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Global Constraint Catalog: Cscalar_product
<< 5.341. same_sign5.343. sequence_folding >>
Arithmetic constraint.
\mathrm{𝚜𝚌𝚊𝚕𝚊𝚛}_\mathrm{𝚙𝚛𝚘𝚍𝚞𝚌𝚝}\left(\mathrm{𝙻𝙸𝙽𝙴𝙰𝚁𝚃𝙴𝚁𝙼},\mathrm{𝙲𝚃𝚁},\mathrm{𝚅𝙰𝙻}\right)
\mathrm{𝚎𝚚𝚞𝚊𝚝𝚒𝚘𝚗}
\mathrm{𝚕𝚒𝚗𝚎𝚊𝚛}
\mathrm{𝚜𝚞𝚖}_\mathrm{𝚠𝚎𝚒𝚐𝚑𝚝}
\mathrm{𝚠𝚎𝚒𝚐𝚑𝚝𝚎𝚍𝚂𝚞𝚖}
\mathrm{𝙻𝙸𝙽𝙴𝙰𝚁𝚃𝙴𝚁𝙼}
\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚌𝚘𝚎𝚏𝚏}-\mathrm{𝚒𝚗𝚝},\mathrm{𝚟𝚊𝚛}-\mathrm{𝚍𝚟𝚊𝚛}\right)
\mathrm{𝙲𝚃𝚁}
\mathrm{𝚊𝚝𝚘𝚖}
\mathrm{𝚅𝙰𝙻}
\mathrm{𝚍𝚟𝚊𝚛}
\mathrm{𝚛𝚎𝚚𝚞𝚒𝚛𝚎𝚍}
\left(\mathrm{𝙻𝙸𝙽𝙴𝙰𝚁𝚃𝙴𝚁𝙼},\left[\mathrm{𝚌𝚘𝚎𝚏𝚏},\mathrm{𝚟𝚊𝚛}\right]\right)
\mathrm{𝙲𝚃𝚁}\in \left[=,\ne ,<,\ge ,>,\le \right]
Constraint a linear term defined as the sum of products of coefficients and variables. More precisely, let
𝚂
denote the sum of the product between a coefficient and its variable of the different items of the
\mathrm{𝙻𝙸𝙽𝙴𝙰𝚁𝚃𝙴𝚁𝙼}
collection. Enforce the following constraint to hold:
𝚂\mathrm{𝙲𝚃𝚁}\mathrm{𝚅𝙰𝙻}
\left(\begin{array}{c}〈\mathrm{𝚌𝚘𝚎𝚏𝚏}-1\mathrm{𝚟𝚊𝚛}-1,\mathrm{𝚌𝚘𝚎𝚏𝚏}-3\mathrm{𝚟𝚊𝚛}-1,\mathrm{𝚌𝚘𝚎𝚏𝚏}-1\mathrm{𝚟𝚊𝚛}-4〉,=,8\hfill \end{array}\right)
\mathrm{𝚜𝚌𝚊𝚕𝚊𝚛}_\mathrm{𝚙𝚛𝚘𝚍𝚞𝚌𝚝}
constraint holds since the condition
1·1+3·1+1·4=8
|\mathrm{𝙻𝙸𝙽𝙴𝙰𝚁𝚃𝙴𝚁𝙼}|>1
\mathrm{𝚛𝚊𝚗𝚐𝚎}
\left(\mathrm{𝙻𝙸𝙽𝙴𝙰𝚁𝚃𝙴𝚁𝙼}.\mathrm{𝚌𝚘𝚎𝚏𝚏}\right)>1
\mathrm{𝚛𝚊𝚗𝚐𝚎}
\left(\mathrm{𝙻𝙸𝙽𝙴𝙰𝚁𝚃𝙴𝚁𝙼}.\mathrm{𝚟𝚊𝚛}\right)>1
\mathrm{𝙲𝚃𝚁}\in \left[=,<,\ge ,>,\le \right]
\mathrm{𝙻𝙸𝙽𝙴𝙰𝚁𝚃𝙴𝚁𝙼}
\mathrm{𝙻𝙸𝙽𝙴𝙰𝚁𝚃𝙴𝚁𝙼}
\left(\mathrm{𝚌𝚘𝚎𝚏𝚏},\mathrm{𝚟𝚊𝚛}\right)
(permutation not necessarily applied to all items).
\mathrm{𝙻𝙸𝙽𝙴𝙰𝚁𝚃𝙴𝚁𝙼}
\mathrm{𝙲𝚃𝚁}\in \left[<,\le \right]
\mathrm{𝚖𝚒𝚗𝚟𝚊𝚕}\left(\mathrm{𝙻𝙸𝙽𝙴𝙰𝚁𝚃𝙴𝚁𝙼}.\mathrm{𝚌𝚘𝚎𝚏𝚏}\right)\ge 0
\mathrm{𝚖𝚒𝚗𝚟𝚊𝚕}\left(\mathrm{𝙻𝙸𝙽𝙴𝙰𝚁𝚃𝙴𝚁𝙼}.\mathrm{𝚟𝚊𝚛}\right)\ge 0
\mathrm{𝙻𝙸𝙽𝙴𝙰𝚁𝚃𝙴𝚁𝙼}
\mathrm{𝙲𝚃𝚁}\in \left[\ge ,>\right]
\mathrm{𝚖𝚒𝚗𝚟𝚊𝚕}\left(\mathrm{𝙻𝙸𝙽𝙴𝙰𝚁𝚃𝙴𝚁𝙼}.\mathrm{𝚌𝚘𝚎𝚏𝚏}\right)\ge 0
\mathrm{𝚖𝚒𝚗𝚟𝚊𝚕}\left(\mathrm{𝙻𝙸𝙽𝙴𝙰𝚁𝚃𝙴𝚁𝙼}.\mathrm{𝚟𝚊𝚛}\right)\ge 0
Aggregate:
\mathrm{𝙻𝙸𝙽𝙴𝙰𝚁𝚃𝙴𝚁𝙼}\left(\mathrm{𝚞𝚗𝚒𝚘𝚗}\right)
\mathrm{𝙲𝚃𝚁}\left(\mathrm{𝚒𝚍}\right)
\mathrm{𝚅𝙰𝙻}\left(+\right)
\mathrm{𝚜𝚌𝚊𝚕𝚊𝚛}_\mathrm{𝚙𝚛𝚘𝚍𝚞𝚌𝚝}
\mathrm{𝚕𝚒𝚗𝚎𝚊𝚛}
in Gecode (http://www.gecode.org/). It is called
\mathrm{𝚜𝚞𝚖}_\mathrm{𝚠𝚎𝚒𝚐𝚑𝚝}
in JaCoP (http://www.jacop.eu/). In the 2008 CSP solver competition the
\mathrm{𝚜𝚌𝚊𝚕𝚊𝚛}_\mathrm{𝚙𝚛𝚘𝚍𝚞𝚌𝚝}
\mathrm{𝚠𝚎𝚒𝚐𝚑𝚝𝚎𝚍𝚂𝚞𝚖}
and required
\mathrm{𝚅𝙰𝙻}
to be fixed.
Most filtering algorithms first merge multiple occurrences of identical variables in order to potentially make more deductions. When
\mathrm{𝙲𝚃𝚁}
corresponds to the less than or equal to constraint, a filtering algorithm achieving bound-consistency for the
\mathrm{𝚜𝚌𝚊𝚕𝚊𝚛}_\mathrm{𝚙𝚛𝚘𝚍𝚞𝚌𝚝}
constraint with large numbers of variables is described in [HarveySchimpf02].
equation in Choco, linear in Gecode, sumweight in JaCoP, scalar_product in SICStus.
\mathrm{𝚜𝚞𝚖}_\mathrm{𝚌𝚝𝚛}
(arithmetic constraint where all coefficients are equal to 1).
characteristic of a constraint: sum.
constraint type: predefined constraint, arithmetic constraint.
filtering: duplicated variables.
|
Multidirectional slowness vector for computing angle gathers from reverse time migrationMultidirectional slowness vector | Geophysics | GeoScienceWorld
, Center for Lithospheric Studies, Richardson, Texas,
. E-mail: chen.tang.diary@gmail.com; mcmec@utdallas.edu.
Errata: To: “Multidirectional slowness vector for computing angle gathers from reverse time migration,” Chen Tang and George A. McMechan, Geophysics, 81, no. 2, S55–S68, doi.org/10.1190/geo2015-0134.1.
Chen Tang, George A. McMechan; Multidirectional slowness vector for computing angle gathers from reverse time migration. Geophysics 2016;; 81 (2): S55–S68. doi: https://doi.org/10.1190/geo2015-0134.1
Angle gathers are important for true-amplitude migration, migration velocity analysis, and angle-dependent inversion. Among existing methods, calculating the
t-x
direction vector is efficient, but it can give only one direction per grid point and fails to give multiple directions for overlapping wavefields associated with multipaths and reflections. The slowness vectors (SVs) in
t-x
ω-k
can be connected by Fourier transforms (FTs); the forward FT from
t-x
ω-k
decomposes the wavefields into different vector components, and the inverse FT sums these components into a unique direction. Therefore, the SV has multiple directions in
ω-k
, but it has only one direction in
t-x
. Based on this relation, we have separated the computation of propagation direction into two steps: First, we used the forward FT,
k/ω
binning, and several inverse FTs to separate the wavefields into vector subsets with different approximate propagation angles, which contained much less wave overlapping; then, we computed
t-x
SVs for each separated wavefield, and the set of these single-direction SVs constituted a multidirectional SV (MSV). In this process, the FTs between
t
ω
domains required a large input/output (I/O) time. We prove the conjugate relation between the decomposition results using positive- and negative-frequency wavefields, and we use complex-valued modeling to obtain the positive-frequency wavefields. Thus, we did wavefield decomposition in
t-k
ω-k
, and avoided the huge I/O caused by the FT between the
t
ω
domains. Our tests demonstrated that the MSV can give multiple directions for overlapping wavefields and improve the quality of angle gathers.
The optimized expansion based low-rank method for wavefield extrapolation
Tilted transverse isotropic reverse time migration with angle gathers: Implementation and efficiency
|
Global Constraint Catalog: Cdisjoint
<< 5.122. disj5.124. disjoint_sboxes >>
\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}
\mathrm{𝚍𝚒𝚜𝚓𝚘𝚒𝚗𝚝}\left(\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{1},\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{2}\right)
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{1}
\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚟𝚊𝚛}-\mathrm{𝚍𝚟𝚊𝚛}\right)
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{2}
\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚟𝚊𝚛}-\mathrm{𝚍𝚟𝚊𝚛}\right)
\mathrm{𝚛𝚎𝚚𝚞𝚒𝚛𝚎𝚍}
\left(\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{1},\mathrm{𝚟𝚊𝚛}\right)
\mathrm{𝚛𝚎𝚚𝚞𝚒𝚛𝚎𝚍}
\left(\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{2},\mathrm{𝚟𝚊𝚛}\right)
Each variable of the collection
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{1}
should take a value that is distinct from all the values assigned to the variables of the collection
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{2}
\left(〈1,9,1,5〉,〈2,7,7,0,6,8〉\right)
In this example, values
1,5,9
are used by the variables of
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{1}
0,2,6,7,8
by the variables of
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{2}
. Since there is no intersection between the two previous sets of values the
\mathrm{𝚍𝚒𝚜𝚓𝚘𝚒𝚗𝚝}
\mathrm{𝚍𝚒𝚜𝚓𝚘𝚒𝚗𝚝}
{U}_{1}\in \left[0..2\right]
{U}_{2}\in \left[1..2\right]
{U}_{3}\in \left[1..2\right]
{V}_{1}\in \left[0..1\right]
{V}_{2}\in \left[1..2\right]
\mathrm{𝚍𝚒𝚜𝚓𝚘𝚒𝚗𝚝}
\left(〈{U}_{1},{U}_{2},{U}_{3}〉,〈{V}_{1},{V}_{2}〉\right)
\mathrm{𝚍𝚒𝚜𝚓𝚘𝚒𝚗𝚝}
|\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{1}|>1
|\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{2}|>1
Arguments are permutable w.r.t. permutation
\left(\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{1},\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{2}\right)
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{1}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{2}
An occurrence of a value of
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{1}.\mathrm{𝚟𝚊𝚛}
can be replaced by any value of
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{1}.\mathrm{𝚟𝚊𝚛}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{2}.\mathrm{𝚟𝚊𝚛}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{2}.\mathrm{𝚟𝚊𝚛}
All occurrences of two distinct values in
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{1}.\mathrm{𝚟𝚊𝚛}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{2}.\mathrm{𝚟𝚊𝚛}
can be swapped; all occurrences of a value in
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{1}.\mathrm{𝚟𝚊𝚛}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{2}.\mathrm{𝚟𝚊𝚛}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{1}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{2}
Despite the fact that this is not an uncommon constraint, it can not be modelled in a compact way neither with a disequality constraint (i.e., two given variables have to take distinct values) nor with the
\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}
constraint. The
\mathrm{𝚍𝚒𝚜𝚓𝚘𝚒𝚗𝚝}
constraint can bee seen as a special case of the
\mathrm{𝚌𝚘𝚖𝚖𝚘𝚗}
\left(\mathrm{𝙽𝙲𝙾𝙼𝙼𝙾𝙽}\mathtt{1},\mathrm{𝙽𝙲𝙾𝙼𝙼𝙾𝙽}\mathtt{2},\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{1},\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{2}\right)
constraint where
\mathrm{𝙽𝙲𝙾𝙼𝙼𝙾𝙽}\mathtt{1}
\mathrm{𝙽𝙲𝙾𝙼𝙼𝙾𝙽}\mathtt{2}
are both set to 0.
MiniZinc (http://www.minizinc.org/) has a
\mathrm{𝚍𝚒𝚜𝚓𝚘𝚒𝚗𝚝}
constraint between two set variables rather than between two collections of variables.
{n}_{1}
the minimum number of distinct values taken by the variables of the collection
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{1}
{n}_{2}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{2}
{n}_{12}
the maximum number of distinct values taken by the union of the variables of
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{1}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{2}
One invariant to maintain for the
\mathrm{𝚍𝚒𝚜𝚓𝚘𝚒𝚗𝚝}
constraint is
{n}_{1}+{n}_{2}\le {n}_{12}
. A lower bound of
{n}_{1}
{n}_{2}
can be obtained by using the algorithms provided in [Beldiceanu01], [BeldiceanuCarlssonThiel02]. An exact upper bound of
{n}_{12}
can be computed by using a bipartite matching algorithm.
𝚔_\mathrm{𝚍𝚒𝚜𝚓𝚘𝚒𝚗𝚝}
\mathrm{𝚍𝚒𝚜𝚓𝚘𝚒𝚗𝚝}_\mathrm{𝚝𝚊𝚜𝚔𝚜}
\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎}
\mathrm{𝚝𝚊𝚜𝚔}
\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}_\mathrm{𝚘𝚗}_\mathrm{𝚒𝚗𝚝𝚎𝚛𝚜𝚎𝚌𝚝𝚒𝚘𝚗}
\mathrm{𝚕𝚎𝚡}_\mathrm{𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}
system of constraints:
𝚔_\mathrm{𝚍𝚒𝚜𝚓𝚘𝚒𝚗𝚝}
characteristic of a constraint: disequality, automaton, automaton with array of counters.
constraint type: value constraint.
modelling: empty intersection.
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{1}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{2}
\mathrm{𝑃𝑅𝑂𝐷𝑈𝐶𝑇}
↦\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}\mathtt{1},\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}\mathtt{2}\right)
\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}\mathtt{1}.\mathrm{𝚟𝚊𝚛}=\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}\mathtt{2}.\mathrm{𝚟𝚊𝚛}
\mathrm{𝐍𝐀𝐑𝐂}
=0
\mathrm{𝑃𝑅𝑂𝐷𝑈𝐶𝑇}
is used in order to generate the arcs of the graph between all variables of
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{1}
and all variables of
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{2}
. Since we use the graph property
\mathrm{𝐍𝐀𝐑𝐂}
=
0 the final graph will be empty. Figure 5.123.2 shows the initial graph associated with the Example slot. Since we use the
\mathrm{𝐍𝐀𝐑𝐂}
=
0 graph property the final graph is empty.
Figure 5.123.2. Initial graph of the
\mathrm{𝚍𝚒𝚜𝚓𝚘𝚒𝚗𝚝}
constraint (the final graph is empty)
Since 0 is the smallest number of arcs of the final graph we can rewrite
\mathrm{𝐍𝐀𝐑𝐂}
=
\mathrm{𝐍𝐀𝐑𝐂}
\le
0. This leads to simplify
\underline{\overline{\mathrm{𝐍𝐀𝐑𝐂}}}
\underline{\mathrm{𝐍𝐀𝐑𝐂}}
Figure 5.123.3 depicts the automaton associated with the
\mathrm{𝚍𝚒𝚜𝚓𝚘𝚒𝚗𝚝}
constraint. To each variable
\mathrm{𝚅𝙰𝚁}{\mathtt{1}}_{i}
of the collection
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{1}
corresponds a signature variable
{S}_{i}
that is equal to 0. To each variable
\mathrm{𝚅𝙰𝚁}{\mathtt{2}}_{i}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{2}
{S}_{i+|\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{1}|}
that is equal to 1.
Figure 5.123.3. Automaton of the
\mathrm{𝚍𝚒𝚜𝚓𝚘𝚒𝚗𝚝}
\left(\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{1},\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{2}\right)
constraint, where state
s
handles variables of the collection
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{1}
and state
t
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\mathtt{2}
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↵In a quadrilateral ABCD , equal diagonals AC BD intersect at P, such that AP=PC BP=PD - Maths - Quadrilaterals - 9727783 | Meritnation.com
↵In a quadrilateral ABCD , equal diagonals AC & BD intersect at P, such that AP=PC & BP=PD ,also angle BPC=90 degreee, then quadrilateral is exactly (a) a parallelogram (b) a square (c) a rhombus (d) a rectangle
\mathrm{In} \mathrm{quadrilateral} ABCD, AC \mathrm{and} BD \mathrm{are} \mathrm{equal} \mathrm{diagonals} \mathrm{and}\phantom{\rule{0ex}{0ex}} AP=PC & BP=OD\phantom{\rule{0ex}{0ex}}\therefore AP=BP=CP=DP=a \left(\mathrm{say}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\angle BPC=90°\phantom{\rule{0ex}{0ex}}⇒\angle APD=90° \left(\mathrm{vertically} \mathrm{opposite} \mathrm{angles}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Here},\phantom{\rule{0ex}{0ex}}\angle APD+\angle CPD=180° \left(\mathrm{vertically} \mathrm{opposite} \mathrm{angles}\right)\phantom{\rule{0ex}{0ex}}⇒90°+\angle CPD=180°\phantom{\rule{0ex}{0ex}}⇒\angle CPD=90°\phantom{\rule{0ex}{0ex}}\mathrm{Again}\phantom{\rule{0ex}{0ex}}\angle APB+\angle BPC=180° \left(\mathrm{vertically} \mathrm{opposite} \mathrm{angles}\right)\phantom{\rule{0ex}{0ex}}⇒\angle APB+90°=180°\phantom{\rule{0ex}{0ex}}⇒\angle APB=90°\phantom{\rule{0ex}{0ex}}\mathrm{Therefore}, \angle APB=\angle BPC=\angle CPD=\angle APD=90°\phantom{\rule{0ex}{0ex}}\mathrm{In} \mathrm{all} \mathrm{four} \mathrm{triangles}; △APB, △BPC, △CPD \mathrm{and} △APD,\phantom{\rule{0ex}{0ex}}\mathrm{two} \mathrm{sides} \mathrm{equal} \mathrm{to} \mathrm{a} \mathrm{and} \mathrm{angles} \mathrm{between} \mathrm{these} \mathrm{two} \mathrm{sides} \mathrm{are} 90°\phantom{\rule{0ex}{0ex}}\mathrm{hence}, \mathrm{all} \mathrm{four} \mathrm{triangles} \mathrm{are} \mathrm{congruent}.\phantom{\rule{0ex}{0ex}}\therefore Ab=BC=CD=DA\phantom{\rule{0ex}{0ex}}\therefore ABCD \mathrm{is} \mathrm{a} \mathrm{rhobus}.\phantom{\rule{0ex}{0ex}}\mathrm{Now}, \mathrm{a} \mathrm{rhombus} \mathrm{whose} \mathrm{both} \mathrm{diagonals} \mathrm{are} \mathrm{equal} \mathrm{is} \mathrm{square}.\phantom{\rule{0ex}{0ex}}\therefore ABCD \mathrm{is} \mathrm{a} \mathrm{square}.\phantom{\rule{0ex}{0ex}}\mathrm{Hence}, \left(\mathrm{b}\right) \mathrm{is} \mathrm{correct} \mathrm{option}.
Hope this information will clear your doubts about quadrilateral.
Abhishek Soni answered this
(c) - answer
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Global Constraint Catalog: Kbusiness_rules
<< 3.7.38. Bound-consistency3.7.40. Centered cyclic(1) constraint network(1) >>
\mathrm{𝚌𝚢𝚌𝚕𝚎}
\mathrm{𝚍𝚒𝚏𝚏𝚗}
\mathrm{𝚐𝚎𝚘𝚜𝚝}
Denotes that a dedicated language was introduced within an argument of a global constraint for directly specifying a specific type of business rules:
\mathrm{𝚌𝚢𝚌𝚕𝚎}
constraint was extended in order to accept rules specifying forbidden sequences of vertices within each cycle [Bourreau99].
\mathrm{𝚍𝚒𝚏𝚏𝚗}
constraint was extended in order to accept calendar rules specifying the way tasks can be interrupted or not on each resource [Beldiceanu98]. This was done since many real scheduling problems have not only to consider disjunctive and assignment constraints, but also operational rules expressing how tasks can be interrupted.
\mathrm{𝚐𝚎𝚘𝚜𝚝}
constraint was extended in order to directly accept a great variety of packing and placement rules [CarlssonBeldiceanuMartin08].
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SPM Maths – user's Blog!
Home › Archive for SPM Maths
(a) Diagram 9.1 shows point P (5, 1) on a Cartesian plane.
Transformation T is a translation
\left(\begin{array}{l}\text{ }4\\ -3\end{array}\right)
Transformation S is an enlargement about the centre (–5, 2) with a scale factor 2.
State the coordinates of the image of point P under the following transformations:
(i) T2,
(ii) TS.
(b) Diagram 9.2 shows geometrical shapes KLMNP, KSRQP and KTUVW drawn on a Cartesian plane.
(i) KTUVW is the image of KLMNP under the combined transformation YZ.
Describe, in full, the transformation:
(a) Z,
(b) Y.
(ii) It is given that KSRQP represents a region of area 30 m2.
Calculate the area, in m2, of the shaded region.
(i) TT = P(–3, 3) → T1 → P’(1, 0) ) → T2 → P’’(5, –3)
(ii) TS = P (–3, 3) → S → P’(–1, 4) → T → P’’(3, 1)
Z: Reflection in the line x = 0.
Y: Enlargement with the centre at (0, 0) and a scale factor of 2.
Area of KTUVW = (Scale factor)2 × Area of object
= Area KTUVW – Area KSRQP
Posted in Transformations III
A(25o N, 35o E), B(25o N, 40o W), C and D are four points which lie on the surface of the earth. AD is the diameter of the common parallel latitude 25o N .
(a) Find the longitude of D.
(b) C lies 3300 nautical miles due south of A measured along the surface of the earth.
Calculate the latitude of C.
(c) Calculate the shortest distance, in nautical mile, from A to D measured along the surface of the earth.
(d) An aeroplane took off from C and flew due north to point A.
The total time taken for the whole flight was 12 hours 24 minutes.
(i) Calculate the distance, in nautical mile, from A due west to B measured along the common parallel of latitude.
(ii) Calculate the average speed, in knot, of the whole flight.
Longitude of D = (180o – 35o)W
= 145oW
\begin{array}{l}\angle AOC=\frac{3300}{60}={55}^{o}\\ \text{Latitude of }C\\ ={\left(55-25\right)}^{o}S\\ ={30}^{o}S\end{array}
Shortest distance of A to D
= (35o + 40o) × 60’ × cos 25o
= 75o × 60’ × cos 25o
= 4078.4 nautical miles
\begin{array}{l}\text{Total distance travelled}\\ =CA+AB\\ =3300+4078.4\\ =7378.4\text{ nautical miles}\\ \\ \text{Average speed}=\frac{\text{Total distance}}{\text{Total time}}\\ =\frac{7378.4}{12.4}\text{ knot}\\ =595.0\text{ knot}\end{array}
Posted in Earth as a Sphere
Diagram 8 shows four points, G, H, I and J on the surface of the Earth. JI is the diameter of the parallel of latitude 50o N. O is the centre of the Earth.
(a) State the location of G.
(b) Calculate the shortest distance, in nautical mile, from I to J measured along the surface of the Earth.
(c) Calculate the shortest distance, in nautical mile, from G to H measured along the common parallel of latitude.
(d) An aeroplane took off from I and flew due south to point P. The average speed of the journey was 800 knots. The time taken for the flight was 5.25 hours.
Calculate the latitude of P.
Location of G = (70o S, 20o W)
Distance of I to J
Distance of G to H
= (20o + 120o) × 60’ × cos 70o
= 140o × 60’ × cos 70o
\begin{array}{l}\text{Average speed}=\frac{\text{Total distance travelled}}{\text{Total time taken}}\\ 800=\frac{x}{5.25}\\ x=4200\text{ nautical miles}\\ I\text{ to }P=4200\\ \\ \text{Difference between parallel}=y\\ y×60=4200\\ y={70}^{o}\\ \\ \text{Thus, latitude of }P\\ ={70}^{o}-{50}^{o}\\ ={20}^{o}S\end{array}
Diagram 7 in the answer space shows the locations of points J, L and M, which lie on the surface of the earth. O is the centre of the earth. The longitude of M is 30o W. K is another point on the surface of the earth such that KJ is the diameter of the common parallel of latitude 45o S.
(a)(i) Mark and label point K on Diagram 7 in the answer space.
(ii) Hence, state the longitude of point K.
(b) L lies due north of M and the shortest distance from M to L measured along the surface of the earth is 7500 nautical miles.
Calculate the latitude of L.
(c) Calculate the distance, in nautical mile, from K due east to M measured along the common parallel of latitude.
(d) An aeroplane took off from K and flew due east to M along the common parallel of latitude. The average speed of the aeroplane for the flight was 750 knots.
Calculate the total time, in hour, taken for the whole flight.
Longitude of point K = 130oW
\begin{array}{l}\angle \text{ }LOM×60=7500\\ \angle \text{ }LOM=\frac{7500}{60}\\ \angle \text{ }LOM={125}^{o}\\ \\ \text{Latitude of }L={125}^{o}-{45}^{o}\\ ={80}^{o}N\end{array}
KM = (130o – 30o) × 60 × cos 45o
\begin{array}{l}\text{Time}=\frac{\text{Distance}}{\text{speed}}\\ \text{}=\frac{4242.64}{750}\\ \text{}=5.66\text{ hours}\end{array}
You are not allowed to use graph paper to answer this question.
(a) Diagram 7.1 shows a solid prism with a rectangular base JKLM on a horizontal plane. The surface MQRSTL is the uniform cross section of the prism. Triangle STU and PQR are horizontal planes. Edges PJ, RS and UK are vertical. PQ = TU = 3 cm.
Draw to full scale, the elevation of the solid on a vertical plane parallel to KL as viewed from X.
(b) Another solid cuboid with rectangle base JMDE is combined to the prism in Diagram 7.1 at the vertical plane JMQP. The composite solid is as shown in Diagram 7.2. The base EJKLMD lies on a horizontal plane.
Draw to full scale,
(i) the elevation of the composite solid on a vertical plane parallel to EJK as viewed from Y.
(ii) the plan of the composite solid.
Posted in Plans and Elevations
(a) Diagram 6.1 shows two solid right prisms joined at the vertical plane EGLK. The planes JKL and MNP are the uniform cross sections of the prism HEGLJK and prism EFGMNP respectively. The base EFGH is a rectangle which lies on a horizontal plane. Edges HJ and EK are vertical.
Draw, to full scale, the plan of the composite solid.
(b) Another solid half cylinder with a diameter of 4 cm is joined to the prism in Diagram 9.1 at the vertical plane EFQR. The composite solid is as shown in Diagram 6.2. The base HETFG lies on a horizontal plane.
(i) the elevation of the composite solid on a vertical plane parallel to HE as viewed from X.
(ii) the elevation of the composite solid on a vertical plane parallel to EF as viewed from Y.
(a) Diagram 5.1 shows a pyramid with rectangular base ABCD on a horizontal plane. Vertex E is vertically above C. Triangles BCE and DCE are vertical planes. Triangles ABE and ADE are inclined planes.
(b) Another solid cuboid with rectangle base BLKC is joined to the pyramid in Diagram 5.1 at the vertical plane BCFM. The composite solid is as shown in Diagram 5.2. The base ABLKCD lies on a horizontal plane
(i) the elevation of the composite solid on a vertical plane parallel to ABL as viewed from X.
(ii) the elevation of the composite solid on a vertical plane parallel to LK as viewed from Y.
Question 10 (12 marks):
Diagram shows a histogram which represents the mass, in kg, for a group of 100 students.
(a) Based on the Diagram, complete the Table in the answer space.
(b) Calculate the estimated mean mass of a student.
(c) For this part of the question, use graph paper. You may use a flexible curve ruler.
Using a scale of 2 cm to 10 kg on the horizontal axis and 2 cm to 10 students on the vertical axis, draw an ogive for the data.
(d) Based on the ogive drawn in 10(c), state the third quartile.
\begin{array}{l}\text{Estimated mean mass}\\ =\frac{8110}{100}\\ =81.1\text{ kg}\end{array}
= 75th student
Posted in Statistics III
The data in the Diagram shows the mass, in g, of 30 strawberries plucked by a tourist from a farm.
(a) Based on the Diagram, complete Table 3 in the answer space.
(b) Based on Table, calculate the estimated mean mass of a strawberry.
(c) For this part of the question, use graph paper.
By using the scale of 2 cm to 10 g on the horizontal axis and 2 cm to 1 strawberry on the vertical axis, draw a histogram for the data.
(d) Based on the histogram drawn in 14(c), state the number of strawberries with the mass of more than 50 g.
\begin{array}{l}\text{Estimated mean mass}\\ =\frac{\text{Total }\left[\text{midpoint}×\text{frequency}\right]}{\text{Total frequencies}}\\ =\frac{\begin{array}{l}2\left(14.5\right)+5\left(24.5\right)+10\left(34.5\right)+8\left(44.5\right)+\\ \text{3}\left(54.5\right)+2\left(64.5\right)\end{array}}{2+5+10+8+3+2}\\ =\frac{1145}{30}\\ =38.17\end{array}
Number of strawberries with the mass more than 50 g
Diagram shows the marks obtained by a group of 36 students in a Mathematics test.
(a) Based on the data in Diagram, complete Table in the answer space.
(b) Based on the Table, calculate the estimated mean mark of a student.
By using the scale of 2 cm to 5 marks on the horizontal axis and 2 cm to 1 student on the vertical axis, draw a frequency polygon for the data.
(d) Based on the frequency polygon in 8(c), state the number of students who obtained more than 40 marks.
\begin{array}{l}\text{Estimated mean mark}\\ =\frac{\text{Total }\left[\text{Frequency}×\text{midpoint}\right]}{\text{Total frequency}}\\ =\frac{\begin{array}{l}2\left(28\right)+5\left(33\right)+7\left(38\right)+10\left(43\right)+\\ \text{ }8\left(48\right)+4\left(53\right)\end{array}}{2+5+7+10+8+4}\\ =\frac{1513}{36}\\ =42.03\end{array}
Number of students who obtained more than 40 marks
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Home : Support : Online Help : Programming : Logic : Boolean : verify : plot3d
verify(P, Q, plot3d)
verify(P, Q, 'plot3d'(opts))
anything, assumed to be PLOT3D data structures
equation(s) of the form option=value where option is one of curves, display, display_floats, feature_floats, features, grid, isosurface, mesh, points, polygons, text, or traversal; specify options for the comparison
The verify(P, Q, plot3d) and verify(P, Q, 'plot3d'(opts)) calling sequences return true if the two PLOT3D data structures compare using the default comparator.
The default comparator determines that two PLOT3D data structures are equal if:
Given two functions, they must have the same name, the same number of arguments, and their arguments must each compare using the default comparator,
Given either two lists or Arrays, their corresponding entries must be equal,
Given two ranges, their endpoints must be compare using the default comparator,
0
All other objects are compared by using evalb.
If either P or Q is not a PLOT3D data structure then false is returned.
The data structures within a 3-D plot object are separated into two classes, display objects, and feature objects. The data structures, which fall into the first class are CURVES, GRID, ISOSURFACE, MESH, POINTS, POLYGONS, TEXT, and VIEW. All other data structures are considered to be feature data structures.
At each stage, display and feature objects are compared separately, and any features are always compared using the default comparator with any options given by the features option.
By default, the display objects of a 3-D plot are compared. This can be turned off by setting this option to false.
display_floats = nonnegative or verification
0
Common verifications are float(10) or neighborhood.
feature_floats = nonnegative or verification
0
By default, the feature objects of a 3-D plot are compared. This can be turned off by setting this option to false.
grid = verification
If this option is given, the set of all GRID objects are compared using this verification.
isosurface = verification
If this option is given, the set of all ISOSURFACE objects are compared using this verification.
mesh = verification
If this option is given, the set of all MESH objects are compared using this verification.
If this option is given, the set of all POLYGONS objects are compared using this verification. Otherwise, the default comparator is used to compare POLYGONS objects.
By default, this option is set to ordered, in which case objects are compared in the same order in both plots. If this option is set to unordered, then it is only necessary that a match be found for each object in the other PLOT3D data structure. If this option is set to one_to_one, then a one-to-one correspondence must be found between data structures in both objects.
The option one_to_one is significantly more expensive than either other test. The option ordered is the least expensive, but is subject to ordering problems, which affect the structure, but not the final 3-D plot.
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Web crawler - CodeDocs
A Web crawler, sometimes called a spider or spiderbot and often shortened to crawler, is an Internet bot that systematically browses the World Wide Web, typically operated by search engines for the purpose of Web indexing (web spidering).[1]
Crawlers consume resources on visited systems and often visit sites without approval. Issues of schedule, load, and "politeness" come into play when large collections of pages are accessed. Mechanisms exist for public sites not wishing to be crawled to make this known to the crawling agent. For example, including a robots.txt file can request bots to index only parts of a website, or nothing at all.
A Web crawler starts with a list of URLs to visit, called the seeds. As the crawler visits these URLs, it identifies all the hyperlinks in the pages and adds them to the list of URLs to visit, called the crawl frontier. URLs from the frontier are recursively visited according to a set of policies. If the crawler is performing archiving of websites (or web archiving), it copies and saves the information as it goes. The archives are usually stored in such a way they can be viewed, read and navigated as if they were on the live web, but are preserved as 'snapshots'.[5]
Semantic focused crawler
{\displaystyle F_{p}(t)={\begin{cases}1&{\rm {if}}~p~{\rm {~is~equal~to~the~local~copy~at~time}}~t\\0&{\rm {otherwise}}\end{cases}}}
{\displaystyle A_{p}(t)={\begin{cases}0&{\rm {if}}~p~{\rm {~is~not~modified~at~time}}~t\\t-{\rm {modification~time~of}}~p&{\rm {otherwise}}\end{cases}}}
Further information: List of search engine software, each of which typically has an associated web crawler
Historical web crawlers
Frontera is web crawling framework implementing crawl frontier component and providing scalability primitives for web crawler applications.
Trandoshan, a free, open source distributed web-crawler designed for the deep-web.
^ Kobayashi, M. & Takeda, K. (2000). "Information retrieval on the web". ACM Computing Surveys. 32 (2): 144–173. CiteSeerX . doi:10.1145/358923.358934. S2CID 3710903.
^ Edwards, J., McCurley, K. S., and Tomlin, J. A. (2001). "An adaptive model for optimizing performance of an incremental web crawler". Proceedings of the tenth international conference on World Wide Web - WWW '01. In Proceedings of the Tenth Conference on World Wide Web. pp. 106–113. CiteSeerX . doi:10.1145/371920.371960. ISBN 978-1581133486. S2CID 10316730. CS1 maint: multiple names: authors list (link)
^ Steve Lawrence; C. Lee Giles (8 July 1999). "Accessibility of information on the web". Nature. 400 (6740): 107–9. Bibcode:1999Natur.400..107L. doi:. PMID 10428673. S2CID 4347646.
^ Paolo Boldi; Bruno Codenotti; Massimo Santini; Sebastiano Vigna (2004). "UbiCrawler: a scalable fully distributed Web crawler" (PDF). Software: Practice and Experience. 34 (8): 711–726. CiteSeerX . doi:10.1002/spe.587. S2CID 325714. Archived from the original (PDF) on 20 March 2009. Retrieved 23 March 2009.
^ Paolo Boldi; Massimo Santini; Sebastiano Vigna (2004). "Do Your Worst to Make the Best: Paradoxical Effects in PageRank Incremental Computations" (PDF). Algorithms and Models for the Web-Graph. Lecture Notes in Computer Science. 3243. pp. 168–180. doi:10.1007/978-3-540-30216-2_14. ISBN 978-3-540-23427-2. Retrieved 23 March 2009.
^ Pant, Gautam; Srinivasan, Padmini; Menczer, Filippo (2004). "Crawling the Web" (PDF). In Levene, Mark; Poulovassilis, Alexandra (eds.). Web Dynamics: Adapting to Change in Content, Size, Topology and Use. Springer. pp. 153–178. ISBN 978-3-540-40676-1.
^ Cothey, Viv (2004). "Web-crawling reliability" (PDF). Journal of the American Society for Information Science and Technology. 55 (14): 1228–1238. CiteSeerX . doi:10.1002/asi.20078.
^ Menczer, F. (1997). ARACHNID: Adaptive Retrieval Agents Choosing Heuristic Neighborhoods for Information Discovery. In D. Fisher, ed., Machine Learning: Proceedings of the 14th International Conference (ICML97). Morgan Kaufmann
^ Menczer, F. and Belew, R.K. (1998). Adaptive Information Agents in Distributed Textual Environments. In K. Sycara and M. Wooldridge (eds.) Proc. 2nd Intl. Conf. on Autonomous Agents (Agents '98). ACM Press
^ Dong, Hai; Hussain, Farookh Khadeer; Chang, Elizabeth (2009). "State of the Art in Semantic Focused Crawlers". Computational Science and Its Applications – ICCSA 2009. Lecture Notes in Computer Science. 5593. pp. 910–924. doi:10.1007/978-3-642-02457-3_74. hdl:20.500.11937/48288. ISBN 978-3-642-02456-6.
^ a b E. G. Coffman Jr; Zhen Liu; Richard R. Weber (1998). "Optimal robot scheduling for Web search engines". Journal of Scheduling. 1 (1): 15–29. CiteSeerX . doi:10.1002/(SICI)1099-1425(199806)1:1<15::AID-JOS3>3.0.CO;2-K.
^ a b Junghoo Cho; Hector Garcia-Molina (2003). "Estimating frequency of change". ACM Transactions on Internet Technology. 3 (3): 256–290. CiteSeerX . doi:10.1145/857166.857170. S2CID 9362566.
^ Koster, M. (1993). Guidelines for robots writers.
^ Heydon, Allan; Najork, Marc (26 June 1999). "Mercator: A Scalable, Extensible Web Crawler" (PDF). Archived from the original (PDF) on 19 February 2006. Retrieved 22 March 2009. Cite journal requires |journal= (help)
^ Michael L Nelson; Herbert Van de Sompel; Xiaoming Liu; Terry L Harrison; Nathan McFarland (24 March 2005). "mod_oai: An Apache Module for Metadata Harvesting": cs/0503069. arXiv:. Bibcode:2005cs........3069N. Cite journal requires |journal= (help)
^ Sun, Yang (25 August 2008). "A COMPREHENSIVE STUDY OF THE REGULATION AND BEHAVIOR OF WEB CRAWLERS. The crawlers or web spiders are software robots that handle trace files and browse hundreds of billions of pages found on the Web. Usually, this is determined by tracking the keywords that make the searches of search engine users, a factor that varies second by second: according to Moz, only 30% of searches performed on search engines like Google, Bing or Yahoo! corresponds generic words and phrases. The remaining 70% are usually random". Retrieved 11 August 2014. Cite journal requires |journal= (help)
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Lewis Symbols and Structures | Introductory Chemistry - Lecture & Lab | Course Hero
As previously mentioned, when a pair of atoms shares one pair of electrons, we call this a single bond. However, a pair of atoms may need to share more than one pair of electrons in order to achieve the requisite octet. A double bond forms when two pairs of electrons are shared between a pair of atoms, as between the carbon and oxygen atoms in CH2O (formaldehyde) and between the two carbon atoms in C2H4 (ethylene): A triple bond forms when three electron pairs are shared by a pair of atoms, as in carbon monoxide (CO) and the cyanide ion (CN–):
Let us determine the Lewis structures of SiH4,
{\text{CHO}}_{2}^{-},
NO+, and OF2 as examples in following this procedure:
\begin{array}{l}\\ \phantom{\rule{0.8em}{0ex}}{\text{SiH}}_{4}\\ \phantom{\rule{0.8em}{0ex}}\text{Si: 4 valence electrons/atom}\times \text{1 atom}=4\\ \underline{+\text{H: 1 valence electron/atom}\times \text{4 atoms}=4}\\ \\ \phantom{\rule{15.95em}{0ex}}=\text{8 valence electrons}\end{array}
For a negative ion, such as
{\text{CHO}}_{2}^{-},
we add the number of valence electrons on the atoms to the number of negative charges on the ion (one electron is gained for each single negative charge):
\begin{array}{l}\\ {\text{CHO}}_{2}^{-}\\ \phantom{\rule{0.48em}{0ex}}\text{C: 4 valence electrons/atom}\times \text{1 atom}=4\\ \phantom{\rule{0.8em}{0ex}}\text{H: 1 valence electron/atom}\times \text{1 atom}=1\\ \phantom{\rule{0.05em}{0ex}}\text{O: 6 valence electrons/atom}\times \text{2 atoms}=12\\ \underline{+\phantom{\rule{6.5em}{0ex}}\text{1 additional electron}=1}\\ \\ \phantom{\rule{15.45em}{0ex}}=\text{18 valence electrons}\end{array}
\begin{array}{l}\\ \\ {\text{NO}}^{+}\\ \text{N: 5 valence electrons/atom}\times \text{1 atom}=5\\ \\ \phantom{\rule{0.4em}{0ex}}\text{O: 6 valence electron/atom}\times \text{1 atom}=6\\ \phantom{\rule{0.35em}{0ex}}\underline{+{-1 electron (positive charge)}\phantom{\rule{1.8em}{0ex}}=-1}\\ \\ \phantom{\rule{15.02em}{0ex}}=\text{10 valence electrons}\end{array}
\begin{array}{l}\\ \phantom{\rule{0.8em}{0ex}}{\text{OF}}_{\text{2}}\\ \phantom{\rule{1.25em}{0ex}}\text{O: 6 valence electrons/atom}\times \text{1 atom}=6\\ \underline{+\text{F: 7 valence electrons/atom}\times \text{2 atoms}=14}\\ \phantom{\rule{16.28em}{0ex}}=\text{20 valence electrons}\end{array}
When several arrangements of atoms are possible, as for
{\text{CHO}}_{2}^{-},
we must use experimental evidence to choose the correct one. In general, the less electronegative elements are more likely to be central atoms. In
{\text{CHO}}_{2}^{-},
the less electronegative carbon atom occupies the central position with the oxygen and hydrogen atoms surrounding it. Other examples include P in POCl3, S in SO2, and Cl in
{\text{ClO}}_{4}^{-}.
An exception is that hydrogen is almost never a central atom. As the most electronegative element, fluorine also cannot be a central atom.
For SiH4,
{\text{CHO}}_{2}^{-},
and NO+, there are no remaining electrons; we already placed all of the electrons determined in Step 1.
{\text{CHO}}_{2}^{-}:
{\text{ICl}}_{4}^{-}.
{\text{NH}}_{4}^{+}
{\text{BF}}_{4}^{-}
{\text{C}}_{2}^{\text{2+}}
{\text{PF}}_{6}^{-}
{\text{SeCl}}_{3}^{+}
{\text{PO}}_{4}^{\text{3-}}
{\text{IC}}_{4}^{-}
{\text{SO}}_{3}^{2-}
5. 7. The Lewis structures are as follows:
{\text{NH}}_{4}^{+}
{\text{BF}}_{4}^{-}
{\text{C}}_{2}^{\text{2+}}
{\text{SeCl}}_{3}^{+}:
\begin{array}{l}\frac{85.7\text{g}}{12.011{\text{g mol}}^{-1}}=7.14\text{mol C}\\ \frac{14.3\text{g}}{1.00794{\text{g mol}}^{-1}}=14.19\text{mol H}\end{array}
This is a ratio of 2 H to 1 C, or an empirical formula of CH2 with a formula mass of approximately 14. As
\frac{42}{14}=3,
the formula is 3 × CH2 or C3H6. The Lewis structure is:
Lab10 Lewis Structures PowerPoint Lecture.pptx
CHEM 1311 • Lamar University
C9_Lec_8_Basic_Concepts_of_Chemical_Bonding.docx
CHEM 9 • Mindanao State University Main Campus
Lecture-7_Concepts-of-Chemical-Bonding.docx
CHEMISTRY 43 • Fr. Saturnino Urios University
Lec10-Chapter3b.pdf
JAVA 123 • Johns Hopkins University
CHEM 104 • University of Regina
Chapter 10 Solutions.pdf
CHEM 101 • Madison Area Technical College, Madison
Lab 1 Prelab Assignment.pdf
Organic Chemistry Lecture Notes.docx
CHEM 18A • Yuba College
Introductory Chemistry.pdf
Lecture 5 Bonding in Chemistry Ionic Bonds.pdf
CHEM 1003 • The University of Western Australia
M03 Quiz - Chapters 9 & 10_ 0AJ-Spring 2020-Introductory Chemistry I.pdf
2016 Lect 2 Chemical Bonding Pt. 1 (2) (1).pptx
CHEM 191 • University of Canterbury
Introductory Chemistry Computer Workshop Practice - Week 5.docx
CHEM 1009 • Western Sydney University
Spring 2021 Chemical Principles II Chp 7 Part 4.pdf
CHEMISTRY 50:160:348 • Rutgers University, Camden
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pH of two solutions A and B are 2 and 5 respectively This means that : (A) Solution A - Science - Acids Bases and Salts - 13063767 | Meritnation.com
pH of two solutions A and B are 2 and 5 respectively. This means that :
(A) Solution A is 3 times more basic than B
(B) Solution A is 3 times less basic than B
(C) Solution A is 1000 times more basic than B
(D) Solution A is 1000 times less basic than B
\phantom{\rule{0ex}{0ex}}From the formula,\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}pH=-\mathrm{log}\left[{H}^{+}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Given pH of solutions A, B as 2 and 5 respectively then,based on above formula\phantom{\rule{0ex}{0ex}}we get\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left[{H}^{+}\right] for solution A ={10}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Where as for Solution B is\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left[{H}^{+}\right] ={10}^{5}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Hence this shows that Solution A is {10}^{3} or 1000 times less basic than Solution B
Option 'D' is correct
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Global Constraint Catalog: Kabstract_interpretation
<< 3.7.2. 3-SAT3.7.4. Acyclic >>
\mathrm{𝚐𝚌𝚍}
\mathrm{𝚙𝚘𝚠𝚎𝚛}
Denotes that abstract interpretation was used for deriving a filtering algorithm for a constraint
C
from a polynomial algorithm describing a checker for a ground instance of
C
. Abstract interpretation [CousotCousot77] executes an algorithm on abstract values in order to deduce some information about that algorithm.
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Algebra Problem on Telescoping Series - Sum: Playing with Square Root - Christopher Boo | Brilliant
Playing with Square Root
\displaystyle \frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+\dots+\frac{1}{100\sqrt{99}+99\sqrt{100}}
\frac{a}{b},
and
b
a+b
This problem is adapted from a past year KMC Contest.
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Written by Sarah Stecher published 1 month ago
When we talk with other teachers, the prevailing themes of our conversations are the desire to teach our students well--and by that we mean deeply, and engagingly--and the time constraint that makes reaching that goal so difficult.
\text{\textquotedblleft}
How can we get students engaged in X?
\text{\textquotedblright}
\text{\textquotedblleft}
How do you have time for X?
\text{\textquotedblright}
The reason that the traditional lecture-style instructional model has persisted so long is because it is so efficient! But efficient at what? It may be efficient at transferring information, or at getting students to mimic worked examples, or at avoiding any dead space during the class period. After all, if the teacher does all of the talking and students don’t have to negotiate for meaning among themselves, then little time is
\text{\textquotedblleft}
\text{\textquotedblright}
. As teachers, we know we can present or summarize ideas more succinctly and more articulately than a student who is exploring it for the first time. When students in my class have a question, they prefer to get an answer from me, rather than from one of their group mates because they say it’s easier to understand. While that's of course a little flattering, I realize that granting their request works against other messages I am trying to convey about students owning the mathematical knowledge and the power of collaboration in moving our thinking forward. The lecture-style instructional model is not efficient at getting students to become problem solvers and flexible thinkers. As a result, I’ve had to make decisions about when I intentionally choose a more inefficient route and when I can incorporate a strategy or tool to save time in my teaching.
To motivate the need for a future lesson that will provide a new strategy or shortcut: sometimes we purposefully do things the harder, longer, less efficient way, so that students will almost be begging for the new learning. As Dan Meyer says, “If math is the aspirin, how do we create the headache?”
To regulate our workload so we can focus our energy on what’s important
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A Mathematical Model of Multi-Hop HF Radio Propagation
A Mathematical Model of Multi-Hop HF Radio Propagation ()
Yaru Chen, Lu Han, Junrun Huang, Yufeng Gui*
It is an eternal topic to study the minimum loss under the maximum transmission distance and the practicality of radio propagation in the navigation of ships. The transmission loss by dividing it to Median medium attenuation and Media attenuation correction factor is discussed. The Longley-Rice Model is introduced and improved; the correlation proportion of our model is work out. Then the reflection correction factor is calculated in the case of silent and turbulent ocean, which comes to the total loss and strength of the first reflection. The signal to noise ratio is explained and the relationship between various angle of incidence and correction factor are explored. According to fixed signal to noise ratio 14, the highest bounce frequency is got.
Longley-Rice, Radio Propagation, Ship, Mathematical Model, Random Waves
Chen, Y. , Han, L. , Huang, J. and Gui, Y. (2018) A Mathematical Model of Multi-Hop HF Radio Propagation. Applied Mathematics, 9, 779-788. doi: 10.4236/am.2018.96054.
Radio waves can travel long distances by multiple reflections off the ionosphere and off the earth with a high frequencies (HF, defined to be 3 - 30 mHz). HF radio waves from the ground whose frequencies are under maximum usable frequency (MUF) travel further and further with each successive hop by the reflections between the earth and the ionosphere again and again. And MUF has something to do with the season, time of day, and solar conditions. There is no reflection or refraction when the frequencies are over MUF. The characteristics of the reflecting surface determine the strength of the reflected wave and how far the signal will ultimately travel while maintaining useful signal integrity. The state of ocean also influences the attenuation of reflections. Ocean turbulence will affect the electromagnetic gradient of seawater, alter the local permittivity and permeability of the ocean, and chang the height and angle of the reflection surface.
We build a mathematical model based on Longley-Rice Model pinciple [1] for this signal reflection off the ocean [2] and then we simulate the three dimensional model of the waves [3] .
For a 100-watt HF constant-carrier signal, below the MUF, from a point source on land, it determines the strength of the first reflection off a turbulent ocean and we can compare it with the strength of a first reflection off a calm ocean [4] to support that additional reflections (2 through n) taking place off calm oceans [5] and working out the maximum number of hops the signal can take before its strength falls below a usable signal-to-noise ratio (SNR) threshold of 10 Db. Last we optimize the angle and find the maximum number of hops [4] [6] .
Last but not least, the highlight of this paper is that the model draws on all the advantages of the previous semi-empirical model and more comprehensive consideration of various types of attenuation in transmission. Meanwhile [7] [8] , according to the scope of the model we amend other factors, establish a highly generalized model of transmission loss and remain much to improve for more factors, which are continuous improvements of the growth mode [9] [10] . The model accurately describes the various types of losses in radio transmission acceptance, with great significance to the communication in the route [11] .
2. The Establishment of a Model
2.1. Radio Propagation Receiver
The Longley-Rice radio wave transmission model is a radio wave transmission model proposed by Longley and Rice, which is a statistical model based on the radio wave propagation theory. And it combines many real-world measurement data, so the model is called semi-empirical prediction mode [7] .
This model has a certain scope of application, so the decay needs to be corrected when the Medium attenuation, the introduction of correction factor. On the other hands, this model investigates radio waves as spherical emissions so we should sum each radio wave. As a result, our outcome should multiply by 4.
2.2. The Establish of Sealed Radio Propagation Receiver
First of all, assuming that the frequency is 20 mHz, the attenuation of the radio wave in the first reflection is calculated by two parts of the medium median value and the medium attenuation correction factor [8] .
The transmission loss of the Longley-Rice model can be roughly divided into three cases: 1) line-of-sight propagation loss:
{d}_{\mathrm{min}}\le d\le {d}_{s}
2) diffraction propagation loss:
{d}_{Ls}\le d\le {d}_{x}
3) scattering propagation loss:
{d}_{x}\le d
The propagation loss of radio waves in a calm sea surface is:
{L}_{b}={L}_{ref}+{L}_{free}
{L}_{ref}\left(d\right)=\left\{\begin{array}{l}\mathrm{max}\left(0,{L}_{be}+{k}_{1}d+{k}_{2}\mathrm{lg}d\right)\text{}{d}_{\mathrm{min}}\le d\le {d}_{x}\\ {L}_{bed}+{m}_{d}d\text{}{d}_{ls}\le d\le {d}_{x}\\ {L}_{bes}+{m}_{s}d\text{}{d}_{x}\le d\end{array}
d is the propagation distance in km;
f is the radio wave frequency in MHz;
dLs is the Smooth ground distance;
dx, where the diffraction loss and scattering loss are equal;
{L}_{be},{L}_{bed},{L}_{bes}
represents the value of propagation loss in line-of-sight, diffraction, and scattering under free space;
k1 and k2 are the propagation loss coefficients;
md and ms are the diffraction and scattering loss coefficients, respectively.
Since this formula is a semi-empirical and the unknown coefficient is complicated, a large amount of data is collected according to the actual situation to be fitted.
L=\left\{\begin{array}{l}0.8677d+93.31\text{}d\le 56\\ 0.6073d+108.9\text{}56\le d\le 136\\ 0.1363d+172.9\text{}d\ge 136\end{array}
Then we come to multi-hop: Known elevation can be based on geometric knowledge to get the relationship between the hopping angle and angle of incidence as shown in Figure 1, we take the ionosphere height h = 100 km and the Earth’s radius R = 6371 km.
d=2R\left(\mathrm{arccos}\left(\frac{R\mathrm{cos}\partial }{h+R}\right)-\partial \right)
2.3. Correction of Attenuation in Medium
1) Since radio waves propagate in the line of sight range, the radio wave propagation mode is mainly diffractive propagation, and the subject requires the study of reflection. Therefore, it is not necessary to discuss the diffraction correction of the sea surface within the viewing distance range.
2) The following picture suggest that when a point light emits signals in all directions, the energy of all the electric waves that can reach ③ only comes from the loss of electric wave propagation in the area ②, and the energy loss of the electric waves emitted in the range ① is not considered. For that case the coefficients need to be corrected with correction ratio which is mean of angle, see Figure 2.
Since the corresponding propagation distance in scattering is infinite, all
Figure 1. Earth and atmosphere.
Figure 2. Relation of angles.
signal losses are taken into account, so there is no need to correct the scattering losses.
3) The calm sea surface reflection coefficient by the law of refraction refraction. Because of the law of refraction refraction:
{n}_{1}{\theta }_{i}={n}_{2}{\theta }_{i}
We calculate the reflection coefficient: with:
{Q}_{1}=L\times 4\text{π}{d}^{2}\times {R}_{1}
{R}_{1}=\frac{1}{2}\left({r}_{1}^{2}+{r}_{2}^{2}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{r}_{1}=\frac{{\lambda }_{2}\mathrm{cos}{\theta }_{i}-{\lambda }_{1}\mathrm{cos}{\theta }_{t}}{{\lambda }_{2}\mathrm{cos}{\theta }_{i}+{\lambda }_{1}\mathrm{cos}{\theta }_{t}}
{r}_{\text{2}}=\frac{{\lambda }_{\text{1}}\mathrm{cos}{\theta }_{i}-{\lambda }_{2}\mathrm{cos}{\theta }_{t}}{{\lambda }_{\text{1}}\mathrm{cos}{\theta }_{i}+{\lambda }_{2}\mathrm{cos}{\theta }_{t}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\lambda }_{i}=\sqrt{\frac{{\mu }_{i}}{{\epsilon }_{i}}}\text{\hspace{0.17em}}\left(i=1,2\right)
The relationship between the reflection coefficient and the incident angle is shown in Figure 3.
Based on the empirical formula in the longley-Rice model, we can deduce that the attenuation of the first reflection at calm sea surface is:
4) Rough turbulent ocean surface reflection coefficient correction
① The establishment of 3D turbulent ocean wave model based on ocean wave spectrum.
Because there is sea slope in turbulent sea compared to the calm sea surface, a three-dimensional turbulent ocean wave model needs to be established to study it.
② Create a wave model
The wave is described by a stationary stochastic process with ergodicity. The wave is viewed as a superposition of waves and swells in a simple cosine wave of infinitely different amplitudes, of varying frequency and of an incipient phase. That is, for the composition of wave propagation direction relative to the wind direction angle. θ is the angle which wave spread by in x. We use the wave spectrum function in reference:
S\left(\rho ,\theta \right)=\frac{8.1\times {10}^{-3}{g}^{2}}{{w}^{5}}\mathrm{exp}\left[-0.74{\left(\frac{g}{uw}\right)}^{4}\right]\frac{2}{\text{π}}{\mathrm{cos}}^{2}(\theta )
Figure 3. The relationship between angle and reflection confficient.
S\left(\rho ,\theta \right)=\frac{8.1\times {10}^{-3}{g}^{2}}{{w}^{5}}\mathrm{exp}\left[-0.74{\left(\frac{g}{uw}\right)}^{4}\right]\frac{\text{8}}{\text{3π}}{\mathrm{cos}}^{2}(\theta )
Create a random wave model shown in Figure 4.
From this model, we get the slope distribution function of each discrete sea surface randomly according to the wave spectrum (Figure 5).
③ Set up the sea coordinate system
Set up the sea coordinate system as shown. Let the origin be located on the sea surface where the study object is located. The X and Y axes are located on the horizontal plane of the coordinate system. The positive direction of Y axis is the position of the detector and the positive direction of Z axis is upwar, see Figure 6.
We change the coordinates, maintaining the Z axis direction unchanged. The coordinate system is rotated clockwise:
\left\{\begin{array}{l}{z}_{u}={S}_{x}\mathrm{cos}\gamma +{S}_{y}\mathrm{sin}\gamma \\ {z}_{v}=-{S}_{x}\mathrm{cos}\gamma +{S}_{y}\mathrm{sin}\gamma \end{array}
Figure 4. Wave mode.
Figure 5. Spectrum.
Roughness correction factor
\xi
\xi \text{=}\frac{\text{π}P\left({z}_{x},{z}_{y}\right)}{\text{4}\mathrm{cos}{\omega }_{i}{\mathrm{cos}}^{4}\beta }
Rough correction factor and the relationship between the incidence angle shown in Figure 7.
Because: The reflection coefficient on the turbulent sea is the reflection coefficient on calm sea surface with rough correction factor.
The reflection coefficient on the turbulent sea is:
{R}_{2}=\frac{\text{π}P\left({z}_{x},{z}_{y}\right){R}_{1}}{4\mathrm{cos}{\omega }_{i}{\mathrm{cos}}^{4}\beta }
Figure 6. Relation of angle.
Figure 7. The relationship between angle of incidence and correction factor.
3. Model Calculation and Result Analysis
3.1. Optimize the Angle and Find the Maximum Number of Hops
As for S/N ratio explanation [5] , the main parameters of the world profile of atmospheric radio noise reported by CCIR-3 22 are the effective noise figure Fa. It is defined as the ratio of the external radio noise power received by the non-directional short vertical antenna to the noise power generated by the heat source at temperature T0 in the unit bandwidth that is the title of the signal to noise ratio [6] :
{F}_{a}=\frac{{P}_{n}}{k{T}_{0}B}
Pn is the noise power received by the non-directional short vertical antenna.
K is boltzmann constant; B is the effective noise bandwidth of the receiver.
W is initial signal power.
{F}_{a}=10\mathrm{lg}\frac{{P}_{n}}{W}
, the critical condition is when the noise power is equal to the signal power，so according to the problem (what is the maximum number of hops the signal can take before its strength falls below a usable signal-to-noise ratio (SNR) threshold of 10 dB):
W{\left(L\times {R}_{\text{1}}\right)}^{n}>{P}_{n}
With W = 100W, we can get the remaining power after the reflection with
{W}_{1}=W\left(L\times {R}_{1}\right),\text{\hspace{0.17em}}{W}_{\text{2}}=W{\left(L\times {R}_{1}\right)}^{\text{2}},\text{\hspace{0.17em}}{W}_{\text{3}}=W{\left(L\times {R}_{1}\right)}^{\text{3}},\cdots ,\text{\hspace{0.17em}}{W}_{n}=W{\left(L\times {R}_{1}\right)}^{n}
In the case of the incident angle is determined, the maximum number of hops is uniquely determined, we make an interval traversal of the angle and get angle of incidence and the maximum number of hops in Figure 8.
And maximum number is 14.
The size of the incident angle can affect the radio wave propagation time in the air and thus the reflection intensity, so we can get the relationship between the first reflection intensity and the incident angle in Figure 9.
3.2. Intensities Comparison of the First Reflections in Calm and Turbulent
The angle is 57˚ from the figure and we can get the first reflection strenghth [8] :
{P}_{1}=W{Q}_{1}=7,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{P}_{2}=W{Q}_{2}=3,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{P}_{1}>{P}_{2}
By using original Longley-Rice formula, the minimum loss under the maximum transmission distance and the practicality of radio propagation in the navigation of ships are discussed. The multiple correction parameters is found, and the loss of radio propagation is calculated, which can be used to estimated loss and save
Figure 8. The relationship between angle of incidence and max-hop.
Figure 9. The relationship between angle of incidence and first strength.
energy. The model accurately describes the various types of losses in radio transmission acceptance, with great significance to the communication in the route. In addition to that, our paper is not only used to the situation above, but widely applies to kinds of fields such as marine radar communication, mountain exploration, etc. In more details, the correction factor we propose can also apply to many problems which have a traditional formula without matching situation.
[1] Li, J.-S. (2014) Study on AIS Sea Wave Propagation Model. Dalian Maritime University, Dalian, 32-45.
[2] Yi, Q.D. (2015) Research on Electromagnetic Wave Propagation model in Sea Area. Hainan University, Hainan.
[3] Fang, H. (2015) Research on Propagation Characteristics and Channel Modeling of Maritime Radio Waves. Hainan University, Hainan, 67-98.
[4] Yu, W.Z., Chi, X. and Ren, J. (2014) Transmission Model of Maritime Mobile Channel Based on ITM. Journal of Electronics, 40, 106-111.
