Tuesday, December 1, 2009

Cannibalistic Truth

Blog post # 20:

First, a little politics.

You know what Al Qaeda needs? They need LOBBYISTS, that's what. Poor AQ. They get such a bad rap from the US government and from the US media. Sure, they mass-murder innocent people. But think of all the other mass-murdering organizations that have received "legitimacy" as a result of they having lobbyists who pressure and bribe (pay campaign contributions to) American politicians: The defense industry, the US insurance industry, the prescription drug industry, the tobacco industry, the gun lobby. On and on.

---------------------------

The rest of today's post my repulse some people. There is nothing too dark, but there is a (fun and easy) math trick, as well as a poem of mine. (Gads!)

----



Okay, some fun. A math trick! Yay! Yes, the following is math, but it only involves adding, subtracting, and counting.
Even a kid can do it.
(I tried this on my non-math friend, and she liked it. Be sure to look at the examples.)

First, write down a list of any number of positive integers (whole numbers at least as big as 1), from left to right, where each
integer is greater than the integer to the left of it.
(So, you have, say, 5, 6, 7, or 8 numbers in your list)

Example: 2, 3, 5, 7, 8, 11

Second, make another list below the first list, where 0 is subtracted from the first list's first integer, 1 is subtracted from the first list's second number, 2 is subtracted from the 3rd number, 3 from the 4th number, etc.

Example:
First list: 2, 3, 5, 7, 8, 11
subtract: 0, 1, 2, 3, 4, 5
Second list: 2, 2, 3, 4, 4, 6

Next, make a 3rd list where the first term of the 3rd list gives the total number of 1's in your 2nd list, the 2nd term of your 3rd list gives the total number of 2's in your 2nd list, the 3rd term of the 3rd list gives the total number of 3's in your 2nd list, the 4th term gives the number of 4's, etc. Do this until you reach the highest number in the 2nd list.
So, the 3rd list will have, when it is complete, the number of terms which is equal to the highest number of the 2nd list.
Some of the numbers in the 3rd list may be 0's.
(Just to be clear, by "total number of...'s", I am asking about numbers themselves, not about the digits of multi-digit numbers.)

Example: 3rd list: 0, 2, 1, 2, 0, 1
(We have zero 1's in our second list,
two 2's, one 3, two 4's, zero 5's, and one 6.)

Next we make a 4th list of the same number of terms as is in the 3rd list, where the first term is the same first term as is in the 3rd list.
The following terms are each equal to
{the previous term in the fourth list}
plus
{the corresponding term in the 3rd list}.
(So, we want the fourth list to be a running-total of the terms in the
3rd list.)

Example:
3rd list again: 0, 2, 1, 2, 0, 1
0,0+2,2+1,3+2,5+0,5+1
4th list: 0, 2, 3, 5, 5, 6
(In middle row above, each number before + is previous number in 4th
list, each number after + is corresponding number in 3rd list.)

Finally, make a 5th list where 1 is added to the first term of the 4rd list, 2 is added to the 2nd term of the 3rd list, 3 is added to the 3rd term, 4 to the 4th term, etc.

Example:
4th list: 0, 2, 3, 5, 5, 6
Add: 1, 2, 3, 4, 5, 6
5th list: 1, 4, 6, 9, 10, 12

So, comparing your first list with your last list, we should have every integer from 1 to {1 more than in your first list's biggest number}, where each integer is in one list or the other, but not in both.

Example: (*'s placed at missing integers' positions to help show what I mean.)
First list:
*, 2, 3, *, 5, *, 7, 8, *, * , 11, *
5th list:
1, *, *, 4, *, 6, *, *, 9, 10, * , 12
And you should have this result (each integer in one list or the other, but not in both) no matter which integers are in your first list! (as long as they are positive and increasing)
By the way, you probably do not want your first list's biggest number to be that big. As I had in my example, you probably want to have each integer in the first list to be only 1, 2, 3, or 4 bigger than each previous integer.
This is because the 3rd list has as many terms as the highest number in the second list.
(So, if the second list's highest number is 110875, for example, you might have a problem...)

-----------------------------




I wrote this just yesterday.


Cannibalistic Herbivore
------------------------
(Into Tautological Antecedents)
--------------------------------

In such parallelism of coincidental tangents, I prod
The tapering annulus, and I rectify its juxtapositions.
In
Such parameterizations of convincing tableaux, I protrude
Into
Tautological antecedents, and I repulse their judgment.

Yes, I am as the head of an insect's corpse. And
My pincers are those strangely shaped concentricities.
Oh, I spin and fly and feast on concurrent dreams
Of such parallelism and parameters constrained by this
Cannibalistic truth, certainly. In such ropes, each
Grasped and odorless, the moist blood tastes of sediment.
Yet, by one millimeter, the lines converge...
Almost, always, and again; and they spit upon
My mandible. They spiral within my sharp
Antennae aglow. And I coordinate each odd
Process with its vastness. For, I am but
A vacuous specimen, ha. I am but myself,
Somewhat equal to myself. I am but
Existential, and I wiggle in this gap.
Yes, I am a cannibalistic herbivore. And I
Am surely the equations that I multiply
By my tirades, by my solitary abutment.




Thanks,
Leroy Quet

No comments: