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> (define r .8)
> (for/last ([p '(a b c d)][f '(.2 .5 .7 1)] #:break (< r f)) p)
'c

Or WORSE: 

> (define r (random))
> r
0.011105628290672482
> (for/last ([p '(a b c d)][f '(.2 .5 .7 1)] #:break (< r f)) p)
#f




On Oct 13, 2015, at 3:37 PM, Nguyen Linh Chi <nguyenlinhch...@gmail.com> wrote:

> the list 0 isnt a bug. I add it because #:break will break the fitness vector 
> at r < f.
> 
> For example, 
> Population : a b c d
> Cumulative fitness .2 .5 .7 1
> 
> If the random r = .8, it means that the lottery points to the interval of 
> automaton d. But #:break will breaks at .7, and the associated automaton is 
> falsely identified as automaton c.
> 
> (Hm, right?)
> 
> On 13/ott/2015, at 20:21, Nguyen Linh Chi <nguyenlinhch...@gmail.com> wrote:
> 
>> Hi Mathias, thank you so much for helping me a lot.
>> You can use the code as you want, i still have tons of them on my github. 
>> (Just dont use it against me ok :D )
>> 
>> I'd try to see through your list of suggestions.
>> 
>> About the automata payoff.
>> 
>> There are 2 different things that are always mixed up: round and cycle. In 
>> each cycle, automata are matched in pair and play a game for r rounds per 
>> match. At the end of each cycle, there is a replacement period (respawning 
>> the fittest). In what im doing, the payoff after r rounds is transformed 
>> into the fitness, and the strongest automata (biggest payoff) survive. In 
>> some sense they already carry on their relative payoff with them just 
>> because they survive.
>> 
>> Make the automaton carrying their absolute payoff history over cycles would 
>> be intereting to see. I've never thought of that and the coding effort is 
>> also a reason i havent tried.
>> 
>> On 13/ott/2015, at 19:59, Matthias Felleisen <matth...@ccs.neu.edu> wrote:
>> 
>>> 
>>> p.s. One more question. Why do you 'reset' the payoff for automata after 
>>> each round? Shouldn't they carry along their overall, historical payoff? 
>>> 
>>> 
>>> 
>>> 
>>> On Oct 13, 2015, at 1:40 PM, Matthias Felleisen <matth...@ccs.neu.edu> 
>>> wrote:
>>> 
>>>> 
>>>> 1. Your code is a good example for something that we could use as a 
>>>> benchmark, and working on it has helped me find a couple of problems in 
>>>> drracket and typed racket. Thanks for triggering my activity. I'll stop 
>>>> soon. 
>>>> 
>>>> 2. I figured out that your code was general so you could accommodate more 
>>>> situations than the ones you explored. My interest diverges from yours. To 
>>>> me it is now just a cute benchmark. You will be able to use my insights to 
>>>> speed up your performance: 
>>>> 
>>>> -- factor the program into separate modules/files, 
>>>> -- make the modules independent of the representations, 
>>>> -- change automata and population representations to use vectors instead 
>>>> of lists, 
>>>> -- avoid mutation, 
>>>> -- eliminate all situations where you use (append l (list x)); instead use 
>>>> cons and reverse the result at the end. 
>>>> 
>>>> That way you can run the program 6 times faster. 
>>>> 
>>>> 3. Otherwise I think your program needs real testing and I think there are 
>>>> two bugs: 
>>>> 
>>>> a: I created a pull request for this one: 
>>>> 
>>>> 
>>>> (define (accumulated-payoff-percentages payoff-list)
>>>> (define payoff-sum (apply + payoff-list))
>>>> (define-values (accumulated _)
>>>>  (for/fold ([accumulated (list 0)] ;; change to '(); otherwise you get a 
>>>> list that's too long; your other for loop accidentally compensates for 
>>>> this bug 
>>>>             [init 0])
>>>>            ([y (in-list payoff-list)])
>>>>    [define next-init (+ init (/ y payoff-sum))]
>>>>    (values (cons next-init accumulated) next-init)))
>>>> (reverse accumulated))
>>>> 
>>>> b: I don't know how to fix this one properly: 
>>>> 
>>>> (define (randomise-over-fitness population fitness speed)
>>>> (for/list ([n (in-range speed)])
>>>> [define r (random)]
>>>> (define candidate
>>>>  (for/and ([p (in-vector population)]
>>>>             [f (in-list fitness)]
>>>>             #:break (< r f))
>>>>    p))
>>>> candidate))
>>>> 
>>>> replace last line with something like this: 
>>>> (or candidate (vector-ref population 0))))
>>>> 
>>>> Otherwise there is a non-0 chance otherwise that you get a boolean value 
>>>> eventually. I am not sure which default value you want to throw in. 
>>>> 
>>>> 4. You're right:
>>>> 
>>>> -- I re-created the match-pair function when I looked over my code. 
>>>> -- using in-list and in-vector and in-range speeds up for loops. 
>>>> -- types are irrelevant to your project. They helped me. 
>>>> 
>>>> I have pushed the last changes for now. Can we have your permission to use 
>>>> this code for a benchmark in our research world? Thanks. 
>>>> 
>>>> 
>>>> 
>>>> 
>>>> 
>>>> 
>>>> 
>>>> On Oct 13, 2015, at 12:00 PM, Linh Chi Nguyen <nguyenlinhch...@gmail.com> 
>>>> wrote:
>>>> 
>>>>> Matthias you work like a machine, too fast.
>>>>> 
>>>>> Anyway, I have no idea what is type or not type.
>>>>> 
>>>>> But I have some general remarks:
>>>>> 
>>>>> 1. A general automaton can have many states, say, 10 states.
>>>>> 
>>>>> so the action in each state can be either cooperate or defect (this is a 
>>>>> deterministic automaton, for a probabilistic one, they can even randomise 
>>>>> between cooperate or defect, but the case of deterministic and 
>>>>> probabilistic should be separated far).
>>>>> 
>>>>> it means that, even for a 2 state automaton, it can happen that both 
>>>>> state use the same cooperate action.
>>>>> 
>>>>> 
>>>>> other small notices
>>>>> - the all defect automaton start with defecting and always defecting so 
>>>>> its code is : defect defect defect... not cooperate defect defect... 
>>>>> (same for the all cooperates one)
>>>>> 
>>>>> - i'm considering to add a mutation phase at the end of each cycle (ie 
>>>>> choose some random automaton and mutate their code) but im working very 
>>>>> slow.
>>>>> 
>>>>> - why dont you define a (match-pair ...) function separately? you insert 
>>>>> it in the (match-population ...). doesnt it deserve to be outside?
>>>>> 
>>>>> - why you use [i (in-range 10)] in all for loop? what's the difference 
>>>>> with [i 10]. they both produce stream, right?
>>>>> 
>>>>> 
>>>>> 
>>>>> On Tuesday, 13 October 2015 01:46:04 UTC+2, Matthias Felleisen  wrote:
>>>>>> Take a crack at it. 
>>>>>> 
>>>>>> I am pretty sure I have introduced the proper level of representation 
>>>>>> independence (I can hear Ben laugh all the way now — Typed Racket
>>>>>> doesn’t have types) and the typed documentation is pretty solid. (I’ll
>>>>>> do types later to validate.)
>>>>> 
>>>>> -- 
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>>> 

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