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LEARNING OBJECTIVES
By the end of this section, you will be able to:
- Explain what a continuous source charge distribution is and how it is related to the concept of quantization of charge
- Describe line charges, surface charges, and volume charges
- Calculate the field of a continuous source charge distribution of either sign
The charge distributions we have seen so far have been discrete: made up of individual point particles. This is in contrast with a continuous charge distribution, which has at least one nonzero dimension. If a charge distribution is continuous rather than discrete, we can generalize the definition of the electric field. We simply divide the charge into infinitesimal pieces and treat each piece as a point charge.
Note that because charge is quantized, there is no such thing as a “truly” continuous charge distribution. However, in most practical cases, the total charge creating the field involves such a huge number of discrete charges that we can safely ignore the discrete nature of the charge and consider it to be continuous. This is exactly the kind of approximation we make when we deal with a bucket of water as a continuous fluid, rather than a collection of (ce{H2O}) molecules.
Our first step is to define a charge density for a charge distribution along a line, across a surface, or within a volume, as shown in Figure (PageIndex{1}).
Definitions: Charge Densities
Definitions of charge density:
- linear charge density: (lambda equiv ) charge per unit length (Figure (PageIndex{1a})); units are coulombs per meter ((C/m))
- surface charge density: (sigma equiv ) charge per unit area (Figure (PageIndex{1b})); units are coulombs per square meter ((C/m^2))
- volume charge density: (rho equiv ) charge per unit volume (Figure (PageIndex{1c})); units are coulombs per square meter ((C/m^3))
For a line charge, a surface charge, and a volume charge, the summation in the definition of an Electric field discussed previously becomes an integral and (q_i) is replaced by (dq = lambda dl), (sigma dA), or (rho dV), respectively:
[ begin{align} vec{E}(P) &= underbrace{dfrac{1}{4pi epsilon_0} sum_{i=1}^N left(dfrac{q_i}{r^2}right)hat{r}}_{text{Point charges}} label{eq1} [4pt] vec{E}(P) &= underbrace{dfrac{1}{4pi epsilon_0} int_{line} left(dfrac{lambda , dl}{r^2}right) hat{r}}_{text{Line charge}} label{eq2} [4pt] vec{E}(P) &= underbrace{dfrac{1}{4pi epsilon_0} int_{surface} left(dfrac{sigma ,dA}{r^2}right) hat{r} }_{text{Surface charge}}label{eq3} [4pt] vec{E}(P) &= underbrace{dfrac{1}{4pi epsilon_0} int_{volume} left(dfrac{rho ,dV}{r^2}right) hat{r}}_{text{Volume charge}} label{eq4} end{align}]
![Disk Disk](https://upload.wikimedia.org/wikipedia/commons/thumb/b/be/Unit_disk_graph.svg/1200px-Unit_disk_graph.svg.png)
The integrals in Equations ref{eq1}-ref{eq4} are generalizations of the expression for the field of a point charge. They implicitly include and assume the principle of superposition. The “trick” to using them is almost always in coming up with correct expressions for (dl), (dA), or (dV), as the case may be, expressed in terms of r, and also expressing the charge density function appropriately. It may be constant; it might be dependent on location.
Note carefully the meaning of (r) in these equations: It is the distance from the charge element ((q_i, , lambda , dl, , sigma , dA, , rho , dV)) to the location of interest, (P(x, y, z)) (the point in space where you want to determine the field). However, don’t confuse this with the meaning of (hat{r}); we are using it and the vector notation (vec{E}) to write three integrals at once. That is, Equation ref{eq2} is actually
[ begin{align} E_x (P) &= dfrac{1}{4pi epsilon_0} int_{line} left(dfrac{lambda , dl}{r^2}right)_x, [4pt] E_y(P) &= dfrac{1}{4pi epsilon_0} int_{line} left(dfrac{lambda , dl}{r^2}right)_y, [4pt] E_z(P) &= dfrac{1}{4pi epsilon_0} int_{line} left(dfrac{lambda , dl}{r^2}right)_z end{align} ]
Example (PageIndex{1}): Electric Field of a Line Segment
Find the electric field a distance (z) above the midpoint of a straight line segment of length (L) that carries a uniform line charge density (lambda).
