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kelalaka
kelalaka
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Phi-hiding assumption can be simply stated as (wrt hardness)

It is difficult to findingfind small factors of φ(m)$\varphi(m)$ where m$m$ is a number whose whose factorization is unknown, and φ$\varphi$ is Euler's totient function.

Is the hardness due to this assumption comparatively higher than than the hardness of integer factorization?

My intuition says that finding prime factors of φ(m)$\varphi(m)$ is simpler than finding the prime factors of m$m$. So I am believe that the hardness of the phi-hiding assumption is at most equal to the hardness of integer factorization.

Phi-hiding assumption can be simply stated as (wrt hardness)

It is difficult to finding small factors of φ(m) where m is a number whose factorization is unknown, and φ is Euler's totient function.

Is the hardness due to this assumption comparatively higher than than the hardness of integer factorization?

My intuition says that finding prime factors of φ(m) is simpler than finding the prime factors of m. So I am believe that the hardness of the phi-hiding assumption is at most equal to the hardness of integer factorization.

Phi-hiding assumption can be simply stated as (wrt hardness)

It is difficult to find small factors of $\varphi(m)$ where $m$ is a number whose factorization is unknown and $\varphi$ is Euler's totient function.

Is the hardness due to this assumption comparatively higher than than the hardness of integer factorization?

My intuition says that finding prime factors of $\varphi(m)$ is simpler than finding the prime factors of $m$. So I believe that the hardness of the phi-hiding assumption is at most equal to the hardness of integer factorization.

Grammar and made question clearer
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edit approved Aug 3, 2020 at 11:15
Aman Grewal
edit approved Aug 3, 2020 at 11:15
Aman Grewal
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Phi-hiding assumption, as a layman, I can be simply state itstated as (wrt hardness)

It is difficult to finding small factors of φ(m) where m is a number whose factorization is unknown, and φ is Euler's totient function.

Is the hardness due to this assumption is comparatively higher than than the hardness of integer factorisationfactorization?

My intuition says that finding prime factors euler-totientof φ(m) is simpler than finding the prime factorfactors of a numberm. So I am believingbelieve that the hardness of Phithe phi-hiding assumption is atmostat most equal to the hardness of integer factorization.

Phi-hiding assumption, as a layman, I can be simply state it as (wrt hardness)

It is difficult to finding small factors of φ(m) where m is a number whose factorization is unknown, and φ is Euler's totient function.

Is the hardness due to this assumption is comparatively higher than than the hardness of integer factorisation?

My intuition says that finding prime factors euler-totient is simpler than finding the prime factor of a number. So I am believing that the hardness of Phi-hiding assumption is atmost equal to the hardness of integer factorization.

Phi-hiding assumption can be simply stated as (wrt hardness)

It is difficult to finding small factors of φ(m) where m is a number whose factorization is unknown, and φ is Euler's totient function.

Is the hardness due to this assumption comparatively higher than than the hardness of integer factorization?

My intuition says that finding prime factors of φ(m) is simpler than finding the prime factors of m. So I am believe that the hardness of the phi-hiding assumption is at most equal to the hardness of integer factorization.

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Fateh
Fateh
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Is phi-hiding assumption as hard as integer factorization?

Phi-hiding assumption, as a layman, I can be simply state it as (wrt hardness)

It is difficult to finding small factors of φ(m) where m is a number whose factorization is unknown, and φ is Euler's totient function.

Is the hardness due to this assumption is comparatively higher than than the hardness of integer factorisation?

My intuition says that finding prime factors euler-totient is simpler than finding the prime factor of a number. So I am believing that the hardness of Phi-hiding assumption is atmost equal to the hardness of integer factorization.