compressibility and probabilistic proofs
play

Compressibility and probabilistic proofs alexander.shen@lirmm.fr, - PowerPoint PPT Presentation

Compressibility and probabilistic proofs alexander.shen@lirmm.fr, www.lirmm.fr/~ashen LIRMM CNRS & University of Montpellier CiE 2017 alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs Probabilistic


  1. Same proof using the compression language n × n matrix can be encoded as a n 2 -bit string most strings are incompressible (cannot be described by fewer bits) if matrix with a k × k monochromatic minor for k ≫ 2 log n is compressible why? it has a short description: each of 2 k rows/columns of the minor requires log n bits, 2 k log n in total alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  2. Same proof using the compression language n × n matrix can be encoded as a n 2 -bit string most strings are incompressible (cannot be described by fewer bits) if matrix with a k × k monochromatic minor for k ≫ 2 log n is compressible why? it has a short description: each of 2 k rows/columns of the minor requires log n bits, 2 k log n in total one bit for the color of the minor alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  3. Same proof using the compression language n × n matrix can be encoded as a n 2 -bit string most strings are incompressible (cannot be described by fewer bits) if matrix with a k × k monochromatic minor for k ≫ 2 log n is compressible why? it has a short description: each of 2 k rows/columns of the minor requires log n bits, 2 k log n in total one bit for the color of the minor the rest of the matrix ( n 2 − k 2 bits) alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  4. Same proof using the compression language n × n matrix can be encoded as a n 2 -bit string most strings are incompressible (cannot be described by fewer bits) if matrix with a k × k monochromatic minor for k ≫ 2 log n is compressible why? it has a short description: each of 2 k rows/columns of the minor requires log n bits, 2 k log n in total one bit for the color of the minor the rest of the matrix ( n 2 − k 2 bits) replacing k 2 by 2 k log n + 1: compression if k ≫ 2 log n alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  5. So what? alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  6. So what? may be the compression language is more intuitive alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  7. So what? may be the compression language is more intuitive but not very impressive. . . alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  8. So what? may be the compression language is more intuitive but not very impressive. . . more interesting examples alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  9. So what? may be the compression language is more intuitive but not very impressive. . . more interesting examples Lovasz local lemma instead of the union bound alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  10. So what? may be the compression language is more intuitive but not very impressive. . . more interesting examples Lovasz local lemma instead of the union bound algorithmic version due to Moses-Tardos alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  11. So what? may be the compression language is more intuitive but not very impressive. . . more interesting examples Lovasz local lemma instead of the union bound algorithmic version due to Moses-Tardos do not need to know what is LL and MT algorithm alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  12. So what? may be the compression language is more intuitive but not very impressive. . . more interesting examples Lovasz local lemma instead of the union bound algorithmic version due to Moses-Tardos do not need to know what is LL and MT algorithm scheme: we try to most natural randomized algorithm alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  13. So what? may be the compression language is more intuitive but not very impressive. . . more interesting examples Lovasz local lemma instead of the union bound algorithmic version due to Moses-Tardos do not need to know what is LL and MT algorithm scheme: we try to most natural randomized algorithm it succeeds with high probability. . . alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  14. So what? may be the compression language is more intuitive but not very impressive. . . more interesting examples Lovasz local lemma instead of the union bound algorithmic version due to Moses-Tardos do not need to know what is LL and MT algorithm scheme: we try to most natural randomized algorithm it succeeds with high probability. . . because if it fails, the random bits used are compressible alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  15. So what? may be the compression language is more intuitive but not very impressive. . . more interesting examples Lovasz local lemma instead of the union bound algorithmic version due to Moses-Tardos do not need to know what is LL and MT algorithm scheme: we try to most natural randomized algorithm it succeeds with high probability. . . because if it fails, the random bits used are compressible A: forbidden factors (Ochem, Gon¸ calves) alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  16. So what? may be the compression language is more intuitive but not very impressive. . . more interesting examples Lovasz local lemma instead of the union bound algorithmic version due to Moses-Tardos do not need to know what is LL and MT algorithm scheme: we try to most natural randomized algorithm it succeeds with high probability. . . because if it fails, the random bits used are compressible A: forbidden factors (Ochem, Gon¸ calves) B: CNF with bounded neighborhood (Moser, Fortnow) alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  17. Forbidden factors alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  18. Forbidden factors F 1 , . . . , F k : binary strings (“forbidden strings”) alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  19. Forbidden factors F 1 , . . . , F k : binary strings (“forbidden strings”) is there an infinite bit sequence that does not have any of F i as a substring? alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  20. Forbidden factors F 1 , . . . , F k : binary strings (“forbidden strings”) is there an infinite bit sequence that does not have any of F i as a substring? infinite ⇔ arbitrarily long alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  21. Forbidden factors F 1 , . . . , F k : binary strings (“forbidden strings”) is there an infinite bit sequence that does not have any of F i as a substring? infinite ⇔ arbitrarily long the answer depends on the list: 0 , 11 does not exist; alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  22. Forbidden factors F 1 , . . . , F k : binary strings (“forbidden strings”) is there an infinite bit sequence that does not have any of F i as a substring? infinite ⇔ arbitrarily long the answer depends on the list: 0 , 11 does not exist; 0 , 00 does exist alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  23. Forbidden factors F 1 , . . . , F k : binary strings (“forbidden strings”) is there an infinite bit sequence that does not have any of F i as a substring? infinite ⇔ arbitrarily long the answer depends on the list: 0 , 11 does not exist; 0 , 00 does exist for a fixed list we get a regular expression / finite automaton alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  24. Forbidden factors F 1 , . . . , F k : binary strings (“forbidden strings”) is there an infinite bit sequence that does not have any of F i as a substring? infinite ⇔ arbitrarily long the answer depends on the list: 0 , 11 does not exist; 0 , 00 does exist for a fixed list we get a regular expression / finite automaton quantitative results: “if there are not too many forbidden strings of each length, then there are long sequences without forbidden strings” alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  25. Forbidden factors F 1 , . . . , F k : binary strings (“forbidden strings”) is there an infinite bit sequence that does not have any of F i as a substring? infinite ⇔ arbitrarily long the answer depends on the list: 0 , 11 does not exist; 0 , 00 does exist for a fixed list we get a regular expression / finite automaton quantitative results: “if there are not too many forbidden strings of each length, then there are long sequences without forbidden strings” Let a i be the number of forbidden strings of length i. If a i t i < mt − 1 for some t > 0 � then there exist arbitrarily long strings without forbidden factors. (For the case of m letters) alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  26. Tetris algorithm Forbidden strings: 01, 110 Random bits are added one by one; if a forbidden string appears (at the end), it vanishes, and the process continues alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  27. Tetris algorithm Forbidden strings: 01, 110 0 0 Random bits are added one by one; if a forbidden string appears (at the end), it vanishes, and the process continues alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  28. Tetris algorithm Forbidden strings: 01, 110 0 0 00 Random bits are added one by one; if a forbidden string appears (at the end), it vanishes, and the process continues alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  29. Tetris algorithm Forbidden strings: 01, 110 0 0 1 001 Random bits are added one by one; if a forbidden string appears (at the end), it vanishes, and the process continues alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  30. Tetris algorithm Forbidden strings: 01, 110 0 0 1 001 Random bits are added one by one; if a forbidden string appears (at the end), it vanishes, and the process continues alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  31. Tetris algorithm Forbidden strings: 01, 110 0 001 Random bits are added one by one; if a forbidden string appears (at the end), it vanishes, and the process continues alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  32. Tetris algorithm Forbidden strings: 01, 110 0 1 0011 Random bits are added one by one; if a forbidden string appears (at the end), it vanishes, and the process continues alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  33. Tetris algorithm Forbidden strings: 01, 110 0 1 0011 Random bits are added one by one; if a forbidden string appears (at the end), it vanishes, and the process continues alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  34. Tetris algorithm Forbidden strings: 01, 110 0011 Random bits are added one by one; if a forbidden string appears (at the end), it vanishes, and the process continues alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  35. Tetris algorithm Forbidden strings: 01, 110 1 00111 Random bits are added one by one; if a forbidden string appears (at the end), it vanishes, and the process continues alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  36. Tetris algorithm Forbidden strings: 01, 110 1 1 001111 Random bits are added one by one; if a forbidden string appears (at the end), it vanishes, and the process continues alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  37. Tetris algorithm Forbidden strings: 01, 110 1 1 0 0011110 Random bits are added one by one; if a forbidden string appears (at the end), it vanishes, and the process continues alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  38. Tetris algorithm Forbidden strings: 01, 110 1 