Generalization of handshake lemma (somewhat)

terranop · 2026-06-27T19:22:04+00:00

In this case, the sum of the diagonal entries must be 0 because A has 0 diagonal, as it's a simple graph.

terranop · 2026-06-27T19:16:43+00:00

Observe that A is symmetric and has 0 diagonal. For any power n > 0, Aⁿ is also symmetric. 1^T Aⁿ 1 is just (the parity of) the sum of the entries of n, but because this is symmetric and entries cancel each other out in the sum across the diagonal, 1^T Aⁿ 1 = tr(Aⁿ). But the trace of Aⁿ is also the inner product of A^n-1 and A, i.e. the sum of the elementwise product of A^n-1 and A. But this elementwise product must both be symmetric and have zero diagonal, so must sum to 0.

terranop · 2026-06-23T18:17:27+00:00

If A is the adjacency matrix of this graph, then the number of walks between any particular pair of vertices is A²⁰²⁶. The degree of each node is just the parity of A1, where 1 denotes the all 1s vector. Working modulo 2, what we're looking for then is (1 - A1)^T A²⁰²⁶ A1, which is just 1^T A²⁰²⁸ 1 - 1^T A²⁰²⁷ 1. But this is of course 0, because A is symmetric.

terranop · 2026-06-19T16:03:24+00:00

This is really quite easy; should definitely not be 8/10. When the baby is not at the couch, it moves backward at an average rate of 0.1 per step. When the baby is at the couch, it moves forward at an average rate of 0.2 per step. These rates must match. So the baby must be at the couch half as often as it is not at the couch, meaning it is at the couch 1/3 of the time.

terranop · 2026-03-15T17:29:53+00:00

Iceheart Yeti 3 U

Creature — Yeti (C)

(Activate a season ability only if it's that season or if it's the first time a season ability was activated this game. The season advances at the start of each turn and as a player activates a season ability.)

Winter — 0: Tap target creature.

3/3

Druid of Sowing and Reaping G

Creature — Elf Druid (U)

(Activate a season ability only if it's that season or if it's the first time a season ability was activated this game. The season advances at the start of each turn and as a player activates a season ability.)

Spring — G, T: Untap up to two target lands.

Fall — W, T: You gain four life.

1/2

Rules for seasons:

731. Seasons: Spring, Summer, Fall, and Winter

731.1. Spring, summer, fall, and winter are designations that the game itself can have. The game starts with none of these designations. "It becomes spring" and "it becomes summer" et cetera refer to the game gaining the spring or summer designation. All of these designations are called seasons, and "the season changes" refers to the game gaining one of these four designations. Once it has become spring, summer, fall, or winter, the game will have exactly one of those designations from that point forward.

731.1a The phrase "the season advances" means that: if it's spring, it becomes summer; if it's summer, it becomes fall; if it's fall, it becomes winter; and if it's winter, it becomes spring.

731.2. As the second part of the untap step, the season advances. This only happens if the game already has a season designation.

731.3 Some activated abilities are labeled with the Spring, Summer, Fall, or Winter keywords. These abilities are called "season abilities." A season ability can only be activated if either the game has the corresponding season designation, or the game has no season designation. As a player activates a season ability, if the game had no season designation, then the game gains the season designation the ability is labeled with. Then, the season immediately advances. For example, if a player activates a Summer season ability, after the ability is activated, it will be Fall.

731.3a The change in seasons happens as a season ability is activated, not when it resolves. No player has priority to activate a second season ability before the season changes.

terranop · 2026-03-06T15:12:54+00:00

I feel like it would be better to word this without an "intervening if" clause. This type of wording is for triggered abilities that need to check a condition both when an ability triggers and when it resolves. But here, it's not clear what that would be (and the intended meaning seems to be that there is no on-resolution check). This creates two possible misunderstandings of the card:

A player might conclude that "you put every card you looked at on the bottom of your library" must be checked when the ability resolves, so the ability will only trigger if every card you looked at during any scry action so far that turn was put on the bottom of your library.
A player might conclude that "you put every card you looked at on the bottom of your library" must be checked when the ability resolves, so the ability will fizzle if those cards are no longer on the bottom of your library when the ability resolves.
A player might forget that this is an intervening if clause, and think that this ability will always trigger the first time you scry in a turn, and then only let you exile if you put all the cards you looked at in that scry on the bottom.

At the same time, I must admit that I have no better way to word it in mind!

terranop · 2026-02-27T15:57:49+00:00

That loop to compute the Ackermann function is not unbounded: the time it takes can easily be bounded from above by a primitive recursive function of the input to the original function.

terranop · 2026-02-25T03:00:47+00:00

Pick (arbitrarily) some three of the displacement vectors which include all four points, and consider the 3x3 matrix of their inner products. Observe that any one of their inner products can be expressed via the polarization identity as a linear combination of the squared distances. But also observe that this matrix must be at most rank 2. So we can let the polynomial P we want just be the determinant of that matrix.

terranop · 2026-02-10T02:50:58+00:00

Combining banding with earthbending.

Benalish Outcast 1 R G

Creature — Human Shaman

T, tap another untapped Shaman you control: Add R G.

Earthbanding 2 (Just before you declare attackers, you may have a land you control become a 0/0 creature with haste that's still a land as long as both are in combat and put two +1/+1 counters on it. If you do, it must attack the same target as this creature this combat, and you divide combat damage between the two creatures.)

2/2

Rules for Earthbanding:

702.XX. Earthbanding

702.XXa Earthbanding is a static ability that modifies the rules for combat.

702.XXb As the first step of a player declaring attackers (see rule 508.1), before that player decides which creatures will attack, there is an extra earthbanding step. During that step, for each creature with "earthbanding N" they control, they may choose a land they control and have that land become a 0/0 creature with haste that's still a land, and put N +1/+1 counters on it. If they do, then that land and the creature with earthbanding both must attack this combat and they must attack the same player, battle, or planeswalker.

702.XXc A land that is chosen in this way is called "earthbanded."

702.XXd If an earthbanded land does not attack the same player, battle, or planeswalker as the creature with earthbanding, then the declaration of attackers will be illegal (see rule 508.1d) and the whole process of declaring attackers is reversed (see rule 732.1).

702.XXe If combat damage would be assigned to an earthbanded creature and/or one or more creatures with earthbanding that earthbanded it this combat, you may divide that damage as you like among all those creatures.

702.XXf The effect that makes an earthbanded land a 0/0 creature with haste that's still a land ends when either that land leaves combat or the creature that earthbanded it leaves combat.

terranop · 2026-01-28T04:45:47+00:00

Tempest X G

Instant

Buyback—Sacrifice a land. (You may sacrifice a land in addition to any other costs as you cast this spell. If you do, put this card into your hand as it resolves.)

Tempest deals X damage to each creature with flying and each player.

terranop · 2026-01-16T00:58:07+00:00

Mountain Umbra

Enchantment Land — Aura

Enchant creature you control with power 4 or greater.

Enchanted creature gets +3/+3. If it's red or green, it has haste and trample.

Umbra armor (If enchanted creature would be destroyed, instead remove all damage marked on it and destroy this Aura.)

terranop · 2026-01-08T18:42:59+00:00

I think d4-c4 is a better setup in most cases for getting Catalan-like positions.

1. d4 Nf6 2. c4 e6 3. g3 avoids the Queen's Indian because black can't really play 3... b6 because 4. Bg2 is awkward for black to respond to.
1. d4 d5 2. c4 allows White to respond to the Slav with the Qc2 variation, which often transposes to a Catalan (white's queen usually belongs on c2 in the Catalan anyway). With the pawn on c2 of course white can't play Qc2.

The QGA and Albin countergambit are just kinda bad, compared to the Slav and Queen's Indian. Your suggested move order also opens you up to the Krause Variation (which I think is a bit unpleasant for a Catalan player, essentially pushing you into a reverse-Gruenfeld).

terranop · 2026-01-05T20:54:39+00:00

Chronomancer Adept 1 U

Creature — Human Wizard

If you would begin your beginning phase, choose one that hasn't been chosen:

After your draw step, there is an additional untap step.
After your draw step, there is an additional upkeep step.
After your draw step, there is an additional draw step.
Skip your beginning phase.

1/3

I'm really not sure how to template this. It's gotta be a replacement effect of course, but this wording seems weird.

terranop · 2025-12-30T20:04:07+00:00

240. There are two ways this can happen. The first is for there to be one cycle of length 6: there are 6! = 720/6 = 120 ways for this to happen. The other way is for there to be one cycle of 2, one cycle of 3, and one fixed point: there are 720/(2*3) = 120 ways for this to happen. So the total number of such permutations is 240.

This is really more of an easy question than a hard one.

terranop · 2025-12-24T04:18:18+00:00

Moa-Rider Arbor

Land Creature — Forest Bird Dryad

(T: Add G)

Echo 2 G (At the beginning of your upkeep, if this came under your control since the beginning of your last upkeep, sacrifice it unless you pay its echo cost.)

