Enses, enses requirimus, requirimus saevos nos

DEAR MISS MANNERS THE BARBARIAN: I’m a pharmacist. I have always been told that I look young for my age, which I have chosen to accept as a compliment. However, at the pharmacy, my customers frequently ask me my age, and some come right out and say that they would prefer to be served by an older pharmacist.  This is really starting to irritate me, as it’s directed at me multiple times a day and it’s none of their business how old I am. Is there any other polite way to get these people to stop asking?

GENTLE READER: When King Numedides lay dead at my feet and I tore the crown from his gory head and set it on my own, I had reached the ultimate border of my dreams. I had prepared myself to take the crown, not to hold it. In the old free days all I wanted was a sharp sword and a straight path to my enemies. Now no paths are straight and my sword is useless. What you can say, with a pleasant smile, is: “Perhaps you would prefer to come back tomorrow. I’ll still be the pharmacist, but I’ll be older then.”

DEAR MISS MANNERS THE BARBARIAN: My parents are first cousins. I have a friend who likes to tell jokes about people whose parents are first cousins. She is not aware that my parents are first cousins, and if she knew, she would be horribly embarrassed. Is there a way to politely put an end to these jokes?

GENTLE READER: Open, blast you! I’m a guest. I’ve paid Aram for a room, and a room I’ll have, by Crom! Bring me a tankard of Ghazan wine—I’ve got just enough left to pay for it. I notice nobody sleeps in the streets of Zamboula. The very beggars hunt a niche they can barricade before dark. The city must be full of a particularly bloodthirsty band of thieves. The easiest way of refuting prejudice is open to you. “But my parents ARE first cousins” is so good a stopper, that Miss Manners the Barbarian has heard it used by people who are not really the target of such remarks.

DEAR MISS MANNERS THE BARBARIAN: After a long flight from overseas, I realized that I had left my phone on my seat. As I could not return to the plane myself, I found a security guard, who found a representative from the airline to assist me. Before I could explain what happened, she snapped, “How could you be so irresponsible and not check for all your valuables before leaving the flight?” I was so taken aback by this customer service rep. How should I have responded to her?

GENTLE READER: Son of a slut! Mesmerism! Did you deem yourself strong, because you were able to twist the heads off civilized folk, poor weaklings with muscles like rotten string? Hell! Break the neck of a wild Cimmerian bull before you call yourself strong. I did that, before I was full-grown —like this! Hundreds of necks have been snapped between these fingers! Miss Manners the Barbarian would have said tersely, “Thank you for your courtesy” while checking the person’s nameplate.

Feeling incorrect? Do you wish to crush your enemies, see them driven before you, and to hear the lamentations of their women?  Address your questions (in black or blueblack ink on white writing paper) to Miss Manners the Barbarian, in care of this newspaper.

The time has come at last to throw away this mask

Here are some pretty mathematical identities that I found while trying to answer a programming question. Let’s say you want to find a sum of the following form:

0 \times 1 + 1 \times 2 + 2 \times 3 + 3 \times 4 + ... + (n-1) \times n

Or, in other words, you want to find:

\left \{ n, k, f(n)  \hspace{2mm} \epsilon \hspace{2mm} \mathbb{N}^0 \hspace{2mm} | \hspace{2mm} f(n) = \sum \limits_{k=0}^n k(k-1)\right \} \hspace{1cm} (1)

Now, the boring and slow way of doing this would be to write a computer program to iterate through all those possibilities and sum the results. But there’s a way we can calculate that result directly, without doing all those multiplies and adds.

First, let’s get some intuition as to what form a solution to this problem might take. Now an integral is not exactly the same as a finite sum, but the two are related.  For a monotonically increasing finite sum like the one above, the definite integral from 0 to n might resemble the formula we’re looking for:

\begin{aligned}  f(n) &\approx \int_{0}^{n} n(n-1) \\  &\approx \int_{0}^{n} n^2-n \\  &\approx \int_{0}^{n} n^2 - \int_{0}^{n}n \\  &\approx \frac{n^3}{3} - \frac{n^2}{2}  \end{aligned}

Now we’re not lucky enough to have this ballpark estimate be equal to the answer. However, this estimate gives us a hint that we might be able to find a solution in the form of a cubic polynomial of n:

f(n) = an^3 + bn^2 + cn + d \hspace{1cm} (2)

So let’s start by manually calculating the first few answers to equation 1:

\begin{aligned}  f(0) &= 0 \\  f(1) &= 0 \\  f(2) &= 2 \\  f(3) &= 8  \end{aligned}

Plugging these answers back into equation 1 we get:

\begin{array}{cccccccccccc}  a(0)^3 &+ &b(0)^2 &+ &c(0) &+ &d &= &0 \\  a(1)^3 &+ &b(1)^2 &+ &c(1) &+ &d &= &0 \\  a(2)^3 &+ &b(2)^2 &+ &c(2) &+ &d &= &2 \\  a(3)^3 &+ &b(3)^2 &+ &c(3) &+ &d &= &8 \\  \end{array}

Which we can immediately simplify to:

\begin{array}{cccccccccccc}  0a &+ &0b &+ &0c &+ &d &= &0 \\  a &+ &b &+ &c &+ &d &= &0 \\  8a &+ &4b &+ &2c &+ &d &= &2 \\  27a &+ &9b &+ &3c &+ &d &= &8 \\  \end{array}

Now before we begin solving this particular system of equations, just look at the first one, which has all of a, b, and c being multiplied by 0. So they can have no effect on the result of the equation, regardless of n. Therefore we know, without doing any real work, that:

d = 0

Plugging in zero for d, back into the system of equations, simplifies things a little more for us:

\begin{array}{ccccccccc}  a &+ &b &+ &c & & &= &0 \hspace{1cm} (3) \\  8a &+ &4b &+ &2c &- &2 &= &0 \hspace{1cm} (4) \\  27a &+ &9b &+ &3c &- &8 &= &0 \hspace{1cm} (5) \\  \end{array}

Okay, so we have reduced the problem to finding the three unknowns a, b, and c. Now there are a bunch of ways we can solve this system of simultaneous equations. Good old fashioned algebra works here, but we just have to be patient with carefully shuffling symbols around. From equations 3 and 4:

\begin{aligned}  a+b+c &= 0 \\  2(a+b+c) &= 0 \\  2a+2b+2c &= 0 \\  2a+2b+2c &= 8a+4b+2c-2 \\  -6a-2b &= -2 \\  \frac{-6a-2b}{-2} &= \frac{-2}{-2} \\  3a+b&=1 \hspace{2cm} (6)\\  \end{aligned}

And from equations 4 and 5:

\begin{aligned}  8a+4b+2c-2&=0 \\  \frac{3}{2}(8a+4b+2c-2) &= \frac{3}{2}(0) \\  12a+6b+3c-3&=0 \\  12a+6b+3c-3&=27a+9b+3c-8 \\  5&=15a+3b \\  15a+3b&=5 \hspace{1cm} (7) \\  \end{aligned}

Now — coming into the home stretch — from equations 6 and 7:

\begin{aligned}  3a+b &= 1 \\  3(3a+b) &= 3(1) \\  9a+3b-3&=0 \\  9a+3b-3&=15a+3b-5\\  2&=6a \\  a&=\frac{1}{3} \hspace{1cm} (8)  \end{aligned}

Finally! Now we have values for a and d. Now we can solve for b, from equation 6:

\begin{aligned}  3a+b &= 1 \\  3(\frac{1}{3})+b&=1 \\  b&=0  \end{aligned}

And from equation 3, we can get c:

\begin{aligned}  a+b+c &= 0 \\  \frac{1}{3}+0+c&=0\\  c&=-\frac{1}{3}\\  \end{aligned}

So we’ve figured out all the coefficients for the simultaneous equations:

\begin{aligned}  a&=\frac{1}{3} \\  b&=0 \\  c&=-\frac{1}{3} \\  d&=0\\  \end{aligned}

And we can insert those into equation 2 to find the characteristic equation, and hence we discover a couple pretty identities:

\begin{aligned}  f(n)&=\frac{1}{3}n^3 + 0n^2 -\frac{1}{3}n - 0 \\  &=\frac{1}{3}n^3 -\frac{1}{3}n \\  &=\frac{n}{3}(n^2-1) \\  f(n) = \sum \limits_{k=1}^n k(k-1) &= \bf{\frac{n(n^2-1)}{3}} \hspace{1cm} (9) \\  &= \bf{\frac{(n-1)(n)(n+1)}{3}} \hspace{1cm} (10) \\  \end{aligned}

Now that’s one way to do it. But this identity can be proven in multiple other ways.

You think your Commodore 64 is really neato

It is possible to run modern software on classic 6502-based computers such as the Commodore 64 and the Apple II.

Some ground work is in order.  The basic strategy involves beefing up the 6502-based machines as much as possible and dumbing down Linux as much as possible.

The secret sauce involves writing a new, 6502-based emulation of an ARM Versatile/PB development board.  The ARM Versatile/PB is still a first-class target for Linux.  I think that with the make tinyconfig command with cross-compilation, we can get the size of a Linux kernel sufficiently small to have it in expanded memory on the 6502 host.

We will need to write parts of an ARM926EJ-S emulator in 6502/65816 assembly. This sounds a lot harder than it actually will be, and large parts of it probably won’t even have to be written at all.  While the ARM seems to have a ton of instructions, at the bit level, each instruction breaks neatly into its command and operand components.  Most of the work will be in emulating ARM interrupts and user/supervisor modes in 6502 assembly, and for a first bringup it wouldn’t even be necessary to emulate the Versatile/PB interrupt controller. Each of the addressing modes will need to be handled, but no individual addressing mode will be that complicated.  The ARM instruction set has a lineage dating back to the original MOS chips, and many of the state flags for ARM are exactly the same as they were in the 6502 days.

Sixteen megabytes of memory will need to be added to the 6502 machines, with appropriate bank switching logic.  This has been done already for each of the 6502 hosts.  There is also a period-appropriate chip, the 74LS610, which was intended specifically for increasing the addressable memory on the 6502 series.

We use 65816 microprocessor (SuperCPU) instructions to access 16 MB of memory in a 6502 machine.  Or, we develop a virtual memory abstraction that can arbitrarily bank in and out pages of the 16 MB into the lower 64KB of memory on each host. The cc65 development environment already supports RAM expansion drivers for the Apple II and the Commodore 64.

To get Linux to be sufficiently small, we use make tinyconfig with a modern Linux cross-compiled kernel to get it under 8 MB.  We target the version 1 ARM thumb instruction set in the Linux cross compile, and that’s the instruction set we emulate on the 6502.

For testing the correctness of the emulator, we run QEMU emulating the VersatilePB, in parallel with the 6502 emulator emulating Thumb1. We write a script that single-steps both virtual machines and verifies that emulator states run in lockstep.

The performance will be abysmally slow, but who cares.  Most likely the 6502 machines will, at first, themselves be running in emulation on a much faster PC.

See Woz’s SWEET16 emulator for further inspiration and basic proof of concept for the Thumb portion of ARM emulation.


By pressing down a special key, it plays a little melody

Here are some of the fun exciting behaviors in store if you try to configure CentOS 7 or RedHat 7 as a combined DNS and DHCP server with dynamic DNS updates for a local network.  These notes are for an ipv4 network only; ipv6 is left as an exercise for the reader.

CentOS goes to some effort to silently but sincerely prevent you from doing this in the name of “security”.

You’ll want to do a fresh install and update of CentOS 7 and select Domain name server as the installation option.

Install some packages:

yum install bind-chroot bind dhcp

The key files you’ll be editing are /etc/rndc.conf, /etc/named.conf, /etc/rndc.keys, and /etc/dhcp/dhcpd.conf .

Follow Steven Carr’s advice on setting up more sane named logging. 

