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Python & EDA

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Python & EDA

  1. 1. dokelung (Ko-Lung Yuan)
  2. 2. About Me https://github.com/dokelung/ dokelung@gmail.com http://dokelung-blog.logdown.com/
  3. 3. ⼩小⼩小的⼼心得分享 今天講些什麼?
  4. 4. • EDA • script-king • msat • minickt Agenda
  5. 5. 常有⼈人問我這樣⼀一個問題... 什麼是電⼦子設計⾃自動化?
  6. 6. IC Design Flow
  7. 7. In EDA World … 我在這裡
  8. 8. 在最⼀一開始,我也是學過⼀一點 功夫的... C / C++ / Java … 三 腳 貓
  9. 9. 直到我遇⾒見了... 他! 直到我遇⾒見了...
  10. 10. Python可以拿來幹⿇麻呢? 除了吃飯和睡覺之外我都想⽤用Python啊
  11. 11. 拿來寫 script 好了 真是送禮⾃自⽤用兩相宜啊!
  12. 12. Script King https://github.com/dokelung/script-king
  13. 13. 接著幹嘛好呢? 來寫個 solver 好了 於是乎msat 就誕⽣生了!
  14. 14. https://github.com/dokelung/msat msat
  15. 15. Multiset Constraint Solving { 1, 3, 3, 1 } { 2, 3, 3 } Subset-Sum Problem { 1, 2, 3, 4, 5 } { 6 }= >= Element set Target setRelation What Is this?
  16. 16. subset-sum problem k-partition problem bin-packing problem knapsack problem pseudo Boolean constraint symmetry encoding What?
  17. 17. { 1, 2, 3, 4, 5 } k = 3 { 1, 4 } { 2, 3 } { 5 } Let E = { 1, 2, 3, 4, 5 } Let T = { 5, 5, 5 } Relation “=” k-partition Problem
  18. 18. { 2, 2, 3, 4 } bin size = 5 { 2 } { 2, 3 } { 4 } Let E = { 2, 2, 3, 4 } with all use rule Try T = { 5 }, { 5, 5 }, { 5, 5, 5 } … Relation “<=” Bin-packing Problem
  19. 19. size = { 1, 2, 1, 2, 1 } value = { 2, 3, 1, 2, 2 } knapsack capacity = 5 Let E1 = { 1, 2, 1, 2, 1 } T1 = { 5 } Relation “<=“ Let E2 = { 2, 3, 1, 2, 2 } T2 = { Vmax } Relation “=“ Try Vmax … Knapsack Problem
  20. 20. 3x + 5y - 2z >= 5 6x + 2y + 2z >= 2 x = 1, y = 1, z =0 Let E1 = { 3, 5, -2 } T1 = { 5 } Relation “>=“ Let E2 = { 6, 2, 2 } T2 = { 2 } Relation “>=“ PBC
  21. 21. How to Solve Let E = { 1, 1, 3, 3 } Try T = { 2, 3, 3 } Relation “>=” !2 !1 0 1 2 3 0 True True True False False False 1 True True True True False False 1 True True True True True False 3 True True True True True True 3 True True True True True True
  22. 22. minickt https://github.com/dokelung/minickt AND OR XOR a b c d e f out
  23. 23. –dokelung 「⼯工商服務,真⼼心不誤。」
  24. 24. 6/9(二) 天瓏買的到
  25. 25. 謝謝⼤大家 Q & A

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