Striiker
Former Flyers Fan
I hate The Wire.
NHL Mega-Mock Draft Reboot - Discussion / Draft Thread – DDU-DU DDU-DU PHASE TWENTY-TWO!
- Thread starter Captain Dave Poulin
- Start date
BiggE
SELL THE DAMN TEAM
The Wire was weird. Liked season 1 a lot, thought season 2 was crap, loved season 3, thought 4 was ok but season 5 was horrid.
These ads are pissing me off. I shouldn't be seeing these ads.
We start the day with @Strawberry Fields on the clock, @Hollywood Cannon on deck, myself on the lido deck, and Toe on the lido afterdeck
RIP @Asnito.
I just now remembered about the Academy Awards, and I saw that "Everything Everywhere All at Once" won for Best Picture. I haven't seen all the nominees, so I have no context established which I can use to get upset. It's just not at all the kind of thing that should be winning Best Picture. The plot is typically "Asian Fantasy" (for lack of a better term) in the sense that they make up enough of the backstory to carry the story forward, but not quite enough for it to make sense. The plot isn't the point of the film, of course - it's about family, and Chinese families in particular (and Chinese American families in even more particular), and especially the relationship between a husband and wife. It's about sacrifice, and love. In those aspects, it is great. But a "great" film should be great in every aspect. I have seen "The Banshees of Inisherin," and that is unquestionably a great film.
At the same time, I'm not upset that it won. I like the fact that they awarded creativity and uniqueness, though I think social issues really drove the decision (which is fine with me, since I share the same social values they were basing the decision on). There's also the fact that these awards lost any cachet they may have had when they were given to "Ordinary People" over "Raging Bull" in 1981 and "Forrest Gump" over "Pulp Fiction" in 1995. So they don't matter.
We start the day with @Strawberry Fields on the clock, @Hollywood Cannon on deck, myself on the lido deck, and Toe on the lido afterdeck
RIP @Asnito.
I just now remembered about the Academy Awards, and I saw that "Everything Everywhere All at Once" won for Best Picture. I haven't seen all the nominees, so I have no context established which I can use to get upset. It's just not at all the kind of thing that should be winning Best Picture. The plot is typically "Asian Fantasy" (for lack of a better term) in the sense that they make up enough of the backstory to carry the story forward, but not quite enough for it to make sense. The plot isn't the point of the film, of course - it's about family, and Chinese families in particular (and Chinese American families in even more particular), and especially the relationship between a husband and wife. It's about sacrifice, and love. In those aspects, it is great. But a "great" film should be great in every aspect. I have seen "The Banshees of Inisherin," and that is unquestionably a great film.
At the same time, I'm not upset that it won. I like the fact that they awarded creativity and uniqueness, though I think social issues really drove the decision (which is fine with me, since I share the same social values they were basing the decision on). There's also the fact that these awards lost any cachet they may have had when they were given to "Ordinary People" over "Raging Bull" in 1981 and "Forrest Gump" over "Pulp Fiction" in 1995. So they don't matter.
I'm almost positive it was the dildo fingers that pushed "Everything Everywhere All at Once" over the top.
Hollywood Cannon
I'm Away From My Desk
Are you still seeing them?
No. You rule.
Hollywood Cannon
I'm Away From My Desk
Asnito
Blood Rival to a Briere Simp
- Mar 2, 2017
- 6,965
- 15,604
Picks incoming. I was off the grid for a couple of days and just found out cuck was shitcanned
Beef Invictus
Revolutionary Positivity
Picks incoming. I was off the grid for a couple of days and just found out cuck was shitcanned
What a day this is for you
Asnito
Blood Rival to a Briere Simp
- Mar 2, 2017
- 6,965
- 15,604
I feel like I just won the lottery
Asnito
Blood Rival to a Briere Simp
- Mar 2, 2017
- 6,965
- 15,604
For our second video game we are going to select "Mutant League Football".
