Matthew van Eerde's web log

Thanks to Gerry F. in Tampa for raising this interesting question.  Given: a perfect watch with three sweep hands (hour, minute and second).  (By "sweep", I mean that the hands move continuously and do not "jump" every hour, minute, or second.)

Find: a list of all times at which the three hands divide the circle of the clock into three equal sections - that is to say, any pair of hands will form a 120° angle, or, in radians, an angle of 2π/3.  (For the rest of the post I will use radians.)

The remainder of this post is my proof of the surprising result.  If you'd like to try solving the problem yourself, stop reading now...

OK.  First, let's define some terms.

Let t be the time.  I'm going to make a rather strange choice for the units.  I have the obvious choices of hours, minutes, or seconds - instead, I'm going to choose a unit of half-days.  So 12:00 corresponds to t = 0 or 1; 4:00 corresponds to t = 1/3; 9:00 corresponds to t = 3/4, etc.  Note the clock has periodic behavior with a period of 1 (in units of t.)

Let H be the directed angle between 12 and the hour hand, increasing from 0 (at 12:00, t = 0) through π/2 (at 3:00) and all the way 'round to 2π at 12:00 again.  Note that at 9:00 H is defined to be 3π/2 and not merely π/2.  Note that H is also periodic with period 1.

Let M be the directed angle between 12 and the minute hand similarly.  Note M is periodic with period 1/12: the minute hand goes all the way 'round in an hour, which is 1/12 of a half-day.

Let S be the directed angle between 12 and the second hand.  Note S is periodic with period 1/720: the second hand goes 'round 720 times in a half-day.

We can now write explicit formulas for H, M, and S in terms of t:

Define θ_xy as the directed angle between hand x and hand y.  In particular:

The key point is that the angles between the hands are periodic.  The angle between the minute and the hour hands cycles from 0 through 2π exactly 11 times every half-day; the angle between the minute and the hour hands, exactly 12*59 times; and the angle between the hour and the second hands, exactly (12*60 - 1) = 719 times.

Now that we have formulas for the three relevant angles, we can answer the question "when are these angles 2π/3 or 4π/3?"  In particular, define {t_xy} as the set of times at which the x and y hands make either a 2π/3 angle or a 4π/3 angle.

Let's consider the angle between the hour and minute hand first.

So the angle is 2π/3 (or 4π/3) at multiples of t = 1/33 of a half-day, or every 12*60/33 = 21 and 9/11 minutes... except that at every third multiple of 21 and 9/11 minutes, the hands coincide, so those don't count.

But rather than think of it as "21 and 9/11 minutes", I'd like to urge you to think of it as "1/(3*11) of a cycle."

Now consider the angle between the minute and the second hand:

The angle between the minute and the second hand is 2π/3 or 4π/3 at multiples of 1/(3*12*59) of a half-day (except that, as before, at every third multiple of 1/2124 the hands coincide instead of making a suitable angle.)

The following is a demonstration that {t_HM} and {t_MS} have no elements in common.  This implies that the hour/minute angle and the minute/second angle are never 2π/3 or 4π/3 simultaneously.  As this is a necessary condition for the three hands to divide the clock evenly, the three hands never divide the clock evenly.

Write {t_HM} and {t_MS} in lowest terms.  How many factors of 3 are there in the numerator of each fraction?  In the denominator?

The numerator is easy.  The numerator is never a multiple of 3 to begin with, and cancelation of common terms in the numerator and denominator will never add a factor, so after cancelation there are (still) no factors of 3.

The denominator for each t_HM has exactly one factor of 3; the denominator for each t_MS has exactly two factors of 3.

Thus, the two sets have no elements in common, and the clock is never evenly divided.  QED.

To illustrate the importance of the numerator being a nonmultiple of 3, let's calculate {t_HS}:

719 happens to be prime.

It is tempting to put forth the following fallacious argument:

Fallacious argument: "{t_HS} consists of a bunch of fractions whose denominator contains the large prime factor 719.  Neither of the other sets {t_HM} or {t_MS} contain this large prime factor in their denominators... therefore {t_HS} contains no elements in common with either {t_HM} or {t_MS}, QED."

The problem with this argument is that it attempts to prove something that is false.  True, the intersection of {t_HS} and {t_MS} is empty, but the intersection of {t_HS} and {t_HM} is not empty - in fact, it consists of precisely the two elements { 1/3, 2/3 } (which correspond to 4:00:00 and 8:00:00 respectively.)

What happened to the factor of 719 in the denominator?  It got canceled out by a multiple of 719 in the numerator.  In particular, 1/3 = (3*239 + 2)/(3*719), and 2/3 = (3*479 + 1)/(3*719).

For as long as there have been humans, we have looked at the skies - and around us on Earth as well.  Things move in the skies; things move on Earth.  The way things move in the skies is called "celestial mechanics"; the way things move on Earth is called "terrestrial mechanics."

The ancient Greeks thought there were five elements, corresponding to the five regular (Platonic) solids: the four terrestrial elements (air, fire, earth, and water) and the fifth celestial element which the Romans called quintessence.

All kinds of rules were discovered/proposed for how things move on Earth.  Galileo demonstrated that falling objects accelerate at the same rate regardless of their mass.  Similar rules were discovered/proposed for how things move in the heavens.  Kepler demonstrated that the planets move in ellipses around the sun, that they swept out equal areas in equal times, etc.  But everybody knew that things in the heavens were different than they were on Earth.

Then Newton, building on Hooke, came up with the crazy idea that perhaps things fall on Earth for the same reason that heavenly objects go in ellipses... that all objects, celestial or terrestrial, act on each other from a distance in a uniform fashion.  (He himself wrote that this was such a crazy idea that no-one should take it seriously.)

That one body may act upon another at a distance through a vacuum without the mediation of anything else, by and through which their action and force may be conveyed from one another, is to me so great an absurdity that, I believe, no man who has in philosophic matters a competent faculty of thinking could ever fall into it. -- Isaac Newton

But he had done some calculations and it seemed to work out... specifically, he worked out how much acceleration you would expect the Earth to exert on an object as far away as the Moon, and how much acceleration the Moon would need to stay in its orbit (and not wander off or spiral down into the Earth.)

I... compared the force requisite to keep the moon in her orb with the force of gravity at the surface of the earth and found them to answer pretty nearly. -- Isaac Newton

Note Newton's use of the word "orb" to describe the motion of the moon, a direct reference to celestial mechanics.

Let's see how compelling those calculations were, shall we?

His first idea was that there was this thing called "force" which was equal to the mass of an object multiplied by the acceleration it was undergoing.  F = m a.  Fine.

His second idea (which he stole, at least partially, from Hooke) was that any two objects have a force that attracts them, which is proportional to both of their masses, and inversely proportional to the square of the distance between them (that's the part he stole from Hooke.)  F = G m₁ m₂ / r².  Fine.

His third idea was that an object that goes around a circle of radius r at a constant speed v is undergoing a constant centripetal acceleration of a = v² / r.  This follows pretty quickly from the definitions with a little geometry, trigonometry, and a dash of calculus.  Fine.

He also knew some facts.  I'm going to state them in metric because, hey, this is the twenty-first century.  He knew:

• The distance from the center of the Earth to the surface (where the apple trees are) is 3,960 miles: 6370 km.
• Falling objects on the surface of the Earth accelerate at 32.2 ft/s²: 9.81 m/s².
  
• The distance from the center of the Earth to the center of the Moon is 239,000 miles: 384,000 km.
• The Moon takes 27.3 days to make a single sidereal orbit around the Earth: 236,000 s.

Coarse estimate time...

If the Moon is roughly 50 times further away from the center of the Earth than an apple is, and "gravitational" acceleration is inversely proportional to the square of the distance between two objects, then the Moon should be accelerated roughly 1/2500th as much as the apple is: 0.004 m/s² instead of 10 m/s²,

More precisely...

Consider the apple case (m_E is the mass of the Earth, m_A the mass of the apple:)

F_A = m_A a_A = G m_E m_A / r_A²
a_A r_A² = G m_E
9.81 m/s² (6.37e6 m)² = G m_E
G m_E = 3.98e14 m³/s²

Now consider the moon case:

F_M = m_M a_M = G m_E m_M / r_M²
a_M = G m_E / r_M²
a_M = 3.98e14 m³/s² / (3.84e5 m)²
a_M = 0.00269 m/s²

So Newton's proposed formula predicts that the Earth's gravity causes the Moon to accelerate at 0.00269 m/s².  This is, indeed, roughly 0.004 m/s² so it jives with our coarse estimate.

Let's see how that compares to the necessary centripetal acceleration to achieve a closed orbit: a = v² / r.

The distance the moon travels in its sidereal orbit is 2π * 3.84e5 m = 2.42e9 m, and it does so in 27.3 days; so its velocity is 2π * 3.84e5 m / (27.3 days * 24 (hours / day) * 60 (minutes / hour) * 60 (s / minute)) = 1020 m/s.  This is roughly Mach 3 in our atmosphere.

The centripetal acceleration is thus measured to be (1020 m/s)² / 3.84e8 m = 0.00273 m/s². Newton's prediction is about 1.5% off, which isn't bad, considering.