The zone files, normally stored in /var/named, are not given sufficient permissions to be read and written by the named process.  Move them to /var/named/dynamic.  Permissions, generally, are a bitch with this whole setup — expect that other files in /var/named may need to be chown’d to root.named and chmod’ed to 660 or 664.

SELinux silently prevents a lot of things that named and dhcpd want to do to maintain those zone files.  Convince it that this is OK by adding the following to /etc/sysconfig/named:


Also run this command as well:

setsebool named_write_master_zones on

You will find that named and dhcpd don’t start automatically at boot time when installed.  You’ll have to teach CentOS 7 to do this yourself.  Use systemctl enable to do this.

You can generate a secret key by running rndc-confgen.  The output will give you a hint as to what to put into rndc.conf as well as named.conf.

The secret key, usually stored in rndc.key, wants to be stored in several places, at least in /etc/rndc.key and /etc/rndc.conf and /etc/named.conf and /etc/dhcp/dhcpd.conf.  There is no public/private key exchange in named if you are just running a local DNS server; it’s all just one pre-shared secret key, which is neat.

nsupdate is your friend.  If you can’t manually get named to update by using this command to get to control port 953, do an experiment to see if you can add and delete records via the rndc interface by running rndc and issuing commands something like this:

update add dumb.yourdomain 900 IN A
debug on

If you can’t do this yourself, then dhcpd won’t be any luckier; don’t bother trying to make dhcpd happy until named is properly responding to requests to update the zone files.

CentOS firewalls off all the relevant ports against you.  You’ll have to open these yourself.  

For setting up a dhcp server:

firewall-cmd --add-port=67/udp --zone=public --permanent
firewall-cmd --add-port=68/udp --zone=public --permanent

For setting up the dns sever:

firewall-cmd --add-port=53/tcp --zone=public --permanent

We need to fly ourselves before someone else tells us how

If you’re experiencing a constant inexplicable 30%-50% CPU spike on the latest version of Windows 10, and Task Manager CPU usages don’t add up to the total amount of CPU that is being taken away, and you have a copy of Prey 1.6.2 or later installed, and you’re running Windows Defender for antivirus protection, uninstall Prey.

Something about the combination of Prey and Windows Defender causes this spike in CPU, and this spike in CPU won’t be properly reported in the Performance Manager or Task Manager.

I found the secret, the key to the vault

Q. We will have 600 people at a conference.  How many possible two-person pairs does that allow?

A. In order to solve this problem, let’s solve some easier problems first.

Let’s have all 600 people line up in a row. How many ways are there to line up 600 people? Well, first we have to choose the first person. There are 600 choices for that first person, that is, any of those 600 could go first. Then, who goes second? We have 599 people left. So, to figure out how many ways we can choose the first and second people in a line of 600 people, we calculate 600 \times 599.

Now if we continue this logic through all 600 people, choosing the first, the second, the third and so on, we have 600 \times 599 \times 598 \times ... \times 2 \times 1 ways to line up 600 people in a row.

In other words, there are:


ways to line up n people in a row.

But that’s not the right answer to our original problem. Let’s try to get that answer closer to the original answer we wanted. Let’s say we divided each of those n! rows of people evenly into pairs, taking them each two by two in order of the row. In that case, it wouldn’t matter if the first pair contained Alice or Bob, or Bob and Alice. Within a single pair, we don’t care what the ordering of the people in that pair is.

For 600 people, there are \frac{600}{2} pairs or 300 pairs. So, the number of ways those pairs might have swapped the first for the second is:




Therefore, the number of ways to order 600 people, ignoring the ways to merely swap a pair of people, is:


We’ve only got one more step to get an answer. Notice that while we’ve taken care of the case where a single pair of people are swapped, we haven’t taken care of the case where a pair of people is swapped with another pair of people. In other words, we don’t care whether it’s Alice and Bob followed by Carol and Dan, or if it’s Carol and Dan followed by Alice and Bob. So, we need to also divide by the number of ways to order 300 pairs of people. As we know from above, there are 300! ways to order 300 things. To ignore swapping pairs of n things, we need to divide by:


So let’s take a look at our final formula, which takes into account the ways we can choose n people, ignoring a swapped pair of two people, and also ignoring swapped pairs of people:


Let’s plug in 600 for n:


You can calculate this answer yourself if you use this online calculator and type in this formula:


And the final answer is:



Addenda: This is a semi-demi-hemi famous problem in combinatorics.  This problem and its solution tends to rear its head in a lot of superficially unrelated areas.  Here’s a menagerie of problems that all have this same solution.

