From 27e90b5f3b84e58577f411d56a1e1582deb62850 Mon Sep 17 00:00:00 2001 From: t0wbo2t <52655804+t0wbo2t@users.noreply.github.com> Date: Thu, 15 May 2025 10:18:15 +0530 Subject: [PATCH 1/6] Update factorization.md [Update Powersmooth Definition] The definition of powersmooth was a bit confusing so I added a formal definition inspired by a number theory book. --- src/algebra/factorization.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/src/algebra/factorization.md b/src/algebra/factorization.md index 11bf4049c..51f206482 100644 --- a/src/algebra/factorization.md +++ b/src/algebra/factorization.md @@ -160,7 +160,7 @@ By looking at the squares $a^2$ modulo a fixed small number, it can be observed ## Pollard's $p - 1$ method { data-toc-label="Pollard's method" } It is very likely that at least one factor of a number is $B$**-powersmooth** for small $B$. -$B$-powersmooth means that every prime power $d^k$ that divides $p-1$ is at most $B$. +$B$-powersmooth means that every prime power $d^k$ that divides $p-1$ is at most $B$. Formally, let $\mathrm{B} \geqslant 1$ and $n \geqslant 1$ with prime factorization $n = \prod {p_i}^{e_i},$ then $n$ is $\mathrm{B}$-powersmooth if, for all $i,$ ${p_i}^{e_i} \leqslant \mathrm{B}$. E.g. the prime factorization of $4817191$ is $1303 \cdot 3697$. And the factors are $31$-powersmooth and $16$-powersmooth respectably, because $1303 - 1 = 2 \cdot 3 \cdot 7 \cdot 31$ and $3697 - 1 = 2^4 \cdot 3 \cdot 7 \cdot 11$. In 1974 John Pollard invented a method to extracts $B$-powersmooth factors from a composite number. From 422fb19bdd7c39a56d8fb51caefa419251b37fab Mon Sep 17 00:00:00 2001 From: t0wbo2t <52655804+t0wbo2t@users.noreply.github.com> Date: Wed, 21 May 2025 17:33:44 +0530 Subject: [PATCH 2/6] Update factorization.md [Update powersmooth definition with suggestions] Commas are now outside $...$ and definiton is for (p - 1) for consistency. --- src/algebra/factorization.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/src/algebra/factorization.md b/src/algebra/factorization.md index 51f206482..997a11be2 100644 --- a/src/algebra/factorization.md +++ b/src/algebra/factorization.md @@ -160,7 +160,7 @@ By looking at the squares $a^2$ modulo a fixed small number, it can be observed ## Pollard's $p - 1$ method { data-toc-label="Pollard's method" } It is very likely that at least one factor of a number is $B$**-powersmooth** for small $B$. -$B$-powersmooth means that every prime power $d^k$ that divides $p-1$ is at most $B$. Formally, let $\mathrm{B} \geqslant 1$ and $n \geqslant 1$ with prime factorization $n = \prod {p_i}^{e_i},$ then $n$ is $\mathrm{B}$-powersmooth if, for all $i,$ ${p_i}^{e_i} \leqslant \mathrm{B}$. +$B$-powersmooth means that every prime power $d^k$ that divides $p-1$ is at most $B$. Formally, let $\mathrm{B} \geqslant 1$ and let $p$ be a prime such that $(p - 1) \geqslant 1$. Suppose the prime factorization of $(p - 1)$ is $(p - 1) = \prod {q_i}^{e_i}$, where each $q_i$ is a prime and $e_i \geqslant 1$ then $(p - 1)$ is $\mathrm{B}$-powersmooth if, for all $i$, ${q_i}^{e_i} \leqslant \mathrm{B}$. E.g. the prime factorization of $4817191$ is $1303 \cdot 3697$. And the factors are $31$-powersmooth and $16$-powersmooth respectably, because $1303 - 1 = 2 \cdot 3 \cdot 7 \cdot 31$ and $3697 - 1 = 2^4 \cdot 3 \cdot 7 \cdot 11$. In 1974 John Pollard invented a method to extracts $B$-powersmooth factors from a composite number. From 1a7db31f720455b647a929596ff70936875f7532 Mon Sep 17 00:00:00 2001 From: t0wbo2t <52655804+t0wbo2t@users.noreply.github.com> Date: Thu, 22 May 2025 09:24:40 +0530 Subject: [PATCH 3/6] Update factorization.md [Update Pollard's (p - 1) Method] Updated materials related to powersmoothness. Corrected some minor mistakes. --- src/algebra/factorization.md | 13 ++++++------- 1 file changed, 6 insertions(+), 7 deletions(-) diff --git a/src/algebra/factorization.md b/src/algebra/factorization.md index 997a11be2..84bdd356d 100644 --- a/src/algebra/factorization.md +++ b/src/algebra/factorization.md @@ -159,11 +159,10 @@ By looking at the squares $a^2$ modulo a fixed small number, it can be observed ## Pollard's $p - 1$ method { data-toc-label="Pollard's method" } -It is very likely that at least one factor of a number is $B$**-powersmooth** for small $B$. -$B$-powersmooth means that every prime power $d^k$ that divides $p-1$ is at most $B$. Formally, let $\mathrm{B} \geqslant 1$ and let $p$ be a prime such that $(p - 1) \geqslant 1$. Suppose the prime factorization of $(p - 1)$ is $(p - 1) = \prod {q_i}^{e_i}$, where each $q_i$ is a prime and $e_i \geqslant 1$ then $(p - 1)$ is $\mathrm{B}$-powersmooth if, for all $i$, ${q_i}^{e_i} \leqslant \mathrm{B}$. +It is very likely that a number $n$ has at least one prime factor $p$ such that $p - 1$ is $\mathrm{B}$**-powersmooth** for small $\mathrm{B}$. An integer $m$ is said to be $\mathrm{B}$-powersmooth if every prime power dividing $m$ is at most $\mathrm{B}$. Formally, let $\mathrm{B} \geqslant 1$ and let $m$ be any positive integer. Suppose the prime factorization of $m$ is $m = \prod {q_i}^{e_i}$, where each $q_i$ is a prime and $e_i \geqslant 1$. Then $m$ is $\mathrm{B}$-powersmooth if, for all $i$, ${q_i}^{e_i} \leqslant \mathrm{B}$. E.g. the prime factorization of $4817191$ is $1303 \cdot 3697$. -And the factors are $31$-powersmooth and $16$-powersmooth respectably, because $1303 - 1 = 2 \cdot 3 \cdot 7 \cdot 31$ and $3697 - 1 = 2^4 \cdot 3 \cdot 7 \cdot 11$. -In 1974 John Pollard invented a method to extracts $B$-powersmooth factors from a composite number. +And the values, $1303 - 1$ and $3697 - 1$, are $31$-powersmooth and $16$-powersmooth respectively, because $1303 - 1 = 2 \cdot 3 \cdot 7 \cdot 31$ and $3697 - 1 = 2^4 \cdot 3 \cdot 7 \cdot 11$. +In 1974 John Pollard invented a method to extracts $\mathrm{B}$-powersmooth factors from a composite number. The idea comes from [Fermat's little theorem](phi-function.md#application). Let a factorization of $n$ be $n = p \cdot q$. @@ -180,7 +179,7 @@ This means that $a^M - 1 = p \cdot r$, and because of that also $p ~|~ \gcd(a^M Therefore, if $p - 1$ for a factor $p$ of $n$ divides $M$, we can extract a factor using [Euclid's algorithm](euclid-algorithm.md). -It is clear, that the smallest $M$ that is a multiple of every $B$-powersmooth number is $\text{lcm}(1,~2~,3~,4~,~\dots,~B)$. +It is clear, that the smallest $M$ that is a multiple of every $\mathrm{B}$-powersmooth number is $\text{lcm}(1,~2~,3~,4~,~\dots,~B)$. Or alternatively: $$M = \prod_{\text{prime } q \le B} q^{\lfloor \log_q B \rfloor}$$ @@ -189,11 +188,11 @@ Notice, if $p-1$ divides $M$ for all prime factors $p$ of $n$, then $\gcd(a^M - In this case we don't receive a factor. Therefore, we will try to perform the $\gcd$ multiple times, while we compute $M$. -Some composite numbers don't have $B$-powersmooth factors for small $B$. +Some composite numbers don't have $\mathrm{B}$-powersmooth factors for small $\mathrm{B}$. For example, the factors of the composite number $100~000~000~000~000~493 = 763~013 \cdot 131~059~365~961$ are $190~753$-powersmooth and $1~092~161~383$-powersmooth. We will have to choose $B >= 190~753$ to factorize the number. -In the following implementation we start with $B = 10$ and increase $B$ after each each iteration. +In the following implementation we start with $\mathrm{B} = 10$ and increase $\mathrm{B}$ after each each iteration. ```{.cpp file=factorization_p_minus_1} long long pollards_p_minus_1(long long n) { From 2597e0558304678a7fa7f92c66a19f9f7913c974 Mon Sep 17 00:00:00 2001 From: t0wbo2t <52655804+t0wbo2t@users.noreply.github.com> Date: Thu, 29 May 2025 11:19:43 +0530 Subject: [PATCH 4/6] Update src/algebra/factorization.md Co-authored-by: Oleksandr Kulkov --- src/algebra/factorization.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/src/algebra/factorization.md b/src/algebra/factorization.md index 84bdd356d..bc606607f 100644 --- a/src/algebra/factorization.md +++ b/src/algebra/factorization.md @@ -188,8 +188,8 @@ Notice, if $p-1$ divides $M$ for all prime factors $p$ of $n$, then $\gcd(a^M - In this case we don't receive a factor. Therefore, we will try to perform the $\gcd$ multiple times, while we compute $M$. -Some composite numbers don't have $\mathrm{B}$-powersmooth factors for small $\mathrm{B}$. -For example, the factors of the composite number $100~000~000~000~000~493 = 763~013 \cdot 131~059~365~961$ are $190~753$-powersmooth and $1~092~161~383$-powersmooth. +Some composite numbers don't have factors $p$ s.t. $p-1$ is $\mathrm{B}$-powersmooth for small $\mathrm{B}$. +For example, for the composite number $100~000~000~000~000~493 = 763~013 \cdot 131~059~365~961$, values $p-1$ are $190~753$-powersmooth and $1~092~161~383$-powersmooth correspondingly. We will have to choose $B >= 190~753$ to factorize the number. In the following implementation we start with $\mathrm{B} = 10$ and increase $\mathrm{B}$ after each each iteration. From 54fec62526c6454ecd43b38544c6a56880031544 Mon Sep 17 00:00:00 2001 From: t0wbo2t <52655804+t0wbo2t@users.noreply.github.com> Date: Thu, 29 May 2025 11:19:54 +0530 Subject: [PATCH 5/6] Update src/algebra/factorization.md Co-authored-by: Oleksandr Kulkov --- src/algebra/factorization.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/src/algebra/factorization.md b/src/algebra/factorization.md index bc606607f..9d9ab7ed7 100644 --- a/src/algebra/factorization.md +++ b/src/algebra/factorization.md @@ -162,7 +162,7 @@ By looking at the squares $a^2$ modulo a fixed small number, it can be observed It is very likely that a number $n$ has at least one prime factor $p$ such that $p - 1$ is $\mathrm{B}$**-powersmooth** for small $\mathrm{B}$. An integer $m$ is said to be $\mathrm{B}$-powersmooth if every prime power dividing $m$ is at most $\mathrm{B}$. Formally, let $\mathrm{B} \geqslant 1$ and let $m$ be any positive integer. Suppose the prime factorization of $m$ is $m = \prod {q_i}^{e_i}$, where each $q_i$ is a prime and $e_i \geqslant 1$. Then $m$ is $\mathrm{B}$-powersmooth if, for all $i$, ${q_i}^{e_i} \leqslant \mathrm{B}$. E.g. the prime factorization of $4817191$ is $1303 \cdot 3697$. And the values, $1303 - 1$ and $3697 - 1$, are $31$-powersmooth and $16$-powersmooth respectively, because $1303 - 1 = 2 \cdot 3 \cdot 7 \cdot 31$ and $3697 - 1 = 2^4 \cdot 3 \cdot 7 \cdot 11$. -In 1974 John Pollard invented a method to extracts $\mathrm{B}$-powersmooth factors from a composite number. +In 1974 John Pollard invented a method to extract factors $p$, s.t. $p-1$ is $\mathrm{B}$-powersmooth, from a composite number. The idea comes from [Fermat's little theorem](phi-function.md#application). Let a factorization of $n$ be $n = p \cdot q$. From 53639d500284ae160ca2e9f968957c360b6f2040 Mon Sep 17 00:00:00 2001 From: t0wbo2t <52655804+t0wbo2t@users.noreply.github.com> Date: Thu, 29 May 2025 11:20:00 +0530 Subject: [PATCH 6/6] Update src/algebra/factorization.md Co-authored-by: Oleksandr Kulkov --- src/algebra/factorization.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/src/algebra/factorization.md b/src/algebra/factorization.md index 9d9ab7ed7..14715605f 100644 --- a/src/algebra/factorization.md +++ b/src/algebra/factorization.md @@ -190,7 +190,7 @@ Therefore, we will try to perform the $\gcd$ multiple times, while we compute $M Some composite numbers don't have factors $p$ s.t. $p-1$ is $\mathrm{B}$-powersmooth for small $\mathrm{B}$. For example, for the composite number $100~000~000~000~000~493 = 763~013 \cdot 131~059~365~961$, values $p-1$ are $190~753$-powersmooth and $1~092~161~383$-powersmooth correspondingly. -We will have to choose $B >= 190~753$ to factorize the number. +We will have to choose $B \geq 190~753$ to factorize the number. In the following implementation we start with $\mathrm{B} = 10$ and increase $\mathrm{B}$ after each each iteration. pFad - Phonifier reborn

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