[5] Jin, F. (2015) Radio Tracking Based on the Law of Radio Propagation in the Hilly Areas. Nanjing University of Posts and Telecommunications, Nanjing, 42-51.
[6] Xu C., Qiu, C.-C. and Li, L. (2015) Status and Analysis of Optimum Frequency Selection for Maritime HF Communications. Communications Technology, 1, 17-23.
[7] Zhao, Y.C. (2009) Research and Implementation of Radio Wave Propagation Prediction and Interference Analysis. National University of Defense Technology, Changsha.
[8] Peng, F.F. and Zhou, X.J. (2017) Prediction of Maritime Shortwave Telecommunication Links. Ship Electronic Engineering, 31, 125-127.
[9] Li, Z. (2018) Analysis of Maritime Radio Communication Time Based on Longley-Rice Model. Science and Technology Innovation and Application, 13, 16-17.
[10] Liu, J. (2018) On the Development Trend of Anti-Interference Technology for Ultra-Shortwave Radio Communication. Shandong Industrial Technology, 2, 117-118.
[11] Zhang, X.G. (2018) “Internet + Radio” to Achieve Efficient Radio Supervision. Science and Technology Economics Guide, 26, 192.
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Global Constraint Catalog: Call_differ_from_at_least_k_pos
<< 5.1. abs_value5.3. all_differ_from_at_most_k_pos >>
Inspired by [Frutos97].
\mathrm{𝚊𝚕𝚕}_\mathrm{𝚍𝚒𝚏𝚏𝚎𝚛}_\mathrm{𝚏𝚛𝚘𝚖}_\mathrm{𝚊𝚝}_\mathrm{𝚕𝚎𝚊𝚜𝚝}_𝚔_\mathrm{𝚙𝚘𝚜}\left(𝙺,\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁𝚂}\right)
\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁}
\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚟𝚊𝚛}-\mathrm{𝚍𝚟𝚊𝚛}\right)
𝙺
\mathrm{𝚒𝚗𝚝}
\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁𝚂}
\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚟𝚎𝚌}-\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁}\right)
\mathrm{𝚛𝚎𝚚𝚞𝚒𝚛𝚎𝚍}
\left(\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁},\mathrm{𝚟𝚊𝚛}\right)
|\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁}|\ge 1
|\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁}|\ge 𝙺
𝙺\ge 0
\mathrm{𝚛𝚎𝚚𝚞𝚒𝚛𝚎𝚍}
\left(\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁𝚂},\mathrm{𝚟𝚎𝚌}\right)
\mathrm{𝚜𝚊𝚖𝚎}_\mathrm{𝚜𝚒𝚣𝚎}
\left(\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁𝚂},\mathrm{𝚟𝚎𝚌}\right)
Enforce all pairs of distinct vectors of the
\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁𝚂}
collection to differ from at least
𝙺
\left(2,〈\mathrm{𝚟𝚎𝚌}-〈2,5,2,0〉,\mathrm{𝚟𝚎𝚌}-〈3,6,2,1〉,\mathrm{𝚟𝚎𝚌}-〈3,6,1,0〉〉\right)
\mathrm{𝚊𝚕𝚕}_\mathrm{𝚍𝚒𝚏𝚏𝚎𝚛}_\mathrm{𝚏𝚛𝚘𝚖}_\mathrm{𝚊𝚝}_\mathrm{𝚕𝚎𝚊𝚜𝚝}_𝚔_\mathrm{𝚙𝚘𝚜}
The first and second vectors differ from 3 positions, which is greater than or equal to
𝙺=2
The first and third vectors differ from 3 positions, which is greater than or equal to
𝙺=2
The second and third vectors differ from 2 positions, which is greater than or equal to
𝙺=2
𝙺>0
|\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁𝚂}|>1
\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁𝚂}
\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁𝚂}.\mathrm{𝚟𝚎𝚌}
\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁𝚂}
\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁𝚂}.\mathrm{𝚟𝚎𝚌}
\mathrm{𝚊𝚕𝚕}_\mathrm{𝚍𝚒𝚏𝚏𝚎𝚛}_\mathrm{𝚏𝚛𝚘𝚖}_\mathrm{𝚎𝚡𝚊𝚌𝚝𝚕𝚢}_𝚔_\mathrm{𝚙𝚘𝚜}
\ge
𝙺
=
𝙺
\mathrm{𝚍𝚒𝚏𝚏𝚎𝚛}_\mathrm{𝚏𝚛𝚘𝚖}_\mathrm{𝚊𝚝}_\mathrm{𝚕𝚎𝚊𝚜𝚝}_𝚔_\mathrm{𝚙𝚘𝚜}
\mathrm{𝚍𝚒𝚏𝚏𝚎𝚛}_\mathrm{𝚏𝚛𝚘𝚖}_\mathrm{𝚊𝚝}_\mathrm{𝚕𝚎𝚊𝚜𝚝}_𝚔_\mathrm{𝚙𝚘𝚜}
application area: bioinformatics.
characteristic of a constraint: disequality, vector.
constraint type: system of constraints, decomposition.
final graph structure: no loop, symmetric.
\mathrm{𝚊𝚕𝚕}_\mathrm{𝚍𝚒𝚏𝚏𝚎𝚛}_\mathrm{𝚏𝚛𝚘𝚖}_\mathrm{𝚊𝚝}_\mathrm{𝚕𝚎𝚊𝚜𝚝}_𝚔_\mathrm{𝚙𝚘𝚜}\left(𝙺,\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁𝚂}\right)
𝙺\le |\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁𝚂}|
\mathrm{𝚊𝚝𝚕𝚎𝚊𝚜𝚝}_\mathrm{𝚗𝚟𝚎𝚌𝚝𝚘𝚛}
\left(\mathrm{𝙽𝚅𝙴𝙲},\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁𝚂}\right)
\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁𝚂}
\mathrm{𝐶𝐿𝐼𝑄𝑈𝐸}
\left(\ne \right)↦\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚟𝚎𝚌𝚝𝚘𝚛𝚜}\mathtt{1},\mathrm{𝚟𝚎𝚌𝚝𝚘𝚛𝚜}\mathtt{2}\right)
\mathrm{𝚍𝚒𝚏𝚏𝚎𝚛}_\mathrm{𝚏𝚛𝚘𝚖}_\mathrm{𝚊𝚝}_\mathrm{𝚕𝚎𝚊𝚜𝚝}_𝚔_\mathrm{𝚙𝚘𝚜}
\left(𝙺,\mathrm{𝚟𝚎𝚌𝚝𝚘𝚛𝚜}\mathtt{1}.\mathrm{𝚟𝚎𝚌},\mathrm{𝚟𝚎𝚌𝚝𝚘𝚛𝚜}\mathtt{2}.\mathrm{𝚟𝚎𝚌}\right)
\mathrm{𝐍𝐀𝐑𝐂}
=|\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁𝚂}|*|\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁𝚂}|-|\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁𝚂}|
•
\mathrm{𝙽𝙾}_\mathrm{𝙻𝙾𝙾𝙿}
•
\mathrm{𝚂𝚈𝙼𝙼𝙴𝚃𝚁𝙸𝙲}
The Arc constraint(s) slot uses the
\mathrm{𝚍𝚒𝚏𝚏𝚎𝚛}_\mathrm{𝚏𝚛𝚘𝚖}_\mathrm{𝚊𝚝}_\mathrm{𝚕𝚎𝚊𝚜𝚝}_𝚔_\mathrm{𝚙𝚘𝚜}
constraint defined in this catalogue.
Parts (A) and (B) of Figure 5.2.1 respectively show the initial and final graph associated with the Example slot. Since we use the
\mathrm{𝐍𝐀𝐑𝐂}
graph property, the arcs of the final graph are stressed in bold. The previous constraint holds since exactly
3·\left(3-1\right)=6
arc constraints hold.
Figure 5.2.1. Initial and final graph of the
\mathrm{𝚊𝚕𝚕}_\mathrm{𝚍𝚒𝚏𝚏𝚎𝚛}_\mathrm{𝚏𝚛𝚘𝚖}_\mathrm{𝚊𝚝}_\mathrm{𝚕𝚎𝚊𝚜𝚝}_𝚔_\mathrm{𝚙𝚘𝚜}
\mathrm{𝐶𝐿𝐼𝑄𝑈𝐸}\left(\ne \right)
arc generator on the items of the
\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁𝚂}
collection, the expression
|\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁𝚂}|·|\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁𝚂}|-|\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁𝚂}|
corresponds to the maximum number of arcs of the final graph. Therefore we can rewrite the graph property
\mathrm{𝐍𝐀𝐑𝐂}
=
|\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁𝚂}|·|\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁𝚂}|-|\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁𝚂}|
\mathrm{𝐍𝐀𝐑𝐂}
\ge
|\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁𝚂}|·|\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁𝚂}|-|\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁𝚂}|
. This leads to simplify
\underline{\overline{\mathrm{𝐍𝐀𝐑𝐂}}}
\overline{\mathrm{𝐍𝐀𝐑𝐂}}
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2.2.2. Compound data types >>
We provide the following basic data types:
\mathrm{𝚊𝚝𝚘𝚖}
corresponds to an atom. Predefined atoms are
\mathrm{𝙼𝙸𝙽𝙸𝙽𝚃}
\mathrm{𝙼𝙰𝚇𝙸𝙽𝚃}
, which respectively correspond to the smallest and to the largest integer.
\mathrm{𝚒𝚗𝚝}
corresponds to an integer value.
\mathrm{𝚍𝚟𝚊𝚛}
corresponds to a domain variable. A domain variable
V
is a variable that will be assigned an integer value taken from an initial finite set of integer values denoted by
\mathrm{𝑑𝑜𝑚}\left(V\right)
\underline{V}
\overline{V}
respectively denote the minimum and the maximum values of
\mathrm{𝑑𝑜𝑚}\left(V\right)
\mathrm{𝚏𝚍𝚟𝚊𝚛}
corresponds to a possibly unbounded domain variable. A possibly unbounded domain variable is a variable that will be assigned an integer value from an initial finite set of integer values denoted by
\mathrm{𝑑𝑜𝑚}\left(V\right)
or from interval minus infinity, plus infinity. This type is required for declaring the domain of a variable. It is also required by some systems in the context of specific constraints like arithmetic or
\mathrm{𝚎𝚕𝚎𝚖𝚎𝚗𝚝}
\mathrm{𝚜𝚒𝚗𝚝}
corresponds to a finite set of integer values.
\mathrm{𝚜𝚟𝚊𝚛}
corresponds to a set variable. A set variable
V
is a variable that will be assigned to a finite set of integer values. Its lower bound
\underline{V}
denotes the set of integer values that for sure belong to
V
, while its upper bound
\overline{V}
denotes the set of integer values that may belong to
V
\mathrm{𝑑𝑜𝑚}\left(V\right)=\left\{{𝐯}_{\mathbf{1}},\cdots ,{𝐯}_{𝐧},{v}_{n+1},\cdots ,{v}_{m}\right\}
is a shortcut for combining the lower and upper bounds of
V
in a single notation:
Bold values designate those values that only belong to
\underline{V}
Plain values indicate those values that belong to
\overline{V}
and not to
\underline{V}
\mathrm{𝚖𝚒𝚗𝚝}
corresponds to a multiset of integer values.
\mathrm{𝚖𝚟𝚊𝚛}
corresponds to a multiset variable. A multiset variable is a variable that will be assigned to a multiset of integer values.
\mathrm{𝚛𝚎𝚊𝚕}
corresponds to a real number.
\mathrm{𝚛𝚟𝚊𝚛}
corresponds to a real variable. A real variable is a variable that will be assigned a real number taken from an initial finite set of intervals. A real number is usually represented by an interval of two floating point numbers.
Beside domain, set, multiset and float variables we have not yet introduced graph variables [Dooms06]. A graph variable is currently simulated by using one set variable for each vertex of the graph (see the third example of type declaration of 2.2.2).
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Implicit Differentiation - Related Rates Practice Problems Online | Brilliant
In a fairy tale, a wizard rides a cloud which is moving to the right at a speed of
15\text{ m/s.}
The wizard throws a ball vertically upward with a speed of
3\text{ m/s}
relative to the cloud. When the cloud has moved to the right
30 \text{ m}
following the throw, what is the velocity of the ball as seen by people on the ground?
Assume that air resistance is negligible and ignore the effect of gravity.
3
\displaystyle{\frac{\sqrt{936}}{4}}
\displaystyle{\frac{\sqrt{936}}{2}}
9
As shown in the above diagram, a plane taxis off a runway at a speed of
a \text{ m/s} = 42 \text{ m/s}
in order to take off. After taking off, the plane continues flying at
42 \text{ m/s}
, with vertical speed
b \text{ m/s} = 24 \text{ m/s}.
What is the horizontal speed of the plane
5
seconds after take-off?
\displaystyle{\frac{\sqrt{32580}}{7}}
\displaystyle{\frac{\sqrt{29700}}{7}}
\displaystyle{\frac{\sqrt{29700}}{5}}
\displaystyle{\frac{\sqrt{32580}}{5}}
A
B
start off at the point
(0,0)
at the same time and walk east and north at
10\text{ m/min}
6\text{ m/min},
respectively. What is the rate at which their distance increases after
3\text{ min}?
\frac{408}{\sqrt{1214}}
\frac{408}{\sqrt{1234}}
\frac{408}{\sqrt{1229}}
\frac{408}{\sqrt{1224}}
As shown in the above digram, a thick layer of snow accumulates on a vertically symmetrical peaked roof, which causes the roof to slowly collapse downward under the weight of the snow. The length of the peaked roof is
x=28 \text{ m}
on each side and the length of the base of the roof is
2a.
If the collapse of the roof is such that
a
increases at a rate of
5
meters per hour, how fast is the height
b
of the roof collapsing when
a=18?
\displaystyle{-\frac{90}{\sqrt{460}}}
m/hour
\displaystyle{-\frac{95}{\sqrt{460}}}
\displaystyle{-\frac{90}{\sqrt{470}}}
\displaystyle{-\frac{95}{\sqrt{470}}}
An aircraft is flying 8000 feet above the ground. It is moving due east at a velocity of 900 feet per second and is located due east of an observation tower. At a certain moment, the angle of elevation is
\frac{ \pi}{4}
. At that time, what is the rate of increase of the angle of elevation (in radians per second)?
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December 2021 Universality for Langevin-like spin glass dynamics
Amir Dembo, Eyal Lubetzky, Ofer Zeitouni
Amir Dembo,1 Eyal Lubetzky,2 Ofer Zeitouni3
1Mathematics Department and Statistics Department, Stanford University
2Courant Institute of Mathematical Sciences, New York University
3Department of Mathematics, Weizmann Institute of Science
We study dynamics for asymmetric spin glass models, proposed by Hertz et al. and Sompolinsky et al. in the 1980’s in the context of neural networks: particles evolve via a modified Langevin dynamics for the Sherrington–Kirkpatrick model with soft spins, whereby the disorder is i.i.d. standard Gaussian rather than symmetric. Ben Arous and Guionnet (Probab. Theory Related Fields 102 (1995) 455–509), followed by Guionnet (Probab. Theory Related Fields 109 (1997) 183–215), proved for Gaussian interactions that as the number of particles grows, the short-term empirical law of this dynamics converges a.s. to a nonrandom law
{\mathit{\mu }}_{\star }
of a “self-consistent single spin dynamics,” as predicted by physicists. Here we obtain universality of this fact: For asymmetric disorder given by i.i.d. variables of zero mean, unit variance and exponential or better tail decay, at every temperature, the empirical law of sample paths of the Langevin-like dynamics in a fixed time interval has the same a.s. limit
{\mathit{\mu }}_{\star }
A.D. was supported in part by NSF grant DMS- 1954337 and E.L. was supported in part by NSF Grant DMS-1812095. This research was further supported in part by BSF Grant 2018088.
We thank G. Ben Arous and A. Guionnet for a valuable feedback on our preliminary draft and for pointing our attention to the references [4, 15, 17, 26].
Amir Dembo. Eyal Lubetzky. Ofer Zeitouni. "Universality for Langevin-like spin glass dynamics." Ann. Appl. Probab. 31 (6) 2864 - 2880, December 2021. https://doi.org/10.1214/21-AAP1665
Received: 1 January 2020; Revised: 1 August 2020; Published: December 2021
Secondary: 60F10 , 60H10 , 82C31 , 82C44
Keywords: Interacting random processes , Langevin dynamics , SDEs , Universality
Amir Dembo, Eyal Lubetzky, Ofer Zeitouni "Universality for Langevin-like spin glass dynamics," The Annals of Applied Probability, Ann. Appl. Probab. 31(6), 2864-2880, (December 2021)
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Global Constraint Catalog: Copen_alldifferent
<< 5.293. nvisible_from_start5.295. open_among >>
[HoeveRegin06]
\mathrm{𝚘𝚙𝚎𝚗}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}\left(𝚂,\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\right)
\mathrm{𝚘𝚙𝚎𝚗}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏}
\mathrm{𝚘𝚙𝚎𝚗}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚜𝚝𝚒𝚗𝚌𝚝}
\mathrm{𝚘𝚙𝚎𝚗}_\mathrm{𝚍𝚒𝚜𝚝𝚒𝚗𝚌𝚝}
𝚂
\mathrm{𝚜𝚟𝚊𝚛}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚟𝚊𝚛}-\mathrm{𝚍𝚟𝚊𝚛}\right)
𝚂\ge 1
𝚂\le |\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}|
\mathrm{𝚛𝚎𝚚𝚞𝚒𝚛𝚎𝚍}
\left(\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂},\mathrm{𝚟𝚊𝚛}\right)
𝒱
be the variables of the collection
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
for which the corresponding position belongs to the set
𝚂
. Positions are numbered from 1. Enforce all variables of
𝒱
to take distinct values.
\left(\left\{2,3,4\right\},〈9,1,9,3〉\right)
\mathrm{𝚘𝚙𝚎𝚗}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}
constraint holds since the last three (i.e.,
𝚂=\left\{2,3,4\right\}
) values of the collection
〈9,1,9,3〉
|\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}|>2
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}.\mathrm{𝚟𝚊𝚛}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}.\mathrm{𝚟𝚊𝚛}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
In their article [HoeveRegin06], W.-J. van Hoeve and J.-C. Régin motivate the
\mathrm{𝚘𝚙𝚎𝚗}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}
constraint by the following scheduling problem. Consider a set of activities (where each activity has a fixed duration 1 and a start variable) that can be processed on two factory lines such that all the activities that will be processed on a given line must be pairwise distinct. This can be modelled by using one
\mathrm{𝚘𝚙𝚎𝚗}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}
constraint for each line, involving all the start variables as well as a set variable whose final value specifies the set of activities assigned to that specific factory line.
Note that this can also be directly modelled by a single
\mathrm{𝚍𝚒𝚏𝚏𝚗}
constraint. This is done by introducing an assignment variable for each activity. The initial domain of each assignment variable consists of two values that respectively correspond to the two factory lines.
A slight adaptation of the flow model that handles the original
\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}
constraint [Regin96] is described in [HoeveRegin06]. The rightmost part of Figure 3.7.28 illustrates this flow model.
\mathrm{𝚜𝚒𝚣𝚎}_\mathrm{𝚖𝚊𝚡}_\mathrm{𝚜𝚎𝚚}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}
\mathrm{𝚜𝚒𝚣𝚎}_\mathrm{𝚖𝚊𝚡}_\mathrm{𝚜𝚝𝚊𝚛𝚝𝚒𝚗𝚐}_\mathrm{𝚜𝚎𝚚}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}
(all different,disequality).
\mathrm{𝚘𝚙𝚎𝚗}_\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}
(control the number of occurrence of each active valueAn active value corresponds to a value occuring at a position mentionned in the set
𝚂
. with a counter variable),
\mathrm{𝚘𝚙𝚎𝚗}_\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}_\mathrm{𝚕𝚘𝚠}_\mathrm{𝚞𝚙}
(control the number of occurrence of each active value with an interval).
\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}
\mathrm{𝚒𝚗}_\mathrm{𝚜𝚎𝚝}
constraint type: open constraint, soft constraint, value constraint.
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
\mathrm{𝐶𝐿𝐼𝑄𝑈𝐸}
↦\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}\mathtt{1},\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}\mathtt{2}\right)
•\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}\mathtt{1}.\mathrm{𝚟𝚊𝚛}=\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}\mathtt{2}.\mathrm{𝚟𝚊𝚛}
•
\mathrm{𝚒𝚗}_\mathrm{𝚜𝚎𝚝}
\left(\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}\mathtt{1}.\mathrm{𝚔𝚎𝚢},𝚂\right)
•
\mathrm{𝚒𝚗}_\mathrm{𝚜𝚎𝚝}
\left(\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}\mathtt{2}.\mathrm{𝚔𝚎𝚢},𝚂\right)
\mathrm{𝐌𝐀𝐗}_\mathrm{𝐍𝐒𝐂𝐂}
\le 1
\mathrm{𝙾𝙽𝙴}_\mathrm{𝚂𝚄𝙲𝙲}
We generate a clique with an equality constraint between each pair of vertices (including a vertex and itself) and state that the size of the largest strongly connected component should not exceed one. Variables for which the corresponding position does not belong to the set
𝚂
are removed from the final graph by the second and third conditions of the arc-constraint.
\mathrm{𝐌𝐀𝐗}_\mathrm{𝐍𝐒𝐂𝐂}
graph property we show one of the largest strongly connected components of the final graph. The
\mathrm{𝚘𝚙𝚎𝚗}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}
holds since all the strongly connected components have at most one vertex: a value is used at most once.
\mathrm{𝚘𝚙𝚎𝚗}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}
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From first principles, expand
{\left(x+y\right)}^{5}
by the Binomial theorem.
The Binomial theorem:
{\left(x+y\right)}^{n}=\underset{k=0}{\overset{n}{∑}}\left(\genfrac{}{}{0}{}{n}{k}\right){x}^{n-k} {y}^{k}
\left(\genfrac{}{}{0}{}{n}{k}\right)=\frac{n!}{k! \left(n-k\right)!}
is the binomial coefficient, and
n! =n\cdot \left(n-1\right)\cdot ⋯ \cdot 1
is read "n factorial." By definition, 0!=1.
{\left(x+y\right)}^{5}
=\underset{k=0}{\overset{5}{∑}}\left(\genfrac{}{}{0}{}{5}{k}\right) {x}^{5-k} {y}^{k}
=\left(\genfrac{}{}{0}{}{5}{0}\right){x}^{5}{y}^{0}+\left(\genfrac{}{}{0}{}{5}{1}\right){x}^{4}{y}^{1}+\left(\genfrac{}{}{0}{}{5}{2}\right){x}^{3}{y}^{2}+\left(\genfrac{}{}{0}{}{5}{3}\right){x}^{2}{y}^{3}+\left(\genfrac{}{}{0}{}{5}{4}\right){x}^{1}{y}^{4}+\left(\genfrac{}{}{0}{}{5}{5}\right){x}^{0}{y}^{5}
=\frac{5!}{0! 5!}{x}^{5}+\frac{5!}{1! 4!}{x}^{4}y+\frac{5!}{2! 3!}{x}^{3}{y}^{2}+\frac{5!}{3! 2!}{x}^{2}{y}^{3}+\frac{5!}{4! 1!}{\mathrm{xy}}^{4}+\frac{5!}{5! 0!}{y}^{5}
={x}^{5}+5{x}^{4}y+10{x}^{3}{y}^{2}+10{x}^{2}{y}^{3}+5x{y}^{4}+{y}^{5}
Expression palette: Summation template
\underset{\textcolor[rgb]{0.784313725490196,0,0.784313725490196}{i}=\textcolor[rgb]{0.784313725490196,0,0.784313725490196}{k}}{\overset{\textcolor[rgb]{0.784313725490196,0,0.784313725490196}{n}}{∑}}\textcolor[rgb]{0,0.627450980392157,0.313725490196078}{f}
and binomial coefficient
\left(\genfrac{}{}{0}{}{\textcolor[rgb]{0,0.627450980392157,0.313725490196078}{a}}{\textcolor[rgb]{0.784313725490196,0,0.784313725490196}{b}}\right)
{\left(x+y\right)}^{5}=\underset{k=0}{\overset{5}{∑}}\left(\genfrac{}{}{0}{}{5}{k}\right) {x}^{5-k} {y}^{k}={x}^{5}+5{x}^{4}y+10{x}^{3}{y}^{2}+10{x}^{2}{y}^{3}+5x{y}^{4}+{y}^{5}
Use the Sum and binomial commands to write the unevaluated form of the sum.
q≔\mathrm{Sum}\left(\mathrm{binomial}\left(5,k\right)\cdot {x}^{5-k}\cdot {y}^{k},k=0..5\right)
Evaluate the sum with the value command.
\mathrm{value}\left(q\right)
<< Previous Example Section A-4 Next Section of Appendix >>
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LeadingRank - Maple Help
Home : Support : Online Help : Mathematics : Differential Equations : Differential Algebra : Tools : LeadingRank
returns the leading rank of a differential polynomial
LeadingRank(ideal, opts)
LeadingRank(p, R, opts)
LeadingRank(L, R, opts)
fullset = boolean. In the case of the function call LeadingRank(ideal), applies the function also over the differential polynomials which state that the derivatives of the parameters are zero. Default value is false. This option is incompatible with the diff notation.
The function call LeadingRank(p,R) returns the leading rank of p with respect to the ranking of R, or of its embedding ring, if R is an ideal.
The function is extended to numeric polynomials: the leading rank of
0
0
. The leading rank of any nonzero numerical polynomial is
1
. It is also extended to differential polynomials which involve independent variables only.
The function call LeadingRank(L,R) returns the list or the set of the leading ranks of the elements of L with respect to the ranking of R.
If ideal is a regular differential chain, the function call LeadingRank(ideal) returns the list of the leading ranks of the chain elements. If ideal is a list of regular differential chains, the function call LeadingRank(ideal) returns a list of lists of leading ranks.
This command is part of the DifferentialAlgebra:-Tools package. It can be called using the form LeadingRank(...) after executing the command with(DifferentialAlgebra:-Tools). It can also be directly called using the form DifferentialAlgebra[Tools][LeadingRank](...).
\mathrm{with}\left(\mathrm{DifferentialAlgebra}\right):
\mathrm{with}\left(\mathrm{Tools}\right):
R≔\mathrm{DifferentialRing}\left(\mathrm{derivations}=[x,y],\mathrm{blocks}=[[v,u],p],\mathrm{parameters}=[p]\right)
\textcolor[rgb]{0,0,1}{R}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{differential_ring}}
\mathrm{LeadingRank}\left(u[x,y]v[y]-u+p,R\right)
{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}}
\mathrm{ideal}≔\mathrm{RosenfeldGroebner}\left([{u[x]}^{2}-4u,u[x,y]v[y]-u+p,v[x,x]-u[x]],R\right)
\textcolor[rgb]{0,0,1}{\mathrm{ideal}}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{\mathrm{regular_differential_chain}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{regular_differential_chain}}]
\mathrm{Equations}\left(\mathrm{ideal}[1]\right)
[{\textcolor[rgb]{0,0,1}{v}}_{\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{p}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{u}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{u}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{v}}_{\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{u}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{u}]
\mathrm{LeadingRank}\left(\mathrm{ideal}[1]\right)
[{\textcolor[rgb]{0,0,1}{v}}_{\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{x}}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{v}}_{\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{2}}]
\mathrm{LeadingRank}\left([0,421],R\right)
[\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}]
|
Global Constraint Catalog: Cordered_global_cardinality
<< 5.307. ordered_atmost_nvector5.309. ordered_nvector >>
[PetitRegin09]
\mathrm{𝚘𝚛𝚍𝚎𝚛𝚎𝚍}_\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}\left(\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂},\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}\right)
\mathrm{𝚘𝚛𝚍𝚐𝚌𝚌}
\mathrm{𝚘𝚛𝚍𝚎𝚛𝚎𝚍}_\mathrm{𝚐𝚌𝚌}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚟𝚊𝚛}-\mathrm{𝚍𝚟𝚊𝚛}\right)
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}
\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚟𝚊𝚕}-\mathrm{𝚒𝚗𝚝},\mathrm{𝚘𝚖𝚊𝚡}-\mathrm{𝚒𝚗𝚝}\right)
\mathrm{𝚛𝚎𝚚𝚞𝚒𝚛𝚎𝚍}
\left(\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂},\mathrm{𝚟𝚊𝚛}\right)
|\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}|>0
\mathrm{𝚛𝚎𝚚𝚞𝚒𝚛𝚎𝚍}
\left(\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂},\left[\mathrm{𝚟𝚊𝚕},\mathrm{𝚘𝚖𝚊𝚡}\right]\right)
\mathrm{𝚒𝚗𝚌𝚛𝚎𝚊𝚜𝚒𝚗𝚐}_\mathrm{𝚜𝚎𝚚}
\left(\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂},\left[\mathrm{𝚟𝚊𝚕}\right]\right)
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}.\mathrm{𝚘𝚖𝚊𝚡}\ge 0
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}.\mathrm{𝚘𝚖𝚊𝚡}\le |\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}|
i\in \left[1,|\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}|\right]
, the values of the corresponding set of values
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}\left[j\right].\mathrm{𝚟𝚊𝚕}
\left(i\le j\le |\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}|\right)
should be taken by at most
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}\left[i\right].\mathrm{𝚘𝚖𝚊𝚡}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
From that previous definition, the
\mathrm{𝚘𝚖𝚊𝚡}
attributes are decreasing.
\left(\begin{array}{c}〈2,0,1,0,0〉,\hfill \\ 〈\mathrm{𝚟𝚊𝚕}-0\mathrm{𝚘𝚖𝚊𝚡}-5,\mathrm{𝚟𝚊𝚕}-1\mathrm{𝚘𝚖𝚊𝚡}-3,\mathrm{𝚟𝚊𝚕}-2\mathrm{𝚘𝚖𝚊𝚡}-1〉\hfill \end{array}\right)
\mathrm{𝚘𝚛𝚍𝚎𝚛𝚎𝚍}_\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}
constraint holds since the values of the three sets of values
\left\{0,1,2\right\}
\left\{1,2\right\}
\left\{2\right\}
are respectively used no more than 5, 3 and 1 times within the collection
〈2,0,1,0,0〉
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}
\mathrm{𝚘𝚛𝚍𝚎𝚛𝚎𝚍}_\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}
can be used in order to restrict the way we assign the values of the
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}
collection to the variables of the
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
collection. It expresses the fact that, when we use a value
v
, we implicitly also use all values that are less than or equal to
v
. As depicted by Figure 5.308.1 this is for instance the case for a soft cumulative constraint where we want to control the shape of cumulative profile by providing for each instant
i
{h}_{i}
that gives the height of the cumulative profile at instant
i
. These variables
{h}_{i}
are passed as the first argument of the
\mathrm{𝚘𝚛𝚍𝚎𝚛𝚎𝚍}_\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}
constraint. Then the
\mathrm{𝚘𝚖𝚊𝚡}
j
-th item of the
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}
collection gives the maximum number of instants for which the height of the cumulative profile is greater than or equal to value
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}\left[j\right].\mathrm{𝚟𝚊𝚕}
. In Figure 5.308.1 we should have:
no more than 1 height variable greater than or equal to 2,
no more than 3 height variables greater than or equal to 1,
no more than 5 height variables greater than or equal to 0.
Figure 5.308.1. (A) A cumulative profile wrt two tasks ① and ②, and its corresponding height variables
{h}_{1}
{h}_{2},\cdots ,{h}_{5}
giving at each instant how many resource is used (B) profile of value utilisation of the height variables (e.g., value 1 is assigned to variables
{h}_{3}
{h}_{2}
{h}_{4}
and therefore used three times)
The original definition of the
\mathrm{𝚘𝚛𝚍𝚎𝚛𝚎𝚍}_\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}
constraint mentions a third argument, namely the minimum number of occurrences of the smallest value. We omit it since it is redundant.
An other closely related constraint, the
\mathrm{𝚌𝚘𝚜𝚝}_\mathrm{𝚘𝚛𝚍𝚎𝚛𝚎𝚍}_\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}
constraint was introduced in [PetitRegin09] in order to model the fact that overloads costs may depend of the instant where they occur.
A filtering algorithm achieving arc-consistency in
O\left(|\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}|+|\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}|\right)
is described in [PetitRegin09]. It is based on the equivalence between the following two statements:
\mathrm{𝚘𝚛𝚍𝚎𝚛𝚎𝚍}_\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}
constraint has a solution,
all variables of the
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
collection assigned to their respective minimum value correspond to a solution to the
\mathrm{𝚘𝚛𝚍𝚎𝚛𝚎𝚍}_\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}
\mathrm{𝚘𝚛𝚍𝚎𝚛𝚎𝚍}_\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}
\left(〈\mathrm{𝚟𝚊𝚛}-{V}_{1},\mathrm{𝚟𝚊𝚛}-{V}_{2},\cdots ,\mathrm{𝚟𝚊𝚛}-{V}_{|\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}|}〉,
〈\mathrm{𝚟𝚊𝚕}-{v}_{1}\mathrm{𝚘𝚖𝚊𝚡}-{o}_{1},\mathrm{𝚟𝚊𝚕}-{v}_{2}\mathrm{𝚘𝚖𝚊𝚡}-{o}_{2},\cdots ,\mathrm{𝚟𝚊𝚕}-{v}_{|\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}|}\mathrm{𝚘𝚖𝚊𝚡}-{o}_{|\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}|}〉\right)
constraint can be reformulated into a
\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}
\left(〈\mathrm{𝚟𝚊𝚛}-{V}_{1},\mathrm{𝚟𝚊𝚛}-{V}_{2},\cdots ,\mathrm{𝚟𝚊𝚛}-{V}_{|\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}|}〉,〈\mathrm{𝚟𝚊𝚕}-{v}_{1}\mathrm{𝚗𝚘𝚌𝚌𝚞𝚛𝚛𝚎𝚗𝚌𝚎}-{N}_{1},\mathrm{𝚟𝚊𝚕}-{v}_{2}\mathrm{𝚗𝚘𝚌𝚌𝚞𝚛𝚛𝚎𝚗𝚌𝚎}-{N}_{2},\cdots ,\mathrm{𝚟𝚊𝚕}-{v}_{|\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}|}\mathrm{𝚗𝚘𝚌𝚌𝚞𝚛𝚛𝚎𝚗𝚌𝚎}-{N}_{|\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}|}〉\right)
|\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}|
sliding linear inequalities constraints of the form:
{N}_{1}+{N}_{2}+\cdots +{N}_{|\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}|}\le {o}_{1}
{N}_{2}+\cdots +{N}_{|\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}|}\le {o}_{2}
\cdots \cdots \cdots \cdots \cdots \cdots
{N}_{|\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}|}\le {o}_{|\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}|}
However, with the next example, T. Petit and J.-C. Régin have shown that this reformulation hinders propagation:
{V}_{1}\in \left\{0,1\right\}
{V}_{2}\in \left\{0,1\right\}
{V}_{3}\in \left\{0,1,2\right\}
{V}_{4}\in \left\{2,3\right\}
{V}_{5}\in \left\{2,3\right\}
\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}
\left(〈{V}_{1},{V}_{2},{V}_{3},{V}_{4},{V}_{5}〉,〈\mathrm{𝚟𝚊𝚕}-1\mathrm{𝚗𝚘𝚌𝚌𝚞𝚛𝚛𝚎𝚗𝚌𝚎}-{N}_{1},
\mathrm{𝚟𝚊𝚕}-2\mathrm{𝚗𝚘𝚌𝚌𝚞𝚛𝚛𝚎𝚗𝚌𝚎}-{N}_{2},
\mathrm{𝚟𝚊𝚕}-3\mathrm{𝚗𝚘𝚌𝚌𝚞𝚛𝚛𝚎𝚗𝚌𝚎}-{N}_{3}〉\right)
{N}_{1}+{N}_{2}+{N}_{3}\le 3\wedge {N}_{2}+{N}_{3}\le 2\wedge {N}_{3}\le 2
The previous reformulation does not remove value 2 from the domain of variable
{V}_{3}
\mathrm{𝚌𝚞𝚖𝚞𝚕𝚊𝚝𝚒𝚟𝚎}
(controlling the shape of the cumulative profile for breaking symmetry),
\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}_\mathrm{𝚕𝚘𝚠}_\mathrm{𝚞𝚙}
\mathrm{𝚒𝚗𝚌𝚛𝚎𝚊𝚜𝚒𝚗𝚐}_\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}
(the order is imposed on the main variables, and not on the count variables).
root concept:
\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}
constraint type: value constraint, order constraint.
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
\mathrm{𝑆𝐸𝐿𝐹}
↦\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}\right)
\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}.\mathrm{𝚟𝚊𝚛}\ge \mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}.\mathrm{𝚟𝚊𝚕}
\mathrm{𝐍𝐕𝐄𝐑𝐓𝐄𝐗}
\le \mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}.\mathrm{𝚘𝚖𝚊𝚡}
Since we want to express one unary constraint for each value we use the “For all items of
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}
” iterator. Part (A) of Figure 5.308.2 shows the initial graphs associated with each value 0, 1 and 2 of the
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}
collection of the Example slot. Part (B) of Figure 5.308.2 shows the corresponding final graph associated with value 0. Since we use the
\mathrm{𝐍𝐕𝐄𝐑𝐓𝐄𝐗}
graph property, the vertices of the final graph is stressed in bold.
\mathrm{𝚘𝚛𝚍𝚎𝚛𝚎𝚍}_\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}
|
1 Laboratory of Electronic, Computing, Telecommunication and Renewable Energies (LEITER), UGB, Saint-Louis, Senegal.
2 Renewable Energy Unit, Department of Physics, University of Nouakchott Al-Aasriya, Nouakchott, Mauritania.
3 Laboratory of Applied Energetics, Ecole Polytechnique of Dakar, Dakar, Senegal.
Abstract: This work presents a method of optimization of the photovoltaic generator (PV) based on the electrical model with a diode. The method consists of solving a second degree equation representing the derivative of the power function. The current and the maximum voltage being determined, the maximum power is deduced. Four popular types of photovoltaic panels from different manufacturers were considered for the study: BYD Model (BYD 320P6C-36), Atersa Grupo Model (A-320P GSE), SunPower Model (E19-320) and Model operated in the 50 MW power plant of Nouakchott-Mauritania (JKM320PP-72-V) of JinkoSolar. A comparative study is carried out between the simulated results and the data of the manufacturer of different technologies. The results obtained prove the effectiveness of the proposed method and that the BYD 320P6C-36 model is the most efficient among the four different technologies studied.
Keywords: Model, Photovoltaic Generator, Power Function, Optimization, Maximum Power
{I}_{pv}
{R}_{s}
{I}_{o}
{R}_{sh}
{I}_{ph}
{R}_{s}
{R}_{sh}
{I}_{0}
I={I}_{ph}-{I}_{D}-{I}_{sh}
{I}_{D}={I}_{0}\cdot \left({\text{e}}^{\left(\frac{q\left(V+IRs\right)}{nNk{T}_{c}}\right)}-1\right)
{I}_{sh}=\frac{\left(V+I{R}_{s}\right)}{{R}_{sh}}
I={I}_{ph}-{I}_{0}\cdot \left({\text{e}}^{\left(\frac{q\left(V+IRs\right)}{nNk{T}_{c}}\right)}-1\right)-\frac{\left(V+I{R}_{s}\right)}{{R}_{sh}}
{R}_{sh}
I={I}_{ph}-{I}_{0}\cdot \left({\text{e}}^{\left(\frac{q\left(V+IRs\right)}{nNk{T}_{c}}\right)}-1\right)
{I}_{ph}
{I}_{ph}=\left[{I}_{sc}+\alpha \left({T}_{c}-{T}_{r}\right)\right]\cdot \left(\frac{G}{{G}_{r}}\right)
{I}_{0}={I}_{on}\cdot {\left(\frac{{T}_{c}}{{T}_{r}}\right)}^{3}\cdot {\text{e}}^{\left[\frac{q{E}_{g}\left(\frac{1}{{T}_{r}}-\frac{1}{{T}_{c}}\right)}{nk}\right]}
I=0,V={V}_{oc}
I=0,V={V}_{oc}
0={I}_{sc}-{I}_{0n}\cdot \left({\text{e}}^{\left(\frac{q{V}_{oc}}{nNk{T}_{c}}\right)}-1\right)
{I}_{0n}=\frac{{I}_{sc}}{{\text{e}}^{\left(\frac{q{V}_{oc}}{nkN{T}_{c}}\right)}-1}
{I}_{0}=\frac{{I}_{sc}}{{\text{e}}^{\left(\frac{q{V}_{oc}}{nkN{T}_{c}}\right)}-1}\cdot {\left(\frac{{T}_{c}}{{T}_{r}}\right)}^{3}\cdot {\text{e}}^{\left[\frac{q{E}_{g}\left(\frac{1}{{T}_{r}}-\frac{1}{{T}_{c}}\right)}{nk}\right]}
{R}_{s}
{R}_{s}=\left(-{\frac{\text{d}V}{\text{d}I}|}_{V={V}_{oc}}\right)
{R}_{s}
{R}_{s}=\frac{\frac{{N}_{s}nk{T}_{c}}{q}\cdot \mathrm{ln}\left(1-\frac{{I}_{mp}}{{I}_{sc}}\right)+{V}_{oc}-{V}_{mp}}{{I}_{mp}}
I={I}_{sc}
V=0
{I}_{sc}={I}_{ph,ref}-{I}_{0,ref}\cdot \left({\text{e}}^{\left(\frac{q{R}_{s}{I}_{sc}}{nNk{T}_{c}}\right)}-1\right)
I={I}_{mp}
V={V}_{mp}
{I}_{mp}={I}_{ph,ref}-{I}_{0,ref}\cdot \left({\text{e}}^{\left(\frac{q\left({V}_{mp}+{I}_{mp}{R}_{s}\right)}{nNk{T}_{c}}\right)}-1\right)
{I}_{sc}\approx {I}_{ph,ref}
0={I}_{sc}-{I}_{0,ref}\cdot \left({\text{e}}^{\left(\frac{q{V}_{oc}}{nNk{T}_{r}}\right)}-1\right)
{I}_{mp}={I}_{ph,ref}-{I}_{0,ref}\cdot {\text{e}}^{\left(\frac{q\left({V}_{mp}+{I}_{mp}{R}_{s}\right)}{nNk{T}_{r}}\right)}
n=\frac{q\left(2{V}_{mp}-{V}_{oc}\right)}{Nk{T}_{r}\left(\frac{{I}_{mp}}{{I}_{sc}-{I}_{mp}}+\mathrm{ln}\left(\frac{{I}_{sc}-{I}_{mp}}{{I}_{sc}}\right)\right)}
V=\frac{nNK{T}_{c}}{q}\mathrm{ln}\left(1+\frac{{I}_{ph}-I}{{I}_{o}}\right)\cdot {R}_{s}I
P=V\cdot I
P=\frac{nNK{T}_{c}}{q}\mathrm{ln}\left(1+\frac{{I}_{ph}-I}{{I}_{o}}\right)\cdot {R}_{s}{I}^{2}
P=f\left(I\right)
\frac{\text{d}P}{\text{d}I}=0
\begin{array}{l}2{R}_{s}{I}^{2}-\left[C\mathrm{ln}\left(1+\frac{{I}_{ph}-I}{{I}_{o}}\right)+C+2{R}_{s}\left({I}_{o}+{I}_{ph}\right)\right]\cdot I\\ \text{ }+C\mathrm{ln}\left(1+\frac{{I}_{ph}-I}{{I}_{o}}\right)\left({I}_{o}+{I}_{ph}\right)=0\end{array}
C=\frac{nNk{T}_{c}}{q}
I=0
\mathrm{ln}\left(1+\frac{{I}_{ph}-I}{{I}_{o}}\right)
\mathrm{ln}\left(1+\frac{{I}_{ph}-I}{{I}_{o}}\right)=\mathrm{ln}\left(1+\frac{{I}_{ph}}{{I}_{o}}\right)-\frac{I}{{I}_{ph}+{I}_{o}}
\begin{array}{l}2{R}_{s}{I}^{2}-\left[C\mathrm{ln}\left(1+\frac{{I}_{ph}}{{I}_{o}}\right)-\frac{C{I}_{o}}{{I}_{ph}+{I}_{o}}+C+2{R}_{s}\left({I}_{o}+{I}_{ph}\right)\right]\cdot I\\ \text{ }+C\mathrm{ln}\left(1+\frac{{I}_{ph}}{{I}_{o}}\right)\left({I}_{o}+{I}_{ph}\right)-\frac{CI\left({I}_{o}+{I}_{ph}\right)}{{I}_{ph}+{I}_{o}}=0\end{array}
\begin{array}{l}\left(C+2{R}_{s}\left({I}_{o}+{I}_{ph}\right)\right){I}^{2}-\left({I}_{o}+{I}_{ph}\right)\left[C\mathrm{ln}\left(1+\frac{{I}_{ph}}{{I}_{o}}\right)+2C+2{R}_{s}\left({I}_{o}+{I}_{ph}\right)\right]I\\ \text{ }+C\mathrm{ln}\left(1+\frac{{I}_{ph}}{{I}_{o}}\right){\left({I}_{o}+{I}_{ph}\right)}^{2}\end{array}
{X}_{1}=\left(C+2{R}_{s}\left({I}_{o}+{I}_{ph}\right)\right)
{X}_{2}=\left({I}_{o}+{I}_{ph}\right)\left[C\mathrm{ln}\left(1+\frac{{I}_{ph}}{{I}_{o}}\right)+2C+2{R}_{s}\left({I}_{o}+{I}_{ph}\right)\right]
{X}_{3}=C\mathrm{ln}\left(1+\frac{{I}_{ph}}{{I}_{o}}\right){\left({I}_{o}+{I}_{ph}\right)}^{2}
{X}_{1}{I}^{2}+{X}_{2}I+{X}_{3}=0
{I}_{\mathrm{max}}=\frac{-{X}_{2}\pm \sqrt{{X}_{2}^{2}-4{X}_{1}{X}_{3}}}{2{X}_{1}}
{V}_{\mathrm{max}}=\frac{nNK{T}_{c}}{q}\mathrm{ln}\left(1+\frac{{I}_{ph}-{I}_{\mathrm{max}}}{{I}_{o}}\right)\cdot {R}_{s}{I}_{\mathrm{max}}
{P}_{\mathrm{max}}={V}_{\mathrm{max}}\cdot {I}_{\mathrm{max}}
{I}_{\mathrm{max}}
{V}_{\mathrm{max}}
{P}_{\mathrm{max}}
{E}_{x}=\frac{{x}_{i}-{x}_{mi}}{{x}_{mi}}\cdot 100
{x}_{i}
{x}_{mi}
Cite this paper: Sidibba, A. , Ndiaye, D. , El Bah, M. and Bouhamady, S. (2018) Analytical Modeling and Determination of the Characteristic Parameters of the Different Commercial Technologies of Photovoltaic Modules. Journal of Power and Energy Engineering, 6, 14-27. doi: 10.4236/jpee.2018.63002.
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Lidar Point Cloud Semantic Segmentation Using PointSeg Deep Learning Network - MATLAB & Simulink - MathWorks Deutschland
\mathit{r}=\sqrt{{\mathit{x}}^{2\text{\hspace{0.17em}}}+{\mathit{y}}^{2}+{\mathit{z}}^{2}}
Use the pixelLabelDatastore object to store pixel-wise labels from the label images. The object maps each pixel label to a class name. In this example, cars and trucks are the only objects of interest; all other pixels are the background. Specify these classes (car, truck, and background) and assign a unique label ID to each class.
To see the distribution of class labels in the data set, use the countEachLabel function.
Use the analyzeNetwork (Deep Learning Toolbox) function to display an interactive visualization of the network architecture.
Use the trainNetwork (Deep Learning Toolbox) function to train a PointSeg network if doTraining is true. Otherwise, load the pretrained network.
Use the evaluateSemanticSegmentation function to compute the semantic segmentation metrics from the test set results.
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Global Constraint Catalog: Cstable_compatibility
<< 5.372. sort_permutation5.374. stage_element >>
P. Flener, [BeldiceanuFlenerLorca06]
\mathrm{𝚜𝚝𝚊𝚋𝚕𝚎}_\mathrm{𝚌𝚘𝚖𝚙𝚊𝚝𝚒𝚋𝚒𝚕𝚒𝚝𝚢}\left(\mathrm{𝙽𝙾𝙳𝙴𝚂}\right)
\mathrm{𝙽𝙾𝙳𝙴𝚂}
\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\begin{array}{c}\mathrm{𝚒𝚗𝚍𝚎𝚡}-\mathrm{𝚒𝚗𝚝},\hfill \\ \mathrm{𝚏𝚊𝚝𝚑𝚎𝚛}-\mathrm{𝚍𝚟𝚊𝚛},\hfill \\ \mathrm{𝚙𝚛𝚎𝚌}-\mathrm{𝚜𝚒𝚗𝚝},\hfill \\ \mathrm{𝚒𝚗𝚌}-\mathrm{𝚜𝚒𝚗𝚝}\hfill \end{array}\right)
\mathrm{𝚛𝚎𝚚𝚞𝚒𝚛𝚎𝚍}
\left(\mathrm{𝙽𝙾𝙳𝙴𝚂},\left[\mathrm{𝚒𝚗𝚍𝚎𝚡},\mathrm{𝚏𝚊𝚝𝚑𝚎𝚛},\mathrm{𝚙𝚛𝚎𝚌},\mathrm{𝚒𝚗𝚌}\right]\right)
\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚒𝚗𝚍𝚎𝚡}\ge 1
\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚒𝚗𝚍𝚎𝚡}\le |\mathrm{𝙽𝙾𝙳𝙴𝚂}|
\mathrm{𝚍𝚒𝚜𝚝𝚒𝚗𝚌𝚝}
\left(\mathrm{𝙽𝙾𝙳𝙴𝚂},\mathrm{𝚒𝚗𝚍𝚎𝚡}\right)
\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚏𝚊𝚝𝚑𝚎𝚛}\ge 1
\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚏𝚊𝚝𝚑𝚎𝚛}\le |\mathrm{𝙽𝙾𝙳𝙴𝚂}|
\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚙𝚛𝚎𝚌}\ge 1
\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚙𝚛𝚎𝚌}\le |\mathrm{𝙽𝙾𝙳𝙴𝚂}|
\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚒𝚗𝚌}\ge 1
\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚒𝚗𝚌}\le |\mathrm{𝙽𝙾𝙳𝙴𝚂}|
\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚒𝚗𝚌}>\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚒𝚗𝚍𝚎𝚡}
Enforce the construction of a stably compatible supertree that is compatible with several given trees. The notion of stable compatibility and its context are detailed in the Usage slot.
\left(\begin{array}{c}〈\begin{array}{cccc}\mathrm{𝚒𝚗𝚍𝚎𝚡}-1\hfill & \mathrm{𝚏𝚊𝚝𝚑𝚎𝚛}-4\hfill & \mathrm{𝚙𝚛𝚎𝚌}-\left\{11,12\right\}\hfill & \mathrm{𝚒𝚗𝚌}-\varnothing ,\hfill \\ \mathrm{𝚒𝚗𝚍𝚎𝚡}-2\hfill & \mathrm{𝚏𝚊𝚝𝚑𝚎𝚛}-3\hfill & \mathrm{𝚙𝚛𝚎𝚌}-\left\{8,9\right\}\hfill & \mathrm{𝚒𝚗𝚌}-\varnothing ,\hfill \\ \mathrm{𝚒𝚗𝚍𝚎𝚡}-3\hfill & \mathrm{𝚏𝚊𝚝𝚑𝚎𝚛}-4\hfill & \mathrm{𝚙𝚛𝚎𝚌}-\left\{2,10\right\}\hfill & \mathrm{𝚒𝚗𝚌}-\varnothing ,\hfill \\ \mathrm{𝚒𝚗𝚍𝚎𝚡}-4\hfill & \mathrm{𝚏𝚊𝚝𝚑𝚎𝚛}-5\hfill & \mathrm{𝚙𝚛𝚎𝚌}-\left\{1,3\right\}\hfill & \mathrm{𝚒𝚗𝚌}-\varnothing ,\hfill \\ \mathrm{𝚒𝚗𝚍𝚎𝚡}-5\hfill & \mathrm{𝚏𝚊𝚝𝚑𝚎𝚛}-7\hfill & \mathrm{𝚙𝚛𝚎𝚌}-\left\{4,13\right\}\hfill & \mathrm{𝚒𝚗𝚌}-\varnothing ,\hfill \\ \mathrm{𝚒𝚗𝚍𝚎𝚡}-6\hfill & \mathrm{𝚏𝚊𝚝𝚑𝚎𝚛}-2\hfill & \mathrm{𝚙𝚛𝚎𝚌}-\left\{8,14\right\}\hfill & \mathrm{𝚒𝚗𝚌}-\varnothing ,\hfill \\ \mathrm{𝚒𝚗𝚍𝚎𝚡}-7\hfill & \mathrm{𝚏𝚊𝚝𝚑𝚎𝚛}-7\hfill & \mathrm{𝚙𝚛𝚎𝚌}-\left\{6,13\right\}\hfill & \mathrm{𝚒𝚗𝚌}-\varnothing ,\hfill \\ \mathrm{𝚒𝚗𝚍𝚎𝚡}-8\hfill & \mathrm{𝚏𝚊𝚝𝚑𝚎𝚛}-6\hfill & \mathrm{𝚙𝚛𝚎𝚌}-\varnothing \hfill & \mathrm{𝚒𝚗𝚌}-\left\{9,10,11,12,13,14\right\},\hfill \\ \mathrm{𝚒𝚗𝚍𝚎𝚡}-9\hfill & \mathrm{𝚏𝚊𝚝𝚑𝚎𝚛}-2\hfill & \mathrm{𝚙𝚛𝚎𝚌}-\varnothing \hfill & \mathrm{𝚒𝚗𝚌}-\left\{10,11,12,13\right\},\hfill \\ \mathrm{𝚒𝚗𝚍𝚎𝚡}-10\hfill & \mathrm{𝚏𝚊𝚝𝚑𝚎𝚛}-3\hfill & \mathrm{𝚙𝚛𝚎𝚌}-\varnothing \hfill & \mathrm{𝚒𝚗𝚌}-\left\{11,12,13\right\},\hfill \\ \mathrm{𝚒𝚗𝚍𝚎𝚡}-11\hfill & \mathrm{𝚏𝚊𝚝𝚑𝚎𝚛}-1\hfill & \mathrm{𝚙𝚛𝚎𝚌}-\varnothing \hfill & \mathrm{𝚒𝚗𝚌}-\left\{12,13\right\},\hfill \\ \mathrm{𝚒𝚗𝚍𝚎𝚡}-12\hfill & \mathrm{𝚏𝚊𝚝𝚑𝚎𝚛}-1\hfill & \mathrm{𝚙𝚛𝚎𝚌}-\varnothing \hfill & \mathrm{𝚒𝚗𝚌}-\left\{13\right\},\hfill \\ \mathrm{𝚒𝚗𝚍𝚎𝚡}-13\hfill & \mathrm{𝚏𝚊𝚝𝚑𝚎𝚛}-5\hfill & \mathrm{𝚙𝚛𝚎𝚌}-\varnothing \hfill & \mathrm{𝚒𝚗𝚌}-\left\{14\right\},\hfill \\ \mathrm{𝚒𝚗𝚍𝚎𝚡}-14\hfill & \mathrm{𝚏𝚊𝚝𝚑𝚎𝚛}-6\hfill & \mathrm{𝚙𝚛𝚎𝚌}-\varnothing \hfill & \mathrm{𝚒𝚗𝚌}-\varnothing \hfill \end{array}〉\hfill \end{array}\right)
Figure 5.373.1 shows the two trees we want to merge. Note that the leaves
𝚊
𝚏
occur in both trees.
Figure 5.373.1. The two trees to merge
The left part of Figure 5.373.2 gives one way to merge the two previous trees. This solution corresponds to the ground instance provided by the example. Note that there exist 7 other ways to merge these two trees. They are respectively depicted by Figures 5.373.2 to 5.373.5.
Figure 5.373.2. First solution (corresponding to the ground instance of the example) and second solution on how to merge the two trees
{T}_{1}
{T}_{2}
of Figure 5.373.1
Figure 5.373.3. Third and fourth solutions on how to merge the two trees
{T}_{1}
{T}_{2}
Figure 5.373.4. Fifth and sixth solutions on how to merge the two trees
{T}_{1}
{T}_{2}
Figure 5.373.5. Seventh and eight solutions on how to merge the two trees
{T}_{1}
{T}_{2}
|\mathrm{𝙽𝙾𝙳𝙴𝚂}|>2
\mathrm{𝚛𝚊𝚗𝚐𝚎}
\left(\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚏𝚊𝚝𝚑𝚎𝚛}\right)>1
\mathrm{𝙽𝙾𝙳𝙴𝚂}
One objective of phylogeny is to construct the genealogy of the species, called the tree of life, whose leaves represent the contemporary species and whose internal nodes represent extinct species that are not necessarily named. An important problem in phylogeny is the construction of a supertree [BinindaEmondsGittlemanSteel02] that is compatible with several given trees. There are several definitions of tree compatibility in the literature:
𝒯
is strongly compatible with a tree
{𝒯}^{\text{'}}
{𝒯}^{\text{'}}
is topologically equivalent to a subtree
𝒯
that respects the node labelling. [NgWormald96]
𝒯
is weakly compatible with a tree
{𝒯}^{\text{'}}
{𝒯}^{\text{'}}
𝒯
by a series of arc contractions. [Steel92]
𝒯
is stably compatible with a set
𝒮
of trees if
𝒯
is weakly compatible with each tree in
𝒮
and each internal node of
𝒯
can be labelled by at least one corresponding internal node of some tree in
𝒮
For the supertree problem, strong and weak compatibility coincide if and only if all the given trees are binary [NgWormald96]. The existence of solutions is not lost when restricting weak compatibility to stable compatibility.
Figure 5.373.6. Supertree problem instance and two of its solutions
Figure 5.373.7. Three small phylogenetic trees
For example, the trees
{𝒯}_{1}
{𝒯}_{2}
of Figure 5.373.6 have
𝒯
{𝒯}^{\text{'}}
as supertrees under both weak and strong compatibility. As shown, all the internal nodes of
{𝒯}^{\text{'}}
can be labelled by corresponding internal nodes of the two given trees, but this is not the case for the father of
b
g
𝒯
𝒯
and four other such supertrees are debatable because they speculate about the existence of extinct species that were not in any of the given trees. Consider also the three small trees in Figure 5.373.7:
{𝒯}_{3}
{𝒯}_{4}
{𝒯}_{4}
as a supertree under weak compatibility, as it suffices to contract the arc
\left(3,2\right)
{𝒯}_{3}
{𝒯}_{4}
{𝒯}_{3}
{𝒯}_{4}
have no supertree under strong compatibility, as the most recent common ancestor of
b
c
\mathrm{𝑚𝑟𝑐𝑎}\left(b,c\right)
\mathrm{𝑚𝑟𝑐𝑎}\left(a,b\right)
{𝒯}_{3}
, namely 1, but not the same in
{𝒯}_{4}
\mathrm{𝑚𝑟𝑐𝑎}\left(b,c\right)=3
is an evolutionary descendant of
\mathrm{𝑚𝑟𝑐𝑎}\left(a,b\right)=2
{𝒯}_{4}
{𝒯}_{5}
have neither weakly nor strongly compatible supertrees.
Under strong compatibility, a first supertree algorithm was given in [AhoSagivSzymanskiUllman81], with an application for database management systems; it takes
O\left({l}^{2}\right)
l
is the number of leaves in the given trees. Derived algorithms have emerged from phylogeny, for instance OneTree [NgWormald96]. The first constraint program was proposed in [GentProsserSmithWei03], using standard, non-global constraints. Under weak compatibility, a phylogenetic supertree algorithm can be found in [Steel92] for instance. Under stable compatibility, the algorithm from computational linguistics of [BodirskyKutz07] has supertree construction as special case.
\mathrm{𝚝𝚛𝚎𝚎}
application area: bioinformatics, phylogeny.
final graph structure: tree.
\mathrm{𝙽𝙾𝙳𝙴𝚂}
\mathrm{𝐶𝐿𝐼𝑄𝑈𝐸}
↦\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚗𝚘𝚍𝚎𝚜}\mathtt{1},\mathrm{𝚗𝚘𝚍𝚎𝚜}\mathtt{2}\right)
\mathrm{𝚗𝚘𝚍𝚎𝚜}\mathtt{1}.\mathrm{𝚏𝚊𝚝𝚑𝚎𝚛}=\mathrm{𝚗𝚘𝚍𝚎𝚜}\mathtt{2}.\mathrm{𝚒𝚗𝚍𝚎𝚡}
•
\mathrm{𝐌𝐀𝐗}_\mathrm{𝐍𝐒𝐂𝐂}
\le 1
•
\mathrm{𝐍𝐂𝐂}
=1
•
\mathrm{𝐌𝐀𝐗}_\mathrm{𝐈𝐃}
\le 2
•
\mathrm{𝐏𝐀𝐓𝐇}_\mathrm{𝐅𝐑𝐎𝐌}_\mathrm{𝐓𝐎}
\left(\mathrm{𝚒𝚗𝚍𝚎𝚡},\mathrm{𝚒𝚗𝚍𝚎𝚡},\mathrm{𝚙𝚛𝚎𝚌}\right)=1
•
\mathrm{𝐏𝐀𝐓𝐇}_\mathrm{𝐅𝐑𝐎𝐌}_\mathrm{𝐓𝐎}
\left(\mathrm{𝚒𝚗𝚍𝚎𝚡},\mathrm{𝚒𝚗𝚍𝚎𝚡},\mathrm{𝚒𝚗𝚌}\right)=0
•
\mathrm{𝐏𝐀𝐓𝐇}_\mathrm{𝐅𝐑𝐎𝐌}_\mathrm{𝐓𝐎}
\left(\mathrm{𝚒𝚗𝚍𝚎𝚡},\mathrm{𝚒𝚗𝚌},\mathrm{𝚒𝚗𝚍𝚎𝚡}\right)=0
To each distinct leave (i.e., each species) of the trees to merge corresponds a vertex of the initial graph. To each internal vertex of the trees to merge corresponds also a vertex of the initial graph. Each vertex of the initial graph has the following attributes:
An index corresponding to a unique identifier.
A father corresponding to the father of the vertex in the final tree. Since the leaves of the trees to merge must remain leaves we remove the index value of all the leaves from all the father variables.
A set of precedence constraints corresponding to all the arcs of the trees to merge.
A set of incomparability constraints corresponding to the incomparable vertices of each tree to merge.
The arc constraint describes the fact that we link a vertex to its father variable. Finally we use the following six graph properties on our final graph:
The first graph property
\mathrm{𝐌𝐀𝐗}_\mathrm{𝐍𝐒𝐂𝐂}\le 1
enforces the fact that the size of the largest strongly connected component does not exceed one. This avoid having circuits containing more than one vertex. In fact the root of the merged tree is a strongly connected component with a single vertex.
The second graph property
\mathrm{𝐍𝐂𝐂}=1
imposes having only a single tree.
The third graph property
\mathrm{𝐏𝐀𝐓𝐇}_\mathrm{𝐅𝐑𝐎𝐌}_\mathrm{𝐓𝐎}\left(\mathrm{𝚒𝚗𝚍𝚎𝚡},\mathrm{𝚒𝚗𝚍𝚎𝚡},\mathrm{𝚙𝚛𝚎𝚌}\right)=1
enforces for each vertex
i
a set of precedence constraints; for each vertex
j
of the precedence set there is a path from
i
j
in the final graph.
