Please help me to solve this problem efficiently and calculate its time complexity.
We have an NxM grid there are two animals one is ‘Horse’ and another is ‘Bishop’ which has different moving abilities. A ‘Horse’ can move 2.5 steps and a ‘Bishop’ can move only diagonal but not horizontally or vertically. Some grids are marked as inactive. Return 1 position where these animals can meet at any point. In the following image 0,3 & 2,0 are represented as an inactive grid where none of the animals can visit and ‘Bishop’ is at 3,2 position and possible direction are represented as arrow and ‘Horse’ is available at 6,6 and possible direction where horse can move are represented as arrow and circles. One of the possible point where the Bishop and Horse can meet is 4,5 as represented by yellow color. You need to use your own data structure, data types and test cases asper the requirement. (Hint: use graph algorithm - BFS). PS: Chess knowledge is not required to solve this problem *
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I'm currently working towards a 3D model of this, but I thought I would start with 2D. Basically, I have a grid of longitude and latitude with NO2 concentrations across it. What I want to produce, at least for now, is a total amount of Nitrogen Dioxide between two points. Like so:
2DGrid
Basically, These two points are at different lats and lons and as I stated I want to find the amount of something between them. The tricky thing to me is that the model data I'm working with is gridded so I need to be able to account for the amount of something along a line at the lat and lons at which that line cuts through said grid.
Another approach, and maybe a better one for my purposes, could be visualized like this:3DGrid
Ultimately, I'd like to be able to create a program (within any language honestly) that could find the amount of "something" between two points in a 3D grid. If you would like specfics, the bottom altitude is the surface, the top grid is the top of the atmosphere. The bottom point is a measurement device looking at the sun during a certain time of day (and therefore having a certain zenith and azimuth angle). I want to find the NO2 between that measurement device and the "top of the atmosphere" which in my grid is just the top altitude level (of which there are 25).
I'm rather new to coding, stack exchange, and even the subject matter I'm working with so the sparse code I've made might end up creating more clutter than purely asking the question and seeing what methods/code you might suggest?
Hopefully my question is beneficial!
Best,
Taylor
To traverse all touched cells, you can use Amanatides-Woo algorithm. It is suitable both for 2D and for 3D case.
Implementation clues
To account for quantity used from every cell, you can apply some model. For example, calculate path length inside cell (as difference of enter and exit coordinates) and divide by normalizing factor to get cell weight (for example, byCellSize*Sqrt(3) for 3D case as diagonal length).
By direction I mean for example from a patient's head to bottom or from his bottom to head. The CHEST CT scans I have seen so far indicates that Instance Number 1 slice is usually the first one down from the upper part of the body but I don't know whether this is part of the standard or there are some other tags that I should inspect into to determine.
There is no rule in DICOM that requires the Instance Number to be related to the slice position in a particular way. The link of Bartloimiej shows that there is a rule how the slice coordinates defined by Image Position Patient (0020,0027) and Image Orientation Patient (0020, 0037) are related to directions in the patient's body (head, feet, etc.)
So if you want to apply spatial ordering, these attributes are what you want to use. Slice Location (0020,1041) will not help you as well:
C.7.6.2.1.2 [...] This information is relative to an unspecified implementation specific reference point.
For original (i.e. Image Type (0008,0008) is ORIGINAL\PRIMARY...) CT slices, it is quite safe to assume that some growth in the Z-Direction is always present in a volumetric dataset. But for MRI or for reconstructed CT-slices (MPR), you may find datasets in which slices are parallel to the xz or yz plane. If your application is supposed to handle such images, make sure to avoid division by zero...
Yes, the standard defines it. DICOM PS3.3, part C.7.6.2:
The direction of the axes is defined fully by the patient's orientation.
If Anatomical Orientation Type (0010,2210) is absent or has a value of BIPED, the x-axis is increasing to the left hand side of the patient. The y-axis is increasing to the posterior side of the patient. The z-axis is increasing toward the head of the patient.
There is also a tag (0020,0037), Image Orientation (Patient), which relates actual position of the patient to the global coordinate frame. In trunk CT it is almost always 1 0 0 0 1 0 (no rotation) and you don't need to deal with it. Otherwise, see comments under the link above.
You are correct. The chest CT series are sorted from head to feet. The slice closest to the head should have the lowest Instance Number.
I don't know if this is defined by the DICOM standard or not, but I have seen a lot of DICOM images and the convention is this:
AXIAL - sorted by Z axis high to low (head to feet)
CORONAL - sorted by Y axis high to low (back to front)
SAGITTAL - sorted by X axis low to high (right to left)
Notice in all cases, the first slice in the series will be farthest from the observer.
If you need to generate Instance Number, you should sort the images by the dot product of Image Position Patient and (1,-1,-1) from low to high. In the rare degenerate case (all dot products are the same), I don't know. Pick another direction to sort, but probably (0,-1,-1) would be a good choice.
EDIT: I just discussed this with a friend who is more experienced. He said it varies. Some departments prefer back to front order, some prefer front to back. Also some DICOM viewers will give users the choice of how the slices are sorted (by Instance Number, Slice Location, IPP, Content Time, etc)
Given a 4x4 gridboard (as shown here: http://i.stack.imgur.com/YD2XI.png)
I need to:
1. Select any one of the 16 grids to use as starting point
2. From the starting point, move one grid at a time, and walk on the other 15 grids once without repeating
Example of the pathfinding: http://i.stack.imgur.com/J6lBt.png
Highlighted red grid is the start point
Highlighted green grid is the end point
Sounds like homework... you might want to have a look at the eularian path/circuit (http://en.wikipedia.org/wiki/Eulerian_path) problem.
I have thought some time about writing a program that tells me if three circles with given diameters can fit inside a triangle with given side lengths without overlapping (touching is ok) each other.
How would one think about doing that?
I would try to find some way to enumerate the possible configurations for the three circles, and then test each configuration until one is found where the three circles fit, or until all configurations have been tested and rejected.
(In what follows I am assuming that each circle is known to fit in the triangle by itself. Obviously if any circle fails to fit by itself then it fails to fit in any configuration of three circles.)
Configuration (1) involves putting a circle in each corner of the triangle. (This is the configuration that everyone spotted.)
There are six ways to arrange the circles, and for each arrangement it sufficient to check whether the circles will fit pairwise:
The distance AS₁ is r₁/tan(½α), the distance S₂B is r₂/tan(½β), and the distance S₁S₂ is √((r₁ + r₂)² − (r₁ − r₂)²) = 2√r₁r₂
The circles fit if AS₁ + S₁S₂ + S₂B ≤ AB.
In configuration (2) we place two circles into two of the corners of the triangle, and the third circle between these two and one of the two edges that's not touching both circles:
Figuring out whether these will fit is a bit more complex:
To find the length AS₁ we have to walk round the triangle from corner C via the point T. I'll leave the details of this as an exercise.
There are eighteen ways to arrange the circles into this configuration.
Is there a configuration (3)? I looked but couldn't find one that couldn't be turned into one of the two I gave. For example, if all three circles touch the same side then there's always room to swap the middle circle over to the opposite side, getting configuration (2). However, enumeration of geometric configurations is always tricky and I could easily have missed one.
Just a guess: your problem could be related to that of Appolonius's circles.
I ran into it while trying to fit recursively 3 circles within a 4th one without any intersection for some fractal animation, so it may be worth a try.
You'll find it explained in length at Wolfram (this problem was solved only in 1968):
http://mathworld.wolfram.com/ApolloniusProblem.html
this seems to be a tough and interesting problem. Through the solution of the Marble Problem (related to Malfatti's Circles) by Los and Zalgaller in 1994, it might be possible for you to tediously extract a necessary condition for existence of a configuration of three non-overlapping circles with given radii inside a triangle with prescribed side lengths. If you can place them inside the triangle, the sum of their areas will be at most the maximal possible area for three triangles inside a circle. The Marble Problem is the problem of determining the maximal area of three non-overlapping circles inside a given triangle. Right now, I can't see that this is also sufficient.
Perhaps it would be worth, as a start, to introduce an ε to the problem and then look for an algortihm that in a finite number of steps, could determine if a configuration which is at most "bad by ε" (defined in some sensible way) exists.
Some of the World's top mathematicians participate regularly at stackoverflow's sibling,
mathoverflow.org, so you could try posting this problem over there.
Thanks for posting this.
I think it's sufficient to try all 6 possible permutations (A1 B2 C3, A2 B1 C3, A1 B3 C2, A3 B1 C2, A2 B3 C1, A3 B2 C1). If a circle isn't tangent to two edges of the triangle, then it's placement is suboptimal in some sense and you could make more space for the other two by sliding it into a corner. If it's possible to stick three circles in a triangle without them overlapping then it's probably possible to slide them into the corners (for the case in which they're all jammed against the edges, lift them all up together and rotate by 60 degrees). Of course, this isn't a rigorous proof, but I'm pretty sure it works.
Furthermore, I imagine the solution will always be to place the largest circles at the widest angles, because those are the ones that could potentially take up the most central triangle space.
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In the Communications of the ACM, August 2008 "Puzzled" column, Peter Winkler asked the following question:
On the table before us are 10 dots,
and in our pocket are 10 $1 coins.
Prove the coins can be placed on the
table (no two overlapping) in such a
way that all dots are covered. Figure
2 shows a valid placement of the coins
for this particular set of dots; they
are transparent so we can see them.
The three coins at the bottom are not
needed.
In the following issue, he presented his proof:
We had to show that any 10 dots on a
table can be covered by
non-overlapping $1 coins, in a problem
devised by Naoki Inaba and sent to me
by his friend, Hirokazu Iwasawa, both
puzzle mavens in Japan.
The key is to note that packing disks
arranged in a honeycomb pattern cover
more than 90% of the plane. But how do
we know they do? A disk of radius one
fits inside a regular hexagon made up
of six equilateral triangles of
altitude one. Since each such triangle
has area sqrt(3)/3, the hexagon
itself has area 2*sqrt(3); since the
hexagons tile the plane in a honeycomb
pattern, the disks, each with area π,
cover π /(2*sqrt(3)) ~ .9069 of the
plane's surface.
It follows that if the disks are
placed randomly on the plane, the
probability that any particular point
is covered is .9069. Therefore, if we
randomly place lots of $1 coins
(borrowed) on the table in a hexagonal
pattern, on average, 9.069 of our 10
points will be covered, meaning at
least some of the time all 10 will be
covered. (We need at most only 10
coins so give back the rest.)
What does it mean that the disks cover
90.69% of the infinite plane? The easiest way to answer is to say,
perhaps, that the percentage of any
large square covered by the disks
approaches this value as the square
expands. What is "random" about the
placement of the disks? One way to
think it through is to fix any packing
and any disk within it, then pick a
point uniformly at random from the
honeycomb hexagon containing the disk
and move the disk so its center is at
the chosen point.
I don't understand. Doesn't the probabilistic nature of this proof simply mean that in the majority of configurations, all 10 dots can be covered. Can't we still come up with a configuration involving 10 (or less) dots where one of the dots can't be covered?
I think that I can re-arrange Winkler's argument to make it a little more convincing.
You're given an arrangement of dots on a table. You also have a big template made of coins glued together in a honeycomb pattern. You now do a Monte Carlo simulation, repeatedly throwing the honeycomb on the table at a random location (but always with the same orientation), and counting the number of covered dots. If you get enough samples you will eventually find that the expected average number of dots covered is 9.069 per throw.
The key insight is that if the average is 9.069, there must have been a throw where 10 dots were covered. Because if you never covered 10 dots, the average would be 9 or less.
So now you know that there was at least one throw that covered 10 dots. You duplicate that throw, and remove all the coins that aren't covering a dot. There will be 10 or fewer coins remaining.
A small digression: Is it possible that for some clever arrangement of dots the long range average of covered dots is something other than 9.069? The answer is no, because each of the dots can be considered separately. In other words, in 10000 throws of the honeycomb, the expected number each dot will be covered is 9069 times. So we expect a total of 90690 dots to be covered, for an average of 9.069 per throw.