May 14, 2015 by mchouza

Universality with only two NOT gates

In a previous post we have asked how many NOTs are needed to compute an arbitrary Boolean function. In this post we will see that only two NOT gates are enough.

Building 3 NOT gates starting from 2

If we call the inputs $X$ , $Y$ and $Z$ , we can make a function detecting when no more than one input is active using a single NOT gate:

$\displaystyle f(X, Y, Z) = \overline{XY + YZ + XZ}$ .

Detects when no more than one input is active.

By selecting only the cases where at least one input is present, adding a term to detect when all the inputs are active and using an additional NOT gate, we can detect when exactly zero or two inputs are active:

$\displaystyle g(X, Y, Z) = \overline{f(X, Y, Z)(X + Y + Z) + XYZ}$

$\displaystyle = \overline{\overline{XY + YZ + XZ}(X + Y + Z) + XYZ}$ .

Detects when zero or two inputs are active.

Now we know that if $X$ is not present, we either have:

0 inputs present: we can check that by simultaneously ensuring that we don’t have more than one input present and that we have either zero or two inputs present, $f(X, Y, Z)\cdot g(X, Y, Z)$ .
1 input present: we should have no more than one input present and $Y$ or $Z$ should be present, $f(X, Y, Z)\cdot(Y + Z)$ .
2 inputs present: we can check that by simultaneously ensuring that either zero or two inputs are present and that $Y$ and $Z$ are present, $g(X, Y, Z)\cdot YZ$ .

Putting all together and adding the symmetrical cases:

$\displaystyle \overline{X} = f(X, Y, Z) \cdot (Y + Z) + (f(X, Y, Z) + YZ)\cdot g(X, Y, Z)$

$\displaystyle \overline{Y} = f(X, Y, Z) \cdot (X + Z) + (f(X, Y, Z) + XZ)\cdot g(X, Y, Z)$

$\displaystyle \overline{Z} = f(X, Y, Z) \cdot (X + Y) + (f(X, Y, Z) + XY)\cdot g(X, Y, Z)$ .

Computing NOT X (the other cases are symmetrical).

Checking the solution

To get independent confirmation, let’s check the truth tables with a simple Python script:

from itertools import product

for x, y, z in product((False, True), repeat=3):
	f_xyz = not (x and y or x and z or y and z)
	g_xyz = not (f_xyz and (x or y or z) or x and y and z)
	not_x = f_xyz and (y or z) or (f_xyz or y and z) and g_xyz
	not_y = f_xyz and (x or z) or (f_xyz or x and z) and g_xyz
	not_z = f_xyz and (x or y) or (f_xyz or x and y) and g_xyz
	assert (not x == not_x) and (not y == not_y) and (not z == not_z)

Conclusion

As this technique allows us to expand two NOT gates to three and it can be applied repeatedly, we can compute an arbitrary Boolean function with a circuit containing only two NOT gates.

In a following post we will see how I arrived to the solution by using brute force.

May 2, 2015 by mchouza

How many NOTs are needed?

It’s relatively easy to see that we cannot compute an arbitrary Boolean function using only AND and OR gates. For example, even the NOT function cannot be computed using only those gates (why?).

Can we build a circuit to compute an arbitrary Boolean function using a constant number of NOT gates?

Solution to the “42 code golf” problem

This was my best result:

n=1e6,m,c,d;main(){while(c+=d==42,d=0,m=--n)while(d+=m%10,m/=10);printf("%d\n",c);}

It would have been nice to find a solution under 80 bytes but, after one hour of trying, that was the best I could manage…

April 29, 2015 by mchouza

Hilbert matrices are positive definite

(Argument stolen from Yahoo Answers)

It can be seen by using its integral form:

$\displaystyle H_{ij} = \int_0^1 dx\,x^{i+j-2}$

Then we can express the positive definite condition as

$\displaystyle \sum_{i,j} x_i H_{ij} x_j > 0$

for $\mathbf{x} \ne 0$ .

