1 / 42

Understanding State Machine Transitions: Examples and Patterns

This document explores the intricate mechanics of state machine transitions using formal notation and examples. Each transition is illustrated with clear examples, demonstrating how different inputs lead the state machine through various states. The structured approach lays out each step of transitioning, aiding comprehension of complex state changes. Examples include multiple states and their corresponding expected actions within the system. Readers will gain insight into automata theory and the functioning of state machines through it.

noah-rutledge
Télécharger la présentation

Understanding State Machine Transitions: Examples and Patterns

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. A = {w#w | w in {0,1}*} 1 2 8 3 4 accept 5 6 7

  2. 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR Σ = {0,1,#} Γ = {0,1,#,x,_} #L 0,1,xL 0,1L 7

  3. Example 1 • s = #

  4. s = # 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  5. s = # 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  6. s = # 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  7. Example 2 • s = 0#

  8. s = 0# 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  9. s = 0# 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  10. s = 0# 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  11. Example 3 • s = 01#0

  12. s = 01#0 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  13. s = 01#0 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  14. s = 01#0 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  15. s = 01#0 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  16. s = 01#0 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  17. s = 01#0 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  18. Example 3 • s = 01#01

  19. s = 01#01 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  20. s = 01#01 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  21. s = 01#01 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  22. s = 01#01 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  23. s = 01#01 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  24. s = 01#01 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  25. s = 01#01 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  26. s = 01#01 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  27. s = 01#01 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  28. s = 01#01 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  29. s = 01#01 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  30. s = 01#01 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  31. s = 01#01 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  32. s = 01#01 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  33. s = 01#01 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  34. s = 01#01 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  35. s = 01#01 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  36. s = 01#01 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  37. s = 01#01 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  38. s = 01#01 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  39. s = 01#01 1 1x,R 0x,R #R 0,1R 0,1R 2 8 3 xR #R _R #R xR xR 4 accept 5 0x,R 1x,L 6 xR #L 0,1,xL 0,1L 7

  40. Finding a symbol “-”Move until you hit _ xR q _R

  41. Branch inside of a loop 1 1x,R 0x,R ?R 2 accept 3 6 7

  42. Expanding the Tape Alphabet Σ = {0,1,#} Γ = {0,1,#,x,0,1,#,x,_}

More Related