The fourth graph property
\mathrm{𝐌𝐀𝐗}_\mathrm{𝐈𝐃}\le 2
enforces that the number of predecessors (i.e., arcs from a vertex to itself are not counted) of each vertex does not exceed 2 (i.e., the final graph is a binary tree).
The fifth and sixth graph properties
\mathrm{𝐏𝐀𝐓𝐇}_\mathrm{𝐅𝐑𝐎𝐌}_\mathrm{𝐓𝐎}\left(\mathrm{𝚒𝚗𝚍𝚎𝚡},\mathrm{𝚒𝚗𝚍𝚎𝚡},
\mathrm{𝚒𝚗𝚌}\right)=0
\mathrm{𝐏𝐀𝐓𝐇}_\mathrm{𝐅𝐑𝐎𝐌}_\mathrm{𝐓𝐎}\left(\mathrm{𝚒𝚗𝚍𝚎𝚡},\mathrm{𝚒𝚗𝚌},
\mathrm{𝚒𝚗𝚍𝚎𝚡}\right)=0
i
a set of incomparability constraints; for each vertex
j
of the incomparability set there is neither a path from
i
j
, nor a path from
j
i
Figures 5.373.8 and 5.373.9 respectively show the precedence and the incomparability graphs associated with the Example slot. As it contains too many arcs the initial graph is not shown. Figures 5.373.2 shows the first solution satisfying all the precedence and incomparability constraints.
Figure 5.373.8. Precedence graph associated with the two trees to merge described by Figure 5.373.1
Figure 5.373.9. Incomparability graph associated with the two trees to merge described by Figure 5.373.1; the two cliques respectively correspond to the leaves of the two trees to merge.
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showstop - Maple Help
Home : Support : Online Help : Programming : Debugging : showstop
display information about breakpoints, watchpoints, and error watchpoints
showstop()
The showstop function displays a report of all functions containing breakpoints, all watchpoints, and all error watchpoints currently set.
f := proc(x) local y; if x < 2 then y := x; print(y^2) end if; print(-x); x^3 end proc:
\mathrm{stopat}\left(f\right)
[\textcolor[rgb]{0,0,1}{f}]
\mathrm{stopat}\left(f,2\right)
[\textcolor[rgb]{0,0,1}{f}]
\mathrm{stopat}\left(\mathrm{int}\right)
[\textcolor[rgb]{0,0,1}{f}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{int}}]
\mathrm{stopwhen}\left(f,y\right)
[[\textcolor[rgb]{0,0,1}{f}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}]]
\mathrm{stopwhen}\left(\mathrm{Digits}\right)
[\textcolor[rgb]{0,0,1}{\mathrm{Digits}}\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{f}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}]]
\mathrm{stopwhenif}\left(\mathrm{answer},42\right)
[\textcolor[rgb]{0,0,1}{\mathrm{Digits}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{answer}}\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{f}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}]]
\mathrm{stoperror}\left(\mathrm{`division by zero`}\right)
[\textcolor[rgb]{0,0,1}{"division by zero"}]
\mathrm{showstop}\left(\right)
Breakpoints in:
Watched variables:
y in procedure f
Watched errors:
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Overlap - Maple Help
Home : Support : Online Help : Programming : Names and Strings : StringTools Package : Combinatorics on Words : Overlap
compute the overlap of two strings
Overlap( s, t )
The Overlap(s,t) command computes the length of the overlap between s and t in linear time. This is defined to be the length of the longest suffix of s that is a prefix of t.
Note that, in general, Overlap( s, t ) and Overlap( t, s ) are different.
8
\mathrm{with}\left(\mathrm{StringTools}\right):
\mathrm{Overlap}\left("abc","def"\right)
\textcolor[rgb]{0,0,1}{0}
\mathrm{Overlap}\left("abc","cdef"\right)
\textcolor[rgb]{0,0,1}{1}
\mathrm{Overlap}\left("abcdef","defg"\right)
\textcolor[rgb]{0,0,1}{3}
StringTools[Border]
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Force Analysis of a Vibratory Bowl Feeder for Automatic Assembly | J. Mech. Des. | ASME Digital Collection
Richard Silversides,
Richard Silversides
School of Physical Sciences and Engineering
, King’s College London, Strand, London, WC2R 2LS, UK
e-mail: richard.silversides@kcl.ac.uk
Jian S Dai,
e-mail: lakmal.seneviratne@kcl.ac.uk
Silversides, R., Dai, J. S., and Seneviratne, L. (August 27, 2004). "Force Analysis of a Vibratory Bowl Feeder for Automatic Assembly." ASME. J. Mech. Des. July 2005; 127(4): 637–645. https://doi.org/10.1115/1.1897407
This paper investigates the vibratory bowl feeder for automatic assembly, presents a geometric model of the feeder, and develops force analysis, leading to dynamical modeling of the vibratory feeder. Based on the leaf-spring modeling of the three legs of the symmetrically arranged bowl of the feeder, and equating the vibratory feeder to a three-legged parallel mechanism, the paper reveals the geometric property of the feeder. The effects of the leaf-spring legs are transformed to forces and moments acting on the base and bowl of the feeder. Resultant forces are obtained based upon the coordinate transformation, and the moment analysis is produced based upon the orthogonality of the orientation matrix. This reveals the characteristics of the feeder, that the resultant force is along the
z
-axis and the resultant moment is about the
z
direction and further generates the closed-form motion equation. The analysis presents a dynamic model that integrates the angular displacement of the bowl with the displacement of the leaf-spring legs. Both Newtonian and Lagrangian approaches are used to verify the model, and an industrial case-based simulation is used to demonstrate the results.
assembling, production equipment, vibrations, springs (mechanical), matrix algebra, Force, Dynamics, Analysis, Geometry, Transformation, Modeling, Bowl Feeder, Assembly
Manufacturing, Springs, Equations of motion, Simulation
Simulation of Vibratory Feeders, Proc. of the Symp. on Computer-Aided Engineering
Analog and Digital Analysis and Synthesis of Oscillatory Tracks
Analysis of Vibratory Feeding Where the Track has Directional Characteristics
Numerical Study of a Model of a Vibro-Transporter
ASME J. Manuf. Systems
Feedback Control for Electromagnetic Vibration Feeder (Applications of Two Degrees-of-Freedom Proportional Plus Integral Plus Derivative Controller With Nonlinear Element)
Effects of Driving Signal Forms and Frequency in the Performance of a Vibratory Bowl Feeder
Proc. of International Conference on Manufacturing Automation
Univ. Hong Kong.
Modelling and Control of a Novel Vibratory Feeder
1999 IEEE/ASME Int. Conference on Advanced Intelligent Mechatronics
A Six-Component Contact Force Measurement Device Based on the Stewart Platform
Design and Analysis of a New Six Component Force Transducer for Robotic Grasping
Proc. of Second Biennial European Joint Conference on Engineering Systems Design and Analysis
Lagrangian Formulation of Rotating Beam with Active Constrained Layer Damping in Time Domain Analysis
Dynamic Vibration Absorbers
Mechanical Engineering Publications Ltd
The Hamel Representation: A Diagonalized Poincaré Form
Design and Validation of Electro-Hydraulic Pressure-Control Valves for Closed-Loop Controller Implementation
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Acid/Base Equilibrium | Brilliant Math & Science Wiki
Jordan Calmes, Vishwak Srinivasan, Gautam Sharma, and
Dimitrios Kidonakis
Jalaj Swami
Acids and bases have a chemical equilibrium in solution. At chemical equilibrium, the products and reactants have reached a state of balance. Reactions may still be taking place within the sample, but the forward and reverse reactions are taking place at the same rate, so the concentrations of the products and reactants are not changing with time.
Concepts related to acid base equilibrium include designing buffers and plotting pH curves.
Additionally, determining the acid base equilibrium can help in predicting the products of a reaction and the relative concentration of the products, as well as aiding in the identification of the weaker acid and base.
Weak acids and bases are common in nature. For example, many insects of the order hymenoptera, including bees, wasps, and ants, use a weak acid (formic acid) as a weapon when they bite or sting. As a result, many animals avoid contact with these insects. In order for the weaponized acid to be effective, it must produce enough hydrogen ions to wound the bee's enemy, but not enough to dissolve the bee's stinger and organs. Formic acid
This defense occasionally backfires, as other animals have adapted their own pH in response to formic acid. Anteaters, a species well known for feasting on anthills, do not produce the hydrochloric acid (a strong acid) that most animals rely on for digestion. The anteaters are attracted to the ants, because they need the formic acid to break up their stomach contents.
Giant Anteater [1]
Defining Strong and Weak
A strong Arrhenius acid dissociates completely in aqueous solution to form
\ce{H^+}
A strong Arrhenius base dissociates completely in aqueous solution to form
\ce{OH^-}
An acid showing 100% dissociation in a solution is considered a very very strong acid.
Similarly, a base showing 100% dissociation in a solution is considered a very very strong base.
\ce{HCl}
is considered a strong acid, because it readily and completely dissociates, making
\ce{H^+}
ions, as illustrated by the following chemical equation:
\ce{HCl_{aq} + H_2O_{l} \rightarrow H_3O^{+}_{aq} + Cl^{-}_{aq}}.
\ce{NaOH}
is considered a strong base, because it dissociates completely, forming hydroxide ions
(\ce{OH^-})
. The reaction proceeds as follows:
\ce{NaOH_{aq} \rightarrow Na^{+}_{aq} + OH^{-}_{aq}}.
Common strong acids and strong bases are listed below.
Strong Acids Chemical Structure
\ce{HCl}
\ce{HBr}
\ce{HI}
\ce{HNO3}
\ce{HClO_4}
\ce{H_2SO_4}
Strong Bases Chemical Structure
\ce{LiOH}
\ce{NaOH}
\ce{KOH}
\ce{Ca(OH)2}
\ce{Ba(OH)_2}
True for both True for acid, false for base False for acid, true for base False for both
A solution containing a strong acid will always burn your skin if you spill it on yourself.
A solution containing a strong base will always burn your skin if you spill it on yourself.
Weak acids and bases do not dissociate completely in aqueous solution.
Here is the general format for weak acid dissociation:
\ce{HA_{aq} + H_2O \rightleftharpoons H_3O_{aq}^{+} + A_{aq}^{-}}.
Notice the double arrows,
\rightleftharpoons
, which mean that as the acids is dissociating, it is also re-combining. The reaction can proceed either forward or in reverse, depending on the environmental conditions.
In aqueous solutions, the reaction between an acid and a base can be thought of as the transfer of a proton from the acid to the base.
\ce{HBr + H_2O \rightarrow Br^- + H_3O^+}
Note that the hydrogen from
\ce{HBr}
(the acid) has moved to
\ce{H2O}
(the base), forming the hydronium ion,
\ce{H_3O^+}
. Water and hydronium are called a conjugate acid-base pair.
\ce{OH^-}
\ce{NH_3}
\ce{H_2O}
\ce{NH_{4}^{+}}
Identify the conjugate acid of
\ce{NH_3}
in the following chemical reaction:
\ce{H_2O + NH_3 \rightarrow NH_{4}^{+} + OH^-}.
Water is amphoteric, meaning it can act both as a base and as an acid, depending on its environment.
Common weak acids and their conjugate bases
Weak Acid Chemical Structure Conjugate Base Chemical Structure
\ce{H2O}
\ce{OH^-}
\ce{CH_3COOH}
\ce{CH_3COO^-}
\ce{HCN}
\ce{CN^-}
Common weak bases and their conjugate acids
Weak Base Chemical Structure Conjugate Acid Chemical Structure
\ce{H2O}
\ce{H_3O^+}
\ce{C_5H_5N}
Pyridinium
\ce{C_5H_5NH^+}
\ce{NH_3}
\ce{NH_4^+}
Reactions between acids and bases are very common in nature, and often follow one of the following equations.
(\text{strong acid}) + (\text{strong base}) \rightarrow (\text{salt}) + (\text{water})
\ce{HCl + KOH \to KCl + H_2O}
(\text{stronger acid}) + (\text{stronger base}) \rightleftharpoons (\text{weaker acid}) + (\text{weaker base})
\ce{CH_3CO_2H + H_2O \rightleftharpoons H_3O^+ + CH_3CO_2^-}
Lewis diagram showing the movement of hydrogen in the reaction between acetic acid and water
Acid-base equilibria favor the side of the reaction that has the weaker acid and base. In the above example, the reactants side would be favored and there would be more acetic acid and water in solution, because acetic acid is a weak acid and water is a weak base.
The equilibrium constants of acids (
K_a
) and bases (
K_b
) can be used to determine the relative strengths of the compounds. These constants, which are also called dissociation constants or ionization constants, can be used to calculate the percent ionization of an aqueous solution or to determine pH. Additionally, the equilibrium constant for an acid-base reaction can be calculated from the
K_a
K_b
values for the reactants.
K_a = \dfrac{[\ce{H_3O^+}][\ce{A^-}]}{[\ce{HA}]},
[\ce{HA}]
is the concentration of the acid,
[\ce{A^-}]
is the concentration of the conjugate base, and
[\ce{H_3O^+}]
is the concentration of hydronium.
K_b = \dfrac{[\ce{BH^+}][\ce{OH^-}]}{[\ce{B}]},
[\ce{B}]
is the concentration of the base,
\ce{BH^+}
is the concentration of the conjugate acid, and
\ce{OH^-}
is the concentration of hydroxide.
Flores, F. Oso hormiguero (Myrmecophaga tridactyla). Retrieved from https://www.flickr.com/photos/ferjflores/8696742551/
Cite as: Acid/Base Equilibrium. Brilliant.org. Retrieved from https://brilliant.org/wiki/acidbase-equilibrium/
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ndfUnlock
Unlock an NDF so it can then be locked by another thread
This function unlocks the supplied NDF (both the supplied identifier and the associated base NDF) so that another thread can then lock it for its own use using function ndfLock. After calling this function, an error will be reported if an attempt is made to access the NDF in any way, either through the supplied identifier or any other identifier that refers to the same base NDF. There are however two exceptions to this rule:
An unlocked NDF identifier can be locked using ndfLock, allowing the thread full access to the NDF using the locked identifier.
An unlocked NDF identifier can be annulled using ndfAnnul.
void ndfUnlock( int indf, int
\ast
This function returns without action if the NDF is already unlocked.
This function will report an error if the supplied NDF is locked by any thread other than the currently running thread.
The supplied NDF identifier will be removed from the NDF context associated with the currently running thread, and placed in a " null" context that is ignored by the ndfEnd function. The ndfReport function can be used to determine if any NDF identifiers are currently in this " null" context.
\ast
Building C applications which call the NDF_ library differs very little from building Fortran applications and is covered in §24.
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TRANNDF
Transforms a list of images by resampling
This application performs the arbitrary transformation of a list of images. The output images are calculated by resampling the data of the input images. Output array elements are set to the bad value if their inverse-transformed coordinates lie outside the corresponding input image’s coordinate limits. Many images can be resampled with a single invocation of TRANNDF, but it is the user’s responsibility to ensure that they are resampled into the same coordinate system if they are subsequently to be combined or compared on a pixel-by-pixel basis.
Images processed by CCDPACK are resampled in one of two ways, depending on the value of the USEWCS parameter.
If USEWCS is TRUE then they are resampled from their Pixel coordinates into their Current attached coordinate system (this is the default). Since the resampling means that a 1 x 1 square in the Current coordinates will represent one pixel in the output image, the Current coordinate system must be of an appropriate size (so for instance resampling into SKY coordinates is not suitable because they have units of radians). The Current coordinate system will typically have been added by the CCDPACK REGISTER or WCSREG applications, and be labelled ’CCD_REG’ or ’CCD_WCSREG’ accordingly - if it has another label (domain) a warning will be issued but resampling will proceed. A copy of the original PIXEL coordinate system will be retained in the WCS component of the new image under the name CCD_OLDPIXEL; this can be useful for transforming positions back into the pre-transformation coordinate system.
If USEWCS is set to FALSE, then the resampling will take place according to the TRANSFORM structure stored in the .MORE.CCDPACK extension of the file. This option exists chiefly for compatibility with older versions of CCDPACK.
tranndf in out [method]
If CONSERVE is TRUE, the output values are normalised by the ratio of the output-to-input pixel areas. In other words this conserves flux. If CONSERVE is FALSE, there is no normalisation. Flux can only be conserved if the transformation is linear, so that even if CONSERVE is TRUE, flux will be incorrectly conserved if the transformation is of a non-linear nature. [TRUE]
If TRUE then the transformation which is to be applied to the image is stored in the image’s CCDPACK extension (.MORE.CCDPACK.TRANSFORM). If FALSE then a transformation structure must be supplied via the parameter TRANSFORM. This transformation is then applied to the list of images. [TRUE]
LBOUND() = _INTEGER (Read)
If SHAPE is "SPECIFY" then this parameter specifies the lower pixel-index bounds of all the output images. The number of values should equal the maximum number of dimensions of the input images. The suggested defaults are the lower bounds generated by the SHAPE="AUTO" option for the first image. These bounds are probably small enough to ensure that all the transformed data (of the first image) will appear in the output image. [Dynamic default]
The interpolation method used to resample the input image data arrays. Permitted values are "NEAREST" for nearest-neighbour, and "LININT" for linear interpolation. [NEAREST]
Names of the output – transformed – images. These may be specified as list of comma separated names, using indirection if required, or, as a single modification element (of the input names). The simplest modification element is the asterisk "
\ast
>
\ast
>
\ast
signifies that all the images in the current directory should be used and the output images should have the same names. Other types of modification can also be used, such as,
>
\ast
-TRN
which means call the output images the same as the input images but add -TRN to the end of the names. Replacement of a specified string with another in the output file names can also be used:
>
\ast
|
|
|
this replaces the string RAW with RES in any of the output names.
SHAPE = LITERAL (Read)
The method to be used to determine the SHAPE of the output images. Can take one of the values "AUTO", "SAME", "SPECIFY". With the meanings.
AUTO – automatically determine the bounds of the output images such that all of the input data appears. This is achieved by transforming test points along the current bounds so assumes that the transformation will behave reasonably.
SAME – set the output image bounds to those of the corresponding input images.
SPECIFY – you will specify a single set of bounds for all the output images. (See the LBOUND and UBOUND parameters.) [AUTO]
Title for the output images. [Output from TRANNDF]
If INEXT is FALSE then this parameter specifies the transformation structure. This includes the file name and the HDS object. For example, DISTORT.MAPPING would use the TRANSFORM structure called MAPPING in the HDS file DISTORT. Normally the object name is TRANSFORM. The structure must contain both the forward and inverse mappings. This transform if supplied acts on all the input images.
UBOUND() = _INTEGER (Read)
If SHAPE is "SPECIFY" then this parameter specifies the upper pixel-index bounds of all the output images. The number of values should equal the maximum number of dimensions of the input images. The suggested defaults are the upper bounds generated by the SHAPE="AUTO" option for the first image. These bounds are probably large enough to ensure that all the transformed data (of the first image) will appear in the output image. [Dynamic default]
USEWCS = _LOGICAL (Read)
If TRUE then the transformation which is to be applied to the image is stored in the image’s WCS extension as an attached coordinate system. If FALSE then the transformation is either stored as a TRN structure in the image’s CCDPACK extension (.MORE.CCDPACK.TRANSFORM), or is supplied by the user (see the INEXT parameter). [TRUE]
tranndf ’
\ast
\ast
-resamp’ reset
This transforms all the images in the current directory from pixel coordinates to their Current coordinate system. It uses nearest-neighbour resampling and conserves the flux levels (assuming that the transformation is linear). The output images are of a size such that all the input pixels have contributed.
tranndf curved straight linint shape=same
As above, except linear interpolation is used, and the straight array uses the bounds of curved.
tranndf ’a119
\ast
\ast
s’ inext=false transform=proj.merc shape=bounds lbound=’[1,-20]’ ubound=’[256,172]’
This transforms the images called a119
\ast
, using the transformation structure merc in the HDS file called proj, into images called a119
\ast
s. It uses nearest-neighbour resampling. All the output images have size 256 x 192 pixels and origin (1,-20).
“Data resampling”.
LBOUND – always uses a dynamic default
UBOUND – always uses a dynamic default
TITLE – always "Output from TRANNDF"
Flux conservation can only be applied to constant-determinant or linear transformations. It is currently impossible to tell whether an AST Mapping is linear, but in the expectation that it is (most of them are, and most of the rest very nearly are), it is turned on, without a warning, by default.
The NDF components are processed by this application as follows:
AXES, LABEL, UNITS, HISTORY, and extensions are merely propagated.
QUALITY is not derived from the input NDF for a linearly interpolated NDF. The DATA and VARIANCE arrays are resampled.
If USEWCS is TRUE, then the NDF WCS component is updated and propagated.
Bad pixels, including automatic quality masking, are supported.
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Batch normalization layer - MATLAB - MathWorks ä¸å›½
After normalization, the layer scales the input with a learnable scale factor γ and shifts it by a learnable offset β.
Channel scale factors γ, specified as a numeric vector.
Channel offsets β, specified as a numeric vector.
{\mathrm{μ}}^{*}={\mathrm{λ}}_{\mathrm{μ}}\stackrel{^}{\mathrm{μ}}+\left(1â{\mathrm{λ}}_{\mathrm{μ}}\right)\mathrm{μ},
{\mathrm{μ}}^{*}
{\mathrm{λ}}_{\mathrm{μ}}
\stackrel{^}{\mathrm{μ}}
\mathrm{μ}
{\mathrm{Ï}}^{2}{}^{*}={\mathrm{λ}}_{{\mathrm{Ï}}^{2}}\stackrel{^}{{\mathrm{Ï}}^{2}}+\left(1â{\mathrm{λ}}_{{\mathrm{Ï}}^{2}}\right){\mathrm{Ï}}^{2},
{\mathrm{Ï}}^{2}{}^{*}
{\mathrm{λ}}_{{\mathrm{Ï}}^{2}}
\stackrel{^}{{\mathrm{Ï}}^{2}}
{\mathrm{Ï}}^{2}
The layer first normalizes the activations of each channel by subtracting the mini-batch mean and dividing by the mini-batch standard deviation. Then, the layer shifts the input by a learnable offset β and scales it by a learnable scale factor γ. β and γ are themselves learnable parameters that are updated during network training.
The batch normalization operation normalizes the elements xi of the input by first calculating the mean μB and variance σB2 over the spatial, time, and observation dimensions for each channel independently. Then, it calculates the normalized activations as
\stackrel{^}{{x}_{i}}=\frac{{x}_{i}â{\mathrm{μ}}_{B}}{\sqrt{{\mathrm{Ï}}_{B}^{2}+\mathrm{ϵ}}},
where ϵ is a constant that improves numerical stability when the variance is very small.
{y}_{i}=\mathrm{γ}{\stackrel{^}{x}}_{i}+\mathrm{β},
where the offset β and scale factor γ are learnable parameters that are updated during network training.
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Exiting the Market (Withdrawals) - Premia
LPs can withdraw their free capital at any time and have a few options for withdrawing their active capital.
Liquidity providers can withdraw their un-utilized free capital in a pool at any moment in time (including unused premiums). If an LP would like to remove active (underwritten option) capital from a pool, they have two choices. They can either pay the pool price to sell the option to another underwriter in the pool, or they can use the Gradual Withdrawal feature to prevent their capital from being used to underwrite future options after expiration.
At any given time of withdrawal, a LP can have up to 3 components to consider: 1) Free capital (or intermediate pool capital); 2) accumulated premiums; 3) active positions underwritten.
Users can withdraw free capital or accumulated fees at any time for no fee.
At time of withdrawal request, the amount available for withdrawal is calculated as follows:
WithdrawalAmount = FreeCapital + PremiumsEarned + AvailableActiveCapital
FreeCapital
= amount of free/intermediate capital available
PremiumsEarned
= total premiums accumulated from option sales
AvailableActiveCapital = sum(OptionSize * (1-OutstandingExposure))
OutstandingExposure
= outstanding pool value of underwritten options
LPs can always instantly withdraw their full value from a pool, if they'd like to. However, if a user chooses to withdraw their
AvailableActiveCapital
, the user will need to pay the pool price of their actively underwritten options (the
OutstandingExposure)
to withdraw the remaining collateral value.
Alternatively to the complete withdrawal of active funds (at cost), an LP can select a gradual withdrawal. This will prevent the user's capital from being used to underwrite future option sale. As soon as capital is released from unwound positions, it is blocked from re-converting to free capital and can be withdrawn by the user.
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P (25o N, 35o E), Q (25o N, 40o W), R and S are four points on the surface of the earth. PS is the diameter of the common parallel of latitude 25o N.
(a) Find the longitude of S.
(b) R lies 3300 nautical miles due south of P measured along the surface of the earth.
Calculate the latitude of R.
(c) Calculate the shortest distance, in nautical mile, from P to S measured along the surface of the earth.
(d) An aeroplane took off from R and flew due north to P. Then, it flew due west to Q.
(i) Calculate the distance, in nautical mile, from P due west Q measured along the common parallel of latitude.
Longitude of S = (180o – 35o) W = 145o W
\begin{array}{l}\angle POR=\frac{3300}{60}\\ \text{ }={55}^{o}\\ \text{Latitude of }R={\left(55-25\right)}^{o}\\ \text{ }={30}^{o}S\end{array}
Shortest distance from P to S
= (65 + 65) x 60
Distance of PQ
= 75 x 60 x cos 25o
\begin{array}{l}\left(\text{ii}\right)\\ \text{Total distance travelled}\\ RP+PQ\\ =3300+4078.4\\ =7378.4\text{ nautical miles}\\ \\ \text{Average speed =}\frac{\text{Total distance travelled}}{\text{Time taken}}\\ \text{}=\frac{7378.4}{12.4}←\overline{)\text{12 hours 24 min}=12+\frac{12}{60}=12+0.4}\\ \text{}=595.0\text{ knot}\end{array}
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Custom network function for nonlinear ARX and Hammerstein-Wiener models - MATLAB - MathWorks 日本
Mathematically, idCustomNetwork is a function that maps m inputs X(t) = [x(t1),x2(t),…,xm(t)]T to a scalar output y(t) using the following relationship:
y\left(t\right)={y}_{0}+\mathrm{Χ}{\left(t\right)}^{T}PL+C\left(\mathrm{Χ}\left(t\right)\right)
{F}_{L}\left(t\right)=\mathrm{Χ}{\left(t\right)}^{T}PL
\left(Xâ\stackrel{¯}{X}\right)
C\left(X\right)=\underset{i=1}{\overset{n}{â}}{s}_{i}f\left({X}^{T}Q{b}_{i}+{c}_{i}\right)
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how the formula sin(2npi+x)=sinx came why was 2npi avoided - Maths - Trigonometric Functions - 7779345 | Meritnation.com
how the formula sin(2npi+x)=sinx came.why was 2npi avoided
As every trigonometric function is periodic for 360 degrees , it means after every 360 degrees the values will be repeated again.
Hence for sin (2n
\mathrm{\pi }
+ x ) = sinx , we can take any n here and n is natural number.
Similarly it is true for cos , tan also.
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Elastic_instability Knowpia
Elastic instability is a form of instability occurring in elastic systems, such as buckling of beams and plates subject to large compressive loads.
Elastic instability of a rigid beam supported by an angular spring.
There are a lot of ways to study this kind of instability. One of them is to use the method of incremental deformations based on superposing a small perturbation on an equilibrium solution.
Single degree of freedom-systemsEdit
Consider as a simple example a rigid beam of length L, hinged in one end and free in the other, and having an angular spring attached to the hinged end. The beam is loaded in the free end by a force F acting in the compressive axial direction of the beam, see the figure to the right.
Moment equilibrium conditionEdit
Assuming a clockwise angular deflection
{\displaystyle \theta }
, the clockwise moment exerted by the force becomes
{\displaystyle M_{F}=FL\sin \theta }
. The moment equilibrium equation is given by
{\displaystyle FL\sin \theta =k_{\theta }\theta }
{\displaystyle k_{\theta }}
is the spring constant of the angular spring (Nm/radian). Assuming
{\displaystyle \theta }
is small enough, implementing the Taylor expansion of the sine function and keeping the two first terms yields
{\displaystyle FL{\Bigg (}\theta -{\frac {1}{6}}\theta ^{3}{\Bigg )}\approx k_{\theta }\theta }
which has three solutions, the trivial
{\displaystyle \theta =0}
{\displaystyle \theta \approx \pm {\sqrt {6{\Bigg (}1-{\frac {k_{\theta }}{FL}}{\Bigg )}}}}
which is imaginary (i.e. not physical) for
{\displaystyle FL<k_{\theta }}
and real otherwise. This implies that for small compressive forces, the only equilibrium state is given by
{\displaystyle \theta =0}
, while if the force exceeds the value
{\displaystyle k_{\theta }/L}
there is suddenly another mode of deformation possible.
Energy methodEdit
The same result can be obtained by considering energy relations. The energy stored in the angular spring is
{\displaystyle E_{\mathrm {spring} }=\int k_{\theta }\theta \mathrm {d} \theta ={\frac {1}{2}}k_{\theta }\theta ^{2}}
and the work done by the force is simply the force multiplied by the vertical displacement of the beam end, which is
{\displaystyle L(1-\cos \theta )}
{\displaystyle E_{\mathrm {force} }=\int {F\mathrm {d} x=FL(1-\cos \theta )}}
The energy equilibrium condition
{\displaystyle E_{\mathrm {spring} }=E_{\mathrm {force} }}
now yields
{\displaystyle F=k_{\theta }/L}
as before (besides from the trivial
{\displaystyle \theta =0}
Stability of the solutionsEdit
{\displaystyle \theta }
is stable iff a small change in the deformation angle
{\displaystyle \Delta \theta }
results in a reaction moment trying to restore the original angle of deformation. The net clockwise moment acting on the beam is
{\displaystyle M(\theta )=FL\sin \theta -k_{\theta }\theta }
An infinitesimal clockwise change of the deformation angle
{\displaystyle \theta }
results in a moment
{\displaystyle M(\theta +\Delta \theta )=M+\Delta M=FL(\sin \theta +\Delta \theta \cos \theta )-k_{\theta }(\theta +\Delta \theta )}
{\displaystyle \Delta M=\Delta \theta (FL\cos \theta -k_{\theta })}
{\displaystyle FL\sin \theta =k_{\theta }\theta }
due to the moment equilibrium condition. Now, a solution
{\displaystyle \theta }
is stable iff a clockwise change
{\displaystyle \Delta \theta >0}
results in a negative change of moment
{\displaystyle \Delta M<0}
and vice versa. Thus, the condition for stability becomes
{\displaystyle {\frac {\Delta M}{\Delta \theta }}={\frac {\mathrm {d} M}{\mathrm {d} \theta }}=FL\cos \theta -k_{\theta }<0}
{\displaystyle \theta =0}
is stable only for
{\displaystyle FL<k_{\theta }}
, which is expected. By expanding the cosine term in the equation, the approximate stability condition is obtained:
{\displaystyle |\theta |>{\sqrt {2{\Bigg (}1-{\frac {k_{\theta }}{FL}}{\Bigg )}}}}
{\displaystyle FL>k_{\theta }}
, which the two other solutions satisfy. Hence, these solutions are stable.
Multiple degrees of freedom-systemsEdit
Elastic instability, 2 degrees of freedom
By attaching another rigid beam to the original system by means of an angular spring a two degrees of freedom-system is obtained. Assume for simplicity that the beam lengths and angular springs are equal. The equilibrium conditions become
{\displaystyle FL(\sin \theta _{1}+\sin \theta _{2})=k_{\theta }\theta _{1}}
{\displaystyle FL\sin \theta _{2}=k_{\theta }(\theta _{2}-\theta _{1})}
{\displaystyle \theta _{1}}
{\displaystyle \theta _{2}}
are the angles of the two beams. Linearizing by assuming these angles are small yields
{\displaystyle {\begin{pmatrix}FL-k_{\theta }&FL\\k_{\theta }&FL-k_{\theta }\end{pmatrix}}{\begin{pmatrix}\theta _{1}\\\theta _{2}\end{pmatrix}}={\begin{pmatrix}0\\0\end{pmatrix}}}
The non-trivial solutions to the system is obtained by finding the roots of the determinant of the system matrix, i.e. for
{\displaystyle {\frac {FL}{k_{\theta }}}={\frac {3}{2}}\mp {\frac {\sqrt {5}}{2}}\approx \left\{{\begin{matrix}0.382\\2.618\end{matrix}}\right.}
Thus, for the two degrees of freedom-system there are two critical values for the applied force F. These correspond to two different modes of deformation which can be computed from the nullspace of the system matrix. Dividing the equations by
{\displaystyle \theta _{1}}
{\displaystyle {\frac {\theta _{2}}{\theta _{1}}}{\Big |}_{\theta _{1}\neq 0}={\frac {k_{\theta }}{FL}}-1\approx \left\{{\begin{matrix}1.618&{\text{for }}FL/k_{\theta }\approx 0.382\\-0.618&{\text{for }}FL/k_{\theta }\approx 2.618\end{matrix}}\right.}
For the lower critical force the ratio is positive and the two beams deflect in the same direction while for the higher force they form a "banana" shape. These two states of deformation represent the buckling mode shapes of the system.
Cavitation (elastomers)
Theory of elastic stability, S. Timoshenko and J. Gere
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Global Constraint Catalog: Cminimum
<< 5.261. min_width_valley5.263. minimum_except_0 >>
\mathrm{𝚖𝚒𝚗𝚒𝚖𝚞𝚖}\left(\mathrm{𝙼𝙸𝙽},\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\right)
\mathrm{𝚖𝚒𝚗}
\mathrm{𝙼𝙸𝙽}
\mathrm{𝚍𝚟𝚊𝚛}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚟𝚊𝚛}-\mathrm{𝚍𝚟𝚊𝚛}\right)
|\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}|>0
\mathrm{𝚛𝚎𝚚𝚞𝚒𝚛𝚎𝚍}
\left(\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂},\mathrm{𝚟𝚊𝚛}\right)
\mathrm{𝙼𝙸𝙽}
is the minimum value of the collection of domain variables
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
\left(2,〈3,2,7,2,6〉\right)
\left(7,〈8,8,7,8,7〉\right)
\mathrm{𝚖𝚒𝚗𝚒𝚖𝚞𝚖}
\mathrm{𝙼𝙸𝙽}=2
is set to the minimum value of the collection
〈3,2,7,2,6〉
|\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}|>1
\mathrm{𝚛𝚊𝚗𝚐𝚎}
\left(\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}.\mathrm{𝚟𝚊𝚛}\right)>1
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}.\mathrm{𝚟𝚊𝚛}
One and the same constant can be added to
\mathrm{𝙼𝙸𝙽}
as well as to the
\mathrm{𝚟𝚊𝚛}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
\mathrm{𝙼𝙸𝙽}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
\mathrm{𝙼𝙸𝙽}\left(\mathrm{𝚖𝚒𝚗}\right)
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\left(\mathrm{𝚞𝚗𝚒𝚘𝚗}\right)
In some project scheduling problems one has to introduce dummy activities that correspond for instance to the starting time of a given set of activities. In this context one can use the
\mathrm{𝚖𝚒𝚗𝚒𝚖𝚞𝚖}
constraint to get the minimum starting time of a set of tasks.
\mathrm{𝚖𝚒𝚗𝚒𝚖𝚞𝚖}
is a constraint and not just a function that computes the minimum value of a collection of variables: potential values of
\mathrm{𝙼𝙸𝙽}
influence the variables of
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
, and reciprocally potential values that can be assigned to variables of
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
\mathrm{𝙼𝙸𝙽}
\mathrm{𝚖𝚒𝚗𝚒𝚖𝚞𝚖}
\mathrm{𝚖𝚒𝚗}
in JaCoP (http://www.jacop.eu/).
\mathrm{𝚖𝚒𝚗𝚒𝚖𝚞𝚖}
constraint is described in [Beldiceanu01].
\mathrm{𝚖𝚒𝚗𝚒𝚖𝚞𝚖}
constraint is entailed if all the following conditions hold:
\mathrm{𝙼𝙸𝙽}
At least one variable of
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
is assigned value
\mathrm{𝙼𝙸𝙽}
All variables of
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
have their minimum value greater than or equal to value
\mathrm{𝙼𝙸𝙽}
n
\mathrm{𝚖𝚒𝚗𝚒𝚖𝚞𝚖}
0..n
n
1 3 19 175 2101 31031 543607 11012415
2 1 7 65 781 11529 201811 4085185
3 - 1 15 211 3367 61741 1288991
4 - - 1 31 665 14197 325089
5 - - - 1 63 2059 58975
6 - - - - 1 127 6305
7 - - - - - 1 255
\mathrm{𝚖𝚒𝚗𝚒𝚖𝚞𝚖}
0..n
min in Choco, min in Gecode, min in JaCoP, minimum in MiniZinc, minimum in SICStus.
\mathrm{𝚖𝚒𝚗𝚒𝚖𝚞𝚖}_\mathrm{𝚐𝚛𝚎𝚊𝚝𝚎𝚛}_\mathrm{𝚝𝚑𝚊𝚗}
\mathrm{𝚗𝚎𝚡𝚝}_\mathrm{𝚎𝚕𝚎𝚖𝚎𝚗𝚝}
\mathrm{𝚗𝚎𝚡𝚝}_\mathrm{𝚐𝚛𝚎𝚊𝚝𝚎𝚛}_\mathrm{𝚎𝚕𝚎𝚖𝚎𝚗𝚝}
\mathrm{𝚖𝚊𝚡𝚒𝚖𝚞𝚖}
(order constraint).
comparison swapped:
\mathrm{𝚖𝚊𝚡𝚒𝚖𝚞𝚖}
\mathrm{𝚖𝚒𝚗𝚒𝚖𝚞𝚖}_\mathrm{𝚖𝚘𝚍𝚞𝚕𝚘}
\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎}
\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎}\mathrm{mod}\mathrm{𝚌𝚘𝚗𝚜𝚝𝚊𝚗𝚝}
\mathrm{𝚊𝚗𝚍}
\mathrm{𝚋𝚎𝚝𝚠𝚎𝚎𝚗}_\mathrm{𝚖𝚒𝚗}_\mathrm{𝚖𝚊𝚡}
\mathrm{𝚒𝚗}
\mathrm{𝚖𝚒𝚗𝚒𝚖𝚞𝚖}_\mathrm{𝚎𝚡𝚌𝚎𝚙𝚝}_\mathtt{0}
(value 0 is ignored),
\mathrm{𝚘𝚙𝚎𝚗}_\mathrm{𝚖𝚒𝚗𝚒𝚖𝚞𝚖}
(open constraint).
\mathrm{𝚖𝚒𝚗}_𝚗
(minimum or order
𝚗
replaced by absolute minimum).
uses in its reformulation:
\mathrm{𝚌𝚢𝚌𝚕𝚎}
characteristic of a constraint: minimum, maxint, automaton, automaton without counters, reified automaton constraint.
constraint arguments: reverse of a constraint, pure functional dependency.
filtering: glue matrix, arc-consistency, entailment.
\mathrm{𝚖𝚒𝚗𝚒𝚖𝚞𝚖}\left(\mathrm{𝙼𝙸𝙽},\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\right)
\mathrm{𝚏𝚒𝚛𝚜𝚝}\left(\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}.\mathrm{𝚟𝚊𝚛}\right)>\mathrm{𝙼𝙸𝙽}
\mathrm{𝚕𝚊𝚜𝚝}\left(\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}.\mathrm{𝚟𝚊𝚛}\right)>\mathrm{𝙼𝙸𝙽}
\mathrm{𝚍𝚎𝚎𝚙𝚎𝚜𝚝}_\mathrm{𝚟𝚊𝚕𝚕𝚎𝚢}
\left(\mathrm{𝙳𝙴𝙿𝚃𝙷},\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\right)
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
\mathrm{𝐶𝐿𝐼𝑄𝑈𝐸}
↦\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}\mathtt{1},\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}\mathtt{2}\right)
\bigvee \left(\begin{array}{c}\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}\mathtt{1}.\mathrm{𝚔𝚎𝚢}=\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}\mathtt{2}.\mathrm{𝚔𝚎𝚢},\hfill \\ \mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}\mathtt{1}.\mathrm{𝚟𝚊𝚛}<\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}\mathtt{2}.\mathrm{𝚟𝚊𝚛}\hfill \end{array}\right)
\mathrm{𝐎𝐑𝐃𝐄𝐑}
\left(0,\mathrm{𝙼𝙰𝚇𝙸𝙽𝚃},\mathrm{𝚟𝚊𝚛}\right)=\mathrm{𝙼𝙸𝙽}
\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}\mathtt{1}.\mathrm{𝚔𝚎𝚢}=\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}\mathtt{2}.\mathrm{𝚔𝚎𝚢}
\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}\mathtt{1}
\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}\mathtt{2}
corresponds to the same vertex. It is used in order to enforce to keep all the vertices of the initial graph.
\mathrm{𝐎𝐑𝐃𝐄𝐑}\left(0,\mathrm{𝙼𝙰𝚇𝙸𝙽𝚃},\mathrm{𝚟𝚊𝚛}\right)
refers to the source vertices of the graph, i.e., those vertices that do not have any predecessor.
\mathrm{𝐎𝐑𝐃𝐄𝐑}
graph property, the vertices of rank 0 (without considering the loops) of the final graph are outlined with a thick circle.
\mathrm{𝚖𝚒𝚗𝚒𝚖𝚞𝚖}
Figure 5.262.2 depicts a first counter free deterministic automaton associated with the
\mathrm{𝚖𝚒𝚗𝚒𝚖𝚞𝚖}
{\mathrm{𝚅𝙰𝚁}}_{i}
{i}^{th}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
\left(\mathrm{𝙼𝙸𝙽},{\mathrm{𝚅𝙰𝚁}}_{i}\right)
{S}_{i}
\left(\mathrm{𝙼𝙸𝙽}<{\mathrm{𝚅𝙰𝚁}}_{i}⇔{S}_{i}=0\right)\wedge \left(\mathrm{𝙼𝙸𝙽}={\mathrm{𝚅𝙰𝚁}}_{i}⇔{S}_{i}=1\right)\wedge \left(\mathrm{𝙼𝙸𝙽}>{\mathrm{𝚅𝙰𝚁}}_{i}⇔{S}_{i}=2\right)
Figure 5.262.2. Counter free automaton of the
\mathrm{𝚖𝚒𝚗𝚒𝚖𝚞𝚖}
\mathrm{𝚖𝚒𝚗𝚒𝚖𝚞𝚖}
Figure 5.262.3 depicts a second counter free non deterministic automaton associated with the
\mathrm{𝚖𝚒𝚗𝚒𝚖𝚞𝚖}
constraint, where the argument
\mathrm{𝙼𝙸𝙽}
is also part of the sequence passed to the automaton.
Figure 5.262.4. Counter free non deterministic automaton of the
\mathrm{𝚖𝚒𝚗𝚒𝚖𝚞𝚖}
\left(\mathrm{𝙼𝙸𝙽},\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\right)
constraint assuming that the union of the domain of the variables is the set
\left\{1,2,3,4\right\}
and that the elements of
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
are first passed to the automaton followed by
\mathrm{𝙼𝙸𝙽}
{s}_{i}
means that no value strictly less than value
i
was found and that value
i
was already encountered at least once)
Figure 5.262.5. Hypergraph of the reformulation corresponding to the counter free non deterministic automaton of the
\mathrm{𝚖𝚒𝚗𝚒𝚖𝚞𝚖}
Figure 5.262.6 depicts a third deterministic automaton with one counter associated with the
\mathrm{𝚖𝚒𝚗𝚒𝚖𝚞𝚖}
\mathrm{𝙼𝙸𝙽}
is unified to the final value of the counter.
Figure 5.262.6. Automaton (with one counter) of the
\mathrm{𝚖𝚒𝚗𝚒𝚖𝚞𝚖}
constraint and its glue matrix
Figure 5.262.7. Hypergraph of the reformulation corresponding to the automaton (with one counter) of the
\mathrm{𝚖𝚒𝚗𝚒𝚖𝚞𝚖}
{Q}_{0},{Q}_{1},\cdots ,{Q}_{n}
s
C
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Dynamic range limiter - MATLAB - MathWorks í•œêµ
Specific to limiter
Limit Audio Signal
Tune Limiter Parameters
Sidechain Ducking with Limiter
Dynamic range limiter
The limiter System object™ performs brick-wall dynamic range limiting independently across each input channel. Dynamic range limiting suppresses the volume of loud sounds that cross a given threshold. It uses specified attack and release times to achieve a smooth applied gain curve. Properties of the limiter System object specify the type of dynamic range limiting.
To perform dynamic range limiting:
Create the limiter object and set its properties.
dRL = limiter
dRL = limiter(thresholdValue)
dRL = limiter(___,Name,Value)
dRL = limiter creates a System object, dRL, that performs brick-wall dynamic range limiting independently across each input channel.
dRL = limiter(thresholdValue) sets the Threshold property to thresholdValue.
dRL = limiter(___,Name,Value) sets each property Name to the specified Value. Unspecified properties have default values.
Example: dRL = limiter('AttackTime',0.01,'SampleRate',16000) creates a System object, dRL, with a 10 ms attack time and a sample rate of 16 kHz.
Knee width is the transition area in the limiter characteristic.
y=xâ\frac{{\left(xâT+\frac{W}{2}\right)}^{2}}{\left(2ÃW\right)}
\left(2Ã|xâT|\right)â¤W
Attack time is the time it takes the limiter gain to rise from 10% to 90% of its final value when the input goes above the threshold.
Release time is the time it takes the limiter gain to drop from 90% to 10% of its final value when the input goes below the threshold.
'Auto' –– Make-up gain is applied at the output of the dynamic range limiter such that a steady-state 0 dB input has a 0 dB output.
Make-up gain compensates for gain lost during limiting. It is applied at the output of the dynamic range limiter.
Enable sidechain input, specified as true or false. This property determines the number of available inputs on the limiter object.
false –– Sidechain input is disabled and the limiter object accepts one input: the audioIn data to be limited.
true –– Sidechain input is enabled and the limiter object accepts two inputs: the audioIn data to be limited and the sidechain input used to compute the limiter gain.
audioOut = dRL(audioIn)
[audioOut,gain] = dRL(audioIn)
audioOut = dRL(audioIn)performs dynamic range limiting on the input signal, audioIn, and returns the limited signal, audioOut. The type of dynamic range limiting is specified by the algorithm and properties of the limiter System object, dRL.
[audioOut,gain] = dRL(audioIn)also returns the applied gain, in dB, at each input sample.
audioIn — Audio input to limiter
Audio input to the limiter, specified as a matrix. The columns of the matrix are treated as independent audio channels.
audioOut — Audio output from limiter
Audio output from the limiter, returned as a matrix the same size as audioIn.
gain — Gain applied by limiter (dB)
Gain applied by the limiter, returned as a matrix the same size as audioIn.
The createAudioPluginClass and configureMIDI functions map tunable properties of the limiter System object to user-facing parameters:
Use dynamic range limiting to suppress the volume of loud sounds.
Set up the limiter to have a threshold of -15 dB, an attack time of 0.005 seconds, and a release time of 0.1 seconds. Set make-up gain to 0 dB (default). To specify this value, set the make-up gain mode to 'Property' but do not specify the MakeUpGain property. Use the sample rate of your audio file reader.
dRL = limiter(-15, ...
'ReleaseTime',0.1, ...
Set up a time scope to visualize the original signal and the limited signal.
'Title',['Original vs. Limited Audio (top)' ...
' and Limiter Gain in dB (bottom)']);
[y,g] = dRL(x);
g1 = g(:,1);
scope([x1,y1],g1);
Create a dsp.AudioFileReader to read in audio frame-by-frame. Create a audioDeviceWriter to write audio to your sound card. Create a limiter to process the audio data.
dRL = limiter('SampleRate',fileReader.SampleRate);
Call parameterTuner to open a UI to tune parameters of the limiter while streaming.
parameterTuner(dRL)
Apply dynamic range limiting.
While streaming, tune parameters of the dynamic range limiter and listen to the effect.
audioOut = dRL(audioIn);
Use the EnableSidechain input of a limiter object to limit the amplitude level of a separate audio signal. The sidechain signal controls the dynamic range limiting of the input audio signal. When the sidechain signal exceeds the limiter Threshold, the limiter activates and limits the amplitude of the input signal. When the sidechain signal level falls below the threshold, the audio input returns to its original amplitude. For a detailed comparison of compression and dynamic range limiting, see Compare Dynamic Range Limiter and Compressor.
In this section, you resample and zero-pad a speech file to use as input to the EnableSidechain property of your limiter object.
Pad the beginning of the resampled signal with 10 seconds worth of zeros. This allows the input audio signal to be clearly heard before any limiting is applied.
Construct a limiter object. Use a fast AttackTime, and a moderately slow ReleaseTime. These settings are ideal for voice-over work. The fast attack time ensures that the input audio is limited almost immediately after the sidechain signal surpasses the limiter threshold. The slow release time ensures the limiting on the input audio lasts through any potential short silent regions in the sidechain signal.
iAmYourLimiter = limiter('EnableSidechain',true,...
'ReleaseTime',1.75);
'Title','Original Input Audio - Guitar');
scope.Title = 'Dynamic Range Limited Input Audio - Guitar';
Read in a frame of audio from your input and sidechain signals. Process your input and sidechain signals with your limiter object. Playback your processed audio signals and display the audio data using a timescope object.
The top panel of your timescope displays the unprocessed input audio signal and the middle panel displays the sidechain audio signal. The bottom panel displays the limited input audio signal. Notice the amplitudes of the signals in the top and bottom panels are identical until the sidechain signal begins. Once the sidechain signal activates, the amplitude in the bottom panel decreases. Once the sidechain signal ends, the amplitude of the bottom panel returns to its original level.
limiterOutput = iAmYourLimiter(inputAudioFrame,sideChainAudioFrame);
afw(sideChainAudioFrame+limiterOutput);
scope(inputAudioFrame,sideChainAudioFrame,limiterOutput);
release(iAmYourLimiter)
The limiter System object processes a signal frame by frame and element by element.
{x}_{\text{dB}}\left[n\right]=20Ã{\mathrm{log}}_{10}|x\left[n\right]|
xdB[n] passes through the gain computer. The gain computer uses the static characteristic properties of the dynamic range limiter to brick-wall gain that is above the threshold.
{x}_{\text{sc}}\left({x}_{\text{dB}}\right)=\left\{\begin{array}{cc}{x}_{\text{dB}}& {x}_{\text{dB}}<\left(Tâ\frac{W}{2}\right)\\ {x}_{\text{dB}}â\frac{{\left({x}_{\text{dB}}âT+\frac{W}{2}\right)}^{2}}{2W}& \begin{array}{c}\\ \\ \end{array}\left(Tâ\frac{W}{2}\right)â¤{x}_{\text{dB}}â¤\left(T+\frac{W}{2}\right)\\ T& {x}_{\text{dB}}>\left(T+\frac{W}{2}\right)\end{array}\text{â},
where T is the threshold and W is the knee width.
{x}_{\text{sc}}\left({x}_{\text{dB}}\right)=\left\{\begin{array}{cc}{x}_{\text{dB}}& {x}_{\text{dB}}<T\\ T& {x}_{\text{dB}}â¥T\end{array}
{g}_{\text{c}}\left[n\right]={x}_{\text{sc}}\left[n\right]â{x}_{\text{dB}}\left[n\right].
gc[n] is smoothed using specified attack and release time:
{g}_{\text{s}}\left[n\right]=\left\{\begin{array}{cc}{\mathrm{α}}_{\text{A}}{g}_{\text{s}}\left[nâ1\right]+\left(1â{\mathrm{α}}_{\text{A}}\right){g}_{\text{c}}\left[n\right],& {g}_{\text{c}}\left[n\right]â¤{g}_{\text{s}}\left[nâ1\right]\\ {\mathrm{α}}_{\text{R}}{g}_{\text{s}}\left[nâ1\right]+\left(1â{\mathrm{α}}_{\text{R}}\right){g}_{\text{c}}\left[n\right],& {g}_{\text{c}}\left[n\right]>{g}_{\text{s}}\left[nâ1\right]\end{array}
{\mathrm{α}}_{\text{A}}=\mathrm{exp}\left(\frac{â\mathrm{log}\left(9\right)}{FsÃ{T}_{\text{A}}}\right)\text{â}.
{\mathrm{α}}_{\text{R}}=\mathrm{exp}\left(\frac{â\mathrm{log}\left(9\right)}{FsÃ{T}_{\text{R}}}\right)\text{â}.
If MakeUpGainMode is set to the default 'Auto', the make-up gain is calculated as the negative of the computed gain for a 0 dB input:
M={â{x}_{\text{sc}}|}_{{x}_{\text{dB}}=0}
Given a steady-state input of 0 dB, this configuration achieves a steady-state output of 0 dB. The make-up gain is determined by the Threshold and KneeWidth properties. It does not depend on the input signal.
{g}_{\text{m}}\left[n\right]={g}_{\text{s}}\left[n\right]+M
{g}_{\text{lin}}\left[n\right]={10}^{\left(\frac{{g}_{\text{m}}\left[n\right]}{20}\right)}\text{â}.
The output of the dynamic range limiter is given as
y\left[n\right]=x\left[n\right]Ã{g}_{\text{lin}}\left[n\right].
Limiter | noiseGate | compressor | expander
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Polynomial curve fitting - MATLAB polyfit
p\left(x\right)={p}_{1}{x}^{n}+{p}_{2}{x}^{n-1}+...+{p}_{n}x+{p}_{n+1}.
\stackrel{^}{x}=\frac{x-\overline{x}}{{\sigma }_{x}}\text{\hspace{0.17em}}.
y\left(x\right)=\left(1+x{\right)}^{-1}
\mathit{y}±2\Delta
\left(\begin{array}{cccc}{x}_{1}^{n}& {x}_{1}^{n-1}& \cdots & 1\\ {x}_{2}^{n}& {x}_{2}^{n-1}& \cdots & 1\\ ⋮& ⋮& \ddots & ⋮\\ {x}_{m}^{n}& {x}_{m}^{n-1}& \cdots & 1\end{array}\right)\left(\begin{array}{c}{p}_{1}\\ {p}_{2}\\ ⋮\\ {p}_{n+1}\end{array}\right)=\left(\begin{array}{c}{y}_{1}\\ {y}_{2}\\ ⋮\\ {y}_{m}\end{array}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}},
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Quantile expected shortfall (ES) backtest by Acerbi and Szekely - MATLAB quantile - MathWorks Australia
Run an ES Quantile Test
Quantile expected shortfall (ES) backtest by Acerbi and Szekely
TestResults = quantile(ebts)
[TestResults,SimTestStatistic] = quantile(ebts,Name,Value)
TestResults = quantile(ebts) runs the quantile ES backtest of Acerbi-Szekely (2014).
[TestResults,SimTestStatistic] = quantile(ebts,Name,Value) adds an optional name-value pair argument for TestLevel.
Generate the ES quantile test report.
PortfolioID VaRID VaRLevel Quantile PValue TestStatistic CriticalValue Observations Scenarios TestLevel
___________ _____________ ________ ________ ______ _____________ _____________ ____________ _________ _________
"S&P" "t(10) 95%" 0.95 reject 0.002 -0.10602 -0.055798 1966 1000 0.95
"S&P" "t(10) 97.5%" 0.975 reject 0 -0.15697 -0.073513 1966 1000 0.95
"S&P" "t(10) 99%" 0.99 reject 0 -0.26561 -0.10117 1966 1000 0.95
Example: [TestResults,SimTestStatistic] = quantile(ebts,'TestLevel',0.99)
0.95 (default) | numeric with values between 0 and 1
'Quantile'— Categorical array with categories 'accept' and 'reject' indicating the result of the quantile test
'PValue'— P-value of the quantile test
'TestStatistic'— Quantile test statistic
'CriticalValue'— Critical value for the quantile test
'Scenarios'— Number of scenarios simulated to get the p-values
'TestLevel'— Test confidence level
The quantile test (also known as the third Acerbi-Szekely test) uses a sample estimator of the expected shortfall.
The expected shortfall for a sample Y1,...,YN is:
\stackrel{⌢}{ES}\left(Y\right)=-\frac{1}{⌊N{p}_{VaR}⌋}\sum _{i=1}^{⌊N{p}_{VaR}⌋}{Y}_{\left[i\right]}
Y[1],...,Y[N] are the sorted sample values (from smallest to largest), and
⌊N{p}_{VaR}⌋
{U}_{1}={P}_{1}\left({X}_{1}\right),...,{U}_{N}={P}_{N}\left({X}_{N}\right)
using the cumulative distribution function Pt. If the distribution assumptions are correct, the rank values U1,...,UN are uniformly distributed in the interval (0,1). Then at each time t:
Invert the ranks U = (U1,...,UN) to get N quantiles
{P}_{t}^{-1}\left(U\right)=\left({P}_{t}^{-1}\left({U}_{1}\right),...,{P}_{t}^{-1}\left({U}_{N}\right)\right)
\stackrel{⌢}{ES}\left({P}_{t}^{-1}\left(U\right)\right)
E\left[\stackrel{⌢}{ES}\left({P}_{t}^{-1}\left(V\right)\right)\right]
where V = (V1,...,VN is a sample of N independent uniform random variables in the interval (0,1). This value can be computed analytically.
Define the quantile test statistic as
{Z}_{quantile}=-\frac{1}{N}\sum _{t=1}^{N}\frac{\stackrel{⌢}{ES}\left({P}_{t}^{-1}\left(U\right)\right)}{E\left[\stackrel{⌢}{ES}\left({P}_{t}^{-1}\left(V\right)\right)\right]}+1
E\left[\stackrel{⌢}{ES}\left({P}_{t}^{-1}\left(V\right)\right)\right]=-\frac{N}{⌊{N}_{pVaR}⌋}{\int }_{0}^{1}{I}_{1-p}\left(N-⌊{N}_{pVaR}⌋,⌊{N}_{pVaR}⌋\right){P}_{t}^{-1}\left(p\right)dp
where Ix(z,w) is the regularized incomplete beta function. For more information, see betainc.
Assuming that the distributional assumptions are correct, the expected value of the test statistic Zquantile is 0.
E\left[{Z}_{quantile}\right]=0
Negative values of the test statistic indicate risk underestimation. The quantile test is a one-sided test that rejects the model when there is evidence that the model underestimates risk. (For technical details on the null and alternative hypotheses, see Acerbi-Szekely, 2014). The quantile test rejects the model when the p-value is less than 1 minus the test confidence level.
For more information on simulating the test statistics and computing the p-values and critical values, see simulate.
The quantile test statistic is well-defined when there are no VaR failures in the data.
However, when the expected number of failures NpVaR is small, an adjustment is required. The sample estimator of the expected shortfall takes the average of the smallest Ntail observations in the sample, where
{N}_{tail}=⌊{N}_{pVaR}⌋
. If NpVaR < 1, then Ntail = 0, the sample estimator of the expected shortfall becomes an empty sum, and the quantile test statistic is undefined.
To account for this, whenever NpVaR < 1, the value of Ntail is set to 1. Thus, the sample estimator of the expected shortfall has a single term and is equal to the minimum value of the sample. With this adjustment, the quantile test statistic is then well-defined and the significance analysis is unchanged.
summary | runtests | conditional | unconditional | simulate | minBiasRelative | minBiasAbsolute | esbacktestbysim | esbacktestbyde
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Global Constraint Catalog: Ccardinality_atleast
<< 5.57. calendar5.59. cardinality_atmost >>
\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}
\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}_\mathrm{𝚊𝚝𝚕𝚎𝚊𝚜𝚝}\left(\mathrm{𝙰𝚃𝙻𝙴𝙰𝚂𝚃},\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂},\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}\right)
\mathrm{𝙰𝚃𝙻𝙴𝙰𝚂𝚃}
\mathrm{𝚍𝚟𝚊𝚛}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚟𝚊𝚛}-\mathrm{𝚍𝚟𝚊𝚛}\right)
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}
\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚟𝚊𝚕}-\mathrm{𝚒𝚗𝚝}\right)
\mathrm{𝙰𝚃𝙻𝙴𝙰𝚂𝚃}\ge 0
\mathrm{𝙰𝚃𝙻𝙴𝙰𝚂𝚃}\le |\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}|
\mathrm{𝚛𝚎𝚚𝚞𝚒𝚛𝚎𝚍}
\left(\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂},\mathrm{𝚟𝚊𝚛}\right)
\mathrm{𝚛𝚎𝚚𝚞𝚒𝚛𝚎𝚍}
\left(\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂},\mathrm{𝚟𝚊𝚕}\right)
\mathrm{𝚍𝚒𝚜𝚝𝚒𝚗𝚌𝚝}
\left(\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂},\mathrm{𝚟𝚊𝚕}\right)
\mathrm{𝙰𝚃𝙻𝙴𝙰𝚂𝚃}
is the minimum number of time that a value of
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}
is taken by the variables of the collection
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
\left(1,〈3,3,8〉,〈3,8〉\right)
In this example, values 3 and 8 are respectively used 2, and 1 times. The
\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}_\mathrm{𝚊𝚝𝚕𝚎𝚊𝚜𝚝}
\mathrm{𝙰𝚃𝙻𝙴𝙰𝚂𝚃}=1
is assigned to the minimum number of time that values 3 and 8 occur in the collection
〈3,3,8〉
\mathrm{𝙰𝚃𝙻𝙴𝙰𝚂𝚃}>0
\mathrm{𝙰𝚃𝙻𝙴𝙰𝚂𝚃}<|\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}|
|\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}|>1
|\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}|>0
|\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}|>|\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}|
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}.\mathrm{𝚟𝚊𝚛}
that does not belong to
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}.\mathrm{𝚟𝚊𝚕}
can be replaced by any other value that also does not belong to
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}.\mathrm{𝚟𝚊𝚕}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}.\mathrm{𝚟𝚊𝚛}
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}.\mathrm{𝚟𝚊𝚕}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}.\mathrm{𝚟𝚊𝚛}
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}.\mathrm{𝚟𝚊𝚕}
\mathrm{𝙰𝚃𝙻𝙴𝙰𝚂𝚃}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}
An application of the
\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}_\mathrm{𝚊𝚝𝚕𝚎𝚊𝚜𝚝}
constraint is to enforce a minimum use of values.
This is a restricted form of a variant of an
\mathrm{𝚊𝚖𝚘𝚗𝚐}
constraint and of the
\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}
constraint. In the original
\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}
constraint, one specifies for each value its minimum and maximum number of occurrences.
\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}
[Regin96].
\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}
(single
\mathrm{𝚌𝚘𝚞𝚗𝚝}\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎}
replaced by an individual
\mathrm{𝚌𝚘𝚞𝚗𝚝}\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎}
for each value).
modelling: functional dependency, at least.
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}
\mathrm{𝑃𝑅𝑂𝐷𝑈𝐶𝑇}
↦\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜},\mathrm{𝚟𝚊𝚕𝚞𝚎𝚜}\right)
\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}.\mathrm{𝚟𝚊𝚛}\ne \mathrm{𝚟𝚊𝚕𝚞𝚎𝚜}.\mathrm{𝚟𝚊𝚕}
\mathrm{𝐌𝐀𝐗}_\mathrm{𝐈𝐃}
=|\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}|-\mathrm{𝙰𝚃𝙻𝙴𝙰𝚂𝚃}
•
\mathrm{𝙰𝙲𝚈𝙲𝙻𝙸𝙲}
•
\mathrm{𝙱𝙸𝙿𝙰𝚁𝚃𝙸𝚃𝙴}
•
\mathrm{𝙽𝙾}_\mathrm{𝙻𝙾𝙾𝙿}
Using directly the graph property
\mathrm{𝐌𝐈𝐍}_\mathrm{𝐈𝐃}
=
\mathrm{𝙰𝚃𝙻𝙴𝙰𝚂𝚃}
, and replacing the disequality of the arc constraint by an equality does not work since it ignores values that are not assigned to any variable. This comes from the fact that isolated vertices are removed from the final graph.
Parts (A) and (B) of Figure 5.58.1 respectively show the initial and final graph associated with the Example slot. Since we use the
\mathrm{𝐌𝐀𝐗}_\mathrm{𝐈𝐃}
graph property, the vertex with the maximum number of predecessor (i.e., namely two predecessors) is stressed with a double circle. As a consequence the first argument
\mathrm{𝙰𝚃𝙻𝙴𝙰𝚂𝚃}
\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}_\mathrm{𝚊𝚝𝚕𝚎𝚊𝚜𝚝}
constraint is assigned to the total number of variables 3 minus 2.
Figure 5.58.1. Initial and final graph of the
\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}_\mathrm{𝚊𝚝𝚕𝚎𝚊𝚜𝚝}
Figure 5.58.2 depicts the automaton associated with the
\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}_\mathrm{𝚊𝚝𝚕𝚎𝚊𝚜𝚝}
{\mathrm{𝚅𝙰𝚁}}_{i}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
corresponds a 0-1 signature variable
{S}_{i}
. The following signature constraint links
{\mathrm{𝚅𝙰𝚁}}_{i}
{S}_{i}
{\mathrm{𝚅𝙰𝚁}}_{i}\in \mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}⇔{S}_{i}
Figure 5.58.2. Automaton of the
\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}_\mathrm{𝚊𝚝𝚕𝚎𝚊𝚜𝚝}
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Actor model - CodeDocs
Research on the actor model has been carried out at California Institute of Technology, Kyoto University Tokoro Laboratory, Microelectronics and Computer Technology Corporation (MCC), MIT Artificial Intelligence Laboratory, SRI, Stanford University, University of Illinois at Urbana–Champaign,[7]Pierre and Marie Curie University (University of Paris 6), University of Pisa, University of Tokyo Yonezawa Laboratory, Centrum Wiskunde & Informatica (CWI) and elsewhere.
The actor model can be used as a framework for modeling, understanding, and reasoning about a wide range of concurrent systems. For example:
Objects with locks (e.g., as in Java and C#) can be modeled as a serializer, provided that their implementations are such that messages can continually arrive (perhaps by being stored in an internal queue). A serializer is an important kind of actor defined by the property that it is continually available to the arrival of new messages; every message sent to a serializer is guaranteed to arrive.
In the early 1960s, interrupts began to be used to simulate the concurrent execution of several programs on one processor.[15] Having concurrency with shared memory gave rise to the problem of concurrency control. Originally, this problem was conceived as being one of mutual exclusion on a single computer. Edsger Dijkstra developed semaphores and later, between 1971 and 1973,[16]Tony Hoare[17] and Per Brinch Hansen[18] developed monitors to solve the mutual exclusion problem. However, neither of these solutions provided a programming language construct that encapsulated access to shared resources. This encapsulation was later accomplished by the serializer construct ([Hewitt and Atkinson 1977, 1979] and [Atkinson 1980]).
Composing actor systems
Modeling other concurrency systems
Computational Representation Theorem
{\displaystyle {\mathtt {S}}}
{\displaystyle {\mathtt {S}}}
{\displaystyle \mathbf {Denote} _{\mathtt {S}}\equiv \lim _{i\to \infty }\mathbf {progression} _{{\mathtt {S}}^{i}}(\bot _{\mathtt {S}})}
{\displaystyle \mathbf {Denote} _{\mathtt {S}}}
{\displaystyle {\mathtt {S}}}
Relationship to logic programming
Synthesizing addresses of actors
Early actor programming languages
Later actor programming languages
Actor libraries and frameworks
CloudI Active 2018-12-19[53] MIT C/C++, Elixir/Erlang/LFE, Go, Haskell, Java, Javascript, OCaml, Perl, PHP, Python, Ruby
vlingo Active 2020-07-26 Mozilla Public License 2.0 Java, Kotlin, soon .NET
^ Hewitt, Carl; Bishop, Peter; Steiger, Richard (1973). "A Universal Modular Actor Formalism for Artificial Intelligence". IJCAI. Cite journal requires |journal= (help)
^ a b c d William Clinger (June 1981). "Foundations of Actor Semantics". Mathematics Doctoral Dissertation. MIT. hdl:1721.1/6935. Cite journal requires |journal= (help)
^ a b Irene Greif (August 1975). "Semantics of Communicating Parallel Processes". EECS Doctoral Dissertation. MIT. Cite journal requires |journal= (help)
^ a b Henry Baker; Carl Hewitt (August 1977). "Laws for Communicating Parallel Processes". IFIP. Cite journal requires |journal= (help)
^ a b c Gul Agha (1986). "Actors: A Model of Concurrent Computation in Distributed Systems". Doctoral Dissertation. MIT Press. hdl:1721.1/6952. Cite journal requires |journal= (help)
^ Carl Hewitt (2006-04-27). "What is Commitment? Physical, Organizational, and Social" (PDF). [email protected] Cite journal requires |journal= (help)
^ M. Gaspari; G. Zavattaro (1999). "An Algebra of Actors". Formal Methods for Open Object Based Systems. Cite journal requires |journal= (help)
^ Gul Agha; Prasanna Thati (2004). "An Algebraic Theory of Actors and Its Application to a Simple Object-Based Language" (PDF). From OO to FM (Dahl Festschrift) LNCS 2635. Archived from the original (PDF) on 2004-04-20. Cite journal requires |journal= (help)
^ John Darlington; Y. K. Guo (1994). "Formalizing Actors in Linear Logic". International Conference on Object-Oriented Information Systems. Cite journal requires |journal= (help)
^ Milner, Robin (1993). "Elements of interaction". Communications of the ACM. 36: 78–89. doi:.
^ Henry Lieberman (June 1981). "A Preview of Act 1". MIT AI memo 625. hdl:1721.1/6350. Cite journal requires |journal= (help)
^ Henry Lieberman (June 1981). "Thinking About Lots of Things at Once without Getting Confused: Parallelism in Act 1". MIT AI memo 626. hdl:1721.1/6351. Cite journal requires |journal= (help)
^ Philipp Haller and Martin Odersky (September 2006). "Event-Based Programming without Inversion of Control" (PDF). Proc. JMLC 2006. Cite journal requires |journal= (help)
^ Philipp Haller and Martin Odersky (January 2007). "Actors that Unify Threads and Events" (PDF). Technical report LAMP 2007. Archived from the original (PDF) on 2011-06-07. Retrieved 2007-12-10. Cite journal requires |journal= (help)
Carl Hewitt, et https://link.springer.com/chapter/10.1007/3-540-06859-7_147al.[permanent dead link]Behavioral Semantics of Nonrecursive Control Structure Proceedings of Colloque sur la Programmation, April 1974.
Jean-Pierre Briot. From objects to actors: Study of a limited symbiosis in Smalltalk-80 Rapport de Recherche 88-58, RXF-LITP, Paris, France, September 1988
Carl Hewitt (2006b) What is Commitment? Physical, Organizational, and Social [email protected] April 27, 2006b.
Carl Hewitt (2007b) Large-scale Organizational Computing requires Unstratified Paraconsistency and Reflection [email protected]'07.
Hewitt, Meijer and Szyperski: The Actor Model (everything you wanted to know, but were afraid to ask) Microsoft Channel 9. April 9, 2012.
Akka – actor based library in Scala and Java, from Lightbend Inc..
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Global Constraint Catalog: Csoft_all_equal_min_ctr
<< 5.356. soft_all_equal_max_var5.358. soft_all_equal_min_var >>
[HebrardSullivanRazgon08]
\mathrm{𝚜𝚘𝚏𝚝}_\mathrm{𝚊𝚕𝚕}_\mathrm{𝚎𝚚𝚞𝚊𝚕}_\mathrm{𝚖𝚒𝚗}_\mathrm{𝚌𝚝𝚛}\left(𝙽,\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\right)
\mathrm{𝚜𝚘𝚏𝚝}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏}_\mathrm{𝚖𝚊𝚡}_\mathrm{𝚌𝚝𝚛}
\mathrm{𝚜𝚘𝚏𝚝}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}_\mathrm{𝚖𝚊𝚡}_\mathrm{𝚌𝚝𝚛}
\mathrm{𝚜𝚘𝚏𝚝}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚜𝚝𝚒𝚗𝚌𝚝}_\mathrm{𝚖𝚊𝚡}_\mathrm{𝚌𝚝𝚛}
𝙽
\mathrm{𝚒𝚗𝚝}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚟𝚊𝚛}-\mathrm{𝚍𝚟𝚊𝚛}\right)
𝙽\ge 0
𝙽\le |\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}|*|\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}|-|\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}|
\mathrm{𝚛𝚎𝚚𝚞𝚒𝚛𝚎𝚍}
\left(\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂},\mathrm{𝚟𝚊𝚛}\right)
Consider the equality constraints involving two distinct variables of the collection
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
. Among the previous set of constraints,
𝙽
is less than or equal to the number of equality constraints that hold.
\left(6,〈5,1,5,5〉\right)
〈5,1,5,5〉
six equality constraints holds. Consequently, the
\mathrm{𝚜𝚘𝚏𝚝}_\mathrm{𝚊𝚕𝚕}_\mathrm{𝚎𝚚𝚞𝚊𝚕}_\mathrm{𝚌𝚝𝚛}
constraint holds since the argument
𝙽=6
𝙽>0
𝙽<|\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}|*|\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}|-|\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}|
|\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}|>1
𝙽
\ge 0
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}.\mathrm{𝚟𝚊𝚛}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}.\mathrm{𝚟𝚊𝚛}
It was shown in [HebrardSullivanRazgon08] that, finding out whether the
\mathrm{𝚜𝚘𝚏𝚝}_\mathrm{𝚊𝚕𝚕}_\mathrm{𝚎𝚚𝚞𝚊𝚕}_\mathrm{𝚌𝚝𝚛}
constraint has a solution or not is NP-hard. This was achieved by reduction from 3-dimensional-matching. Hebrard et al. also identify a tractable class when no value occurs in more than two variables of the collection
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
that is equivalent to the vertex matching problem. One year later, [HebrardMarxSullivanRazgon09] shows how to achieve bound-consistency in polynomial time.
\mathrm{𝚜𝚘𝚏𝚝}_\mathrm{𝚊𝚕𝚕}_\mathrm{𝚎𝚚𝚞𝚊𝚕}_\mathrm{𝚖𝚊𝚡}_\mathrm{𝚟𝚊𝚛}
\mathrm{𝚜𝚘𝚏𝚝}_\mathrm{𝚊𝚕𝚕}_\mathrm{𝚎𝚚𝚞𝚊𝚕}_\mathrm{𝚖𝚒𝚗}_\mathrm{𝚟𝚊𝚛}
\mathrm{𝚜𝚘𝚏𝚝}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}_\mathrm{𝚌𝚝𝚛}
\mathrm{𝚜𝚘𝚏𝚝}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}_\mathrm{𝚟𝚊𝚛}
\mathrm{𝚊𝚕𝚕}_\mathrm{𝚎𝚚𝚞𝚊𝚕}
\mathrm{𝚊𝚗𝚍}
\mathrm{𝚋𝚊𝚕𝚊𝚗𝚌𝚎}
\mathrm{𝚎𝚚𝚞𝚒𝚟𝚊𝚕𝚎𝚗𝚝}
\mathrm{𝚗𝚘𝚛}
\mathrm{𝚊𝚝𝚖𝚘𝚜𝚝}_\mathrm{𝚗𝚟𝚊𝚕𝚞𝚎}
complexity: 3-dimensional-matching.
constraint type: soft constraint, value constraint, relaxation, decomposition-based violation measure.
filtering: bound-consistency.
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
\mathrm{𝐶𝐿𝐼𝑄𝑈𝐸}
\left(\ne \right)↦\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}\mathtt{1},\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}\mathtt{2}\right)
\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}\mathtt{1}.\mathrm{𝚟𝚊𝚛}=\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎𝚜}\mathtt{2}.\mathrm{𝚟𝚊𝚛}
\mathrm{𝐍𝐀𝐑𝐂}
\ge 𝙽
We generate an initial graph with binary equalities constraints between each vertex and its successors. We use the arc generator
\mathrm{𝐶𝐿𝐼𝑄𝑈𝐸}\left(\ne \right)
in order to avoid considering equality constraints between the same variable. The graph property states that
𝙽
is less than or equal to the number of equalities that hold in the final graph.
\mathrm{𝐍𝐀𝐑𝐂}
graph property, the arcs of the final graph are stressed in bold. Six equality constraints remain in the final graph.
\mathrm{𝚜𝚘𝚏𝚝}_\mathrm{𝚊𝚕𝚕}_\mathrm{𝚎𝚚𝚞𝚊𝚕}_\mathrm{𝚖𝚒𝚗}_\mathrm{𝚌𝚝𝚛}
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Wheel and axle mechanism in mechanical systems - MATLAB - MathWorks Switzerland
Mechanism orientation
Simscape / Foundation Library / Mechanical / Mechanisms
The Wheel and Axle block represents a wheel and axle mechanism shown in the following schematic.
The wheel and the axle have the same axis, and the axis is assumed to be rigidly connected to the frame, thus making this mechanism an ideal converter of mechanical rotational into mechanical translational motion. The mechanism has two connections: a mechanical rotational port A, which corresponds to the axle, and a mechanical translational port P, which corresponds to the wheel periphery. The mechanism is described with the following equations:
T=r·F·or
v=r·\omega ·or
T is torque on the axle.
F is force on the wheel periphery.
ω is angular velocity.
v is linear velocity on the wheel periphery.
r is wheel radius.
or is mechanism orientation indicator. The value is +1 if axle rotation in the globally assigned positive direction is converted into translational motion in positive direction. The value is –1 if positive rotation results in translational motion in negative direction.
Use the block in simulation of rack-pinions, steering wheels, hoisting devices, windlasses, and so on.
The block positive directions are from A to the reference point and from the reference point to P.
Mechanical rotational conserving port associated with the axle.
P — Wheel periphery
Mechanical translational conserving port associated with the wheel periphery.
Wheel radius — Radius of the wheel
0.05 m (default)
Mechanism orientation — Relationship between axle rotation and direction of translational motion
Drives in positive direction (default) | Drives in negative direction
The value Drives in positive direction specifies a mechanism where axle rotation in the globally assigned positive direction is converted into translational motion in positive direction. The value Drives in negative direction specifies a mechanism where axle rotation in the globally assigned positive direction is converted into translational motion in negative direction.
Gear Box | Lever | Slider-Crank
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Another Train Paradox: May the Myelin Be with You!
Bernard Delalande1, Hirohisa Tamagawa2, Vladimir Matveev3
1NeuroDynamique.fr, La Verpilliere, France.
2Department of Mechanical Engineering, Faculty of Engineering, Gifu University, 1-1 Yanagido, Gifu, Japan.
3Laboratory of Cell Physiology, Institute of Cytology, Russian Academy of Sciences, St. Petersburg, Russia.
For more than 70 years, biologists and biophysicists have been trying to unravel the mystery surrounding the saltatory conduction of so-called myelinated neurons. This conduction is indeed very different and faster than for fibres “without” myelin. Many theories have been developed. Albert Einstein used the metaphor of the train to explain the theory of relativity. It is also possible to use a similar metaphor to better understand this transient functioning of the neuron: the transmission of the action potential in myelinated fibres. By studying the various theories that have been put forward and confronting them with physics, mathematics and microscopic anatomical observations, it is possible to refute or confirm certain hypotheses. It is easy and simple, then, to demonstrate unequivocally that the action potential cannot, in any way, jump from node of Ranvier (noR) to node of Ranvier as has been assumed and taught until now. It is possible to describe that the neuron uses an elegant and very simple method to increase the speed of transmission of the neuronal message. It is also important to conclude that this increase in speed, contrary to common belief, has an energetic cost that is greater than expected and that is proportional to the speed and in perfect agreement with the laws of thermodynamics.
Axon, Neuron, Saltatory Conduction, Action Potential Propagation, Hodgkin and Huxley Model
Delalande, B., Ta- magawa, H. and Matveev, V. (2021) Another Train Paradox: May the Myelin Be with You!. Open Access Library Journal, 8, 1-14. doi: 10.4236/oalib.1107379.
“This nerve impulse propagates electrically” is a common notion among most physiologists [1] [2] , while some propose that it is a mechanical soliton [3] . Finally, some attempt to link all the theories to unify them while others return to more plausible bases [4] . Some hypothesise a waveguide [5] of the order of one or more Mhz, but this has so far escaped all our measuring devices and the geometrical configuration imposes a drastic functional limitation.
To complete their explanations and hypotheses, biologists have separated neurons into two categories: the so-called myelinated neurons and those that are not. Of course, neurocytologists do not entirely agree with this last distinction [6] . All axons (with very, very rare exceptions) are surrounded by a myelin sheath. The myelin sheath may be compact and spiralling around the axon or it may be non-compact and form a simple envelope around the axon. The use of the term unmyelinated would suggest that the membrane of some axons is in direct contact with the external environment which would favor the functioning of certain theories.
l=v\cdot t
These few sentences may take us back to the algebra problems of our childhood, of course, but they contain the very principles of applied science. The
Considering the electrical component of the action potential (see Figure 1): it has a duration that depends on the type of neuron being considered.
{d}_{AP}={t}_{2}-{t}_{0}
{d}_{spike}={t}_{1}-{t}_{0}
Figure 1. Graphic representation of the electrical component of an action potential.
{v}_{unmyel}<{v}_{myel}
Because it has a duration
{d}_{spike}
and also because it has a velocity (
{v}_{myel}
{v}_{unmyel}
), it is possible to say that the spike has a length.
The spike occupies a finite surface on the axon. It can therefore be said that it has a rather cylindrical shape that has a length
{l}_{myel}
{l}_{unmyel}
{l}_{myel}=Const
{l}_{unmyel}={v}_{unmyel}\cdot {d}_{spike}
All axons are myelinated [7] . They are all surrounded by specialized cells that produce a myelin barrier that may or may not be compact [8] . It is interesting to ask the question of the existence of so-called unmyelinated fibers since neuro-cytology dispels this myth through anatomical evidence [6] [9] (Figure 2).
The worst example is the giant squid axon [10] used as a model in the HH model: It is surrounded by the thickest (0.7 to 1.3 μm) myelin layer in the animal kingdom and yet we are taught that it is not myelinated and it is a fusion of multiple axons [11] .
Figure 2. Eight unmyelinated axons surrounded by myelin (in light green). The Schwann nucleus takes the major area. 3D reconstruction from a microscopy by courtesy of Ennio Pannese.
3.2. Unique Properties of the Unmyelinated Neuron
It is therefore possible to know the position at a time t. of the AP by simple computation, with
{x}_{0}
the position of the AP at a time
{t}_{0}
{d}_{t}={v}_{unmyel}\cdot t+{x}_{0}
s=2\pi \cdot {r}_{unmyel}\cdot \left({v}_{unmyel}\cdot {d}_{spike}\right)
s\propto {v}_{unmyel}
The speeds normally observed for this type of neurons are between 0.2 and 3 m∙s−1.
The spike has a more or less constant duration, this length varies between 2 × 10−4 and 3 × 10−3 m.
3.3. Special Properties of the Myelinated Neuron
In myelinated neurons, conduction is called saltatory although this statement is not yet clear-cut [12] [13] [14] [15] . It is stated, without irrefutable evidence, that the action potential seems to jump from noR to the next one [16] [17] . The process would of course be linked to the presence of myelin, which would improve electrical conduction while reducing energy costs [18] .
If we know the internode length, it is perfectly possible to know both the distance covered since the instant
{t}_{0}
but also the number of jumped nodes
no{R}_{t}
Let us not neglect any length because all of them seem to be more crucial than they appear [19] [20] [21] .
{d}_{t}={v}_{myel}\cdot t
no{R}_{t}=\frac{{d}_{t}}{{l}_{inoR}+{l}_{noR}}
This gives us with a
{l}_{inoR}
value of 10−3 and
{l}_{noR}
of 10−6 m and a speed between 4 and 150 m∙s−1:
It is also perfectly possible to state without any doubt with Equation (10) and Equation (11) that the number of noRs crossed is both proportional to the duration t and the conduction speed
{v}_{myel}
4. Let’s Watch the Trains
4.1. The Train Metaphor
Our trains have a length determined by their speed and type. They are more or less long as described above for unmyelinated neurons. A train, on the opposite, appears relatively compressed (it should be even more so) and reflects the length of the AP at a noR (see Figure 3).
4.2. An Observer Looking at His Watch
4.3. Trains Don’t Jump
We have been led to believe for too long that the action potential jumps from noR to noR and that is what explains its increased speed [12] [22] .
Figure 3. 3 action potentials: Axons are shown as rails. 2 APs on unmyelinated fibers at the top, the fastest is the longer one but its duration is similar. The tunnels are like compact myelin. The AP at noR seems compressed and looks quite stationary. Its duration is the same as the others.
The second observer must see the train’s headlights while the first observer must see the rear lights: there would then be an unequivocal saltatory conduction (see Figure 4).
That is not what is being witnessed [23] !
It can be said that the number of APs existing on the axon is proportional to the conduction speed (see Figure 5).
Figure 4. Conventional saltatory conduction: the train jumps at the next node when it ends at the current one.
Figure 5. Numbers of noRs activated vs conduction velocity (4 to 150 m∙s−1) and internode length, and AP duration (0.3 to 2 ms). The energetic cost is higher with myelinated neurons.
A myelinated fiber with an internode length of 2 × 10−3 m, an AP (spike) of 5 × 10−4 s and an average velocity of 40 m∙s−1.
If it is true that it is only at the end of the action potential that the signal jumps to the next noR then we should be able to confirm by computation the speed of 40 m∙s−1.
\frac{1}{{d}_{spike}}\cdot {l}_{inoR}=\frac{1}{5\times {10}^{-4}}\times 2\times {10}^{-3}=4.0\text{\hspace{0.17em}}\text{m}\cdot {\text{s}}^{-1}\ll 40\text{\hspace{0.17em}}\text{m}\cdot {\text{s}}^{-1}
Or by the opposite method of calculation;
{v}_{myel}\cdot {d}_{spike}=40\times 5\times {10}^{-4}=2.0\times {10}^{-2}\text{\hspace{0.17em}}\text{m}\gg 2\times {10}^{-3}\text{\hspace{0.17em}}\text{m}
It is also possible to check that the system reaches its low operating limit of 4.0 m∙s−1 which is well over the length of the internode.
{v}_{myel}\cdot {d}_{spike}=4.0\times 5\times {10}^{-4}=2.0\times {10}^{-3}\text{\hspace{0.17em}}\text{m}
The system therefore only works when;
{v}_{myel}\cdot {d}_{spike}\ge {l}_{inoR}
4.4. Electric Trains without Catenary Cables
While it is true, and why contest it, that the action potential is closely linked to the presence of ion channels, it is also perfectly proven that myelin damage, compact or not (as in multiple sclerosis) leads to a slower conduction rate for all types of fibres. The action potential is even eliminated when the compact myelin disappears consecutively over a tiny length of the axon [24] .
4.5. You Shall not Pass!
There are also many references [25] - [30] showing that the compression of axons, myelinated or not, leads to a slowing down and then the disappearance of APs.
We have enough evidence to understand that the basic propagation system is based on a vertical force associated with horizontal translation (see Figure 6). This process is repeated throughout the axon, which obviously wastes time and energy.
It is quite easy to verify because the available speed can go up to more than 1500 m∙s−1.
The AP that activates the next noR is not transmitted because the ion channels under the myelin activate as soon as the “liquid” wave passes. These firm locks provide better strength and pressure transmission and limit ionic movements at the noR (see Figure 6).
Figure 6. Propagation methods of APs in unmyelinated vs myelinated neurons.
[1] Hodgkin, A.L. (1937) Evidence for Electrical Transmission in Nerve: Part I. The Journal of Physiology, 90, 183-210. https://doi.org/10.1113/jphysiol.1937.sp003507
[2] Hodgkin, A.L. (1937) Evidence for Electrical Transmission in Nerve: Part II. The Journal of Physiology, 90, 211-232. https://doi.org/10.1113/jphysiol.1937.sp003508
[3] Heimburg, T. and Jackson, A.D. (2005) On Soliton Propagation in Biomembranes and Nerves. Proceedings of the National Academy of Sciences of the United States of America, 102, 9790-9795. https://doi.org/10.1073/pnas.0503823102
[4] Akaishi, T. (2018) Saltatory Conduction as an Electrostatic Compressional Wave in the Axoplasm. The Tohoku Journal of Experimental Medicine, 244, 151-161. https://doi.org/10.1620/tjem.244.151
[5] Jacak, J.E. and Jacak, W.A. (2020) New Wave-Type Mechanism of Salutatory Conduction in Myelinated Axons and Micro-Salutatory Conduction in C Fibres. European Biophysics Journal, 49, 343-360. https://doi.org/10.1007/s00249-020-01442-z
[6] Pannese, E. (1994) Neurocytology: Fine Structure of Neurons, Nerve Processes, and Neuroglial Cells. G. Thieme Verlag, Stuttgart; Thieme Medical Publishers, New York.
[7] De Pittà, M. (2019) Myelin and Saltatory Conduction. arXiv: 1708.00534. http://arxiv.org/abs/1708.00534
[8] Bear, R.S., Schmitt, F.O. and Young, J.Z. (1937) The Sheath Components of the Giant Nerve Fibres of the Squid. Proceedings of the Royal Society of London. Series B, Biological Sciences, 123, 496-504. https://doi.org/10.1098/rspb.1937.0065
[9] Fraher, J.P. and O’Sullivan, A.W. (2000) Interspecies Variation in Axon-Myelin Relationships. Cells Tissues Organs, 167, 206-213. https://doi.org/10.1159/000016783
[10] Brown, E.R. and Abbott, N.J. (1993) Ultrastructure and Permeability of the Schwann Cell Layer Surrounding the Giant Axon of the Squid. Journal of Neurocytology, 22, 283-298. https://doi.org/10.1007/BF01187127
[11] Martin, R. (1965) On the Structure and Embryonic Development of the Giant Fibre System of the Squid Loligo vulgaris. Zeitschrift für Zellforschung und Mikroskopische Anatomie, 67, 77-85. https://doi.org/10.1007/BF00339277
[12] Fitzhugh, R. (1962) Computation of Impulse Initiation and Saltatory Conduction in a Myelinated Nerve Fiber. Biophysical Journal, 2, 11-21. https://doi.org/10.1016/S0006-3495(62)86837-4
[13] Giuliodori, M.J. and DiCarlo, S.E. (2004) Myelinated vs. Unmyelinated Nerve Conduction: A Novel Way of Understanding the Mechanisms. Advances in Physiology Education, 28, 80-81. https://doi.org/10.1152/advan.00045.2003
[14] Laporte, Y. (1951) Continuous Conduction of Impulses in Peripheral Myelinated Nerve Fibers. The Journal of General Physiology, 35, 343-360. https://doi.org/10.1085/jgp.35.2.343
[15] Rasminsky, M. and Sears, T.A. (1972) Internodal Conduction in Undissected Demyelinated Nerve Fibres. The Journal of Physiology, 227, 323-350. https://doi.org/10.1113/jphysiol.1972.sp010035
[16] Huxley, A.F. and Stampfli, R. (1951) Effect of Potassium and Sodium on Resting and Action Potentials of Single Myelinated Nerve Fibers. The Journal of Physiology, 112, 496-508. https://doi.org/10.1113/jphysiol.1951.sp004546
[17] Huxley, A.F. and St?mpfli, R. (1949) Evidence for Saltatory Conduction in Peripheral Myelinated Nerve Fibres. The Journal of Physiology, 108, 315-339. https://doi.org/10.1113/jphysiol.1949.sp004335
[18] Sangrey, T. and Levy, W.B. (2005) Conduction Velocity Costs Energy. Neurocomputing, 65-66, 907-913. https://doi.org/10.1016/j.neucom.2004.10.091
[19] Ford, M.C., Alexandrova, O., Cossell, L., Stange-Marten, A., Sinclair, J., Kopp- Scheinpflug, C., et al. (2015) Tuning of Ranvier Node and Internode Properties in Myelinated Axons to Adjust Action Potential Timing. Nature Communications, 6, Article No. 8073. https://doi.org/10.1038/ncomms9073
[20] Foust, A., Popovic, M., Zecevic, D. and McCormick, D.A. (2010) Action Potentials Initiate in the Axon Initial Segment and Propagate through Axon Collaterals Reliably in Cerebellar Purkinje Neurons. Journal of Neuroscience, 30, 6891-6902. https://doi.org/10.1523/JNEUROSCI.0552-10.2010
[21] Foust, A.J., Yu, Y., Popovic, M., Zecevic, D. and McCormick, D.A. (2011) Somatic Membrane Potential and Kv1 Channels Control Spike Repolarization in Cortical Axon Collaterals and Presynaptic Boutons. Journal of Neuroscience, 31, 15490- 15498. https://doi.org/10.1523/JNEUROSCI.2752-11.2011
[22] Goodman, B.E. and Waller, S.B. (2002) Propagation of Action Potentials in Myelinated vs. Unmyelinated Neurons. Advances in Physiology Education, 26, 223. https://doi.org/10.1152/advan.00023.2002
[23] Debanne, D., Campanac, E., Bialowas, A., Carlier, E. and Alcaraz, G. (2011) Axon Physiology. Physiological Reviews, 91, 555-602. https://doi.org/10.1152/physrev.00048.2009
[24] Baraban, M., Mensch, S. and Lyons, D.A. (2016) Adaptive Myelination from Fish to Man. Brain Research, 1641, 149-161. https://doi.org/10.1016/j.brainres.2015.10.026
[25] Court, F.A., Sherman, D.L., Pratt, T., Garry, E.M., Ribchester, R.R., Cottrell, D.F., et al. (2004) Restricted Growth of Schwann Cells Lacking Cajal Bands Slows Conduction in Myelinated Nerves. Nature, 431, 191-195. https://doi.org/10.1038/nature02841
[26] Fern, R. and Harrison, P.J. (1994) The Relationship between Ischaemic Conduction Failure and Conduction Velocity in Cat Myelinated Axons. Experimental Physiology, 79, 571-581. https://doi.org/10.1113/expphysiol.1994.sp003790
[27] Ffrench-Constant, C., Colognato, H. and Franklin, R.J.M. (2004) Neuroscience. The Mysteries of Myelin Unwrapped. Science, 304, 688-689. https://doi.org/10.1126/science.1097851
[28] Grandis, M., Leandri, M., Vigo, T., Cilli, M., Sereda, M.W., Gherardi, G., et al. (2004) Early Abnormalities in Sciatic Nerve Function and Structure in a Rat Model of Charcot-Marie-Tooth Type 1A Disease. Experimental Neurology, 190, 213-223. https://doi.org/10.1016/j.expneurol.2004.07.008
[29] Padrón, R. and Mateu, L. (1982) Repetitive Propagation of Action Potentials Destabilizes the Structure of the Myelin Sheath. A Dynamic X-Ray Diffraction Study. Biophysical Journal, 39, 183-188. https://doi.org/10.1016/S0006-3495(82)84506-2
[30] Williams, A. C. and Brophy, P.J. (2002) The Function of the Periaxin Gene during Nerve Repair in a Model of CMT4F. Journal of Anatomy, 200, 323-330. https://doi.org/10.1046/j.1469-7580.2002.00038.x
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A (53o N, 84o E), B (53o N, 25o W), C and D are four points on the surface of the earth. AC is the diameter of the parallel of latitude 53o N.
(a) State the location of C.
(b) Calculate the shortest distance, in nautical mile, from A to C measured along the surface of the earth.
(c) Calculate the distance, in nautical mile, from A due east B measured along the common parallel of latitude.
(d) An aeroplane took off from B and flew due south to D. The average speed of the flight was 420 knots and the time taken was 6½ hours.
(i) the distance, in nautical mile, from B to D measured along the meridian.
(ii) the latitude of D.
Latitude of C = 53o N
Longitude of C = (180o – 84o) E = 96o E
Therefore location of C = (53o N, 96o E)
Shortest distance from A to C
= (84 – 25) x 60 x cos 53o
\begin{array}{l}\left(\text{i}\right)\\ \text{Distance travel from }B\text{ to }D\\ =420×6\frac{1}{2}←\overline{)\begin{array}{l}\text{ Distance travelled}\\ \text{ = average speed }×\text{ time taken }\end{array}}\\ =2730\text{ nautical miles}\\ \\ \left(\text{ii}\right)\\ \text{Difference in latitude between }B\text{ to }D\\ =\frac{2730}{60}\\ ={45.5}^{\text{o}}\\ \\ \therefore \text{Latitude of }D=\left({53}^{\text{o}}-{45.5}^{\text{o}}\right)N\\ \text{}={7.5}^{\text{o}}N\end{array}
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PT3 Maths – user's Blog!
Home › Archive for PT3 Maths
1.2.2 Angles and Lines II, PT3 Practice
In Diagram below, PQRST is a straight line. Find the value of x.
\begin{array}{l}\text{Interior angle of }R=\left({180}^{o}-{76}^{o}\right)÷2={52}^{o}\\ \angle x=\text{exterior angle of }R\text{ }\left(\text{corresponding angles}\right)\\ \text{Hence, }x={76}^{o}+{52}^{o}={128}^{o}\end{array}
In Diagram below, find the value of y.
\begin{array}{l}\angle ABC=\angle BCD\\ \text{ }={108}^{o}\text{ (alternate angle)}\\ {y}^{o}+{130}^{o}+{108}^{o}={360}^{o}\\ \text{ }{y}^{o}={360}^{o}-{238}^{o}\\ \text{ }{y}^{o}={122}^{o}\end{array}
In Diagram below, PSR and QST are straight lines.
\begin{array}{l}\angle UST+\angle STV={180}^{o}\\ \angle UST={180}^{o}-{116}^{o}={64}^{o}\\ \angle PST=\angle QSR\\ {x}^{o}+\angle UST={135}^{o}\\ {x}^{o}+\angle UST={135}^{o}\\ {x}^{o}={71}^{o}\\ x=71\end{array}
In Diagram below, PWV is a straight line.
(a) Which line is perpendicular to line PWV?
(b) State the value of ∠ RWU.
(b) ∠ RWU = 13o + 29o + 20o = 62o
In Diagram below, UVW is a straight line.
(a) Which line is parallel to line TU?
(b) State the value of ∠ QVS.
(b) ∠ QVS = 8o + 18o = 26o
Posted in Lines and Angles (II), PT3 Maths
12.2.2 Solid Geometry (II), PT3 Focus Practice
Sphere below has a surface area of 221.76 cm2.
Calculate its radius.
\left(\pi =\frac{22}{7}\right)
\begin{array}{l}4\pi {r}^{2}=221.76\\ 4×\frac{22}{7}×{r}^{2}=221.76\\ {r}^{2}=\frac{221.76×7}{4×22}\\ {r}^{2}=17.64\\ r=\sqrt{17.64}\\ r=4.2\text{cm}\end{array}
Diagram below shows a right pyramid with a square base.
Given the height of the pyramid is 4 cm.
Calculate the total surface area, in cm2, of the right pyramid.
h = √25 = 5 cm2
Total surface area of the right pyramid
= Base area + 4 (Area of triangle)
= (6 × 6) + 4 × 4 (½ × 6 × 5)
Diagram below shows a prism.
Draw to full scale, the net of the prism on the grid in the answer space. The grid has equal squares with sides of 1 unit.
Posted in PT3 Maths, Solid Geometry (II)
8.2.2 Coordinates, PT3 Focus Practice
The point M (x, 4), is the midpoint of the line joining straight line Q (-2, -3) and R (14, y).
The value of x and y are
\begin{array}{l}x=\frac{-2+14}{2}\\ x=\frac{12}{2}\\ x=6\\ \\ 4=\frac{-3+y}{2}\\ 8=-3+y\\ y=11\end{array}
In diagram below, PQR is a right-angled triangle. The sides QR and PQ are parallel to the y-axis and the x-axis respectively. The length of QR = 6 units.
Given that M is the midpoint of PR, then the coordinates of M are
x-coordinate of R = 3
y-coordinate of R = 1 + 6 = 7
\begin{array}{l}P\left(1,1\right),R\left(3,7\right)\\ \text{Coordinates of }M\\ =\left(\frac{1+3}{2},\frac{1+7}{2}\right)\\ =\left(2,4\right)\end{array}
Given points P (–2, 8) and Q (10, 8), find the length of PQ.
\begin{array}{l}\text{Length of }PQ\\ =\sqrt{{\left[10-\left(-2\right)\right]}^{2}+{\left(8-8\right)}^{2}}\\ =\sqrt{{\left(14\right)}^{2}+0}\\ =14\text{ units}\end{array}
In diagram below, ABC is an isosceles triangle.
(b) the length of BC.
\begin{array}{l}\left(\text{a}\right)\\ \text{For an isosceles triangle, }\\ y-\text{coordinate of }C\text{ is the midpoint of straight line }AB.\\ \frac{2+k}{2}=-3\\ 2+k=-6\\ \text{ }k=-8\\ \\ \left(\text{b}\right)\\ B=\left(-2,-8\right)\\ BC=\sqrt{{\left[10-\left(-2\right)\right]}^{2}+{\left[-3-\left(-8\right)\right]}^{2}}\\ \text{ }=\sqrt{{12}^{2}+{5}^{2}}\\ \text{ }=13\text{ units}\end{array}
Diagram below shows a rhombus PQRS drawn on a Cartesian plane. PS is parallel to x-axis.
Given the perimeter of PQRS is 40 units, find the coordinates of point R.
\begin{array}{l}\text{All sides of rhombus have the same length,}\\ \text{therefore length of each side}=\frac{40}{4}=10\text{ units}\\ PQ=10\\ {\left(9-{x}_{1}\right)}^{2}+{\left(7-\left(-1\right)\right)}^{2}={10}^{2}\\ 81-18{x}_{1}+{x}_{1}{}^{2}+64=100\\ {x}_{1}{}^{2}-18{x}_{1}+45=0\\ \left({x}_{1}-3\right)\left({x}_{1}-15\right)=0\\ {x}_{1}=3,15\\ {x}_{1}=3\\ Q=\left(3,-1\right),R=\left({x}_{2},-1\right)\\ \\ QR=10\\ {\left({x}_{2}-3\right)}^{2}+{\left[-1-\left(-1\right)\right]}^{2}={10}^{2}\\ {x}_{2}{}^{2}-6{x}_{2}+9+0=100\\ {x}_{2}{}^{2}-6{x}_{2}-91=0\\ \left({x}_{2}+7\right)\left({x}_{2}-13\right)=0\\ {x}_{2}=-7,13\\ {x}_{2}=13\\ \\ \therefore R=\left(13,-1\right)\end{array}
Posted in Coordinates, PT3 Maths
4.2.2 Linear Equations I, PT3 Practice
\begin{array}{l}\text{(a) }5a-16=a\\ \text{(b) }6+\frac{2}{3}\left(-9p+12\right)=p\end{array}
\begin{array}{l}5a-16=a\\ 5a-a=16\\ 4a=16\\ a=4\end{array}
\begin{array}{l}6+\frac{2}{3}\left(-9p+12\right)=p\\ 6-6p+8=p\\ -6p-p=-8-6\\ -7p=-14\\ p=2\end{array}
\begin{array}{l}\text{(a) }a-2=\frac{a}{5}\\ \text{(b) }\frac{b-3}{4}=\frac{2+b}{5}\end{array}
\begin{array}{l}a-2=\frac{a}{5}\\ 5a-2=a\\ 4a=2\\ a=2\end{array}
\begin{array}{l}\frac{b-3}{4}=\frac{2+b}{5}\\ 5b-15=8+4b\\ 5b-4b=8+15\\ b=23\end{array}
\begin{array}{l}\text{(a) }x=-24-x\\ \text{(b) }y+\frac{5}{4}\left(4-2y\right)=-4\end{array}
\begin{array}{l}x=-24-x\\ 2x=-24\\ x=-12\end{array}
\begin{array}{l}y+\frac{5}{4}\left(4-2y\right)=-4\\ 4y+20-10y=-16\text{ }\left(×4\right)\\ -6y=-16-20\\ -6y=-36\\ y=6\end{array}
\begin{array}{l}\text{(a) }5p=8p-9\\ \text{(b) }3q=\frac{20-13q}{4}\end{array}
\begin{array}{l}5p=8p-9\\ 5p-8p=-9\\ -3p=-9\\ p=3\end{array}
\begin{array}{l}3q=\frac{20-13q}{4}\\ 12q=20-13q\\ 12q+13q=20\\ 25q=20\\ q=\frac{4}{5}\end{array}
Posted in Linear Equations (I), PT3 Maths
2.2.2 Number Patterns and Sequences, PT3 Practice
There are 8 prime numbers between 10 and 40.
Diagram below shows a sequence of prime numbers.
\overline{)13,\text{ }17,\text{ }x,\text{ }y,\text{ }29}
Find the value of x + y.
Value of x + y
The following data shows a sequence of prime numbers in ascending order.
\overline{)\text{ }5,\text{ }7,\text{ }x,\text{ }13,\text{ }17,\text{ 19},\text{ 23},\text{ }29,\text{ }y,\text{ }z\text{ }}
Find the value of x, y and of z.
\overline{)\text{ }p,\text{ 6}1,\text{ 6}7,\text{ }q,\text{ 73},\text{ }r,\text{ 83 }}
Find the value of p, q and of r.
p = 59, q = 71, r = 79
Diagram below shows several number cards.
\overline{)\begin{array}{l}\text{ }\overline{)49}\text{ }\overline{)39}\text{ }\overline{)29}\text{ }\\ \text{ }\overline{)87}\text{ }\overline{)97}\text{ }\overline{)77}\text{ }\\ \text{ }\overline{)\text{ }5\text{ }}\text{ }\overline{)15}\text{ }\overline{)\text{2}5}\text{ }\end{array}}
List three prime numbers from the diagram.
Prime number card from the diagram = 5, 29 and 97.
Posted in Number Patterns and Sequences, PT3 Maths
Diagram below is part of a number line.
What is the value of P and of Q?
Each gradation on the number line represents 4 units. Thus P = –12 and Q = 12.
Diagram below shows a sequence of numbers. K and M represent two numbers.
\overline{)\text{ }\overline{)19}\text{ }\overline{)13}\text{ }\overline{)K}\text{ }\overline{)\text{ }1\text{ }}\text{ }\overline{)M}\text{ }}
What are the values of K and M?
K = 13 – 6 = 7
Diagram below shows five integers.
Find the sum of the largest integer and the smallest integer.
The largest integer = 1
The smallest integer = –10
Sum of the largest integer and the smallest integer
1.2.3 Whole Numbers, PT3 Practice
Britney rents a stall to sell cakes. The rental of the stall is RM500 per month. The cost for baking a cake is RM5 per box and she sells it at RM10 per box. What is the minimum box of cakes must Britney sell each month to make a profit?
Rental of the stall = RM500 monthly
Cost of baking a cake per box = RM5
Selling price per box = RM10
= RM10 – RM5
The number of boxes of cakes to accommodate the cost
= RM500 ÷ RM5
Hence, the minimum box of cakes must Britney sell each month to make a profit = 101
Posted in PT3 Maths, Whole Numbers
Solve 4 (8 + 14)(37 – 18) + 46
4 (8 + 14)(37 – 18) + 46
= 4(22)(19) + 46
= 4 × 22 × 19 + 46
113 + 6(47 – 4 × 9)
= 113 + 6[47 – (4 × 9)]
= 113 + 6(47 – 36)
Solve 128 + (9 × 4) – 318 ÷ 6
128 + (9 × 4) – 318 ÷ 6
62 × 15 – 40 + 68 ÷ 17.
(62 × 15) – 40 + (68 ÷ 17)
402 – ( ) = 78 – 6 × 8 ÷ 16.
402 – ( ) = 78 – 6 × 8 ÷ 16
402 – ( ) = 78 – 48 ÷ 16
402 – ( ) = 78 – 3
402 – ( ) = 75
( ) = 402 – 75
( ) = 327
Solve 8 × 2 ÷ 4 + 12.
8 × 2 ÷ 4 + 12
Solve 7(6 + 5) – 3(13 – 9).
7(6 + 5) – 3(13 – 9)
Solve 64 + (136 – 87) × (28 ÷ 4).
64 + (136 – 87) × (28 ÷ 4)
Solve 84 ÷ 4 – 9 + 6 × 7
84 ÷ 4 – 9 + 6 × 7
Solve 92 + 13 × 5 – 42
= 92 + (13 × 5) – 42
15.2.3 Trigonometry, PT3 Focus Practice
A technician needs to climb a ladder to repair a street lamp as shown in Diagram below.
(a) What is the length, in m, of the ladder?
(b) Find the height, in m, of the lamp post.
\begin{array}{l}\text{cos}{70}^{o}=\frac{1}{h}\\ h=\frac{1}{\text{cos}{70}^{o}}\\ h=\frac{1}{0.342}\\ h=2.924\text{ m}\\ \\ \text{Hence, the length of the ladder}\\ =2.924\text{ m}\end{array}
\begin{array}{l}\text{tan}{70}^{o}=\frac{T}{1}\\ T=\text{tan}{70}^{o}×1\\ \text{ }=2.747×1\\ \text{ }=2.747\text{ m}\\ \\ \text{The height of the lamp post}\\ =2.747+1.2\\ =3.947\text{ m}\end{array}
Posted in PT3 Maths, Trigonometry
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Four p -value distributions of (transformed) normalized mutual information values for human GCK and EFGR proteins having PDB-ID 1V4S and 2J6M, respectively. The bar charts illustrate the two steps of our model: i) blue bars show the p-value distribution of the
\mathbb{U}\left(i,j\right)
-scores; ii) red bars display the p-value distribution of the
{\mathbb{U}}_{D\left(1\right)}\left(i,j\right)
-values. The p-values close to zero represent the significant pairs by means of which we assess the individual residue position. As one can see, within [0.25,0.70] these four distributions are approximately uniform.
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Wavelet network function for nonlinear ARX and Hammerstein-Wiener models - MATLAB - MathWorks 日本
Mathematically, idWaveletNetwork is a function that maps m inputs X(t) = [x(t1),x2(t),…,xm(t)]T to a scalar output y(t) using the following relationship:
y\left(t\right)={y}_{0}+\mathrm{Χ}{\left(t\right)}^{T}PL+W\left(\mathrm{Χ}\left(t\right)\right)+S\left(\mathrm{Χ}\left(t\right)\right)
\left(Xâ\stackrel{¯}{X}\right)
W\left(X\right)=\underset{i=1}{\overset{{d}_{w}}{â}}{w}_{i}{f}_{w}\left({b}_{i}{X}^{T}Qâ{c}_{i}\right)
Q is an m-by-q projection matrix, where m ≥ q
w1, w2, …, wdw are scalar coefficients called wavelet coefficients.
b1, b2, …, bdw are scalars called wavelet dilations that multiply the input matrix X.
c1, c2, …, cdw are 1-by-q row vectors called wavelet translations.
{f}_{w}\left(x\right)={e}^{âx{x}^{T}/2}
S\left(X\right)=\underset{i=1}{\overset{{d}_{s}}{â}}{s}_{i}{f}_{s}\left({b}_{i}{X}^{T}Qâ{e}_{i}\right)
s1, s2, …, sds are scalar coefficients called scaling coefficients.
d1, d2, …, dds are scalars called scaling dilations that multiply the input matrix X.
e1, e2, …, eds are 1-by-q row vectors called scaling translations.
{f}_{s}\left(x\right)=\left(mâx{x}^{T}\right){e}^{âx{x}^{T}/2}
[1] Qinghua Zhang. “Using Wavelet Network in Nonparametric Estimation.†IEEE Transactions on Neural Networks 8, no. 2 (March 1997): 227–36. https://doi.org/10.1109/72.557660.
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ASTEXP
Exports coordinate system information from images
This task exports coordinate system information from a set of images, writing it to an AST file. For each image a frameset is written containing information about how to map between a selected Base frame and the image’s Current frame. Each frameset is identified by a key which is derived from the image itself, and matches keys which can be derived from other images to which similar framesets ought to apply. The key should be generated in the same way when the AST file is used for importing the mapping information by ASTIMP or MAKESET. Currently these keys can be generated according to a FITS header card or the order in which the images are presented. Additional information may be written describing what use to make of FITS headers in the images.
Used together, the framesets written out to an AST file can thus contain information about the positioning of images in a set of related images.
AST files written out by this application can be applied to other images of similar origin using the ASTIMP or MAKESET programs, so that registration information present in the WCS components of one set of images (put there for instance by the REGISTER or WCSEDIT applications) can be transferred using ASTIMP and ASTEXP to another similar set. This "similar set" will typically be one from chips in the same mosaic camera instrument.
A 2-frame frameset is output for each image. The Base frame is one selected by the BASEFRAME parameter, and is identical in the exported frameset to the one in the original image. The Current frame in the exported frameset is the same as the Current frame in the original image, but may be given a different Domain name by the OUTDOMAIN parameter.
Under normal circumstances, the Current frames of all the input images should share the same Domain name, and so should the frames identified by the BASEFRAME parameter. A warning will be issued if this is not the case. Warnings will also be issued if the image identifiers are not all unique.
ASTEXP in astfile outdomain baseframe
ASTFILE = LITERAL (Read)
The name of the AST file to be written.
BASEEPOCH = _DOUBLE (Read)
If a "Sky Co-ordinate System" specification is supplied (using parameter BASEFRAME) for a celestial co-ordinate system, then an epoch value is needed to qualify it. This is the epoch at which the supplied sky positions were determined. It should be given as a decimal years value, with or without decimal places ("1996.8" for example). Such values are interpreted as a Besselian epoch if less than 1984.0 and as a Julian epoch otherwise.
BASEFRAME = LITERAL (Read)
This parameter specifies the WCS frame from the images relative to which the Current frames will be defined in the output AST file. To be useful, this must specify a frame which occurs in all the images in the IN list, and can be expected to occur in any image to which the AST file will later be applied using ASTIMP. AXIS is a good choice since this may be applicable to frames which have been modified, for instance by an application like KAPPA’s COMPAVE.
The value of the parameter can be one of the following:
A domain name is usually the most suitable choice.
Unlike the Current frame, the frame selected using this parameter is copied to the AST file unmodified; in particular it retains the same Domain name. [PIXEL]
FITSID = LITERAL (Read)
If the IDTYPE parameter has the value FITSID, this parameter gives the FITS header keyword whose value distinguishes frames with different coordinate system information. If any lower case characters are given, they are converted to upper case. This may be a compound name to handle hierarchical keywords, in which case it has the form keyword1.keyword2 etc. Each keyword must be no longer than 8 characters.
FITSROT = LITERAL (Read)
If this parameter is not null, it gives the name of a FITS header keyword whose value gives a number of degrees to rotate the coordinate system by when it is imported. If any lower case characters are given, they are converted to upper case. This may be a compound name to handle hierarchical keywords, in which case it has the form keyword1.keyword2 etc. Each keyword must be no longer than 8 characters. [!]
IDTYPE = LITERAL (Read)
This parameter destermines the form of the ID value which distinguishes the framesets from each other in the exported AST file. It may have one of the following values:
FITSID – ID is generated from FITS header (see also the FITSID parameter).
INDEX – ID is given by an integer as taken from the INDICES parameter. This normally gives the frameset generated from the N’th image in the IN list an ID with index N.
SET – ID is given by an integer taken from the Set Index attribute of the CCDPACK Set header of each input file.
A list of images from which framesets are to be extracted. The Current frame of each should normally be the same, and should be a frame in which the different images are correctly registered. The image names may be specified using wildcards, or may be specified using an indirection file (the indirection character is "
\text{^}
INDICES( ) = _INTEGER (Read)
If IDTYPE is set to INDEX, then this parameter is a list of integers with as many elements as there are images accessed by the IN parameter. It gives the sequence of indices N to be used for generating the ID values. If set null (!) the images will be considered in the order 1,2,3,… which will normally be appropriate unless the images are being presented in an order different from that in which they are likely to be presented to ASTIMP. [!]
This parameter gives the name of the new alignment domain for the frames written out to the AST file. It is a good idea to choose a value which is not likely to exist previously in the WCS components of the images to which ASTFILE will be applied. A suitable value might be the name of the instrument from which the images are obtained.
Note that the frames which are written to the AST file are always the Current frames of the images supplied; this parameter only gives the name that the frames will have in the AST file, and consequently the name by which they will be known when the WCS information is imported into other images using ASTIMP or MAKESET.
The name is converted to upper case, and whitespace is removed. [CCD_EXPORT]
astexp reg_data
\ast
camera.ast idtype=fitsid fitsid=CHIPNUM outdomain=camera
This will save the information about the relative positioning of the images ’reg_data
\ast
’ to the file ’camera.ast’, calling the alignment domain ’CAMERA’. The file ’camera.ast’ can later be used by the ASTIMP or MAKESET applications to add the same coordinate information to a different set of images from the same instrument. Before running this, the images ’reg_data
\ast
’ should be correctly aligned in their Current domain. CHIPNUM must be the name of a FITS header keyword present in the FITS extension of each image whose value distinguishes the CCDs from each other (presumably present in the unreduced data). The mappings between the pixel coordinates and Current coordinates of the input images are recorded.
astexp "im1,im2,im3" astfile=camera.ast baseframe=axis title="Focal plane alignment" accept
In this case the OUTDOMAIN parameter takes its default value of ’CCD_EXPORT’, but mappings are between the Current coordinates of the input images and their ’AXIS’ coordinates. This could be a good idea if the images had been shrunk using KAPPA’s COMPAVE or something similar, which modifies the PIXEL coordinates but leaves the AXIS coordinates unchanged. No suitable FITS header is available to distinguish the different types of image, so the IDTYPE parameter is allowed to assume its default value of INDEX. When camera.ast is used for importing frameset information, the images from the three different chips must be listed in the same order as when this command was invoked. The title of the output Current frame will be as given.
astexp "r10595[2345]" wfc.ast outdomain=wfc idtype=fitsid fitsid=CHIPNAME fitsrot=ROTSKYPA
This exports the alignment information from the four named images to a file wfc.ast. The CHIPNAME FITS header identifies the source CCD for each, and the ROTSKYPA FITS header gives a number of degrees to rotate each frame additional to the relative alignment information.
“Re-use of coordinate system information with AST files”, ASTIMP.
AST file format
The AST file is designed to be written by ASTEXP and read by ASTIMP or MAKESET, and the user does not need to understand its format. It is however a text file, and if care is taken it may be edited by hand. Removing entire framesets and modifying ID values or domain names may be done fairly easily, but care should be taken (see SUN/210) if any more involved changes are to be undertaken. The format of the file is explained here.
The AST file consists of the following, in order:
<
global modifiers
>
<
frameset 1
>
<
frameset 1 modifiers
>
<
>
<
>
Characters after a ’#’ character are normally ignored. The constituent parts are composed as follows:
A single blank line, which may contain spaces but no comments.
The framesets are written in AST native format, as explained in SUN/210.
Each frameset has an ID, and contains two frames (a Base frame and a Current frame) and a mapping between them. The domains of all the Base frames should normally be the same, and likewise for all the Current frames. For the images to which the file will be applied by ASTIMP, their WCS components should contain frames in the same domain as the AST file’s Base frame.
The ID of each frameset is used to determine, for each image, which of the framesets in the file should be applied to it. This ID is a string which can assume one of the following forms:
"FITSID KEY VALUE" — This will match an image if the first FITS header card with the keyword KEY has the value VALUE. If the value is of type CHARACTER it must be in single quotes. KEY may be compound (of the form keyword1.keyword2 etc) to permit reading of hierarchical keywords.
"INDEX N" — This associates a frameset with an integer N. Usually N will take the values 1,2,3,... for the framesets in the file. Typically the N’th image in a list will match the one with an ID of "INDEX N".
"SET N" — This will match an image if the Set Index attribute iin its CCDPACK Set header is equal to the integer N.
Modifiers describe additional modifications to be made to the framesets on import. They are of the form
USE keyword arguments
Currently the only modifier defined is FITSROT, which defines the name of a FITS header which specifies how many degrees to rotate the image before use. This rotation is carried out after the mapping defined by the frameset itself.
Global modifiers affect all images processed with the AST file. Frameset modifiers affect only those images which correspond to their frameset.
Rigorous error checking of the AST file is not performed, so that unhelpful modifications to the WCS components of the target images may occur if it is not in accordance with these requirements.
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Heavy-Duty Data Transfer · Sam Henri Gold
Medieval soldiers clamping down the arm of a trebuchet. Image credit: Ian V. Hogg via Encyclopædia Britannica.
Every waking minute, even when I’m in private, someone comes up and asks me, “Sam, how fast can we possibly move data from one place to another?” I’m tired of it. I just want to eat alone. But because I am kind, funny, hot, but mostly kind, I will run through the math. Just for you.
The current fastest (recorded) data transfer speed is 319 terabits/second over 30,001 kilometers, or slightly faster than a South Korean internet connection, by using optical fibers.
These fibers are tiny. Teeny tiny. Teeny weeny tiny. I'll be blunt: if I can’t see my internet pipes with m'own peepers, I simply cannot trust that data. So our ideal data transfer must be visible. Preferably, I should be able to construct this infrastructure using standard household materials – nothing more than a dozen baulks of wood harvested from ship hulls and the remains of homes of a nearby pillaged town. Since the internet is everywhere and yet nowhere, we must adapt our new data transfer system to any arbitrary location. This contraption would have to be mobile.
As it turns out, a team of engineers called the Pannonian Avars had solved this problem years ago: with the advent of the Trebuchet in the 6th century AD.
A modern counterweight trebuchet can theoretically yeet a 100 kg projectile 200 meters with a 6,000 kg counterweight. Assuming we use solely 1 TB micro SD cards (250 milligrams), our trebuchet can handle 400,000 cards before snapping in half like a darn-tootin’ toothpick. 400 petabytes!
Math time! So to make our lives easier, we'll normalize all units to gigabytes and meters. This is the fastest optical speed:
\frac{39,880\phantom{\rule{0.22em}{0ex}}\text{gigabytes}}{3,001,000\phantom{\rule{0.22em}{0ex}}\text{m}}\phantom{\rule{0.22em}{0ex}}=\phantom{\rule{0.22em}{0ex}}\frac{0.0132889037\phantom{\rule{0.22em}{0ex}}\text{gigabytes}}{1\phantom{\rule{0.22em}{0ex}}\text{m}}
And our projected trebuchet:
\frac{400,000,000\phantom{\rule{0.22em}{0ex}}\text{gigabytes}}{200\phantom{\rule{0.22em}{0ex}}\text{m}}\phantom{\rule{0.22em}{0ex}}=\phantom{\rule{0.22em}{0ex}}\frac{2,000,000\phantom{\rule{0.22em}{0ex}}\text{gigabytes}}{1\phantom{\rule{0.22em}{0ex}}\text{m}}
To account for speed, I'll make some assumptions about our MicroSD payload. The bundled payload's diameter is 6 feet and we've maxed out the payload at 400K cards (100 kg). We'll also assume the windspeed is 2.68 m/s, which is the low end for average windspeed in the US.
I’m gonna level with you, I barely passed high school science. “Punnett square” this, “stop sniffing the gas out of the Bunsen burner” that. So when I started thinking about how fast this bundle of micro SD cards would be traveling, my brain started to hurt. And I did NOT manifest that for 2022. So here we are. I can only imagine flinging a bunch of cards across land would move significantly faster than fiber. Fiber is tiny, and it's easy for the internet juice to get caught in a tangle.
Trebuchet Data Transfer (TDT for short, since I imagine we'll be seeing this term often) seems promising. Encryption in transit barely holds a candle to the trebuchet method. What could be more secure than being able to see your precious data being flung 200 meters by a medieval siege weapon? That's the Web3-level trust we were promised. If it was good enough for the Romans, it’ll be good enough for you.
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Aliquot sequence - Wikipedia
Do all aliquot sequences eventually end with a prime number, a perfect number, or a set of amicable or sociable numbers? (Catalan's aliquot sequence conjecture)
In mathematics, an aliquot sequence is a sequence of positive integers in which each term is the sum of the proper divisors of the previous term. If the sequence reaches the number 1, it ends, since the sum of the proper divisors of 1 is 0.
2 Catalan–Dickson conjecture
3 Systematically searching for aliquot sequences
The aliquot sequence starting with a positive integer k can be defined formally in terms of the sum-of-divisors function σ1 or the aliquot sum function s in the following way:[1]
sn = s(sn−1) = σ1(sn−1) − sn−1 if sn−1 > 0
sn = 0 if sn−1 = 0 ---> (if we add this condition, then the terms after 0 are all 0, and all aliquot sequences would be infinite sequence, and we can conjecture that all aliquot sequences are convergent, the limit of these sequences are usually 0 or 6)
and s(0) is undefined.
For example, the aliquot sequence of 10 is 10, 8, 7, 1, 0 because:
σ1(10) − 10 = 5 + 2 + 1 = 8,
σ1(8) − 8 = 4 + 2 + 1 = 7,
σ1(7) − 7 = 1,
σ1(1) − 1 = 0.
Many aliquot sequences terminate at zero; all such sequences necessarily end with a prime number followed by 1 (since the only proper divisor of a prime is 1), followed by 0 (since 1 has no proper divisors). See (sequence A080907 in the OEIS) for a list of such numbers up to 75. There are a variety of ways in which an aliquot sequence might not terminate:
A perfect number has a repeating aliquot sequence of period 1. The aliquot sequence of 6, for example, is 6, 6, 6, 6, ...
An amicable number has a repeating aliquot sequence of period 2. For instance, the aliquot sequence of 220 is 220, 284, 220, 284, ...
A sociable number has a repeating aliquot sequence of period 3 or greater. (Sometimes the term sociable number is used to encompass amicable numbers as well.) For instance, the aliquot sequence of 1264460 is 1264460, 1547860, 1727636, 1305184, 1264460, ...
Some numbers have an aliquot sequence which is eventually periodic, but the number itself is not perfect, amicable, or sociable. For instance, the aliquot sequence of 95 is 95, 25, 6, 6, 6, 6, ... . Numbers like 95 that are not perfect, but have an eventually repeating aliquot sequence of period 1 are called aspiring numbers.[2]
n Aliquot sequence of n length (OEIS: A098007) n Aliquot sequence of n length (OEIS: A098007) n Aliquot sequence of n length (OEIS: A098007) n Aliquot sequence of n length (OEIS: A098007)
The lengths of the aliquot sequences that start at n are
1, 2, 2, 3, 2, 1, 2, 3, 4, 4, 2, 7, 2, 5, 5, 6, 2, 4, 2, 7, 3, 6, 2, 5, 1, 7, 3, 1, 2, 15, 2, 3, 6, 8, 3, 4, 2, 7, 3, 4, 2, 14, 2, 5, 7, 8, 2, 6, 4, 3, ... (sequence A044050 in the OEIS)
The final terms (excluding 1) of the aliquot sequences that start at n are
1, 2, 3, 3, 5, 6, 7, 7, 3, 7, 11, 3, 13, 7, 3, 3, 17, 11, 19, 7, 11, 7, 23, 17, 6, 3, 13, 28, 29, 3, 31, 31, 3, 7, 13, 17, 37, 7, 17, 43, 41, 3, 43, 43, 3, 3, 47, 41, 7, 43, ... (sequence A115350 in the OEIS)
Numbers whose aliquot sequence terminates in 1 are
1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 26, 27, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, ... (sequence A080907 in the OEIS)
Numbers whose aliquot sequence known to terminate in a perfect number, other than perfect numbers themselves (6, 28, 496, ...), are
25, 95, 119, 143, 417, 445, 565, 608, 650, 652, 675, 685, 783, 790, 909, 913, ... (sequence A063769 in the OEIS)
Numbers whose aliquot sequence terminates in a cycle with length at least 2 are
220, 284, 562, 1064, 1184, 1188, 1210, 1308, 1336, 1380, 1420, 1490, 1604, 1690, 1692, 1772, 1816, 1898, 2008, 2122, 2152, 2172, 2362, ... (sequence A121507 in the OEIS)
Numbers whose aliquot sequence is not known to be finite or eventually periodic are
276, 306, 396, 552, 564, 660, 696, 780, 828, 888, 966, 996, 1074, 1086, 1098, 1104, 1134, 1218, 1302, 1314, 1320, 1338, 1350, 1356, 1392, 1398, 1410, 1464, 1476, 1488, ... (sequence A131884 in the OEIS)
A number that is never the successor in an aliquot sequence is called an untouchable number.