The game deviates from usual football simulations in several ways. Most notably it takes place in a post-apocalyptic world where radiation has caused the human race to mutate and the dead to rise from the grave. The game instruction manual states that the exact causes of the upheaval have been lost or corrupted, due to (among many things) the chaos of an alien invasion, spin control, a sloppy filing system set up by a temp, and what appears to be barbecue sauce.
en.wikipedia.org
The game deviates from usual football simulations in several ways. Most notably it takes place in a post-apocalyptic world where radiation has caused the human race to mutate and the dead to rise from the grave. The game instruction manual states that the exact causes of the upheaval have been lost or corrupted, due to (among many things) the chaos of an alien invasion, spin control, a sloppy filing system set up by a temp, and what appears to be barbecue sauce.
Mutant League Football - Wikipedia
Asnito
Blood Rival to a Briere Simp
- Mar 2, 2017
- 6,965
- 15,604
Our second college course will be a step up from our previous selection, Abstract Algebra.
Abstract algebra is a broad field of mathematics, concerned with algebraic structures such as groups, rings, vector spaces, and algebras.
On the 12-hour clock, 9+4=19+4=1, rather than 13 as in usual arithmetic
Roughly speaking, abstract algebra is the study of what happens when certain properties of number systems are abstracted out; for instance, altering the definitions of the basic arithmetic operations result in a structure known as a ring, so long as the operations are consistent.
For example, the 12-hour clock is an example of such an object, where the arithmetic operations are redefined to use modular arithmetic (with modulus 12). An even further level of abstraction--where only one operation is considered--allows the clock to be understood as a group. In either case, the abstraction is useful because many properties can be understood without needing to consider the specific structure at hand, which is especially important when considering the relationship(s) between structures; the concept of a group isomorphism is an example.
brilliant.org
Abstract algebra is a broad field of mathematics, concerned with algebraic structures such as groups, rings, vector spaces, and algebras.
Roughly speaking, abstract algebra is the study of what happens when certain properties of number systems are abstracted out; for instance, altering the definitions of the basic arithmetic operations result in a structure known as a ring, so long as the operations are consistent.
For example, the 12-hour clock is an example of such an object, where the arithmetic operations are redefined to use modular arithmetic (with modulus 12). An even further level of abstraction--where only one operation is considered--allows the clock to be understood as a group. In either case, the abstraction is useful because many properties can be understood without needing to consider the specific structure at hand, which is especially important when considering the relationship(s) between structures; the concept of a group isomorphism is an example.
Abstract Algebra | Brilliant Math & Science Wiki
Abstract algebra is a broad field of mathematics, concerned with algebraic structures such as groups, rings, vector spaces, and algebras. Roughly speaking, abstract algebra is the study of what happens when certain properties of number systems are abstracted out; for instance, altering the...
brilliant.org
PROBLEMS ON ABSTRACT ALGEBRA 1 (Putnam 1972 A2).
Let S be a set and let ∗ be a binary operation on S satisfying the laws x ∗ (x ∗ y) = y for all x, y in S, (y ∗ x) ∗ x = y for all x, y in S.
Show that ∗ is commutative but not necessarily associative.
2 (Putnam 1972 B3). Let A and B be two elements in a group such that ABA = BA2B, A3 = 1 and B2n−1 = 1 for some positive integer n. Prove B = 1.
3 (Putnam 2007 A5). Suppose that a finite group has exactly n elements of order p, where p is a prime. Prove that either n = 0 or p divides n + 1.
4 (Putnam 2011 A6). Let G be an abelian group with n elements, and let {g1 = e, g2, . . . , gk} ⊊ G be a (not necessarily minimal) set of distinct generators of G. A special die, which randomly selects one of the elements g1, g2, . . . , gk with equal probability, is rolled m times and the selected elements are multiplied to produce an element g ∈ G. Prove that there exists a real number b ∈ (0, 1) such that �� �2 1 1 lim Prob(g = x) − m→∞ b2m n x∈G is positive and finite.
5 (Putnam 1990 B4). Let G be a finite group of order n generated by a and b. Prove or disprove: there is a sequence g1, g2, g3, . . . , g2n such that (a) every element of G occurs exactly twice, and (b) gi+1 equals gia or gib for i = 1, 2, . . . , 2n. (Interpret g2n+1 as g1.)
6 (Putnam 2016 A5). Suppose that G is a finite group generated by the two elements g and h, where the order of g is odd. Show that every element of G can be written in the form m1 hn1 m2 hn2 mr hnr g g · · · g with 1 ≤ r ≤ |G| and mn, n1, m2, n2, . . . , mr, nr ∈ {1, −1}. (Here |G| is the number of elements of G.)
7 (Putnam 1977 B6). Let H be a subgroup with h elements in a group G. Suppose that G has an element a such that for all x in H, (xa)3 = 1, the identity. In G, let P be the subset of all products x1ax2a · · · xna, with n a positive integer and the xi’s in H. (a) Show that P is a finite set. (b) Show that, in fact, P has no more than 3h2 elements.
8 (Putnam 1984 B3). Prove or disprove the following statement: If F is a finite set with two or more elements, then there exists a binary operation ∗ on F such that for all x, y, z in F, 1 (i) x ∗ z = y ∗ z implies x = y (right cancellation holds), and (ii) x ∗ (y ∗ z) =∕ (x ∗ y) ∗ z (no case of associativity holds).
9 (Putnam 1987 B6). Let F be the field of p2 elements where p is an odd prime. Suppose S is a set of (p2 − 1)/2 distinct nonzero elements of F with the property that for each a =∕ 0 in F, exactly one of a and −a is in S. Let N be the number of elements in the intersection S ∩ {2a : a ∈ S}. Prove that N is even.
10 (Putnam 1989 B2). Let S be a nonempty set with an associative operation that is left and right cancellative (xy = xz implies y = z, and yx = zx implies y = z). Assume that for every a in S the n set {a : n = 1, 2, 3, . . . } is finite. Must S be a group?
11 (Putnam 1992 B6). Let M be a set of real n × n matrices such that (i) I ∈ M, where I is the n × n identity matrix; (ii) if A ∈ M and B ∈ M, then either AB ∈ M or −AB ∈ M, but not both; (iii) if A ∈ M and B ∈ M, then either AB = BA or AB = −BA; (iv) if A ∈ M and A ∈/ I, there is at least one B ∈ M such that AB = −BA. Prove that M contains at most n2 matrices.
12 (Putnam 1996 A4). Let S be a set of ordered triples (a, b, c) of distinct elements of a finite set A. Suppose that (1) (a, b, c) ∈ S if and only if (b, c, a) ∈ S; (2) (a, b, c) ∈ S if and only if (c, b, a) ∈/ S [for a, b, c distinct]; (3) (a, b, c) and (c, d, a) are both in S if and only if (b, c, d) and (d, a, b) are both in S. Prove that there exists a one-to-one function g from A to R such that g(a) < g(b) < g(c) implies (a, b, c) ∈ S.
13 (Putnam 2008 A6). Prove that there exists a constant c > 0 such that in every nontrivial finite group G there exists a sequence of length at most c ln |G| with the property that each element of G equals the product of some subsequence. (The elements of G in the sequence are not required to be distinct. A subsequence of a sequence is obtained by selecting some of the terms, not necessarily consecutive, without reordering them; for example, 4, 4, 2 is a subsequence of 2, 4, 6, 4, 2, but 2, 2, 4 is not.)
14 (Putnam 2009 A5). Is there a finite abelian group G such that the product of the orders of all its elements is 22009?
15 (Putname 2010 A5). Let G be a group, with operation ∗. Suppose that 1. G is a subset of R3 (but ∗ need not be related to addition of vectors); 2. For each a, b ∈ G, either a × b = a ∗ b or a × b = 0 (or both), where × is the usual cross product in R3. Prove that a × b = 0 for all a, b ∈ G. 2
16. Let R be a noncommutative ring with identity. Suppose that x, y are elements of R such that 1 − xy and 1 − yx are invertible. (By the previous problem it suffice to assume that only 1 − xy is invertible, but this is irrelevant.) Show that (1 + x)(1 − yx) −1(1 + y) = (1 + y)(1 − xy) −1(1 + x). (1) This problem illustrates that “noncommutative high school algebra” is a lot harder than ordinary (commutative) high school algebra. Note. Formally we have (1 − yx) −1 = 1 + yx + yxyx + yxyxyx + · · · and similarly for (1−xy)−1. Thus both sides of (1) are formally equal to the sum of all “alternating words” (products of x’s and y’s with no two x’s or y’s appearing consecutively). This makes the identity (1) plausible, but our formal argument is not a proof.