Strategy
Since this is a continuous charge distribution, we conceptually break the wire segment into differential pieces of length (dl), each of which carries a differential amount of charge
[dq = lambda , dl. nonumber]
Then, we calculate the differential field created by two symmetrically placed pieces of the wire, using the symmetry of the setup to simplify the calculation (Figure (PageIndex{2})). Finally, we integrate this differential field expression over the length of the wire (half of it, actually, as we explain below) to obtain the complete electric field expression.
Solution
Before we jump into it, what do we expect the field to “look like” from far away? Since it is a finite line segment, from far away, it should look like a point charge. We will check the expression we get to see if it meets this expectation.
The electric field for a line charge is given by the general expression
[vec{E}(P) = dfrac{1}{4pi epsilon_0} int_{line} dfrac{lambda dl}{r^2}hat{r}. nonumber]
The symmetry of the situation (our choice of the two identical differential pieces of charge) implies the horizontal (x)-components of the field cancel, so that the net field points in the (z)-direction. Let’s check this formally.
The total field (vec{E}(P)) is the vector sum of the fields from each of the two charge elements (call them (vec{E}_1) and (vec{E}_2), for now):
[ begin{align*} vec{E}(P) &= vec{E}_1 + vec{E}_2 [4pt] &= E_{1x}hat{i} + E_{1z}hat{k} + E_{2x} (-hat{i}) + E_{2z}hat{k}. end{align*}]
Disk Graph 2 1 3 0 5
Because the two charge elements are identical and are the same distance away from the point (P) where we want to calculate the field, (E_{1x} = E_{2x}), so those components cancel. This leaves
[ begin{align*} vec{E}(P) &= E_{1z}hat{k} + E_{2z}hat{k} [4pt] &= E_1 , cos , theta hat{k} + E_2 , cos , theta hat{k}. end{align*}]
These components are also equal, so we have
[ begin{align*} vec{E}(P) &= dfrac{1}{4 pi epsilon_0}int dfrac{lambda dl}{r^2} , cos , theta hat{k} + dfrac{1}{4 pi epsilon_0}int dfrac{lambda dl}{r^2} , cos , theta hat{k} [4pt] &= dfrac{1}{4 pi epsilon_0}int_0^{L/2} dfrac{2lambda dx}{r^2} , cos , theta hat{k} end{align*}]
where our differential line element dl is dx, in this example, since we are integrating along a line of charge that lies on the x-axis. (The limits of integration are 0 to (frac{L}{2}), not (-frac{L}{2}) to (+frac{L}{2}), because we have constructed the net field from two differential pieces of charge (dq). If we integrated along the entire length, we would pick up an erroneous factor of 2.)
In principle, this is complete. However, to actually calculate this integral, we need to eliminate all the variables that are not given. In this case, both (r) and (theta) change as we integrate outward to the end of the line charge, so those are the variables to get rid of. We can do that the same way we did for the two point charges: by noticing that
[r = (z^2 + x^2)^{1/2} nonumber]
and
[cos , theta = dfrac{z}{r} = dfrac{z}{(z^2 + x^2)^{1/2}}. nonumber]
Substituting, we obtain
[ begin{align*} vec{E}(P) &= dfrac{1}{4 pi epsilon_0}int_0^{L/2} dfrac{2lambda dx}{(z^2 + x^2)} dfrac{z}{(z^2 + x^2)^{1/2}} hat{k} [4pt] &= dfrac{1}{4 pi epsilon_0}int_0^{L/2} dfrac{2lambda z}{(z^2 + x^2)^{3/2}} dx hat{k} [4pt] &= dfrac{2 lambda z}{4 pi epsilon_0} left[dfrac{x}{z^2sqrt{z^2 + x^2}}right]_0^{L/2} hat{k}. end{align*}]
which simplifies to
[vec{E}(z) = dfrac{1}{4 pi epsilon_0} dfrac{lambda L}{zsqrt{z^2 + dfrac{L^2}{4}}} , hat{k}. label{5.12}]
Significance
Notice, once again, the use of symmetry to simplify the problem. This is a very common strategy for calculating electric fields. The fields of nonsymmetrical charge distributions have to be handled with multiple integrals and may need to be calculated numerically by a computer.
Exercise (PageIndex{1}) Money pro 2 0 9 putter review.
How would the strategy used above change to calculate the electric field at a point a distance (z) above one end of the finite line segment?
We will no longer be able to take advantage of symmetry. Instead, we will need to calculate each of the two components of the electric field with their own integral.
Example (PageIndex{2}): Electric Field of an Infinite Line of Charge
Find the electric field a distance (z) above the midpoint of an infinite line of charge that carries a uniform line charge density (lambda).
Strategy
Disk Graph 2 1 3 0 9
This is exactly like the preceding example, except the limits of integration will be (-infty) to (+infty).
Solution
Again, the horizontal components cancel out, so we wind up with
[vec{E}(P) = dfrac{1}{4 pi epsilon_0} int_{-infty}^{infty} dfrac{lambda dx}{r^2} , cos , theta hat{k} nonumber]
where our differential line element dl is dx, in this example, since we are integrating along a line of charge that lies on the x-axis. Again,
[ begin{align*} cos , theta &= dfrac{z}{r} [4pt] &= dfrac{z}{(z^2 + x^2)^{1/2}}. end{align*} ]
Substituting, we obtain
[ begin{align*} vec{E}(P) &= dfrac{1}{4 pi epsilon_0} int_{-infty}^{infty} dfrac{lambda dx}{(z^2 + x^2)} , dfrac{z}{(z^2 + x^2)^{1/2}} hat{k} [4pt] &= dfrac{1}{4 pi epsilon_0} int_{-infty}^{infty} dfrac{lambda z}{(z^2 + x^2)^{3/2}}dx hat{k} [4pt] &= dfrac{1}{4 pi epsilon_0} left[ dfrac{x}{z^2sqrt{z^2 + x^2}}right]_{-infty}^{infty} , hat{k} end{align*}]
which simplifies to
[vec{E}(z) = dfrac{1}{4 pi epsilon_0} dfrac{2lambda}{z}hat{k}. nonumber]
Significance
Our strategy for working with continuous charge distributions also gives useful results for charges with infinite dimension.
In the case of a finite line of charge, note that for (z gg L), (z^2) dominates the L in the denominator, so that Equation ref{5.12} simplifies to
[vec{E} approx dfrac{1}{4pi epsilon_0} dfrac{lambda L}{z^2} hat{k}.]
If you recall that (lambda L = q) the total charge on the wire, we have retrieved the expression for the field of a point charge, as expected.
In the limit (L rightarrow infty) on the other hand, we get the field of an infinite straight wire, which is a straight wire whose length is much, much greater than either of its other dimensions, and also much, much greater than the distance at which the field is to be calculated:
Shredo 1 2 5 download free. [vec{E}(z) = dfrac{1}{4 pi epsilon_0} dfrac{2lambda}{z}hat{k}. label{infinite straight wire}]
An interesting artifact of this infinite limit is that we have lost the usual (1/r^2) dependence that we are used to. This will become even more intriguing in the case of an infinite plane.
Example (PageIndex{3A}): Electric Field due to a Ring of Charge
A ring has a uniform charge density (lambda), with units of coulomb per unit meter of arc. Find the electric field at a point on the axis passing through the center of the ring.
Strategy
We use the same procedure as for the charged wire. The difference here is that the charge is distributed on a circle. We divide the circle into infinitesimal elements shaped as arcs on the circle and use polar coordinates shown in Figure (PageIndex{3}).
Solution
The electric field for a line charge is given by the general expression
[vec{E}(P) = dfrac{1}{4pi epsilon_0} int_{line} dfrac{lambda dl}{r^2} hat{r}. nonumber]
A general element of the arc between (theta) and (theta + dtheta) is of length (Rdtheta) and therefore contains a charge equal to (lambda R ,dtheta). The element is at a distance of (r = sqrt{z^2 + R^2}) from (P), the angle is (cos , phi = dfrac{z}{sqrt{z^2+R^2}}) and therefore the electric field is
[ begin{align*} vec{E}(P) &= dfrac{1}{4pi epsilon_0} int_{line} dfrac{lambda dl}{r^2} hat{r} = dfrac{1}{4pi epsilon_0} int_0^{2pi} dfrac{lambda Rdtheta}{z^2 + R^2} dfrac{z}{sqrt{z^2 + R^2}} hat{z} [4pt] &= dfrac{1}{4pi epsilon_0} dfrac{lambda Rz}{(z^2 + R^2)^{3/2}} hat{z} int_0^{2pi} dtheta [4pt] &= dfrac{1}{4pi epsilon_0} dfrac{2pi lambda Rz}{(z^2 + R^2)^{3/2}} hat{z} [4pt] &= dfrac{1}{4pi epsilon_0} dfrac{q_{tot}z}{(z^2 + R^2)^{3/2}} hat{z}. end{align*}]
Significance
As usual, symmetry simplified this problem, in this particular case resulting in a trivial integral. Also, when we take the limit of (z gg R), we find that
[vec{E} approx dfrac{1}{4pi epsilon_0} dfrac{q_{tot}}{z^2} hat{z}, nonumber ]
as we expect.
Example (PageIndex{3B}): The Field of a Disk
Find the electric field of a circular thin disk of radius (R) and uniform charge density at a distance (z) above the center of the disk (Figure (PageIndex{4}))
Strategy
The electric field for a surface charge is given by
[vec{E}(P) = dfrac{1}{4 pi epsilon_0} int_{surface} dfrac{sigma dA}{r^2} hat{r}. nonumber]
To solve surface charge problems, we break the surface into symmetrical differential “stripes” that match the shape of the surface; here, we’ll use rings, as shown in the figure. Again, by symmetry, the horizontal components cancel and the field is entirely in the vertical ((hat{k})) direction. The vertical component of the electric field is extracted by multiplying by (theta), so
[vec{E}(P) = dfrac{1}{4pi epsilon_0} int_{surface} dfrac{sigma dA}{r^2} , cos , theta , hat{k}. nonumber]
![Disk graph 2 1 3 0 3 Disk graph 2 1 3 0 3](https://picturethismaths.files.wordpress.com/2018/02/bg4.png)
As before, we need to rewrite the unknown factors in the integrand in terms of the given quantities. In this case,
[dA = 2pi r'dr']
[r^2 = r'^2 + z^2]
[cos , theta = dfrac{z}{(r'^2 + z^2)^{1/2}}.]
(Please take note of the two different “(r)’s” here; (r) is the distance from the differential ring of charge to the point (P) where we wish to determine the field, whereas (r') is the distance from the center of the disk to the differential ring of charge.) Also, we already performed the polar angle integral in writing down (dA).
Solution
Substituting all this in, we get
[ begin{align*} vec{E}(P) &= vec{E}(z) [4pt] &= dfrac{1}{4 pi epsilon_0} int_0^R dfrac{sigma (2pi r' dr')z}{(r'^2 + z^2)^{3/2}} hat{k} [4pt] &= dfrac{1}{4 pi epsilon_0} (2pi sigma z)left(dfrac{1}{z} - dfrac{1}{sqrt{R^2 + z^2}}right) hat{k} end{align*}]
or, more simply,
[vec{E}(z) = dfrac{1}{4 pi epsilon_0} left( 2 pi sigma - dfrac{2 pi sigma z}{sqrt{R^2 + z^2}}right)hat{k}. label{5.14}]
Significance
Again, it can be shown (via a Taylor expansion) that when (z gg R), this reduces to
[vec{E}(z) approx dfrac{1}{4 pi epsilon_0} dfrac{sigma pi R^2}{z^2} hat{k},nonumber]
which is the expression for a point charge (Q = sigma pi R^2).
Exercise (PageIndex{3})
How would the above limit change with a uniformly charged rectangle instead of a disk?
The point charge would be (Q = sigma ab) where (a) and (b) are the sides of the rectangle but otherwise identical.
As (R rightarrow infty), Equation ref{5.14} reduces to the field of an infinite plane, which is a flat sheet whose area is much, much greater than its thickness, and also much, much greater than the distance at which the field is to be calculated:
[ begin{align} vec{E} &= lim_{R rightarrow infty} dfrac{1}{4 pi epsilon_0} left( 2 pi sigma - dfrac{2 pi sigma z}{sqrt{R^2 + z^2}}right)hat{k} [4pt] &= dfrac{sigma}{2 epsilon_0} hat{k}. label{5.15} end{align}]
Note that this field is constant. This surprising result is, again, an artifact of our limit, although one that we will make use of repeatedly in the future. To understand why this happens, imagine being placed above an infinite plane of constant charge. Does the plane look any different if you vary your altitude? No—you still see the plane going off to infinity, no matter how far you are from it. It is important to note that Equation ref{5.15} is because we are above the plane. If we were below, the field would point in the (- hat{k}) direction.
Example (PageIndex{4}): The Field of Two Infinite Planes
Find the electric field everywhere resulting from two infinite planes with equal but opposite charge densities (Figure (PageIndex{5})).
Strategy
We already know the electric field resulting from a single infinite plane, so we may use the principle of superposition to find the field from two.
Solution
The electric field points away from the positively charged plane and toward the negatively charged plane. Since the (sigma) are equal and opposite, this means that in the region outside of the two planes, the electric fields cancel each other out to zero. However, in the region between the planes, the electric fields add, and we get
[vec{E} = dfrac{sigma}{epsilon_0}hat{i} nonumber]
for the electric field. The (hat{i}) is because in the figure, the field is pointing in the +x-direction.
Significance
Systems that may be approximated as two infinite planes of this sort provide a useful means of creating uniform electric fields.
Exercise (PageIndex{4})
What would the electric field look like in a system with two parallel positively charged planes with equal charge densities?
The electric field would be zero in between, and have magnitude (dfrac{sigma}{epsilon_0}) everywhere else.
Contributors and Attributions
- Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).
A collection of unit circles and the corresponding unit disk graph.
In geometric graph theory, a unit disk graph is the intersection graph of a family of unit disks in the Euclidean plane. That is, it is a graph with one vertex for each disk in the family, and with an edge between two vertices whenever the corresponding vertices lie within a unit distance of each other.
They are commonly formed from a Poisson point process, making them a simple example of a random structure.
Definitions[edit]
There are several possible definitions of the unit disk graph, equivalent to each other up to a choice of scale factor:
- A graph formed from a collection of points in the Euclidean plane, in which two points are connected if their distance is below a fixed threshold.
- An intersection graph of equal-radius circles, or of equal-radius disks (see Fig. 1).
- A graph formed from a collection of equal-radius circles, in which two circles are connected by an edge if one circle contains the centre of the other circle.
Properties[edit]
Every induced subgraph of a unit disk graph is also a unit disk graph. An example of a graph that is not a unit disk graph is the star K1,7 with one central node connected to seven leaves: if each of seven unit disks touches a common unit disk, some two of the seven disks must touch each other (as the kissing number in the plane is 6). Therefore, unit disk graphs cannot contain an induced K1,7 subgraph.
Applications[edit]
Beginning with the work of Huson & Sen (1995), unit disk graphs have been used in computer science to model the topology of ad hoc wireless communication networks. In this application, nodes are connected through a direct wireless connection without a base station. It is assumed that all nodes are homogeneous and equipped with omnidirectional antennas. Node locations are modelled as Euclidean points, and the area within which a signal from one node can be received by another node is modelled as a circle. If all nodes have transmitters of equal power, these circles are all equal. Random geometric graphs, formed as unit disk graphs with randomly generated disk centres, have also been used as a model of percolation and various other phenomena.[1]
Computational complexity[edit]
If one is given a collection of unit disks (or their centres) in a space of any fixed dimension, it is possible to construct the corresponding unit disk graph in linear time, by rounding the centres to nearby integer grid points, using a hash table to find all pairs of centres within constant distance of each other, and filtering the resulting list of pairs for the ones whose circles intersect. The ratio of the number of pairs considered by this algorithm to the number of edges in the eventual graph is a constant, giving the linear time bound. However, this constant grows exponentially as a function of the dimension (Bentley, Stanat & Williams 1977).
It is NP-hard (more specifically, complete for the existential theory of the reals) to determine whether a graph, given without geometry, can be represented as a unit disk graph.[2] Additionally, it is probably impossible in polynomial time to output explicit coordinates of a unit disk graph representation: there exist unit disk graphs that require exponentially many bits of precision in any such representation.[3]
However, many important and difficult graph optimization problems such as maximum independent set, graph coloring, and minimum dominating set can be approximated efficiently by using the geometric structure of these graphs,[4] and the maximum clique problem can be solved exactly for these graphs in polynomial time, given a disk representation.[5] Even if a disk representation is not known, and an abstract graph is given as input, it is possible in polynomial time to produce either a maximum clique or a proof that the graph is not a unit disk graph,[6] and to 3-approximate the optimum coloring by using a greedy coloring algorithm.[7]
When a given vertex set forms a subset of a triangular lattice, a necessary and sufficient condition for the perfectness of a unit graph is known.[8] For the perfect graphs, a number of NP-complete optimization problems (graph coloring problem, maximum clique problem, and maximum independent set problem) are polynomially solvable.
See also[edit]
- Barrier resilience, an algorithmic problem of breaking cycles in unit disk graphs
- Indifference graph, a one-dimensional analogue of the unit disk graphs
- Penny graph, the unit disk graphs for which the disks can be tangent but not overlap (contact graph)
- Coin graph, the contact graph of (not necessarily unit-sized) disks
- Vietoris–Rips complex, a generalization of the unit disk graph that constructs higher-order topological spaces from unit distances in a metric space
- Unit distance graph, a graph formed by connecting points that are at distance exactly one rather than (as here) at most a given threshold
Notes[edit]
- ^See, e.g., Dall & Christensen (2002).
- ^Breu & Kirkpatrick (1998); Kang & Müller (2011).
- ^McDiarmid & Mueller (2011).
- ^Marathe et al. (1994); Matsui (2000).
- ^Clark, Colbourn & Johnson (1990).
- ^Raghavan & Spinrad (2003).
- ^Gräf, Stumpf & Weißenfels (1998).
- ^Miyamoto & Matsui (2005).
References[edit]
- Bentley, Jon L.; Stanat, Donald F.; Williams, E. Hollins, Jr. (1977), 'The complexity of finding fixed-radius near neighbors', Information Processing Letters, 6 (6): 209–212, doi:10.1016/0020-0190(77)90070-9, MR0489084.
- Breu, Heinz; Kirkpatrick, David G. (1998), 'Unit disk graph recognition is NP-hard', Computational Geometry: Theory and Applications, 9 (1–2): 3–24, doi:10.1016/s0925-7721(97)00014-x.
- Clark, Brent N.; Colbourn, Charles J.; Johnson, David S. (1990), 'Unit disk graphs', Discrete Mathematics, 86 (1–3): 165–177, doi:10.1016/0012-365X(90)90358-O.
- Dall, Jesper; Christensen, Michael (2002), 'Random geometric graphs', Phys. Rev. E, 66: 016121, arXiv:cond-mat/0203026, Bibcode:2002PhRvE.66a6121D, doi:10.1103/PhysRevE.66.016121.
- Gräf, A.; Stumpf, M.; Weißenfels, G. (1998), 'On coloring unit disk graphs', Algorithmica, 20 (3): 277–293, doi:10.1007/PL00009196, MR1489033.
- Huson, Mark L.; Sen, Arunabha (1995), 'Broadcast scheduling algorithms for radio networks', Military Communications Conference, IEEE MILCOM '95, 2, pp. 647–651, doi:10.1109/MILCOM.1995.483546, ISBN0-7803-2489-7.
- Kang, Ross J.; Müller, Tobias (2011), 'Sphere and dot product representations of graphs', Proceedings of the Twenty-Seventh Annual Symposium on Computational Geometry (SoCG'11), June 13–15, 2011, Paris, France, pp. 308–314.
- Marathe, Madhav V.; Breu, Heinz; Hunt, III, Harry B.; Ravi, S. S.; Rosenkrantz, Daniel J. (1994), Geometry based heuristics for unit disk graphs, arXiv:math.CO/9409226.
- Matsui, Tomomi (2000), 'Approximation Algorithms for Maximum Independent Set Problems and Fractional Coloring Problems on Unit Disk Graphs', Lecture Notes in Computer Science, Lecture Notes in Computer Science, 1763: 194–200, doi:10.1007/978-3-540-46515-7_16, ISBN978-3-540-67181-7.
- McDiarmid, Colin; Mueller, Tobias (2011), Integer realizations of disk and segment graphs, arXiv:1111.2931, Bibcode:2011arXiv1111.2931M
- Miyamoto, Yuichiro; Matsui, Tomomi (2005), 'Perfectness and Imperfectness of the kth Power of Lattice Graphs', Lecture Notes in Computer Science, Lecture Notes in Computer Science, 3521: 233–242, doi:10.1007/11496199_26, ISBN978-3-540-26224-4.
- Raghavan, Vijay; Spinrad, Jeremy (2003), 'Robust algorithms for restricted domains', Journal of Algorithms, 48 (1): 160–172, doi:10.1016/S0196-6774(03)00048-8, MR2006100.
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