2, 5, 52, 88, 96, 120, 124, 146, 162, 188, 206, 210, 216, 238, 246, 248, 262, 268, 276, 288, 290, 292, 304, 306, 322, 324, 326, 336, 342, 372, 406, 408, 426, 430, 448, 472, 474, 498, ... (sequence A005114 in the OEIS)
Catalan–Dickson conjecture
An important conjecture due to Catalan, sometimes called the Catalan–Dickson conjecture, is that every aliquot sequence ends in one of the above ways: with a prime number, a perfect number, or a set of amicable or sociable numbers.[3] The alternative would be that a number exists whose aliquot sequence is infinite yet never repeats. Any one of the many numbers whose aliquot sequences have not been fully determined might be such a number. The first five candidate numbers are often called the Lehmer five (named after D.H. Lehmer): 276, 552, 564, 660, and 966.[4] However, it is worth noting that 276 may reach a high apex in its aliquot sequence and then descend; the number 138 reaches a peak of 179931895322 before returning to 1.
Guy and Selfridge believe the Catalan–Dickson conjecture is false (so they conjecture some aliquot sequences are unbounded above (i.e., diverge)).[5]
As of April 2015[update], there were 898 positive integers less than 100,000 whose aliquot sequences have not been fully determined, and 9190 such integers less than 1,000,000.[6]
Systematically searching for aliquot sequences
The aliquot sequence can be represented as a directed graph,
{\displaystyle G_{n,s}}
, for a given integer
{\displaystyle n}
{\displaystyle s(k)}
denotes the sum of the proper divisors of
{\displaystyle k}
.[7] Cycles in
{\displaystyle G_{n,s}}
represent sociable numbers within the interval
{\displaystyle [1,n]}
. Two special cases are loops that represent perfect numbers and cycles of length two that represent amicable pairs.
^ Weisstein, Eric W. "Aliquot Sequence". MathWorld.
^ Sloane, N. J. A. (ed.). "Sequence A063769 (Aspiring numbers: numbers whose aliquot sequence terminates in a perfect number.)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation.
^ Weisstein, Eric W. "Catalan's Aliquot Sequence Conjecture". MathWorld.
^ Creyaufmüller, Wolfgang (May 24, 2014). "Lehmer Five". Retrieved June 14, 2015.
^ A. S. Mosunov, What do we know about aliquot sequences?
^ Creyaufmüller, Wolfgang (April 29, 2015). "Aliquot Pages". Retrieved June 14, 2015.
^ Rocha, Rodrigo Caetano; Thatte, Bhalchandra (2015), Distributed cycle detection in large-scale sparse graphs, Simpósio Brasileiro de Pesquisa Operacional (SBPO), doi:10.13140/RG.2.1.1233.8640
Current status of aliquot sequences with start term below 2 million
Active research site on aliquot sequences (Jean-Luc Garambois) (in French)
Retrieved from "https://en.wikipedia.org/w/index.php?title=Aliquot_sequence&oldid=1070814075"
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MAKECAL
Produces a dark or pre-flash calibration image
This routine performs the combination of a series of dark count or pre-flash exposure frames. The input images should have been bias subtracted. The input data are divided by the exposure factors before combination into a calibration "master", giving an output image whose data represent one unit of the given exposure time per pixel. Thus the calibration frame should be multiplied by the appropriate factor before subtracting from other frames (i.e. by the dark time or the flash-exposure time). This can be performed by CALCOR and should be done prior to the production of a flatfield and flatfield correction. The data combination methods give a mixture of very robust (median) to very efficient (mean) methods to suit the data.
makecal in expose out method
Either: An exact number of exposure factors for the input images. The values must be in the same order as the input images.
Or: A single value which applies to all the input images.
Indirection through an ASCII file may be used to specify these values. If more than one line is required at prompt time then a continuation line may be requested by adding "-" to the end of the line.
This parameter will not be used if USEEXT is set TRUE.
A list of image names which contain the calibration data. The image names should be separated by commas and may include wildcards.
Whether to keep (i.e. not delete) the input images or not. Deleting the input images has the advantage of saving disk space, but should probably only be used if this program is part of a sequence of commands and the intermediary data used by it are not important.
If METHOD = "THRESH" then this value defines the upper limit for values which can be used when combining the data. This limit applies to the range of the output data (i.e. the values after the exposure factors have been divided into the input data).
TRIMMED – An "alpha trimmed mean" in which a fraction alpha/2 of the values are removed from each extreme
If METHOD = "THRESH" then this value defines the lower limit for values which can be used when combining the data. This limit applies to the range of the output data (i.e. the values after the exposure factors have been divided into the input data).
The minimum number of good (ie. not BAD) pixels required to contribute to the value of an output pixel. Output pixels not meeting this requirement are set BAD. [1]
Name of the output image to contain the calibration data. Note this image will have a type of at least _REAL. If USESET is true and multiple Sets are represented in the IN list then this name will be used as the name of an HDS container file containing one NDF structure for each Set Index value. This name may be specified using indirection through a file.
Number of standard deviations to reject data at. Used for "MODE", "SIGMA" and "CLIPMED" methods. For METHOD = "MODE" the standard deviation is estimated from the population of values, for METHOD = "SIGMA" and "CLIPMED" the variances are used. If no variances exist then the variation of the population is used to estimate one. [4.0]
Title for the output image. [Output from MAKECAL]
TYPE = LITERAL (Read)
The frame types of the input data. This should be a recognised name "FLASH", "DARK" or "NONE". The value of this parameter affects the output image frame type which will be set to "MASTER_FLASH" or "MASTER_DARK" or "MASTER_?". [NONE]
Whether to use Set header information or not. If USESET is false then any Set header information will be ignored. If USESET is true, then input files will be considered in groups; a separate calibration frame will be constructed for each group of corresponding input frames (i.e. those sharing the same Set Index attribute). If this results in multiple output calibration files, they will be written as separate NDF structures into a single HDS container file. If no Set header information is present in the input files, then calibration is done on all the input files together, so USESET can usually be safely set to TRUE.
If TRUE then the EXPOSE parameter of this program will not be used and the required values will be obtained from the CCDPACK extensions of the input images instead. This method can only be used if the images have been "imported" using the programs PRESENT or IMPORT. Typically it is used when processing using CCDPACK’s "automated" methods.
\ast
makecal in=’"f1,f2,f3,f4"’ expose=’"100,200,300,400"’ method=median out=master_flash
This example forms a flash calibration image from the data in images f1,f2,f3 and f4. The data are divided by the relative exposure factors before combination. The combination method used is the (weighted) median, the resultant data are written to the image master_flash.
makecal ’"d1,d2,d3,d4"’ 1 master_dark trimmed alpha=0.2
This example produces a dark-count-calibration frame from the data in images d1,d2,d3 and d4. The exposure factors are given as 1 which probably indicates that the dark-exposure times in these datasets are exactly right to correct any subsequent data frames. The combination mode used is the trimmed mean with trimming fraction 0.2 and the output data are written to image master_dark.
\text{^}
flash_frames
\text{^}
flash_exposures flash_master
In this example a list of frames to be processed is passed to the program by indirection through an ASCII file flash_frames.dat, the corresponding exposure times are passed from the file flash_exposures.dat. This is probably the only safe method for entering images to this routine other than as in the above examples. Using wildcards for the file specifications will mean that the exposures cannot be associated correctly. Thus wildcards should not be used except when the input images have the same exposure times.
“Flash or dark calibration”, CALCOR.
TITLE – always "Output from MAKECAL"
The parameter EXPOSE will not be used if the USEEXT parameter is set TRUE. In this case the necessary values will be extracted from the CCDPACK extensions of the input images.
The routine supports BAD pixels and all data types except COMPLEX. All combinational arithmetic is performed in floating point. The AXIS and TITLE components are correctly propagated. The variances are propagated through the combination processing, assuming that the input data have a normal distribution.
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MIMO State-Space Models - MATLAB & Simulink - MathWorks Deutschland
MIMO Explicit State-Space Models
MIMO Descriptor State-Space Models
State-Space Model of Jet Transport Aircraft
You create a MIMO state-space model in the same way as you create a SISO state-space model. The only difference between the SISO and MIMO cases is the dimensions of the state-space matrices. The dimensions of the B, C, and D matrices increase with the numbers of inputs and outputs as shown in the following illustration.
In this example, you create a state-space model for a rotating body with inertia tensor J, damping force F, and three axes of rotation, related as:
\begin{array}{c}J\frac{d\omega }{dt}+F\omega =T\\ y=\omega .\end{array}
The system input T is the driving torque. The output y is the vector of angular velocities of the rotating body.
To express this system in state-space form:
\begin{array}{c}\frac{dx}{dt}=Ax+Bu\\ y=Cx+Du\end{array}
\begin{array}{c}\frac{d\omega }{dt}=-{J}^{-1}F\omega +{J}^{-1}T\\ y=\omega .\end{array}
Then the state-space matrices are:
A=-{J}^{-1}F,\text{ }B={J}^{-1},\text{ }C=I,\text{ }D=0.
To create this model, enter the following commands:
These commands assume that J is the inertia tensor of a cube rotating about its corner, and the damping force has magnitude 0.2.
sys_mimo is an ss model.
This example shows how to create a continuous-time descriptor (implicit) state-space model using dss.
This example uses the same rotating-body system shown in MIMO Explicit State-Space Models, where you inverted the inertia matrix J to obtain the value of the B matrix. If J is poorly-conditioned for inversion, you can instead use a descriptor (implicit) state-space model. A descriptor (implicit) state-space model is of the form:
\begin{array}{c}E\frac{dx}{dt}=Ax+Bu\\ y=Cx+Du\end{array}
Create a state-space model for a rotating body with inertia tensor J, damping force F, and three axes of rotation, related as:
\begin{array}{c}J\frac{d\omega }{dt}+F\omega =T\\ y=\omega .\end{array}
The system input T is the driving torque. The output y is the vector of angular velocities of the rotating body. You can write this system as a descriptor state-space model having the following state-space matrices:
A=-F,\text{ }B=I,\text{ }C=I,\text{ }D=0,\text{ }E=J.
To create this system, enter:
A = -F;
sys_mimo = dss(A,B,C,D,E)
sys is an ss model with a nonempty E matrix.
This example shows how to build a MIMO model of a jet transport. Because the development of a physical model for a jet aircraft is lengthy, only the state-space equations are presented here. See any standard text in aviation for a more complete discussion of the physics behind aircraft flight.
A = [-0.0558 -0.9968 0.0802 0.0415
0 0.0805 1.0000 0];
B = [ 0.0073 0
C = [0 1 0 0
D = [0 0
Use the following commands to specify this state-space model as an LTI object and attach names to the states, inputs, and outputs.
outputs = {'yaw rate' 'bank angle'};
sys_mimo = ss(A,B,C,D,'statename',states,...
You can display the LTI model by typing sys_mimo.
sys_mimo
beta yaw roll phi
beta -0.0558 -0.9968 0.0802 0.0415
yaw 0.598 -0.115 -0.0318 0
roll -3.05 0.388 -0.465 0
phi 0 0.0805 1 0
rudder aileron
beta 0.0073 0
yaw -0.475 0.0077
roll 0.153 0.143
phi 0 0
yaw rate 0 1 0 0
bank angle 0 0 0 1
yaw rate 0 0
bank angle 0 0
The model has two inputs and two outputs. The units are radians for beta (sideslip angle) and phi (bank angle) and radians/sec for yaw (yaw rate) and roll (roll rate). The rudder and aileron deflections are in degrees.
As in the SISO case, use tf to derive the transfer function representation.
tf(sys_mimo)
Transfer function from input "rudder" to output...
-0.475 s^3 - 0.2479 s^2 - 0.1187 s - 0.05633
yaw rate: ---------------------------------------------------
s^4 + 0.6358 s^3 + 0.9389 s^2 + 0.5116 s + 0.003674
0.1148 s^2 - 0.2004 s - 1.373
bank angle: ---------------------------------------------------
Transfer function from input "aileron" to output...
0.0077 s^3 - 0.0005372 s^2 + 0.008688 s + 0.004523
0.1436 s^2 + 0.02737 s + 0.1104
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Global Constraint Catalog: Clex2
<< 5.218. leq_cst5.220. lex_alldifferent >>
\mathrm{𝚕𝚎𝚡}\mathtt{2}\left(\mathrm{𝙼𝙰𝚃𝚁𝙸𝚇}\right)
\mathrm{𝚍𝚘𝚞𝚋𝚕𝚎}_\mathrm{𝚕𝚎𝚡}
\mathrm{𝚛𝚘𝚠}_\mathrm{𝚊𝚗𝚍}_\mathrm{𝚌𝚘𝚕𝚞𝚖𝚗}_\mathrm{𝚕𝚎𝚡}
\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁}
\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚟𝚊𝚛}-\mathrm{𝚍𝚟𝚊𝚛}\right)
\mathrm{𝙼𝙰𝚃𝚁𝙸𝚇}
\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚟𝚎𝚌}-\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁}\right)
|\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁}|\ge 1
\mathrm{𝚛𝚎𝚚𝚞𝚒𝚛𝚎𝚍}
\left(\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁},\mathrm{𝚟𝚊𝚛}\right)
\mathrm{𝚛𝚎𝚚𝚞𝚒𝚛𝚎𝚍}
\left(\mathrm{𝙼𝙰𝚃𝚁𝙸𝚇},\mathrm{𝚟𝚎𝚌}\right)
\mathrm{𝚜𝚊𝚖𝚎}_\mathrm{𝚜𝚒𝚣𝚎}
\left(\mathrm{𝙼𝙰𝚃𝚁𝙸𝚇},\mathrm{𝚟𝚎𝚌}\right)
Given a matrix of domain variables, enforces that both adjacent rows, and adjacent columns are lexicographically ordered (adjacent rows and adjacent columns can be equal).
\left(〈\mathrm{𝚟𝚎𝚌}-〈2,2,3〉,\mathrm{𝚟𝚎𝚌}-〈2,3,1〉〉\right)
\mathrm{𝚕𝚎𝚡}\mathtt{2}
The first row
〈2,2,3〉
is lexicographically less than or equal to the second row
〈2,3,1〉
The first column
〈2,2〉
is lexicographically less than or equal to the second column
〈2,3〉
The second column
〈2,3〉
is lexicographically less than or equal to the third column
〈3,1〉
|\mathrm{𝚅𝙴𝙲𝚃𝙾𝚁}|>1
|\mathrm{𝙼𝙰𝚃𝚁𝙸𝚇}|>1
\mathrm{𝚟𝚊𝚛}
\mathrm{𝙼𝙰𝚃𝚁𝙸𝚇}.\mathrm{𝚟𝚎𝚌}
The idea of this symmetry-breaking constraint can already be found in the following articles of A. Lubiw [Lubiw85], [Lubiw87].
In block designs you sometimes want repeated blocks, so using the non-strict order would be required in this case.
\mathrm{𝚕𝚎𝚡}\mathtt{2}
constraint can be expressed as a conjunction of two
\mathrm{𝚕𝚎𝚡}_\mathrm{𝚌𝚑𝚊𝚒𝚗}_\mathrm{𝚕𝚎𝚜𝚜𝚎𝚚}
constraints: A first
\mathrm{𝚕𝚎𝚡}_\mathrm{𝚌𝚑𝚊𝚒𝚗}_\mathrm{𝚕𝚎𝚜𝚜𝚎𝚚}
constraint on the
\mathrm{𝙼𝙰𝚃𝚁𝙸𝚇}
argument and a second
\mathrm{𝚕𝚎𝚡}_\mathrm{𝚌𝚑𝚊𝚒𝚗}_\mathrm{𝚕𝚎𝚜𝚜𝚎𝚚}
constraint on the transpose of the
\mathrm{𝙼𝙰𝚃𝚁𝙸𝚇}
argument.
lex2 in MiniZinc.
\mathrm{𝚊𝚕𝚕𝚙𝚎𝚛𝚖}
\mathrm{𝚕𝚎𝚡}_\mathrm{𝚕𝚎𝚜𝚜𝚎𝚚}
(matrix symmetry,lexicographic order).
\mathrm{𝚜𝚝𝚛𝚒𝚌𝚝}_\mathrm{𝚕𝚎𝚡}\mathtt{2}
\mathrm{𝚕𝚎𝚡}_\mathrm{𝚌𝚑𝚊𝚒𝚗}_\mathrm{𝚕𝚎𝚜𝚜𝚎𝚚}
\mathrm{𝚕𝚎𝚡}_\mathrm{𝚌𝚑𝚊𝚒𝚗}_\mathrm{𝚕𝚎𝚜𝚜𝚎𝚚}
constraint type: predefined constraint, system of constraints, order constraint.
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Using Question Structures to Write Assessments
Written by Sarah Stecher published 9 months ago
Finding good sources for assessment questions can be tricky. If you're like us, you've probably spent hours on internet deep dives searching for interesting problems. What you find instead are hundreds of questions that assess basic facts and procedures. Where do we go for deeper, more thoughtful questions? Maybe you already have an online resource that you love, but we're here to propose that writing your own questions may be easier than you think.
Here are our favorite strategies for writing good questions in an intentional but efficient way:
Focus on underlying structure, not just specific examples:
Replace some numbers with parameters so students can't just perform a standard algorithm to get an answer. While students may be able to apply a learned procedure to a specific example and get an answer, they often don’t understand why that method works in general. By stripping away some of the numbers, we ask students to consider how changing the part affects the whole.
f(x)=asin(bx-c)+d
, changing which parameter(s) would affect the range of the function?
Find the value of k such that the linear system has no solution.
Ask the question backward:
Consider turning a standard question on its head.
Standard: What is the horizontal asymptote of
y=\frac{3x^2-5x}{2+2x^2}
Backward: Write an equation of a rational function that has a horizontal asymptote of y=3/2.
Incorporate multiple representations:
Have students reason using graphs and tables, not just equations. For example, instead of always providing an equation (analytical approach), have students evaluate functions using selected values on a table (numerical approach) or by finding ordered pairs on a graph (graphical approach). Students should be able to demonstrate the deeper understanding that “evaluate” doesn’t just mean "plug in", but finding a corresponding output for a particular input.
Use question stems:
We use some versions of these questions in almost every test we write. You don’t have to reinvent the wheel!
Which of the following is false? (Provide four statements that have to do with the current unit)
Find the value of k such that…
Determine if the statement is always, sometimes, or never true.
Give a possible (value, equation, graph) such that…
Interpret the ______ in the context of this problem.
Which of the following is/is NOT equivalent to _____?
Error analysis: In which step did the student make his first mistake? (Label the steps of a sample solution. Make sure one of the answer choices is “there is no mistake.”)
As always, the more you do this, the more efficient you will become. We encourage you to do this work alongside your colleagues and share the workload!
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3.7.120.2. Strategy that gradually creates a compul >>
When the available space is equal to the total area of the rectangles to place (i.e., we have no slack) this is a two-phase search procedure originally introduced in [AggounBeldiceanu93] where we first fix all the
x
-coordinates and then, in the second phase, all the
y
-coordinates. The intuitions behind this heuristics are:
To systematically fill the placement space from right to left in order to avoid creating small holes that cannot be filled.
To decrease the combinatorial aspect of the problem by focussing first on all
x
-coordinates. This stems from the fact that it is usually easy to extend a partial solution, where all
x
-coordinates are fixed, to a full solution.
Fixing the
x
-coordinates is done by:
First, compute the minimum
mi{n}_{x}
over the minimum values of the
x
-coordinates of the rectangles for which the
x
-coordinate is not already fixed.
Second, create a choice point and, in each branch:
Fix the
x
-coordinate of a rectangle
R
x
-coordinates is not already fixed to value
mi{n}_{x}
. Usually rectangles are considered by decreasing height (and decreasing width in case of tie).
On backtracking, enforce that the
x
-coordinate of rectangle
R
mi{n}_{x}
Third, fail when all branches issued from a choice point have been tried (since otherwise we would create a hole at position
mi{n}_{x}
because, on the
x
axis all rectangles that could start at position
mi{n}_{x}
were delayed after
mi{n}_{x}
; in order to not cut valid choices, this third part assumes that the minimum value of the
x
-coordinate of each rectangle is pruned with respect to the compulsory part profile of the corresponding
\mathrm{𝚌𝚞𝚖𝚞𝚕𝚊𝚝𝚒𝚟𝚎}
constraint.).
Since, as we said early on, it is usually easy to extend a partial solution, where all
x
-coordinates are fixed, to a full solution where all
y
-coordinates are also fixed, the search strategy used for fixing the
y
-coordinates is usually not so important, at least when strong filtering algorithms are used [BeldiceanuCarlssonPoder08].
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Speed, Distance, and Time | Brilliant Math & Science Wiki
Andrew Ellinor, Christopher Williams, Tim O'Brien, and
A common set of physics problems ask students to determine either the speed, distance, or travel time of something given the other two variables. These problems are interesting since they describe very basic situations that occur regularly for many people. For example, a problem might say: "Find the distance a car has traveled in fifteen minutes if it travels at a constant speed of
75 \text {km/hr}
." Often in these problems, we work with an average velocity or speed, which simplifies the laws of motion used to calculate the desired quantity. Let's see how that works.
As long as the speed is constant or average, the relationship between speed, distance, and time is expressed in this equation
\mbox{Speed} = \frac{\mbox{Distance}}{\mbox{Time}},
which can also be rearranged as
\mbox{Time} = \frac{\mbox{Distance}}{\mbox{Speed}}
\mbox{Distance} = \mbox{Speed} \times \mbox{Time}.
Speed, distance, and time problems ask to solve for one of the three variables given certain information. In these problems, objects are moving at either constant speeds or average speeds.
Most problems will give values for two variables and ask for the third.
Bernie boards a train at 1:00 PM and gets off at 5:00 PM. During this trip, the train traveled 360 kilometers. What was the train's average speed in kilometers per hour?
In this problem, the total time is 4 hours and the total distance is
360\text{ km},
which we can plug into the equation:
\mbox{Speed} = \frac{\mbox{Distance}}{\mbox{Time}}= \frac{360~\mbox{km}}{4~\mbox{h}} = 90~\mbox{km/h}. \ _\square
When working with these problems, always pay attention to the units for speed, distance, and time. Converting units may be necessary to obtaining a correct answer.
A horse is trotting along at a constant speed of 8 miles per hour. How many miles will it travel in 45 minutes?
The equation for calculating distance is
\mbox{Distance} = \mbox{Speed} \times \mbox{Time},
but we won't arrive at the correct answer if we just multiply 8 and 45 together, as the answer would be in units of
\mbox{miles} \times \mbox{minute} / \mbox{hour}
. To fix this, we incorporate a unit conversion:
\mbox{Distance} = \frac{8~\mbox{miles}}{~\mbox{hour}} \times 45~\mbox{minutes} \times \frac{1~\mbox{hour}}{60~\mbox{minutes}} = 6~\mbox{miles}. \ _\square
Alternatively, we can convert the speed to units of miles per minute and calculate for distance:
\mbox{Distance} = \frac{2}{15}~\frac{\mbox{miles}}{\mbox{minute}} \times 45~\mbox{minutes} = 6~\mbox{miles},
or we can convert time to units of hours before calculating:
\mbox{Distance} = 8~\frac{\mbox{miles}}{\mbox{hour}} \times \frac{3}{4}~\mbox{hours} = 6~\mbox{miles}.
Any of these methods will give the correct units and answer.
_\square
In more involved problems, it is convenient to use variables such as
v
d
t
for speed, distance, and velocity, respectively.
Alice, Bob, Carly, and Dave are in a flying race!
Alice's plane is twice as fast as Bob's plane.
When Alice finishes the race, the distance between her and Carly is
D.
When Bob finishes the race, the distance between him and Dave is
D.
If Bob's plane is three times as fast as Carly's plane, then how many times faster is Alice's plane than Dave's plane?
Albert and Danny are running in a long-distance race. Albert runs at 6 miles per hour while Danny runs at 5 miles per hour. You may assume they run at a constant speed throughout the race. When Danny reaches the 25 mile mark, Albert is exactly 40 minutes away from finishing. What is the race's distance in miles?
Let's begin by calculating how long it takes for Danny to run 25 miles:
\mbox{Time} = \frac{\mbox{Distance}}{\mbox{Speed}}= \frac{25~\mbox{miles}}{5~\mbox{miles/hour}}= 5~\mbox{hours}.
So, it will take Albert
5~\mbox{hours} + 40~\mbox{minutes}
\frac{17}{3}~\mbox{hours}
, to finish the race. Now we can calculate the race's distance:
\begin{aligned} \mbox{Distance} &= \mbox{Speed} \times \mbox{Time} \\ &= (6~\mbox{miles/hour}) \times \left(\frac{17}{3}~\mbox{hours}\right) \\ &= 34~\mbox{miles}.\ _\square \end{aligned}
A cheetah spots a gazelle
300\text{ m}
away and sprints towards it at
100\text{ km/h}.
At the same time, the gazelle runs away from the cheetah at
80\text{ km/h}.
How many seconds does it take for the cheetah to catch the gazelle?
Let's set up equations representing the distance the cheetah travels and the distance the gazelle travels. If we set distance
d
0
as the cheetah's starting point, we have
\begin{aligned} d_\text{cheetah} &= 100t \\ d_\text{gazelle} &= 0.3 + 80t. \end{aligned}
Note that time
t
here is in units of hours, and
300\text{ m}
was converted to
0.3\text{ km}.
The cheetah catches the gazelle when
\begin{aligned} d_\text{cheetah} &=d_\text{gazelle} \\ 100t &= 0.3 + 80t \\ 20t &= 0.3 \\ t &= 0.015~\mbox{hours}. \end{aligned}
Converting that answer to seconds, we find that the cheetah catches the gazelle in
54~\mbox{seconds}
_\square
Two friends are crossing a hundred meter railroad bridge when they suddenly hear a train whistle. Desperate, each friend starts running, one towards the train and one away from the train. The one that ran towards the train gets to safety just before the train passes, and so does the one that ran in the same direction as the train.
If the train is five times faster than each friend, then what is the train-to-friends distance when the train whistled (in meters)?
Cite as: Speed, Distance, and Time. Brilliant.org. Retrieved from https://brilliant.org/wiki/speed-distance-and-time/
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Diagram below shows the locations of points A (34o S, 40o W) and B (34o S, 80o E) which lie on the surface of the earth. AC is a diameter of the common parallel of latitude 34o S.
(a) State the longitude of C.
(b) Calculate the distance, in nautical mile, from A due east to B, measured along the common parallel of latitude 34o S.
(c) K lies due north of A and the shortest distance from A to K measured along the surface of the earth is 4440 nautical miles.
Calculate the latitude of K.
(d) An aeroplane took off from B and flew due west to A along the common parallel of latitude. Then, it flew due north to K. The average speed for the whole flight was 450 knots.
Longitude of C = (180o – 40o) E = 140o E
Distance of AB
= (40 + 80) x 60 x cos 34o
= 120 x 60 x cos 34o
\begin{array}{l}\angle AOK=\frac{4440}{60}\\ \text{ }={74}^{o}\\ \text{Latitude of }K={\left(74-34\right)}^{o}N\\ \text{ }={40}^{o}N\end{array}
\begin{array}{l}\text{Total distance travelled}\\ BA+AK\\ =5969+4440\\ =10409\text{ nautical miles}\\ \\ \text{Total time taken =}\frac{\text{Total distance travelled}}{\text{Average speed}}\\ \text{}=\frac{10409}{450}\\ \text{}=23.13\text{ hours}\end{array}
← 7.5 SPM Practice (Long Questions)
9.6 SPM Practice (Long Questions) →
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PDA_RNEXP
Returns pseudo-random numbers from an exponential distribution
This is a simple random-number generator providing deviates in the from an exponential distribution, with a period of 2
\ast
\ast
RESULT = PDA_RNEXP( X )
PDA_RNEXP = INTEGER
Ahrens, J.H., & Dieter, U. 1972, "Computer Methods for sampling from the exponential and Normal distributions", Comm. ACM 15(10), pp.873–882.
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Econometric Modeler App Overview - MATLAB & Simulink - MathWorks ä¸å›½
H1: Series has at least one nonzero autocorrelation coefficient Ïj, j ∈ {1,…,m}.
On the test type tab, in the Parameters section, adjust parameters for the test. For example, consider performing an Engle's ARCH test. On the ARCH tab, in the Parameters section, select the number of lags in the test statistic using the Number of Lags spin box, or the significance level (that is, the value of α) using the Significance Level spin box.
Multiple testing inflates the false discovery rate. One conservative way to maintain an overall false discovery rate of α is to apply the Bonferroni correction to the significance level of each test. That is, for a total of t tests, set Significance Level value to α/t.
In the Type Model Parameters dialog box, select the lag to constrain by using the AR Coefficients (ϕ) (VAR or VARX) or Short-Run Coefficients (Φ) (for VEC) list.
\stackrel{^}{y}
{\stackrel{^}{\mathrm{Ï}}}_{t}^{2}
\frac{{y}_{t}âc}{{\stackrel{^}{\mathrm{Ï}}}_{t}^{2}}
Consider a SARIMA(0,1,1)×(0,1,1)12 for the monthly international airline passenger numbers from 1949 to 1960 in the Data_Airline data set. To estimate this model using the Econometric Modeler:
\stackrel{^}{y}
Consider performing goodness-of-fit checks on the estimated SARIMA(0,1,1)×(0,1,1)12 model for the airline counts data in Estimating a Univariate Model.
Fit a SARIMA(0,1,q)×(0,1,q12)12 to PSSGLog, where all unknown orders are 0 (see Estimating a Univariate Model).
Iterate steps 4 and 5, but adjust q and q12 to cover the nine permutations of q ∈ {0,1,2} and q12 ∈ {0,1,2}. Econometric Modeler distinguishes subsequent models of the same type by appending consecutive digits to the end of the variable name.
SARIMA(0,1,0)×(0,1,0)12 SARIMA_PSSGLog1 -491.8042
Because it yields the minimal AIC, the SARIMA(0,1,1)×(0,1,1)12 model is the model with the best parsimonious, in-sample fit.
Consider generating a live function that returns SARIMA_PSSGLog, the SARIMA(0,1,1)×(0,1,1)12 model fit to the log of airline passenger data (see Estimating a Univariate Model). This figure shows the generated live function.
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Global Constraint Catalog: Cinverse
<< 5.198. interval_and_sum5.200. inverse_offset >>
\mathrm{𝚒𝚗𝚟𝚎𝚛𝚜𝚎}\left(\mathrm{𝙽𝙾𝙳𝙴𝚂}\right)
\mathrm{𝚊𝚜𝚜𝚒𝚐𝚗𝚖𝚎𝚗𝚝}
\mathrm{𝚌𝚑𝚊𝚗𝚗𝚎𝚕}
\mathrm{𝚒𝚗𝚟𝚎𝚛𝚜𝚎}_\mathrm{𝚌𝚑𝚊𝚗𝚗𝚎𝚕𝚒𝚗𝚐}
\mathrm{𝙽𝙾𝙳𝙴𝚂}
\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚒𝚗𝚍𝚎𝚡}-\mathrm{𝚒𝚗𝚝},\mathrm{𝚜𝚞𝚌𝚌}-\mathrm{𝚍𝚟𝚊𝚛},\mathrm{𝚙𝚛𝚎𝚍}-\mathrm{𝚍𝚟𝚊𝚛}\right)
\mathrm{𝚛𝚎𝚚𝚞𝚒𝚛𝚎𝚍}
\left(\mathrm{𝙽𝙾𝙳𝙴𝚂},\left[\mathrm{𝚒𝚗𝚍𝚎𝚡},\mathrm{𝚜𝚞𝚌𝚌},\mathrm{𝚙𝚛𝚎𝚍}\right]\right)
\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚒𝚗𝚍𝚎𝚡}\ge 1
\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚒𝚗𝚍𝚎𝚡}\le |\mathrm{𝙽𝙾𝙳𝙴𝚂}|
\mathrm{𝚍𝚒𝚜𝚝𝚒𝚗𝚌𝚝}
\left(\mathrm{𝙽𝙾𝙳𝙴𝚂},\mathrm{𝚒𝚗𝚍𝚎𝚡}\right)
\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚜𝚞𝚌𝚌}\ge 1
\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚜𝚞𝚌𝚌}\le |\mathrm{𝙽𝙾𝙳𝙴𝚂}|
\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚙𝚛𝚎𝚍}\ge 1
\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚙𝚛𝚎𝚍}\le |\mathrm{𝙽𝙾𝙳𝙴𝚂}|
Enforce each vertex of a digraph to have exactly one predecessor and one successor. In addition the following two statements are equivalent:
The successor of the
{i}^{th}
node is the
{j}^{th}
node.
The predecessor of the
{j}^{th}
{i}^{th}
\left(\begin{array}{c}〈\begin{array}{ccc}\mathrm{𝚒𝚗𝚍𝚎𝚡}-1\hfill & \mathrm{𝚜𝚞𝚌𝚌}-2\hfill & \mathrm{𝚙𝚛𝚎𝚍}-2,\hfill \\ \mathrm{𝚒𝚗𝚍𝚎𝚡}-2\hfill & \mathrm{𝚜𝚞𝚌𝚌}-1\hfill & \mathrm{𝚙𝚛𝚎𝚍}-1,\hfill \\ \mathrm{𝚒𝚗𝚍𝚎𝚡}-3\hfill & \mathrm{𝚜𝚞𝚌𝚌}-5\hfill & \mathrm{𝚙𝚛𝚎𝚍}-4,\hfill \\ \mathrm{𝚒𝚗𝚍𝚎𝚡}-4\hfill & \mathrm{𝚜𝚞𝚌𝚌}-3\hfill & \mathrm{𝚙𝚛𝚎𝚍}-5,\hfill \\ \mathrm{𝚒𝚗𝚍𝚎𝚡}-5\hfill & \mathrm{𝚜𝚞𝚌𝚌}-4\hfill & \mathrm{𝚙𝚛𝚎𝚍}-3\hfill \end{array}〉\hfill \end{array}\right)
\mathrm{𝚒𝚗𝚟𝚎𝚛𝚜𝚎}
\mathrm{𝙽𝙾𝙳𝙴𝚂}\left[1\right].\mathrm{𝚜𝚞𝚌𝚌}=2⇔\mathrm{𝙽𝙾𝙳𝙴𝚂}\left[2\right].\mathrm{𝚙𝚛𝚎𝚍}=1
\mathrm{𝙽𝙾𝙳𝙴𝚂}\left[2\right].\mathrm{𝚜𝚞𝚌𝚌}=1⇔\mathrm{𝙽𝙾𝙳𝙴𝚂}\left[1\right].\mathrm{𝚙𝚛𝚎𝚍}=2
\mathrm{𝙽𝙾𝙳𝙴𝚂}\left[3\right].\mathrm{𝚜𝚞𝚌𝚌}=5⇔\mathrm{𝙽𝙾𝙳𝙴𝚂}\left[5\right].\mathrm{𝚙𝚛𝚎𝚍}=3
\mathrm{𝙽𝙾𝙳𝙴𝚂}\left[4\right].\mathrm{𝚜𝚞𝚌𝚌}=3⇔\mathrm{𝙽𝙾𝙳𝙴𝚂}\left[3\right].\mathrm{𝚙𝚛𝚎𝚍}=4
\mathrm{𝙽𝙾𝙳𝙴𝚂}\left[5\right].\mathrm{𝚜𝚞𝚌𝚌}=4⇔\mathrm{𝙽𝙾𝙳𝙴𝚂}\left[4\right].\mathrm{𝚙𝚛𝚎𝚍}=5
|\mathrm{𝙽𝙾𝙳𝙴𝚂}|>1
\mathrm{𝙽𝙾𝙳𝙴𝚂}
\mathrm{𝙽𝙾𝙳𝙴𝚂}
\left(\mathrm{𝚒𝚗𝚍𝚎𝚡}\right)
\left(\mathrm{𝚜𝚞𝚌𝚌},\mathrm{𝚙𝚛𝚎𝚍}\right)
\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚜𝚞𝚌𝚌}
\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚒𝚗𝚍𝚎𝚡}
\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚙𝚛𝚎𝚍}
\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚙𝚛𝚎𝚍}
\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚒𝚗𝚍𝚎𝚡}
\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚜𝚞𝚌𝚌}
This constraint is used in order to make the link between the successor and the predecessor variables. This is sometimes required by specific heuristics that use both predecessor and successor variables. In some problems, the
\mathrm{successor}
\mathrm{predecessor}
variables are respectively interpreted as column an row variables (i.e., we have a bijection between the
\mathrm{successor}
variables and their values). This is for instance the case in the
n
-queens problem (i.e., place
n
queens on an
n
chessboard in such a way that no two queens are on the same row, the same column or the same diagonal) when we use the following model: to each column of the chessboard we associate a variable that gives the row where the corresponding queen is located. Symmetrically, to each row of the chessboard we create a variable that indicates the column where the associated queen is placed. Having these two sets of variables, we can now write a heuristics that selects the column or the row for which we have the fewest number of alternatives for placing a queen.
\mathrm{𝚒𝚗𝚟𝚎𝚛𝚜𝚎}
\mathrm{𝚒𝚗𝚍𝚎𝚡}
attribute was not explicitly present. It was implicitly defined as the position of a variable in a list, the first position being 1. This is also the case for SICStus Prolog, JaCoP and Gecode where the variables are respectively indexed from 1, 0 and 0. Within SICStus Prolog and JaCoP (http://www.jacop.eu/), the
\mathrm{𝚒𝚗𝚟𝚎𝚛𝚜𝚎}
\mathrm{𝚊𝚜𝚜𝚒𝚐𝚗𝚖𝚎𝚗𝚝}
. Within Gecode, it is called
\mathrm{𝚌𝚑𝚊𝚗𝚗𝚎𝚕}
(http://www.gecode.org/).
\mathrm{𝚒𝚗𝚟𝚎𝚛𝚜𝚎}
We first normalize the domains of the variables by removing value
i
{j}^{th}
\mathrm{predecessor}
variable if value
j
does not belong to the
{i}^{th}
\mathrm{successor}
variable, and by removing value
j
{i}^{th}
\mathrm{successor}
i
{j}^{th}
\mathrm{predecessor}
Second, one can map solutions to the
\mathrm{𝚒𝚗𝚟𝚎𝚛𝚜𝚎}
constraint to perfect matchings in a so-called variable bipartite graph derived from the domain of the variables of the constraint in the following way: to each successor variable corresponds a vertex; similarly to each predecessor variable corresponds a vertex; there is and edge between the
{i}^{th}
\mathrm{successor}
variable and the
{j}^{th}
\mathrm{predecessor}
variable if and only if value
i
belongs to the domain of the
{j}^{th}
\mathrm{predecessor}
variable and value
j
{i}^{th}
\mathrm{successor}
Third, Dulmage-Mendelsohn decomposition [DulmageMendelsohn58] is used to characterise all edges that do not belong to any perfect matching, and therefore prune the corresponding variables.
inverseChanneling in Choco, channel in Gecode, inverse in MiniZinc, assignment in SICStus.
\mathrm{𝚌𝚢𝚌𝚕𝚎}
\mathrm{𝚜𝚢𝚖𝚖𝚎𝚝𝚛𝚒𝚌}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}
\mathrm{𝚒𝚗𝚟𝚎𝚛𝚜𝚎}_\mathrm{𝚘𝚏𝚏𝚜𝚎𝚝}
(do not assume anymore that the smallest value of the
\mathrm{𝚙𝚛𝚎𝚍}
\mathrm{𝚜𝚞𝚌𝚌}
attributes is equal to 1),
\mathrm{𝚒𝚗𝚟𝚎𝚛𝚜𝚎}_\mathrm{𝚜𝚎𝚝}
\mathrm{𝚍𝚘𝚖𝚊𝚒𝚗}\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎}
\mathrm{𝚜𝚎𝚝}
\mathrm{𝚟𝚊𝚛𝚒𝚊𝚋𝚕𝚎}
\mathrm{𝚒𝚗𝚟𝚎𝚛𝚜𝚎}_\mathrm{𝚠𝚒𝚝𝚑𝚒𝚗}_\mathrm{𝚛𝚊𝚗𝚐𝚎}
(partial mapping between two collections of distinct size).
\mathrm{𝚕𝚎𝚡}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}
heuristics: heuristics.
modelling: channelling constraint, permutation channel, dual model, functional dependency.
modelling exercises: n-Amazons, zebra puzzle.
puzzles: n-Amazons, n-queens, zebra puzzle.
\mathrm{𝙽𝙾𝙳𝙴𝚂}
\mathrm{𝐶𝐿𝐼𝑄𝑈𝐸}
↦\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚗𝚘𝚍𝚎𝚜}\mathtt{1},\mathrm{𝚗𝚘𝚍𝚎𝚜}\mathtt{2}\right)
•\mathrm{𝚗𝚘𝚍𝚎𝚜}\mathtt{1}.\mathrm{𝚜𝚞𝚌𝚌}=\mathrm{𝚗𝚘𝚍𝚎𝚜}\mathtt{2}.\mathrm{𝚒𝚗𝚍𝚎𝚡}
•\mathrm{𝚗𝚘𝚍𝚎𝚜}\mathtt{2}.\mathrm{𝚙𝚛𝚎𝚍}=\mathrm{𝚗𝚘𝚍𝚎𝚜}\mathtt{1}.\mathrm{𝚒𝚗𝚍𝚎𝚡}
\mathrm{𝐍𝐀𝐑𝐂}
=|\mathrm{𝙽𝙾𝙳𝙴𝚂}|
In order to express the binary constraint that links two vertices one has to make explicit the identifier of the vertices. This is why the
\mathrm{𝚒𝚗𝚟𝚎𝚛𝚜𝚎}
constraint considers objects that have three attributes:
One fixed attribute
\mathrm{𝚒𝚗𝚍𝚎𝚡}
that is the identifier of the vertex,
One variable attribute
\mathrm{𝚜𝚞𝚌𝚌}
that is the successor of the vertex,
\mathrm{𝚙𝚛𝚎𝚍}
that is the predecessor of the vertex.
\mathrm{𝐍𝐀𝐑𝐂}
\mathrm{𝚒𝚗𝚟𝚎𝚛𝚜𝚎}
\mathrm{𝚒𝚗𝚍𝚎𝚡}
\mathrm{𝙽𝙾𝙳𝙴𝚂}
collection are distinct and because of the first condition
\mathrm{𝚗𝚘𝚍𝚎𝚜}\mathtt{1}.\mathrm{𝚜𝚞𝚌𝚌}=\mathrm{𝚗𝚘𝚍𝚎𝚜}\mathtt{2}.\mathrm{𝚒𝚗𝚍𝚎𝚡}
of the arc constraint all the vertices of the final graph have at most one predecessor.
\mathrm{𝚒𝚗𝚍𝚎𝚡}
\mathrm{𝙽𝙾𝙳𝙴𝚂}
collection are distinct and because of the second condition
\mathrm{𝚗𝚘𝚍𝚎𝚜}\mathtt{2}.\mathrm{𝚙𝚛𝚎𝚍}=\mathrm{𝚗𝚘𝚍𝚎𝚜}\mathtt{1}.\mathrm{𝚒𝚗𝚍𝚎𝚡}
of the arc constraint all the vertices of the final graph have at most one successor.
From the two previous remarks it follows that the final graph is made up from disjoint paths and disjoint circuits. Therefore the maximum number of arcs of the final graph is equal to its maximum number of vertices
\mathrm{𝙽𝙾𝙳𝙴𝚂}
\mathrm{𝐍𝐀𝐑𝐂}=|\mathrm{𝙽𝙾𝙳𝙴𝚂}|
\mathrm{𝐍𝐀𝐑𝐂}\ge |\mathrm{𝙽𝙾𝙳𝙴𝚂}|
\underline{\overline{\mathrm{𝐍𝐀𝐑𝐂}}}
\overline{\mathrm{𝐍𝐀𝐑𝐂}}
\mathrm{𝚒𝚗𝚟𝚎𝚛𝚜𝚎}
\mathrm{𝙽𝙾𝙳𝙴𝚂}
{S}_{i}
\mathrm{𝚒𝚗𝚟𝚎𝚛𝚜𝚎}
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Global Constraint Catalog: Kn-queens
<< 3.7.163. n-Amazons3.7.165. Non-deterministic automaton >>
\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}
\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}_\mathrm{𝚌𝚜𝚝}
\mathrm{𝚒𝚗𝚟𝚎𝚛𝚜𝚎}
A constraint that can be used for modelling the
n
-queen problem. Place
n
queens on an
n
chessboard in such a way that no queen attacks another. Two queens attack each other if they are located on the same column, on the same row or on the same diagonal. A constructive method for arbitrary
n>3
was first given in [FalkowskiSchmitz86]. An effective heuristics for the
n
-queen problem was given in [Kale90]. It consists of starting to place the queens in the centre of the chessboard so that they eliminate the maximum number of potential positions.
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9.6 SPM Practice (Long Questions) – user's Blog!
Home › SPM Maths › Earth as a Sphere › 9.6 SPM Practice (Long Questions)
\begin{array}{l}\angle POM=\frac{5700}{60}\\ \text{}={95}^{o}\\ \therefore \text{Latitude of}M=\left({95}^{o}-{40}^{o}\right)S\\ \text{}={55}^{o}S\end{array}
\begin{array}{l}\text{Time taken =}\frac{\text{distance from}R\text{to}P}{\text{average speed}}\\ \text{}=\frac{\left(180-25\right)×60×\mathrm{cos}{40}^{o}}{660}\\ \text{}=\frac{155×60×\mathrm{cos}{40}^{o}}{660}\\ \text{}=10.79\text{hours}\end{array}
← 9.4.2 Distance between Two Points along a Parallel of Latitude
10.3 SPM Practice (Long Questions) →
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Chess Abstract: Level 3 Challenges Practice Problems Online | Brilliant
Define a Galloping Queen as a chess piece whose legal move is that of a Knight, and that of a Queen.
What is the minimum value of integer
n > 1
such that you can place
non-attacking Galloping Queens on an
n \times n
Try more questions on Galloping Queens
What is the maximum number of bishops on a
8 \times 8
chessboard, so that they cannot attack each other?
Above is a
3\times3
board with 4 knights two white knights and two black knights. As in a standard game of chess, the knight can move only two steps in the horizontal or vertical direction and then one step in the other direction for one move. Define an action as moving a knight of any color.
The objective of the game is to interchange the position of both the black and white knights while alternately moving a knight of different color. The final state of the board is:
Using only actions, what is the minimum number of actions required to complete the game?
by naitik sanghavi
What is the maximum number of counters you can place on an
8 \times 8
chessboard given that each row, column, and the two main diagonals contain 5 or fewer counters?
by Sam Bealing
A white pawn had been accidentally knocked off the board. Neither player could remember for sure on which square it stood. If neither king has yet moved, where is the pawn?
This problem was inspired by the book Chess Mysteries of Sherlock Holmes by Raymond Smullyan.
This problem is the first one of the set Retrograde Chess.
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SHOWSET
Outputs image Set header information
This routine is used to examine the Set membership attributes of images. It will show the Set Name and Set Index attributes for each image, and whether it contains a CCD_SET coordinate frame in its WCS component. The images are output grouped by Set Name or Set Index. If required, a restricted list of images, those with certain Name and/or Index attributes, may be selected for output; in this case the acceptable Names/Indexes can be given explicitly or as a list of template images whose attributes they have to match. The names of the images selected for output may be written to a list file. SHOWSET can therefore be used to construct files listing those images in a given Set, or corresponding images in different Sets.
showset in
A list of images to examine.
INDEX = LITERAL (Read)
If PICKINDEX=EQUAL this parameter restricts which files will be selected for output. It must be a group expression (a comma-separated list) each member of which is an acceptable INDEX value. Only files with a Set Index value equal to one of these will be selected.
INDEXLIKE = LITERAL (Read)
If PICKINDEX=LIKE this parameter restricts which files will be selected for output. It must be a group expression (a comma-separated list which may employ wildcards or indirection) each member of which represents an image to be used as a template. Only images with a Set Index value matching that of one of the template images will be selected.
LISTBY = LITERAL (Read)
Indicates the way in which images should be grouped for output. It may take the values ’NAME’, ’INDEX’ or ’NONE’. If set to NAME, then all the images in the same Set are grouped together in the output; if set to INDEX then all the corresponding images from different Sets are grouped together, and if set to NONE images will be listed in the same order as the IN parameter. If only images with the same Name or with the same Index are being output, this will have no effect. [NAME]
If PICKNAME=EQUAL this parameter restricts which files will be selected for output. It must be a group expression (a comma-separated list) each member of which is a string. Only files with a Set Name value the same as one of these will be selected.
NAMELIKE = LITERAL (Read)
If PICKNAME=LIKE this parameter restricts which files will be selected for output. It must be a group expression (a comma-separated list which may employ wildcards or indirection) each member of which represents an image to be used as a template. Only images with a Set Name value matching that of one of the template images will be selected.
The name of an output file in which to write the names of the images selected for output. The (non-comment) lines of this file are of the form:
image-name # set-index set-name
since the set-index and set-name values appear to the right of a comment character, the file can thus be used as an indirection file for input to other CCDPACK commands. [showset.lis]
PICKINDEX = LITERAL (Read)
Indicates how images are to be filtered by Set Index attribute for output. Takes one of the following values:
ALL – All Index values are acceptable
EQUAL – Only Index values listed in the INDEX parameter value are acceptable
LIKE – Only Index values the same as those of the images listed in the INDEXLIKE parameter are acceptable.
PICKNAME = LITERAL (Read)
Indicates how images are to be filtered by Set Name attribute for output. Takes one of the following values:
ALL – All Name values are acceptable
EQUAL – Only Name values listed in the NAME parameter value are acceptable
LIKE – Only Name values the same as those of the images listed in the NAMELIKE parameter are acceptable.
SETLESS = _LOGICAL (Read)
If there are no restrictions on which Sets to display, because PICKNAME and PICKINDEX are both set to ALL, this parameter determines what happens to images which have no Set headers. If SETLESS is true, they are selected for output, but if SETLESS is false, they are discarded. [FALSE]
\ast
This will list all the images in the current directory which contain Set header information; the listing will be grouped by the Set Name attribute and Set Index will be shown.
\ast
setless=true
This will do the same as the previous example, except that those images with no Set header information will be displayed as well.
\ast
pickname=like namelike="gc6235a,gc4021a" namelist=gc.lis
This will list all the images in the current directory which are in the same Set as the images gc6235a and gc4021a. As well as showing the Set information of these files on the screen, the names of the files thus selected will be written to the file gc.lis.
showset fdata setless reset
This will just show the Name and Set information of the file fdata. If fdata is a container file, it will show the Set information for all the datasets within it. Since the SETLESS parameter is given, even if it has no Set header output will be written.
showset dat
\ast
pickindex=equal index=3 logto=neither namelist=out.lis
This will write a list of image names to the file out.lis choosing only those which have a Set Index attribute value of 3. There will be no output to the screen or log file.
“Examining Set headers”.
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Diagram below shows the locations of points P, Q, R, A, K and C, on the surface of the earth. O is the centre of the earth.
(a) Find the location of A.
(b) Given the distance QR is 3240 nautical miles, find the longitude of Q.
(c) Calculate the distance, in nautical miles of KA, measured along the common parallel latitude.
(d) An aeroplane took off from A and flew due west to K along the common parallel of latitude. Then, it flew due south to Q. The average speed of the aeroplane was 550 knots.
Longitude of A = (180o – 15o) = 165o E
Latitude of A = 50o N
Therefore, position of A = (50o N, 165o E).
\begin{array}{l}\angle QOR=\frac{3240}{60}\\ \text{}={54}^{o}\\ \therefore \text{Longitude of }Q=\left({165}^{o}-{54}^{o}\right)E\\ \text{ }={111}^{o}E\end{array}
Distance of KA
\begin{array}{l}\text{Total distance}=AK+KQ\\ \text{ }=2082.6+\left(50×60\right)\\ \text{ }=5082.6\text{ nautical miles}\\ \\ \text{Total time }=\frac{5082.6}{550}\\ \text{ }=9.241\text{ hours}\end{array}
4.5 Multiplication of Two Matrices (Sample Question 2) →
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Find Image Rotation and Scale - MATLAB & Simulink Example
Step 3: Select Control Points
Step 6: Recover Original Image
This example shows how to align or register two images that differ by a rotation and a scale change. You can use fitgeotrans to find the rotation angle and scale factor after manually picking corresponding points. You can then transform the distorted image to recover the original image.
distorted = imresize(original,scale); % Try varying the scale factor.
distorted = imrotate(distorted,theta); % Try varying the angle, theta.
Use the Control Point Selection Tool to pick at least two pairs of control points.
movingPoints = [151.52 164.79; 131.40 79.04];
fixedPoints = [135.26 200.15; 170.30 79.30];
You can run the rest of the example with these pre-picked points, but try picking your own points to see how the results vary.
cpselect(distorted,original,movingPoints,fixedPoints);
Save control points by choosing the File menu, then the Save Points to Workspace option. Save the points, overwriting variables movingPoints and fixedPoints.
Fit a nonreflective similarity transformation to your control points.
tform = fitgeotrans(movingPoints,fixedPoints,'nonreflectivesimilarity');
After you have done Steps 5 and 6, repeat Steps 4 through 6 but try using 'affine' instead of 'NonreflectiveSimilarity'. What happens? Are the results as good as they were with 'NonreflectiveSimilarity'?
The geometric transformation, tform, contains a transformation matrix in tform.T. Since you know that the transformation includes only rotation and scaling, the math is relatively simple to recover the scale and angle.
\mathrm{sc}=\mathrm{scale}*\mathrm{cos}\left(\mathrm{theta}\right)
\mathrm{ss}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\mathrm{scale}*\mathrm{sin}\left(\mathrm{theta}\right)
Then, Tinv = invert(tform), and Tinv.T =
\left[\begin{array}{ccc}\mathrm{sc}& -\mathrm{ss}& 0\\ \mathrm{ss}& \mathrm{sc}& 0\\ \mathrm{tx}& \mathrm{ty}& 1\end{array}\right]
tformInv = invert(tform);
Tinv = tformInv.T;
scale_recovered = sqrt(ss*ss + sc*sc)
scale_recovered = 0.7000
theta_recovered = atan2(ss,sc)*180/pi
theta_recovered = 29.3741
The recovered values of scale_recovered and theta_recovered should match the values you set in Step 2: Resize and Rotate the Image.
Recover the original image by transforming distorted, the rotated-and-scaled image, using the geometric transformation tform and what you know about the spatial referencing of original. The 'OutputView' name-value argument is used to specify the resolution and grid size of the resampled output image.
Roriginal = imref2d(size(original));
recovered = imwarp(distorted,tform,'OutputView',Roriginal);
montage({original,recovered})
The recovered (right) image quality does not match the original (left) image because of the distortion and recovery process. In particular, the image shrinking causes loss of information. The artifacts around the edges are due to the limited accuracy of the transformation. If you were to pick more points in Step 3: Select Control Points, the transformation would be more accurate.
imresize | imrotate | cpselect | fitgeotrans | imwarp | imref2d
Find Image Rotation and Scale Using Automated Feature Matching (Computer Vision Toolbox)
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C-level Overview - Capital Pricing and Initial Liquidity - Premia
Every pool has a C-level (price level) at any given time. The higher the C-level, the more expensive the options, the higher the premiums earned on capital by LPs.
C-level is the unique Premia mechanic that regulates the option price levels in a pool, according to the relative supply and demand of capital in the pool.
When the C-level is high, options are expensive and LPs earn a higher return on capital.
As a LP you want to provide more liquidity when the C-level is high.
Each pool is given a high C value on initialization - LPs can get in early for the highest likelihood of above market risk-adjusted returns.
To best understand how the C-level works, let's consider a newly created ETH/DAI put pool. Let's suppose that the initial
C_0
value with which the pool is initiated is
C_0=5
. Intuitively, the C-level is the dynamic multiple to the base Black-Scholes model that adjusts after every trade. This means that initial pool prices are set to
5 * BlackScholesPrice
C_0=5
, the options in that pool are almost guaranteed to be overpriced (by design) - this is good for LPs. As an LP, I want to provide liquidity to such a pool, in case some option buyers are either irrational or have an extremely high willingness to pay for ETH/DAI put options. But while
C_0
is high, this is almost guaranteed to attract more LPs than buyers. Note: The pools are liquidity aware. Every time liquidity is added, the C-level decreases. Every time options are purchased (causing free liquidity to be removed), the C-level increases.
As LPs continue to deposit capital into the new pool, they will drive down the pool price levels to a market clearing price range, after some number of transactions, n. If we assume that, after n transactions, the pool converges to the market clearing C-value,
C_{clearing}
, the LPs will continue earning market-competitive returns on their liquidity.
However, before the C-level converges to
C_{clearing}
, there are very likely to be buyers willing to overpay for the options, thus generating early LPs above market risk-adjusted returns (for the capital that is utilized).
Theoretical pool utilization will reach 100% at the limit
Due to the design of Premia pricing mechanics, each pool will always converge towards a theoretical 100% utilization. For the LPs this is great, because none of their capital will be sitting idle. All the liquidity deposited by an LP will be incentivized to be utilized at the best market return. However, practically, there will always be deviations from 100% utilization for reasons such as buyers needing free liquidity, withdrawals, gradual divestment, etc. and the Liquidity Buffer.
The Liquidity Buffer is the excess liquidity available in the pool, created by the initial C-value convergence. In other words, between the pool being initialized and the C-level reaching
C_{clearing}
, not all of this capital will be utilized, since it's assumed to be overpriced. Simulations show that once
C_{clearing}
is reached, the free capital in the pool will roughly hover around the liquidity buffer levels, though over time this buffer will become an increasingly smaller portion of the total pool size because the buffer is denoted in nominal terms, not percentage terms.
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Global Constraint Catalog: Cdag
<< 5.108. cyclic_change_joker5.110. decreasing >>
[Dooms06]
\mathrm{𝚍𝚊𝚐}\left(\mathrm{𝙽𝙾𝙳𝙴𝚂}\right)
\mathrm{𝙽𝙾𝙳𝙴𝚂}
\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚒𝚗𝚍𝚎𝚡}-\mathrm{𝚒𝚗𝚝},\mathrm{𝚜𝚞𝚌𝚌}-\mathrm{𝚜𝚟𝚊𝚛}\right)
\mathrm{𝚛𝚎𝚚𝚞𝚒𝚛𝚎𝚍}
\left(\mathrm{𝙽𝙾𝙳𝙴𝚂},\left[\mathrm{𝚒𝚗𝚍𝚎𝚡},\mathrm{𝚜𝚞𝚌𝚌}\right]\right)
\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚒𝚗𝚍𝚎𝚡}\ge 1
\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚒𝚗𝚍𝚎𝚡}\le |\mathrm{𝙽𝙾𝙳𝙴𝚂}|
\mathrm{𝚍𝚒𝚜𝚝𝚒𝚗𝚌𝚝}
\left(\mathrm{𝙽𝙾𝙳𝙴𝚂},\mathrm{𝚒𝚗𝚍𝚎𝚡}\right)
\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚜𝚞𝚌𝚌}\ge 1
\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚜𝚞𝚌𝚌}\le |\mathrm{𝙽𝙾𝙳𝙴𝚂}|
G
\mathrm{𝙽𝙾𝙳𝙴𝚂}
collection. Select a subset of arcs of
G
so that the corresponding graph does not contain any circuit.
\left(\begin{array}{c}〈\begin{array}{cc}\mathrm{𝚒𝚗𝚍𝚎𝚡}-1\hfill & \mathrm{𝚜𝚞𝚌𝚌}-\left\{2,4\right\},\hfill \\ \mathrm{𝚒𝚗𝚍𝚎𝚡}-2\hfill & \mathrm{𝚜𝚞𝚌𝚌}-\left\{3,4\right\},\hfill \\ \mathrm{𝚒𝚗𝚍𝚎𝚡}-3\hfill & \mathrm{𝚜𝚞𝚌𝚌}-\varnothing ,\hfill \\ \mathrm{𝚒𝚗𝚍𝚎𝚡}-4\hfill & \mathrm{𝚜𝚞𝚌𝚌}-\varnothing ,\hfill \\ \mathrm{𝚒𝚗𝚍𝚎𝚡}-5\hfill & \mathrm{𝚜𝚞𝚌𝚌}-\left\{6\right\},\hfill \\ \mathrm{𝚒𝚗𝚍𝚎𝚡}-6\hfill & \mathrm{𝚜𝚞𝚌𝚌}-\varnothing \hfill \end{array}〉\hfill \end{array}\right)
\mathrm{𝚍𝚊𝚐}
constraint holds since the
\mathrm{𝙽𝙾𝙳𝙴𝚂}
collection depicts a graph without circuit.
|\mathrm{𝙽𝙾𝙳𝙴𝚂}|>2
\mathrm{𝙽𝙾𝙳𝙴𝚂}
A filtering algorithm for the
\mathrm{𝚍𝚊𝚐}
constraint is given in [Dooms06]. It removes potential arcs that would create a circuit of mandatory arcs.
\mathrm{𝚒𝚗}_\mathrm{𝚜𝚎𝚝}
constraint type: graph constraint.
\mathrm{𝙽𝙾𝙳𝙴𝚂}
\mathrm{𝑆𝐸𝐿𝐹}
↦\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚗𝚘𝚍𝚎𝚜}\right)
\mathrm{𝚒𝚗}_\mathrm{𝚜𝚎𝚝}
\left(\mathrm{𝚗𝚘𝚍𝚎𝚜}.\mathrm{𝚔𝚎𝚢},\mathrm{𝚗𝚘𝚍𝚎𝚜}.\mathrm{𝚜𝚞𝚌𝚌}\right)
\mathrm{𝐍𝐀𝐑𝐂}
=0
\mathrm{𝙽𝙾𝙳𝙴𝚂}
\mathrm{𝐶𝐿𝐼𝑄𝑈𝐸}
↦\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚗𝚘𝚍𝚎𝚜}\mathtt{1},\mathrm{𝚗𝚘𝚍𝚎𝚜}\mathtt{2}\right)
\mathrm{𝚒𝚗}_\mathrm{𝚜𝚎𝚝}
\left(\mathrm{𝚗𝚘𝚍𝚎𝚜}\mathtt{2}.\mathrm{𝚒𝚗𝚍𝚎𝚡},\mathrm{𝚗𝚘𝚍𝚎𝚜}\mathtt{1}.\mathrm{𝚜𝚞𝚌𝚌}\right)
\mathrm{𝐌𝐀𝐗}_\mathrm{𝐍𝐒𝐂𝐂}
\le 1
The first graph constraint removes the loop of each vertex. The second graph constraint forbids the creation of circuits involving more than one vertex.