Note that although Newton knew the value of the product G m_E, he didn't know either G or m_E seperately.  It wasn't until Cavendish took two known masses and measured the gravitational pull between them directly that G was evaluated; we could then calculate the mass of the earth as G m_E / G.

Once G was known we could also evaluate the mass of the Sun based on Earth's orbit.  However, we still could not determine the mass of the Moon... in fact, back in the 1950s, American scientists were unable to determine the mass of Sputnik (which would have been very helpful) even though we could see its orbit.

Which raises the question... how did we know the mass of the Moon?

A body that moves in a circular motion (of radius r) at constant speed (v) is always being accelerated.  The acceleration is at right angles to the direction of motion (towards the center of the circle) and of magnitude v² / r.

The direction of acceleration is deduced by symmetry arguments.  If the acceleration pointed out of the plane of the circle, then the body would leave the plane of the circle; it doesn't, so it isn't.  If the acceleration pointed in any direction other than perpendicular (left or right) then the body would speed up or slow down.  It doesn't.

Now for the magnitude.  Consider the distance traveled by the body over a small time increment Δt:

We can calculate the arc length s as both the distance traveled (distance = rate * time = v Δt) and using the definition of a radian (arc = radius * angle in radians = r Δθ:)

The angular velocity of the object is thus v / r (in radians per unit of time.)

The right half of the diagram is formed by putting the tails of the two v vectors together.  Note that Δθ is the same in both diagrams.

Note the passing from sin to cos is via l'Hôpital's rule.

QED.

It happens a million times a day.  Somebody pays cash for something and gets change.  But there are rules, and Raymond's friend seems to like to break those rules.  Luckily for Raymond's friend, one of the rules is the customer is always right... or as Mr. Krabs would say, the money is always right.  This blog post is about a couple of the more mathematical rules.

Whether to honor requests for non-canonical change is a complicated morass I will not attempt to tackle in this blog post as too much depends on context.  But the initial offering must be canonical.

By "canonical change", I mean the cashier must give change using the smallest possible number of canonical coins and bills.  For example, (one quarter + one nickel) is a rendering of $0.30 using only two coins, while (three dimes) is three coins, so a cashier must never give change that includes three dimes.

By "canonical coins and bills" I mean (in the U.S.) this precise list:

• penny
• nickel
• dime
• quarter
• $1 bill
• $5 bill
• $10 bill
• $20 bill

I discuss usage of non-canonical (unusual) coins and bills later.

Rule 1 for the customer: "Tap, tap, no takebacks"

The tender and canonical change cannot intersect.

Don't give the cashier anything they would just hand back to you directly.

This is the rule that Raymond's friend broke - it just wastes everyone's time to play musical chairs with the money.

Actually, this rule is just a corollary of the more general rule:

Rule 2 for the customer: "No side transactions."

No subset of the tender can have the same value as any subset of canonical change for that tender.

Don't give the cashier anything that they would just hand back to you in coalesced form.

If you want to consolidate your change, go to a bank or use one of those big machines.

Corollary 1: it is always OK to pay with a single large bill.
Corollary 2: it is always OK to give exact change.

Unusual denominations:

The two-dollar bill, the dollar coin (in any of its forms), the half-dollar coin, and bills higher than $20 are "unusual" denominations that complicate the situation slightly.

Rule for the cashier: Unusual denominations stay in the drawer.

Accept them (unless there are security rules at your place of business) but they are explicitly not part of any canonical change.

Rule for the customer: Go right ahead.

You can pay with an unusual denomination provided that you do not break the "no side transactions" rule or any large-denomination security rule at the business in question.

Examples:

In the following examples, "total" is the amount owed; "tender" is the amount the customer gives to the cashier; and "change" is what the cashier gives back to the customer.

In the last row, you can think of the "side transaction" as either five pennies for a nickel or ten pennies for a dime. I prefer the former.

There's an MSDN sample of how to turn on HDCP or SCMS in a playback app.  The sample is loosely based on a test app I wrote, but there are still some rough edges.  For example, the CMFAttributeImpl<T> template is not part of the SDK or the DDK.  Also, there's a leak in the GetDigitalAudioEndpoint implementation.  (Using the private template was my bug.  The leak was someone else's.)

Exercise: find the leak.

There are also rough edges in the test app.  For one thing, it doesn't play audio - the way to use the app is to

1. Start playing audio using your favorite audio playback application, to the digital output on which want to turn on HDCP or SCMS.
2. Run the trustedaudiodrivers2.exe command line to do what you want it to.

Source and binaries (x86 and amd64) attached.

>trustedaudiodrivers2
trustedaudiodrivers2 --list-devices

trustedaudiodrivers2
    --device [ communications | console | multimedia | <device-name>]
    --copy-ok [ 0 | 1 ]
    --digital-output-disable [ 0 | 1 ]
    --test-certificate-enable [ 0 | 1 ]
    --drm-level <drm-level>

--list-devices: displays a list of output devices.

Sets a Trusted Audio Drivers 2 output policy on a given audio endpoint.
--device
    communications: set policy on default communications render endpoint.
    console: set policy on default console render endpoint.
    multimedia: set policy on default multimedia render endpoint.
    <device-name>: set policy on the endpoint with this long name.
--copy-ok: 1 or 0. Set to 0 to turn on "copy protection" (SCMS or HDCP)
--digital-output-disable: 1 or 0. Set to 1 to disable output (or turn on HDCP.) --test-certificate-enable: 1 or 0. Set to 1 to allow the test certificate.
--drm-level: set to a number. 1100 is a good default.

Here's a sample command line which should disable the digital out (or enable HDCP, if the driver supports it.)  This is line-wrapped for readability only.

>trustedaudiodrivers2.exe
    --device "Acer H213H (High Definition Audio Device)"
    --copy-ok 1
    --digital-output-disable 1
    --test-certificate-enable 1
    --drm-level 1300

If all goes well the output should look something like this:

Output Trust Authorities on this endpoint: 2
Processing Output Trust Authority #1 of 2
OTA action is PEACTION_PLAY (1)
GetOriginatorID() called
GetOriginatorID() called
GenerateRequiredSchemas() called
dwAttributes: MFOUTPUTATTRIBUTE_DIGITAL (0x00000001)
guidOutputSubType: MFCONNECTOR_HDMI ({57CD596D-CE47-11D9-92DB-000BDB28FF98})
cProtectionSchemasSupported: 1
MFPROTECTION_TRUSTEDAUDIODRIVERS ({65BDF3D2-0168-4816-A533-55D47B027101})
MFPROTECTION_TRUSTEDAUDIODRIVERS supported.
IMFOutputSchema::GetSchemaType called
IMFOutputSchema::GetConfigurationData called
IMFOutputTrustAuthority::SetPolicy returned 0x00000000
Processing Output Trust Authority #2 of 2
OTA action is PEACTION_EXTRACT (4)
Skipping as the OTA action is not PEACTION_PLAY
Policy successfully applied.  Press any key to release IMFTrustedOutput...

At that point I pressed a key, which redacted the policy, and the music started playing again.

Common causes of failures:

1. No audio is playing to that device.  Setting protection requires an active audio stream.
2. The audio device is not S/PDIF or HDMI.
   
3. The audio driver is not capable of enforcing the protection.
4. The audio driver is test-signed only; try --test-certificate-enable 1
5. Kernel debugging is enabled.
6. Driver verifier is enabled.

This source should compile with just publically available headers and no special voodoo on the part of the developer.

EDIT: 11/13/2009 10:05 AM

A subtle problem was called to my attention.  This line triggers an ATL ASSERT:

hr = pMFOutputTrustAuthority->SetPolicy(&pMFOutputPolicy, 1, &pbTicket, &cbTicket);

The ASSERT is in CComPtr<IMFOutputPolicy>::operator& which ASSERTs if the pointer CComPtr<IMFOutputPolicy>::p is not NULL.

My first instinct was to bypass the assertion like this:

hr = pMFOutputTrustAuthority->SetPolicy(&pMFOutputPolicy.p, 1, &pbTicket, &cbTicket);

But ASSERTs are there for a reason.  The fundamental problem here is that I'm trying to be too clever.

Despite its name, IMFOutputTrustAuthority::SetPolicy takes, not an IMFOutputPolicy *, but an array of IMFOutputPolicy *s.  A little confusing, yes.  As this is effectively a sample, I should be emphasizing this behavior, not covering it up.

So here's the fix I decided on.  Updated source and binaries attached.

// we're only setting a single policy but the API allows setting multiple policies
// (like WaitForMultipleObjects)
IMFOutputPolicy *rMFOutputPolicies[1] = { pMFOutputPolicy };
hr = pMFOutputTrustAuthority->SetPolicy(
    rMFOutputPolicies,
    ARRAYSIZE(rMFOutputPolicies),
    &pbTicket,
    &cbTicket
);

http://programmingpraxis.com/2009/10/09/calculating-pi/

>perl -e"print 4/(map{$n+=rand()**2+rand()**2<1}1..pop)*$n" 5000
3.1336

59 characters, plus arguments.

I was reading László Polgár's excellent book Chess: 5334 Problems, Combinations, and Games and ran across #1071, White to move and mate in 2:

The purported solution is...

1. ♘xf6+ ♚xf6
2. ♕f7#

... which, indeed, works - and there is no other mate in two that I can see - so why blog this?

Because there's a mate in one.

If you're installing Windows via a boot DVD, and you choose Custom, you have the option to rearrange partitions.  I like use this to have each drive be one big partition.

Windows 7 wants to set aside a 100 MB partition for something-or-other.  I'm sure there's a very good reason for this but I am too lazy to look up the team that owns this space and ask them what it is.

So I'm in the "Where do you want to install Windows?" stage, I've gone into "advanced drive setup", and I've deleted all the partitions.  Fine.  I then create a partition that fills the drive, and I get this popup:

Install Windows
To ensure that all Windows features work correctly, Windows might create additional partitions for system files.
OK | Cancel

After letting Windows finish installing, I jump into diskpart.exe and sure enough, Windows has created an additional partition.  A small one, to be sure... but an additional partition (horrors!)

Not being one to let Windows push me around, I decided to experiment, and came up with the following dance to allow me to just have one big partition thankyouverymuch:

1. Boot from the Windows DVD
2. Choose Custom (advanced) as opposed to Upgrade
3. Go into Drive options (advanced)
   
4. Delete all partitions on the drive
5. Create a new partition - you will get the prompt above
6. Click OK
7. There will now be two partitions - a small (System) one and a large (Primary) one.
   
8. Delete the large one.
9. Extend the new one to fill the drive.
10. Install Windows.
11. Open Windows Explorer
12. Right-click the C: drive | Properties
13. Delete the "System Reserved" partition name
    

As preparation for moving one of my machines from Vista to Windows 7, I'm compiling a list of all the little tweaks I like to make to machines that I use a lot:

Boot from the Windows DVD.  Delete all partitions; make each hard drive one big partition.  (Hmm... apparently Windows 7 really wants a second 100 MB partition.  Do the partition dance to force it into installing on a single partition.)

In the "password hint" box, type a misleading hint.

In "Help protect your computer and improve Windows automatically", choose "Ask me later."

Once I'm in, create a new limited user (not a member of the Administrators group) and use that as my primary account.

Control Panel | Hardware and Sound | Mouse
    Pointers | Enable pointer shadow (uncheck)
    Pointer Options
        Enhance pointer precision (uncheck)
        Hide pointer while typing (uncheck)

Right-click taskbar | Properties | Taskbar
    Use small icons (check)
    Taskbar location on screen (change to "Right")
    Taskbar buttons (change to "Never combine")
    Notification area | Customize
        Turn system icons on or off
            Clock | Off (select)
            Volume | Off (select)
            Network | Off (select)
            ... turn everything off, except sometimes.
            (for example, I might leave Power on for a laptop.)
        Always show all icons and notifications on the taskbar (check)
    Use Aero Peek to preview the desktop (uncheck)

Windows Explorer | Organize
    Layout | Details pane (uncheck)
    Folder and search options | View | Advanced settings
        Always show icons, never thumbnails (check)
        Always show menus (check)
        Display file icon on thumbnails (uncheck)
        Display file size information in folder tips (uncheck)
        Display the full path in the title bar (Classic theme only) (check)
        Hidden files and folders | Show hidden files, folders, and drives (select)
        Hide empty drives in the Computer folder (uncheck)
        Hide extensions for known file types (uncheck)
        Hide protected operating system files (Recommended) (uncheck, Yes I'm sure)
        Show pop-up description for folder and desktop items (uncheck)

Control Panel | System and Security | Windows Update | Change settings
    Download updates but let me choose whether to install them (select)
    Allow all users to install updates on this computer (check)

Elevated command prompt | gpedit.msc | Local Computer Policy
    User Configuration | Administrative Templates | Windows Components
        Windows Explorer | Turn off numerical sorting in Windows Explorer (enable)
    Computer Configuration | Administrative Templates
        System
            Power Management | Video and Display Settings
                Turn Off Adaptive Display Timeout (Plugged In) (enable)
                Turn Off Adaptive Display Timeout (On Battery) (enable)
        Windows Components | Windows Update
            Do not display 'Install Updates and Shut Down' ... (enable)
            Do not adjust default option to 'Install Updates and Shut Down' ... (enable)

Control Panel | View by: Small icons (select)
    AutoPlay | Use AutoPlay for all media and devices (uncheck)
    Indexing Options | Modify | Show all locations
        Offline Files (uncheck)
        C:\Users (uncheck)
        C:\ProgramData\Microsoft\Windows\Start Menu (uncheck)
    Troubleshooting | Change settings | Computer Maintenance | Off (select)
    Windows Defender | Tools
        Automatic scanning | Automatically scan my computer (uncheck)
        Real-time protection | Use real-time protection (recommended) (uncheck)
        Administrator | Use this program (uncheck)

Right-click Start Menu | Properties
    Customize
        Computer | Don't display this item (select)
        Connect To (uncheck)
        Control Panel | Don't display this item (select)
        Default Programs (uncheck)
        Devices and Printers (uncheck)
        Documents | Don't display this item (select)
        Enable context menus and dragging and dropping (uncheck)
        Games | Don't display this item (select)
        Help (uncheck)
        Highlight newly installed programs (uncheck)
        Music | Don't display this item (select)
        Open submenus when I pause on them with the mouse pointer (uncheck)
        Personal folder | Don't display this item (select)
        Pictures | Don't display this item (select)
        Search other files and libraries | Don't search (select)
        Search programs and Control Panel (uncheck)
        Use large icons (uncheck)
        Number of recent programs to display (set to 20 to make the menu bigger)
        Number of recent items to display in Jump Lists (set to 20 to make the menu bigger)
    Store and display recently opened programs in the Start menu (uncheck)
    Store and display recently opened items in the Start menu and the taskbar (uncheck)

Right-click everything that is pinned to the taskbar
    Unpin this program from taskbar

Right-click Recycle Bin | Properties
    For each hard drive in turn (select)
        Don't move files to the Recycle Bin. Remove files immediately... (select)

Control Panel | Appearance and Personalization
    Personalization | Change desktop icons | Recycle Bin (uncheck)
    Change desktop background
        Picture Location: Solid Colors (select, choose black)

Control Panel | User Accounts | Change your account picture
    Browse for more pictures
       

Control Panel | Ease of access
    Change how your mouse works
        Mouse pointers | Regular Black (select)
        Prevent windows from being automatically arranged when moved to the edge of the screen (check)
    Change how your keyboard works
        Set up Sticky Keys | Turn on Sticky Keys when SHIFT is pressed five times (uncheck)
    Optimize visual display
        Turn off all unnecessary animations (when possible) (check)

Make a folder on the desktop named "_"
    Open the following folders simultaneously
        _
        C:\ProgramData\Microsoft\Windows\Start Menu
        C:\Users\%username%\AppData\Roaming\Microsoft\Windows\Start Menu
    Copy various shortcuts from Start Menu over to _
        Notepad
        Paint
        Command Prompt
        Calculator
        ... whatever else strikes my fancy
    ... as I install programs, consider adding them here if I use them a lot

Right-click the taskbar | Toolbars | New toolbar... | Desktop | _ (Select Folder)
    Right-click the taskbar | Lock the taskbar (uncheck)
    Drag the thumb of the _ toolbar to the top of the taskbar
    Right-click _
        Show Text (uncheck)
        Show title (uncheck)
        View | Small Icons (check)
    Drag the taskbar to be a tiny bit wider so three small icons fit side-by-side
    Drag the view-active-tasks part of the taskbar to be big
    Right-click the taskbar | Lock the taskbar (check)

Make a 1-pixel-by-1-pixel black .jpg and set it as the LogonUI background

I'm sure I'm forgetting some other things.  I'll add them later when I run into them.

I could probably make a series out of this.  Possible candidates for future posts: "Tweaks I make every time I install Office", "Tweaks I make every time I install Firefox"...

Command Prompt | Alt-Space | Defaults | QuickEdit Mode (check)

Three quarters are better than a dollar because they make noise!
    -- Lilly, Lilly's Purple Plastic Purse

Last time I talked about how to calculate a sine sweep mathematically.  There's a closed-form solution which is quite elegant but, alas, useless in a practical computer implementation due to the very big numbers that are being fed to the sin(...) function.

A computer implementation will care only about the discrete samples that need to be generated.  The integral in the previous post turns out to be counterproductive - we're much more interested in the infinite sequence of φ_i, and more particularly in their residue mod 2π.

Here are Matlab scripts that implement a sine sweep both ways:

First, the naïve mathematical solution that just calculates the exponential:

function signal = sinesweep_nostate( ...
    start, ...
    finish, ...
    seconds, ...
    sample_rate, ...
    amplitude, ...
    initial_phase, ...
    dc ...
)
% sinesweep_nostate returns a single-channel sine logarithmic sweep
%     DO NOT USE THIS FUNCTION IN PRODUCTION
%     THIS IS A PEDAGOGICAL EXERCISE ONLY
%     THE SIGNAL THIS PRODUCES GRADUALLY DISTORTS
%     INSTEAD USE sinesweep
%    
%     start: starting frequency in Hz
%     finish: ending frequency in Hz
%     seconds: duration of sweep in seconds
%     samplerate: samples per second
%     amplitude: amplitude
%     initial_phase: starting phase
%     dc: dc

% echo these interesting intermediate values to the console
R = (finish / start) ^ (1 / seconds)
B = 2 * pi * start / log(R)
A = initial_phase - B

time = 0 : 1 / sample_rate : seconds;
phase = A + B * R .^ time;
signal = amplitude * sin(phase) + dc;

end % sinesweep_nostate

Now, the iterative version that adds up the little Δφs to calculate φ_i iteratively:

function signal = sinesweep( ...
    start, ...
    finish, ...
    seconds, ...
    sample_rate, ...
    amplitude, ...
    initial_phase, ...
    dc ...
)
% sinesweep returns a single-channel sine logarithmic sweep
%     start: starting frequency in Hz
%     finish: ending frequency in Hz
%     seconds: duration of sweep in seconds
%     samplerate: samples per second
%     amplitude: amplitude
%     initial_phase: starting phase
%     dc: dc

time = 0 : 1 / sample_rate : seconds;
frequency = exp( ...
    log(start) * (1 - time / seconds) + ...
    log(finish) * (time / seconds) ...
);

phase = 0 * time;
phase(1) = initial_phase;
for i = 2:length(phase)
    phase(i) = ...
        mod(phase(i - 1) + 2 * pi * frequency(i) / sample_rate, 2 * pi);
end

signal = amplitude * sin(phase) + dc;

end % sinesweep

If anything, one would expect the first to be more accurate, since it is mathematically precise, and the second is only an approximation.  But let's see how the signal produced by each of them holds up.

I calculate, and plot, the same sweep both ways, with a slight dc offset to facilitate seeing the different sweeps on the same plot.  In particular this is a logarithmic sweep from 20 kHz to 21 kHz, over 30 seconds, using 44100 samples per second, an amplitude of 0.2, an initial phase of π, and a dc of 0.25 for one and 0.26 for the other:

plot(...
    x, sinesweep(20000, 21000, 30, 44100, 0.2, pi, 0.25), 'd', ...
    x, sinesweep_nostate(20000, 21000, 30, 44100, 0.2, pi, 0.26), 's' ...
)

(x is just 1:30*44100 - necessary to allow a single plot statement.)

Note that the approximate method is plotted in diamonds, and will be 0.01 lower on the resulting graph than the "exact" method, plotted in squares.

(This takes a while to plot because there are a heckuva lot of points...)

Ah, a nice solid mass of green and blue.  I won't show it here.

OK, let's zoom in to a very early chunk of the signal - I expect these to match up closely, with only the dc offset separating the points:

Yup.  Looks pretty good - give or take a pixel (quantization error in the display.)

Now let's see what happens to the signal towards the end of the 30 seconds.

Ick.  Something's rotten in the state of Denmark.  This distortion is so bad that it's visible - you don't even have to take an FFT to see it.

Exercise: take an FFT and look at the distortion.

Advanced Exercise: demonstrate that the distortion is, in fact, in the "exact" method and not in the iterative method... that is to say, show which clock is right.

EDIT: On second glance it looks like the second picture is just showing a horizontal shift between the two signals, which is expected.  I'll need to dig deeper if I'm to prove my assertion that the iterative method produces less distortion than the "exact" method.

Bad Perl solution to the "print the open lockers" problem:

perl -e"print join', ',map{$_*$_}1..sqrt pop" 100

54 characters.  I prefer this to the 53-character solution obtained by omitting the space after the first comma.

EDIT: 49 characters:

perl -e"print map{$_*$_,' '}1..sqrt pop" 100

EDIT: 48:

perl -e"print map{$_*$_.$/}1..sqrt pop" 100

EDIT: 47:

perl -e"map{print$/.$_*$_}1..sqrt pop" 100

I still think "say" is cheating but it does afford this very short solution:

perl -E"map{say$_*$_}1..sqrt pop" 100

EDIT: Apparently I need to learn how to count. Counts above are off. Anyway, 41:

perl -e"print$_*$_,$/for 1..sqrt pop" 100

Suppose you want to generate a continuous sine-sweep from time t_start to time t_end. You want the starting frequency to be ω_start, and the ending frequency to be ω_end; you want the sweep to be logarithmic, so that octaves are swept out in equal times. (The alternative would be to sweep linearly, but usually logarithmic sweeping is what you want.) For today we're going to have continuous time and sample values, so sample rate and bit depth will not be part of this discussion; in particular, the units of t will be seconds, and our sample values will go from -1 to 1.

Here's a picture of want we want to end up with:

Without loss of generality, set t_start = 0. This simplifies some of the math.

Let ω(t) be the "instantaneous frequency" at time t. Advanced Exercise: Define "instantaneous frequency" mathematically.

Sweeping ω(t) logarithmically means that log ω(t) is swept linearly from log ω_start to log ω_end. So

Quick check: ω(0) = ω_start, ω(t_end) = ω_end. Yup.

Let's define R (the rate of change of the frequency) as (ω_end / ω_start)^{1 / t_end}. This formula reduces to

ω(t) = ω_start R^t

Next step is to look at phase. For a sine wave of constant frequency ω, phase progresses linearly with t on the circular interval [0, 2π):
φ = k ω t
for some constant k. If the frequency of a sine wave was 1 cycle per second, the phase would go from 0 to 2π in a single second - this reveals that the correct value of k is 2π:
φ = 2π ω t

What happens with a variable frequency? A little calculus provides the answer. Consider a small change Δt, small enough so that ω(t) is almost constant over the interval:
Δφ ≈ 2π ω(t) Δt
because over that interval the sine sweep function is well approximated by a sine wave of constant frequency ω(t).

Starting from an initial phase φ_start, let Δt → 0, and:

Substitute:
u = τ log R
τ = u / log R
dτ = du / log R

Define:
B = 2π ω_start / log R
A = φ_start - B
and we have:

φ(t) = A + BR^t

Just for fun, here's the explicit solution without intermediate values:

Now, the payoff: the signal is just the sine of the phase, so our sine sweep is:

or, more explicitly:

Ah, beautiful mathematically.

But useless practically.

Why useless?

Well, note that our equation for phase is an exponential. That expression inside the sin(...) gets very big, very quickly. Any practical implementation of sin(...) is going to be utterly useless once its argument gets beyond a certain threshold - you'll get ugly chunky steps before you're halfway into the sweep.

Nonetheless, there is a practical way to generate (pretty good) sine sweeps. More on that next time.

EDIT: all logs are base e.

Another programming contest asks to solve the Josephus problem.

Bad Perl solution (83 characters... so close...)

>perl -e"@_=(1..$ARGV[0]);++$c%$ARGV[1]?$i++:splice@_,$i%=@_,1while$#_;print@_" 40 3
28

EDIT: got it down to 80.

>perl -e"@_=(1..shift);++$c%$ARGV[0]?$i++:splice@_,$i%=@_,1while$#_;print@_" 40 3
28

EDIT2: 78 dropping the parentheses.

>perl -e"@_=1..shift;++$c%$ARGV[0]?$i++:splice@_,$i%=@_,1while$#_;print@_" 40 3
28

EDIT3: 66, shamelessly cannibalizing others' ideas from the contest (though I refuse to use "say")

>perl -e"$k=pop;@_=1..pop;@_=grep{++$i%$k}@_ while$#_;print@_" 40 3
28

I found this programming contest interesting: here's what I've got.

perl -e "($a,$b)=@ARGV;map{$c+=$_*$b}grep{$a&$_}map{1<<$_}(0..log($a)/log 2);print$c" 7 19

I'm calling this a one-liner because the part between the quotes is less than 80 characters (75, to be exact.)  The full command line goes over :-(

Requires the inputs to be positive whole numbers.

Perfect example of Bad Perl.  Exercise: rewrite in Good Perl.

EDIT: got the whole thing down to 80 characters (with single-digit multiplicands.)

perl -e "($a,$b)=@ARGV;map{$c+=$b<<$_ if$a>>$_&1}(0..log($a)/log 2);print$c" 8 7

There's a bug that got away from me in Windows 7's HD Audio class driver (hdaudio.sys.)

Before I explain the bug, a few caveats:

If you're using Vista, there's no problem.  This only affects Windows 7.

If you use the third-party audio driver, there's no problem.  This only affects the Microsoft HD Audio class driver (hdaudio.sys.)

If your recording devices (mic and line in) are independent (can capture from both at the same time) or muxed (can capture from one or the other but not from their sum) then you're fine; this only affects mixed capture (can capture from one, from the other, or from their sum, but not from both sides of the mix independently.)

If either of your recording devices doesn't support jack presence detection then you're fine; this only affects mixed capture where both sides of the mix support jack presence detection.

Here's the bug.

1. There are various ways to "reset" the system and bring the OS recording device state into synchronization with the hardware state
• reboot
• reset the HD Audio controller
• reinstall the audio driver
• restart the AudioEndpointBuilder service
• unplug all recording devices in the mix
  
2. The first state change (plug or unplug) after a reset works
3. The second and further state changes are not recognized by the OS until a reset.

Here's a sample state diagram assuming a mixed Mic and Line In (click to view full-size:)

Workarounds:

Install the third-party audio driver instead of the HD Audio class driver.

Use audio extension cables to fool the jack presence detection hardware into thinking something is always plugged in.