Holger Petersen FCT 2017, Bordeaux
Defined by T. Rado in 1962 Based on deterministic single-tape Turing machines with binary alphabet, steps characterized by: 1. Current state 2. Symbol scanned 3. Symbol written 4. Head movement (mandatory) 5. New state (including halt-state “for free”) “quintuple Turing machine variant”
Start TM on all- blank (0‘s) tape, if it stops: • Number of steps is act ctivi ivity ty • Number of ones is productivit ductivity Functions defined by Rado: • 𝑇 𝑜 = maximum activity of any 𝑜 -state TM • Σ 𝑜 = maximum productivity of any 𝑜 -state TM Rado showed that 𝑇 𝑜 and Σ 𝑜 grow faster than any computable function 𝑇 𝑜 and Σ 𝑜 are not computable (or recursive)
Generalization to alphabets with 𝑛 ≥ 2 symbols: • 𝑇 𝑜, 𝑛 = max. activity of 𝑜 -state, 𝑛 -symbol TM • Σ 𝑜, 𝑛 = max. productivity 𝑜 -state, 𝑛 -symbol TM Productivity: number of non-blanks Trivial observation: 𝑇 𝑜, 𝑛 ≥ 𝑇 𝑜, 2 and Σ 𝑜, 𝑛 ≥ Σ 𝑜, 2 𝑇 𝑜, 𝑛 and Σ 𝑜, 𝑛 are not computable Goal l of of the game: : Find machines of maximum activity/productivity (Busy Beavers)
On On a s simple ple source rce for non-computabl computable e functions ctions ( Rado‘s contribution to: Symposium on Mathematical Theory of Automata, April 1962) Metamathematical amathematical inter erest: est: Knowing Σ 𝑜 or 𝑇 𝑜 for sufficiently large 𝑜 would settle Goldbach’s conjecture and other conjectures disproved by counterexamples (Chaitin 1987) Concrete value of 𝑜 (Yedidia and Aaronson 2016): 𝑜 = 4888 suffices for Goldbach’s conjecture Improved to 47 and then 31 in Aaronson’s blog
Let‘s compute Σ 1 : 1 0 <symbol written><head movement>
Let‘s compute Σ 1 : 1 loop, machine never stops 0 <symbol written><head movement>
Let‘s compute Σ 1 : 01<head movement> 1 HALT
Let‘s compute Σ 1 : 01<head movement> 1 HALT ⇒ Σ 1 = 1
Re Results ts Author hors Yea ear T. Rado 1962 Σ 2, 2 = 4 R. Hegelman 1962 Σ 3, 2 ≥ 6 Σ 3, 2 = 6, 𝑇 3, 2 = 21 S. Lin, T. Rado 1965 A. H. Brady 1964 Σ 4, 2 ≥ 13, 𝑇 4, 2 ≥ 107 A.H. Brady 1974 Σ 4, 2 = 13, 𝑇 4, 2 = 1983 107 R. J. Kopp 1981
Early Results: Author hor Year Sco cores es reported by M. W. 1964 17 ones Green D.S. Lynn 1972 22 ones 435 steps B. Weimann 1973 40 ones D.S. Lynn 1974 112 ones 7,707 steps
Results after c ompetition initiated at 6th GI- Conference on TCS, Dortmund : Author hor Year Sco cores es U. Schult (winner out 1983 501 ones of 133 submissions) 134,467 steps G. Uhing 1985 1,915 ones 2,358,063 steps H. Marxen, 1989 4,098 ones J. Buntrock 47,176,870 steps
Adding a state increases scores: x1R HALT
Adding a state increases scores: x1R 01R HALT HALT n+1 yyR ( y ≠ 0)
Adding a state increases scores: x1R 01R HALT HALT n+1 yyR ( y ≠ 0) ⇒ Σ 𝑜 + 1, 𝑛 > Σ 𝑜, 𝑛 , S 𝑜 + 1, 𝑛 > S 𝑜, 𝑛
Let 𝑁 be a 𝑙 -halting* Turing machine with 𝑜 states and 𝑛 symbols for some 𝑙 ≥ 1 with finite activity. Then there is a 𝑙 -halting 𝑜 -state (𝑛 + 1) -symbol Turing machine 𝑁′ with finite activity such that: activity( 𝑁′ ) > activity( 𝑁 ) productivity( 𝑁′ ) > productivity( 𝑁 ) * 𝑙 transitions to the halt state
Σ 2, 2 = 4 < Σ 2, 3 = 9 < 2,050 < Σ 2, 4 S 2, 2 = 6 < S 2, 3 = 38 < 3,932 < S 2, 4 Σ 3, 2 = 6 < 374,676,383 ≤ Σ 3, 3 S 3, 2 = 21 < 119,112,334,170,342,540 ≤ S 3, 3
Σ 2, 2 = 4 < Σ 2, 3 = 9 < 2,050 < Σ 2, 4 S 2, 2 = 6 < S 2, 3 = 38 < 3,932 < S 2, 4 Σ 3, 2 = 6 < 374,676,383 ≤ Σ 3, 3 S 3, 2 = 21 < 119,112,334,170,342,540 ≤ S 3, 3 one hundred nineteen quadrillion… Results of Rado, Lin, Lafitte, Papazian, T. Ligocki and S. Ligocki
For every 𝑛 ≥ 2 and 𝑙 ≥ 1 there is an 𝑂 𝑛, 𝑙 such that for every 𝑙 -halting Turing machine 𝑁 with 𝑜 ≥ 𝑂 𝑛, 𝑙 states and 𝑛 symbols with finite activity there is an 𝑜 -state, (𝑛 + 1) -symbol 𝑙 -halting Turing machine 𝑁′ with finite activity such that activity( 𝑁′ ) > activity( 𝑁 ) productivity( 𝑁′ ) > productivity( 𝑁 ).
m symbols n states
m symbols n states additional state increases scores
m symbols additional symbol keeps area constant area corresponds to descritional complexity of TM n states additional state increases scores
m symbols additional symbol keeps area constant area corresponds to descritional complexity of TM n states Extractor / simulator additional state treating the blue area increases scores as a ROM
11R 00R 00R 00R 00R 00R 00R 00R f 11R 11R State f reached after reading: 1000000 bits 110 encoded • 0100 bits 010 • 0010000000 bits 111 • Transition encodes log. number of bits
For every Turing machine 𝑁 with 𝑜 ≥ 2 states and two symbols having finite activity there is an 𝑜 -state, 3 -symbol Turing machine 𝑁 with finite activity such that activity( 𝑁′ ) > activity( 𝑁 ).
22R 22R 22R 22R 22R HALT Add halting transitions on new symbol „2“
Consider halting transition: Q X Y Z H X W Z Modify transition depending on X, Z
X = 0, w.l.o.g. initial state 1 moves right on 0: Q 0 Y Z 1 0 2 Z S 0 2 Z H 0 2 Z
X = 1, some state R moves right on 1: Q 1 Y Z R 1 2 Z H S W 2 Z H W 2 Z
X = 1, Z = 1, all states move left on 1: Q 1 Y 1 L 1 2 1 H S 1 2 W H 1 2 W
X = 1, Z = 0, some state L moves left on 0: Q 1 Y 0 L 1 2 0 H S 1 2 W H 1 2 W
If all states move right on 0, then activity is bounded by 𝑜 . For all 𝑜 ≥ 2 we have S 𝑜, 2 > 𝑜 Take 𝑜 – state Busy Beaver for 2 symbols
For every Turing machine 𝑁 with 𝑜 ≥ 2 states and 𝑛 ≥ 2 symbols having finite activity there is a 2 -state, (4𝑜𝑛 + 5𝑛) -symbol Turing machine 𝑁′ with finite activity such that activity( 𝑁′ ) > activity( 𝑁 ) productivity( 𝑁′ ) > productivity( 𝑁 ).
By monotonicity in number of states, 𝑁 with 𝑜 + 1 states and 𝑛 symbols increases activity and productivity. Classical construction of Shannon transforms it into equivalent 2 -state machine with 4𝑛 𝑜 + 1 + 𝑛 = 4𝑜𝑛 + 5𝑛 symbols.
For every Turing machine M with 𝑜 ≥ 2 states and 𝑛 ≥ 3 symbols having finite activity there is an 𝑜 -state, 3𝑛 -symbol Turing machine 𝑁′ with finite activity such that activity( 𝑁′ ) > activity( 𝑁 ).
1 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 Idea ea: Let head bounce on additional symbols with subscripts L, R
1 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 1 L 0 0 0 0 0 1 0 0 0 0 1 L 1 R 0 0 0 0 1 0 0 0 0 1 1 R 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0
Attribution: By Steve from washington, dc, usa (American Beaver) [CC BY-SA 2.0 (http://creativecommons.org/licenses/by-sa/2.0)], via Wikimedia Commons
COMPUTING THE BUSY BEAVER FUNCTION (Chaitin 1987): „… to information theorists it is clear that the correct measure is bits, not states. .. …to deal with Σ (number of bits) one would need a model of a binary computer as simple and compelling as the Turing machine model, and no obvious natural choice is at hand .”
Reasonable programming languages encode 𝑜 -state Turing machines (fixed 𝑛 ) in 𝑃(𝑜 log 𝑜) bits (not sure about Befunge and Ook!) Turing machines encode 𝑜 log 𝑜 bits of programme code in 𝑃(𝑜) states by introspective computing
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