2
play

2 3 4 HALT HALT 5 For all n >2 there are no - PowerPoint PPT Presentation

2 3 4 HALT HALT 5 For all n >2 there are no natural a,b,c such that a n +b n = c n . I have a truly marvelous demonstration of this proposition which this margin is too narrow to contain. 6 FERMAT()


  1. 2

  2. Σ • Σ Σ 3

  3. 4

  4. HALT HALT • • 5

  5. For all n >2 there are no natural a,b,c such that a n +b n = c n . I have a truly marvelous demonstration of this proposition which this margin is too narrow to contain. 6

  6. FERMAT() 𝑢 ← 3 while true for all 𝑜 ∈ 3, … , 𝑢 and 𝑦, 𝑧, 𝑨 ∈ 1, … , 𝑢 if 𝑦 𝑜 + 𝑧 𝑜 = 𝑨 𝑜 then return 𝑦, 𝑧, 𝑨, 𝑜 end for 𝑢 ← 𝑢 + 1 end while 7

  7. 8

  8. • HALT 𝑞𝑠𝑝𝑕𝑠𝑏𝑛, 𝑗𝑜𝑞𝑣𝑢 • Turing 𝑞𝑠𝑝𝑕𝑠𝑏𝑛 if HALT (𝑞𝑠𝑝𝑕𝑠𝑏𝑛, 𝑞𝑠𝑝𝑕𝑠𝑏𝑛) then loop forever else return true • Halt(Turing,Turing) Halt(Turing,Turing) Turing(Turing) o Halt(Turing,Turing) Turing(Turing) o 9

  9. • HALT = { 𝑁, 𝑦 ∶ 𝑁 is a TM that halts on 𝑦} 𝑁 𝐼𝐵𝑀𝑈 HALT • 𝑁 𝑈𝑉𝑆𝐽𝑂𝐻 • 〈𝑁〉 𝑁 𝑁 𝐼𝐵𝑀𝑈 〈𝑁, 𝑁〉 10

  10. • HALT = { 𝑁, 𝑦 ∶ 𝑁 is a TM that halts on 𝑦} 𝑁 𝐼𝐵𝑀𝑈 HALT • 𝑁 𝑈𝑉𝑆𝐽𝑂𝐻 • 𝑁 𝑈𝑉𝑆𝐽𝑂𝐻 ✕ ✔ 〈𝑁〉 〈𝑁, 𝑁〉 𝑁 𝐼𝐵𝑀𝑈 ✔ ∞ 11

  11. 〈𝑁 𝑈𝑉𝑆𝐽𝑂𝐻 〉 𝑁 𝑈𝑉𝑆𝐽𝑂𝐻 ? 𝑁 𝑈𝑉𝑆𝐽𝑂𝐻 ✕ ✔ 〈𝑁〉 〈𝑁, 𝑁〉 𝑁 𝐼𝐵𝑀𝑈 ✔ ∞ 12

  12. 〈𝑁 1 〉 〈𝑁 2 〉 〈𝑁 3 〉 〈𝑁 4 〉 〈𝑁 5 〉 ⋯ 𝑁 1 ⋯ 𝑁 2 ⋯ 𝑁 3 ⋯ 𝑁 4 𝑁 5 ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ 𝑁 𝑈𝑉𝑆𝐽𝑂𝐻 13

  13. 14

  14. 𝐵 ≤ 𝐶 𝐶 • 𝐵 𝐵 𝐶 • 𝐶 ⟹ 𝐵 o 𝐵 ⟹ 𝐶 o 15

  15. 𝐵 𝐶 • 𝐵 𝐶 𝐵 ≤ 𝐶 𝐵 𝐶 𝑁 𝐵 𝑦 𝑧 𝑁 𝐶 16

  16. 𝐶 HALT ≤ 𝐶 17

  17. • ACCEPTS = { 𝑁, 𝑦 ∶ 𝑁 is a TM that accepts 𝑦} • 𝑁, 𝑦 ∈ ACCEPTS ⟹ 𝑦 o 𝑁 𝑁, 𝑦 ∉ ACCEPTS ⟹ 𝑦 o 𝑁 ACCEPTS • 18

  18. 𝑁 𝐼𝐵𝑀𝑈 ✔ ✔ 〈𝑁, 𝑦〉 𝑁 𝐵𝐷𝐷𝐹𝑄𝑈𝑇 ✕ 〈𝑁, 𝑦〉 〈𝑁〉 〈𝑁′〉 ✔ 〈𝑁 ′ , 𝑦〉 ✔ 𝑁 𝐵𝐷𝐷𝐹𝑄𝑈𝑇 ✕ ✕ 19

  19. HALT ≤ ACCEPTS • 𝑁 ACCEPTS • 𝐵𝐷𝐷𝐹𝑄𝑈𝑇 HALT • 〈𝑁, 𝑦〉 𝑁 𝐵𝐷𝐷𝐹𝑄𝑈𝑇 〈𝑁, 𝑦〉 o o 𝑁 𝑁′ o 𝑁 𝐵𝐷𝐷𝐹𝑄𝑈𝑇 〈𝑁′, 𝑦〉 o o • 𝑁, 𝑦 ∈ HALT o 𝑁, 𝑦 ∉ HALT ∎ o 20

  20. • EMPTY = { 𝑁 ∶ 𝑁 is a TM that accepts nothing} EMPTY • E F 21

  21. ACCEPTS ≤ EMPTY • 〈𝑁, 𝑦〉 𝑁 𝑦 𝑧 • 𝑁(𝑦) 𝑁 𝐵𝐷𝐷𝐹𝑄𝑈𝑇 𝑁 𝑦 • 𝑁 𝐹𝑁𝑄𝑈𝑍 (〈𝑁 𝑦 〉) • 𝑦 ⟹ 𝑀 𝑁 𝑦 = Σ ∗ ⟹ 𝑁 𝐹𝑁𝑄𝑈𝑍 𝑁 〈𝑁 𝑦 〉 o 𝑁 𝑦 𝑦 ⟹ 𝑀 𝑁 𝑦 = ∅ o ⟹ 𝑁 𝐹𝑁𝑄𝑈𝑍 𝑁 𝑦 ∎ 22

  22. 23

  23. 𝑏 𝑏 𝑐𝑑𝑑 𝑏 𝑐𝑑𝑑 𝑏 𝑐𝑑𝑑 𝑏𝑐 𝑑𝑏𝑐𝑑 𝑑 𝑏𝑐 𝑑 𝑑𝑏𝑐𝑑 𝑑 24

  24. 25

  25. • https://youtu.be/RG2uPLG5K48 o o ¬∃𝑦, 𝑧, 𝑨, 𝑜 ∈ ℕ: 𝑜 ≥ 3 ∧ 𝑦 𝑜 + 𝑧 𝑜 = 𝑨 𝑜 o o • o 3𝑦 2 − 2𝑦𝑧 − 𝑧 2 𝑨 − 7 = 0 (𝑦 = 1, 𝑧 = 2, 𝑨 = −2) o o o 26

  26. • 𝑁, 𝑁 ′ ∶ 𝑁, 𝑁 ′ TMs, 𝑀 𝑁 = 𝑀(𝑁 ′ ) EQ = 1. GRAVITON = ∅ {1} 2. 27

  27. 28

  28. • HALT, ACCEPTS, EMPTY o o • o HALT, ACCEPTS, EMPTY o • o 29

Recommend


More recommend