17. Let G be a group of order 4n + 2, n ≥ 1. Prove that G is not a simple group, i.e., G has a proper normal subgroup.
18. Let R satisfy all the axioms of a ring except commutativity of addition. Show that ax + by = by + ax for all a, b, x, y ∈ R.
19. Let G denote the set of all infinite sequences (a1, a2, . . .) of integers ai. We can add elements of G coordinate-wise, i.e., (a1, a2, . . .) + (b1, b2, . . .) = (a1 + b1, a2 + b2, . . .). Let Z denote the set of integers. Suppose f : → Z is a function satisfying f(x + y) = f(x) + f(y) for all x, y ∈ G. Let ei be the element of G with a 1 in position i and 0’s elsewhere. (a) Suppose that f(ei) = 0 for all i. Show that f(x) = 0 for all x ∈ G. (b) Show that f(ei) = 0 for all but finitely many i.
20. Let G be a finite group, and set f(G) = #{(u, v) ∈ G × G : uv = vu}. Find a formula for f(G) in terms of the order of G and the number k(G) of conjugacy classes of G. (Two elements x, y ∈ G are conjugate if y = axa−1 for some a ∈ G. Conjugacy is an equivalence relation whose equivalence classes are called conjugacy classes.)
21 (difficult). Let n be an odd positive integer. Show that the number of ways to write the identity permutation ι of 1, 2, . . . , n as a product uvw = ι of three n-cycles is 2(n − 1)!2/(n + 1).
22. Let G be any finite group, and let w ∈ G. Find the number of pairs (u, v) ∈ G × G satisfying 2 w = uvu vuv.
23. Show that the number of ways to write the cycle (1, 2, . . . , n) as a product of n−1 transpositions is nn−2. For instance, when n = 3 we have (multiplying permutations left-to-right) three ways: (1, 2, 3) = (1, 3)(2, 3) = (1, 2)(1, 3) = (2, 3)(1, 2).
24 (difficult). Let si = (i,i + 1) ∈ Sn, i.e., si is the permutation of 1, 2, . . . , n that transposes i and i+1 and fixes all other j. Let f(n) be the number of ways to write the permutation n, n−1, . . . , 1 in � �n the form si1 si2 · · · sip , where p = . For instance, 321 = s1s2s1 = s2s1s2, so f(3) = 2. Moreover, 2 f(4) = 16. Show that f(n) is the number of sequences a1, . . . , ap of n − 1 1’s, n − 2 2’s, . . . , one n − 1, such that in any prefix a1, a2, . . . , ak, the number of i + 1’s does not exceed the number of i’s. For instance, when n = 3 there are the two sequences 112 and 121. 3 Note. An explicit formula is known for f(n), but this is irrelevant here.
25 (difficult). In the notation of the previous problem, show that � i1i2 · · ·ip = p!, i1,i2,...,ip where the sum is over all sequences i1, . . . , ip for which n, n − 1, . . . , 1 = si1 si2 · · · sip . For instance, when n = 3 we get 1 · 2 · 1 + 2 · 1 · 2 = 3!. Note. The only known proofs are algebraic. It would be interesting to give a combinatorial proof.
Let S be a set and let ∗ be a binary operation on S satisfying the laws x ∗ (x ∗ y) = y for all x, y in S, (y ∗ x) ∗ x = y for all x, y in S.
Show that ∗ is commutative but not necessarily associative.
2 (Putnam 1972 B3). Let A and B be two elements in a group such that ABA = BA2B, A3 = 1 and B2n−1 = 1 for some positive integer n. Prove B = 1.
3 (Putnam 2007 A5). Suppose that a finite group has exactly n elements of order p, where p is a prime. Prove that either n = 0 or p divides n + 1.
