Two of the classes (second and fourth period) are Earth Science, and two of the classes (third and fifth) are called "Integrated Science." Oddly enough, the Earth Science classes are for freshmen, while most of the Integrated Science students are juniors. So once again, I don't know how schools are organizing science classes these days.
I will say what the main classroom management issues are, even though this isn't DITL:
- The students are required to sit in assigned seats. No seating chart is provided, but I do make the students sit in actual seats rather than the lab stations.
- The students are not allowed to eat or drink. One girl eats a lollipop that a teacher in the office has given her, and so I write her name down.
- The students must write 15 notes from the DVD's that I'm playing for them. Several students refuse to take notes. One guy is summoned to the office for half the period. Upon his return, I tell him that he only needs to take half as many notes (so about 7-8) for the remaining time, but he doesn't. I never catch his name -- he's already holding the summons slip he received in the previous class -- so I go to my backup plan and describe his backpack. It has a very distinctive design -- the California Bear Flag. Once again, maybe students with easily noticed backpacks or other features shouldn't break the rules.
Chapter 4 of Wayne Wickelgren's How to Solve Problems is "Classification of Action Sequences." It opens with:
"The first thing that most people do when confronted with a problem is to start applying the allowable operations to the givens in the problem. Call this random trial and error [equivalent to random sampling with replacement, in probability theory]."
But random trial and error usually isn't effective, so the author suggests a better method:
"Most desirable of all random trial-and-error schemes would be a generation method that automatically produced a mutually exclusive and exhaustive listing of all sequences of actions up to some maximum length. Call this systematic trial and error (equivalent to random sampling without replacement)."
Wicklegren now gives the best, most powerful trial-and-error method:
"Such classificatory trial and error only works for problems where sequences of actions that are equivalent with respect to a solution of the problem, but most of the problems people solve probably exhibit some such equivalences."
As the author explains, one way in which two actions may be considered equivalent is if they are commutative, as is often the case in algebra:
"For example, in solving for x, given the equation 5x + 17 = 3x + 21, we could subtract 3x from both sides of the equation as the first step, then subtract 17 from both sides as the second step; but the same result is achieved by performing the actions in the reverse order."
Wicklegren's first example of a problem in this chapter is the six-arrow problem. Unfortunately, many of the problems in this chapter aren't available on other websites, so now I'm forced to type the statement of the problem into this blog post:
You are given six arrows in a row, the left three pointing up, and the right three pointing down. The goal is to transform these arrows into an alternating sequence such that the leftmost arrow points up, the next arrow to it points down, the next up, then down, then up, and then down. The actions allowed are to simultaneously invert (turn upside down) any two adjacent arrows. Note that you cannot invert one arrow at a time but must invert two arrows at a time, and the two arrows must be adjacent. Achieve the solution using the minimum number of actions (inversion of adjacent pairs.
"Before reading any further, try to solve this problem by determining the very small number of different equivalence classes of action sequences."
Wicklegren gives us several tricks to finding these equivalences:
- First, there are only five possible actions -- invert 12, 23, 34, 45, or 56.
- All the actions commute -- inverting 34 then 45 is equivalent to inverting 45 then 34.
- Each action is its own inverse -- inverting 23 then 23 again is equivalent to the identity.
- Inverting 12 or 56 is useless because arrows 1 and 6 don't need to change at all -- if we were to invert 12 we'd have to do it again, and this is the identity so we might as well not invert.
So there are only three possible inversions left -- 23, 34, and 45. We see that performing all three of them produces our goal -- but these three can be performed in any order, so there are six solutions.
We skip to the next problem that I don't have to type -- the cheap-necklace problem. In fact, the following link actually provides a video of the problem:
In the video, viewers are asked to find the cheapest possible circular chain. Wickelgren, on the other hand, tells us the final cost of the chain -- 15 cents -- and asks us how to achieve this cost.
"Stop reading and try to solve the problem by defining equivalence classes of action sequences based on the achievement of equivalent states."
There are several ways to solve this problem. The author tells us that the most obvious possibility to consider -- joining the four chains end-to-end -- requires four opens and four closes and so would end up costing 20 cents. The correct solution must not be equivalent to this one -- indeed, since we know that any link that's opened must eventually be closed, our final solution must have only three opens and three closes in order to make it 15 cents.
The solution is what Wicklegren calls destroying a chain. The three opens are done to the three links of a single chain. The opened links are then used to connect the other three chains, one open link between two adjoining chains. Finally, the three joining links are closed.
Now Wicklegren describes the concepts of microaction and macroaction:
"Now consider a sequence of actions that starts with some given state and achieves some other state. Call that a microaction sequence."
He then goes on to define a macroaction to be an equivalence class of microaction sequences. His example comes from algebra -- solve ax + b = cx + d. Using microactions:
"One could subtract b from both sides, then subtract cx from both sides, and then divide both sides by (a - c), but one could also add -cx to both sides, then multiply both sides by 1/(a - c), and then subtract b/(a - c) from both sides, and so on."
