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Chess: mobility: opening files?

The tutor discusses the (dis)advantage of opening files.

I suffered some chess burnout earlier this holiday season, but hope I’m making a recovery. I still lose more than I win, but computers aren’t prone to error the way a human might be.

Rooks need open files to operate, or course. In a recent game I had a chance to snatch a centre pawn from the computer; I didn’t, because it would have opened a file. The computer is better with its rooks than I am.

In early training about chess, one hears about the advantage of opening the board to mobilize pieces. The heavy pieces – rook and queen – benefit most from an open field. However, the computer has a much greater awareness of the board than most humans can.

As often as not, I make moves to try keep the computer’s pieces bottled up. For instance, if a bishop is in front of a rook, I try to leave that bishop alone, hoping it will stay there. Often, a player’s own pawns stall movement of other pieces. Even when I can snatch such a pawn, I’ll often let it be. This may represent a beginning towards understanding positional play.

HTH:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

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English: Homonyms (or homophones): peel and peal

Call them homonyms, homophones, or what you will; the tutor introduces another pair.

My coverage of homonyms began with my post of August 21, 2013. Searching up homophone and/or homonym on this site reveals my love of them.

Playing the Parker Brothers game Boggle (our version is from 1977), homonyms often surface. Here is a recent pair:

peel: (noun) the skin of a fruit.

peal: (noun) sound audible over a long distance.  As the storm began, he heard a peal of thunder.

Often, with a pair of homonyms, one is very familiar, while the other is rarely used.

HTH:)

Source:

Collins Essential Canadian English Dictionary & Thesaurus. Glasgow: HarperCollins,
   2006.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

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English: plurals: -oes

The tutor visits a favourite potential hazard of English.

Famously, potato and tomato require the addition of -es to make the plural:

potato → potatoes

tomato → tomatoes

Unknown to me before today, veto and embargo follow the same pattern:

veto → vetoes

embargo → embargoes

Perhaps surprisingly, with motto, mango, and mosquito, the e is optional:

mosquito → mosquitoes or mosquitos

motto → mottoes or mottos

mango → mangoes or mangos

Interesting, eh?

HTH:)

Source:

Collins Essential Canadian English Dictionary and Thesaurus. Glasgow: HarperCollins,
   2006.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

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Mobile phone apps: Nexus 4 onboard calculator: trig functions

The tutor offers a hint about using the Nexus 4 onboard calculator for sin, cos, and tan.

One of my New Year’s resolutions is to get acquainted with my mobile phone. Believe it or not, I’ve been lounging on the couch, panning through its menus to discover how to use its features.

My phone is a Nexus 4, running Android 5.1.1. Last night, among the apps, I discovered its onboard calculator. I love the simple design. At the right is a green tab that you can “pull out”, revealing scientific functions.

On mine, anyway, sin, cos, and tan expect the angle to be in rads. For instance, sin30° is 0.5. However, if I enter sin30 on the Nexus 4 calculator, it replies -0.9880…., which is sin(30rads).

If you’re in the same situation, you can get the answer for degrees by entering, for example,

sin((angle in degrees)xπ÷180)

You’ll probably see the π button on the green pull-out, perhaps below sin, cos, and tan.

Therefore, if you want tan45°, just key in

tan(45xπ÷180)

The answer should say 1.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

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Math: linear diophantine equations: integer solutions to Ax+By=C

The tutor delves deeper into when integer solutions can be expected for Ax+By=C.

In my dec 16 post I discussed finding integer coordinates for a linear equation in the form Ax+By=C. I pointed out that not always can integer solutions be found.

An equation of form Ax+By=C can be referred to as a linear diophantine equation. Furthermore, there is an easy way to tell if it has integer solutions. Specifically, if the greatest common factor of A and B is also a factor of C, the equation does have integer solutions.

Example 1: Determine if 2x-3y=11 has integer solutions; if so, find one.

Solution:

The greatest common factor of 2 and -3 is 1, which is also a factor of 11. Therefore,
2x-3y=11 does indeed have integer solutions, one being (4,-1): 2(4)-3(-1)=8+3=11.

Example 2: Determine if 4x-8y=6 has integer solutions; if so, find one.

Solution:

The greatest common factor of 4 and -8 is 4, which is not a factor 6. This equation has no integer solution.

Note that, for graphing purposes, convenient coordinates can still be found for 4x-8y=6. Despite its not having integer solutions, it does have (-2.5,-2) and (1.5,0), which are quite convenient for graphing.

I’ll be further discussing linear diophantine equations in a future post:)

Source:

Dudley, Underwood. Elementary Number Theory. New York: W H Freeman and
   Company, 1978.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

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Winter ice: heat of fusion

As the ice melts away in the sun, the tutor entertains the heat involved.

The heat of fusion is the amount of energy, per gram, that is required to melt solid to liquid without increasing its temperature. For ice, it’s 334J/g.

The tutor wonders how quickly the ice on the pavement melts in the sunlight on a day like today.

We’ll start with some assumptions. Let’s imagine we have a layer of ice 0.1cm (aka 1mm or 0.001m) thick over the pavement. Then the volume of ice per square metre is 1mx1mx0.001m = 0.001m³. The density of ice is 917kg/m³, so the mass of ice per square metre of pavement would be 0.917kg = 917g. Furthermore, we’ll assume the temperature is hovering around 0°C.

To find the amount of heat required to melt 917g of ice, we multiply it by ice’s heat of fusion, earlier said to be 334J/g. 917gx334J/g=306278J.

The wattage of sunlight reaching the pavement, on a day like today, might be 341W/m².

The time for the ice to melt under the assumptions above is t=306278÷341=898s: 15 minutes.