Replacing and operating:

$\displaystyle \sum_{i,j} x_i \int_0^1 du\,u^{i+j-2} x_j = \int_0^1 du \sum_{i,j} x_i u^{i+j-2} x_j$

$\displaystyle = \int_0^1 du \sum_{i,j} x_i u^{i-1} x_j u^{j-1}$

$\displaystyle = \int_0^1 du \sum_i x_i u^{i-1} \sum_j x_j u^{j-1}$

$\displaystyle = \int_0^1 du \left( \sum_i x_i u^{i-1} \right) \left( \sum_j x_j u^{j-1} \right)$

$\displaystyle = \int_0^1 du \left( \sum_i x_i u^{i-1} \right)^2$

The term inside the parentheses is just a (polynomial) function of $u$ :

$\displaystyle = \int_0^1 du\,p(u)^2$

The only way the integral can be zero is if $p(u)$ is identically zero:

$\displaystyle \sum_i x_i u^{i-1} = 0 \implies \forall i: x_i = 0$ .

QED

April 17, 2015 by mchouza

Solving the viral Singaporean math problem

The following problem has become very popular in social media:

In this blog we have solved similar problems before, but this is one that can be easily solved by hand. We only need to be careful not to confuse our knowledge state with the one of Albert & Bernard.

Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is. Cheryl gives them a list of 10 possible dates.

May 15 May 16 May 19

June 17 June 18

July 14 July 16

August 14 August 15 August 17

Cheryl then tells Albert and Bernard separately the month and the day of her birthday, respectively.

We will describe a list of the possible knowledge states of Albert and Bernard after being given that information:

Albert

May 15 or May or May 19
June 17 or June 18
July 14 or July 16
August 14 or August 15 or August 17

Bernard

July 14 or August 14
May 15 or August 15
May 16 or July 16
June 17 or August 17
June 18
May 19

Albert: I don’t know when Cheryl’s birthday is, but I know that Bernard does not know, too.

We already knew that Albert wouldn’t know the day based just on being given the month, but he is giving us additional information by telling us that he knows that Bernard doesn’t know. Bernard would know the date if the day were 18 or 19, so Albert knows that those days could not be the right ones. That excludes options 1 and 2 from our knowledge of Albert knowledge:

Albert

July 14 or July 16
August 14 or August 15 or August 17

Bernard can do the same deductions we have and eliminate the options that are not possible from his state of knowledge (all the options with months different from July and August).

Bernard

July 14 or August 14
August 15
July 16
August 17

Bernard: At first, I didn’t know when Cheryl’s birthday is, but I know now.

Updating our knowledge of Bernard knowledge:

Bernard

August 15
July 16
August 17

As Albert also knows what we know about Bernard knowledge…

Albert

July 16
August 15 or August 17

Albert: Then I also know when Cheryl’s birthday is.

Now we know the right date:

Albert

July 16

Bernard

July 16

January 7, 2015 by mchouza

Force over a dipole

In an arbitrary field

The easiest way to get the force over a dipole is to consider it as the limit of two oppositely charged monopoles that are closely spaced. If the dipole has moment $\mathbf{m}$ and is at position $\mathbf{r_0}$ , it can be considered as the limit of two monopoles, one with charge $-\epsilon^{-1}$ at position $\mathbf{r_0} - (\epsilon / 2) \mathbf{m}$ and the other with charge $\epsilon^{-1}$ at position $\mathbf{r_0} + (\epsilon / 2) \mathbf{m}$ , when $\epsilon$ goes to zero.

If we consider the finite size dipole immersed in a (let’s say magnetic) field $\mathbf{B}(\mathbf{r})$ , the total force will be

$\displaystyle \mathbf{F} = -\epsilon^{-1}\mathbf{B}(\mathbf{r_0} - (\epsilon / 2) \mathbf{m}) +\epsilon^{-1}\mathbf{B}(\mathbf{r_0} + (\epsilon / 2) \mathbf{m})$

$\displaystyle \mathbf{F} = \frac{\mathbf{B}(\mathbf{r_0} + (\epsilon / 2) \mathbf{m}) - \mathbf{B}(\mathbf{r_0} - (\epsilon / 2) \mathbf{m})}{\epsilon}$ .

We can get the force for the infinitesimal dipole by taking the limit when $\epsilon$ goes to zero

$\displaystyle \mathbf{F} = \lim_{\epsilon \to 0}\frac{\mathbf{B}(\mathbf{r_0} + (\epsilon / 2) \mathbf{m}) - \mathbf{B}(\mathbf{r_0} - (\epsilon / 2) \mathbf{m})}{\epsilon}$

$\displaystyle \mathbf{F} = \mathbf{m} \cdot (\mathbf{\nabla B})(\mathbf{r_0})$ ,

where $\mathbf{\nabla B}$ is the (tensor) derivative of the magnetic field.