The single macroaction that solves the equation is just to write x = (d - b)/(a - c). A more famous example of an algebraic macroaction is the Quadratic Formula.
Wickelgren now returns to the construction problem from yesterday. It's natural to consider micro- and macroactions in constructions -- indeed, the U of Chicago text does exactly this. In Lesson 3-6, the U of Chicago text combines the microactions used to construct perpendicular lines into what the text calls a "subroutine" -- a macroaction used in other constructions.
Here's the problem from yesterday:
This is what I wrote last year about today's lesson:
Let me at least post the answers to the review worksheet.
1a. 16x
1b. 16xsqrt(2)
2a. 7sqrt(2)
2b. 7
2c. 7sqrt(3)
2d. 14
3. 12/13
4. 5/12
5. 0.406
6. 1
7. 1/2
8. 1
9-10. There is an error in the way I wrote this questions last year. Teachers may fix the error, and then the answers will vary.
11. AC/AB
12. BD, BC
13. DAB, DCA
14. Draw a vector the same direction as AB, but pointing in the opposite direction.
15. (-2, -9)
16. about 9 yards
17-20. In years past, I combined this with an activity, but this year I don't -- in fact, I won't post that activity until Monday. Remember that the digit pattern naturally leads to test day on Monday, so I changed it to Friday. Teachers can decide what to do about these questions.
There are several ways to solve this problem. The author tells us that the most obvious possibility to consider -- joining the four chains end-to-end -- requires four opens and four closes and so would end up costing 20 cents. The correct solution must not be equivalent to this one -- indeed, since we know that any link that's opened must eventually be closed, our final solution must have only three opens and three closes in order to make it 15 cents.
The solution is what Wicklegren calls destroying a chain. The three opens are done to the three links of a single chain. The opened links are then used to connect the other three chains, one open link between two adjoining chains. Finally, the three joining links are closed.
Now Wicklegren describes the concepts of microaction and macroaction:
"Now consider a sequence of actions that starts with some given state and achieves some other state. Call that a microaction sequence."
He then goes on to define a macroaction to be an equivalence class of microaction sequences. His example comes from algebra -- solve ax + b = cx + d. Using microactions:
"One could subtract b from both sides, then subtract cx from both sides, and then divide both sides by (a - c), but one could also add -cx to both sides, then multiply both sides by 1/(a - c), and then subtract b/(a - c) from both sides, and so on."
The single macroaction that solves the equation is just to write x = (d - b)/(a - c). A more famous example of an algebraic macroaction is the Quadratic Formula.
Wickelgren now returns to the construction problem from yesterday. It's natural to consider micro- and macroactions in constructions -- indeed, the U of Chicago text does exactly this. In Lesson 3-6, the U of Chicago text combines the microactions used to construct perpendicular lines into what the text calls a "subroutine" -- a macroaction used in other constructions.
Here's the problem from yesterday:
- Given an acute angle UVW and a point P within the angle, use a compass and straightedge to construct a segment
QRpassing through P, such that QP and PR stand in the ratio 2:1, Q and R lying onUVandVWrespectively.
The macroaction needed to solve this problem isn't a construction from the U of Chicago text, but it does appear in other Geometry texts. We must construct the parallel to UV through P. This line will intersect VW in some point M. Then we find R on VW such that M divides VR is in the goal ratio 2:1, which is a task that itself requires other macroactions (such as the U of Chicago's favorite perpendicular bisector). Then draw PR, which intersects UV at Q.
The reason that this works is the Side-Splitting Theorem -- in Triangle QRV, PM | | UV and so PM must divide both VR and QR in the same 2:1 ratio.
OK, now that we've reached the Geometry example I wish to end this discussion. So let me link to Wicklegren's next example to illustrate "getting out of loops" -- the nine-dot four-line problem:
He wraps up the chapter with one final problem-solving technique -- incubation, or in laymen's terms, sleeping on it. He's not sure why it seems to work, but:
"In any event, incubation often works, whatever the mechanism."
This is what I wrote last year about today's lesson:
Let me at least post the answers to the review worksheet.
1a. 16x
1b. 16xsqrt(2)
2a. 7sqrt(2)
2b. 7
2c. 7sqrt(3)
2d. 14
3. 12/13
4. 5/12
5. 0.406
6. 1
7. 1/2
8. 1
9-10. There is an error in the way I wrote this questions last year. Teachers may fix the error, and then the answers will vary.
11. AC/AB
12. BD, BC
13. DAB, DCA
14. Draw a vector the same direction as AB, but pointing in the opposite direction.
15. (-2, -9)
16. about 9 yards
17-20. In years past, I combined this with an activity, but this year I don't -- in fact, I won't post that activity until Monday. Remember that the digit pattern naturally leads to test day on Monday, so I changed it to Friday. Teachers can decide what to do about these questions.
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