Go

https://blog.csdn.net/john_f_lau/article/details/55254324?utm_source=itdadao&utm_medium=referral
https://swtch.com/~rsc/thread/#2
https://www.cnblogs.com/foxmailed/p/3509359.html
https://www.zhihu.com/question/26192499
https://golang.org/test/chan/powser1.go
https://codegolf.stackexchange.com/questions/2403/implement-lazy-lists-preferably-in-a-language-you-dont-know-well
https://en.wikipedia.org/wiki/CAR_and_CDR

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https://en.wikipedia.org/wiki/Cons
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Power Series
https://books.google.com/books?id=dXRlaoQ7wBkC&pg=PA247&lpg=PA247&dq=psderiv+power+series&source=bl&ots=77T9CKl2oW&sig=FnsjUUPfO4QOa74GQ29Zrc9Dk3o&hl=en&sa=X&ved=2ahUKEwj6yZPanPLcAhWNEnwKHdQRD0AQ6AEwAnoECAcQAQ#v=onepage&q&f=false

https://github.com/jessehzj/golang-playground/blob/master/powerSeries/powerSeries.go

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split
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Basic splitter
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Optimized splitter
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添加

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两个ones相乘:
只加上F0 <- f的情况

加上

case <-wait:
.......
case f = <- F:
  case F0 <-= f:
    begin do_split_new(F, F0, F1, signal);
    become do_split_mid(f, F1, wait, signal);
  case F1 <-= f:
    begin do_split_new(F, F1, F0, signal);
    become do_solit_mid(f, F0, wait, signal);

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split的问题:

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一直卡着:证明在一直递归调用do_split_new

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基本方法的进程数，可以看到大概是n^2

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这个是没有疑问的
其他的差别不大，基本呈线性
下图是做了优化之后的，在250到370左右

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下图是增加了demand channel之后的:大概在290~500之间，没有上面的好？？？
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下面是官方版本的在230~240之间

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https://stackoverflow.com/questions/29775836/no-op-explicitly-do-nothing-in-go

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https://courses.cs.washington.edu/courses/cse341/08au/haskell/InfiniteDataStructures.txt
http://www.cs.cornell.edu/courses/cs3110/2011fa/supplemental/lec24-streams/streams.htm
http://crypto.stanford.edu/~blynn/c/haskell.html

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这样是有bug的，我还是改回去吧。。。改成被注释掉的部分，goroutine数变最多了。。。WHY！！！

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下一个任务是

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1.分清发信号和收信号的区别，不然死锁了
2.第二个错误

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怎么越变越多了？？？

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增加lazy evaluation以及把do_split_new改成下面那样取得最小的goroutine数(产生50个数时最大为261，一般为215)：
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power series 的Scala版本
https://www.cs.dartmouth.edu/~doug/powser.html

interesting things about exponetial:
因为积分后的第一项的计算不依赖于输入，所以不会死锁，用论文的话来说就是，Deadlock is avoided because the integral operator produces the first term of its output（the constant of integration ）before it needs any input. This result gets fed onto stream X to help calculate the next term, and so on.

scala的lazy list天生有splitting和demand channel的特性，但是却没有memorization,有这个就更好了，
不对是有的吧，看multiplication的公式，
https://en.m.wikibooks.org/wiki/Haskell/Pattern_matching
https://stackoverflow.com/questions/36771251/lazy-evaluation-for-list-generating-functions
https://www.cs.dartmouth.edu/~doug/powser1.html
https://kseo.github.io/posts/2017-01-14-memoization-in-hasekll.html
http://okmij.org/ftp/ML/index.html
论文power series, power serious

https://www.scala-exercises.org/scala_tutorial/lazy_evaluation
https://stackoverflow.com/questions/8566532/scala-streams-and-their-memory-usage