Between two antialigned dipoles

The general field of a magnetic dipole of moment $\mathbf{m}$ at position $\mathbf{r_0}$ is

$\displaystyle \mathbf{B}_d(\mathbf{r}) = \frac{\mu_0}{4\pi}\left(\frac{3\mathbf{m}\cdot(\mathbf{r}-\mathbf{r_0})(\mathbf{r}-\mathbf{r_0})-\mathbf{m}|\mathbf{r}-\mathbf{r_0}|^2}{|\mathbf{r}-\mathbf{r_0}|^5}\right)$ .

If we assume we have one dipole at $x = 0$ with its moment $+m\mathbf{\hat{x}}$ and the other at $x = +R$ with its moment $-m\mathbf{\hat{x}}$ , we get a field at $x = +R$ of

$\displaystyle \mathbf{B}(+R\mathbf{\hat{x}}) = \frac{\mu_0}{4\pi}\left(\frac{3m\mathbf{\hat{x}}\cdot(+R\mathbf{\hat{x}})(+R\mathbf{\hat{x}})-m\mathbf{\hat{x}}|+R\mathbf{\hat{x}}|^2}{|+R\mathbf{\hat{x}}|^5}\right)$

$\displaystyle \mathbf{B}(+R\mathbf{\hat{x}}) = \frac{\mu_0}{4\pi}\left(\frac{3mR^2-mR^2}{R^5}\right)\mathbf{\hat{x}}$

$\displaystyle \mathbf{B}(+R\mathbf{\hat{x}}) = \frac{\mu_0 m}{2\pi R^3}\mathbf{\hat{x}}$ .

By symmetry, we are only interested in the x-component of the x-derivative of the field,

$\displaystyle \left(\frac{\partial\mathbf{B}}{\partial x}\right)_x = -\frac{3\mu_0 m}{2\pi R^4}$ .

And the force on the dipole at $x = +R$ will be

$\displaystyle \mathbf{F} = -m\mathbf{\hat{x}} \cdot (\mathbf{\nabla B})(+R\mathbf{\hat{x}})$

$\displaystyle \mathbf{F} = -m \left(-\frac{3\mu_0 m}{2\pi R^4}\right)\mathbf{\hat{x}}$

$\displaystyle \mathbf{F} = \frac{3\mu_0 m^2}{2\pi R^4}\mathbf{\hat{x}}$ ,

a repulsive force as expected of antialigned dipoles.

January 16, 2014 by mchouza

No increasing functions in the long line

For our purposes, let’s define the “long line” as the set $L = (0, 1) \times [0, 1)$ together with a lexicographic ordering:

$\displaystyle (a, b) \prec (c, d) \Leftrightarrow (a < c) \vee (a = c) \wedge (b < d)$ .

We want to prove that we cannot have a strictly increasing real-valued function in $L$ , meaning that there is no function $f: L \to \mathbb{R}$ such that

$\displaystyle f((a, b)) < f((c, d)) \Leftrightarrow (a, b) \prec (c, d)$ .

To start with the proof we are going to require a lemma.

Between two different real numbers there is at least one rational number

The basic idea (lifted from this math.SE answer) is to define “a grid” of rationals that is thick enough to be sure we have at least one rational between the two reals. So, if $a < b$ are the two real numbers, we can use as denominator

$\displaystyle n = \left\lceil \frac{2}{b - a}\right\rceil$

and as numerator

$\displaystyle m = \lfloor n a + 1 \rfloor$ .

Now we need to prove that $m/n$ is always strictly between $a$ and $b$ .

It’s easy to see that is bigger than $a$ ,

$\displaystyle \frac{m}{n} = \frac{\lfloor n a + 1 \rfloor}{n} > \frac{n a}{n} = a$ ,

and that $(m-1)/n$ is less than or equal to $a$ :

$\displaystyle \frac{m-1}{n} = \frac{\lfloor n a + 1 \rfloor - 1}{n} \le \frac{n a + 1 - 1}{n} = a$ .

As we know that $1/n$ is smaller than $b - a$ because

$\displaystyle \frac{b - a}{1 / n} = n(b - a) = \left\lceil \frac{2}{b - a}\right\rceil (b - a) > \frac{2}{b - a}(b - a) = 2$ ,

we have

$\displaystyle \frac{m}{n} = \frac{m-1}{n} + \frac{1}{n} < a + (b - a) = b$ .

