maths.free › Abstract Algebra › Cyclic Groups › The Method of Repeated Squares

The Method of Repeated Squares

Computing large powers can be very time-consuming. Just as anyone can compute 2^2 or 2^8, everyone knows how to compute \[\begin{aligned}\end{aligned}\].

The Method of Repeated Squares

Computing large powers can be very time-consuming. Just as anyone can compute \(2^2\) or \(2^8\), everyone knows how to compute \[\begin{aligned}\end{aligned}\]. However, such numbers are so large that we do not want to attempt the calculations; moreover, past a certain point the computations would not be feasible even if we had every computer in the world at our disposal. Even writing down the decimal representation of a very large number may not be reasonable. It could be thousands or even millions of digits long. However, if we could compute something like \[\begin{aligned}\end{aligned}\], we could very easily write the result down since it would be a number between \(0\) and \(46{,}388\). If we want to compute powers modulo \(n\) quickly and efficiently, we will have to be clever. The results in this section are needed only in

The first thing to notice is that any number \(a\) can be written as the sum of distinct powers of \(2\); that is, we can write \[\begin{aligned}\end{aligned}\], where \(k_1 \lt k_2 \lt \cdots \lt k_n\). This is just the binary representation of \(a\). For example, the binary representation of \(57\) is \(111001\), since we can write \(57 = 2^0 + 2^3 + 2^4 + 2^5\).

The laws of exponents still work in \({\mathbb Z}_n\); that is, if \(b \equiv a^x \pmod{ n}\) and \(c \equiv a^y \pmod{ n}\), then \(bc \equiv a^{x+y} \pmod{ n}\). We can compute \(a^{2^k} \pmod{ n}\) in \(k\) multiplications by computing \[\begin{aligned}a^{2^0} \pmod{ n} \\ a^{2^1} \pmod{ n } \\ \vdots \\ a^{2^k} \pmod{ n}\end{aligned}\]. Each step involves squaring the answer obtained in the previous step, dividing by \(n\), and taking the remainder.

The method of repeated squares will prove to be a very useful tool when we explore RSA cryptography in . To encode and decode messages in a reasonable manner under this scheme, it is necessary to be able to quickly compute large powers of integers mod \(n\).

Sage support for cyclic groups is a little spotty but we can still make effective use of Sage and perhaps this situation could change soon.

Condensed: the full section is in Judson, Abstract Algebra: Theory and Applications.

Jetzt du Kein Rechner löst diese Frage, aber ihre Teile sind berechenbar. Probieren Sie unten eine aus oder geben Sie Ihre eigene ein.

Behalten Sie Ihre eigenen Lösungswege

Ein kostenloses Konto fügt Notizen zu jeder Lektion, eine Aufzeichnung von dem, was Sie beendet haben, Ihre gelösten Probleme an einem Ort, und ein Tutor können Sie über diese Seite fragen. Die Mathematik selbst ist offen für alle, angemeldet oder nicht.

Registrieren Anmelden

Hier verwendete Symbole

Tippen Sie auf jedes Symbol für die vollständige Definition, ein Bild und was jeder Buchstabe darin bedeutet.

Fragen, die die Leute stellen

What is a group, in plain words?

A set with one operation that is associative, has an identity, and lets every element be undone. Symmetries of any object form a group. That is where the idea came from.

What is the difference between a ring and a field?

A ring has addition and multiplication that behave like the integers (you cannot always divide); a field is a ring where every non-zero element has a reciprocal, like the rationals or the reals.

Teile dieser Seite sind adaptiert aus Judson, Abstract Algebra: Theory and Applications (GFDL 1.3). Hier gekürzt und neu erklärt; Fehler gehen auf uns.

Mehr in Abstract Algebra