# CS代考 EECS 70 Discrete Mathematics and Probability Theory Fall 2021 – cscodehelp代写

EECS 70 Discrete Mathematics and Probability Theory Fall 2021
Polynomials
Polynomials constitute a rich class of functions which are both easy to describe and widely applicable in topics ranging from Fourier analysis, cryptography and communication, to control and computational geom- etry. You’ve seen them earlier in many contexts like Taylor approximation and other contexts in EECS16B, and so they are familiar to you. In this note, we will discuss further properties of polynomials which make them so useful. The key idea here is to extend what you already know about polynomials over the real and complex numbers to modulo arithmetic. We will then describe how to take advantage of these properties to develop a secret-sharing scheme.
Recall that a polynomial in a single variable is an expression that has an associated function.1 The poly- nomial expression p(x) = adxd +ad−1xd−1 +···+a1x+a0. Here the variable x and the coefficients ai are usually real numbers. For example, p(x) = 5×3 + 2x + 1 is a polynomial of degree d = 3. Its coefficients are a3 = 5, a2 = 0, a1 = 2, and a0 = 1. Polynomials have some remarkably simple, elegant and powerful properties, which we will explore in this note.
The polynomial function can be viewed as a function in some domain with multiplication and addition operations, by evaluating the expression in the natural manner. For example, p(x) = 5×3 + 2x + 1 can be evaluated at x = 3 and to obtain p(3) = 5(3)3 +2(3)+1 = 142.
To proceed, a familiar definition: we say that a is a root of the polynomial p(x) if p(a) = 0. For example, the degree 2 polynomial p(x) = x2 − 4 has two roots, namely 2 and −2, since p(2) = p(−2) = 0. If we plot the polynomial p(x) in the x-y plane, then the roots of the polynomial are just the places where the curve crosses the x axis:
We now state two fundamental properties of polynomials that we will prove in due course. Property 1: A non-zero polynomial of degree d has at most d roots.
Property2:Givend+1pairs(x1,y1),…,(xd+1,yd+1),withallthexi distinct,thereisauniquepolynomial p(x)ofdegree(atmost)dsuchthatp(xi)=yi for1≤i≤d+1.
Let us consider what these two properties say in the case that d = 1. A plot over the reals of a linear (degree
1In the reals, the association is unique. In modular arithmetic modulo a prime p, it isn’t necessarily unique as modulo p, by Fermat’s Little Theorem, xp = x mod p for a prime p. Still, there is a uniquely associated polynomial expression to a function where the expression does not have terms with exponents larger than p − 1 under arithmetic modulo a prime p. Our applications generally assume that this is the expression we are working with.
EECS 70, Fall 2021, Note 8 1
1) polynomial y = a1 x + a0 is a line that may not go through the origin. Property 1 says that if a line is not the x-axis (i.e., if the polynomial is not y = 0), then it can intersect the x-axis in at most one point.
Property 2 says that two points uniquely determine a line.
Polynomial Interpolation
Property 2 says that two points uniquely determine a degree 1 polynomial (a line), three points uniquely determine a degree 2 polynomial, four points uniquely determine a degree 3 polynomial, and so on. Given d+1pairs(x1,y1),…,(xd+1,yd+1),howdowedeterminethepolynomialp(x)=adxd+···+a1x+a0 such that p(xi)=yi fori=1tod+1? Wewillgiveanalgorithmforreconstructingthecoefficientsa0,…,ad,and therefore the polynomial p(x). Another standard algorithm that is more clearly linear-algebraic is described in the next note, because it will be important as a stepping stone.
The method here is called Lagrange interpolation and it should remind you of the Chinese Remainder Theorem: Let us start by solving an easier problem. Suppose that we are told that y1 = 1 and y j = 0 for 2 ≤ j≤d+1. Nowcanwereconstruct p(x)? Yes,thisiseasy! Considerq(x)=(x−x2)(x−x3)···(x−xd+1). This is a polynomial of degree d (the xi’s are constants, and x appears d times). Also, we clearly have q(xj)=0for2≤ j≤d+1. Butwhatisq(x1)? Well,q(x1)=(x1−x2)(x1−x3)···(x1−xd+1),whichis some constant not equal to 0 (since x1 is not equal to any of the other xi). Thus if we let p(x) = q(x)/q(x1) (dividing is ok since q(x1) ̸= 0), we have the polynomial we are looking for. For example, suppose you were given the pairs (1, 1), (2, 0), and (3, 0). Then we can construct the degree d = 2 polynomial p(x) by letting q(x) = (x−2)(x−3) = x2 −5x+6, and q(x1) = q(1) = 2. Thus, we can now construct p(x) = q(x)/q(x1) =
(x2 − 5x + 6)/2.
Of course, the problem is no harder if we single out some arbitrary index i instead of 1: i.e. yi = 1 and y j = 0
for j ̸= i. Let us introduce some notation: denote by ∆i(x) the degree d polynomial that goes through these
d+1points.Then∆i(x)= Πj̸=i(x−xj). Πj̸=i(xi−xj)
Wenowreturntotheoriginalproblem. Givend+1pairs(x1,y1),…,(xd+1,yd+1),wefirstconstructthed+1 polynomials ∆1(x),…,∆d+1(x) as described above. Now we claim that we can write p(x) = ∑d+1 yi∆i(x).
Why does this work? First notice that p(x) is a polynomial of degree d, as required, since it is the sum of
EECS 70, Fall 2021, Note 8 2
polynomials of degree d. And when it is evaluated at xi, d of the d + 1 terms in the sum evaluate to 0 and the i-th term evaluates to yi times 1, as required.
In the above construction, we can think of the polynomials ∆i(x) as a “natural basis” for all polynomials whose values are specified at the points {xj}. Note that these basis polynomials depend only on the xj, and not on the values yj at the points. We then sum the basis polynomials ∆i, with coefficients equal to the values yi, to construct the desired polynomial p(x).
As an example, suppose we want to find the degree-2 polynomial p(x) that passes through the three points (x1,y1) = (1,1), (x2,y2) = (2,2) and (x3,y3) = (3,4). The three polynomials ∆i are as follows:
∆1(x) = (x−2)(x−3) = (x−2)(x−3) = 1×2 − 5x+3; (1−2)(1−3) 2 2 2
∆2(x) = (x−1)(x−3) = (x−1)(x−3) = −x2 +4x−3; (2−1)(2−3) −1
∆3(x) = (x−1)(x−2) = (x−1)(x−2) = 1×2 − 3x+1. (3−1)(3−2) 2 2 2
The polynomial p(x) is therefore given by
p(x) = 1·∆1(x)+2·∆2(x)+4·∆3(x) = 12×2 − 12x+1.
You should verify that this polynomial does indeed pass through the above three points.
Proof of Property 2
We are now in a position to prove Property 2 stated earlier, namely: Property2:Givend+1pairs(x1,y1),…,(xd+1,yd+1),withallthexi distinct,thereisauniquepolynomial
p(x)ofdegree(atmost)dsuchthatp(xi)=yi for1≤i≤d+1.
Wehaveshownabovehowtofindapolynomial p(x)suchthat p(xi)=yi ford+1pairs(x1,y1),…,(xd+1,yd+1). This proves part of Property 2 (the existence of the polynomial). How do we prove the second part, that the polynomial is unique? Suppose for contradiction that there is another polynomial q(x) such that q(xi) = yi for all d + 1 pairs above. Now consider the polynomial r(x) = p(x) − q(x). Since we are assuming that q(x) and p(x) are different polynomials, r(x) must be a non-zero polynomial of degree at most d. Therefore, Prop- erty 1 (to be proved below) implies it can have at most d roots. But on the other hand r(xi) = p(xi)−q(xi) = 0
at d + 1 distinct points xi, so r(x) has at least d + 1 roots. Contradiction. Therefore, p(x) is the unique poly- nomial that satisfies the d + 1 conditions.
