As much as I have understood,for any query f(x)
, we need to take maximum of |f(x)-f(y)|
over all neighboring databases.
please explain how to find global sensitivity of queries like average height or maximum height.
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As much as I have understood,for any query f(x)
, we need to take maximum of |f(x)-f(y)|
over all neighboring databases.
please explain how to find global sensitivity of queries like average height or maximum height.
I’m 6 feet 3 and I wanted to know if being tall is a disadvantage when red teaming. Does it make it easier to get spotted?
I was searching for answers to the Question:
Show that there are at most $ \lceil n / 2^{h + 1} \rceil$ nodes of height $ h$ in any $ n$ -element heap.
Recently I asked a related question and found out the solution was flawed so I looked for another one.
So I took over to another answer and found this
It took over to prove by induction and it is quite understood on the first read except for the statement:
Note that the nodes at height $ h$ in $ T$ would be at height $ h − 1$ in tree $ T’$ .
Preface:
Let $ N_h$ be the number of nodes at height $ h$ in the n-node tree $ T$ . Consider the tree $ T’$ formed by removing the leaves of $ T$ .
Please help me clarify my doubts. Thank you.
The spell description for rope trick (PHB, pg. 272) states:
You touch a length of rope that is up to 60 feet long. One end of the rope then rises into the air until the whole rope hangs perpendicular to the ground.
What happens if I cast rope trick on a 60 foot rope in a 12 foot high room? The ceiling is not high enough for the whole rope to hang perpendicular to the ground; conceivably, it either stops rising when it hits the ceiling or it begins to pile up on the ceiling until 12 feet of it dangle to the floor. Either way, the rope does not satisfy the bolded condition in the spell description.
Does the spell fail? Do I need to cut down my rope to be less than the height of the room before casting the spell?
Given an arbitrary binary tree on $ n$ nodes, choose an assignment $ A$ from each parent to one of its children (the "favored child" as it were). We define the skew height of the tree as $ H_A(\mathsf{nil})=0$ and $ H_A(\mathsf{node}\;a\;b)=\max(H_A(a), H_A(b)+1)$ if $ A(\mathsf{node}\;a\;b)=a$ is the favored child and symmetrically $ \max(H_A(a)+1, H_A(b))$ if $ b$ is favored.
The question is: For a fixed tree $ T$ , what is the minimum skew height over all assignments? I would like to get an asymptotic bound on $ f(n)=\max_{|T|=n}\min_AH_A(T)$ .
Other variations on this problem I am interested in are when the trees are not binary (but there is still one favored child and all others add one to the height), and when there is sharing (i.e. it is a dag), which doesn’t affect the height computation but allows for much wider "trees" while staying under the $ n$ node bound.
The obvious bounds are $ f(n)=\Omega(\log n)$ and $ f(n)=O(n)$ . My guess is that $ f(n)=\Theta(\log n)$ for binary trees, and $ f(n)=\Theta(\sqrt n)$ for dags (with some kind of grid graph as counterexample).
I’m plotting a 3d function and looking at it from above. here is the function definition:
lagz[n_, m_, z_, zbar_] := 1/Sqrt[Pi*a^2*n!*m!]* E^(z*zbar/(2.*a^4)) * (a)^(m + n)* D[ E^(-(z*zbar)/a^4), {z, n}, {zbar, m}] /. {a -> 1} lagnlz[n_, l_, z_, zbar_] := Sqrt[n!/(Pi*a^2*(n + l)!)]* (z/a)^l LaguerreL[n, l, z*zbar/a^2]* E^(-z*zbar/(2.*a^2)) /. {a -> 1} lag[n_, l_, r_, \[Theta]_] := lagnlz[n, l, z, zbar] /. {z -> r*E^(I*\[Theta]), zbar -> r*E^(-I*\[Theta])} lagcc[n_, l_, r_, \[Theta]_] := lagnlz[n, l, z, zbar] /. {z -> r*E^(-I*\[Theta]), zbar -> r*E^(I*\[Theta])}
And here is the code I’m using to plot:
RevolutionPlot3D[ lag[1, 1, r, \[Theta]]*lagcc[1, 1, r, \[Theta]], {r, 0, 3}, {\[Theta], 0, 2 Pi}, ViewPoint -> Above]
From the “Above” view, it is not at all clear (at least to me) that the height of the inner ring is larger than the height of the outer ring. Is there some setting or coloring I can use that will make the height difference much more apparent from this “Above” view?
I am pretty new in the field of DP. My friend recently gave me a question from one of his old assignments to solve and practice and I am stuck at it.
The question says:
Given n blocks of same height and width but the different thickness and list B = (b_1, . . . , b_n) of the thickness of the block, define a DP algorithm to fit all the blocks in k boxes where each box can fit t thickness of blocks at most.
I was thinking of putting as many blocks in the first 2 boxes and the third one will be filled with the left blocks. But I have no idea how to break it down into the subproblems and create the base cases to define the time required. Any help will be appreciated. The given time complexity is O(nt^2).
Thanks!
If a caster has spent the duration of the Levitate spell moving upward to 2000′ is all that distance covered by the included feather fall component. A falling creature’s rate of descent slows to 60 feet per round until the spell ends and Levitate is a 2nd level spell. Also, how long would it take to float back down again?
I know that the height of a red-black tree is at most 2 lg(n + 1). However, what is the mathematical proof of this? I searched various sites, however I couldn’t find a good proof. I already know the properties of a red-black tree.
As a flying character I’m trying to do some aerial recon/scouting and wanted to know how a perception check works (passive and active) while flying.
Aside from lightly/heavily obscured (PHB p. 183), what other penalties/factors are there?
For example:
Does height play a factor? (See note 2)
Does speed play a factor? (See note 1. If yes, assume moving at base fly speed (50ft)).
I’m planning on doing daytime scouting for mobs (aka targets/enemy units), both hidden and wandering (seeing if the group’s path is clear or not), and anything out of the ordinary (campsites, lairs, caves, huts, anything that might require further investigation).
At night I plan to sky patrol (using ground level light spell, dancing lights spell). Also looking for other light sources (campfires, torches) that may be in the area/on approach.
If height makes a difference/you need a number, assume aerial scouting at just over 600 ft (say 625 ft). This is the magical number where long bows (all normal ranged weapons) cannot hit during the day.
For night patrols, assume just over 120ft (say 125ft), which is outside most night vision sight.
If possible please include any RAW that relates to this question.
Note 1: On PHB p. 182 there are penalties to passive perception checks for traveling at a “fast” pace (see table below). However this is for ground not flying (and I believe only passive checks not active checks).
Pace Distance Traveled per... Effect Minute Hour Day Fast 400 ft 4 miles 30 miles -5 penalty to passive Wisdom (Perception) Normal 300 ft 3 miles 24 miles — Slow 200 ft 2 miles 18 miles Able to use stealth
Note 2: My understanding (and would like input/confirmation of this) of RAW is that only lightly obscured (disadvantage on perception checks), heavily obscured (blocks vision), and possibly travel pace are the only factors in perception checks (both ground and air). Only LOS matters, thus the higher up I go the farther I can see/get perception (passive or active) checks on any items of interest (mobs, structures, etc.) not heavily obscured. Any limitation on this would be a DM house rule.