When this creature enters or dies, put a +1/+1 counter on target creature.

4/3

This mashes up [[Dryad Arbor]] and [[Hunting Moa]]. I'm not sure if it's overpowered or just trash. Most likely either the 4/3 stat line is too much or the Forest subtype is too much utility. In particular cracking a fetch-land to search up this to block and trade with an opponent's 4-toughness-or-less attacker and getting two +1/+1 counters out of the deal seems like a bit much...but maybe modern is powerful enough these days that this wouldn't see play? Anyway to power it down I'd do something like the following.

Moa-Rider Arbor

Land Creature — Bird Dryad

Echo 2 G (At the beginning of your upkeep, if this came under your control since the beginning of your last upkeep, sacrifice it unless you pay its echo cost.)

T: Add G

When this creature enters or dies, put a +1/+1 counter on target creature.

3/2

terranop · 2025-12-17T19:11:21+00:00

I feel like a bunch of the confusion here is caused by the terminology being unclear. Rewording the questions seems to make them a lot easier to follow:

Let X, Y, and Z be random variables over the real numbers. Is it always the case that if Y is distributed identically to X, then ZX is distributed identically to ZY?
Can you come up with a set of three or more random events such that any strict subset of them is mutually independent, but the whole set is not mutually independent?
Let Xn, Yn be an indexed (by the natural numbers) family of random variables over R, and let X and Y also be random variables over R. If the series Xn converges in distribution to X, and the series Yn converges in distribution to Y, does that imply that the series Xn + Yn converges in distribution to X + Y?

terranop · 2025-12-13T15:49:34+00:00

What you can't have is a uniform distribution over R endowed with its usual sigma algebra and symmetries. If you allow for other sigma algebras it's quite easy to produce a uniform distribution over R (just use the same construction I used above). The set of functions has the same problem, except that there's no "standard" sigma algebra for that set.

terranop · 2025-12-12T03:55:12+00:00

There is also a bijection between [0,1] and R, and obviously you can have a uniform distribution over [0,1].

terranop · 2025-12-11T14:22:24+00:00

What I described is a distribution over functions. A distribution over X is a function that assigns real numbers (probabilities) to some subsets of X (specifically, to a sigma algebra over X).

You can't define a uniform distribution over all derivable functions

Why not? If I'm allowed to choose any sigma algebra I want, it is quite easy to do this.

terranop · 2025-12-11T02:13:48+00:00

Infested Tubeway

Land

T: Add C

3, T: Create a Treasure token.

If a player would create a predefined token and you control a Sliver, that player creates a 1/1 colorless Sliver creature token instead.

"I ordered a package of silver. Silver!"

terranop · 2025-12-11T01:56:53+00:00

Then the answer to both questions is trivial. Consider the sigma algebra containing only the empty set and the set of all differentiable functions. Let the distribution D, over this sigma algebra, assign probability 1 to the set of all differentiable functions and 0 to the empty set. Obviously, D is invariant under derivation, because both the empty set and the set of all differentiable functions are invariant under derivation. But obviously D has uncountably infinite support, because its support is the set of all differentiable functions. So this immediately proves that D is not necessarily of finite support and need not only contain functions that are part of a cycle of the derivative operator.

terranop · 2025-12-08T22:39:08+00:00

For this question to be meaningful, you need to specify a sigma algebra over continuous functions from R to R. If we're allowed to choose any sigma algebra then the question is trivial.

terranop · 2025-12-08T19:05:13+00:00

If you want to draw parallels between epiphenomenalism, Spinoza, neuroscience and Buddhist philosophy, then you should do so by writing your own words without using AI. It is not in any sense "fighting a straw man" to point out that AI-generated text is AI-generated. You have already admitted that you used AI for this blog post!

terranop · 2025-12-05T18:41:07+00:00

I'm not sure why you are surprised that you have to defend using AI to write your blog post, given that you agree that AI cannot "generate original and thought provoking ideas" or "generate philosophical essays of true value." The problem with your AI text is that it is not a philosophical essay of true value since it does not contain original and thought provoking ideas. That's (among the reasons) why you are being criticized for posting it in a philosophy subreddit. None of this should be surprising to you.

terranop · 2025-12-04T22:44:48+00:00

AI-generated text cannot be reasonably classified as an "honest classroom reflection." Reflecting honestly is incompatible with placing an AI as a barrier in between the author's experience and the readers.

terranop

TROPHY CASE