Heavy Metal History/Evolution Chart

Whoa! When googling for more information on Blue Cheer after seeing Mudhoney the other night, I found this crazy heavy metal history/evolution chart on this Turkish? message board site. Hmm….I wonder who made this? It’s sort of like the world history timeline posters but for heavy metal! Oh, I just found out, I guess it’s a chart from a film “Metal: A Headbanger’s Journey”!

Probably missing from the chart is the Godfather of Metal, at least Prog/Technical Metal, Bela Bartok...

Here’s some of the text on the site from the chart:

Early metal (1966 – 1971) – Jimi Hendrix – Blue Cheer – Deep Purple – Iron Butterfly – Led Zeppelin – MC5 – Mountain – The Stooges – Black Sabbath

Power metal (1976 – Present)
– Scorpions – Judas Priest – Rainbow – Accept – Manowar – Dio – Yngwie J. Malmsteen – Helloween – Blind Guardian – HammerFall – Primal Fear

New Wave of British Heavy Metal (1979 – 1983) – Motörhead – Saxon – Iron Maiden – Angel Witch – Girlschool – Tygers of Pan Tang – Diamond Head

Progressive metal (1970 – Present) – Uriah Heep – Rush – Queensrÿche – Savatage – Fates Warning – Voivod – Dream Theater – Meshuggah – Symphony X – Evergrey – Tool

Death metal (1985 – Present)
– Possessed – Sodom – Death – Morbid Angel – Obituary – Deicide – Cannibal Corpse – Immolation – Autopsy – Nile

This calls for some…Def Leppard!!!

Def Leppard: Foolin

I wonder what category Def Leppard is in, pop metal? Yep! How many Bud Lights total do you think people have drunk to this song since it came out?

I thought I’d throw in this picture from the movie Clash of the Titans, that’s pretty metal looking, and Def Leppard-era, right?

My favorite vote for the best early metal song is the Troggs: Come Now.


Lil’ Wayne, Jay-Z, Just Blaze: Mr. Carter

Another new Just Blaze track! Mr. Carter – Lil’ Wayne featuring Jay-Z, that’s pretty big.

Big album too–Tha Carter III is the first album to sell a million copies the first week in three years!

MP3 courtesy of Passion of the Weiss

Lil Wayne etch a sketch by etchasketchist

Making Change from a Dollar

I was trying to figure out this problem, “How many different ways can we make change of $ 1.00, given half-dollars, quarters, dimes, nickels, and pennies?” This problem is featured in this Ramblings of a Geek blog post and is also in section 1.2.2 of the book Structure and Interpretation of Computer Programs (PDF version).

I think it involves the pigeonhole principle when trying to figure out how the recursive formula works. Here’s part of the set up from the Ramblings of a Geek blog post:

Let An be the number of ways to make n cents with pennies.
Let Bn be the number of ways to make n cents with pennies and nickels.
Let Cn be the number of ways to make n cents with pennies, nickels, and dimes.
Let Dn be the number of ways to make n cents with pennies, nickels, dimes, and quarters.

The question is asking us to find D100

So if you’re trying to figure out Dn=100 as in the post, where n=the amount, and D is a combination of quarters, dimes, nickels and pennies, you use the “pigeonhole principle” and split the choices into combinations using quarters and combinations without using quarters first.

So, if you use 0 quarters, that could be C100 combinations, where Cn is the number of combinations using dimes, nickels, and pennies but no quarters. That’s your “no quarters” box. Then your other box, to begin with, for a combination to belong in the “using quarters” box, you know automatically that the combination must be using at least one quarter to belong in that box. So since the any combination is for sure using at least one quarter (25 cents), you will have a remaining 75cents (the beginning 100 cents minus the quarter you’re sure exists in the combination) that you’re unsure of the composition of, that is, you have 75 cents that will be made up of a combination of 0-3 quarters, 0-7 dimes, 0-15 nickels, and 0-75 pennies. Since it could be a combination of quarters, nickels, and pennies, that can fit in category “D”, and since the amount is 75 cents, you can label that as D75.