1 0 0011110 Random bits are added one by one; if a forbidden string appears (at the end), it vanishes, and the process continues alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  39. Tetris algorithm Forbidden strings: 01, 110 0011110 Random bits are added one by one; if a forbidden string appears (at the end), it vanishes, and the process continues alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  40. Tetris algorithm Forbidden strings: 01, 110 1 00111101 Random bits are added one by one; if a forbidden string appears (at the end), it vanishes, and the process continues alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  41. Tetris algorithm Forbidden strings: 01, 110 1 0 001111010 Random bits are added one by one; if a forbidden string appears (at the end), it vanishes, and the process continues alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  42. Tetris algorithm Forbidden strings: 01, 110 1 0 0 0011110100 Random bits are added one by one; if a forbidden string appears (at the end), it vanishes, and the process continues alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  43. Tetris algorithm Forbidden strings: 01, 110 1 0 0 0 00111101000 Random bits are added one by one; if a forbidden string appears (at the end), it vanishes, and the process continues alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  44. Tetris algorithm Forbidden strings: 01, 110 1 0 0 0 1 001111010001 Random bits are added one by one; if a forbidden string appears (at the end), it vanishes, and the process continues alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  45. Tetris algorithm Forbidden strings: 01, 110 1 0 0 0 1 001111010001 Random bits are added one by one; if a forbidden string appears (at the end), it vanishes, and the process continues alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  46. Tetris algorithm Forbidden strings: 01, 110 1 0 0 001111010001 Random bits are added one by one; if a forbidden string appears (at the end), it vanishes, and the process continues alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  47. Tetris algorithm Forbidden strings: 01, 110 1 0 0 0 0011110100010 Random bits are added one by one; if a forbidden string appears (at the end), it vanishes, and the process continues alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  48. Will it grow indefinitely? alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  49. Will it grow indefinitely? alphabet size m alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  50. Will it grow indefinitely? alphabet size m a n is the number of forbidden strings of length n ≥ 2 alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  51. Will it grow indefinitely? alphabet size m a n is the number of forbidden strings of length n ≥ 2 n a n t n < mt − 1 for some t > 0 assume that � alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  52. Will it grow indefinitely? alphabet size m a n is the number of forbidden strings of length n ≥ 2 n a n t n < mt − 1 for some t > 0 assume that � Claim: if the string remains short forever, then the sequence of random bits is compressible alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  53. Will it grow indefinitely? alphabet size m a n is the number of forbidden strings of length n ≥ 2 n a n t n < mt − 1 for some t > 0 assume that � Claim: if the string remains short forever, then the sequence of random bits is compressible log file: sequence of signs like +, + 01 , + 110 for adding the new bit (not indicated) without or with cancelled string alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  54. Will it grow indefinitely? alphabet size m a n is the number of forbidden strings of length n ≥ 2 n a n t n < mt − 1 for some t > 0 assume that � Claim: if the string remains short forever, then the sequence of random bits is compressible log file: sequence of signs like +, + 01 , + 110 for adding the new bit (not indicated) without or with cancelled string going backwards: + means deletion of the last bit, + u means adding u and then deleting the last bit alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  55. Will it grow indefinitely? alphabet size m a n is the number of forbidden strings of length n ≥ 2 n a n t n < mt − 1 for some t > 0 assume that � Claim: if the string remains short forever, then the sequence of random bits is compressible log file: sequence of signs like +, + 01 , + 110 for adding the new bit (not indicated) without or with cancelled string going backwards: + means deletion of the last bit, + u means adding u and then deleting the last bit current sequence + log file → random bits used alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  56. Will it grow indefinitely? alphabet size m a n is the number of forbidden strings of length n ≥ 2 n a n t n < mt − 1 for some t > 0 assume that � Claim: if the string remains short forever, then the sequence of random bits is compressible log file: sequence of signs like +, + 01 , + 110 for adding the new bit (not indicated) without or with cancelled string going backwards: + means deletion of the last bit, + u means adding u and then deleting the last bit current sequence + log file → random bits used few forbidden strings ⇒ few symbols in log file ⇒ efficient encoding alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  57. More details alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  58. More details current string + log file → sequence of random bits alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  59. More details current string + log file → sequence of random bits current string is O (1) if it doesn’t grow indefinitely alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  60. More details current string + log file → sequence of random bits current string is O (1) if it doesn’t grow indefinitely the length of log file is the number of random bits alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  61. More details current string + log file → sequence of random bits current string is O (1) if it doesn’t grow indefinitely the length of log file is the number of random bits so we need to encode the log file efficiently ( < 1 bit/symbol) alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  62. More details current string + log file → sequence of random bits current string is O (1) if it doesn’t grow indefinitely the length of log file is the number of random bits so we need to encode the log file efficiently ( < 1 bit/symbol) large alphabet + x but most symbols are + alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  63. More details current string + log file → sequence of random bits current string is O (1) if it doesn’t grow indefinitely the length of log file is the number of random bits so we need to encode the log file efficiently ( < 1 bit/symbol) large alphabet + x but most symbols are + arithmetic coding: use less that 1 bit for + alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  64. More details current string + log file → sequence of random bits current string is O (1) if it doesn’t grow indefinitely the length of log file is the number of random bits so we need to encode the log file efficiently ( < 1 bit/symbol) large alphabet + x but most symbols are + arithmetic coding: use less that 1 bit for + the savings due to +’s are used for encoding + x letters alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  65. More details current string + log file → sequence of random bits current string is O (1) if it doesn’t grow indefinitely the length of log file is the number of random bits so we need to encode the log file efficiently ( < 1 bit/symbol) large alphabet + x but most symbols are + arithmetic coding: use less that 1 bit for + the savings due to +’s are used for encoding + x letters amortized analysis: + increases the length by 1 and + x decreases the length by | x | − 1 alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  66. More details current string + log file → sequence of random bits current string is O (1) if it doesn’t grow indefinitely the length of log file is the number of random bits so we need to encode the log file efficiently ( < 1 bit/symbol) large alphabet + x but most symbols are + arithmetic coding: use less that 1 bit for + the savings due to +’s are used for encoding + x letters amortized analysis: + increases the length by 1 and + x decreases the length by | x | − 1 so there couldn’t be many + x unless there are many + alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  67. More details current string + log file → sequence of random bits current string is O (1) if it doesn’t grow indefinitely the length of log file is the number of random bits so we need to encode the log file efficiently ( < 1 bit/symbol) large alphabet + x but most symbols are + arithmetic coding: use less that 1 bit for + the savings due to +’s are used for encoding + x letters amortized analysis: + increases the length by 1 and + x decreases the length by | x | − 1 so there couldn’t be many + x unless there are many + role of t : parameter for amortized analysis of the encoding efficiency alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  68. Technical details alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  69. Technical details arithmetic coding: each symbol z has some weight p z > 0 alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  70. Technical details arithmetic coding: each symbol z has some weight p z > 0 � z p z ≤ 1 alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  71. Technical details arithmetic coding: each symbol z has some weight p z > 0 � z p z ≤ 1 encoding z by log(1 / p z ) bits alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  72. Technical details arithmetic coding: each symbol z has some weight p z > 0 � z p z ≤ 1 encoding z by log(1 / p z ) bits allocate weight q 0 for + and total weight q n for all + u where u are forbidden strings of length n . alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  73. Technical details arithmetic coding: each symbol z has some weight p z > 0 � z p z ≤ 1 encoding z by log(1 / p z ) bits allocate weight q 0 for + and total weight q n for all + u where u are forbidden strings of length n . code lengths: − log q 0 , for “+” − log q n + log a n , for each “+ u ” with | u | = n alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  74. Technical details arithmetic coding: each symbol z has some weight p z > 0 � z p z ≤ 1 encoding z by log(1 / p z ) bits allocate weight q 0 for + and total weight q n for all + u where u are forbidden strings of length n . code lengths: − log q 0 , for “+” − log q n + log a n , for each “+ u ” with | u | = n each log symbol corresponds to one random symbol, so we want to encode log symbols with less than log m bits alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

  75. Technical details arithmetic coding: each symbol z has some weight p z > 0 � z p z ≤ 1 encoding z by log(1 / p z ) bits allocate weight q 0 for + and total weight q n for all + u where u are forbidden strings of length n . code lengths: − log q 0 , for “+” − log q n + log a n , for each “+ u ” with | u | = n each log symbol corresponds to one random symbol, so we want to encode log symbols with less than log m bits + increases the length by 1, and + u decreases the length by n − 1 for | u | = n alexander.shen@lirmm.fr, www.lirmm.fr/~ashen Compressibility and probabilistic proofs

Recommend


More recommend