Am I misreading the effect-cancelling part of Warforged Juggernaut’s Crag of Steel power, or is it meant to be extremely limited?

Warforged Juggernaut‘s Level 20 Daily Power is Crag of Steel; a stance that gives you a bunch of stuff, but the part of the power I’m interested in here is:

“Until the stance ends, you gain resist 5 to all damage, and whenever this reduces an attack’s damage to 0, you also negate that attack’s effect on you.”


The way I’m reading this, in order for this to kick in: an attack must do damage, which must be less than 5, AND also deal an effect.

An attack that just deals an effect isn’t ignored because the stance doesn’t “reduce the attack’s damage to 0″ – it already was 0 (or rather, there was no damage/null/etc.)

I can’t envision any other way of triggering the text in question;

  • Resistance doesn’t stack.
  • If a different resistance is higher and negates the attack it doesn’t count because it’s “whenever THIS [Resist 5] reduces an attack”.
  • Temporary hit points are just me taking the damage somewhere else…
  • ???

At best, I can see it synergising with a handful of specific powers that otherwise reduce attack damage by mechanics other than resist, but those are few and far between (and generally reduce the damage by a not significant enough amount to negate 90% of a reasonable opponent’s damage.)(And my party hasn’t picked any of the dozen specific paragon paths/backgrounds that get them.)

Overall, this means, I can’t really see an occasion where this text would actually kick in. It’s good for negating level 1 creature’s attack effects if they happen to roll a 1 on their damage dice… as a level 20 Daily.

(Don’t get me wrong, the rest of the power is fine. Resist 5 is fine. Free damage is good. Resist forced movement is good. It’s just an unlikely enough confluence of events for that power to trigger that.. it feels like I’m missing something?)

TL;DR: Am I mis-reading this part of the Crag of Steel power or is it just not going to come up in normal level-appropriate fights?

How to generate a CSR (certificate signing request) for creating a limited CA (Certificate Authority) with LibreSSL?

Related to this (too broad) question: How to implement my PKI?

I have a self-signed CA (ca0)

I would like to create a CA (ca1) with limited power derived from that first CA. ca1 should only be able to sign certificates for *.foo.com and for foo.com.

From this question, I found out that the Name Constraints extension is probably what I want.

The key for ca1 is already created and is ca1.foo.key.pem.

I already have an incomplete command for creating the request:

libressl req -new -sha512 -key ca1.foo.key.pem -out ca1.foo.csr.pem 

What should I add to that line to limit ca1’s power to what I want?

Are NP proofs limited to polynomial length?


In computational complexity theory, NP (nondeterministic polynomial time) is a complexity class used to classify decision problems. NP is the set of decision problems for which the problem instances, where the answer is “yes”, have proofs verifiable in polynomial time by a deterministic Turing machine.

The proofs for an NP decision problem are verified in polynomial time.

Does this imply the proofs are at most polynomial length?

“Well you have to read the whole input. If the input is longer than polynomial, then the time is greater than polynomial.”

The decision problem “Is the first bit of the input a 0?” can be solved in constant time and space – without reading the whole input.

Therefore, perhaps some NP problem has candidate proofs that are longer than polynomial length but checked in polynomial time.

Does falling damage by weight get limited to 20d6?

I’m trying to build a character that does damage by dropping on people. (It’s a baleen whale with grafted wings.) I was reading the SRD on falling damage and it has this very confusing line about falling objects:

For each 200 pounds of an object’s weight, the object deals 1d6 points of damage, provided it falls at least 10 feet. Distance also comes into play, adding an additional 1d6 points of damage for every 10-foot increment it falls beyond the first (to a maximum of 20d6 points of damage).

Is the total falling damage limited to 20d6 or just the falling damage from height (which would make more sense since falling damage is limited because of terminal velocity)?

Min-coin change problem with limited coins

I have been assigned the min-coin change problem for homework. I have to calculate the least number of coins needed to make change for a certain amount of cents in 2 scenarios: we have an infinite supply of coins and also where we only have 1 of each coin. I was able to quickly implement the solution for the first scenario using a recursive formula from here but was not able to find a recursive formula for the second case and I instead came up with an iterative solution.

We can just sum every possible combination of coins, check and see if the sum is equal to the target wanted and compare the number of coins used to the current lowest number of coins. I also realized that I can use binary counting to simulate all possible combinations using the bits in the iterator as a coefficient (either 1 or 0) for each coin and then calculate the Hamming Weight to check how many coins was used in the sum.

To visualize an example with 3 coins:

C1 C2 C3 | S                        | W  0  0  0 | 0 * C1 + 0 * C2 + 0 * C3 | 0  0  0  1 | 0 * C1 + 0 * C2 + 1 * C3 | 1  0  1  0 | 0 * C1 + 1 * C2 + 0 * C3 | 1  0  1  1 | 0 * C1 + 1 * C2 + 1 * C3 | 2  1  0  0 | 1 * C1 + 0 * C2 + 0 * C3 | 1  1  0  1 | 1 * C1 + 0 * C2 + 1 * C3 | 2  1  1  0 | 1 * C1 + 1 * C2 + 0 * C3 | 2  1  1  1 | 1 * C1 + 1 * C2 + 1 * C3 | 3 

Obviously we can omit the case where all the coefficients are 0, but for completeness (when target = 0) and simplicity I’ve kept it in the solution.

My implementation in Python is:

import math   def least_coins(coins, target):     least = math.inf      # loop all 2^n possible sums of coins     for i in range(2 ** len(coins)):         sum = 0          # add all the coins multiplied by their coefficient in this iteration         for j in range(len(coins)):             # get the j-th bit in i             coefficient = i >> j & 1             sum += coefficient * coins[j]          if sum == target:             count = 0             # calculate the Hamming Weight of i to determine the number of coins used             while i:                 i &= i - 1                 count += 1             least = min(least, count)      return least   coins = [3, 1, 1, 3, 2, 4] target = 10 print(least_coins(coins, target)) >>> 3 # 3 + 3 + 4 = 10 

As you can see the coins are not necessarily unique, but can only be used once. How can I turn this into a recursive formula (if possible and for the reason of possibly being able to optimize it using dynamic programming) or even introduce coefficients other than 0 and 1 so the problem can be more general where we have a certain amount of each coin?

Can limited wish be used to cast spells from prohibited schools of magic?

I’m playing a silverbrow human sorcerer in D&D 3.5e and a prestige class I plan on taking in the future requires me to select a school of magic to prohibit. I’m having trouble deciding which school to choose, so I’m wondering if limited wish would allow me to duplicate the effect of a spell from the prohibited school.

D&D 5e: Can the Wish spell overcome limited magic immunity?

I always wondered how the two interacted. What if I cast Wish, copying the spell Contagion (5th level), at a rakshasa? The creature is immune to spells of 6th level or under, but does my spell count as 9th level because I used wish, or 5th level because the effect is that of contagion? Perhaps it counts as 9th for counterspell purposes, but 5th level for immunity purposes? If anyone has an answer, I would love to hear it.