4 (Putnam 2011 A6). Let G be an abelian group with n elements, and let {g1 = e, g2, . . . , gk} ⊊ G be a (not necessarily minimal) set of distinct generators of G. A special die, which randomly selects one of the elements g1, g2, . . . , gk with equal probability, is rolled m times and the selected elements are multiplied to produce an element g ∈ G. Prove that there exists a real number b ∈ (0, 1) such that �� �2 1 1 lim Prob(g = x) − m→∞ b2m n x∈G is positive and finite.
5 (Putnam 1990 B4). Let G be a finite group of order n generated by a and b. Prove or disprove: there is a sequence g1, g2, g3, . . . , g2n such that (a) every element of G occurs exactly twice, and (b) gi+1 equals gia or gib for i = 1, 2, . . . , 2n. (Interpret g2n+1 as g1.)
6 (Putnam 2016 A5). Suppose that G is a finite group generated by the two elements g and h, where the order of g is odd. Show that every element of G can be written in the form m1 hn1 m2 hn2 mr hnr g g · · · g with 1 ≤ r ≤ |G| and mn, n1, m2, n2, . . . , mr, nr ∈ {1, −1}. (Here |G| is the number of elements of G.)
7 (Putnam 1977 B6). Let H be a subgroup with h elements in a group G. Suppose that G has an element a such that for all x in H, (xa)3 = 1, the identity. In G, let P be the subset of all products x1ax2a · · · xna, with n a positive integer and the xi’s in H. (a) Show that P is a finite set. (b) Show that, in fact, P has no more than 3h2 elements.
8 (Putnam 1984 B3). Prove or disprove the following statement: If F is a finite set with two or more elements, then there exists a binary operation ∗ on F such that for all x, y, z in F, 1 (i) x ∗ z = y ∗ z implies x = y (right cancellation holds), and (ii) x ∗ (y ∗ z) =∕ (x ∗ y) ∗ z (no case of associativity holds).
9 (Putnam 1987 B6). Let F be the field of p2 elements where p is an odd prime. Suppose S is a set of (p2 − 1)/2 distinct nonzero elements of F with the property that for each a =∕ 0 in F, exactly one of a and −a is in S. Let N be the number of elements in the intersection S ∩ {2a : a ∈ S}. Prove that N is even.
10 (Putnam 1989 B2). Let S be a nonempty set with an associative operation that is left and right cancellative (xy = xz implies y = z, and yx = zx implies y = z). Assume that for every a in S the n set {a : n = 1, 2, 3, . . . } is finite. Must S be a group?
11 (Putnam 1992 B6). Let M be a set of real n × n matrices such that (i) I ∈ M, where I is the n × n identity matrix; (ii) if A ∈ M and B ∈ M, then either AB ∈ M or −AB ∈ M, but not both; (iii) if A ∈ M and B ∈ M, then either AB = BA or AB = −BA; (iv) if A ∈ M and A ∈/ I, there is at least one B ∈ M such that AB = −BA. Prove that M contains at most n2 matrices.
12 (Putnam 1996 A4). Let S be a set of ordered triples (a, b, c) of distinct elements of a finite set A. Suppose that (1) (a, b, c) ∈ S if and only if (b, c, a) ∈ S; (2) (a, b, c) ∈ S if and only if (c, b, a) ∈/ S [for a, b, c distinct]; (3) (a, b, c) and (c, d, a) are both in S if and only if (b, c, d) and (d, a, b) are both in S. Prove that there exists a one-to-one function g from A to R such that g(a) < g(b) < g(c) implies (a, b, c) ∈ S.
13 (Putnam 2008 A6). Prove that there exists a constant c > 0 such that in every nontrivial finite group G there exists a sequence of length at most c ln |G| with the property that each element of G equals the product of some subsequence. (The elements of G in the sequence are not required to be distinct. A subsequence of a sequence is obtained by selecting some of the terms, not necessarily consecutive, without reordering them; for example, 4, 4, 2 is a subsequence of 2, 4, 6, 4, 2, but 2, 2, 4 is not.)
14 (Putnam 2009 A5). Is there a finite abelian group G such that the product of the orders of all its elements is 22009?