Part (A) of Figure 5.109.1 shows the initial graph associated with the second graph constraint of the Example slot. This initial graph from which we start is derived from the set associated with each vertex. Each set describes the potential values of the
\mathrm{𝚜𝚞𝚌𝚌}
attribute of a given vertex. Part (B) of Figure 5.109.1 gives the final graph associated with the Example slot.
\mathrm{𝚍𝚊𝚐}
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Global Constraint Catalog: Cint_value_precede_chain
<< 5.195. int_value_precede5.197. interval_and_count >>
\mathrm{𝚒𝚗𝚝}_\mathrm{𝚟𝚊𝚕𝚞𝚎}_\mathrm{𝚙𝚛𝚎𝚌𝚎𝚍𝚎}_\mathrm{𝚌𝚑𝚊𝚒𝚗}\left(\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂},\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\right)
\mathrm{𝚙𝚛𝚎𝚌𝚎𝚍𝚎}
\mathrm{𝚙𝚛𝚎𝚌𝚎𝚍𝚎𝚗𝚌𝚎}
\mathrm{𝚟𝚊𝚕𝚞𝚎}_\mathrm{𝚙𝚛𝚎𝚌𝚎𝚍𝚎}_\mathrm{𝚌𝚑𝚊𝚒𝚗}
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}
\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚟𝚊𝚛}-\mathrm{𝚒𝚗𝚝}\right)
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚟𝚊𝚛}-\mathrm{𝚍𝚟𝚊𝚛}\right)
\mathrm{𝚛𝚎𝚚𝚞𝚒𝚛𝚎𝚍}
\left(\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂},\mathrm{𝚟𝚊𝚛}\right)
\mathrm{𝚍𝚒𝚜𝚝𝚒𝚗𝚌𝚝}
\left(\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂},\mathrm{𝚟𝚊𝚛}\right)
\mathrm{𝚛𝚎𝚚𝚞𝚒𝚛𝚎𝚍}
\left(\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂},\mathrm{𝚟𝚊𝚛}\right)
n
denotes the number of items of the
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}
collection, the following condition holds for every
i\in \left[1,n-1\right]
: When it is defined, the first occurrence of the
{\left(i+1\right)}^{th}
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}
collection should be preceded by the first occurrence of the
{i}^{th}
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}
\left(〈4,0,1〉,〈4,0,6,1,0〉\right)
\mathrm{𝚒𝚗𝚝}_\mathrm{𝚟𝚊𝚕𝚞𝚎}_\mathrm{𝚙𝚛𝚎𝚌𝚎𝚍𝚎}_\mathrm{𝚌𝚑𝚊𝚒𝚗}
constraint holds since within the sequence 4, 0, 6, 1, 0:
The first occurrence of value 4 occurs before the first occurrence of value 0.
|\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}|>1
\mathrm{𝚜𝚝𝚛𝚒𝚌𝚝𝚕𝚢}_\mathrm{𝚒𝚗𝚌𝚛𝚎𝚊𝚜𝚒𝚗𝚐}
\left(\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}\right)
|\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}|>|\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}|
\mathrm{𝚛𝚊𝚗𝚐𝚎}
\left(\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}.\mathrm{𝚟𝚊𝚛}\right)>1
\mathrm{𝚞𝚜𝚎𝚍}_\mathrm{𝚋𝚢}
\left(\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂},\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}\right)
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}.\mathrm{𝚟𝚊𝚛}
that does not occur in
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}.\mathrm{𝚟𝚊𝚛}
can be replaced by any other value that also does not occur in
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}.\mathrm{𝚟𝚊𝚛}
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}\left(\mathrm{𝚒𝚍}\right)
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}\left(\mathrm{𝚞𝚗𝚒𝚘𝚗}\right)
\mathrm{𝚒𝚗𝚝}_\mathrm{𝚟𝚊𝚕𝚞𝚎}_\mathrm{𝚙𝚛𝚎𝚌𝚎𝚍𝚎}_\mathrm{𝚌𝚑𝚊𝚒𝚗}
constraint is useful for breaking symmetries in graph colouring problems. We set a
\mathrm{𝚒𝚗𝚝}_\mathrm{𝚟𝚊𝚕𝚞𝚎}_\mathrm{𝚙𝚛𝚎𝚌𝚎𝚍𝚎}_\mathrm{𝚌𝚑𝚊𝚒𝚗}
constraint on all variables
{V}_{1},{V}_{2},\cdots ,{V}_{n}
associated with the vertices of the graph to colour, where we state that the first occurrence of colour
i
should be located before the first occurrence of colour
i+1
within the sequence
{V}_{1},{V}_{2},\cdots ,{V}_{n}
Figure 5.196.1 illustrates the problem of colouring earth and mars from Thom Sulanke. Part (A) of Figure 5.196.1 provides a solution where the first occurrence of each value of
i
\left(i\in \left\{1,2,\cdots ,8\right\}\right)
is located before the first occurrence of value
i+1
. This is obtained by using the following constraints:
\left\{\begin{array}{c}𝙰\ne 𝙱,𝙰\ne 𝙴,𝙰\ne 𝙵,𝙰\ne 𝙶,𝙰\ne 𝙷,𝙰\ne 𝙸,𝙰\ne 𝙹,𝙰\ne 𝙺,\hfill \\ 𝙱\ne 𝙰,𝙱\ne 𝙲,𝙱\ne 𝙵,𝙱\ne 𝙶,𝙱\ne 𝙷,𝙱\ne 𝙸,𝙱\ne 𝙹,𝙱\ne 𝙺,\hfill \\ 𝙲\ne 𝙱,𝙲\ne 𝙳,𝙲\ne 𝙵,𝙲\ne 𝙶,𝙲\ne 𝙷,𝙲\ne 𝙸,𝙲\ne 𝙹,𝙲\ne 𝙺,\hfill \\ 𝙳\ne 𝙲,𝙳\ne 𝙴,𝙳\ne 𝙵,𝙳\ne 𝙶,𝙳\ne 𝙷,𝙳\ne 𝙸,𝙳\ne 𝙹,𝙳\ne 𝙺,\hfill \\ 𝙴\ne 𝙰,𝙴\ne 𝙳,𝙴\ne 𝙵,𝙴\ne 𝙶,𝙴\ne 𝙷,𝙴\ne 𝙸,𝙴\ne 𝙹,𝙴\ne 𝙺,\hfill \\ 𝙵\ne 𝙰,𝙵\ne 𝙱,𝙵\ne 𝙲,𝙵\ne 𝙳,𝙵\ne 𝙴,𝙵\ne 𝙶,𝙵\ne 𝙷,𝙵\ne 𝙸,𝙵\ne 𝙹,𝙵\ne 𝙺,\hfill \\ 𝙶\ne 𝙰,𝙶\ne 𝙱,𝙶\ne 𝙲,𝙶\ne 𝙳,𝙶\ne 𝙴,𝙶\ne 𝙵,𝙶\ne 𝙷,𝙶\ne 𝙸,𝙶\ne 𝙹,𝙶\ne 𝙺,\hfill \\ 𝙷\ne 𝙰,𝙷\ne 𝙱,𝙷\ne 𝙲,𝙷\ne 𝙳,𝙷\ne 𝙴,𝙷\ne 𝙵,𝙷\ne 𝙶,𝙷\ne 𝙸,𝙷\ne 𝙹,𝙷\ne 𝙺,\hfill \\ 𝙸\ne 𝙰,𝙸\ne 𝙱,𝙸\ne 𝙲,𝙸\ne 𝙳,𝙸\ne 𝙴,𝙸\ne 𝙵,𝙸\ne 𝙶,𝙸\ne 𝙷,𝙸\ne 𝙹,𝙸\ne 𝙺,\hfill \\ 𝙹\ne 𝙰,𝙹\ne 𝙱,𝙹\ne 𝙲,𝙹\ne 𝙳,𝙹\ne 𝙴,𝙹\ne 𝙵,𝙹\ne 𝙶,𝙹\ne 𝙷,𝙹\ne 𝙸,𝙹\ne 𝙺,\hfill \\ 𝙺\ne 𝙰,𝙺\ne 𝙱,𝙺\ne 𝙲,𝙺\ne 𝙳,𝙺\ne 𝙴,𝙺\ne 𝙵,𝙺\ne 𝙶,𝙺\ne 𝙷,𝙺\ne 𝙸,𝙺\ne 𝙹,\hfill \\ \mathrm{𝚒𝚗𝚝}_\mathrm{𝚟𝚊𝚕𝚞𝚎}_\mathrm{𝚙𝚛𝚎𝚌𝚎𝚍𝚎}_\mathrm{𝚌𝚑𝚊𝚒𝚗}\left(〈1,2,3,4,5,6,7,8,9〉,〈𝙰,𝙱,𝙲,𝙳,𝙴,𝙵,𝙶,𝙷,𝙸,𝙹,𝙺〉\right).\hfill \end{array}\right\
Part (B) provides a symmetric solution where the value precedence constraints between the pairs of values
\left(1,2\right)
\left(2,3\right)
\left(4,5\right)
\left(7,8\right)
\left(8,9\right)
are all violated (each violation is depicted by a dashed arc).
Figure 5.196.1. Using the
\mathrm{𝚒𝚗𝚝}_\mathrm{𝚟𝚊𝚕𝚞𝚎}_\mathrm{𝚙𝚛𝚎𝚌𝚎𝚍𝚎}_\mathrm{𝚌𝚑𝚊𝚒𝚗}
constraint for breaking symmetries in graph colouring problems; there is an arc between the first occurrence of value
v
\left(1\le v\le 8\right)
in the sequence of variables
𝙰
𝙱
𝙲
𝙳
𝙴
𝙵
𝙶
𝙷
𝙸
𝙹
𝙺
, and the first occurrence of value
v+1
(a plain arc if the corresponding value precedence constraint holds, a dashed arc otherwise)
When we have more than one class of interchangeable values (i.e., a partition of interchangeable values) we can use one
\mathrm{𝚒𝚗𝚝}_\mathrm{𝚟𝚊𝚕𝚞𝚎}_\mathrm{𝚙𝚛𝚎𝚌𝚎𝚍𝚎}_\mathrm{𝚌𝚑𝚊𝚒𝚗}
constraint for breaking value symmetry in each class of interchangeable values. However it was shown in [Walsh07] that enforcing arc-consistency for such a conjunction of
\mathrm{𝚒𝚗𝚝}_\mathrm{𝚟𝚊𝚕𝚞𝚎}_\mathrm{𝚙𝚛𝚎𝚌𝚎𝚍𝚎}_\mathrm{𝚌𝚑𝚊𝚒𝚗}
constraints is NP-hard.
The 2004 reformulation [Beldiceanu04] associated with the automaton of the Automaton slot achieves arc-consistency since the corresponding constraint network is a Berge-acyclic constraint network. Later on, another formulation into a sequence of ternary sliding constraints was proposed by [Walsh06]. It also achieves arc-consistency for the same reason.
precede in Gecode, value_precede_chain in MiniZinc.
\mathrm{𝚒𝚗𝚝}_\mathrm{𝚟𝚊𝚕𝚞𝚎}_\mathrm{𝚙𝚛𝚎𝚌𝚎𝚍𝚎}
\mathrm{𝚜𝚎𝚚𝚞𝚎𝚗𝚌𝚎}
\mathrm{𝚟𝚊𝚕𝚞𝚎𝚜}
\mathrm{𝚜𝚎𝚚𝚞𝚎𝚗𝚌𝚎}
\mathrm{𝚟𝚊𝚕𝚞𝚎𝚜}
\mathrm{𝚒𝚗𝚝}_\mathrm{𝚟𝚊𝚕𝚞𝚎}_\mathrm{𝚙𝚛𝚎𝚌𝚎𝚍𝚎}_\mathrm{𝚌𝚑𝚊𝚒𝚗}
n
m
respectively denote the number of variables of the
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
collection and the number of values of the
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}
{\mathrm{𝚅𝙰𝚁}}_{i}
{i}^{th}
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
{\mathrm{𝚟𝚊𝚕}}_{v}
\left(1\le v\le m\right)
{v}^{th}
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}
\mathrm{𝚒𝚗𝚝}_\mathrm{𝚟𝚊𝚕𝚞𝚎}_\mathrm{𝚙𝚛𝚎𝚌𝚎𝚍𝚎}_\mathrm{𝚌𝚑𝚊𝚒𝚗}
{s}_{i}
means that (1) each value
{\mathrm{𝚟𝚊𝚕}}_{1},{\mathrm{𝚟𝚊𝚕}}_{2},\cdots ,{\mathrm{𝚟𝚊𝚕}}_{i}
was already encountered at least once, and that (2) value
{\mathrm{𝚟𝚊𝚕}}_{i+1}
was not yet found)
\mathrm{𝚒𝚗𝚝}_\mathrm{𝚟𝚊𝚕𝚞𝚎}_\mathrm{𝚙𝚛𝚎𝚌𝚎𝚍𝚎}_\mathrm{𝚌𝚑𝚊𝚒𝚗}
constraint (since all states of the automaton are accepting there is no restriction on the last variable
{Q}_{n}
We now show how to construct such an automaton systematically. For this purpose let us first introduce some notations:
Without loss of generality we assume that we have at least two values (i.e.,
m\ge 2
𝒞
be the set of values that can be potentially assigned to a variable of the
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
collection, but which do not belong to the values of the
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}
collection (i.e.,
𝒞=\left(\mathrm{𝑑𝑜𝑚}\left({\mathrm{𝚅𝙰𝚁}}_{1}\right)\cup \mathrm{𝑑𝑜𝑚}\left({\mathrm{𝚅𝙰𝚁}}_{2}\right)\cup \cdots \cup \mathrm{𝑑𝑜𝑚}\left({\mathrm{𝚅𝙰𝚁}}_{n}\right)-\left\{{\mathrm{𝚟𝚊𝚕}}_{1},{\mathrm{𝚟𝚊𝚕}}_{2},\cdots ,{\mathrm{𝚟𝚊𝚕}}_{m}\right\}=\left\{{w}_{1},{w}_{2},\cdots ,{w}_{|𝒞|}\right\}
The states and transitions of the automaton are respectively defined in the following way:
m+1
states labelled
{s}_{0},{s}_{1},\cdots ,{s}_{m}
{s}_{0}
is the initial state. All states are accepting states.
We have the following three sets of transitions:
v\in \left[0,m-1\right]
, a transition from
{s}_{v}
{s}_{v+1}
labelled by value
{\mathrm{𝚟𝚊𝚕}}_{v+1}
. Each transition of this type will be triggered on the first occurrence of value
{\mathrm{𝚟𝚊𝚕}}_{v+1}
within the variables of the
\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}
v\in \left[1,m\right]
w\in \left[1,v\right]
, a self loop on
{s}_{v}
{\mathrm{𝚟𝚊𝚕}}_{w}
. Such transitions encode the fact that we stay in the same state as long as we have a value that was already encountered.
𝒞
is not empty, then for all
v\in \left[0,m\right]
a self loop on
{s}_{v}
labelled by the fact that we take a value not in
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}
(i.e., a value in
𝒞
). This models the fact that, encountering a value that does not belong to the set of values of the
\mathrm{𝚅𝙰𝙻𝚄𝙴𝚂}
collection, leaves us in the same state.
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14.2.1 Ratio, Rates and Proportions II, PT3 Focus Practice – user's Blog!
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14.2.1 Ratio, Rates and Proportions II, PT3 Focus Practice
The distance from town A to town B is 120 km. A car leaves town A for town B at 11.00 a.m. The average speed is 80 km h-1 .
At what time does the car arrive at town B.
\begin{array}{l}\text{Time taken}=\frac{\text{distance travelled}}{\text{average speed}}\\ =\frac{120\text{ km}}{80{\text{ km h}}^{-1}}\\ =\frac{3}{2}\text{ hours}\\ \text{= 1 hour 30 minutes}\\ \text{1 hour 30 minutes after 11}\text{.00 a}\text{.m}\text{. is 12}\text{.30 p}\text{.m}\text{.}\end{array}
Kenny drives his car from town P to town Q at a distance of 180 km in 3 hours.
Faisal takes 30 minutes less than Kenny for the same journey.
Calculate the average speed, in km/h, of Faisal’s car.
\begin{array}{l}\text{Time taken by Faisal}\\ \text{= 3 hours}-30\text{ minutes}\\ \text{= 3 hours}-\frac{1}{2}\text{ hour}\\ \text{= 2}\frac{1}{2}\text{ hours}\\ \\ \text{Average speed of Faisal's car}\\ =\frac{\text{distance travelled}}{\text{time taken}}\\ =\frac{180\text{ km}}{\text{2}\frac{1}{2}\text{ hours}}\\ =72\text{ km/h}\end{array}
Rafidah drives her car from town L to town M at an average speed of 90 km/h for 2 hours 40 minutes. She continues her journey for a distance of 100 km from town M to town N and takes 1 hour 20 minutes.
Calculate the average speed, in km/h, for the journey from L to M.
\begin{array}{l}\text{Distance}=\text{average speed}×\text{time taken}\\ \text{Distance from }L\text{ to }M=90×2\frac{40}{60}\\ \text{ }=90×2\frac{2}{3}\\ \text{ }=90×\frac{8}{3}\\ \text{ }=240\text{ km}\\ \\ \text{Total distance from }L\text{ to }N=240+100\\ \text{ }=340\text{ km}\\ \\ \text{Total time taken = 2 hours 40 minutes + 1 hour 20 minutes}\\ \text{ = 4 hours}\\ \text{Average speed for the journey from }L\text{ to }N=\frac{340\text{ km}}{4\text{ h}}\\ \text{ }=85\text{ km/h}\end{array}
Susan drives at an average speed of 105 km/h from town F to town G.
Susan takes 30 minutes longer for her return journey from town G to town F. Calculate the average speed, in km/h, for the return journey.
\begin{array}{l}\text{Distance from }F\text{ to }G\\ \text{= 105 km/h}×\text{3 hours}\\ \text{=315 km}\\ \\ \text{Average speed for return journey}\\ =\frac{\text{distance travelled}}{\text{time taken}}\\ =\frac{315\text{ km}}{3\frac{1}{2}\text{ hours}}\\ =90\text{ km/h}\end{array}
Table below shows the distances travelled and the average speeds for four vehicles.
Vehicle Distance (km) Average speed (km/h)
Which vehicles took the same time to complete the journey?
\begin{array}{l}\text{Time taken for vehicle }A=\frac{230}{115}=2\text{ hours}\\ \text{Time taken for vehicle }B=\frac{250}{100}=2\frac{1}{2}\text{ hours}\\ \text{Time taken for vehicle }C=\frac{170}{85}=2\text{ hours}\\ \text{Time taken for vehicle }D=\frac{245}{60}=3\frac{1}{2}\text{ hours}\end{array}
Thus, the vehicles A and C took the same time to complete the journey.
Posted in PT3 Maths, Ratio, Rates and Proportions II
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14.2.2 Ratio, Rates and Proportions II, PT3 Focus Practice →
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Global Constraint Catalog: Ccycle_resource
<< 5.105. cycle_or_accessibility5.107. cyclic_change >>
\mathrm{𝚌𝚢𝚌𝚕𝚎}_\mathrm{𝚛𝚎𝚜𝚘𝚞𝚛𝚌𝚎}\left(\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴},\mathrm{𝚃𝙰𝚂𝙺}\right)
\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}
\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚒𝚍}-\mathrm{𝚒𝚗𝚝},\mathrm{𝚏𝚒𝚛𝚜𝚝}_\mathrm{𝚝𝚊𝚜𝚔}-\mathrm{𝚍𝚟𝚊𝚛},\mathrm{𝚗𝚋}_\mathrm{𝚝𝚊𝚜𝚔}-\mathrm{𝚍𝚟𝚊𝚛}\right)
\mathrm{𝚃𝙰𝚂𝙺}
\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚒𝚍}-\mathrm{𝚒𝚗𝚝},\mathrm{𝚗𝚎𝚡𝚝}_\mathrm{𝚝𝚊𝚜𝚔}-\mathrm{𝚍𝚟𝚊𝚛},\mathrm{𝚛𝚎𝚜𝚘𝚞𝚛𝚌𝚎}-\mathrm{𝚍𝚟𝚊𝚛}\right)
\mathrm{𝚛𝚎𝚚𝚞𝚒𝚛𝚎𝚍}
\left(\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴},\left[\mathrm{𝚒𝚍},\mathrm{𝚏𝚒𝚛𝚜𝚝}_\mathrm{𝚝𝚊𝚜𝚔},\mathrm{𝚗𝚋}_\mathrm{𝚝𝚊𝚜𝚔}\right]\right)
\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}.\mathrm{𝚒𝚍}\ge 1
\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}.\mathrm{𝚒𝚍}\le |\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}|
\mathrm{𝚍𝚒𝚜𝚝𝚒𝚗𝚌𝚝}
\left(\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴},\mathrm{𝚒𝚍}\right)
\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}.\mathrm{𝚏𝚒𝚛𝚜𝚝}_\mathrm{𝚝𝚊𝚜𝚔}\ge 1
\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}.\mathrm{𝚏𝚒𝚛𝚜𝚝}_\mathrm{𝚝𝚊𝚜𝚔}\le |\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}|+|\mathrm{𝚃𝙰𝚂𝙺}|
\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}.\mathrm{𝚗𝚋}_\mathrm{𝚝𝚊𝚜𝚔}\ge 0
\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}.\mathrm{𝚗𝚋}_\mathrm{𝚝𝚊𝚜𝚔}\le |\mathrm{𝚃𝙰𝚂𝙺}|
\mathrm{𝚛𝚎𝚚𝚞𝚒𝚛𝚎𝚍}
\left(\mathrm{𝚃𝙰𝚂𝙺},\left[\mathrm{𝚒𝚍},\mathrm{𝚗𝚎𝚡𝚝}_\mathrm{𝚝𝚊𝚜𝚔},\mathrm{𝚛𝚎𝚜𝚘𝚞𝚛𝚌𝚎}\right]\right)
\mathrm{𝚃𝙰𝚂𝙺}.\mathrm{𝚒𝚍}>|\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}|
\mathrm{𝚃𝙰𝚂𝙺}.\mathrm{𝚒𝚍}\le |\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}|+|\mathrm{𝚃𝙰𝚂𝙺}|
\mathrm{𝚍𝚒𝚜𝚝𝚒𝚗𝚌𝚝}
\left(\mathrm{𝚃𝙰𝚂𝙺},\mathrm{𝚒𝚍}\right)
\mathrm{𝚃𝙰𝚂𝙺}.\mathrm{𝚗𝚎𝚡𝚝}_\mathrm{𝚝𝚊𝚜𝚔}\ge 1
\mathrm{𝚃𝙰𝚂𝙺}.\mathrm{𝚗𝚎𝚡𝚝}_\mathrm{𝚝𝚊𝚜𝚔}\le |\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}|+|\mathrm{𝚃𝙰𝚂𝙺}|
\mathrm{𝚃𝙰𝚂𝙺}.\mathrm{𝚛𝚎𝚜𝚘𝚞𝚛𝚌𝚎}\ge 1
\mathrm{𝚃𝙰𝚂𝙺}.\mathrm{𝚛𝚎𝚜𝚘𝚞𝚛𝚌𝚎}\le |\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}|
G
To each item of the
\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}
\mathrm{𝚃𝙰𝚂𝙺}
collections corresponds one vertex of
G
. A vertex that was generated from an item of the
\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}
\mathrm{𝚃𝙰𝚂𝙺}
) collection is called a resource vertex (respectively task vertex).
There is an arc from a resource vertex
r
to a task vertex
t
t\in \mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}\left[r\right].\mathrm{𝚏𝚒𝚛𝚜𝚝}_\mathrm{𝚝𝚊𝚜𝚔}
There is an arc from a task vertex
to a resource vertex
r
r\in \mathrm{𝚃𝙰𝚂𝙺}\left[t\right].\mathrm{𝚗𝚎𝚡𝚝}_\mathrm{𝚝𝚊𝚜𝚔}
{t}_{1}
{t}_{2}
{t}_{2}\in \mathrm{𝚃𝙰𝚂𝙺}\left[{t}_{1}\right].\mathrm{𝚗𝚎𝚡𝚝}_\mathrm{𝚝𝚊𝚜𝚔}
There is no arc between two resource vertices.
Enforce to cover
G
in such a way that each vertex belongs to a single circuit. Each circuit is made up from a single resource vertex and zero, one or more task vertices. For each resource-vertex a domain variable indicates how many task-vertices belong to the corresponding circuit. For each task a domain variable provides the identifier of the resource that can effectively handle that task.
\left(\begin{array}{c}〈\begin{array}{ccc}\mathrm{𝚒𝚍}-1\hfill & \mathrm{𝚏𝚒𝚛𝚜𝚝}_\mathrm{𝚝𝚊𝚜𝚔}-5\hfill & \mathrm{𝚗𝚋}_\mathrm{𝚝𝚊𝚜𝚔}-3,\hfill \\ \mathrm{𝚒𝚍}-2\hfill & \mathrm{𝚏𝚒𝚛𝚜𝚝}_\mathrm{𝚝𝚊𝚜𝚔}-2\hfill & \mathrm{𝚗𝚋}_\mathrm{𝚝𝚊𝚜𝚔}-0,\hfill \\ \mathrm{𝚒𝚍}-3\hfill & \mathrm{𝚏𝚒𝚛𝚜𝚝}_\mathrm{𝚝𝚊𝚜𝚔}-8\hfill & \mathrm{𝚗𝚋}_\mathrm{𝚝𝚊𝚜𝚔}-2\hfill \end{array}〉,\hfill \\ 〈\begin{array}{ccc}\mathrm{𝚒𝚍}-4\hfill & \mathrm{𝚗𝚎𝚡𝚝}_\mathrm{𝚝𝚊𝚜𝚔}-7\hfill & \mathrm{𝚛𝚎𝚜𝚘𝚞𝚛𝚌𝚎}-1,\hfill \\ \mathrm{𝚒𝚍}-5\hfill & \mathrm{𝚗𝚎𝚡𝚝}_\mathrm{𝚝𝚊𝚜𝚔}-4\hfill & \mathrm{𝚛𝚎𝚜𝚘𝚞𝚛𝚌𝚎}-1,\hfill \\ \mathrm{𝚒𝚍}-6\hfill & \mathrm{𝚗𝚎𝚡𝚝}_\mathrm{𝚝𝚊𝚜𝚔}-3\hfill & \mathrm{𝚛𝚎𝚜𝚘𝚞𝚛𝚌𝚎}-3,\hfill \\ \mathrm{𝚒𝚍}-7\hfill & \mathrm{𝚗𝚎𝚡𝚝}_\mathrm{𝚝𝚊𝚜𝚔}-1\hfill & \mathrm{𝚛𝚎𝚜𝚘𝚞𝚛𝚌𝚎}-1,\hfill \\ \mathrm{𝚒𝚍}-8\hfill & \mathrm{𝚗𝚎𝚡𝚝}_\mathrm{𝚝𝚊𝚜𝚔}-6\hfill & \mathrm{𝚛𝚎𝚜𝚘𝚞𝚛𝚌𝚎}-3\hfill \end{array}〉\hfill \end{array}\right)
\mathrm{𝚌𝚢𝚌𝚕𝚎}_\mathrm{𝚛𝚎𝚜𝚘𝚞𝚛𝚌𝚎}
constraint holds since the graph corresponding to the vertices described by its arguments consists of the following 3 disjoint circuits:
The first circuit involves the resource vertex 1 as well as the task vertices 5, 4 and 7.
The second circuit is limited to the resource vertex 2.
Finally the third circuit is made up from the remaining vertices, namely the resource vertex 3 and the task vertices 8 and 6.
|\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}|>1
|\mathrm{𝚃𝙰𝚂𝙺}|>1
|\mathrm{𝚃𝙰𝚂𝙺}|>|\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}|
\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}
\mathrm{𝚃𝙰𝚂𝙺}
\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}.\mathrm{𝚒𝚍}
\mathrm{𝚃𝙰𝚂𝙺}.\mathrm{𝚛𝚎𝚜𝚘𝚞𝚛𝚌𝚎}
This constraint is useful for some vehicles routing problem where the number of locations to visit depends of the vehicle type that is actually used. The resource attribute allows expressing various constraints such as:
The compatibility or incompatibility between tasks and vehicles,
The fact that certain tasks should be performed by the same vehicle,
The pre-assignment of certain tasks to a given vehicle.
This constraint could be expressed with the
\mathrm{𝚌𝚢𝚌𝚕𝚎}
constraint of CHIP by using the following optional parameters:
The resource node parameter [Bourreau99],
The circuit weight parameter [Bourreau99],
The name parameter [Bourreau99].
\mathrm{𝚌𝚢𝚌𝚕𝚎}
constraint type: graph constraint, resource constraint, graph partitioning constraint.
final graph structure: connected component, strongly connected component.
\mathrm{𝚌𝚘𝚕}\left(\begin{array}{c}\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}_\mathrm{𝚃𝙰𝚂𝙺}-\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\begin{array}{c}\mathrm{𝚒𝚗𝚍𝚎𝚡}-\mathrm{𝚒𝚗𝚝},\hfill \\ \mathrm{𝚜𝚞𝚌𝚌}-\mathrm{𝚍𝚟𝚊𝚛},\hfill \\ \mathrm{𝚗𝚊𝚖𝚎}-\mathrm{𝚍𝚟𝚊𝚛}\hfill \end{array}\right),\hfill \\ \left[\begin{array}{c}\mathrm{𝚒𝚝𝚎𝚖}\left(\begin{array}{c}\mathrm{𝚒𝚗𝚍𝚎𝚡}-\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}.\mathrm{𝚒𝚍},\hfill \\ \mathrm{𝚜𝚞𝚌𝚌}-\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}.\mathrm{𝚏𝚒𝚛𝚜𝚝}_\mathrm{𝚝𝚊𝚜𝚔},\hfill \\ \mathrm{𝚗𝚊𝚖𝚎}-\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}.\mathrm{𝚒𝚍}\hfill \end{array}\right),\hfill \\ \mathrm{𝚒𝚝𝚎𝚖}\left(\begin{array}{c}\mathrm{𝚒𝚗𝚍𝚎𝚡}-\mathrm{𝚃𝙰𝚂𝙺}.\mathrm{𝚒𝚍},\hfill \\ \mathrm{𝚜𝚞𝚌𝚌}-\mathrm{𝚃𝙰𝚂𝙺}.\mathrm{𝚗𝚎𝚡𝚝}_\mathrm{𝚝𝚊𝚜𝚔},\hfill \\ \mathrm{𝚗𝚊𝚖𝚎}-\mathrm{𝚃𝙰𝚂𝙺}.\mathrm{𝚛𝚎𝚜𝚘𝚞𝚛𝚌𝚎}\hfill \end{array}\right)\hfill \end{array}\right]\hfill \end{array}\right)
\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}_\mathrm{𝚃𝙰𝚂𝙺}
\mathrm{𝐶𝐿𝐼𝑄𝑈𝐸}
↦\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚛𝚎𝚜𝚘𝚞𝚛𝚌𝚎}_\mathrm{𝚝𝚊𝚜𝚔}\mathtt{1},\mathrm{𝚛𝚎𝚜𝚘𝚞𝚛𝚌𝚎}_\mathrm{𝚝𝚊𝚜𝚔}\mathtt{2}\right)
•\mathrm{𝚛𝚎𝚜𝚘𝚞𝚛𝚌𝚎}_\mathrm{𝚝𝚊𝚜𝚔}\mathtt{1}.\mathrm{𝚜𝚞𝚌𝚌}=\mathrm{𝚛𝚎𝚜𝚘𝚞𝚛𝚌𝚎}_\mathrm{𝚝𝚊𝚜𝚔}\mathtt{2}.\mathrm{𝚒𝚗𝚍𝚎𝚡}
•\mathrm{𝚛𝚎𝚜𝚘𝚞𝚛𝚌𝚎}_\mathrm{𝚝𝚊𝚜𝚔}\mathtt{1}.\mathrm{𝚗𝚊𝚖𝚎}=\mathrm{𝚛𝚎𝚜𝚘𝚞𝚛𝚌𝚎}_\mathrm{𝚝𝚊𝚜𝚔}\mathtt{2}.\mathrm{𝚗𝚊𝚖𝚎}
•
\mathrm{𝐍𝐓𝐑𝐄𝐄}
=0
•
\mathrm{𝐍𝐂𝐂}
=|\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}|
•
\mathrm{𝐍𝐕𝐄𝐑𝐓𝐄𝐗}
=|\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}|+|\mathrm{𝚃𝙰𝚂𝙺}|
\mathrm{𝙾𝙽𝙴}_\mathrm{𝚂𝚄𝙲𝙲}
\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}
\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}_\mathrm{𝚃𝙰𝚂𝙺}
\mathrm{𝐶𝐿𝐼𝑄𝑈𝐸}
↦\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚛𝚎𝚜𝚘𝚞𝚛𝚌𝚎}_\mathrm{𝚝𝚊𝚜𝚔}\mathtt{1},\mathrm{𝚛𝚎𝚜𝚘𝚞𝚛𝚌𝚎}_\mathrm{𝚝𝚊𝚜𝚔}\mathtt{2}\right)
•\mathrm{𝚛𝚎𝚜𝚘𝚞𝚛𝚌𝚎}_\mathrm{𝚝𝚊𝚜𝚔}\mathtt{1}.\mathrm{𝚜𝚞𝚌𝚌}=\mathrm{𝚛𝚎𝚜𝚘𝚞𝚛𝚌𝚎}_\mathrm{𝚝𝚊𝚜𝚔}\mathtt{2}.\mathrm{𝚒𝚗𝚍𝚎𝚡}
•\mathrm{𝚛𝚎𝚜𝚘𝚞𝚛𝚌𝚎}_\mathrm{𝚝𝚊𝚜𝚔}\mathtt{1}.\mathrm{𝚗𝚊𝚖𝚎}=\mathrm{𝚛𝚎𝚜𝚘𝚞𝚛𝚌𝚎}_\mathrm{𝚝𝚊𝚜𝚔}\mathtt{2}.\mathrm{𝚗𝚊𝚖𝚎}
•\mathrm{𝚛𝚎𝚜𝚘𝚞𝚛𝚌𝚎}_\mathrm{𝚝𝚊𝚜𝚔}\mathtt{1}.\mathrm{𝚗𝚊𝚖𝚎}=\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}.\mathrm{𝚒𝚍}
\mathrm{𝐍𝐕𝐄𝐑𝐓𝐄𝐗}
=\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}.\mathrm{𝚗𝚋}_\mathrm{𝚝𝚊𝚜𝚔}+1
The graph model of the
\mathrm{𝚌𝚢𝚌𝚕𝚎}_\mathrm{𝚛𝚎𝚜𝚘𝚞𝚛𝚌𝚎}
constraint illustrates the following points:
How to differentiate the constraint on the length of a circuit according to a resource that is assigned to a circuit? This is achieved by introducing a collection of resources and by asking a different graph property for each item of that collection.
How to introduce the concept of name that corresponds to the resource that handles a given task? This is done by adding to the arc constraint associated with the
\mathrm{𝚌𝚢𝚌𝚕𝚎}
constraint the condition that the name variables of two consecutive vertices should be equal.
Part (A) of Figure 5.106.1 shows the initial graphs (of the second graph constraint) associated with resources 1, 2 and 3 of the Example slot. Part (B) of Figure 5.106.1 shows the corresponding final graphs (of the second graph constraint) associated with resources 1, 2 and 3. Since we use the
\mathrm{𝐍𝐕𝐄𝐑𝐓𝐄𝐗}
graph property, the vertices of the final graphs are stressed in bold. To each resource corresponds a circuit of respectively 3, 0 and 2 task-vertices.
\mathrm{𝚌𝚢𝚌𝚕𝚎}_\mathrm{𝚛𝚎𝚜𝚘𝚞𝚛𝚌𝚎}
Since the initial graph of the first graph constraint contains
|\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}|+|\mathrm{𝚃𝙰𝚂𝙺}|
vertices, the corresponding final graph cannot have more than
|\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}|+|\mathrm{𝚃𝙰𝚂𝙺}|
vertices. Therefore we can rewrite the graph property
\mathrm{𝐍𝐕𝐄𝐑𝐓𝐄𝐗}
=
|\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}|+|\mathrm{𝚃𝙰𝚂𝙺}|
\mathrm{𝐍𝐕𝐄𝐑𝐓𝐄𝐗}
\ge
|\mathrm{𝚁𝙴𝚂𝙾𝚄𝚁𝙲𝙴}|+|\mathrm{𝚃𝙰𝚂𝙺}|
\underline{\overline{\mathrm{𝐍𝐕𝐄𝐑𝐓𝐄𝐗}}}
\overline{\mathrm{𝐍𝐕𝐄𝐑𝐓𝐄𝐗}}
|
Parallel Computation - Maple Help
Home : Support : Online Help : System : Information : Updates : Maple 2016 : Parallel Computation
Thread-Safety in Parallel Computation
CodeTools:-ThreadSafetyCheck
A new option, lock, can be added to any procedure. There can be only one thread running a procedure with option lock at a time. If a second thread tries to run a procedure with option lock, then the second thread will block until the first thread's procedure is done. Other threads are free to run any other procedure.
option lock;
printf("Running p %d\n",n);
printf("Done p %d\n",n);
Threads:-Task:-Start(()->NULL,Task=[p,1],Task=[p,2]);
Done p 2
Notice that the first procedure that begins to run starts and finishes before the second procedure can begin. Without option lock, you would see two "Running" statements printed before two "Done" statements.
Option lock is now used on over 10,000 procedures in Maple's library of mathematical algorithms. This is a small fraction of the total number of procedures, the wider number already being thread-safe.
The new ThreadSafetyCheck command in the CodeTools package can be used to help identify global state in a procedure or module. This is helpful in determining if your package is thread-safe or not.
Here is a module that implements a counter. The Jump procedure simply increments the counter by a given increment, although it does it by ones in a loop. This example is meant to be a simplification of a procedure that iterates through a longer process.
local count := 0;
export Reset := proc() count := 0; end proc;
export Jump := proc(n)
for j from 1 to 1000 do (* delay *) od;
Running this in serial we see that the internal counter is incremented by the given stride.
Counter:-Jump(10);
\textcolor[rgb]{0,0,1}{10}
\textcolor[rgb]{0,0,1}{100}
Let's Jump by 1000 in two different threads.
Counter:-Reset();
\textcolor[rgb]{0,0,1}{0}
Threads:-Task:-Start((a,b)->[a,b], Task=[Counter:-Jump,1000],Task=[Counter:-Jump,1000]);
\left[\textcolor[rgb]{0,0,1}{1894}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1994}\right]
The counter should be 1000 + 1000 = 2000, but the results above are strange numbers. The two threads interfered with each other. A quick test using the ThreadSafetyCheck command in the CodeTools package warns us of the presence of a global variable.
CodeTools:-ThreadSafetyCheck( Counter );
Warning, proc Jump uses lexical count
\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}
One way to fix this is to add option lock. This prevents two instances of Jump from running at the same time.
CounterLock := module()
Threads:-Task:-Start((a,b)->[a,b], Task=[CounterLock:-Jump,1000],Task=[CounterLock:-Jump,1000]);
\left[\textcolor[rgb]{0,0,1}{2000}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1000}\right]
Locking in the above example is unsatisfying because it limits parallelism. Another way to fix this algorithm is to make the counter be a thread-local variable -- a different variable on each thread.
CounterThreadLocal := module()
local count::thread_local := 0;
Threads:-Task:-Start((a,b)->[a,b], Task=[CounterThreadLocal:-Jump,1000],Task=[CounterThreadLocal:-Jump,1000]);
\left[\textcolor[rgb]{0,0,1}{1000}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1000}\right]
CodeTools:-ThreadSafetyCheck does analysis on the given procedure or module and warns about the following:
global variables declared in a procedure's global statement
global variables appearing on the left side of an assignment statement
lexically scoped variables appearing on the left side of an assignment statement
global G1, G2, G3;
local L1, L2, L3, P1, P2;
P1 := proc()
L2 := 2*L3*G2;
CodeTools:-ThreadSafetyCheck( M );
Warning, proc P1 uses lexical L2
Warning, proc P1 uses globals [G3]
Warning, proc P2 has declared globals [G4]
\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}
Notice the following variables are not flagged:
G1 - globals declared in a module are not flagged
G2 - globals declared in a module are not flagged even if they are used in a procedure (as in P1)
L1 - locals and exports assigned at the top-level of a module get that assignment when the module is created and thus are not flagged
L3 - locals that are not found on the lhs of an assignment are not flagged
P1, P2 - same as L1
Analysis of each procedure is done based on just the statements inside the present procedure, not on it's children. Therefore, it is insufficient to simply apply option lock to any procedure flagged during CodeTools:-ThreadSafetyCheck.
For example, here is a situation that would not be thread-safe:
local data, FetchData, ProcessData;
export Compute;
FetchData := proc( file )
data := ImportMatrix(file);
ProcessData := proc()
data[..,1];
Compute := proc( file )
FetchData(file);
Warning, proc ModuleApply uses lexical i
Warning, proc FetchData uses lexical data
Warning, proc ProcessData uses lexical data
\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{71}
(Note: the ModuleApply here is actually ReadBinaryL5File:-ModuleApply -- something used by ImportMatrix.)
Adding option lock to FetchData and ProcessData is not sufficient. The parent function, Compute, also needs option lock (or some other mechanism) in order for this to be thread-safe. Otherwise, if threads A and B call Compute simultaneously, the call to FetchData in thread A may run in between the calls to FetchData and ProcessData in thread B: in that case, ProcessData in thread B will see the data from thread A.
How to Make your code Thread-Safe
Pattern 1: Read-only Constant
This is a situation where the code actually is thread-safe, but you want to resolve the warning.
global `debugger/max_width`;
CodeTools:-ThreadSafetyCheck( p );
Warning, proc p has declared globals [`debugger/max_width`]
\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}
The appropriate fix is to add option threadsafe; to this procedure. This is a declaration that has no runtime significance, but does prevent ThreadSafetyCheck from warning about this procedure.
option threadsafe;
\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}
Pattern 2: Initialization of Constants
A package module requires some initialization to set read-only constants.
m1 := module()
local init := proc()
export user := proc()
CodeTools:-ThreadSafetyCheck(m1);
Warning, proc user uses lexical a
Warning, proc init uses lexical a
\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}
There are several good options to solve this:
Move your init code to ModuleLoad.
local ModuleLoad := proc()
\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}
Initialize your code in the module body.
local a := 4;
\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}
Use the protect command to mark the variable as protected -- this declares it as read-only.
protect('a');
\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}
Declare the variable as thread_local.
local a::thread_local;
\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}
Declare all procedures that use the variable as option threadsafe.
\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}
Pattern 3: The code is NOT thread-safe
In this situation you know the code is not thread-safe, but you want to get rid of the warning. Just mark it with option lock.
CodeTools:-ThreadSafetyCheck(p);
\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}
Pattern 4: define_external
The define_external command allows you to link to compiled procedures in external DLLs or shared libraries. In order to use an external procedure in a package that may be run on different operating systems, it was often the case that a "trampoline" pattern was used. The first invocation would create the link and then reassign itself.
pkg := module()
export RandPerm := proc()
unprotect(RandPerm);
RandPerm := define_external("mstring_randperm",MAPLE,LIB=ExternalLibraryName("mstring"));
protect(RandPerm);
procname( args )
CodeTools:-ThreadSafetyCheck(pkg);
Warning, proc RandPerm uses lexical RandPerm
\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}
In a multi-threaded environment, this pattern could lead to the external routine being linked in a second time. The define_external command now will accept the base name of the external library file, and apply standard transformations as ExternalLibraryName would. So, LIB="mstring" will map to "mstring.dll" on Windows, and to "libmstring.so" on Mac OS X and Linux. Saving such procedures will remember the original definitions so they can still be used in a platform-independent way.
export RandPerm := define_external("mstring_randperm",MAPLE,LIB="mstring");
\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}
|
9.2.3 Loci in Two Dimensions, PT3 Focus Practice
Diagram below in the answer space shows a circle with centre O drawn on a grid of equal squares with sides of 1 unit. POQ is a diameter of the circle.
W, X and Y are three moving points inside the circle.
(a) W is the point which moves such that it is constantly 4 units from the point O. Describe fully the locus of W.
(b) On the diagram, draw,
(i) the locus of the point X which moves such that its distance is constantly 3 units from the line PQ,
(ii) the locus of the point Y which moves such that it is equidistant from the point P and the point Q.
(c) Hence, mark with the symbol
\otimes
the intersection of the locus of X and the locus of Y.
(b)(i),(ii) and (c)
(a) The locus of W is a circle with the centre O and a radius of 4 units.
The diagram in the answer space shows two squares ABCD and CDEF each of sides 4 cm. K is a point on the line CD. W, X and Y are three moving points in the diagram.
(a) Point W moves such that it is always equidistant from the straight lines AB and EF. By using the letters in diagram, state the locus of W.
(b) On the diagram, draw
(i) the locus X such that it is always 2 cm from the straight line ACE,
(ii) the locus of Y such that KY = KC.
\otimes
(b)(i), (ii) and (c)
(a) The locus of W is the line CD.
Diagram below shows a Cartesian plane.
Draw the locus of X, Y and Z.
(a) X is a point which moves such that its distance is constantly 4 units from the origin.
(b) Y is a point which moves such that it is always equidistant from point K and point L.
(c) Z is a point which moves such that it is always equidistant from lines KM and y-axis.
Posted in Loci in Two Dimensions, PT3 Maths
Diagram below in the answer space shows a quadrilateral ABCD drawn on a grid of equal squares with sides of 1 unit.
X, Y and Z are three moving points inside the quadrilateral ABCD.
(a) X is the point which moves such that it is always equidistant from point B and point D.
By using the letters in diagram, state the locus of X.
(i) the locus of the point Y such that it is always 6 units from point A,
(ii) the locus of the point Z which moves such that its distance is constantly 3 units from the
line AB.
(c) Hence, mark with the symbol ⊗ the intersection of the locus of Y and the locus of Z.
(a) The locus of X is the line AC.
Diagram in the answer space below shows a square ABCD. E, F, G and H are the midpoints of straight lines AD, AB, BC and CD respectively. W, X and Y are moving points in the square.
(a) draw the locus of the point W which moves such that it is always equidistant from point AD and BC.
(b) draw the locus of the point X which moves such that XM = MG.
(c) draw the locus of point Y which moves such that its distance is constantly 6 cm from point C.
(d) Hence, mark with the symbol ⊗ the intersection of the locus of W and the locus of Y.
Diagram in the answer space below shows a polygon ABCDEF drawn on a grid of squares with sides of 1 unit. X, Y and Z are the points that move in the polygon.
(a) X is a point which moves such that it is equidistant from point B and point F.
(i) the locus of the point Y which moves such that it is always parallel and equidistant from the straight lines BA and CD.
(ii) the locus of point Z which moves such that its distance is constantly 6 units from the point A.
\otimes
the intersection of the locus of Y and the locus of Z.
(a) The locus of X is AD.
Diagram below in the answer space shows a square PQRS with sides of 6 units drawn on a grid of equal squares with sides of 1 unit. W, X and Y are three moving points inside the square.
(a) W is the point which moves such that it is always equidistant from point P and point R.
By using the letters in diagram, state the locus of W.
(i) the locus of the point X which moves such that it is always equidistant from the straight lines PQ and PS,
(ii) the locus of the point Y which moves such that its distance is constantly 2 units from point K.
(c) Hence, mark with the symbol ⊗ the intersection of the locus of X and the locus of Y.
(b)(i),(ii)
Diagram in the answer space below, shows a regular pentagon PQRST. W, X and Y are moving points which move in the pentagon.
(a) draw the locus of the point W which moves such that it is always equidistant from point R and S.
(b) draw the locus of the point X which moves such that XR = RS.
(c) draw the locus of point Y which moves such that its distance is constantly 3 cm from the line SR.
(d) hence, mark with the symbol ⊗ the intersection of the locus of W and the locus of X.
Diagram in the answer space below shows a polygon. X and Y are two moving points in the polygon.
(a) On the diagram, draw
(i) the locus of the point X such that XQ = XR.
(ii) the locus of the point Y such that YQ = QR.
(b) Hence, mark with the symbol
\otimes
(a)(i), (ii) and (b)
Diagram below shows the route of a Negaraku Run.
The route of male participants is PQUT while the route for female participants is QRST. QR is parallel to UT whereas UQ is parallel to SR.
(a) Given the distance between point Q and point R is 9 km, state the coordinates of point R.
(b) It is given that the male participants start the run at point P and the female participants at point Q.
Find the difference of distance, in km, between the male and female participants.
Route of male participants
= PQ + QU + UT
Route of female participants
= QR + RS + ST
Difference in distance
Diagram below shows a parking lot in the shape of trapezium, RSTU. Two coordinates from the trapezium vertices are R(–30, –4) and S(20, –4).
(a) Given the distance between vertex S and vertex T is 22 units.
State the coordinates of vertex T.
(b) Given the area of the parking lot is 946 unit2, find the coordinates of U.
y-coordinates of T = –4 + 22 = 18
\begin{array}{l}\text{Area of trapezium}=946{\text{ unit}}^{2}\\ \frac{1}{2}×\left(UT+RS\right)×ST=946\\ \frac{1}{2}×\left(UT+50\right)×22=946\\ UT+50=\frac{946×2}{22}\\ UT+50=86\\ UT=36\\ \\ x-\text{coordinate of point }U=20-36=-16\\ y-\text{coordinate of point }U=18\\ U=\left(-16,18\right)\end{array}
In diagram below, Q is the midpoint of the straight line PR.
\begin{array}{l}\frac{2+m}{2}=5\\ 2+m=10\\ \text{}m=8\end{array}
In diagram below, P and Q are points on a Cartesian plane.
If M is the midpoint of PQ, then the coordinates of M are
\begin{array}{l}P\left(-4,8\right),\text{ }Q\left(6,-2\right)\\ \text{Coordinates of }M\\ =\left(\frac{-4+6}{2},\frac{8+\left(-2\right)}{2}\right)\\ =\left(1,3\right)\end{array}
Find the distance between P (–4, 6) and Q (20, –1).
\begin{array}{l}\text{Distance of }PQ\\ =\sqrt{{\left(-4-20\right)}^{2}+{\left[6-\left(-1\right)\right]}^{2}}\\ =\sqrt{{\left(-24\right)}^{2}+{7}^{2}}\\ =\sqrt{576+49}\\ =\sqrt{625}\\ =25\text{ units}\end{array}
Diagram shows a straight line PQ on a Cartesian plane.
Calculate the length, in unit, of PQ.
PS = 15 – 3 = 12 units
SQ = 8 – 3 = 5 units
PQ2 = PS2 + SQ2
The diagram shows an isosceles triangle STU.
Given that ST = 5 units, the coordinates of point S are
For an isosceles triangle STU, M is the midpoint of straight line TU.
\begin{array}{l}x-\text{coordinate of }M\\ =\frac{-2+4}{2}=1\end{array}
Point M = (1, 0)
MT = 4 – 1 = 3 units
SM2 = ST2 – MT2
SM = √16
Therefore, point S = (1, 4).
1. The Cartesian coordinate system is a number plane with a horizontal line (x-axis) drawn at right angles to a vertical line (y-axis), intersecting at a point called origin.
2. It is used to locate the position of a point in reference to the x-axis and y-axis.
3. The coordinate of any point are written as an ordered pair (x, y). The first number is the x-coordinate and the second number is the y-coordinate of the point.
The coordinates of points A and B are (3, 4) and (–5, –2) respectively.
This means that point A is located 3 units from they-axis and 4 units from the x-axis, whereas point B is located 5 units on the left from the y-axis and 2 units from the x-axis.
4. The coordinate of the origin O is (0, 0).
8.1.2 Scales for the Coordinate Axes
1. The scale for an axis is the number of units represented by a specific length along the axes.
2. The scale on a coordinate is usually written in the form of a ratio.
A scale of 1 : 2 means one unit on the graph represents 2 units of the actual length.
3. Both coordinate axes on the Cartesian plane may have
(a) the same scales, or
(b) different scales.
1 unit on the x-axis represents 2 units.
1 unit on the y-axis represents 1 unit.
Therefore the scale for x-axis is 1 : 2 and the scale for y-axis is 1 : 1.
P (4, 3) and Q (10, 5).
8.1.3 Distance between Two Points
1. Finding the distance between two points on a Cartesian plane is the same as finding the length of the straight line joining them.
2. The distance between two points can be calculated by using Pythagoras’ theorem.
AB = 2 – (–4) = 2 + 4 = 6 units
BC = 5 – (–3) = 5 + 3 = 8 units
AC2 = AB2 + AC2
3. Distance is always a positive value.
8.1.4 Midpoint
The midpoint of a straight line joining two points is the middle point that divides the straight line into two equal halves.
\text{Midpoint,}M=\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right)
The coordinate of the midpoint of (7, –5) and (–3, 11) are
\begin{array}{l}\left(\frac{7+\left(-3\right)}{2},\frac{-5+11}{2}\right)\\ =\left(\frac{4}{2},\frac{6}{2}\right)\\ =\left(2,3\right)\end{array}
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Condition number for inversion - MATLAB cond - MathWorks India
Condition Number of Matrix
1-Norm Condition Number
cond returns NaN for nonfinite inputs
C = cond(A,p)
C = cond(A) returns the 2-norm condition number for inversion, equal to the ratio of the largest singular value of A to the smallest.
C = cond(A,p) returns the p-norm condition number, where p can be 1, 2, Inf, or 'fro'.
Calculate the condition number of a matrix and examine the sensitivity to the inverse calculation.
A = [4.1 2.8;
Calculate the 2-norm condition number of A.
Since the condition number of A is much larger than 1, the matrix is sensitive to the inverse calculation. Calculate the inverse of A, and then make a small change in the second row of A and calculate the inverse again.
A2 = [4.1 2.8;
9.671 6.608]
invA2 = inv(A2)
invA2 = 2×2
The results indicate that making a small change in A can completely change the result of the inverse calculation.
Calculate the 1-norm condition number of a matrix.
A = [1 0 -2;
-1 5 7];
Calculate the 1-norm condition number of A. The value of the 1-norm condition number for an m-by-n matrix is
{\kappa }_{1}\left(A\right)=||A|{|}_{1}\phantom{\rule{0.16666666666666666em}{0ex}}\phantom{\rule{0.16666666666666666em}{0ex}}\phantom{\rule{0.16666666666666666em}{0ex}}||{A}^{-1}|{|}_{1}
where the 1-norm is the maximum absolute column sum of the matrix given by
||A|{|}_{1}=\underset{1\le j\le n}{\text{max}}\sum _{i=1}^{m}|{a}_{ij}|.
C = cond(A,1)
For this matrix the condition number is not too large, so the matrix is not particularly sensitive to the inverse calculation.
2 (default) | 1 | 'fro' | Inf
Norm type, specified as one of the values shown in this table. cond computes the condition number using norm(A,p) * norm(inv(A),p) for values of p other than 2. See the norm page for additional information about these norm types.
Infinity norm condition number
Frobenius norm condition number
Example: cond(A,1) calculates the 1-norm condition number.
C — Condition number
Condition number, returned as a scalar. Values of C near 1 indicate a well-conditioned matrix, and large values of C indicate an ill-conditioned matrix. Singular matrices have a condition number of Inf.
A condition number for a matrix and computational task measures how sensitive the answer is to changes in the input data and roundoff errors in the solution process.
The condition number for inversion of a matrix measures the sensitivity of the solution of a system of linear equations to errors in the data. It gives an indication of the accuracy of the results from matrix inversion and the linear equation solution. For example, the 2-norm condition number of a square matrix is
\kappa \left(A\right)=‖{A}^{}‖‖{A}^{-1}‖\text{\hspace{0.17em}}.
In this context, a large condition number indicates that a small change in the coefficient matrix A can lead to larger changes in the output b in the linear equations Ax = b and xA = b. The extreme case is when A is so poorly conditioned that it is singular (an infinite condition number), in which case it has no inverse and the linear equation has no unique solution.
rcond is a more efficient, but less reliable, method of estimating the condition of a matrix compared to cond.
The algorithm for cond has three pieces:
If p = 2, then cond uses the singular value decomposition provided by svd to find the ratio of the largest and smallest singular values.
If p = 1, Inf, or 'fro', then cond calculates the condition number using the appropriate norm of the input matrix and its inverse with norm(A,p) * norm(inv(A),p).
If the input matrix is sparse, then cond ignores any specified p value and calls condest.
cond does not support sparse matrices.
R2021b: cond returns NaN for nonfinite inputs
cond returns NaN when the input contains nonfinite values (Inf or NaN). Previously, cond threw an error when the input contained nonfinite values.
condeig | condest | norm | normest | rank | rcond | svd
Cleve's Corner: What is the Condition Number of a Matrix?
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Determines transformations between lists of positions
This routine determines the transformations between (labelled) position lists. Six different types of transformation are available. The first 5 are based on the linear transformation, the sixth being a function defined by you. The linear transformations are based on the mappings:
+
\ast
+
\ast
+
\ast
+
\ast
and allow:
shift of origin and rotation
shift of origin and magnification
shift of origin, rotation and magnification (solid body)
or a full six parameter fit
The self defined transform can be any mapping given as an algebraic expression (including functions) using the methods allowed by TRANSFORM (SUN/61).
When determining linear transformations REGISTER allows many lists to be processed at once performing a simultaneous registration of all the lists. When using a self defined transform only two lists may be registered at any time.
The results from REGISTER are reported via the logging system and then coded as new coordinate systems attached to images. Normally, the new coordinate systems will be attached to the images with which the lists are associated, but if the lists are not associated with images then they can be attached to a named list of images, or a single named one. The new coordinate system is a copy of the Pixel coordinate system of the refernce image, and so is guaranteed to be a sensible one in which to resample. The resampling can be done by TRANNDF.
register inlist fittype refpos
These parameters supply the values of "sub-expressions" used in the expressions XFOR, YFOR, XINV and YINV. These parameters should be used when repeated expressions are present in complex transformations. Sub-expressions may contain references to other sub-expressions and the variables (PA-PZ). An example of using sub-expressions is:
>
\ast
\ast
>
\ast
\ast
>
\ast
\ast
>
\ast
\ast
>
\ast
+
\ast
>
\ast
+
\ast
The type of fit which should be used when determining the transformation between the input position lists. This may take the values
If more than two position lists are provided, then only the values 1–5 may be used. [5]
FULL = _LOGICAL (Read)
If FITTYPE=6 is chosen then this parameter value determines if a full transformation is to be performed or not. If FALSE then you will only be prompted for expressions for XFOR and YFOR and the inverse transformation will remain undefined.
If TRUE then you will also be prompted for XINV and YINV in response to which the inverse mappings for X’ and Y’ are required. Not performing a full fit will affect the later uses of the transformation. At present not providing an inverse mapping means that image resampling (TRANNDF) may not be performed. [FALSE]
If NDFNAMES is FALSE and PLACEIN is "EACH" then a list of image names in which to store the WCS frames is required. This list of names must correspond exactly to the order of the associated input lists. A listing of the order of inputs is shown before this parameter is accessed.
The image names may (although this is probably not advisable) be specified using wildcards, or may be specified using an indirection file (the indirection character is "
\text{^}
This parameter is used to access the names of the lists which contain the positions and, if NDFNAMES is TRUE, the names of the associated images. If NDFNAMES is TRUE the names of the position lists are assumed to be stored in the extension of the images (in the CCDPACK extension item CURRENT_LIST) and the names of the images themselves should be given (and may include wildcards).
If NDFNAMES is FALSE then the actual names of the position lists should be given. These may not use wildcards but may be specified using indirection (other CCDPACK position list processing routines will write the names of their results files into files suitable for use in this manner) the indirection character is "
\text{^}
This parameter specifies whether the names of the input position lists are stored in the CCDPACK extensions of NDFs. If TRUE then the INLIST parameter accesses a list of images which are used to get the associated position lists. If FALSE then INLIST just accesses the position list names directly.
If the names of the lists are stored in the CCDPACK NDF extension then the new coordinate system is attached to the associated image.
The transformation information is written as a new coordinate system attached to the image. This parameter gives the label (domain) of the new coordinate system. When the new coordinate system is added, any previously existing one with the same Domain will be removed.
If PLACEIN is "SINGLE", then the new coordinate systems are all attached to a single image. In this case the domains are OUTDOMAIN_1, OUTDOMAIN_2, ....
The name is converted to upper case, and whitespace is removed. [CCD_REG]
PA-PZ = LITERAL (Read)
When FITTYPE=6 these parameters are used for supplying initial guesses at the values of the fit parameters. Normally the values of these parameters are not critical, but occasionally the minimization routine fails due to numeric problems (these are usually caused by trig functions etc. which are given invalid values (outside
+
/-1 etc.)). [1.0D0]
PLACEIN = LITERAL (Read)
If NDFNAMES is FALSE then this parameter specifies where you would like to store the final transformation structures. The options are:
EACH – attach them one per image in a set of images
SINGLE – attach them all to a single image
If the EACH option is chosen then you will have the option of supplying the image names via the parameter IN. If the SINGLE option is chosen then the name of an image should be given in response to the WCSFILE parameter; if no NDF by this name exists, a new dummy one will be created. [EACH]
The position within the list of inputs which corresponds to the list to be used as the reference set. [1]
If FITTYPE=6 and FULL=TRUE, this gives the value of the mapping’s SimpFI attribute (whether it is legitimate to simplify the forward followed by the inverse transformation to a unit transformation). [TRUE]
If FITTYPE=6 and FULL=TRUE this gives the value of the mapping’s SimpIF attribute (whether it is legitimate to simplify the inverse followed by the forward transformation to a unit transformation). [TRUE]
The RMS tolerance in positions which is used to determine the best fit. Adjust this value only if the input positions are specified in coordinates with a higher accuracy or smaller units. [0.001]
This parameter determines whether Set header information should be used in the registration. If USESET is true, then REGISTER will try to group position lists according to the Set Name attribute of the images to which they are attached. All lists coming from images which share the same (non-blank) Set Name attribute, and which have a CCD_SET coordinate frame in their WCS component, will be grouped together and treated by the program as a single position list. Images which have no associated position list but are in the same Set as ones which are successfully registered will have a suitable registration frame added too, based on their Set alignment relation to the registered Set member. Thus the assumption is made that the relative alignment of images within a Set is already known and has been fixed.
If USESET is false, all Set header information is ignored. If NDFNAMES is false, USESET will be ignored. If the input images have no Set headers, or if they have no CCD_SET frame in their WCS components, the setting of USESET will make no difference.
This parameter specifies whether the coordinates in the position lists should be transformed from Pixel coordinates into the Current coordinate system of the associated image before use. It should normally be set TRUE, in which case the transformation type set by the FITTYPE parameter is the type which will be fit between the Current coordinate systems of the NDFs. Otherwise the fit will be between the positions in pixel coordinates.
This parameter is ignored if NDFNAMES is not TRUE. [TRUE]
If PLACEIN is "SINGLE" then the value of this parameter gives the the name of an image which will have the new coordinate systems attached to it. They will be added with domains given by the OUTDOMAIN parameter with ’_1’, ’_2’, … appended. If the image named by this parameter does not exist, a dummy one will be created.
If FITTYPE=6 then this parameter specifies the parameterised algebraic expression to be used as the forward X transformation. The expression may use all the functions specified in SUN/61 (TRANSFORM) as well as the usual mathematical operators (
+
,-,
\ast
,/,
\ast
\ast
). Functions are parameterised by the strings PA,PB,PC...PZ which are the values which will be determined. The string must contain at least one reference to either X or Y. So a possible return is:
+
\ast
which is the same as the linear X transformation which just applies an offset and a scale factor.
If FITTYPE=6 and FULL=TRUE then this parameter specifies the inverse X transformation. The expression may use all the functions specified in SUN/61 (TRANSFORM) as well as the usual mathematical operations (
+
\ast
\ast
\ast
). Functions are parameterised by the strings PA,PB,PC...PZ which are the values which will be determined. This expression must contain a reference to either XX or YY. So a possible return is:
(XX-PA)/PB
which is the same as the inverse linear X transformation for an offset and scale.
If FITTYPE=6 then this parameter specifies the parameterised algebraic expression to be used as the forward Y transformation. The expression may use all the functions specified in SUN/61 (TRANSFORM) as well as the usual mathematical operators (
+
\ast
\ast
\ast
+
\ast
which is the same as the linear Y transformation which just applies an offset and a scale factor.
If FITTYPE=6 and FULL=TRUE then this parameter specifies the inverse Y transformation. The expression may use all the functions specified in SUN/61 (TRANSFORM) as well as the usual mathematical operations (
+
\ast
\ast
\ast
(YY-PC)/PD
which is the same as the inverse linear Y transformation for an offset and scale.
register inlist=’
\ast
’ fittype=1
In this example all the images in the current directory are accessed and their associated position lists are opened. A global fit between all the datasets is then performed which results in estimates for the offsets from the first input image’s position. These offsets are between the Current coordinate systems of the images. The results are then attached as new coordinate systems, labelled ’CCD_REG’, in the WCS component of the images. Actual registration of the images can then be achieved by aligning all the images in the CCD_REG domain using TRANNDF.
\ast
’ trtype=5 outdomain=result-set1
This example works as above but this time the global transformations are derived for a full 6-parameter linear fit (which allows offset, rotation, magnification and shear). The results are coded as attached coordinate systems labelled ’RESULT-SET1’.
register inlist=’"myimage1,myimage2"’ fittype=4 refpos=2
In this example a solid body fit is performed between the position lists associated with the images myimage1 and myimage2. The reference positions are chosen to be those associated with myimage2, so that the CCD_REG frame coordinates will be the same as the pixel coordinates in image myimage2.
register inlist=’"one,two"’ fittype=6 xfor=’pa
+
\ast
x’ yfor=’pa
+
\ast
In this example the position lists associated with the images one and two are said to be related by the algebraic expressions "pa
+
\ast
x" and "pa
+
\ast
y", which indicates that a single offset applies in both directions and a single scale factor. A solution for the values PA and PB is found using a general least-squares minimization technique. Starting values for PA and PB can be given using the parameters PA and PB. Since the fittype is 6, only two position lists may be registered in the same run.
register inlist=’"image1,image2"’ fittype=6 xfor=’pa
+
\ast
+
\ast
+
\ast
\ast
y’ yfor=’pe
+
\ast
+
\ast
+
\ast
\ast
In this example a non-linear transformation is fit between the positions associated with the images image1 and image2. This analysis may help in determining whether a 6-parameter fit is good enough, or if you just want to transform positions. A problem with proceeding with this transformation in a general fashion is deriving the inverse as this is required if you want to perform image resampling using TRANNDF (though the more specialised, and less efficient, DRIZZLE can resample with only the forward transformation).
register ndfnames=false inlist=’"list1.acc,list2.acc,list3.acc"’ fittype=3 placein=each in=’"image1,image2,image3"’
In this example the input position lists are not associated with images (ndfnames=false) and have to be specified by name (no wildcards allowed). Since the position lists are not associated with images there is no natural home for the new coordinate systems. In this example it has been decided to attach the coordinate systems to a set of images anyway. PLACEIN could also be given as "SINGLE" in which case the coordinate systems would be attached to a single image with Domain names CCD_REG_1, CCD_REG_2, ...
“Determining transformation parameters”.
In all casese, the coordinates in position lists are pixel coordinates.
Files with EXTERNAL format may be used with this application but all positions have to be present in all lists, no missing positions are allowed.
If NDFNAMES is TRUE then the item "CURRENT_LIST" of the .MORE.CCDPACK structure of the input NDFs will be located and assumed to contain the names of the lists whose positions are to be used for registration.
On exit, a new coordinate frame with a Domain as given by the OUTDOMAIN parameter will be inserted in the WCS component of the input images. Taken together these contain the registration information and can be inspected using WCSEDIT.
Certain parameters (LOGTO, LOGFILE, NDFNAMES and USESET) have global values. These global values will always take precedence, except when an assignment is made on the command line. Global values may be set and reset using the CCDSETUP and CCDCLEAR commands.
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How simplifying and flexible is the simplifying assumption in pair-copula constructions – analytic answers in dimension three and a glimpse beyond
2021 How simplifying and flexible is the simplifying assumption in pair-copula constructions – analytic answers in dimension three and a glimpse beyond
Thomas Mroz, Sebastian Fuchs, Wolfgang Trutschnig
Thomas Mroz,1 Sebastian Fuchs,1 Wolfgang Trutschnig1
1Department for Mathematics, University of Salzburg, Hellbrunnerstrasse 34, A-5020 Salzburg, Austria. Tel.: +43 662 8044 5326, Fax: +43 662 8044 137
Motivated by the increasing popularity and the seemingly broad applicability of pair-copula constructions underlined by numerous publications in the last decade, in this contribution we tackle the unavoidable question on how flexible and simplifying the commonly used ‘simplifying assumption’ is from an analytic perspective and provide answers to two related open questions posed by Nagler and Czado in 2016. Aiming at a simplest possible setup for deriving the main results we first focus on the three-dimensional setting. We prove that the family of simplified copulas is flexible in the sense that it is dense in the set of all three-dimensional copulas with respect to the uniform metric
{d}_{\mathrm{\infty }}
– considering stronger notions of convergence like the one induced by the metric
{D}_{1}
, by weak conditional convergence, by total variation, or by Kullback-Leibler divergence, however, the family even turns out to be nowhere dense and hence insufficient for any kind of flexible approximation. Furthermore, returning to
{d}_{\mathrm{\infty }}
we show that the partial vine copula is never the optimal simplified copula approximation of a given, non-simplified copula C, and derive examples illustrating that the corresponding approximation error can be strikingly large and extend to more than 28% of the diameter of the metric space. Moreover, the mapping ψ assigning each three-dimensional copula its unique partial vine copula turns out to be discontinuous with respect to
{d}_{\mathrm{\infty }}
(but continuous with respect to
{D}_{1}
and to weak conditional convergence), implying a surprising sensitivity of partial vine copula approximations. The afore-mentioned main results concerning
{d}_{\mathrm{\infty }}
are then extended to the general multivariate setting.
The second and the third author gratefully acknowledge the support of the WISS 2025 project ’IDA-lab Salzburg’ (20204-WISS/225/197-2019 and 20102-F1901166-KZP).
Thomas Mroz. Sebastian Fuchs. Wolfgang Trutschnig. "How simplifying and flexible is the simplifying assumption in pair-copula constructions – analytic answers in dimension three and a glimpse beyond." Electron. J. Statist. 15 (1) 1951 - 1992, 2021. https://doi.org/10.1214/21-EJS1832
Keywords: conditional distribution , Dependence , Markov kernel , pair-copula construction , simplifying assumption
Thomas Mroz, Sebastian Fuchs, Wolfgang Trutschnig "How simplifying and flexible is the simplifying assumption in pair-copula constructions – analytic answers in dimension three and a glimpse beyond," Electronic Journal of Statistics, Electron. J. Statist. 15(1), 1951-1992, (2021)
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Global Constraint Catalog: Klinear_programming
<< 3.7.135. Limited discrepancy search3.7.137. Line segments intersection >>
\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}
\mathrm{𝚊𝚖𝚘𝚗𝚐}_\mathrm{𝚜𝚎𝚚}
\mathrm{𝚌𝚒𝚛𝚌𝚞𝚒𝚝}
\mathrm{𝚌𝚞𝚖𝚞𝚕𝚊𝚝𝚒𝚟𝚎}
\mathrm{𝚍𝚒𝚜𝚓𝚞𝚗𝚌𝚝𝚒𝚟𝚎}
\mathrm{𝚍𝚘𝚖𝚊𝚒𝚗}_\mathrm{𝚌𝚘𝚗𝚜𝚝𝚛𝚊𝚒𝚗𝚝}
\mathrm{𝚎𝚕𝚎𝚖𝚎𝚗𝚝}_\mathrm{𝚐𝚛𝚎𝚊𝚝𝚎𝚛𝚎𝚚}
\mathrm{𝚎𝚕𝚎𝚖𝚎𝚗𝚝}_\mathrm{𝚕𝚎𝚜𝚜𝚎𝚚}
\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}_\mathrm{𝚕𝚘𝚠}_\mathrm{𝚞𝚙}
𝚔_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}
𝚔_\mathrm{𝚌𝚞𝚝}
\mathrm{𝚕𝚒𝚗𝚔}_\mathrm{𝚜𝚎𝚝}_\mathrm{𝚝𝚘}_\mathrm{𝚋𝚘𝚘𝚕𝚎𝚊𝚗𝚜}
\mathrm{𝚙𝚊𝚝𝚑}_\mathrm{𝚏𝚛𝚘𝚖}_\mathrm{𝚝𝚘}
\mathrm{𝚛𝚎𝚐𝚞𝚕𝚊𝚛}
\mathrm{𝚜𝚕𝚒𝚍𝚒𝚗𝚐}_\mathrm{𝚜𝚞𝚖}
\mathrm{𝚜𝚝𝚛𝚘𝚗𝚐𝚕𝚢}_\mathrm{𝚌𝚘𝚗𝚗𝚎𝚌𝚝𝚎𝚍}
\mathrm{𝚜𝚞𝚖}
\mathrm{𝚝𝚘𝚞𝚛}
A constraint for which a reference provides a linear relaxation (see, e.g., the
\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}
\mathrm{𝚌𝚒𝚛𝚌𝚞𝚒𝚝}
\mathrm{𝚌𝚞𝚖𝚞𝚕𝚊𝚝𝚒𝚟𝚎}
\mathrm{𝚜𝚞𝚖}
\mathrm{𝚛𝚎𝚐𝚞𝚕𝚊𝚛}
[CoteGendronRousseau07] constraints) or a constraint for which the flow model was derived by reformulating the constraint as a linear program (see, e.g., the
\mathrm{𝚊𝚖𝚘𝚗𝚐}_\mathrm{𝚜𝚎𝚚}
\mathrm{𝚜𝚕𝚒𝚍𝚒𝚗𝚐}_\mathrm{𝚜𝚞𝚖}
constraints), or a constraint that was also proposed within the context of linear programming (see, e.g., the
\mathrm{𝚌𝚒𝚛𝚌𝚞𝚒𝚝}
\mathrm{𝚍𝚘𝚖𝚊𝚒𝚗}_\mathrm{𝚌𝚘𝚗𝚜𝚝𝚛𝚊𝚒𝚗𝚝}
constraints). In the context of linear programming the book of John N. Hooker [Hooker07book] provides a significant set of relaxations for a number of global constraints.
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Interpolate numerical solution of PDE - MATLAB pdeval - MathWorks América Latina
Interpolate PDE Solution
Interpolate numerical solution of PDE
[u,dudx] = pdeval(m,xmesh,usol,xq)
[u,dudx] = pdeval(m,xmesh,usol,xq) interpolates a numerical solution returned by pdepe at new query points xq, and returns the interpolated values of the solution u and their partial derivative dudx. The m, xmesh, and usol arguments are reused from a previous call to pdepe:
The numerical solution produced by sol = pdepe(m,@pdefun,@pdeic,@pdebc,xmesh,tspan) uses the coordinate symmetry m and spatial mesh xmesh to return a 3-D matrix of the solution values sol. Reuse the m and xmesh inputs used to calculate the solution when you call pdeval.
The input vector usol = sol(i,:,k) is the value of component k of the solution at time tspan(i). When there is only one solution component, usol is a row extracted from the solution matrix usol = sol(i,:).
Use pdepe to solve a partial differential equation, and then use pdeval to evaluate the solution at additional points.
Use pdepe to solve the pdex1 example problem. You can type edit pdex1 to see more details on the problem, or see pdepe for details on the PDE solution process. The required functions to solve the pdex1 problem are included at the end of this example as local functions.
Interpolate Solution
The solution sol generated by pdepe uses 20 points for x, evenly spaced between 0 and 1. Create a vector of query points that are located midway between the points used by pdepe.
xq = x;
xq(1:end-1) = xq(1:end-1) + diff(xq)./2;
Use pdeval to interpolate the solution at the query points. Since there is only one solution component, you can extract a row from sol to operate on, such as sol(2,:).
[u,dudx] = pdeval(m,x,sol(2,:),xq);
Plot the solution computed by pdepe, as well as the interpolated solution and its partial derivative computed by pdeval.
plot(x,sol(2,:),'r*')
plot(xq,u,'-o')
plot(xq,dudx,'.')
legend('PDEPE Solution', 'PDEVAL Interpolation', 'PDEVAL Partial Derivative')
Listed here are the local helper functions that the PDE solver pdepe calls to calculate the solution.
m — Coordinate symmetry used with pdepe
Coordinate symmetry used with pdepe, specified as one of the values in this table. Specify the same coordinate symmetry you used in the initial call to pdepe.
Slab/Cartesian
xmesh — Spatial mesh used with pdepe
Spatial mesh used with pdepe, specified as a vector [x0 x1 ... xn] containing the points at which a numerical solution was computed. Specify the same spatial mesh you used in the initial call to pdepe.
usol — Extracted solution component
Extracted solution component, specified as a vector of values computed by pdepe for one solution component at a particular time.
{\mathit{u}}_{\mathit{k}}
evaluated at time t(i) and spatial point xmesh(j). You can create the input usol with the command usol = sol(i,:,k), where sol(i,:,k) is the value of component k of the solution at time tspan(i), evaluated on the entire spatial mesh xmesh. When there is only one solution component, usol is a row extracted from the solution matrix usol = sol(i,:).
Example: usol = sol(10,:,2) extracts the second solution component calculated at time tspan(10).
Example: usol = sol(5,:) extracts the solution calculated at time tspan(5).
Query points, specified as a scalar or vector of x-coordinates. The points specified in xq must lie within the interval [min(xmesh) max(xmesh)].
Typically, the points in xq are specified between the values in xmesh to enable pdeval to evaluate the solution produced by pdepe on a finer mesh.
Example: xq = linspace(0,1,100) specifies 100 evenly spaced query points between 0 and 1.
u — Interpolated solution
Interpolated solution, returned as a vector with the same size as xq.
dudx — Partial derivative of interpolated solution
Partial derivative of interpolated solution, returned as a vector with the same size as xq.
pdeval evaluates the partial derivative
\frac{\partial u}{\partial x}
rather than the flux
f\left(x,t,u,\frac{\partial u}{\partial x}\right)
. Although the flux is continuous, the partial derivative can have a jump at a material interface.
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Electric Circuit Analysis/Resistors in Parallel/Answers - Wikiversity
< Electric Circuit Analysis | Resistors in Parallel
{\displaystyle R_{eq}=2.5\Omega }
{\displaystyle R_{eq}=1.024\Omega }
{\displaystyle I_{1}=7.68A}
{\displaystyle I_{2}=5.12A}
{\displaystyle I_{3}=2.19A}
{\displaystyle V_{s}=15.36Volts}
{\displaystyle R_{eq}=3.377\Omega }
4 Voltage Sources can be connected in parallel if they have the same voltage output.
Retrieved from "https://en.wikiversity.org/w/index.php?title=Electric_Circuit_Analysis/Resistors_in_Parallel/Answers&oldid=1714000"
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Beaufort scale - Wikipedia
(Redirected from Beaufort Scale)
Empirical measure describing wind speed based on observed conditions
A ship in Force 12 storm at sea
1.1 Extended scale
2 Modern scale
3 Weather scale
The scale was devised in 1805 by the Irish hydrographer Francis Beaufort (later Rear Admiral), a Royal Navy officer, while serving on HMS Woolwich. The scale that carries Beaufort's name had a long and complex evolution from the previous work of others (including Daniel Defoe the century before) to when Beaufort was Hydrographer of the Navy in the 1830s, when it was adopted officially and first used during the voyage of HMS Beagle under Captain Robert FitzRoy, who was later to set up the first Meteorological Office (Met Office) in Britain giving regular weather forecasts.[1] In the 18th century, naval officers made regular weather observations, but there was no standard scale and so they could be very subjective – one man's "stiff breeze" might be another's "soft breeze". Beaufort succeeded in standardising the scale.
The scale was made a standard for ship's log entries on Royal Navy vessels in the late 1830s and was adapted to non-naval use from the 1850s, with scale numbers corresponding to cup anemometer rotations. In 1853, the Beaufort scale was accepted as generally applicable at the First International Meteorological Conference in Brussels.[3]
In 1916, to accommodate the growth of steam power, the descriptions were changed to how the sea, not the sails, behaved and extended to land observations. Rotations to scale numbers were standardized only in 1923. George Simpson, CBE (later Sir George Simpson), director of the UK Meteorological Office, was responsible for this and for the addition of the land-based descriptors.[1] The measures were slightly altered some decades later to improve its utility for meteorologists. Nowadays, meteorologists typically express wind speed in kilometres or miles per hour or, for maritime and aviation purposes, knots; but Beaufort scale terminology is still sometimes used in weather forecasts for shipping[4] and the severe weather warnings given to the public.[5]
Extended scale[edit]
The Beaufort scale was extended in 1946 when forces 13 to 17 were added.[3] However, forces 13 to 17 were intended to apply only to special cases, such as tropical cyclones. Nowadays, the extended scale is only used in Taiwan and mainland China, which are often affected by typhoons. Internationally, WMO Manual on Marine Meteorological Services (2012 edition) defined the Beaufort Scale only up to force 12 and there was no recommendation on the use of the extended scale.[6]
Data graphic showing Beaufort wind force in scale units, knots and metres/second
v = 1.625 B3/2 knots (
{\displaystyle ={\frac {13}{8}}{\sqrt {B^{3}}}}
where v is the equivalent wind speed at 10 metres above the sea surface and B is Beaufort scale number. For example, B = 9.5 is related to 24.5 m/s which is equal to the lower limit of "10 Beaufort". Using this formula the highest winds in hurricanes would be 23 in the scale. F1 tornadoes on the Fujita scale and T2 TORRO scale also begin roughly at the end of level 12 of the Beaufort scale, but are independent scales – although the TORRO scale wind values are based on the 3/2 power law relating wind velocity to Beaufort force.[8]
Modern scale[edit]
Beaufort scale[9][10][11][12]
Sea conditions (photo)
Calm < 1 knot
Light air 1–3 knots
Light breeze 4–6 knots
Gentle breeze 7–10 knots
Moderate breeze 11–16 knots
Fresh breeze 17–21 knots
Strong breeze 22–27 knots
near gale 28–33 knots
fresh gale 34–40 knots
Strong/severe gale 41–47 knots
Storm,[13]
whole gale 48–55 knots
Violent storm 56–63 knots
Hurricane force[13] ≥ 64 knots
≥ 118 km/h
≥ 32.7 m/s ≥ 46 ft
≥ 14 m The air is filled with foam and spray; sea is completely white with driving spray; visibility very seriously affected Devastation.
This scale is also widely used in the Netherlands, Germany,[14] Greece, China, Taiwan, Hong Kong, Malta, and Macau, although with some differences between them. Taiwan uses the Beaufort scale with the extension to 17 noted above. China also switched to this extended version without prior notice on the morning of 15 May 2006,[15] and the extended scale was immediately put to use for Typhoon Chanchu. Hong Kong and Macau retain force 12 as the maximum.
In the United States, winds of force 6 or 7 result in the issuance of a small craft advisory, with force 8 or 9 winds bringing about a gale warning, force 10 or 11 a storm warning ("a tropical storm warning" being issued instead of the latter two if the winds relate to a tropical cyclone), and force 12 a hurricane-force wind warning (or hurricane warning if related to a tropical cyclone). A set of red warning flags (daylight) and red warning lights (night time) is displayed at shore establishments which coincide with the various levels of warning.[citation needed]
In Canada, maritime winds forecast to be in the range of 6 to 7 are designated as "strong"; 8 to 9 "gale force"; 10 to 11 "storm force"; 12 "hurricane force". Appropriate wind warnings are issued by Environment Canada's Meteorological Service of Canada: strong wind warning, gale (force wind) warning, storm (force wind) warning and hurricane-force wind warning. These designations were standardized nationally in 2008, whereas "light wind" can refer to 0 to 12 or 0 to 15 knots and "moderate wind" 12 to 19 or 16 to 19 knots, depending on regional custom, definition or practice. Prior to 2008, a "strong wind warning" would have been referred to as a "small craft warning" by Environment Canada, similar to US terminology. (Canada and the USA have the Great Lakes in common.)[citation needed]
Weather scale[edit]
Beaufort's name was also attached to the Beaufort scale for weather reporting:
b blue sky
c detached clouds
d drizzling rain
g dark, gloomy
m misty (hazy)
o overcast
p passing showers
q squally
u ugly (threatening)
v visibility (unusual transparency)
w wet, dew
In this scale the weather could be reported as "s.c." for snow and detached cloud or "g.r.q." for dark, rain and squally.[16]
^ a b "National Meteorological Library and Archive Fact sheet 6 – The Beaufort Scale" (PDF). Met Office. Archived from the original (PDF) on 2 October 2012. Retrieved 13 May 2011.
^ a b Saucier, Walter Joseph (1955). Principles of Meteorological Analysis. Chicago: The University of Chicago Press. OCLC 1082907714. , reprinted in 2003 by Dover Publications.
^ McIlveen, Robin (1991). Fundamentals of Weather and Climate. Cheltenham, England: Stanley Thornes. p. 40. ISBN 978-0-7487-4079-6.
^ Hay, William W. (2016). Experimenting on a Small Planet: A History of Scientific Discoveries, a Future of Climate Change and Global Warming (second ed.). Cham, Switzerland: Springer Verlag. p. 26. ISBN 978-3-319-27402-7.
^ Manual on Marine Meteorological Services: Volume I – Global Aspect (PDF). World Meteorological Organization. 2012. Archived from the original (PDF) on 5 May 2017.
^ Maiden, Terence. "T-Scale: Origins and Scientific Basis". TORRO. Archived from the original on 5 February 2012. Retrieved 4 January 2012.
^ "The Beaufort Scale". RMetS. Retrieved 6 July 2021.
^ "Beaufort wind force scale". Met Office. Retrieved 27 November 2015.
^ "Beaufort Scale". Royal Meteorological Society. Retrieved 27 November 2015.
^ "Beaufort Scale". Encyclopædia Britannica. Retrieved 27 November 2015.
^ a b The names "storm" and "hurricane" on the Beaufort scale refer only to wind strength, and do not necessarily mean that other severe weather (for instance, a thunderstorm or tropical cyclone) is present. To avoid confusion, strong wind warnings will often speak of e.g. "hurricane-force winds".
^ "Wetterlexikon - Beaufort-Skala" (in German). Deutscher Wetterdienst. Archived from the original on 12 December 2013. Retrieved 14 February 2014.
^ "昨日实行新标准"珍珠"属强台风_新闻中心_新浪网". news.sina.com.cn.
^ "The Times". The Times. 29 April 1873. p. 10. ISSN 0140-0460. Retrieved 3 July 2020.
Wikisource has the text of the 1911 Encyclopædia Britannica article "Beaufort Scale".
National Meteorological Library and Archive Archived 13 November 2017 at the Wayback Machine fact sheet on the history of the Beaufort Scale, including various scales and photographic depictions of the sea state.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Beaufort_scale&oldid=1080733582"
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Oil immersion - Wikipedia
Principle of immersion microscopy. Path of rays with immersion medium (yellow) (left half) and without (right half). Rays (black) coming from the object (red) at a certain angle and going through the cover-slip (orange, as is the slide at the bottom) can enter the objective (dark blue) only when immersion is used. Otherwise, the refraction at the cover-slip-air interface causes the ray to miss the objective and its information is lost.
Two Leica oil immersion objective lenses. Oil immersion objective lenses look superficially identical to non-oil immersion lenses.
Without oil, light waves reflect off the slide specimen through the glass cover slip, through the air, and into the microscope lens (see the colored figure to the right). Unless a wave comes out at a 90-degree angle, it bends when it hits a new substance, the amount of bend depending on the angle. This distorts the image. Air has a very different index of refraction from glass, making for a larger bend compared to oil, which has an index more similar to glass. Specially manufactured oil can have nearly exactly the same refractive index as glass, making an oil immersed lens nearly as effective as having entirely glass to the sample (which would be impractical).
Immersion oils are transparent oils that have specific optical and viscosity characteristics necessary for use in microscopy. Typical oils used have an index of refraction of around 1.515.[1] An oil immersion objective is an objective lens specially designed to be used in this way. Many condensers also give optimal resolution when the condenser lens is immersed in oil.
2 Oil immersion objectives
Lenses reconstruct the light scattered by an object. To successfully achieve this end, ideally, all the diffraction orders have to be collected. This is related to the opening angle of the lens and its refractive index. The resolution of a microscope is defined as the minimum separation needed between two objects under examination in order for the microscope to discern them as separate objects. This minimum distance is labelled δ. If two objects are separated by a distance shorter than δ, they will appear as a single object in the microscope.
A measure of the resolving power, R.P., of a lens is given by its numerical aperture, NA:
{\displaystyle \delta ={\frac {\lambda }{\mathrm {2NA} }}}
where λ is the wavelength of light. From this it is clear that a good resolution (small δ) is connected with a high numerical aperture.
The numerical aperture of a lens is defined as
{\displaystyle \mathrm {NA} =n\sin \alpha _{0}\;}
where α0 is half the angle spanned by the objective lens seen from the sample, and n is the refractive index of the medium between the lens and specimen (≈1 for air).
State of the art objectives can have a numerical aperture of up to 0.95. Because sin α0 is always less than or equal to unity (the number "1"), the numerical aperture can never be greater than unity for an objective lens in air. If the space between the objective lens and the specimen is filled with oil however, the numerical aperture can obtain values greater than unity. This is because oil has a refractive index greater than 1.
Oil immersion objectives[edit]
From the above it is understood that oil between the specimen and the objective lens improves the resolving power by a factor 1/n. Objectives specifically designed for this purpose are known as oil immersion objectives.
Oil immersion objectives are used only at very large magnifications that require high resolving power. Objectives with high power magnification have short focal lengths, facilitating the use of oil. The oil is applied to the specimen (conventional microscope), and the stage is raised, immersing the objective in oil. (In inverted microscopes the oil is applied to the objective).
The refractive indices of the oil and of the glass in the first lens element are nearly the same, which means that the refraction of light will be small upon entering the lens (the oil and glass are optically very similar). The correct immersion oil for an objective lens has to be used to ensure that the refractive indices match closely. Use of an oil immersion lens with the incorrect immersion oil, or without immersion oil altogether, will suffer from spherical aberration. The strength of this effect depends on the size of the refractive index mismatch.
Oil immersion can generally only be used on rigidly mounted specimens otherwise the surface tension of the oil can move the coverslip and so move the sample underneath. This can also happen on inverted microscopes because the coverslip is below the slide.
Immersion oil[edit]
Before the development of synthetic immersion oils in the 1940s, cedar tree oil was widely used. Cedar oil has an index of refraction of approximately 1.516. The numerical aperture of cedar tree oil objectives is generally around 1.3. Cedar oil has a number of disadvantages however: it absorbs blue and ultraviolet light, yellows with age, has sufficient acidity to potentially damage objectives with repeated use (by attacking the cement used to join lenses), and diluting it with solvent changes its viscosity (and refraction index and dispersion). Cedar oil must be removed from the objective immediately after use before it can harden, since removing hardened cedar oil can damage the lens.[2]
In modern microscopy synthetic immersion oils are more commonly used, as they eliminate most of these problems.[2] NA values of 1.6 can be achieved with different oils. Unlike natural oils synthetic ones do not harden on the lens and can typically be left on the objective for months at a time, although to best maintain a microscope it is best to remove the oil daily. Over time oil can enter for the front lens of the objective or into the barrel of the objective and damage the objective. There are different types of immersion oils with different properties based on the type of microscopy you will be performing. Type A and Type B are both general purpose immersion oils with different viscosities. Type F immersion oil is best used for fluorescent imaging at room temperature (23 °C), while type N oil is made to be used at body temperature (37 °C) for live cell imaging applications. All have a nD of 1.515, quite similar to the original cedar oil.[3]
Index-matching material
^ "Microscope Objectives: Immersion Media" by Mortimer Abramowitz and Michael W. Davidson, Olympus Microscopy Resource Center (website), 2002.
^ a b Cargille, John (1985) [1964], "Immersion Oil and the Microscope", New York Microscopical Society Yearbook, archived from the original on 2011-09-11, retrieved 2008-01-21
^ Labs, Cargille. "About Immersion Oils". Cargille Labs. Retrieved 2019-12-04.
Practical Microscopy by L.C. Martin and B.K. Johnson, Glasgow (1966).
Light Microscopy by J.K. Solberg, Tapir Trykk (2000).
"Microscope Objectives: Immersion Media" by Mortimer Abramowitz and Michael W. Davidson, Olympus Microscopy Resource Center (website), 2002.
"Immersion Oil Microscopy" by David B. Fankhauser, Biology at University of Cincinnati, Clermont College (website), December 30, 2004.
"History of Oil Immersion Lenses" by Jim Solliday, Southwest Museum of Engineering, Communications, and Computation (website), 2007.
"Immersion Oil and the Microscope" by John J. Cargille, New York Microscopical Society Yearbook, 1964 (revised, 1985). (Archived at Cargille Labs (website).)
Retrieved from "https://en.wikipedia.org/w/index.php?title=Oil_immersion&oldid=1075678968"
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\begin{array}{l}\text{The speed of the bicycle from Town }P\text{ to Town }Q\\ =\frac{\text{Distance}}{\text{Time}}\\ =\frac{10}{2}\\ =5\text{ km/h}\end{array}
\begin{array}{l}\text{Speed}=\frac{\text{Distance}}{\text{Time}}\\ \\ \text{Rahim took 30 minutes rest at Town }Q\text{.}\\ \text{Time taken when his journey to Town }R\text{ three times}\\ \text{faster than his earlier speed}\text{.}\\ =\frac{25}{5×3}\\ =\frac{5}{3}\\ =1\frac{2}{3}\text{ hours}\\ =1\text{ hour 40 minutes }←\overline{)\begin{array}{l}\frac{2}{3}×60\\ =40\text{ minutes}\end{array}}\\ \\ \text{Total time taken from Town }P\text{ to Town }Q\text{ and Town }Q\text{ to Town }R\\ =2\text{ hours }+30\text{ minutes}+1\text{ hour 40 minutes}\\ =4\text{ hour 10 minutes}\\ \\ \text{The time he reached Town }R\text{ at 1}\text{.10 p}\text{.m}\text{.}\end{array}
\begin{array}{l}\text{Mass of concrete pipe on the trailer}\\ =500\text{ kg}×8\\ =4000\text{ kg}\\ =\frac{4000}{1000}\\ =4\text{ tonnes}\\ \\ \text{Total mass of the trailer and its load}\\ =1.5+4.0\\ =5.5\text{ tonnes}\end{array}
\begin{array}{l}\text{Speed}=\frac{\text{Distance}}{\text{Time}}\\ \\ \text{Time taken to travel from the Factory to Location }P\\ =10.00\text{ a}\text{.m}\text{.}-8.00\text{ a}\text{.m}\text{.}\\ =2\text{ hours}\\ \\ \text{Speed of the trailer}\\ =\frac{80}{2}\\ =40\text{ km/h}\\ \\ \text{It stopped for }1\frac{1}{2}\text{ hours to unload half of the concrete pipes}\text{.}\\ \\ \text{Time taken to continue its journey to Location }Q\\ =\frac{\text{Distance}}{\text{Speed}}\\ =\frac{200}{40×2}\\ =2\frac{1}{2}\text{ hours}\\ \\ \text{The time the trailer reached at Location }Q\\ =1400\text{ hours or }2.00\text{ p}\text{.m}\text{.}\end{array}
\begin{array}{l}\text{Speed}=\frac{\text{Distance}}{\text{Time}}\\ \text{Total distance from Town }A\text{ to Town }B\\ =\left(60×3\right)+\left(75×2\right)\\ =180+150\\ =330\text{ km}\\ \\ \text{Time taken by Mary}\\ =\frac{330}{100}\\ =3.3\text{ hours}\\ =3\text{ hours }18\text{ minutes }←\overline{)\begin{array}{l}0.3×60\\ =18\text{ minutes}\end{array}}\\ \\ \text{The time Mary started her journey from Town }A=9.42\text{ a}\text{.m}\text{.}\end{array}
15.1 Trigonometry →
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\left(2,0,-3\right)
\left(4,5,-2\right)
f\left(x,y,z\right)=x {e}^{y}+y {z}^{2}
\mathbf{v}=\mathbf{Q}-\mathbf{P}
\left[\begin{array}{c}4\\ 5\\ -2\end{array}\right]-\left[\begin{array}{c}2\\ 0\\ -3\end{array}\right]
\left[\begin{array}{c}2\\ 5\\ 1\end{array}\right]
be the vector from point P to point Q.
\mathbf{u}=\mathbf{v}/∥\mathbf{v}∥
\mathbf{v}/\sqrt{30}
be the unit vector from point P in the direction of point Q,
\mathbf{R}=\mathbf{P}+t \mathbf{u}
x=2+\frac{2 t}{\sqrt{30}},y=\frac{5 t}{\sqrt{30}},z=-3+\frac{t}{\sqrt{30}}
Along this line the function values of
w\left(t\right)=f\left(x\left(t\right),y\left(t\right),z\left(t\right)\right)=2{ⅇ}^{\frac{1}{6}t\sqrt{30}}+\frac{1}{15}{ⅇ}^{\frac{1}{6}t\sqrt{30}}t\sqrt{30}+\frac{3}{2}t\sqrt{30}-{t}^{2}+\frac{1}{180}{t}^{3}\sqrt{30}
w\prime \left(0\right)=\frac{19}{10}\sqrt{30}
\genfrac{}{}{0}{}{\left(∇f\right)}{\phantom{x=a}}|\genfrac{}{}{0}{}{\phantom{\mathrm{f\left(x\right)}}}{\mathrm{P}}
\genfrac{}{}{0}{}{{f}_{x} \mathbf{i}+{f}_{y} \mathbf{j}+{f}_{z} \mathbf{k}}{\phantom{x=a}}|\genfrac{}{}{0}{}{\phantom{\mathrm{f\left(x\right)}}}{\mathrm{P}}
\left[\begin{array}{r}1\\ 11\\ 0\end{array}\right]
{\mathrm{D}}_{\mathbf{u}}f\left(\mathrm{P}\right)=\left[\begin{array}{r}1\\ 11\\ 0\end{array}\right]·\left(\frac{1}{\sqrt{30}}\left[\begin{array}{r}2\\ 5\\ 1\end{array}\right]\right)=\frac{19}{10}\sqrt{30}
f=x {ⅇ}^{y}+y {z}^{2}
\stackrel{\text{assign}}{\to }
\mathbf{v}=〈2,5,1〉
\stackrel{\text{assign}}{\to }
\mathbf{u}=\mathbf{v}/{∥\mathbf{v}∥}_{2}
\stackrel{\text{assign}}{\to }
f
f
\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{y}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{z}}^{\textcolor[rgb]{0,0,1}{2}}
\stackrel{\text{directional derivative}}{\to }
\frac{\textcolor[rgb]{0,0,1}{19}}{\textcolor[rgb]{0,0,1}{10}}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}
\genfrac{}{}{0}{}{\left(∇f\right)}{\phantom{x=a}}|\genfrac{}{}{0}{}{\phantom{\mathrm{f\left(x\right)}}}{\mathrm{P}}
x {ⅇ}^{y}+y {z}^{2}
\stackrel{\text{gradient}}{\to }
\left[\left[\begin{array}{r}1\\ 11\\ 0\end{array}\right]\right]
\stackrel{\text{select entry 1}}{\to }
\left[\begin{array}{r}1\\ 11\\ 0\end{array}\right]
\stackrel{\text{assign to a name}}{\to }
\textcolor[rgb]{0,0,1}{\mathrm{Gf}}
{\mathrm{D}}_{\mathbf{u}}f\left(\mathrm{P}\right)=(\genfrac{}{}{0}{}{\left(∇f\right)}{\phantom{x=a}}|\genfrac{}{}{0}{}{\phantom{\mathrm{f\left(x\right)}}}{\mathrm{P}})·\mathbf{u}
\mathbf{Gf}·\mathbf{u}
\frac{\textcolor[rgb]{0,0,1}{19}}{\textcolor[rgb]{0,0,1}{10}}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}
\mathbf{R}=\mathbf{P}+t \mathbf{u}
L
\left[2,0,-3\right],\mathbf{u}
\stackrel{\text{make line}}{\to }
\textcolor[rgb]{0,0,1}{\mathrm{Student}}\textcolor[rgb]{0,0,1}{:-}\textcolor[rgb]{0,0,1}{\mathrm{MultivariateCalculus}}\textcolor[rgb]{0,0,1}{:-}\textcolor[rgb]{0,0,1}{\mathrm{Line}}\textcolor[rgb]{0,0,1}{}\left(\left[\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{3}\right]\textcolor[rgb]{0,0,1}{,}\left[\begin{array}{c}\frac{1}{15}\textcolor[rgb]{0,0,1}{}\sqrt{30}\\ \frac{1}{6}\textcolor[rgb]{0,0,1}{}\sqrt{30}\\ \frac{1}{30}\textcolor[rgb]{0,0,1}{}\sqrt{30}\end{array}\right]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{variables}}\textcolor[rgb]{0,0,1}{=}\left[\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{z}\right]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{parameter}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{id}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{1}\right)
\stackrel{\text{representation}}{\to }
[\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}}{\textcolor[rgb]{0,0,1}{15}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{=}\frac{\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}}{\textcolor[rgb]{0,0,1}{6}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}}{\textcolor[rgb]{0,0,1}{30}}]
\stackrel{\text{assign to a name}}{\to }
\textcolor[rgb]{0,0,1}{L}
w\left(t\right)=f\left(x\left(t\right),y\left(t\right),z\left(t\right)\right)
{\mathrm{D}}_{\mathbf{u}}f\left(\mathrm{P}\right)=w\prime \left(0\right)
t
t=0
\genfrac{}{}{0}{}{f}{\phantom{x=a}}|\genfrac{}{}{0}{}{\phantom{\mathrm{f\left(x\right)}}}{L}
\left(\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{15}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}\right)\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{6}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{6}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}\textcolor[rgb]{0,0,1}{}{\left(\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{30}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}\right)}^{\textcolor[rgb]{0,0,1}{2}}
\stackrel{\text{simplify}}{=}
\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{6}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{15}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{6}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{3}}{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{t}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{180}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{t}}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}
\stackrel{\text{differentiate w.r.t. t}}{\to }
\frac{\textcolor[rgb]{0,0,1}{2}}{\textcolor[rgb]{0,0,1}{5}}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{6}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{6}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{3}}{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{60}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{t}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}
\stackrel{\text{evaluate at point}}{\to }
\frac{\textcolor[rgb]{0,0,1}{19}}{\textcolor[rgb]{0,0,1}{10}}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}
\mathrm{with}\left(\mathrm{Student}:-\mathrm{MultivariateCalculus}\right):
f
f≔x {ⅇ}^{y}+y {z}^{2}:
\mathbf{v}≔\left[2,5,1\right]:
\mathrm{DirectionalDerivative}\left(f,\left[x,y,z\right]=\left[2,0,-3\right],\mathbf{v}\right)
\frac{\textcolor[rgb]{0,0,1}{19}}{\textcolor[rgb]{0,0,1}{10}}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}
{\mathrm{D}}_{\mathbf{u}}f\left(\mathrm{P}\right)=(\genfrac{}{}{0}{}{\left(∇f\right)}{\phantom{x=a}}|\genfrac{}{}{0}{}{\phantom{\mathrm{f\left(x\right)}}}{\mathrm{P}})·\mathbf{u}
\mathbf{Gf}\mathbf{≔}\mathrm{Gradient}\left(f,\left[x,y,z\right]=\left[2,0,-3\right]\right)\left[\right]
\left[\begin{array}{r}\textcolor[rgb]{0,0,1}{1}\\ \textcolor[rgb]{0,0,1}{11}\\ \textcolor[rgb]{0,0,1}{0}\end{array}\right]
\mathbf{V}≔\mathrm{convert}\left(\mathbf{v},\mathrm{Vector}\right):\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\mathbf{u}≔\mathrm{Normalize}\left(\mathbf{V}\right)
\left[\begin{array}{c}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{15}}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}\\ \frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{6}}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}\\ \frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{30}}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}\end{array}\right]
\mathrm{DotProduct}\left(\mathbf{Gf},\mathbf{u}\right)
\frac{\textcolor[rgb]{0,0,1}{19}}{\textcolor[rgb]{0,0,1}{10}}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}
L≔\mathrm{GetRepresentation}\left(\mathrm{Line}\left(\left[2,0,-3\right],\mathbf{u}\right),\mathrm{form}=\mathrm{parametric}\right)
\left[\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{15}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{=}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{6}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{30}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}\right]
Use the eval command to obtain the value of
L
w≔\mathrm{simplify}\left(\mathrm{eval}\left(f,L\right)\right)
\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{6}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{15}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{6}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{3}}{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{t}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{t}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{180}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{t}}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}
w\left(t\right)
w\prime \left(0\right)
\mathrm{eval}\left(\mathrm{diff}\left(w,t\right),t=0\right)
\frac{\textcolor[rgb]{0,0,1}{19}}{\textcolor[rgb]{0,0,1}{10}}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{30}}
|
\mathbf{F}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}
R
x+y+z=1
∇·\mathbf{F}={∂}_{x}\left(x {y}^{3}\right)+{∂}_{y}\left(y z\right)+{∂}_{z}\left({x}^{2}z\right)={y}^{3}+z+{x}^{2}
∇·\mathbf{F}
R
{∫}_{0}^{1}{∫}_{0}^{1-x}{∫}_{0}^{1-x-y}\left({y}^{3}+z+{x}^{2}\right) \mathrm{dz} \mathrm{dy} \mathrm{dx}
\frac{1}{15}
R
, note that there are four boundaries, the coordinate planes and the plane
z=1-x-y
. Table 9.8.10(a) lists the faces, the unit normal on that face, the expression for
\mathbf{F}·\mathbf{N} \mathrm{dσ}
\mathbf{F}·\mathbf{N} \mathrm{dσ}
on that face.
\mathbf{F}·\mathbf{N} \mathrm{dσ}
\mathbf{F}·\mathbf{N} \mathrm{dσ}
x=0
-\mathbf{i}
-x {y}^{3} \mathrm{dA}
0
y=0
-\mathbf{j}
-y z \mathrm{dA}
0
z=0
-\mathbf{k}
-{x}^{2}z \mathrm{dA}
0
z=1-x-y
\left(\mathbf{i}+\mathbf{j}+\mathbf{k}\right)/\sqrt{3}
\frac{\left(x {y}^{3}+y z+{x}^{2}z\right)}{\sqrt{3}}\left(\sqrt{3} \mathrm{dA}\right)
x {y}^{3}+\left(y+{x}^{2}\right)\left(1-x-y\right)
Table 9.8.10(a)
\mathbf{F}·\mathbf{N} \mathrm{dσ}
on the four faces of
R
The only face through which there is nonzero flux is the plane
z=1-x-y
. This flux is given by
{∫}_{0}^{1}{∫}_{0}^{1-x}\left(x {y}^{3}+\left(y+{x}^{2}\right)\left(1-x-y\right)\right) \mathrm{dy} \mathrm{dx}
\frac{1}{15}
The total flux through the boundaries of
R
matches the volume integral of the divergence.
〈x {y}^{3},y z,{x}^{2}z〉
\left[\begin{array}{c}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{3}}\\ \textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{z}\\ {\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{z}\end{array}\right]
\stackrel{\text{to Vector Field}}{\to }
\left[\begin{array}{c}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{3}}\\ \textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{z}\\ {\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{z}\end{array}\right]
\stackrel{\text{assign to a name}}{\to }
\textcolor[rgb]{0,0,1}{F}
∇·\mathbf{F}
\mathrm{divF}
∇·\mathbf{F}
{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{z}
\stackrel{\text{assign to a name}}{\to }
\textcolor[rgb]{0,0,1}{\mathrm{divF}}
Write the name given to
∇·\mathbf{F}
Complete the dialogs as shown in the two figures below.
\mathrm{divF}
{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{z}
\stackrel{\text{MultiInt}}{\to }
{\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{∫}}_{\textcolor[rgb]{0,0,1}{0}}^{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{∫}}_{\textcolor[rgb]{0,0,1}{0}}^{\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{x}}{\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{∫}}_{\textcolor[rgb]{0,0,1}{0}}^{\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{y}}\left({\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{z}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{ⅆ}\textcolor[rgb]{0,0,1}{z}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{ⅆ}\textcolor[rgb]{0,0,1}{y}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{ⅆ}\textcolor[rgb]{0,0,1}{x}
=
\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{15}}
There are four parts to the boundary of
R
, the coordinate planes and the plane
z=1-x-y
For the flux through the upper surface, use a task template.
Calculus - Vector≻Integration≻Flux≻3-D≻Through a Surface Defined over a Triangle
Flux through a Surface Defined over a Triangle
On the three coordinate planes,
\mathbf{F}·\mathbf{N} \mathrm{dσ}
vanishes, so all the flux is accounted for. Hence, the flux through
R
\mathrm{with}\left(\mathrm{Student}:-\mathrm{VectorCalculus}\right):
\mathrm{BasisFormat}\left(\mathrm{false}\right):
\mathbf{F}≔\mathrm{VectorField}\left(〈x {y}^{3},y z,{x}^{2}z〉\right):
∇·\mathbf{F}
\mathrm{divF}≔\mathrm{Divergence}\left(\mathbf{F}\right)
{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{z}
R
\mathrm{int}\left(\mathrm{divF},\left[x,y,z\right]=\mathrm{Region}\left(0..1,0..1-x,0..1-x-y\right),\mathrm{output}=\mathrm{integral}\right)
{\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{∫}}_{\textcolor[rgb]{0,0,1}{0}}^{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{∫}}_{\textcolor[rgb]{0,0,1}{0}}^{\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{x}}{\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{∫}}_{\textcolor[rgb]{0,0,1}{0}}^{\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{y}}\left({\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{z}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{ⅆ}\textcolor[rgb]{0,0,1}{z}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{ⅆ}\textcolor[rgb]{0,0,1}{y}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{ⅆ}\textcolor[rgb]{0,0,1}{x}
\mathrm{int}\left(\mathrm{divF},\left[x,y,z\right]=\mathrm{Region}\left(0..1,0..1-x,0..1-x-y\right)\right)
\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{15}}
Use the Flux command to obtain the flux of F through the plane
z=1-x-y
\mathrm{Flux}\left(\mathbf{F},\mathrm{Surface}\left(〈x,y,1-x-y〉,\left[x,y\right]=\mathrm{Triangle}\left(〈0,0〉,〈0,1〉,〈1,0〉\right)\right),\mathrm{output}=\mathrm{integral}\right)
{\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{∫}}_{\textcolor[rgb]{0,0,1}{0}}^{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{∫}}_{\textcolor[rgb]{0,0,1}{0}}^{\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{x}}\left(\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{y}}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{y}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{y}\right)\textcolor[rgb]{0,0,1}{+}{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{y}\right)\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{ⅆ}\textcolor[rgb]{0,0,1}{y}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{ⅆ}\textcolor[rgb]{0,0,1}{x}
\mathrm{Flux}\left(\mathbf{F},\mathrm{Surface}\left(〈x,y,1-x-y〉,\left[x,y\right]=\mathrm{Triangle}\left(〈0,0〉,〈0,1〉,〈1,0〉\right)\right)\right)
\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{15}}
\mathbf{F}·\mathbf{N} \mathrm{dσ}
R
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Binomial Theorem | Brilliant Math & Science Wiki
Patrick Corn, Hua Zhi Vee, Kishlaya Jaiswal, and
The binomial theorem (or binomial expansion) is a result of expanding the powers of binomials or sums of two terms. The coefficients of the terms in the expansion are the binomial coefficients
\binom{n}{k}
. The theorem and its generalizations can be used to prove results and solve problems in combinatorics, algebra, calculus, and many other areas of mathematics.
The binomial theorem generalizes special cases which are common and familiar to students of basic algebra:
\begin{aligned} (x+y)^1 &= x+y \\ (x+y)^2 &= x^2 + 2xy + y^2 \\ (x+y)^3 &= x^3 + 3x^2y+3xy^2+y^3 \\ (x+y)^4 &= x^4 + 4x^3y + 6x^2y^2+4xy^3+y^4 \\ &\vdots \\ \end{aligned}
The binomial theorem also helps explore probability in an organized way:
A friend says that she will flip a coin 5 times. Each time the coin comes up heads, she will give you $10, but each time the coin comes up tails, she gives nothing. What is the probability that you will win $30 playing this game?
The binomial theorem inspires something called the binomial distribution, by which we can quickly calculate how likely we are to win $30 (or equivalently, the likelihood the coin comes up heads 3 times). The binomial theorem tells us that
{5 \choose 3} = 10
2^5 = 32
possible outcomes of this game have us win $30. Therefore, the probability we seek is
\frac{5 \choose 3}{2^5} = \frac{10}{32} = 0.3125.\ _\square
n
x
y
real numbers (or complex numbers, or polynomials). The coefficient of
x^k y^{n-k}
k^\text{th}
(x+y)^n
\binom{n}{k}
\binom{n}{k}=\frac{n!}{(n-k)!k!}.
(x+y)^n = \sum_{r=0}^n {n \choose r} x^{n-r} y^r = \sum_{r=0}^n {n \choose r} x^r y^{n-r}.\ _\square
The above expansion is known as binomial expansion.
The binomial theorem states that for any positive integer
n
\begin{aligned} (x+y)^n &= \binom{n}{0}x^n+\binom{n}{1}x^{n-1}y+ \cdots +\binom{n}{n-1}xy^{n-1}+\binom{n}{n}y^n \\ \\ &= \sum\limits_{k=0}^{n}\binom{n}{k}x^{n-k}y^k. \end{aligned}
We can prove it by combinatorics:
n
n
a^{n-k}b^k
k
n
a^{n-k}b^k
\binom{n}{k}
_\square
Or we can also prove it by induction:
n = 1
is immediate. Now suppose the theorem is true for
(x+y)^{n-1}
\begin{aligned} (x+y)^n &= (x+y)(x+y)^{n-1} \\ &= (x+y)\bigg(\binom{n-1}{0} x^{n-1} + \binom{n-1}{1} x^{n-2}y + \cdots + \binom{n-1}{n-1}y^{n-1}\bigg) \\ &= x^n + \left( \binom{n-1}{0} + \binom{n-1}{1} \right) x^{n-1}y + \left( \binom{n-1}{1} + \binom{n-1}{2} \right) x^{n-2}y^2 \phantom{=} + \cdots + \left(\binom{n-1}{n-2} + \binom{n-1}{n-1} \right) xy^{n-1} + y^n \\ \end{aligned}
and now Pascal's identity applies:
\binom{n-1}{k-1}+\binom{n-1}{k} = \binom{n}{k}.
So the right side simplifies to
x^n + \binom{n}{1} x^{n-1}y + \binom{n}{2} x^{n-2}y^2 + \cdots + \binom{n}{n-1}xy^{n-1} + y^n
_\square
x^4
(x+1)^9
4^\text{th}
term is equal to
\binom{9}{4}=\frac{9!}{(9-4)!4!}=126
. Therefore, the
4^\text{th}
term of the expansion is
126\cdot x^4\cdot 1 = 126x^4
, where the coefficient is
126
_\square
(2x+\frac{k}{x})^8
k
x
700000
k.
2^n = \sum_{k=0}^n {n\choose k}.
x=y=1
in the binomial series to get
(1+1)^n = \sum_{k=0}^n {n\choose k} (1)^{n-k}(1)^k \Rightarrow 2^n = \sum_{k=0}^n {n\choose k}.\ _\square
The following problem has a similar solution. Hint: try
x=1
y = i
\sum_{k = 0}^{49} (-1)^k {99 \choose 2k}
a^b
a, b
b
is as large as possible, what is
a+b?
Now try the following problem:
x^{3}y^{13}
(x+y)^{16}?
x^2y^2z^2
(x+y+z)^6?
The power rule in differential calculus can be proved using the limit definition of the derivative and the binomial theorem.
\big(
To find the derivative of
x^n
, expand the expression
\frac{(x+h)^n-x^n}{h} = \binom{n}{1}x^{n-1} + \binom{n}{2} x^{n-2}h + \cdots + \binom{n}{n} h^{n-1}
and take the limit as
h \to 0
. All the terms except the first term vanish, so the answer is
n x^{n-1}.\big)
The general proof of the principle of inclusion and exclusion involves the binomial theorem. Recall that the principle states that for finite sets
A_i \ (i = 1,\ldots,n)
\begin{aligned} \left| \bigcup_{i=1}^n A_i \right| &= \sum |A_i| - \sum |A_i \cap A_j| + \sum |A_i \cap A_j \cap A_k| \phantom{=} - \cdots + (-1)^{n-1} |A_1 \cap A_2 \cap \cdots \cap A_n|, \end{aligned}
where the sums on the right side are taken over all possible intersections of distinct sets.
Suppose an element in the union appears in
d
A_i
. Then it contributes
d
to the first sum,
-\binom{d}{2}
to the second sum, and so on, so the total contribution is
\sum_{i=1}^d (-1)^{i-1} \binom{d}{i} = 1 - \sum_{i=0}^d (-1)^i \binom{d}{i},
but the last sum is equal to
(1-1)^d = 0
by the binomial theorem. So each element in the union is counted exactly once.
The fact that the Möbius function
\mu
is the Dirichlet inverse of the constant function
\mathbf{1}(n) = 1
is a consequence of the binomial theorem; see here for a proof.
p
p
divides all the binomial coefficients
\binom{p}{k}
1 \le k \le p-1
. (There is a
p
in the numerator but none in the denominator.) So
(x+y)^p \equiv x^p + y^p \pmod p.
This fact is quite useful and has some rather fruitful generalizations to the theory of finite fields, where the function
x \mapsto x^p
is called the Frobenius map. This fact (and its converse, that the above equation is always true if and only if
p
is prime) is the fundamental underpinning of the celebrated polynomial-time AKS primality test.
\left ( \sqrt {71} +1 \right )^{71} - \left ( \sqrt {71} -1 \right )^{71}
The theorem as stated uses a positive integer exponent
n
. It turns out that there are natural generalizations of the binomial theorem in calculus, using infinite series, for any real exponent
\alpha
(1+x)^\alpha = \sum_{k=0}^{\infty} \binom{\alpha}{k} x^k
|x|<1
\binom{\alpha}{k} = \frac{\alpha(\alpha-1)\cdots(\alpha-k+1)}{k!}.
Some special cases of this result are examined in greater detail in the Negative Binomial Theorem and Fractional Binomial Theorem wikis.
(x+y)^n
n
\begin{aligned} (x+y)^0 &=& 1 \\ (x+y)^1 &=& x+y \\ (x+y)^2 &=& x^2 + 2xy + y^2 \\ (x+y)^3 &=& x^3 + 3x^2y + 3xy^2 + y^3 \\ (x+y)^4 &=& x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4 \\ &\vdots \end{aligned}
When we look at the coefficients in the expressions above, we will find the following pattern:
1\\ 1\quad 1\\ 1\quad 2 \quad 1\\ 1\quad 3 \quad 3 \quad 1\\ 1\quad 4 \quad 6 \quad 4 \quad 1\\ 1 \quad 5 \quad 10 \quad 10 \quad 5 \quad 1\\ \vdots
The theorem identifies the coefficients of the general expansion of
(x+y)^n
as the entries of Pascal's triangle.
x^{3}y^{13}
(x+y)^{16}?
x^2y^2z^2
(x+y+z)^6?
Cite as: Binomial Theorem. Brilliant.org. Retrieved from https://brilliant.org/wiki/binomial-theorem-n-choose-k/
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Reporting Merchandise Inventory in Financial Statements - Course Hero
Principles of Accounting/Inventory/Reporting Merchandise Inventory in Financial Statements
Lower-of-Cost-or-Market Value of Inventory
The principle of conservatism is important as it applies to determining the carrying value of the inventory account. Under conservatism, assets and revenues are not overstated, and liabilities and expenses are not understated. To ensure that financial information is reported fairly and is a true representation of a company's position, businesses review amounts to determine if they are current.
Regarding the inventory account, a business must report the value of its inventory each reporting period, as values can change due to factors such as obsolescence, deterioration, or price drop. The accountant determines if the original cost of the inventory is worth at least its carrying costs, or the costs of keeping and maintaining the inventory. If it is not, then an adjustment is needed.
The lower-of-cost-or-market (LCM) method compares the value or cost of inventory versus current market price, and the lower price is recorded. The LCM valuation method is used to determine if the carrying value or purchase cost of the inventory is too high compared to the current market price. The accountant must determine if the inventory could be acquired at the same or higher cost, or if the value of the inventory item has decreased in the marketplace. If the market price is lower than the cost paid for the item, then a journal entry must be made to lower the cost down to the current marketplace cost. However, if the market's cost is higher, nothing is done.
Determining LCM involves calculating the net realizable value (NRV), which is the current value of the inventory in the market as it relates to the estimated selling price minus all direct costs of disposing of the items. As a simplified example, the original cost of an item is $10, however, the current market price of the item is $8, and the estimated selling expenses are $1.50. Thus, the net realizable value is $6.50
{\left(\$8.00 - \$1.50 = \$6.50\right)}
. So, when comparing the original cost of $10 with the market value—net realizable value of $6.50—the company must make an adjustment to decrease the cost of its inventory down to $6.50 per item. The $6.50 is the lower-of-cost-versus-market. If the figures were reversed where the cost was $6.50, but the market price was $10, then the company would leave the inventory at the cost at $6.50.
Another factor in determining LCM involves comparing the replacement cost of an item in inventory with the net realizable value (NRV) and carrying value. The current market price is the replacement cost if it falls within the market ceiling and the market floor. The market ceiling is the NRV, and the market floor is the NRV minus the ordinary profit of the product. Once the market price is selected, the lower of the carrying value or market price is chosen. If the market price is lower, the inventory value is written down to the market price.
Errors on the Income Statement and Balance Sheet
Because the inventory account is updated in real time and all purchases and sales are recorded in the inventory account when using a perpetual inventory system, an organization can discover errors as they occur. While the goal is always to minimize and mitigate errors, a company will want to have a system in place to address these errors if they occur. Any errors in the inventory account will affect both the balance sheet (inventory) and the income statement (COGS). Remember: an error is an accident, while intentional malicious misconduct is fraud. There are several common errors:
Miscounting occurs during the physical inventory.
The FIFO, LIFO, or weighted average cost valuation method is incorrectly used.
Inventory still in transit is included or excluded incorrectly.
Consigned inventory, which is inventory shipped to another seller who acts as selling agent, is included or excluded incorrectly.
As a review, the COGS account has an inverse relationship with the ending inventory account. Therefore, if ending inventory is overstated, then COGS will be understated, and if ending inventory is understated, then COGS will be overstated. If COGS is too low, then gross profit will be too high. If COGS is too high, then gross profit will be too low. One error can create a domino effect on the financial statements.
Effect of COGS on Income Statement
If Ending
Overstated …
Understated …
Sales Same Same
Cost of Goods Sold Understated Overstated
Gross Profit/Net Income Overstated Understated
Similarly, if the ending inventory is overstated, total assets would also be overstated on the balance sheet. Because the net income flows into the equity section of the balance sheet, the equity section would be overstated as well.
Effect of Ending Inventory on Balance Sheet
Cash Same Same
Inventory Overstated Understated
Current Assets/Total Assets Overstated Understated
Owner's Equity Overstated Understated
<Comparing Inventory Costing Methods>Suggested Reading
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<< 3.7.116. Hamiltonian3.7.118. Heuristics and Berge-acyclic constraint >>
\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}
\mathrm{𝚍𝚒𝚜𝚌𝚛𝚎𝚙𝚊𝚗𝚌𝚢}
\mathrm{𝚒𝚗𝚟𝚎𝚛𝚜𝚎}
\mathrm{𝚒𝚗𝚟𝚎𝚛𝚜𝚎}_\mathrm{𝚘𝚏𝚏𝚜𝚎𝚝}
\mathrm{𝚒𝚗𝚟𝚎𝚛𝚜𝚎}_\mathrm{𝚠𝚒𝚝𝚑𝚒𝚗}_\mathrm{𝚛𝚊𝚗𝚐𝚎}
A constraint that was introduced for expressing a heuristics or a constraint (
\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}
) for which an algorithm that evaluate the number of solutions was proposed.
Remark: when we do not have good bounds on the cost variable of a constrained optimisation problem, skewed binary search was introduced in [SellmannKadioglu08] in order to take advantage of the fact that it is usually easier to improve the current solution cost's than to prove that a problem is not feasible.
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Arithmetic Progressions | Brilliant Math & Science Wiki
Sandeep Bhardwaj, Ram Mohith, Mei Li, and
Abhineet Goel
Lucerne O' Brannan
An arithmetic progression (AP), also called an arithmetic sequence, is a sequence of numbers which differ from each other by a common difference. For example, the sequence
2, 4, 6, 8, \dots
is an arithmetic sequence with the common difference
2
We can find the common difference of an AP by finding the difference between any two adjacent terms.
The following sequence is an AP with common difference 5 and initial term 0:
\LARGE \color{#3D99F6}{0} \underbrace{\quad \quad }_{5} \color{#D61F06}{5} \underbrace{\quad \quad }_{5} \color{#20A900}{10} \underbrace{\quad \quad }_{5} \color{cyan}{15} \underbrace{\quad \quad }_{5} \color{orangered}{20} \underbrace{\quad \quad }_{5} \color{grey}{25}
Describing Arithmetic Progressions
Sums of Arithmetic Progressions
Properties of Arithmetic Progressions
Initial term: In an arithmetic progression, the first number in the series is called the "initial term."
Common difference: The value by which consecutive terms increase or decrease is called the "common difference."
We can describe an arithmetic sequence with a recursive formula, which specifies how each term relates to the one before. Since in an arithmetic sequence, each term is given by the previous term with the common difference added, we can write a recursive description as follows:
\text{Term} = \text{Previous term} + \text{Common Difference.}
More concisely, with the common difference
d
a_n=a_{n-1}+d.
If we know the initial term, the following terms are related to it by repeated addition of the common difference. Thus, the explicit formula is
\text{Term} = \text{Initial Term} + \text{Common Difference} \times \text{Number of steps from the initial term}.
We can write this with common difference
d
a_n = a_1 + d(n-1).
What is the sequence described by
a_n = 2 + 4(n-1)
2, 6, 10, 14, \dots
We can see from the explicit formula that the initial term is 2 and the common difference is 4.
What is the explicit formula for the arithmetic progression
3, 6, 9, 12, \dots
Using the form given above, we have an initial term,
a_1=3
, and a common difference,
d
, of 3. Thus,
a_n = 3 + 3(n-1)
Note that we can simplify this expression to
a_n=3+3n-3=3n
What is the seventh term of the arithmetic progression
2, 7, 12, 17, \dots
5^\text{th}
6^\text{th}
-10
15
15^{\text{th}}
Sum of terms: The sum of the first
n
terms of an AP with initial term
and common difference
d
S_n=\frac n2 \big[ 2a+(n-1)d\big] \qquad \text{or} \qquad S_n = \dfrac{n}{2} \big[T_1 + T_n\big] \qquad \text{or} \qquad S_n = n \times (\text{middle term}).