Building an impossible one-to-one function

Using $g$ as a name for the previously defined function that takes a pair of reals and returns a rational between them and $f$ for an increasing function in $L$ , we can define $h: (0, 1) \to \mathbb{Q}$ by the following expression:

$\displaystyle h(x) = g(f((x, 0)), f((x, 0.5)))$ .

We know that is one-to-one because if $x \ne y$ , assuming WLOG that $x < y$ ,

$\displaystyle h(x) = g(f((x, 0)), f((x, 0.5)))$
$\displaystyle < f((x, 0.5))$
$\displaystyle < f((y, 0))$
$\displaystyle < g(f((y, 0)), f((y, 0.5))) = h(y)$ .

But this result would imply that the cardinality of $(0, 1)$ is smaller than the cardinality of a subset of $\mathbb{Q}$ and that is absurd.

Conclusions

This result would seem a random trivia, but it has an important consequence for microeconomics: lexicographic preferences cannot be described by (real-valued) utility functions.

August 25, 2013 by mchouza

Kaufman decimals

Ben Orlin posted an interesting extension of the normal repeating decimals called “Kaufman decimals” (because they were developed jointly with Jeff Kaufman). In this post I will try to define them more precisely and show they can be totally ordered.

Introduction

The Kaufman digit sequences (the decimals without the initial “0.” prefix) are formed by starting from single digit sequences (“0”, “1”, …, “9”) and applying the following operations a finite number of times:

Concatenation: taking two digit sequences and putting one after the other.
Repetition: taking a digit sequence and repeating it an infinite number of times.

As it’s difficult to start directly with the infinite case, let’s start by defining formally a more simple case.

Finite repetitions

If we replace the infinite repetition by repeating the sequence $K$ times, it’s easy to analyze the resulting sequences. For example, if $K = 3$ ,

$\displaystyle \overline{\overline{\strut 9}\,1}\,2 = 9991999199912$ .

So now we can define concatenation in the following way:

$\displaystyle \mathcal{C}(a_{i<n}, b_{j<m})_{k<n+m} = \begin{cases}a_k & \text{if }k < n \\ b_{k-n} & \text{otherwise}\end{cases}$ .

Repetition is also quite easy:

$\displaystyle \mathcal{R}_K(a_{i<n})_{j < n \cdot K} = a_{k \bmod n}$ .

We can check the definition with the following Python code:

def l(a):
    from itertools import count
    for i in count():
        if a(i) is None:
            return i
 
def r(k, a):
    l_a = l(a)
    return lambda i: a(i % l_a) if i < l_a * k else None
 
def c(a, b):
    l_a, l_b = l(a), l(b)
    return lambda i: a(i) if i < l_a else b(i - l_a) if i < l_a + l_b else None
 
def s(c):
    assert len(c) == 1
    return lambda i: c if i == 0 else None
 
def p(a):
    print ''.join(a(i) for i in range(l(a)))
 
if __name__ == '__main__':
    p(c(r(3,c(r(3,s('9')),s('1'))),s('2')))

that gives us the expected output:

9991999199912

Infinite repetitions

For the real Kaufman digit sequences we can promote the ordinary digit sequences to transfinite ones (we use a slightly non-standard definition, allowing sequences of any length), using ordinal indices. Then the concatenation and repetition operations can be defined in essentially the same way:

Concatenation: $\mathcal{C}(a_{\alpha < \delta}, b_{\beta < \zeta})_{\gamma < \delta + \zeta} = \begin{cases}a_\gamma & \text{if }\gamma < \delta \\ b_{\gamma - \delta} & \text{otherwise}\end{cases}$
Repetition: $\mathcal{R}(a_{\alpha < \gamma})_{\beta < \gamma \cdot \omega} = a_{\beta \bmod \gamma}$ .

Ordinal modulus is easily defined by using the division theorem for ordinals.

We can construct the set containing all the Kaufman sequences by starting with the one digit “sequences”,

$\displaystyle K_0 = \{"0", "1", ..., "9"\}$ (where “0” is the length 1 sequence having the single digit 0),

and defining the next step based on the previous one,

$\displaystyle K_{n+1} = K_n \cup \{\mathcal{C}(a_{\alpha < \delta}, b_{\beta < \zeta}) | a_{\alpha < \delta}, b_{\beta < \zeta} \in K_n\} \cup \{\mathcal{R}(a_{\alpha < \delta}) | a_{\alpha < \delta} \in K_n\}$ .