Polynomial Division
Let’s take a short digression to discuss polynomial division, which will be useful in the proof of Property 1. If we have a polynomial p(x) of degree d, we can divide by a polynomial q(x) of degree ≤ d by using long division. The result will be:
p(x) = q′(x)q(x) + r(x)
where q′(x) is the quotient and r(x) is the remainder. The degree of r(x) must be smaller than the degree of q(x). We can compute the quotient and remainder using long division for polynomials, as illustrated in the following example.
Example. We wish to divide p(x) = x3 +x2 −1 by q(x) = x−1:
EECS 70, Fall 2021, Note 8 3
• First we subtract a factor x2(x − 1) to write: p(x) = x2(x − 1) + (2×2 − 1).
• Then we subtract a factor 2x(x−1) to write the remainder as 2×2 −1 = 2x(x−1)+(2x−1). • Then we subtract a factor 2(x−1) to write the remainder as 2x−1 = 2(x−1)+1.
• Finally,puttingtheabovethreelinestogether,wegetthatp(x)=(x2+2x+2)(x−1)+1.
Written out using notation that you might be familiar with: X2 +2X +2
X−1􏰁 X3 +X2 −1 −X3 +X2
2X2 −2X2 +2X
2X−1 −2X+2
Therefore, the quotient is q′(x) = x2 + 2x + 2, and the remainder is r(x) = 1.
Proof of Property 1
Now let us prove Property 1, which we restate from earlier:
Property 1: A non-zero polynomial of degree d has at most d roots.
The idea of the proof is as follows. We will prove the following two claims:
Claim 1 If a is a root of a polynomial p(x) with degree d ≥ 1, then p(x) = (x − a)q(x) for a polynomial q(x) with degree d − 1.
Claim 2 A polynomial p(x) of degree d with distinct roots a1,…,ad can be written as p(x) = c(x− a1)···(x−ad), where c is a real number.
Note that Claim 2 immediately implies Property 1: we just need to show that a ̸= ai for i = 1, . . . d cannot be a root of p(x). But this follows from Claim 2, since p(a) = c(a−a1)···(a−ad) ̸= 0.
Proof of Claim 1
Dividing p(x) by (x − a) gives p(x) = (x − a)q(x) + r(x), where q(x) is the quotient (of degree d − 1) and r(x) is the remainder. The degree of r(x) is necessarily smaller than the degree of the divisor (x − a). Therefore r(x) must have degree 0 and therefore is some constant c. But now substituting x = a, we get p(a) = c. But since a is a root, p(a) = 0. Thus c = 0 and therefore p(x) = (x − a)q(x).
Proof of Claim 2
Proof by induction on d.
• Base Case: d = 0. If p(x) is a polynomial of degree 0 then it is simply a constant c, so it can trivially
be written in the desired form.
EECS 70, Fall 2021, Note 8 4
• Inductive Hypothesis: For some d ≥ 0, any polynomial of degree d with distinct roots a1,…,ad can be written as p(x) = c(x−a1)···(x−ad).
• InductiveStep: Let p(x)beapolynomialofdegreed+1withdistinctrootsa1,···,ad+1. ByClaim1, p(x) = (x − ad+1)q(x) for some polynomial q(x) of degree d. Since 0 = p(ai) = (ai − ad+1)q(ai) for alli̸=d+1,andai−ad+1̸=0,q(ai)mustbeequalto0. Thusq(x)isapolynomialofdegreed with distinct roots a1,…,ad. We can now apply the inductive hypothesis to q(x) to write q(x) = c(x−a1)···(x−ad). Substituting in p(x) = (x−ad+1)q(x), we obtain p(x) = c(x−a1)···(x−ad+1), as desired.