So, D100=C100 + D75. So technically, like in the SICP book and the Ramblings of a Geek blog post, you can think of a recursive formula:

The number of ways to change amount a using n kinds of coins equals the number of ways to change amount a using all but the first kind of coin, plus the number of ways to change amount a – d using all n kinds of coins, where d is the denomination of the first kind of coin.

So we’ve determined above, that D100=C100 + D75. You can use the same logic as above on the D75 to reduce that to a C. For example, for D75, you might get 75 cents by using zero quarters, so you could categorize that as C75…and if you were using at least one quarter, you would know for sure what 25 cents of that 75 cents was made out of (that one quarter), that would leave you with 50 cents you were not sure about the composition of, that might be made up of quarters, dimes, nickels, and pennies, which could be categorized as D50, so D75=C75 + D50. You can do that with the recursive formula, you can then examine D75, that would be equal to D75=C75+D50. And on down the chain until the Ds disappear completely, replaced by all Cs, then replace the Cs with Bs and As.

And with As, there will always only be one way to make anything out of pennies; $1 is 100 pennies, 50 cents is 50 pennies, etc. So when you examine the Bs, the formula is n/5 + 1. The 1 is for one way you can make anything using all pennies, and the n/5 is how many nickels fit in the amount n. So, say n=15. Here n/5 is 15/5=3. B15=4, using the formula n/5 + 1. That corresponds here to these four situations: you use one nickel and ten pennies, or two nickels and five pennies, or three nickels and no pennies, or no nickels and 15 pennies!

So starting at Dn, you reduce down to Cs, then to Bs and As. Here’s the rest of the Rambling Geek’s example:

We want D100, let’s use some substitution here:
D100 = C100+D75
D100 = C100+C75+D50
D100 = C100+C75+C50+D25
D100 = C100+C75+C50+C25+1

C100 = B100+B90+B80+B70+B60+B50+B40+B30+B20+B10+1
C75 = B75+B65+B55+B45+B35+B25+B15+B5 (note no +1, since C5 = B5)
C50 = B50+B40+B30+B20+B10+1
C25 = B25+B15+B5

Substituting with the values of Bn in the formula above we see:
C100 = 21+19+17+15+13+11+9+7+5+3+1 = 121
C75 = 16+14+12+10+8+6+4+2 = 72
C50 = 11+9+7+5+3+1 = 36
C25 = 6+4+2 = 12

D100 = C100+C75+C50+C25+1 = 121+72+36+12+1 = 242

So there are 242 ways to uniquely make change for $1 using pennies, nickels, dimes, and quarters.

Another way of thinking about part of the problem, is, say the amount n is 100, and you want to use quarters, dimes, nickels, and pennies, so you’re looking at D100, is to think about the quarters in the following way: we decided with Bs, you look at how many times nickels divide into n, that would be n/5, and then add 1 for the number of ways you can make n with pennies for combinations where you don’t use any nickels at all. With Ds, we can look at how many times quarters can divide into n, so that would be n/25, and then add C100 for the combinations possible for which you use no quarters, and you make the $1 out of combinations of 0-10 dimes, 0-20 nickels, and 0-100 pennies. So if you use no quarters, that’s C100. Add that to if you use one quarter, then you have 75 cents you don’t know the consistency of, possibly made out of dimes, nickels, and pennies, C75. Then add all that to the combinations if you use two quarters, then you have 50 cents you don’t know about, that could be C50. Then add that all to if you use three quarters, then you have 25 cents which could be made out of dimes, nickels, and pennies, C25. Then add that all to the number of combinations where you use four quarters, in which case you would have no combinations made of dimes, nickels, and pennies left over (C0), so if you use four quarters, there’s only one possible combination for n=100, using those four quarters. So that’s D100=C100 + C75 + C50 + C25 + 1.

Each time you subtract the denomination of the type of coin being examined from the amount, that’s because you’re setting up the case where you are using that type of coin, such as quarters; so C100 is all the ways to make 100 cents made of 0 quarters and 0-10 dimes, 0-20 nickels, and 0-100 pennies; C75 is all the ways to make 100 cents using 1 quarter, 0-7 dimes, 0-15 nickels, and 0-75 pennies; C50 is all the ways to make 100 cents using two quarters, 0-5 dimes, 0-10 nickels, and 0-50 pennies; C25 is all the ways to make 100 cents using three quarters, 0-2 dimes, 0-5 nickels, and 0-25 pennies; C0 is all the ways to make 100 cents using four quarters, namely, 1.

Disclaimer, I hope that helps explain the logic behind the problem and the solution and hope that my explanation is correct, it might not be! 😉

internal tag: math

(Why doesn’t the search function on wordpress.com blogs search post tags too? I can’t believe that when I search my blog with the term “math,” only posts with “math” in the actual post text are retrieved; if the post is tagged “math” but doesn’t contain the term “math” in the post text, the post won’t be retrieved…hence the “internal tag” I had to add. Bah.)

See these other math related posts including:

Mudhoney Live

Just saw Mudhoney live. Saw them with FiestaRed back in 97 too, great show as usual. They have so many hits you probably don’t even realize from all time periods of their career, including some immediate classics from their newest albums. There were segments where just Steve Turner plays guitar, and it’s just as awesome as when they both play guitar. One of the best live acts ever! Love the solos as usual, like a bluesy punk/metal band channeling Blue Cheer and Jimi Hendrix. The audience goes crazy, especially for songs like Hate Police, Sweet Young Thing, Suck You Dry, etc.

In tribute, Mudhoney have a song Magnolia Caboose Babyshit a cover of Magnolia Caboose Babyfinger by Blue Cheer.

Check out this awesome Blue Cheer/Mudhoney-like Grand Funk Railroad song — Paranoid (live).

I heard some Moody Blues recently with some really great psychadelic guitar solos…

Check out this amazing Blue Cheer song, crazy guitar solos, and parts of it are almost like some indie post-punk song, mixed with Jimi Hendrix of course: Blue Cheer — Doctor Please

Best New New Wave Band Ever: Cut Copy…or Sebastien Tellier?

Ah…sort of like For Against meets Depeche Mode etc, this is great.

HOWEVER maybe the great Sebastien Tellier has edged out Cut Copy, which impressed me initially, but seem to lack a certain depth…maybe they sound too much like Depeche Mode? Sebastien Tellier in contrast is much more of a songwriter, has more depth, his sound is more polished, complex, and diverse, out of a pan-60s/70s/80s tradition…

Madvillainy Original Samples

Yeah, there are some samples floating around out there. BUT THIS DOESN’T ANSWER THE MOST IMPORTANT QUESTION, WHAT IS THE SAMPLE FOR MADVILLAIN’S SICKFIT Supervillain??? THAT’S SUCH A GREAT PROG ROCK SONG! Aha, it’s O Terco: Adormeceu!!!

O Terco: Adormeceu (sampled for Madvillain: Supervillain)
Also see: The Best Hip-hop Samples Ever

Logarithms Explained, and the Associative Property of Multiplication

So much advanced math relies on a firm grasp of basic Algebra and Algebra II.

Today, lets take a look at logarithms!

So what are logarithms? Well, first let’s look at exponential equations, such as 2^x = y where the 2 is a base. We all know that for example, 2^3 = 8. A general form is b^x = y where b is the base. Well, with logarithms, the format is log_b y = x. So for 2^3 = 8, we would express that with logarithms as log_2 8=3. Fun, isn’t it! The logarithm is the number that the base is raised to a power by to equal a given number; in the example above, the base 2 is raised by the power 3 to equal the number 8.

