Scholz ’ s First Conjecture : A Brief Demonstration

This paper presents a brief demonstration of Schulz’s first conjecture, which sets the upper and lower limits on the length of the shortest chain of addition. Two methods of the upper limit are demonstrated; the second one is based on the algorithm of one of the most popular methods for obtaining addition chains of a number, known as the binary method.


Introduction
With the development of the Internet and adding chain applications in the development of cryptography-which permits safe handling of data over the Internet-the theme began with the publication in 1937 of Arnold Scholz's paper [1], which defines a minimal addition chain, along with his three famous conjectures.In 1939 [2], Alfred Brauer gave a strong impetus to this issue, gaining importance in this area.During the last decades of the last century deterministic methods flourished.The most popular were the binary method and the window method [3]- [5].Heuristic methods began to emerge in the 70s and toward the end of the century, began to dominate in the literature [6] [7].This is the second paper we write on the theme [8] and in both present simple demonstrations, the third and first conjecture, we use deterministic algorithms.Our intention is to build a framework for the development of intelligent methods for generating addition chains.
In addition to setting maximum and lower bounds for the minimum length of addition chains, this conjecture induces a partition of our study space, in this case the natural numbers.The bounded sets defined by Schulz 1 2 1 2 m m n + + ≤ ≤ for 1 m ≥ , exclude the numbers 1 and 2. For 1 m = the possible values of n correspond to the set { } 3, 4 ; from there, if we increase m by one, the possible values of n are doubled.The point of this partition is that we can delimit our study space, in this case the natural numbers, to set general properties on addition chains.
We will begin our discussion with the following definitions: be a finite sequence of natural numbers.We will call it an addition chain of a natural number if it satisfies: I.
a a a = + , for some ( ) The set i G will be called ith generation of natural numbers.Definition 2.5.For every ith generation with 2 i ≥ , we will define: Definition 2.6.For every n N ∈ , we will define the nth dominant chain as:

Important Properties
Proposition 3.1.The dominant chains are of maximum growth.That is to say, for every x if then there exists 1 j r < ≤ (since the first two values of any addition chain are the same) in such a way that j j b a ≠ as < − because if the first term of the sum had an index less than 1 j − , at most it could be 1 j b − , which would imply that the sequence would not be an addition chain because it does not strictly increase, and as 1 k j < − and the sequence strictly increases we have that: j j b a < .If it was the last term of the sequence then the inequality is true, and if it was not the last term from there j i j i b a at each increment of the difference between the terms of the sequences increases.Hence 2 r r r x b a = < = .Proposition 3.2.Dominant chains are addition chains defined on the numbers of the form 2 n e = , of length ( ) From the definition of dominant chain: 0 0 2 1 a = = , from where its first element is 1and the last one is 2 n n a = , in addition for every i the following is true: from where 0 i n < < , therefore it in- creases and ends in 2 n n a = , which proves the first property of addition chain.Let us have a look at the second property, that is: clearly 0 k j i ≤ ≤ < , with 0 i r < ≤ , so it satisfies the second property and therefore they are addition chains.Finally, since 2 n n a = is an addition chain of 2 n , the Proposition (3.1) assures that any other chain of that length is of a number 2 n x < , which implies that it is the only chain of length n de 2 n .This proves that ( ) ( ) = implies that there exists x S in such a way that ( ) Proof.If we assume that (3.5) is not true, then there exists i x G ∈ in such a way that ( ) S in such a way that ( ) , and this implies that i x G ∉ , which contradicts our hypothesis, from which we conclude that if be a minimum length chain of x, from this one we will build an addition chain of z.
As x is an even number and lower than the upper limit of i G , that is 2 i x < ; then 2 2 be a minimum length chain ( ) l x , then its first term is 1 and its last term is x, from where if we add as the last term 1 z x = + to that sequence, now this sequence is an addition chain of z, that is: , it is an addition chain of Z of length ( ) 1 l x + , from where ( ) ( ) ; , from these numbers only the first predecessor in i G , an addition chain of this number is given by , whose length is ( ) this length is minimum since if there exists another chain of z with a length lower than n according to the Corollary (3.4), we have that , from where we conclude that it is minimum, that is ( ) For the rest of the elements of n i G , according to the Proposition (3.7), there exists a is a partition of i G .Proof.We have to demonstrate that for every 2 i ≥ we will always have: a) Proof.The only addition chain of the number 3 is: , we will demonstrate that it is an addition chain and that it is unique.
It is an addition chain since it begins with 1, it has increasing terms and it ends with the number 3, from where it satisfies the property I of the definition of the chain addition.
Clearly 1 0 0 1 1 2 a a a = + = + = and 2 1 0 2 1 3 a a a = + = + = , which proves part II of the definition ;0 a a a i j k = + ≤ ≤ < .We will demonstrate now that it is unique.Let us suppose that it is not, then there exists 3 be an addition chain different from 3 S , the defini- tion of chain forces 0 1 b = and the property II when 1 k = , and 0 1 i j k ≤ ≤ < =, the only possible value of i and j is zero, from where 1 2 b = .For 2 b the possible values of ( ) , i j are three, which are presented in Table 1.
The only possible value for 2 b is 3, according to the definition of addition chain it would be the last term of the sequence.However, it is equal to 3 S , which contradicts the fact that 3 S + is different from 3 S , from where we conclude that it is unique.As ( )

