Integer Complexity: Experimental and Analytical Results II Juris - PowerPoint PPT Presentation

Integer Complexity: Experimental and Analytical Results II Juris ˇ nenoks 1 anis Iraids 1 s Opmanis 2 Cer¸ J¯ M¯ arti¸ nˇ Rihards Opmanis 2 arlis Podnieks 1 K¯ 1 University of Latvia 2 Institute of Mathematics and Computer Science, University of Latvia DCFS 2015, University of Waterloo, Friday 26 th June, 2015

Introduction Main Results Conclusion Integer Complexity Definition of Integer Complexity Integer complexity Integer complexity of a positive integer n , denoted by � n � , is the least amount of 1’s in an arithmetic expression for n consisting of 1’s, +, · and brackets. For example, � 1 � = 1 � 2 � = 2; 2 = 1 + 1 � 3 � = 3; 3 = 1 + 1 + 1 � 6 � = 5; 6 = (1 + 1) · (1 + 1 + 1) � 8 � = 6; 8 = (1 + 1) · (1 + 1) · (1 + 1) � 11 � = 8; 11 = (1 + 1 + 1) · (1 + 1 + 1) + 1 + 1 http://oeis.org/A005245

Introduction Main Results Conclusion Integer Complexity Lower and Upper Bounds Theorem � n � ∈ Θ(log n ) Sketch of proof. 1 � n � ≤ 3 log 2 n – Horner’s rule Expand n in binary: n = a k a k − 1 · · · a 1 a 0 . Express as a 0 + (1 + 1) · ( a 1 + (1 + 1) · . . . ( a k − 1 + (1 + 1) · a k ) . . . ) . 2 � n � ≥ 3 log 3 n Idea: denote by E ( k ) the largest number having complexity k .

Introduction Main Results Conclusion Integer Complexity Lower and Upper Bounds Theorem � n � ∈ Θ(log n ) Sketch of proof. We will show that E (3 k + 2) = 2 · 3 k ; E (3 k + 3) = 3 · 3 k ; E (3 k + 4) = 4 · 3 k .

Introduction Main Results Conclusion Integer Complexity Largest Number of Complexity k Theorem For all k ≥ 0 : E (3 k + 2) = 2 · 3 k ; E (3 k + 3) = 3 · 3 k ; E (3 k + 4) = 4 · 3 k . Proof (by H. Altman). The value of an expression does not decrease if we: Replace all x · 1 by x + 1; Replace all x · y + 1 by x · ( y + 1); Replace all x · y + u · v by x · y · u · v ; If x = 1 + 1 + . . . + 1 > 3, split it into product of (1 + 1)’s and (1 + 1 + 1)’s; Replace all (1 + 1) · (1 + 1) · (1 + 1) by (1 + 1 + 1) · (1 + 1 + 1).

Introduction Main Results Conclusion Integer Complexity Complexity of Powers From E (3 k ) = 3 k we arrive at � � 3 k � � = 3 k . � � � � n k � What about powers of other numbers � ? � � n k � � < k · � n � : There exist n with � 5 � = 5 � = 10 � � 5 2 � . . . � = 25 � 5 5 � � � = 29; 5 6 = (3 3 · 2 3 + 1) · 3 2 · 2 3 + 1 � � 5 6 �

Introduction Main Results Conclusion Integer Complexity Complexity of 2 a Richard K. Guy, “Unsolved Problems in Number Theory”, problem F26 � = 2 a + 3 b for all ( a , b ) � = (0 , 0)? � 2 a 3 b � � Is In particular, is � 2 a � = 2 a for all a ? [Attributed to Selfridge, Hypothesis H1] Having computed � n � for n up to 10 12 hypothesis H1 holds for all a ≤ 39 [2010]. Recently Harry Altman showed H1 holds for all ( a , b ) with a ≤ 48 (See the PhD thesis of Altman “Integer Complexity, Addition Chains, and Well-Ordering” for excellent introduction to integer complexity.)

Introduction Main Results Conclusion Logarithmic Complexity Logarithmic Complexity Let the logarithmic complexity of n be denoted by � n � � n � log = log 3 n . 3 ≤ � n � log ≤ 3 log 2 3 ≈ 4 . 755 Richard K. Guy, “Unsolved Problems in Number Theory”, problem F26 As n → ∞ does � n � log → 3? [Hypothesis H2] For all n up to 10 12 : � n � log ≤ � 1439 � log ≈ 3 . 928 . In 2014 Arias de Reyna and van de Lune showed that for most n : � n � log < 3 . 635 .

Introduction Main Results Conclusion Logarithmic Complexity Distribution of Logarithmic Complexity [1] 30 20 10 0 3 3 . 2 3 . 4 3 . 6 3 . 8 4 Figure : Distribution of logarithmic complexity of numbers with � n � = 30

Introduction Main Results Conclusion Logarithmic Complexity Distribution of Logarithmic Complexity [2] 800 600 400 200 0 3 3 . 2 3 . 4 3 . 6 3 . 8 4 Figure : Distribution of logarithmic complexity of numbers with � n � = 40

Introduction Main Results Conclusion Logarithmic Complexity Distribution of Logarithmic Complexity [3] 3 · 10 7 2 · 10 7 1 · 10 7 0 3 3 . 2 3 . 4 3 . 6 3 . 8 4 Figure : Distribution of logarithmic complexity of numbers with � n � = 70

Introduction Main Results Conclusion Open Problems Relation of the Open Problems Richard K. Guy, “Unsolved Problems in Number Theory”, problem F26 Is � 2 a � = 2 a for all a ? [Hypothesis H1] As n → ∞ does � n � log → 3? [Hypothesis H2] H 1 = ⇒ ¬ H 2, because 2 a � 2 a � log = log 3 2 a ≈ 3 . 170; hence H2 should be easier to settle. We have not succeeded to prove or disprove either of them.

Introduction Main Results Conclusion Sum of Digits Base-3 Representations of 2 n Observation � 8 � = 6; 8 = (1 + 1)(1 + 1)(1 + 1) � 9 � = 6; 9 = (1 + 1 + 1)(1 + 1 + 1) Base-3 representation of powers of 2: (2) 3 = 2 (2 2 ) 3 = 11 (2 3 ) 3 = 22 (2 10 ) 3 = 1101221 (2 30 ) 3 = 2202211102201212201 (2 50 ) 3 = 12110122110222110100112122112211 The digits seem “random, uniformly distributed”.

Introduction Main Results Conclusion Sum of Digits Pseudorandomness of Powers Let S q ( p n ) denote the sum of digits of p n in base q . If the digits were to be independent, uniformly distributed random variables then the pseudo expectation would be: E n ≈ n log q p · q − 1 2 and pseudo variance V n ≈ n log q p · q 2 − 1 ; 12 and the corresponding normed and centered variable s q ( p n ) should behave as the standard normal distribution . We can try to verify this experimentally...

Introduction Main Results Conclusion Sum of Digits Distribution of Normalized Digit Sums The results for n up to 10 5 : 0 . 4 0 . 2 0 − 4 − 3 − 2 − 1 0 1 2 3 4 Figure : Histogram of the centered and normed variable s 3 (2 n )

Introduction Main Results Conclusion Sum of Digits Related Theoretical Results Conjecture by Paul Erd˝ os For n > 8, the base-3 representation of 2 n contains digit “2”. Corollary of a theorem by C. L. Stewart There exists a constant C p , q > 0 such that: log n S q ( p n ) > − 1 . log log n + C p , q Our result If H1 holds, i.e., if indeed � 2 n � = 2 n , then S 3 (2 n ) > 0 . 107 n . Does this mean proving H1 is very difficult?

Introduction Main Results Conclusion Sum of Digits H1 Implies Linear Sum of Digits Theorem (ˇ Cer¸ nenoks et al.) If, for a prime p, ∃ ǫ > 0 ∀ n > 0 : � p n � log ≥ 3 + ǫ , then S 3 ( p n ) ≥ ǫ n log 3 p . Proof. Write p n in base q : a m a m − 1 · · · a 0 . Using Horner’s rule we obtain an arithmetic expression for p n : � p n � ≤ qm + S q ( p n ) . Since m ≤ log q p n , � p n � ≤ q log q p n + S q ( p n ) .

Introduction Main Results Conclusion Sum of Digits H1 Implies Linear Sum of Digits Theorem (ˇ Cer¸ nenoks et al.) If, for a prime p, ∃ ǫ > 0 ∀ n > 0 : � p n � log ≥ 3 + ǫ , then S 3 ( p n ) ≥ ǫ n log 3 p . Proof. � p n � ≤ q log q p n + S q ( p n ) . When q = 3: S 3 ( p n ) ≥ � p n � log log 3 p n − 3 log 3 p n ≥ ≥ (3 + ǫ ) n log 3 p − 3 n log 3 p = = ǫ n log 3 p .

Introduction Main Results Conclusion Integer Complexity in Basis { 1 , + , · , −} Definition Integer complexity in basis { 1 , + , · , −} Integer complexity (in basis { 1 , + , · , −} ) of a positive integer n , denoted by � n � − , is the least amount of 1’s in an arithmetic expression for n consisting of 1’s, +, · , − and brackets. The corresponding logarithmic complexity is denoted by � n � − log . Having computed � n � − for n up to 2 · 10 11 we present our observations.

Introduction Main Results Conclusion Integer Complexity in Basis { 1 , + , · , −} Experimental Results in Basis { 1 , + , · , −} [1] Smallest number with � n � − < � n � : � 23 � − = 10; � 23 � = 11; 23 = 2 3 · 3 − 1 = 2 2 · 5 + 2 . There are numbers for which subtraction of 6 is necessary: � n � − = 75; n = 55 659 409 816 = (2 4 · 3 3 − 1)(3 17 − 1) − 2 · 3; � n � − = 77; n = 111 534 056 696 = (2 5 · 3 4 − 1)(3 16 + 1) − 2 · 3; � n � − = 78; n = 167 494 790 108 = (2 4 · 3 4 + 1)(3 17 − 1) − 2 · 3 .

Introduction Main Results Conclusion Integer Complexity in Basis { 1 , + , · , −} Experimental Results in Basis { 1 , + , · , −} [2] “Worst” numbers Let e ( n ) denote min { k | � k � = n } and e − ( n ) denote min { k | � k � − = n } . 4 � e ( n ) � log 3 . 8 � e − ( n ) � − log 3 . 6 3 . 4 3 . 2 3 0 10 20 30 40 50 60 70 80 90 n Figure : Logarithmic complexities of the numbers e ( n ) and e − ( n )

Introduction Main Results Conclusion Integer Complexity in Basis { 1 , + , · , −} Upper Bound Theorem (ˇ Cer¸ nenoks et al.) � n � − log ≤ 3 . 679 + 5 . 890 log 3 n Sketch of proof. n = 6 k ; write n as 3 · 2 · k ; n = 6 k + 1; write n as 3 · 2 · k + 1; n = 6 k + 2; write n as 2 · (3 · k + 1); n = 6 k + 3; write n as 3 · (2 · k + 1); n = 6 k + 4; write n as 2 · (3 · ( k + 1) − 1); n = 6 k + 5; write n as 3 · 2 · ( k + 1) − 1;