15 (Putname 2010 A5). Let G be a group, with operation ∗. Suppose that 1. G is a subset of R3 (but ∗ need not be related to addition of vectors); 2. For each a, b ∈ G, either a × b = a ∗ b or a × b = 0 (or both), where × is the usual cross product in R3. Prove that a × b = 0 for all a, b ∈ G. 2
16. Let R be a noncommutative ring with identity. Suppose that x, y are elements of R such that 1 − xy and 1 − yx are invertible. (By the previous problem it suffice to assume that only 1 − xy is invertible, but this is irrelevant.) Show that (1 + x)(1 − yx) −1(1 + y) = (1 + y)(1 − xy) −1(1 + x). (1) This problem illustrates that “noncommutative high school algebra” is a lot harder than ordinary (commutative) high school algebra. Note. Formally we have (1 − yx) −1 = 1 + yx + yxyx + yxyxyx + · · · and similarly for (1−xy)−1. Thus both sides of (1) are formally equal to the sum of all “alternating words” (products of x’s and y’s with no two x’s or y’s appearing consecutively). This makes the identity (1) plausible, but our formal argument is not a proof.
17. Let G be a group of order 4n + 2, n ≥ 1. Prove that G is not a simple group, i.e., G has a proper normal subgroup.
18. Let R satisfy all the axioms of a ring except commutativity of addition. Show that ax + by = by + ax for all a, b, x, y ∈ R.
19. Let G denote the set of all infinite sequences (a1, a2, . . .) of integers ai. We can add elements of G coordinate-wise, i.e., (a1, a2, . . .) + (b1, b2, . . .) = (a1 + b1, a2 + b2, . . .). Let Z denote the set of integers. Suppose f : → Z is a function satisfying f(x + y) = f(x) + f(y) for all x, y ∈ G. Let ei be the element of G with a 1 in position i and 0’s elsewhere. (a) Suppose that f(ei) = 0 for all i. Show that f(x) = 0 for all x ∈ G. (b) Show that f(ei) = 0 for all but finitely many i.
20. Let G be a finite group, and set f(G) = #{(u, v) ∈ G × G : uv = vu}. Find a formula for f(G) in terms of the order of G and the number k(G) of conjugacy classes of G. (Two elements x, y ∈ G are conjugate if y = axa−1 for some a ∈ G. Conjugacy is an equivalence relation whose equivalence classes are called conjugacy classes.)
21 (difficult). Let n be an odd positive integer. Show that the number of ways to write the identity permutation ι of 1, 2, . . . , n as a product uvw = ι of three n-cycles is 2(n − 1)!2/(n + 1).
22. Let G be any finite group, and let w ∈ G. Find the number of pairs (u, v) ∈ G × G satisfying 2 w = uvu vuv.
23. Show that the number of ways to write the cycle (1, 2, . . . , n) as a product of n−1 transpositions is nn−2. For instance, when n = 3 we have (multiplying permutations left-to-right) three ways: (1, 2, 3) = (1, 3)(2, 3) = (1, 2)(1, 3) = (2, 3)(1, 2).
24 (difficult). Let si = (i,i + 1) ∈ Sn, i.e., si is the permutation of 1, 2, . . . , n that transposes i and i+1 and fixes all other j. Let f(n) be the number of ways to write the permutation n, n−1, . . . , 1 in � �n the form si1 si2 · · · sip , where p = . For instance, 321 = s1s2s1 = s2s1s2, so f(3) = 2. Moreover, 2 f(4) = 16. Show that f(n) is the number of sequences a1, . . . , ap of n − 1 1’s, n − 2 2’s, . . . , one n − 1, such that in any prefix a1, a2, . . . , ak, the number of i + 1’s does not exceed the number of i’s. For instance, when n = 3 there are the two sequences 112 and 121. 3 Note. An explicit formula is known for f(n), but this is irrelevant here.
25 (difficult). In the notation of the previous problem, show that � i1i2 · · ·ip = p!, i1,i2,...,ip where the sum is over all sequences i1, . . . , ip for which n, n − 1, . . . , 1 = si1 si2 · · · sip . For instance, when n = 3 we get 1 · 2 · 1 + 2 · 1 · 2 = 3!. Note. The only known proofs are algebraic. It would be interesting to give a combinatorial proof.
- Sep 28, 2014
- 78,072
- 125,470
Hot dog fingers* but there was also a dildo nunchuk scene lol
I know they were hot dogs, but.... c'mon.
Strawberry Fields
12x Calder Cup Champs
College Course II
Politics in the Twilight Zone
This was my first year seminar class in college 8 years ago now. I can't find a syllabus or online evidence it was offered, but I assure you it existed and I took it.
Basically we watched an episode of the show every week or so, then had supplemental historical readings our prof deemed relevant to the episode. Then the rest of the week was dedicated to discussing the episode and readings and tying them all together.
Didn't know it going in but apparently the professor who taught it had a rep of being one of the toughest graders at the school. Well, he was. He spent about five minutes before handing back our first paper telling us how awful they were (first year, first semester students, mind you). I got mine back and unclenched when I saw the A- grade on it. He even came up to me after and told me it was one of the best in the class! Thanks for the heart attack, doc. I wound up getting A-minuses on two of the three papers, but he was a tough grader and pretty unapologetic about it.
Politics in the Twilight Zone
This was my first year seminar class in college 8 years ago now. I can't find a syllabus or online evidence it was offered, but I assure you it existed and I took it.
Basically we watched an episode of the show every week or so, then had supplemental historical readings our prof deemed relevant to the episode. Then the rest of the week was dedicated to discussing the episode and readings and tying them all together.
Didn't know it going in but apparently the professor who taught it had a rep of being one of the toughest graders at the school. Well, he was. He spent about five minutes before handing back our first paper telling us how awful they were (first year, first semester students, mind you). I got mine back and unclenched when I saw the A- grade on it. He even came up to me after and told me it was one of the best in the class! Thanks for the heart attack, doc. I wound up getting A-minuses on two of the three papers, but he was a tough grader and pretty unapologetic about it.
Lord Defect
Secretary of Blowtorching
- Nov 13, 2013
- 18,802
- 34,852
Loved this game. Great pick.For our second video game we are going to select "Mutant League Football".
View attachment 668300
The game deviates from usual football simulations in several ways. Most notably it takes place in a post-apocalyptic world where radiation has caused the human race to mutate and the dead to rise from the grave. The game instruction manual states that the exact causes of the upheaval have been lost or corrupted, due to (among many things) the chaos of an alien invasion, spin control, a sloppy filing system set up by a temp, and what appears to be barbecue sauce.
Mutant League Football - Wikipedia
mja
Everything was beautiful, and nothing hurt
These ads are pissing me off. I shouldn't be seeing these ads.
We start the day with @Strawberry Fields on the clock, @Hollywood Cannon on deck, myself on the lido deck, and Toe on the lido afterdeck
RIP @Asnito.
I just now remembered about the Academy Awards, and I saw that "Everything Everywhere All at Once" won for Best Picture. I haven't seen all the nominees, so I have no context established which I can use to get upset. It's just not at all the kind of thing that should be winning Best Picture. The plot is typically "Asian Fantasy" (for lack of a better term) in the sense that they make up enough of the backstory to carry the story forward, but not quite enough for it to make sense. The plot isn't the point of the film, of course - it's about family, and Chinese families in particular (and Chinese American families in even more particular), and especially the relationship between a husband and wife. It's about sacrifice, and love. In those aspects, it is great. But a "great" film should be great in every aspect. I have seen "The Banshees of Inisherin," and that is unquestionably a great film.
At the same time, I'm not upset that it won. I like the fact that they awarded creativity and uniqueness, though I think social issues really drove the decision (which is fine with me, since I share the same social values they were basing the decision on). There's also the fact that these awards lost any cachet they may have had when they were given to "Ordinary People" over "Raging Bull" in 1981 and "Forrest Gump" over "Pulp Fiction" in 1995. So they don't matter.
I feel like the multiverse conceit, the action sequences, and the maximalist humor are really entertaining on the initial watch but don’t hold up as well on repeat viewings. Once the element of surprise is gone, it all becomes a bit much. The family drama emotional core of the story does hold up though. I absolutely adore that aspect of the film, and Waymond in particular, but it’s a relatively small part of the running time even if it's what the film is really about.