As we have defined $K_n$ for all integer $n$ , now we can define $K_\omega$ , the set of Kaufman sequences, as

$\displaystyle K_\omega = \bigcup_{n < \omega} K_n$ .

Total order

It’s easy to see that this sequences can be totally ordered. Let’s define the extension operation for the Kaufman digit sequences:

$\displaystyle \mathcal{E}(a_{\alpha < \gamma})_{\beta \in \mathrm{On}} = \begin{cases}a_\beta & \text{if }\beta < \gamma \\ 0 & \text{otherwise}\end{cases}$

Then we define two sequences are equal if their extensions match:

$\displaystyle a_{\alpha < \delta} = b_{\beta < \zeta} \Leftrightarrow \forall \gamma \in \mathrm{On} : \mathcal{E}(a_{\alpha < \delta})_\gamma = \mathcal{E}(b_{\beta < \zeta})_\gamma$ .

If they don't match, they must differ in some digit and we can order them based on that digit:

$\displaystyle a_{\alpha < \delta} < b_{\beta < \zeta} \Leftrightarrow \exists \gamma \in \mathrm{On} : \left(\mathcal{E}(a_{\alpha < \delta})_\gamma < \mathcal{E}(b_{\beta < \zeta})_\gamma \right)\wedge \left(\forall \lambda < \gamma : \mathcal{E}(a_{\alpha < \delta})_\lambda < \mathcal{E}(b_{\beta < \zeta})_\lambda\right)$ .

Conclusions

The Kaufman decimals can be formally defined and totally ordered by using transfinite sequences. It wouldn’t be too hard to adapt the previous Python code to accept ordinal indices in Cantor normal form, but I’m still not sure if there is an efficient procedure to order Kaufman decimals.

May 27, 2013 by mchouza

Baez’s “42” – Solving the first puzzle

In a blog post, John Baez proposes the following puzzle:

Consider solutions of

$\displaystyle \frac{1}{p} + \frac{1}{q} + \frac{1}{r} = \frac{1}{2}$

with positive integers $p \le q \le r,$ and show that the largest possible value of $r$ is $42$ .

We can start by writing the following simple Python program to solve the problem, using integer arithmetic to avoid floating point problems:

from itertools import count
for r in count(1):
    for q in xrange(1, r + 1):
        for p in xrange(1, q + 1):
            # 1/p + 1/q + 1/r = 1/2
            # qr + pr + pq = pqr/2
            # 2(qr + pr + pq) = pqr
            if 2*(q*r + p*r + p*q) == p*q*r:
                print p, q, r

This program shows us ten possible values (including 42 in the last triple),

but, as it’s trying to enumerate all integer triples, it doesn’t halt.

To be sure not to miss any interesting integer triple, we must explore them more carefully:

from itertools import count
for p in count(1):
    # checks if we are leaving room for 1/q + 1/r
    # 1/p >= 1/2
    if 2 >= p:
        continue # we need smaller values of 1/p
    # checks if we can reach 1/2 with the biggest legal values for 1/q and 1/r
    # 1/p + 1/p + 1/p < 1/2
    if 3 * 2 < p: 
        break
    for q in count(p):
        # checks if we are leaving room for 1/r
        # 1/p + 1/q >= 1/2
        if 2 * (q + p) >= p * q: 
            continue
        # checks if we can reach 1/2 with the biggest legal value for 1/r
        # 1/p + 1/q + 1/q < 1/2
        if 2 * (q + 2 * p) < p * q:
            break
        for r in count(q):
            lhs = 2 * (q * r + p * r + p * q)
            rhs = p * q * r
            # 1/p + 1/q + 1/r > 1/2
            if lhs > rhs:
                continue
            # 1/p + 1/q + 1/r < 1/2
            elif lhs < rhs:
                break
            # 1/p + 1/q + 1/r = 1/2
            else:
                print p, q, r

The results are the same, but now we can be sure of having the whole solution.

May 23, 2013 by mchouza

A simple combinatorics problem

To break the long hiatus, I’m going to solve a short combinatorics problem extracted from “A Walk Through Combinatorics”:

Find all positive integers $a < b < c$ such that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1$ .

As $a$ is the smallest of the three integers, we know that $\frac{1}{a}$ will be the bigger of the three fractions. The three fractions must add to one and they must be different, so $\frac{1}{a}$ must be equal to $\frac{1}{2}$ . Rewriting the restriction using this newly found knowledge: