynomial,3) – Bob discounts it by strengthening it to (Polynomial,2); and he can successfully support this hypothesis 11/28/201572SCG Innovation

73 I claim I can solve this problem with one program that runs in time t on a single core machine and that runs in time 1.2 * t/c on a machine with c>1 cores. 11/28/2015SCG Innovation73

74 SCG Many kinds of hypotheses. They are defined by – Problems, Solutions – Discounting protocol Refuting protocol Strengthening protocol Problems and solutions to be exchanged in protocols 11/28/2015SCG Innovation74

75 Hypotheses Alice constructive claims I can solve problems of kind k – with quality q – close to your quality – better than you I claim statement S of the form – ForAllExists – ExistsForAll 11/28/2015SCG Innovation75

76 11/28/2015SCG Innovation76

77 Extra: too complex 11/28/2015SCG Innovation77

78 RegExpToAutomata Problem Problems and Solutions Problems: p=(function in two variables f(t,b); interval for t; interval for b). n defines a niche of regular expressions Solutions: min max solution. Quality of solution: Number of states of d. 11/28/201578SCG Innovation

79 Minimizing and Maximizing Functions Problems and Solutions Problems: minimizing and maximizing functions. Solutions: correct values. 11/28/201579SCG Innovation

80 Minimizing and Maximizing Functions Hypotheses Alice claims the hypothesis: function in two variables f(t,b); interval for t; interval for b. min [t] max [b] < h. H = (f(t,b),It,Ib,h) I can solve any problem p=(r,n) with quality q or less: abbreviated H = (n,q) Problems to be delivered for H = (n,q) are of the form p=(r,n). Important: A hypothesis defines a family of problems. Propose: Hypotheses H1 = (5,11), H2 = (5,10) 11/28/201580SCG Innovation

81 11/28/2015SCG Innovation81 Calculus Alice claims the hypothesis: min t max b f(t,b) < 0.8. t and b are vectors in a subset of some vector space. Bob opposes Alice' hypothesis by strengthening it: min t max b f(t,b) < 0.7. Alice opposes Bob's hypothesis by strengthening it further: min t max b f(t,b) < 0.6. Bob opposes Alice' hypothesis by challenging it. Alice provides t=t0 and Bob finds b=b0 and it turns out that f(t0,b0)=0.65. Therefore Bob wins reputation from Alice.

82 Highest Safe Rung Problem Problems and Solutions Problems: p=((r,b),secret hsr), secret hsr in [0,r], r,b natural numbers r = number of rungs b = number of jars that are allowed to break (r,b) is called a niche Solutions: sequence of queries of the form n? to find hsr. Responses: yes/no. Quality of solution: q = length of sequence of queries 11/28/201582SCG Innovation p=((r,b),floating)


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