Banshees was phenomenal. A film you can watch again and pick apart and see a different way each time. I saw 3 of the others as well but I don’t even think they should have been nominated. Was I the only one annoyed at All Quiet on the Western Front’s heavy-handed (and needlessly changed) ending? I’m ok with EEAAO taking home the prize, if only to encourage more inventiveness out of Hollywood. I just hope that they don't take the wrong lesson and start making nothing but whacked-out multiverse action movies, which already felt like it was happening anyway.
Hollywood Cannon
I'm Away From My Desk
Oops.
Will make pick in a little. TV decided to throw up again.
So if you wanna go you can.
Will make pick in a little. TV decided to throw up again.
So if you wanna go you can.
This is totally out of the blue, but every time I see that LSU helmet I like it a little more. Same goes for that white Bengals helmet.
We start the day with my sibling @Hollywood Cannon on the clock, myself on deck, my foot friend @BigToe on the lido deck, and my pharmacist @DancingPanther on the lido afterdeck.
You know how the rocketest of the rockets on that show held my hand in that dream the other night? Last night, on the show, they started their latest trip, and she succumbed to the charms of a douche. Earlier she had been sort of coupled off with the one comedian who is WAY shorter than her (she is very tall) and nerdy and kind of shy, at least when it comes to romance, and she was receptive to him. Even though that description of him doesn't really apply to me in any particular, he makes for a decent cipher to stand in for me and serve as a proxy through which I can live my romance with her vicariously. But last night the douche put on the full-court press. There's always one douche on these shows, and this is a megadouche. He's a total player, doing all the things a player does to make a lady fall - small touches, wiping something off her lips with a napkin (usually when there is nothing on her lips), giving her his coat, etc. Every single one of these moves is orchestrated and telegraphed, but of course the lady never sees it in the moment. It's infuriating. All the rest of the guys (except for the one, who is incapable of being himself and not acting up for the show) are perfectly fine and decent. It's just this one turd in the punchbowl.
I'll tell you something else - they have six guys and four girls. This imbalance is another thing these shows do, and I have no idea why. First-world problems, I guess, even though I am in the middle of Buttf***, Egypt.
We start the day with my sibling @Hollywood Cannon on the clock, myself on deck, my foot friend @BigToe on the lido deck, and my pharmacist @DancingPanther on the lido afterdeck.
You know how the rocketest of the rockets on that show held my hand in that dream the other night? Last night, on the show, they started their latest trip, and she succumbed to the charms of a douche. Earlier she had been sort of coupled off with the one comedian who is WAY shorter than her (she is very tall) and nerdy and kind of shy, at least when it comes to romance, and she was receptive to him. Even though that description of him doesn't really apply to me in any particular, he makes for a decent cipher to stand in for me and serve as a proxy through which I can live my romance with her vicariously. But last night the douche put on the full-court press. There's always one douche on these shows, and this is a megadouche. He's a total player, doing all the things a player does to make a lady fall - small touches, wiping something off her lips with a napkin (usually when there is nothing on her lips), giving her his coat, etc. Every single one of these moves is orchestrated and telegraphed, but of course the lady never sees it in the moment. It's infuriating. All the rest of the guys (except for the one, who is incapable of being himself and not acting up for the show) are perfectly fine and decent. It's just this one turd in the punchbowl.
I'll tell you something else - they have six guys and four girls. This imbalance is another thing these shows do, and I have no idea why. First-world problems, I guess, even though I am in the middle of Buttf***, Egypt.
Hollywood Cannon
I'm Away From My Desk
TV has not improved. Please skip.
BiggE
SELL THE DAMN TEAM
Hope the little dude feels better soon. Did you tell him that Cuck was fired? That should at least cheer him up.
Mine is pretty self-explanatory.
Team College Course II - Japanese Science Fiction and Planetary Possible Futures
@BigToe
Team College Course II - Japanese Science Fiction and Planetary Possible Futures
@BigToe
Lord Defect
Secretary of Blowtorching
- Nov 13, 2013
- 18,802
- 34,852
A yo-yo or even space yo-yo seems like a terrible choice for a weaponMine is pretty self-explanatory.
Team College Course II - Japanese Science Fiction and Planetary Possible Futures
@BigToe
Beef Invictus
Revolutionary Positivity
What am I beating up
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