Demonstration of Scholz's First Conjecture
In terms of the given definitions, Scholz's first conjecture is equivalent to: ( )  x G − ∈ .The Proposition (3.9) proves that , we have that for every i, , that is ( ) = , its minimum addition chain is ( ) , according to the Corollary (3.3) and the 3 according to the Proposition (3.10) ( ) = , from where the upper bound of 2 G is 2.This is 2 2 k = , and substituting this value of 2 k in (2) we will have: l x i ≤ − ; for 2 i ≥ .This demonstrates the second part of the inequality.An alternative way of deriving the upper bound established in Schulz's first conjecture is obtainable by using the binary analysis methods for constructing the addition chains [3].It is expressed in the following algorithm.
Table 1.Sequence of addition chain for possible values of ( ) The sequence would not be strictly increasing The last value must be 3, there can't be higher numbers

Binary Method
Input: an integer: e e e e − =  Output: Addition chain Therefore the length of the obtained chain depends on the length of the binary expression of the number and the quantity of ones that it has, as for each one, without taking into account the first one, is added another number to the output chain; for example see Table 2

{ }
The output chain has the same number of elements that the number of bytes of the expression, in base 2, of the input number e, plus the quantity of ones of the binary expression, minus one, which is at the beginning of the binary expression.In this case: 77 = 1001101, the length of the binary expression is of 7 bytes and the number of ones minus the first one is 3. Hence the length of the output chain is 7 + 3 = 10; U = {1, 2, 4, 8, 9, 18, 19, 38, 76, 77}.Therefore the length of the addition chain is 9.This fact is expressed in the next proposition, as follows: Proposition 5.1.The length of the addition chain generated by the binary method of a number e is equal to the last sub-index of the binary expression The algorithm starts with a one and eliminates 0 e , it is equal to one, and adds it to the addition chain U.For each element from 1 e to m e that is for m elements we have added one more to U.
, , , , , , , where p is quantifies the number of e i 's equal to one.Therefore, according to Definition 2.2 the length of the addition chain is equal to the last sub-index of the chain, in this case 1 m p + − .As m is the last sub-index of the binary expression, and p is the number of ones in the expression.Note that we have that Then from Definition 2.2, we derive that the length of the chain is the last of the sub-indexes, which is 1 m p + − .Q.E.D. Note that the numbers belonging to the generation G i , for 2 i > have has binary expression with length equal to n, minus the superior limit, which has 1 n + elements with 1 and n with zeros, which corresponds to the su- perior limit of the generation, as it is observed in the following Table 3.
Note that the upper bound of the generation is given by 2 n ; its minimum addition chain is n, because of Proposition 3.2.Then we do not take i into account.The maximum expected length of a chain, belonging to the generation n, is given by the number whose binary expression is equal to n, corresponding to 2 1 n e = − , one less than the superior limit, with binary expression: ( ) . The obtained addition chain will have n − 1 components, as the number of ones of the expression is n, where the addition chain is given by { } Hence, the length of this chain is 2 2 n − , which is the longest gener- ated by the method in that generation.Then is proved the following result: Corollary 5.1.The length of the longest addition chain generated by the binary method in i G corresponds to 2 1 i e = − that has as length ( )  e is equal to 1, because if all are zeros, it will be the superior limit of 1 i G − .The generation i has 1 i − elements whose binary expression has two ones, for these elements, according to Proposition 5.1, its generated addition chain is of length equal to i, as we add only another value and the first value of the generated chain has as sub-index 1 1 i i = − + .The rest will have longer chains.The longest corresponding to a 2 1 i n = − has length 2 2 i − due to the results of corollary.This fact proofs that all the chains generated with this method are not larger than 2 2 i − .Hence, the mini- mum chain must be smaller or equal to these values.This last result completes the proof.

3 . 3 .SCorollary 3 . 4 .
last fact we will state the following in this corollary: Corollary The numbers of the form 2 is addition chain of 2 n , and the Proposition (3.1) states that any other chain of that length different from If ( ) not have an even numbered = ≤ ≤ = = , from where we clearly see that a) and b) are true.Proposition 3.10.The number 3 only has an addition chain of length equal to 2.
Proof.The Proposition (3.5) guarantees the first part of the inequality, that is: if demonstrate the second part of the inequality.The Proposition (3.6) guarantees us that if the quantity of ones it has, minus one.That is 1 m p + − , where p is the number of ones that the binary expression e has.Proof: Take e ∈  ; and its binary expression The minimum length of all addition chains of a natural number e is denoted by( ) e l S .Definition 2.3.

Table 3 .
Example binary sequence.Theorem 5.1.The binary method for the generation of addition chains proofs the right hand side of the First Schulz's Conjecture that is: If