Finite Fields
Both Property 1 and Property 2 also hold when the values of the coefficients and the variable x are chosen from the complex numbers, or indeed the rational numbers, rather than from the real numbers. However, the proofs2 do not go through if the values are restricted to being natural numbers or integers. Let us try to understand these facts a little more closely. The only properties of numbers that we used in polynomial interpolation and in the proofs of Properties 1 and 2 are that we can add, subtract, multiply and divide any pair of numbers as long as we are not dividing by 0. (You should go back and check this claim!) Therefore, everything holds just as well for the complex numbers and the rational numbers. On the other hand, we cannot subtract two natural numbers and guarantee that the result is a natural number, and dividing two integers does not generally result in an integer, so everything falls apart for natural numbers and integers.
However, if we work with numbers modulo a prime3 m, then we can add, subtract, multiply and divide (by any non-zero number modulo m). To check this, recall that x has an inverse mod m if gcd(m, x) = 1, so if m is prime all the numbers {1, . . . , m − 1} have an inverse mod m. So both Property 1 and Property 2 hold if the coefficients and the variable x are restricted to take on values modulo m. This remarkable fact—that these properties hold even when we restrict ourselves to a finite set of values—is the key to several applications that we will presently see.
Let us consider an example of degree d = 1 polynomials modulo 5. Let p(x) = 2x + 3 (mod 5). The roots of this polynomial are all values x such that 2x + 3 ≡ 0 (mod 5) holds. Solving for x, we get that 2x = −3 ≡ 2 (mod 5), and thus x = 1 (mod 5). Note that this is consistent with Property 1 since we got only one root of a degree-1 polynomial.
Now consider the polynomials p(x) = 2x + 3 (mod 5) and q(x) = 3x − 2 (mod 5). We can plot the values y of each polynomial as a function of x in the x-y plane. Since we are working modulo 5, there are only 5 possible choices for x, and only 5 possible choices for y:
Notice that these two “lines” intersect in exactly one point, (0,3), even though the picture looks nothing at all like lines in the Euclidean plane! Looking at these graphs it might seem remarkable that both Property 1 and Property 2 hold when we work modulo m for any prime number m. But as we stated above, all that was
2However, the result of property 1 does hold for polynomials with integer or natural number coefficients with the domain restricted to integers or natural numbers. That is because we have proved something stronger by going to the rationals or reals. If a polynomial has no more than d real roots, it certainly has no more than d integer roots. After all, every integer is a real number!
Meanwhile, property 2 does not hold at all if we restrict all the numbers involved to be integers. Try finding a straight line that goes through (0,0) and (2,1). It doesn’t have integer coefficients.
3Again, it is useful to consider what happens if the m is not a prime. What happens to property 1? Consider m = 8 and the equation x3 = 0. How many roots? 0 is a root for sure, but so are 2, 4, 6. That is 4 distinct roots to a third-degree polynomial. This means that property 1 does not hold and neither does property 2.
Unlike in the case of the integers, we cannot simply embed the numbers mod 8 into a larger field for which the result holds. This goes to show that we must be careful. It turns out that there is a way construct a finite field with 8 elements, but it is not simply the integers mod 8.
EECS 70, Fall 2021, Note 8 5
required for the proofs of Properties 1 and 2 was the ability to add, subtract, multiply and divide any pair of numbers (as long as we are not dividing by 0). Therefore, they hold whenever we work with integers modulo a prime m.
When we work with numbers modulo a prime m, we say that we are working over a finite field, denoted by Fm or GF(m) (for Galois Field). In order for a set to be called a field, it must satisfy certain axioms which are the building blocks that allow for these properties and others to hold. If you would like to learn more about fields and the axioms they satisfy, you can visit Wikipedia’s site and read the article on fields: http://en.wikipedia.org/wiki/Field_\%28mathematics\%29. While you are there, if you are interested, you can also read the article on Galois Fields and learn more about some of their applications and elegant properties which will not be covered in this course: http://en.wikipedia. org/wiki/Galois_field. (Take Math 113 and Math 114 to really get into this stuff, or take EECS 229B which also covers many of these properties as it builds the fundamentals of error-correcting codes.)