All the best this holiday season:)

Source:

hypertextbook.com

itacanet.org

Hebden, James A. Chemistry: Theory and Problems, book two. Toronto: McGraw-Hill,
   1980.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

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English: what is a barnburner?

The tutor shares another term from the farming culture he grew up in.

I don’t often hear the word “barnburner,” but of course I don’t live in the culture it’s probably from. Neither my Webster’s, nor my Collins, contains it, but my Oxford Canadian Dictionary of Current English does.

A barnburner is a positively sensational event. For example, a party at which everyone had a memorably fun and exciting time might be considered a “barnburner.”

May the term apply to holiday festivities that are intended to be thrilling.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

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Statistics: Proof of V(X)=E(X²)-[E(X)]²

The tutor offers proof of a formula he recalls from Stats.

In my university Stats courses, a formula referred to often was

    \[V(X)=E(X^2)-[E(X)]^2\]

in which

V(X)=population variance

E(X^2)=expected value of X^2

[E(X)]^2=square of the expected value of X

Here’s how I believe one might prove it:

By definition,

    \[V(X)=\frac{\Sigma_i^n(x_i-\mu)^2}{n}\]

By expansion,

    \[(x_i-\mu)^2=(x_i-\mu)(x_i-\mu)=x_i^2-2x_i\mu+\mu^2\]

Therefore,

    \[V(X)=\frac{\Sigma_i^n(x_i-\mu)^2}{n}=\frac{\Sigma_i^n(x_i^2-2x_i\mu+\mu^2)}{n}\]

which further equals, by taking separate sums,

    \[V(X)=\frac{\Sigma_i^nx_i^2}{n}-2\frac{\Sigma_i^nx_i\mu}{n}+\frac{\Sigma_i^n\mu^2}{n}\]

Now, using summation rules, we can rewrite the above as

    \[V(X)=\frac{\Sigma_i^nx_i^2}{n}-2\mu\frac{\Sigma_i^nx_i}{n}+\frac{n\mu^2}{n}\]

which next becomes

    \[V(X)=\frac{\Sigma_i^nx_i^2}{n}-2\mu\frac{\Sigma_i^nx_i}{n}+\mu^2\]

Since we are calculating for the entire population,

    \[E(X^2)=\frac{\Sigma_i^nx_i^2}{n}\]

and also

    \[E(X)=\frac{\Sigma_i^nx_i}{n}\]

Substituting the above definitions into our expanded, then simplified, formula, we arrive at

    \[V(X)=E(X^2)-2\mu E(X)+[E(X)]^2\]

By definition, E(X)=\mu. Using that idea, our formula becomes

    \[V(X)=E(X^2)-2E(X)E(X)+[E(X)]^2\]

then

    \[V(X)=E(X^2)-2[E(X)]^2+[E(X)]^2\]

and finally

    \[V(X)=E(X^2)-[E(X)]^2\]

I’ll be verifying this variance formula with an actual list of values in a coming post.

HTH:)

Source:

Ross, Sheldon A. A First Course in Probability, 3rd ed. New York: Macmillan, 1988.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

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Chess: queening a pawn

The tutor introduces a resource to help with endgame play.

Some tough games, I’ve won by queening a pawn. (I mainly play the computer, and I lose more often than I win.) Towards the end of a long, tiring game, knowing how to queen a pawn can really help.

A very useful resource that can teach you just how to queen a pawn is by chessvideos on YouTube. It’s less than five minutes long. My understanding of it is to keep your king ahead of your pawn; you’re best to have one or two squares between them. The key is to get your king to one of the the six squares on ranks 7 or 8, left, directly ahead, or right of your pawn.

I highly recommend the video. I’ve used its tips myself, enabling me to queen the pawn with much less effort.

HTH:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

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Home computer use: Computer security: Bitdefender re-install

Amid a busy life and Bitdefender’s solid reliability, it wins again.

Coincidentally, on this first day of Christmas break, my Bitdefender 2015 subscription finally expired. Back in my Dec 6 post, I told of plans to get out to check the computer security software options. While I did so briefly online, I never got to the stores. Bitdefender seems still to be at the top anyway, so I don’t feel remiss about sticking with it.

The package we have is Bitdefender Total Security. It comes with a license for 3 PCs, which is fortuitous because that’s how many Windows ones we have. Today I renewed Bitdefender on all three of them. For those curious, here’s the renewal process as I recall it:

  1. If you renew online, you get an email from Bitdefender. It includes your license key (which likely is the same one you had before, if you’re renewing) and links to the renewal downloads. One link is for Windows 7 and up, while the other is for XP and Vista.
  2. You copy the link (whichever you need), paste it to the address bar of your browser, then go there. In my experience (I use Google Chrome), the download starts immediately.
  3. I clicked on the downloaded file, then chose Open. When asked if I wanted to allow the program to run, I chose Yes.
  4. First, Bitdefender needs to uninstall the previous version. After allowing it to do so, you need to let your computer restart.
  5. Interesting: After the restart, on two of my computers (one Windows 7 and one XP), Bitdefender reappeared front and centre, prompting to start the new installation. However, on one of them, (the other Windows 7 one) it didn’t; I had to go back into the Downloads folder, reopen the program I’d downloaded from the link, and run it again. From then, it went fine.
  6. On one of the computers, I had to tell it to use the installation key detected onboard. For the other two, I just had to tell it I had one – or else just click Finish.
  7. With two of the computers, I had to login to Bitdefender to complete the install. However, with the other one (which was the same one for which I had to re-open the file after restarting), I wasn’t asked to.

Although re-installing Bitdefender is easy, it takes time. I spent around 15 minutes per PC.

In my experience, it’s unwise to do any computer-related task in a hurry:)

HTH:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.