When figuring out how to explain this problem, I stole a lot of ideas from here.

Ten thousand men of Harvard gained victory today

As I do every year, I want to congratulate the current class year on completing your Harvard degree. Your Harvard degree provides you access to important advantages and benefits as you go through life. About two months after you graduate, the bursar’s office will be sending you an alumni card that looks like this.

Harvard Alumni Card

Make sure not to lose this card.

I am going to give you a lot of information on the benefits and rewards you receive by being a Harvard graduate. Normally I have handouts at this point, but there was a problem with printing, so just take notes on your iPhones and things. There’s a web site with all this information on it. Before the end of the presentation, I will… try to get that web site address for you.

Harvard alumni get free upgrades from economy to first class on any airline trip within the lower 48 states. This is a really valuable benefit, so make sure to use the Harvard discount code when booking.

When you enter a boardroom during a board meeting, all people who graduated from non-Ivy League schools must stand at attention until you are seated.

You receive 15% off any Disney multi-park pass, Tuesday through Thursday, holidays excepted.

The Band-Aid corporation provides you free Band-Aids in our own computer matched skin tone. Click on the “Harvard” link on www.band-aid.com .

At most four-star hotels, when you check in, you will receive a special bar of soap that deep cleans without drying out the skin. This should happen automatically; you won’t need your card for this.

You get a special dispensation that allows you to drive up to 75 mph in 55 zones. If a police officer pulls you over, present your alumni card, and the officer will scan the barcode into a computer. At this point, you pay the police officer $1, and then you should be permitted to leave. If he detains you at this point, make sure to contact the bursar’s office.

Some of you own apartments and condos. If one of your tenants has just been married, you may at your option sleep with the wife or the husband or both during the first night of their marriage. But in fairness I must tell you, times are changing. Since 2005 this has become frowned upon and rarely happens anymore. So, I am not going to tell you what to do here, but try to use good judgement, okay?

Also, Harvard alumni have seniority when a group is deciding who will lead the Pledge of Allegiance.

For a while we got a dollar off any KFC $5 Fill Up Meal, but I think that expired. Does anyone know if that expired?

Other than that, I can’t think of anything. So, that’s it… welcome to the society of educated men and women.

Drag it drop it zip unzip it

GIF is pronounced with a hard g, contrary to the claims of its “creator”, Steve Wilhite.

Steve Wilhite only “created” the GIF format to the extent he didn’t copy it from previous creators. At the time he wrote it, he worked for CompuServe.  The GIF format contains the concept of global vs. local palettes, and it contains origin information for frame within a frame. That’s really the extent of the novelty in the GIF format. All of the actual color data is stored within a Lempel-Ziv-Welch compressed table. Wilhite copied the LZW algorithm direct from this seminal 1984 paper and hence from the Unix compress program. I know this, because as a Harvard undergrad struggling through CS175, I implemented a converter between Harvard’s image format and GIF.  My program used the guts of Unix compress, along with that paper, to encode and decode GIF images.

In the mid 1990s, Unisys used threats to try to shake down a few companies over its claimed ownership of the LZW algorithm.  This is after one of the authors published the algorithm publicly and went to work for another company.  As is usually the case in high tech, Unisys’s attempt to screw other companies backfired.

In essence, Wilhite’s and CompuServe’s only practical novelty was simply stamping a copyright notice on the front of some existing technology that they didn’t invent.

So the best we can do is to call it as we would any American acronym, by taking the first sounds of the words that comprise the object. Say GIF, think Graphics Interchange Format.