So the tricky part is that you get rules like log_b y + log_b z = log_b (yz) Huh? How’s that? Well, it has to do with the Associative Property of Multiplication and the rules for multiplying exponents. Basically, when you multiply exponents with the same base, such as b^m * b^n, you add the two powers; so the answer is b^{m+n}. So for example, 2^2 * 2^3 = 2^{2+3} = 2^5 = 25 . Keep in mind that the Associate Rule of Multiplication says that (ab)c=a(bc); so 2^2 = (2*2), 2^3 = (2*2*2), and 2^2 * 2^3 = (2*2) * (2*2*2)=(2*2*2*2*2)=2^5. These rules are important for working with logarithms.

So, why is the following true: log_b y  + log_b z = log_b (yz)? What it’s saying is if you add what you
raise b by to get y plus what you raise b by to get z, you’ll get what you raise b by to get yz. Now, imagine an expression log_2 8 + log_2 16. Why would the equation log_b y  + log_b z = log_b (yz) work here? Well, log_2 8=3 and log_2 16=4. By log_b y  + log_b z = log_b (yz), the left hand side, log_2 8 + log_2 16=7, should equal the right hand side log_b(yz) . Here y=8 and z=16, so yz=128. If you work out the math, log_2 8 + log_2 16=7, and with these numbers, log_b (yz)= log_2 (8*16)=log_2 (128)=7,

, so indeed log_2 8 + log_2 16=log_2 (8*16)=log_2 (128)=7. Wow, so log_b y  + log_b z = log_b (yz) works! Why? A way of looking at this is, log_2 8=3 is equivalent to 2^3 = 8, and
log_2 16=4 is equivalent to 2^4=16 . Since y=8 and 2^3= 8, and z=16 and 2^4=16 , yz = 2^3 * 2^4 = 8*16 which by the rule for multiplying exponents = 2^7=128. And 2^7= log_2 (128). And viewing this by the associative rule, 2^3= 2*2*2 and 2^4 = 2*2*2*2 , so 2^3 *2^4= 8*16 = (2*2*2)*(2*2*2*2) = 2^7 =128. Then
log_2 8 + log_2 16=  log_2(2* 2* 2) + log_2(2* 2* 2* 2) = log_2 (2* 2* 2* 2* 2* 2* 2 )= log_2 128,
the general form of which is log_b y  + log_b z = log_b (yz). I think.

Logarithms are great because you can solve exponential equations with them. You can do this with this log rule: log_b (m^n) = n * log_b m. So for example, if you had the equation 2^x = 50, you could start taking the log of both sides, log (2^x) = log 50 , which equals x* log 2 = log 50, which equals x= log 50/log 2 . Then you can just figure out log 50 and log 2 with a calculator, divide, and you’ve got your answer. You sometimes might want to use the change of base formula too. It’s important to remember that b^{log_b m}= m, this will come in handy when solving some equations. For example if you had the equation x^2 = 50, you could take the log of both sides, log (x^2) = log 50, which equals 2* log (x) = log 50, which equals log (x) = log 50/2. Now, how to isolate the x? Here we’re using base 10. So raise en to the power of each side of the equation, 10^{log (x)} = 10^{log 50/2}. That equals x = 10^{log 50/2}, you can solve the right hand side with a calculator to find out log 50 and divide that by 2, and raise ten to the power of the result. Basically when you have log_b x = y what that means is b^y = x. So if log_b x = y, if you raise the base b to y which equals log_b x, you will get simply x.

**These equations were designed in Microsoft Equation Editor (comes free with Microsoft Word)–but then to my chagrin I realized I don’t know if or how I can export the document with the equations to a format such as text plus image files where the formatting of equations (subscripts, superscripts) would be preserved in a blog post. Until I get scribd/ipaper working or figure out how they put those equations in Wikipedia (do people insert the jpgs of the equations one by one by hand, or do they use a program that can convert the text and equations into text and images all at once, and can Microsoft Word do this? Here’s the original PDF file; I still want to be able to use iPaper for ease, I then figured out how to render the equations in Latex for WordPress, but it seems kind of manually laborious? Is there some easy way to do this all? Here’s some more info on using Latex with WordPress.com