How many polynomials of degree (at most) 2 are there modulo m? This is easy: there are 3 coefficients, each of which can take on one of m values for a total of m3. Writing p(x) = adxd +ad−1xd−1 +···+a0 by specifying its d + 1 coefficients ai is known as the coefficient representation of p(x). Is there any other way to specify p(x)?
Yes, there is! Our polynomial of degree (at most) 2 is uniquely specified by its values at any three points, say x = 0, 1, 2. Once again, the polynomial can take any one of m values at each of these three points, for a total of m3 possibilities. In general, we can specify a degree d polynomial p(x) by specifying its values at d + 1 points, say 0,1,…,d, for a total of md+1 possibilities. These d+1 values, (y0,y1,…,yd), are called the value representation of p(x). The coefficient representation can be converted to the value representation by evaluating the polynomial at 0,1,…,d. And, as we’ve seen, Lagrange interpolation can be used to convert the value representation to the coefficient representation.
So if we are given three pairs (x1,y1),(x2,y2),(x3,y3) then there is a unique polynomial of degree 2 such that p(xi) = yi. But now, suppose we were only given two pairs (x1,y1),(x2,y2); how many distinct degree-2 polynomials are there that go through these two points? Notice that there are exactly m choices for y3, and for each choice there is a unique (and distinct) polynomial of degree two that goes through the three points (x1,y1),(x2,y2),(x3,y3). It follows that there are exactly m polynomials of degree at most 2 that go through two points (x1,y1),(x2,y2), as shown below:
What if you were only given one point (x1,y1)? Well, there are m2 choices for y2 and y3, yielding m2 poly- nomials of degree at most 2 that go through the point given. A pattern begins to emerge, as is summarized in the following table:
EECS 70, Fall 2021, Note 8 6
Polynomials of degree ≤ d over Fm # of points # of polynomials
d+1 1 dm d−1 m2
d−k mk+1 . .
Note that the reason that we can count the number of polynomials in this setting is because we are working over a finite field. If we were working over an infinite field such as the reals, there would be infinitely many polynomials of degree d that can go through d points! (Think of a line, which has degree 1. If you were just given one point, there would be infinitely many possibilities for the second point, each of which uniquely defines a line.)
Secret Sharing
In the late 1950’s and into the 1960’s, during the Cold War, President Dwight D. Eisenhower approved instructions and authorized top commanding officers for the use of nuclear weapons under very serious emergency conditions. Such measures were set up in order to defend the United States in case of an attack in which there was not enough time to confer with the President and decide on an appropriate response. This would allow for a rapid response in case of a Soviet attack on U.S. soil. This is a perfect situation in which a secret sharing scheme could be used to ensure that a certain number of officials must come together in order to successfully launch a nuclear strike, so that for example no single person has the power and control over such a devastating and destructive weapon. Suppose the U.S. government decides that a nuclear strike can be initiated only if at least k > 1 major officials agree to it. We want to devise a scheme such that both of the following properties hold:
1. Any group of k of these officials can pool their information to figure out the launch code and initiate the strike.
2. Nogroupofk−1orfewerofficialshaveanyinformationaboutthelaunchcode,eveniftheypooltheir knowledge. For example, they should not learn whether the secret is odd or even, a prime number, divisible by some number a, or the secret’s least significant bit.
How can we accomplish this4?
4This is a note where polynomials are playing a starring role and so we are going to use polynomials to do this in a nicely structured way. However, your background with solving systems of linear equations should give you a hint as to how you could
EECS 70, Fall 2021, Note 8 7
Suppose that there are n officials indexed from 1 to n and the launch code is some natural number s. Let q be a prime number larger than n and s. We will work over GF(q) from now on.
Now pick a random polynomial P(x) of degree k − 1 such that P(0) = s and give P(1) to the first official, P(2) to the second,. . . , P(n) to the nth. Then we have:
1. Any k officials, having the values of the polynomial at k points, can use Lagrange interpolation to find P, and once they know what P is, they can compute P(0) = s to learn the secret. Another way to say this is that any k officials have between them a value representation of the polynomial, which they can convert to the coefficient representation, which allows t