Let us assume that the sum of the first 100 positive integers is
S
S=1+2+3+ \cdots +98+99+100.
Now writing this expression in reverse order gives
S=100+99+98+ \cdots +3+2+1.
On adding the above two values, we get
\begin{aligned} 2S&=(1+100)+(2+99)+(3+98)+\cdots +(98+3)+(99+2)+(100+1)\\ &=(101)+(101)+(101)+ \cdots + (101)+(101)+(101)\\ &=101 \times 100\\ \\ \Rightarrow S&=\dfrac{101 \times 100}{2}\\&=101 \times 50\\&=5050.\ _\square \end{aligned}
From the example above, notice that the sums of corresponding terms in
S=1+2+3+ \cdots +98+99+100
S=100+99+98+ \cdots +3+2+1
are all the same, i.e.
1+100=2+99=3+98= \cdots =50+51=51+50=\cdots =98+3=99+2=100+1.
The sum of the terms of an AP can be found manually by adding all the terms, but this can be a very tedious process. Based on the above property possessed by an AP, there is a generalized formula for the sum of an AP.
For an arithmetic progression with initial term
a_1
d
n
S_n = \frac{n}{2}\big[2a_1+(n-1)d\big].
What is the sum of the first 50 odd positive integers?
We recognize that this is an arithmetic sequence with common difference
2
1.
We can then use the formula to obtain
S_n = \frac{1}{2}n\big(2a_1+(n-1)d\big) \implies S_{50} =25\times(2+49\times2) = 2500. \ _\square
Note: We can generalize that the sum of the first
n
odd positive integers is
\begin{aligned} S_n &= \frac{ a_1 + a_n } { 2} \times n \\ &= \frac{ 1 + \big(1 + (n-1)\cdot 2\big)} { 2} \times n \\ &= n^2. \end{aligned}
\large 13 + 28 + 43 + \cdots + a_n = 68210
n
n?
Increasing/Decreasing Sequence:
If the common difference is positive, i.e.
d>0,
then the arithmetic progression is an increasing sequence and satisfies the condition
a_{n-1}<a_n:
a_1 < a_2 < a_3 < \cdots .
For example, the arithmetic progression with initial term
2
3,
2,5,8,11,14, \dots,
is an increasing sequence.
If the common difference is negative, i.e.
d<0,
then the arithmetic progression is a decreasing sequence and satisfies the condition
a_{n-1}>a_n:
a_1 > a_2 > a_3 > \cdots .
As an example, the arithmetic progression with initial term
1
-3,
1, -2, -5, -8, -11, \dots,
is a decreasing sequence.
a,b,c
are in AP, then
2b = a + c.
If each term of an AP is increased, decreased, multiplied, or divided by a constant non-zero number, then the resulting sequence is also in AP.
n^\text{th}
term of any sequence is of the form
an + b
, then the sequence is in AP where the common difference is
a
Geometrical Interpretations:
a_n
n^{\text{th}}
term of a the sequence
1, 2, 3, 4, 5, \dots
, corresponding points can be drawn
(n,a_n)
in the Cartesian coordinate system as follows:
All the points
(n,a_n)
are collinear, meaning a straight line can be formed by joining all the points. The result looks like this:
Slope of line: The slope of the line is equal to the common difference of the AP, i.e.
d
. In the above case, the slope of the line
1
is the same as its common difference.
S_{1729 } = S_{29} ,
S_n
n
S_{1758}.
\frac{100001+100003+100005+\cdots+199999}{1+3+5+7+\cdots+99999} = \, ?
99
99!
100
100!
Consider an arithmetic progression whose first term and common difference are both 100. If the
n^\text{th}
term of this progression is equal to
100!
n.
!
8! = 1\times2\times3\times\cdots\times8
5.05 \text{ km}
9.9 \text{ km}
10.1 \text{ km}
20.2 \text{ km}
You are standing next to a bucket and are tasked with collecting
100
potatoes, but can only carry one potato at a time. The potatoes are in a line in front of you, with the first potato
1
meter away and each subsequent potato is located an additional one meter away.
How much distance would you cover while performing this task?
Cite as: Arithmetic Progressions. Brilliant.org. Retrieved from https://brilliant.org/wiki/arithmetic-progressions/
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Machine Learning A-Z · A Blogger who is a Data Science, Web Developer, Game Developer and an enthusiast for anything Tech related.
1.1 Applications of ML
1.2 ML is the future
2.1 Get the data sets
2.4 Importing the dataset
2.4.1 Setting the working directory in spyder
Taking care of the missing data
2.8 Train set/Test set splitting
Facebook image/face tagging.
Kinect motion detection.
VR headset movement.
Speech to text, text to speech.
Swipe keyboard prediction.
IOT robot dogs.
Amazon netflix use ml to recommender systems.
Used in the field of medicine for detection.
Used to explore new areas through space satellites.
Explore new areas like mars.
Since the dawn of time up until 2005 humans have created 130 Exabytes of data.
2005 - 2010 that has become 1200 Exabytes of data.
2010 - 2015 it is 7900 Exabytes.
By 2020, 40900 Exabytes of data will be created.
Install Anaconda 4.2.0 if you are facing some compatibility issues.
The first data set contains 4 columns country, age, salary and purchased
So it’s a data set of the customers of a company with the customers information and whether or not the purchased their product. The first 3 are the independent variables/features. And the last column is the dependent variable/label which is what we need to predict.
Let’s begin by importing necessary libraries.
import numpy as np # n-dim array math library
import matplotlib.pyplot as plt # plotting/charting library
import pandas as pd # importing data sets and managing data sets
Go to file explorer - click on a button to set a folder as working dir or save and run the py file from the same folder as CSV file to set it as working dir (F5 to run)
dataset = pd.read_csv(‘data.csv’) # importing the CSV file
You can verify the above import statement by looking at the variable explorer in spyder. Now let’s create our matrix of features.
X = dataset.iloc[:, :-1].values # the left side of the comma is the rows to include and the right side it the columns and we have excluded the last colum as its the y and not x
Y = dataset.iloc[:, 3] #indexes in python start at 0 so we need the 4th column so we enter 3
We cant remove the rows containing missing data because it may contain other crucial data that we might need the other safer method is to fill the missing data using any of the 3 methods Mean , Median or Mode if you cant see the full array in the variable explorer just add the below code
np.set_printoptions(threshold = np.nan)
imputer = Imputer(missing_values = NaN, strategy=“mean”, axis = 0) # missing values is NaN because that is what shows up for the missing values in the variable explorer, axis 0 means along the columns
imputer = imputer.fit(X[:, 1:3]) # its 3 because the upper bound is excluded here
X[:, 1:3] = imputer.transform(X[:, 1:3]) # Now we take the transformed values and replace into the actual data
To inspect any imported method or class, keep the cursor over it and press cmd + i to open up the docs for that method/class. There are 3 strategies available to fill the missing values. mean, median and most_frequent. axis = 0 means fill along the columns, axis = 1 means fill along the rows. We fit the X data to the imputer for the columns where there are missing data. Hence X[:, 1:3] which means all the rows and columns 1 and 2. To run just a lock of code in spyder just select the lock of new code and hit cmd + enter.
We use the fit_transform method because we want to fit and transform at the same time. Now that we have converted the categorical values of france, germany and spain into numbers 0, 1, 2, We now need to convert it to one hot encoding. As currently the ml algorithm will think Germany is greater than France and Spain is greater than Germany which is not the case.
So lets import the OneHotEncoder from sklearn.preprocessing
from sklearn.preprocessing import Imputer, OneHotEncoder
onehotencoder = OneHotEncoder(categorical_features = [0]) # The columns which needs to be transformed
Now lets do the Label encoding to Y
# Splitting the dataset into train set and test set
X_{\text{stand}} = {{X - \text{mean}(X)} \over \text{standard deviation}(X)}
X_{\text{norm}} = {X - \text{min}(X) \over \text{max}(X) - \text{min}(X)}
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3 Ways to Calculate Interest Expense - wikiHow
3 Posting an Accounting Entry
Interest expense is an expense you incur when you borrow money. Your lender charges you a specific interest rate that is stated in your loan document. As time passes, you are charged interest on the amount that you borrowed. You may need to calculate simple interest or compound interest on your loan, depending on how the loan is structured. If your loan is for business, you will post interest expense to your accounting records.
Understand the interest expense formula. The formula to calculate interest is Interest = Prt where "P" equals Principal, or the amount of the loan outstanding, "r" equals the rate of interest charged, and "t" equals the amount of time that the loan will be outstanding. Your principal is the loan balance that is still owed to the lender. Your rate of interest will be stated in the loan agreement.[1] X Research source
Gather your loan information. Assume that the remaining balance on your loan is $10,000. The interest rate stated on your loan agreement is 12% annually. Your interest expense for a 12 month or 1 year period would be $10,000 multiplied by 12%, or $1,200.
Adjust the period of time in your calculation. You may need to calculate interest expense for a fraction of a year. Use the same interest rate formula and adjust the time period.[2] X Research source
Determine the period of time to properly recognize interest expense. If you need to calculate three months of interest expense, divide the 12 months by the number of months in the desired period. For example, t = 3/12 or 0.25. If the principal is $10,000 and the interest rate is 12 percent (0.12), the calculation would be Interest = $10,000 x 0.12 x 0.25 or $300.
Use a spreadsheet to calculate interest. You can speed up this process by using a spreadsheet application. label columns A,B.C and D "Principal", "Rate", "Time", and "Interest". In cell A2, insert the Principal amount or $10,000. In cell B2, insert the rate of interest or .12. In cell C2, insert the time where a year or 12 months equals 1. Use multiples and fractions where the period is less or greater than one year. In cell D2, enter the formula (A2*B2*C2).
Understand compound interest. Compound interest is the combination of interest earned on the principal during the current period and all previous interest earned but not paid for previous periods. In other words, you earn "interest on interest". With simple interest, the lender only earns interest on the principal amount per annum. For periods of less than year, the lender will receive a pro rata share of the annual interest. For compound interest, all prior interest is added to the principal amount, effectively increases the annual rate earned by the lender. The use of compound interest and the compounding frequency are established in the loan documents.
Learn the compound interest formula. The formula to calculate compound interest in a year is:
{\displaystyle compoundinterest=(P(1+i)^{n}-P)}
. In the formula, the variables stand for the following:
P: Principal, the initial amount borrowed or deposited.
i: Interest rate in percentage. The interest rate must be adjusted for the number of compounding periods in a year by dividing the interest rate by the number of periods. For example, a 12 percent interest rate compounded monthly would be adjusted to 1 percent (12/12).
n: Number of Compounding periods in a year multiplied by the loan period in years. Semi-annual compounding is 2, quarterly is 4, monthly is 12, and daily is 360 or 365 depending upon the financial institution.
Find the interest rate. Solve for i in the equation. This is the interest rate adjusted for the compounding periods per year. For example, a 10 percent loan compounded monthly would have an interest rate of 10/12 or 0.00833. (1 + i) in the formula above would be (1 + .00833), or 1.00833.
Solve for the number of compound periods over the term of the loan. Multiply the number of periods per year by the duration of the loan. For example, a 10 year year loan would have 120 periods (10*12).
Solve for the future value of the loan. This is represented by the figure
{\displaystyle P(1+i)^{n}}
in the formula. Raise the figure is parentheses to the power of n (the number of periods), then multiply by the principal. This will give you the future value of the account.
For example, the future value of a $10,000 loan compounded monthly at 10% interest would be $27,059.
Subtract the principal to get interest expense. Subtract the present value of the loan (the principal) from the future value to arrive at the interest paid over the full ten year period: $27,059 - $10,000 = $17,059.Note that the actual amount may vary depending upon the number of decimal places in the interest rate.
Calculating the compound interest rate for a single year uses the same formula adjusting for the principal amount each year. For example, the first year’s interest on a $10,000 loan compounded monthly at 10% would be $1,047. To calculate the second year’s interest, add the previous year’s interest ($1,047) to the principal amount ($10,000) and recalculate. Interest for the second year would be $1,156.
There are a number of free compound interest calculators on the Internet.
Posting an Accounting Entry Download Article
Determine if your interest expense is business related. If you borrowed money to operate a business, interest expense should be posted to your accounting records. This applies to both simple interest and compound interest calculations.
Use the accrual method of accounting. The accrual method recognizes revenue when it is earned and expenses when they are incurred. This accounting method matches revenue with expenses. It is a better indicator of company profit than the cash method of accounting. The cash method recognizes revenue when cash is received. Expenses are posted when cash is paid.[3] X Research source
Generally Accepted Accounting Principles (GAAP) requires most businesses to use the accrual method of accounting. Companies of all sizes can benefit from conforming with GAAP.
Posting interest expense entries. Assume, for example, that you need to post $100 of interest for the month of December. Also assume that you will pay the December interest on January 15th of the following year.
On the last day of the month (December 31st), you would post one month of interest as interest expense.
Your accounting entry on 12/31 is to debit (increase) interest expense $100 and to credit (increase) interest payable $100.
On January 15th, when interest is paid, you would debit (reduce) interest payable $100 and credit (reduce) cash $100.
The cash method would require you to post the interest expense when cash is paid (January 15th). The $100 of interest expense should be posted in December, since that is the time period when the interest was incurred.
Interest expense is posted to the income statement with other expenses. The income statement formula is (revenue less expenses equals net income).
Your business may have separate interest payable accounts set up for each loan account. Check with the manager of the accounting department to learn the specific policies and procedures.
To calculate interest expense, locate the balance of your loan and the interest rate specified on your loan agreement, which will be a percentage. Next, convert the percentage to decimal format by putting a decimal after the number and moving the decimal 2 places to the left. Then, multiply the loan amount by the decimal to get the total interest. Finally, divide your answer by 12 to calculate monthly interest. For tips on using a spreadsheet to make your calculations, read on!
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Find the co-ordinates of the orthocentre of triangle ABC whose vertices are A(1,2), B(2,3) and C(4,3) - Maths - Coordinate Geometry - 6931128 | Meritnation.com
Find the co-ordinates of the orthocentre of triangle ABC whose vertices are A(1,2), B(2,3) and C(4,3)
Prasanta answered this
Slope of AB=
\frac{3-2}{2-1}=1
Slope of BC=
\frac{3-3}{4-2}=0
Slope of CA=
\frac{3-2}{4-1}=\frac{1}{3}
Let AD, BE and CF be the altitudes in
△ABC
Slope of AD=
-\frac{1}{Slope\quad of\quad BC}=-\frac{1}{0}
Slope of BE=
-\frac{1}{Slope\quad of\quad CA}=-\frac{1}{{\displaystyle \frac{1}{3}}}=-3
Slope of CF=
-\frac{1}{Slope\quad of\quad AB}=-\frac{1}{1}=-1
We now have the vertices and slopes of AD, BE and CF. Let (x, y) be the orthocentre. So, by the equation
y-{y}_{1}=m\left(x-{x}_{1}\right)
, we have:-
AD:\quad y-2=-\frac{1}{0}\left(x-1\right)\quad \quad \Rightarrow \quad x=1\quad \quad \quad \quad [the\quad line\quad is\quad vertical]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}BE:\quad y-3=-3\left(x-2\right)\quad \quad \Rightarrow y-3=-3x+6\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}CF:\quad y-3=-1\left(x-4\right)\quad \quad \quad \Rightarrow y-3=-x+4
Solving these equations, we find x=1, y=6
Thus, the required orthocentre is (1, 6).
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Global Constraint Catalog: Csymmetric_alldifferent_except_0
<< 5.395. symmetric_alldifferent5.397. symmetric_alldifferent_loop >>
\mathrm{𝚜𝚢𝚖𝚖𝚎𝚝𝚛𝚒𝚌}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}
\mathrm{𝚜𝚢𝚖𝚖𝚎𝚝𝚛𝚒𝚌}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}_\mathrm{𝚎𝚡𝚌𝚎𝚙𝚝}_\mathtt{0}\left(\mathrm{𝙽𝙾𝙳𝙴𝚂}\right)
\mathrm{𝚜𝚢𝚖𝚖𝚎𝚝𝚛𝚒𝚌}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏}_\mathrm{𝚎𝚡𝚌𝚎𝚙𝚝}_\mathtt{0}
\mathrm{𝚜𝚢𝚖𝚖𝚎𝚝𝚛𝚒𝚌}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚜𝚝𝚒𝚗𝚌𝚝}_\mathrm{𝚎𝚡𝚌𝚎𝚙𝚝}_\mathtt{0}
\mathrm{𝚜𝚢𝚖𝚖}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}_\mathrm{𝚎𝚡𝚌𝚎𝚙𝚝}_\mathtt{0}
\mathrm{𝚜𝚢𝚖𝚖}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏}_\mathrm{𝚎𝚡𝚌𝚎𝚙𝚝}_\mathtt{0}
\mathrm{𝚜𝚢𝚖𝚖}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚜𝚝𝚒𝚗𝚌𝚝}_\mathrm{𝚎𝚡𝚌𝚎𝚙𝚝}_\mathtt{0}
\mathrm{𝙽𝙾𝙳𝙴𝚂}
\mathrm{𝚌𝚘𝚕𝚕𝚎𝚌𝚝𝚒𝚘𝚗}\left(\mathrm{𝚒𝚗𝚍𝚎𝚡}-\mathrm{𝚒𝚗𝚝},\mathrm{𝚜𝚞𝚌𝚌}-\mathrm{𝚍𝚟𝚊𝚛}\right)
\mathrm{𝚛𝚎𝚚𝚞𝚒𝚛𝚎𝚍}
\left(\mathrm{𝙽𝙾𝙳𝙴𝚂},\left[\mathrm{𝚒𝚗𝚍𝚎𝚡},\mathrm{𝚜𝚞𝚌𝚌}\right]\right)
\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚒𝚗𝚍𝚎𝚡}\ge 1
\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚒𝚗𝚍𝚎𝚡}\le |\mathrm{𝙽𝙾𝙳𝙴𝚂}|
\mathrm{𝚍𝚒𝚜𝚝𝚒𝚗𝚌𝚝}
\left(\mathrm{𝙽𝙾𝙳𝙴𝚂},\mathrm{𝚒𝚗𝚍𝚎𝚡}\right)
\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚜𝚞𝚌𝚌}\ge 0
\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚜𝚞𝚌𝚌}\le |\mathrm{𝙽𝙾𝙳𝙴𝚂}|
Enforce the following three conditions:
\forall i\in \left[1,|\mathrm{𝙽𝙾𝙳𝙴𝚂}|\right]
\forall j\in \left[1,|\mathrm{𝙽𝙾𝙳𝙴𝚂}|\right]
j\ne i
\mathrm{𝙽𝙾𝙳𝙴𝚂}\left[i\right].\mathrm{𝚜𝚞𝚌𝚌}=0\vee \mathrm{𝙽𝙾𝙳𝙴𝚂}\left[j\right].\mathrm{𝚜𝚞𝚌𝚌}=0\vee \mathrm{𝙽𝙾𝙳𝙴𝚂}\left[i\right].\mathrm{𝚜𝚞𝚌𝚌}\ne \mathrm{𝙽𝙾𝙳𝙴𝚂}\left[j\right].\mathrm{𝚜𝚞𝚌𝚌}
\forall i\in \left[1,|\mathrm{𝙽𝙾𝙳𝙴𝚂}|\right]:\mathrm{𝙽𝙾𝙳𝙴𝚂}\left[i\right].\mathrm{𝚜𝚞𝚌𝚌}\ne i
\mathrm{𝙽𝙾𝙳𝙴𝚂}\left[i\right].\mathrm{𝚜𝚞𝚌𝚌}=j\wedge j\ne i\wedge j\ne 0⇔\mathrm{𝙽𝙾𝙳𝙴𝚂}\left[j\right].\mathrm{𝚜𝚞𝚌𝚌}=i\wedge i\ne j\wedge i\ne 0
\left(\begin{array}{c}〈\begin{array}{cc}\mathrm{𝚒𝚗𝚍𝚎𝚡}-1\hfill & \mathrm{𝚜𝚞𝚌𝚌}-3,\hfill \\ \mathrm{𝚒𝚗𝚍𝚎𝚡}-2\hfill & \mathrm{𝚜𝚞𝚌𝚌}-0,\hfill \\ \mathrm{𝚒𝚗𝚍𝚎𝚡}-3\hfill & \mathrm{𝚜𝚞𝚌𝚌}-1,\hfill \\ \mathrm{𝚒𝚗𝚍𝚎𝚡}-4\hfill & \mathrm{𝚜𝚞𝚌𝚌}-0\hfill \end{array}〉\hfill \end{array}\right)
\mathrm{𝚜𝚢𝚖𝚖𝚎𝚝𝚛𝚒𝚌}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}_\mathrm{𝚎𝚡𝚌𝚎𝚙𝚝}_\mathtt{0}
\mathrm{𝙽𝙾𝙳𝙴𝚂}\left[1\right].\mathrm{𝚜𝚞𝚌𝚌}=3⇔\mathrm{𝙽𝙾𝙳𝙴𝚂}\left[3\right].\mathrm{𝚜𝚞𝚌𝚌}=1
\mathrm{𝙽𝙾𝙳𝙴𝚂}\left[2\right].\mathrm{𝚜𝚞𝚌𝚌}=0
and value 2 is not assigned to any variable.
\mathrm{𝙽𝙾𝙳𝙴𝚂}\left[4\right].\mathrm{𝚜𝚞𝚌𝚌}=0
Given 3 successor variables that have to be assigned a value in interval
\left[0,3\right]
, the solutions to the
\mathrm{𝚜𝚢𝚖𝚖𝚎𝚝𝚛𝚒𝚌}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}_\mathrm{𝚎𝚡𝚌𝚎𝚙𝚝}_\mathtt{0}
\left(〈\mathrm{𝚒𝚗𝚍𝚎𝚡}-1\mathrm{𝚜𝚞𝚌𝚌}-{s}_{1},\mathrm{𝚒𝚗𝚍𝚎𝚡}-2\mathrm{𝚜𝚞𝚌𝚌}-{s}_{2},\mathrm{𝚒𝚗𝚍𝚎𝚡}-3\mathrm{𝚜𝚞𝚌𝚌}-{s}_{3}〉\right)
constraint are
〈10,20,30〉
〈10,23,32〉
〈12,21,30〉
〈13,20,31〉
\left[0,3\right]
\mathrm{𝚜𝚢𝚖𝚖𝚎𝚝𝚛𝚒𝚌}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}_\mathrm{𝚎𝚡𝚌𝚎𝚙𝚝}_\mathtt{0}
\left(〈\mathrm{𝚒𝚗𝚍𝚎𝚡}-1\mathrm{𝚜𝚞𝚌𝚌}-{s}_{1},\mathrm{𝚒𝚗𝚍𝚎𝚡}-2\mathrm{𝚜𝚞𝚌𝚌}-{s}_{2},\mathrm{𝚒𝚗𝚍𝚎𝚡}-3\mathrm{𝚜𝚞𝚌𝚌}-{s}_{3},\mathrm{𝚒𝚗𝚍𝚎𝚡}-4\mathrm{𝚜𝚞𝚌𝚌}-{s}_{4}〉\right)
〈10,20,30,40〉
〈10,20,34,43〉
〈10,23,32,40〉
〈10,24,30,42〉
〈12,21,30,40〉
〈12,21,34,43〉
〈13,20,31,40〉
〈13,24,31,42〉
〈14,20,30,41〉
〈14,23,32,41〉
\mathrm{𝚜𝚢𝚖𝚖𝚎𝚝𝚛𝚒𝚌}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}_\mathrm{𝚎𝚡𝚌𝚎𝚙𝚝}_\mathtt{0}
{S}_{1}\in \left[0..5\right]
{S}_{2}\in \left[1..3\right]
{S}_{3}\in \left[1..4\right]
{S}_{4}\in \left[0..3\right]
{S}_{5}\in \left[0..2\right]
\mathrm{𝚜𝚢𝚖𝚖𝚎𝚝𝚛𝚒𝚌}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}_\mathrm{𝚎𝚡𝚌𝚎𝚙𝚝}_\mathtt{0}
\left(〈1{S}_{1},2{S}_{2},3{S}_{3},4{S}_{4},5{S}_{5}〉\right)
\mathrm{𝚜𝚢𝚖𝚖𝚎𝚝𝚛𝚒𝚌}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}_\mathrm{𝚎𝚡𝚌𝚎𝚙𝚝}_\mathtt{0}
constraint of the All solutions slot (the
\mathrm{𝚒𝚗𝚍𝚎𝚡}
attribute is displayed as indices of the
\mathrm{𝚜𝚞𝚌𝚌}
attribute)
|\mathrm{𝙽𝙾𝙳𝙴𝚂}|\ge 4
\mathrm{𝚖𝚒𝚗𝚟𝚊𝚕}
\left(\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚜𝚞𝚌𝚌}\right)=0
\mathrm{𝚖𝚊𝚡𝚟𝚊𝚕}
\left(\mathrm{𝙽𝙾𝙳𝙴𝚂}.\mathrm{𝚜𝚞𝚌𝚌}\right)>0
\mathrm{𝙽𝙾𝙳𝙴𝚂}
Within the context of sport scheduling,
\mathrm{𝙽𝙾𝙳𝙴𝚂}\left[i\right].\mathrm{𝚜𝚞𝚌𝚌}=j
i\ne 0,j\ne 0,i\ne j
) is interpreted as the fact that team
i
plays against team
j
\mathrm{𝙽𝙾𝙳𝙴𝚂}\left[i\right].\mathrm{𝚜𝚞𝚌𝚌}=0
i\ne 0
i
does not play at all.
An arc-consistency filtering algorithm for the
\mathrm{𝚜𝚢𝚖𝚖𝚎𝚝𝚛𝚒𝚌}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}_\mathrm{𝚎𝚡𝚌𝚎𝚙𝚝}_\mathtt{0}
constraint is described in [Cymer13], [CymerPhD13]. The algorithm is based on the following facts:
First, one can map solutions to the
\mathrm{𝚜𝚢𝚖𝚖𝚎𝚝𝚛𝚒𝚌}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}_\mathrm{𝚎𝚡𝚌𝚎𝚙𝚝}_\mathtt{0}
constraint to perfect
\left(g,f\right)
-matchings in a non-bipartite graph derived from the domain of the variables of the constraint where
g\left(x\right)=0
f\left(x\right)=1
for vertices
x
which have 0 in their domain, and
g\left(x\right)=f\left(x\right)=1
for all the remaining vertices. A perfect
\left(g,f\right)
-matching
ℳ
of a graph is a subset of edges such that every vertex
is incident with the number of edges in
ℳ
g\left(x\right)
f\left(x\right)
Second, Gallai-Edmonds decomposition [Gallai63], [Edmonds65] allows to find out all edges that do not belong to any perfect
\left(g,f\right)
-matchings, and therefore prune the corresponding variables.
n
Solutions 2 4 10 26 76 232 764
\mathrm{𝚜𝚢𝚖𝚖𝚎𝚝𝚛𝚒𝚌}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}_\mathrm{𝚎𝚡𝚌𝚎𝚙𝚝}_\mathtt{0}
0..n
\mathrm{𝚜𝚢𝚖𝚖𝚎𝚝𝚛𝚒𝚌}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}
implies (items to collection):
𝚔_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}
\mathrm{𝚕𝚎𝚡}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}
application area: sport timetabling.
characteristic of a constraint: joker value.
combinatorial object: matching.
constraint type: predefined constraint, timetabling constraint.
\mathrm{𝚜𝚢𝚖𝚖𝚎𝚝𝚛𝚒𝚌}_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}_\mathrm{𝚎𝚡𝚌𝚎𝚙𝚝}_\mathtt{0}\left(\mathrm{𝙽𝙾𝙳𝙴𝚂}\right)
\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}_\mathrm{𝚎𝚡𝚌𝚎𝚙𝚝}_\mathtt{0}
\left(\mathrm{𝚅𝙰𝚁𝙸𝙰𝙱𝙻𝙴𝚂}:\mathrm{𝙽𝙾𝙳𝙴𝚂}\right)
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El principio de tiempo mínimo - SEG Wiki
El principio de tiempo mínimo
This page is a translated version of the page The principle of least time and the translation is 42% complete.
The gradient is a concept that is familiar to everyone who has climbed a mountain (Figure 8). As a climber goes up the mountain, he experiences the rate of change of elevation. the gradient is the vector (because it has both magnitude and direction) that points in the steepest direction (i.e., the direction in which the rate of change is greatest). The magnitude of the gradient vector is the rate of change of the elevation along this path. Thus, if the climber continually follows the gradient, he will take the steepest path to the top of the mountain. This steepest path is called a flow line.
Using another analogy, a skier wants to descend on the steepest path available to him at his present position on a hill. The negative gradient is the vector that points in the steepest direction down the hill. This steepest path down is called a fall line. The fall line is always in the direction of the negative gradient of the function whose graph represents the surface of the hill. The fall line and the corresponding flow line coincide. The fall line will change direction as the skier goes downhill from one position to another. A skier feels the fall line and gets visual hints about its location. The skier’s eyes can detect local changes in the steepness of the hill. The skier’s legs indicate whether they are on the fall line or off it. When a skier is skiing the fall line, he feels equal pressure on both legs. All forces are channeled in a direction parallel to the skis. Skiing in any other direction results in pressure on one leg that differs from the pressure on the other. In other words, skiing the fall line means that the skier follows the direction of the negative gradient all the way down the hill.
Figure 8. The steepest paths are the flow lines. The level paths are the contour lines. The two sets are orthogonal (i.e., at right angles).
The Mississippi River provides a geologic example. By depositing sand and silt, the Mississippi has created most of Louisiana. The river’s purpose is to get to the Gulf of Mexico in the least time possible. The river wants to follow the direction of the negative gradient, which is the path of steepest descent. In other words, like all rivers, the Mississippi is destined to follow a flow line. If the river had kept to one channel, southern Louisiana would be a long narrow peninsula reaching into the Gulf of Mexico. However, the river has deposited sediment over a wide area. Over time, as it continues to carry and deposit more sediment, the river lengthens and its mouth advances southward. The steepness of its path declines, the current slows, and sediment builds up the riverbed. Eventually, the bed builds up so much that the river surges over the left or right bank and goes off in a new direction, to follow what has newly become the steepest way down.
Southern Louisiana exists in its present form because periodically the Mississippi River has radically changed course by jumping here and there within an arc about 200 miles wide. Major shifts of that nature have tended to occur about once a millennium. About 1000 years ago, the Mississippi’s channel shifted to the river’s present course. Today, the Mississippi River has advanced far past New Orleans and out into the Gulf of Mexico. The Mississippi River wants to change its course again to follow a shorter and steeper route, but engineers have built a levee system to keep the river from jumping its banks and changing course.
Pierre de Fermat (1601–1665) formulated the rule known as Fermat’s principle of least time. In his original statement, Fermat asserted that the raypath taken by light traveling between any two points is such that the time taken is a minimum. In other words, the ray-paths are the flow lines. Eventually, Fermat’s original statement underwent some expansion. Fermat’s principle is expressed now as: The raypath taken by light traveling between any two points is such that the time taken is stationary with respect to variations of that path. Stationary means that the traveltime can be a minimum or can be a maximum or can be a point of inflection having a horizontal tangent. More specifically, the traveltime of the true trajectory (i.e., the raypath) will equal, to a first approximation, the traveltime of paths in the immediate vicinity. Energy traveling along these neighboring paths will arrive at the destination at about the same time by routes that differ only slightly. Thus, these neighboring paths will tend to reinforce one another. Energy taking other paths arrives out of phase and therefore tends to cancel out. The net result is that energy effectively propagates along the raypaths (i.e., the paths that satisfy Fermat’s principle). In this way, Fermat’s principle helps explain why light is so clever in its meanderings.
A lifeguard on the beach at A sees a drowning person in the water at E (Figure 9a). Which path should the lifeguard take to rescue the drowning person in the least time? The lifeguard might be tempted to take the straight-line path ACE because it represents the minimum distance he must travel. However, he knows that he can run faster on the beach than he can swim in the water. As a result, it pays to go a longer distance AD on the dry land to go a reduced distance DE in the water. Thus, by running to D and then swimming to E, the lifeguard does not minimize the distance he must travel, but he does minimize the time required to reach the drowning person. On the other hand, suppose the lifeguard were a seal. The seal waddles slowly on land but swims rapidly. The seal would take the path ABE. In fact, by instinct, a person in water will make a beeline for land, whereas a seal on land will make a beeline for water.
Figure 9. (a) For a human, the least-time path is ADE. For a seal, the least-time path is ABE. (b) The raypath is orthogonal to the wavefronts. The test path is skewed to the wavefronts.
Let us take an isotropic medium in which the velocity varies continuously so that there is no interface at which reflection and/or refraction could occur. We now show that in such a case, Fermat’s principle holds in its original formulation (i.e., traveltime is a minimum). Let an arbitrary path between two wavefronts be given (Figure 9b). This arbitrary path is called the test path. Let us first figure out what happens between two closely spaced wavefronts. Because the wavefronts are so close together, we might consider the parts of them within a small region to be two parallel straight lines. The test path AC would be a straight line between the two wavefronts, and the flow line AB would be a straight line orthogonal to both wavefronts. If
{\displaystyle \theta }
is the angle between the test path and the flow line, if ds is equal to the length of the flow line, and if
{\displaystyle d\sigma }
is the length of the test path, then
{\displaystyle d\sigma {\rm {=}}{ds/cos}\theta .}
We now use the fact that the medium is isotropic, i.e., that the velocity at any point is the same in all directions. Thus, the time that energy takes to traverse the flow line is
{\displaystyle dt{\ =\ }nds}
, where the slowness n is defined as the reciprocal of velocity. Likewise, the time it takes to traverse the test path is
{\displaystyle {\begin{aligned}d\tau {\ =}\ n\ d\sigma {\ =}\ n{\textit {ds/cos}}\theta .\end{aligned}}}
The traveltime along the flow line between the two wavefronts is
{\displaystyle t{\rm {=}}\int {n}ds}
. The traveltime along the test path is
{\displaystyle \tau {\rm {=}}\int {\left(n/{\rm {\ cos\ }}\theta \right)}ds}
. Because cos
{\displaystyle \theta }
at any point is always less than or equal to one, it follows that
{\displaystyle t\leq \tau }
. Thus, the traveltime along the flow line is less than the traveltime along any test path except along the flow line itself. Finally, the flow line - that is, the line whose direction at any point coincides with the direction of the gradient - is the least-time path.
La derivada direccional La ecuación de Eikonal
Movimiento de ondas Visualización
Sismología de Reflexión
Procesamiento digitales
Realce de señales
Vector unitario tangente
La ecuación de Eikonal
Ecuación del rayo
Ecuacón del rayo para velocidad lineal con la profundidad
Camino del rayo para velocidad lineal con la profundidad
Tiempo de viaje para velocidad lineal con la profundidad
Punto de máxima profundidad
Frente de onda para velocidad lineal con la profundidad
Dos conjuntos ortogonales de círculos
Migración en el caso de velocidad constante
Implementación de la migración
Apéndice B: Ejercicios
Retrieved from "https://wiki.seg.org/index.php?title=The_principle_of_least_time/es&oldid=171880"
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Ground-State Energy and Entropy for One-Dimensional Heisenberg Chain with Alternating D-Term
Ground-State Energy and Entropy for One-Dimensional Heisenberg Chain with Alternating D-Term
Chunhuan Xiang1, Honglei Wang2*
2College of Medical Informatics, Chongqing Medical University, Chongqing, China
We study the ground-state information of one-dimensional Heisenberg chain with alternating D-term. Given the ground-state phase diagram, the ground-state energy and the entanglement entropy are obtained by tensor-net work algorithm. The phase transition points are shown in the entanglement entropy figure. The results are agreed with the phase diagram.
Ground-State Energy, Entropy, One-Dimensional, Heisenberg Chain
Recent improvements in experimental techniques [1] [2] [3] [4] have given access to the quantum mechanical simulation of the dynamics of isolated, interacting quantum many-body systems with a variety of platforms, such as cold atoms in optical lattices and ion traps [5] [6] . Among them, the Haldane ground state of the S = 1 anti-ferromagnetic Heisenberg chain with a gap to the first excited state and the ground-state [7] [8] [9] [10] has been extensively studied by many authors, which is known to have a gapless ground-state for half-integer spin. The gaped state is destroyed by various types of perturbations such as exchange anisotropy, bond alternation and single-ion-type anisotropy. On the other hand, understanding the collective behavior of quantum many-body systems has long presented a formidable challenge due to the exponential growth of Hilbert space dimension with system size N, however, the numerical algorithms are made great progress, such as the matrix product states [11] [12] [13] [14] [15] in one spatial dimension and the projected entanglement-pair states [16] [17] [18] in two and higher spatial dimensions, which is a variational algorithm to compute the ground-state wave-function for transitionally invariant quantum systems on an infinite-size lattice. In this paper, we obtained the approximation ground-state wave-function by the matrix product states, meanwhile, the ground-state energy and the entanglement entropy are also given.
This paper is organized as follows: in the second section, the model Hamiltonian and the ground-state phase diagram are presented. The matrix product state algorithm is simply introduced. The figure for ground-state energy and entanglement entropy for left and right section are shown. The final section is devoted to a summary and discussion.
2. The Hamiltonian and Ground-State Phase Diagram
The anti-ferromagnetic Heisenberg chain with alternating D-term for spin-1 [19] is given as the follow
H=\underset{l=1}{\overset{N}{\sum }}J{S}_{l}{S}_{l+1}+D+\underset{l=1}{\overset{N/2}{\sum }}{S}_{2l-1}^{z2}+{D}_{-}\underset{l=1}{\overset{N/2}{\sum }}{S}_{2l}^{z2},\text{}\left(J>0\right)
where J is the exchange coupling,
{D}_{+}={D}_{0}+\delta D
{D}_{-}={D}_{0}+\delta D
, and S is the spin-1 operator.
{S}^{x}=\frac{1}{\sqrt{2}}\left(\begin{array}{ccc}0& 1& 0\\ 1& 0& 1\\ 0& 1& 0\end{array}\right)
{S}^{y}=\frac{1}{\sqrt{2}}\left(\begin{array}{ccc}0& i& 0\\ i& 0& -i\\ 0& -i& 0\end{array}\right)
{S}^{z}=\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& -1\end{array}\right)
The parameters D0 and dD represent uniform and alternating components of uniaxial single-ion anisotropy, respectively. In what follows we set J = 1 to fix the energy scale. The ground-state phase diagram is given in Ref. [19] , which consists of the Haldane phase, the Large-D phase, udud and u0d0 phases. The Gaussian transition occurred between the Haldane phase and the Large-D phase, which is gapful phase to gapful phase transition. From symmetry consideration, these transitions between the Haldane phase, u0d0 phase and udud phase are Ising type transitions. The ground-state phase diagram is shown in Figure 1. In this paper, we set the D0 = 2, and dD as the control parameter.
Figure 1. Ground state phase diagram of the Hamiltonian (1) [19] . The phases are separated with the symbols. The solid lines are the guide for eye. The broken and dotted lines represent the approximate phase boundary, respectively. The dash-dotted line is the line dD = −D0 + 2J.
The conformal central charge is an important content in field theory, which gives the type of the phase transition in theory. With conformal central charge c = 1, the transition line between the Larged-D and Haldane is expected to be described by the conformal field theory. The phases u0d0 and udud are expected as gapless system.
3. The Matrix Product State Algorithm
The matrix product state algorithm is given in Ref [15] , which exploits two facts, namely invariance under translations of the system and parallelizability of local updates in time-evolving block decimation algorithm. With time evolution for a quantum spin chain in the thermodynamic limit, the approximation ground-states are obtained. For a given wave-function
|{\Psi }_{t}〉=\mathrm{exp}\left(-iHt\right)|{\Psi }_{0}〉
the Schmidt decomposition of
|{\Psi }_{t}〉
|\Psi 〉=\underset{\alpha =\text{1}}{\overset{\chi }{\sum }}{\lambda }_{\alpha }^{\left[r\right]}|{\Phi }_{\alpha }^{\left[⊲r\right]}〉\otimes |{\Phi }_{\alpha }^{\left[r+1⊳\right]}〉
The Equation (3) can be rewritten as
|\Psi 〉=\underset{\alpha ,\beta =1}{\overset{\chi }{\sum }}\underset{i=1}{\overset{d}{\sum }}{\lambda }_{\alpha }^{\left[r\right]}{\Gamma }_{i\alpha \beta }^{\left[r+1\right]}{\lambda }_{\beta }^{\left[r+1\right]}|{\Phi }_{\alpha }^{\left[⊲r\right]}〉|{i}^{\left[r\right]}〉|{\Phi }_{\alpha }^{\left[r+1⊳\right]}〉
where χ is the truncation dimension, d is the Hilbert space, λ is diagonal matrix, the G is three index tensor. By using the two-site Hamiltonian, which is expanded through a Suzuki-Trotter decomposition, the imaginary time evolution for the given wave-function is shown. The approximation ground-state wave-function is obtained until the approximation ground-state energy is lower enough.
4. Ground-State Energy and Entanglement Entropy
The simulation results of the approximation ground-state energy for Equation (1) are shown in Figure 2 with truncation dimension χ = 8, 12, 16, 20 in different label, respectively.
The approximation ground-state energies with different control parameter dD agree with each other.
During the numerical simulation, two phase transitions are obtained, which are shown in entanglement entropy. The figure for entanglement entropy of the left one and the right one are given in Figure 3 and Figure 4. The parameters for the platform of the computer system are given as CPU: Intel(R) Core(TM) i5-64002.7 GHz; memory (RAM): 8.00 GB.
The anti-ferromagnetic Heisenberg chain with alternating D-term for spin-1 is investigated by using matrix product states. The approximation ground-state
Figure 2. The approximation ground-state energy for the Hamiltonian (1). dD as the control parameter, the truncation dimensions are shown with χ = 8, 12, 16, 20. The solid lines are the guide for eye.
Figure 3. The entanglement entropy of the Hamiltonian (1) with truncation dimension χ = 8, 12, 16, 20, the transition points are dD = 0.287, 0.320, 0.361, 0.400. The solid lines are the guide for eye. The peak is bigger and bigger with larger and larger truncation dimension.
Figure 4. The entanglement entropy of the Hamiltonian (1) with truncation dimension χ = 8, 12, 16, 20, the transition points are dD = 0.695, 0601, 0.592, 0.561. The solid lines are the guide for eye. The peak is also bigger and bigger with larger and larger truncation dimension.
energy and the entanglement entropy are shown in this paper. We simply analyzed the results, however, as we have been unable to determine the local order parameter. This is the next direction of research. Besides, we will use alternative techniques beyond the MPS paradigm to yield the scaling behavior of physical observable, which may be more suitable.
Xiang, C.H. and Wang, H.L. (2019) Ground-State Energy and Entropy for One-Dimensional Heisenberg Chain with Alternating D-Term. Journal of Applied Mathematics and Physics, 7, 1220-1225. https://doi.org/10.4236/jamp.2019.75082
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Global Constraint Catalog: KBerge-acyclic_constraint_network
<< 3.7.28. Balanced tree3.7.30. Binary constraint >>
\mathrm{𝚊𝚖𝚘𝚗𝚐}
\mathrm{𝚊𝚗𝚍}
\mathrm{𝚊𝚛𝚒𝚝𝚑}
\mathrm{𝚊𝚛𝚒𝚝𝚑}_\mathrm{𝚘𝚛}
\mathrm{𝚌𝚑𝚊𝚗𝚐𝚎}
\mathrm{𝚌𝚑𝚊𝚗𝚐𝚎}_\mathrm{𝚟𝚎𝚌𝚝𝚘𝚛𝚜}
\mathrm{𝚌𝚕𝚊𝚞𝚜𝚎}_\mathrm{𝚊𝚗𝚍}
\mathrm{𝚌𝚕𝚊𝚞𝚜𝚎}_\mathrm{𝚘𝚛}
\mathrm{𝚌𝚘𝚗𝚍}_\mathrm{𝚕𝚎𝚡}_\mathrm{𝚌𝚘𝚜𝚝}
\mathrm{𝚌𝚘𝚗𝚍}_\mathrm{𝚕𝚎𝚡}_\mathrm{𝚐𝚛𝚎𝚊𝚝𝚎𝚛}
\mathrm{𝚌𝚘𝚗𝚍}_\mathrm{𝚕𝚎𝚡}_\mathrm{𝚐𝚛𝚎𝚊𝚝𝚎𝚛𝚎𝚚}
\mathrm{𝚌𝚘𝚗𝚍}_\mathrm{𝚕𝚎𝚡}_\mathrm{𝚕𝚎𝚜𝚜}
\mathrm{𝚌𝚘𝚗𝚍}_\mathrm{𝚕𝚎𝚡}_\mathrm{𝚕𝚎𝚜𝚜𝚎𝚚}
\mathrm{𝚌𝚘𝚗𝚜𝚎𝚌𝚞𝚝𝚒𝚟𝚎}_\mathrm{𝚐𝚛𝚘𝚞𝚙𝚜}_\mathrm{𝚘𝚏}_\mathrm{𝚘𝚗𝚎𝚜}
\mathrm{𝚎𝚕𝚎𝚖𝚎𝚗𝚝𝚗}
\mathrm{𝚎𝚚𝚞𝚒𝚟𝚊𝚕𝚎𝚗𝚝}
\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚘𝚗𝚝𝚒𝚐𝚞𝚒𝚝𝚢}
\mathrm{𝚒𝚖𝚙𝚕𝚢}
\mathrm{𝚒𝚗}_\mathrm{𝚒𝚗𝚝𝚎𝚛𝚟𝚊𝚕}
\mathrm{𝚒𝚗𝚌𝚛𝚎𝚊𝚜𝚒𝚗𝚐}_\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}
\mathrm{𝚒𝚗𝚌𝚛𝚎𝚊𝚜𝚒𝚗𝚐}_\mathrm{𝚗𝚟𝚊𝚕𝚞𝚎}
\mathrm{𝚒𝚗𝚝}_\mathrm{𝚟𝚊𝚕𝚞𝚎}_\mathrm{𝚙𝚛𝚎𝚌𝚎𝚍𝚎}
\mathrm{𝚒𝚗𝚝}_\mathrm{𝚟𝚊𝚕𝚞𝚎}_\mathrm{𝚙𝚛𝚎𝚌𝚎𝚍𝚎}_\mathrm{𝚌𝚑𝚊𝚒𝚗}
\mathrm{𝚕𝚎𝚡}_\mathrm{𝚋𝚎𝚝𝚠𝚎𝚎𝚗}
\mathrm{𝚕𝚎𝚡}_\mathrm{𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}
\mathrm{𝚕𝚎𝚡}_\mathrm{𝚎𝚚𝚞𝚊𝚕}
\mathrm{𝚕𝚎𝚡}_\mathrm{𝚐𝚛𝚎𝚊𝚝𝚎𝚛}
\mathrm{𝚕𝚎𝚡}_\mathrm{𝚐𝚛𝚎𝚊𝚝𝚎𝚛𝚎𝚚}
\mathrm{𝚕𝚎𝚡}_\mathrm{𝚕𝚎𝚜𝚜}
\mathrm{𝚕𝚎𝚡}_\mathrm{𝚕𝚎𝚜𝚜𝚎𝚚}
\mathrm{𝚗𝚊𝚗𝚍}
\mathrm{𝚗𝚘𝚛}
\mathrm{𝚘𝚛}
\mathrm{𝚙𝚊𝚝𝚝𝚎𝚛𝚗}
\mathrm{𝚜𝚖𝚘𝚘𝚝𝚑}
\mathrm{𝚜𝚝𝚛𝚎𝚝𝚌𝚑}_\mathrm{𝚙𝚊𝚝𝚑}
\mathrm{𝚜𝚝𝚛𝚎𝚝𝚌𝚑}_\mathrm{𝚙𝚊𝚝𝚑}_\mathrm{𝚙𝚊𝚛𝚝𝚒𝚝𝚒𝚘𝚗}
\mathrm{𝚝𝚠𝚘}_\mathrm{𝚘𝚛𝚝𝚑}_\mathrm{𝚊𝚛𝚎}_\mathrm{𝚒𝚗}_\mathrm{𝚌𝚘𝚗𝚝𝚊𝚌𝚝}
\mathrm{𝚝𝚠𝚘}_\mathrm{𝚘𝚛𝚝𝚑}_\mathrm{𝚍𝚘}_\mathrm{𝚗𝚘𝚝}_\mathrm{𝚘𝚟𝚎𝚛𝚕𝚊𝚙}
\mathrm{𝚡𝚘𝚛}
A constraint for which the decomposition associated with its usually counter-free deterministic automatonAll the above constraints, except
\mathrm{𝚊𝚖𝚘𝚗𝚐}
\mathrm{𝚌𝚑𝚊𝚗𝚐𝚎}
\mathrm{𝚜𝚖𝚘𝚘𝚝𝚑}
have a deterministic counter-free automaton. The
\mathrm{𝚊𝚖𝚘𝚗𝚐}
constraint has an automaton involving one counter and a single state, see Figure 5.23.3, while the
\mathrm{𝚌𝚑𝚊𝚗𝚐𝚎}
\mathrm{𝚜𝚖𝚘𝚘𝚝𝚑}
constraints have a counter-free non deterministic automaton, see Figures 5.61.5 and 5.355.4. is Berge-acyclic. Arc-consistency for a Berge-acyclic constraint network is achieved by making each constraint of the corresponding network arc-consistent [BeeriFaginMaierYannakakis83]. A constraint network for which the corresponding intersection graph does not contain any cycle and such that, for any pair of constraints, the two sets of involved variables share at most one variable is Berge-acyclic, where Berge-acyclic is defined by the following two conditions:
There is no more than one shared variable between any pair of constraints,
The hypergraph corresponding to the constraint network does not contain any cycle. Within [Berge87] a cycle of an hypergraph
H
is defined as “Let
H
be an hypergraph on a finite set
X
. A cycle of length
k
k\ge 2
\left({x}_{1},{E}_{1},{x}_{2},{E}_{2},{x}_{3},\cdots ,{E}_{k},{x}_{1}\right)
{E}_{1},{E}_{2},\cdots ,{E}_{k}
are distinct edges of
H
{x}_{1},{x}_{2},\cdots ,{x}_{k}
are distinct vertices of
H
{x}_{i},{x}_{i+1}\in {E}_{i}
i=1,2,\cdots ,k-1
{x}_{k},{x}_{1}\in {E}_{k}
The intersection graph of a constraint network is built in the following way: to each vertex corresponds a constraint and there is an edge between two vertices if and only if the sets of variables involved in the two corresponding constraints intersect.
Figure 3.7.7. (A) and (D): Berge-acyclic constraint networks; (B) and (C): non Berge-acyclic constraint networks; (E), (F), (G), (H): corresponding intersection graphs.
Parts (A), (B), (C) and (D) of Figure 3.7.7 provide four examples of constraint networks, while parts (E), (F), (G) and (F) give their corresponding intersection graphs.
The constraint network corresponding to part (A) is Berge-acyclic since its corresponding intersection graph (E) does not contain any cycle and since there is no more than one shared variable between any pair of constraints.
The constraint network corresponding to part (B) is not Berge-acyclic since its hypergraph (B) contains a cycle.
The constraint network corresponding to (C) is also not Berge-acyclic since its third and fourth constraints share more than one variable.
Finally, the constraint network corresponding to (D) is Berge acyclic, even though its intersection graph (H) has a cycle, since its hypergraph (D) does not contain any cycle and since there is no more than one shared variable between any pair of constraints.
If we execute the filtering algorithm of each constraint of a Berge-acyclic constraint network
𝒩
in an appropriate order then each constraint needs only to be waken twice in order to reach the fix-point. A static ordering for waking the constraints of
𝒩
Consider the intersection graph
{𝒢}_{𝒩}
associated with the constraint network
𝒩
. We perform a topological sort on
{𝒢}_{𝒩}
, which always first selects in the remaining part of
{𝒢}_{𝒩}
a vertex (i.e., a constraint) which has only a single neighbour. Let
{C}_{1},{C}_{2},\cdots ,{C}_{n}
be the constraints successively removed by the topological sort.
Then, the static ordering for reaching a fix-point is given by the sequence
{C}_{1},{C}_{2},\cdots ,{C}_{n-1},{C}_{n},{C}_{n-1},\cdots ,{C}_{2},{C}_{1}
, where each constraint is woken at most twice. This can be done by using the notion of propagator group [LagerkvistSchulte09]. This facility allows the user of a solver controlling the order of execution of a group of constraints. Propagator groups are useful, both to guaranty the theoretical worst case complexity of a decomposition, and for accelerating convergence to the fix-point in practice.
If we consider the Berge-acyclic constraint network given by Part (D) of Figure 3.7.7 an appropriate order for waking the constraints could for instance be
{\mathrm{CTR}}_{1}
{\mathrm{CTR}}_{4}
{\mathrm{CTR}}_{2}
{\mathrm{CTR}}_{3}
{\mathrm{CTR}}_{2}
{\mathrm{CTR}}_{4}
{\mathrm{CTR}}_{1}
For heuristics that try creating a Berge-acyclic constraint network see also the keyword heuristics and Berge-acyclic constraint network.
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Imports FITS information into images
This routine imports FITS information into the CCDPACK extension of a list of images. FITS information (probably provided by the instrument/telescope control systems) can be used to specify certain parameters which are required by CCDPACK to perform "automated" reductions. These might cover such items as the type of data (target, flatfield, bias frame etc.), the Analogue-to-Digital Conversion factor, the nominal readout noise, the position of any bias strips (over-scan regions) etc.
The import is controlled by a "table" which specifies how FITS keyword values should be interpreted. This allows the evaluation of functions containing many FITS keywords as well as the mapping of CCDPACK recognised character items to arbitrary strings.
import in table
TABLE = LITERAL (Read)
The name of the table containing the description of how FITS keyword values are to be translated into CCDPACK extension items. See the topic "Table Format" for information on how to create a translation table. [’import.tab’]
import in=’
\ast
’ table=$CCDPACK_DIR/WHTSKY.DAT
This example shows all the images in the current directory being processed using the import control table $CCDPACK_DIR/WHTSKY.DAT.
The import control (translation) table is an ordinary text file which contains instructions on how to transfer FITS information from the FITS extension to the CCDPACK extension of an image. "Translation" is required since no standard interpretation of FITS keywords can be made and because the items which may be required can be compounds of single FITS keyword values.
In its most simple format a FITS control table is just a series of lines which contain the names of CCDPACK extension items and the names of the FITS keywords to which they map.
Extension-item FITS-keyword
If the HDS type of the destination Extension-item is known this may be included.
Extension-item _HDS-type FITS-keyword
Normally this is inferred. This is the format used by the KAPPA application FITSIMP (as of KAPPA version 0.8-6U). Extension items are the names of CCDPACK items (such as FTYPE, FILTER etc.). These may be heirarchical, e.g. TIMES.DARK. Note that they exclude the "NDF_NAME.MORE.CCDPACK." part of the extension path name.
To allow functions of FITS-keywords to be possible a second "declarative" form of statement is necessary.
_HDS-type FITS-keyword
So for instance if you wanted to derive an exposure time for an observation which was given in milliseconds and which you wanted to convert into seconds you would use this sequence of commands:
_INTEGER EXPOSURE
TIMES.DARK _DOUBLE 1000.0D0
\ast
The "_INTEGER EXPOSURE" tells this application to find a FITS keyword of EXPOSURE and extract its value as an integer. If you wanted to estimate the dark time from a knowledge of the start and end times (TAI0 and TAI1):
_DOUBLE TAI0
TIMES.DARK _DOUBLE (TAI1-TAI0)
The function may use any of the usual Fortran operators; +, -, *, /, ** and many others that are supported by the TRANSFORM package (SUN/61).
Functions are allowed to not contain any FITS-keywords in which case the extension item will be assigned to the value, so for instance numerical constants may be given:
EXTENT.MINX _INTEGER 1
EXTENT.MINY _INTEGER 1
In this way import tables could actually be used to set all the required values in the CCDPACK extension (but see PRESENT also).
Characters strings cannot be manipulated by functions so a single special format for translating their values is provided. The name of the destination extension item and (optionally) its type are given as usual followed by a FITS-keyword which contains the string to be translated. This is then followed by statements which translate an "input" string into an "output" string, i.e.:
FITS1 = Ext1 FITS2 = Ext2 FITS3 = Ext3 ... FITSn = Extn
So for instance if you wanted to translate frame types to those recognised by CCDPACK you might use something like:
FTYPE _CHAR OBSTYPE OBJECT=TARGET -
FF=FLAT -
ZERO=BIAS
Which would compare the value of the FITS-keyword OBSTYPE with the strings "OBJECT", "FF" and "ZERO" (case insensitive) and convert these into the values in the right-hand side of the equals sign.
Logical data types are restricted to a single keyword whose value must be "YES", "TRUE", "T", "Y" for TRUE or "NO", "FALSE", "N", "F".
The FITS keywords may be hierarchical, and on the whole are specified simply by giving their name in the normal way. However, there is one special case: if the value of a FITS header is known to be a string of the form ’[A:B,C:D]’ the numbers A, B, C and D may be extracted individually by appending ’<X1>’, ’<X2>’, ’<Y1>’ or ’<Y2>’ respectively to the name of the keyword. Hence:
EXTENT.MINX TRIMSEC<X1>
EXTENT.MAXX TRIMSEC<X2>
would set the extents from the first two fields of a suitably formatted TRIMSEC header.
Fields in the table may be separated by commas if desired, any amount of white space and tabs are also allowed. Comments may be placed anywhere and should start with the characters "#" or "!". Continuation onto a new line is indicated by use of "-".
CCDPACK extension items
The CCDPACK extension of an image may contain the following items. The names and types of the extension items are those as used in import tables. More complete descriptions of the items can be found with the applications that use these values.
Name HDS data type Description
ADC _DOUBLE The analogue to digital conversion factor.
BOUNDS.END1 _INTEGER The end row or column of the first bias strip region.
BOUNDS.END2 _INTEGER The end row or column of the second bias strip region.
BOUNDS.START1 _INTEGER The first row or column of the first bias strip region.
BOUNDS.START2 _INTEGER The first row or column of the second bias strip region.
DEFERRED _DOUBLE The deferred charge.
DIRECTION _CHAR The "readout" direction (X or Y).
EXTENT.MAXX _INTEGER Maximum X coordinate of useful region.
EXTENT.MAXY _INTEGER Maximum Y coordinate of useful region.
EXTENT.MINX _INTEGER Minimum X coordinate of useful region.
EXTENT.MINY _INTEGER Minimum Y coordinate of useful region.
FILTER _CHAR Filter name.
FTYPE _CHAR Frame type (TARGET, BIAS, FLAT, DARK or FLASH)
RNOISE _DOUBLE Readout noise (ADUs)
SATURATION _DOUBLE Pixel saturation count.
TIMES.DARK _DOUBLE Dark count time.
TIMES.FLASH _DOUBLE Pre-flash time.
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Global Constraint Catalog: Kbound-consistency
<< 3.7.37. Border3.7.39. Business rules >>
\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}
\mathrm{𝚊𝚕𝚕}_\mathrm{𝚖𝚒𝚗}_\mathrm{𝚍𝚒𝚜𝚝}
\mathrm{𝚊𝚝𝚖𝚘𝚜𝚝}\mathtt{1}
\mathrm{𝚊𝚝𝚖𝚘𝚜𝚝}_\mathrm{𝚗𝚟𝚊𝚕𝚞𝚎}
\mathrm{𝚗𝚟𝚊𝚕𝚞𝚎}
\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}
\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}_\mathrm{𝚕𝚘𝚠}_\mathrm{𝚞𝚙}
\mathrm{𝚒𝚗𝚌𝚛𝚎𝚊𝚜𝚒𝚗𝚐}_\mathrm{𝚜𝚞𝚖}
𝚔_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}
\mathrm{𝚖𝚞𝚕𝚝𝚒}_\mathrm{𝚒𝚗𝚝𝚎𝚛}_\mathrm{𝚍𝚒𝚜𝚝𝚊𝚗𝚌𝚎}
\mathrm{𝚜𝚊𝚖𝚎}
\mathrm{𝚜𝚊𝚖𝚎}_\mathrm{𝚊𝚗𝚍}_\mathrm{𝚐𝚕𝚘𝚋𝚊𝚕}_\mathrm{𝚌𝚊𝚛𝚍𝚒𝚗𝚊𝚕𝚒𝚝𝚢}_\mathrm{𝚕𝚘𝚠}_\mathrm{𝚞𝚙}
\mathrm{𝚜𝚕𝚒𝚍𝚒𝚗𝚐}_\mathrm{𝚜𝚞𝚖}
\mathrm{𝚜𝚘𝚏𝚝}_\mathrm{𝚊𝚕𝚕}_\mathrm{𝚎𝚚𝚞𝚊𝚕}_\mathrm{𝚖𝚊𝚡}_\mathrm{𝚟𝚊𝚛}
\mathrm{𝚜𝚘𝚏𝚝}_\mathrm{𝚊𝚕𝚕}_\mathrm{𝚎𝚚𝚞𝚊𝚕}_\mathrm{𝚖𝚒𝚗}_\mathrm{𝚌𝚝𝚛}
\mathrm{𝚜𝚘𝚛𝚝}
\mathrm{𝚜𝚞𝚖}_\mathrm{𝚏𝚛𝚎𝚎}
\mathrm{𝚜𝚞𝚖}_\mathrm{𝚘𝚏}_\mathrm{𝚒𝚗𝚌𝚛𝚎𝚖𝚎𝚗𝚝𝚜}
\mathrm{𝚞𝚜𝚎𝚍}_\mathrm{𝚋𝚢}
Denotes that, for a given constraint, there is a filtering algorithm or a reformulation in term of other constraints that ensures bound-consistency for its domain variables.In the context of the
\mathrm{𝚗𝚟𝚊𝚕𝚞𝚎}
constraint, bound-consistency is only achieved if and only if, the minimum of the variable that denotes the number of distinct values is not constrained at all. In the context of the
𝚔_\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}
constraint, bound-consistency is only achieved when we have two overlapping
\mathrm{𝚊𝚕𝚕𝚍𝚒𝚏𝚏𝚎𝚛𝚎𝚗𝚝}
constraints, see [BessiereKatsirelosNarodytskaQuimperWalsh10] for more details. A filtering algorithm or a reformulation ensure bound-consistency for a given constraint ctr using distinct domain variables if and only if for every domain variable
V
of ctr:
There exists at least one solution to ctr such that
V=\underline{V}
and every other domain variable
W
of ctr is assigned to a value located in its range
\left[\underline{W},\overline{W}\right]
V=\overline{V}
W
\left[\underline{W},\overline{W}\right]
This consistency is called bound(Z) consistency in [Bessiere06]. One of its interest is that it sometimes gives the opportunity to come up with a filtering algorithm that has a lower complexity than the algorithm that achieves arc-consistency. Discarding holes from the domain variables usually leads to graphs with a specific structure for which one can take advantage in order to derive more efficient graph algorithms. Filtering algorithms that achieve bound-consistency can also be used in a pre-processing phase before applying a more costly filtering algorithm that achieves arc-consistency.
Note that there is a second definition of bound-consistency, called bound(D) consistency in [Bessiere06], where the range
\left[\underline{W},\overline{W}\right]
is replaced by the domain of the variable
W
. However within the context of global constraints most filtering algorithms do not refer to this second definition.
Finally, within the context of constraints involving only set variables, bound-consistency is defined in the following way. A constraint ctr defined on distinct set variables is bound-consistent if and only if for every pair
\left(V,v\right)
V
is a set variable of ctr and
v
an integer value, if
v\in \underline{V}
v
belongs to the set assigned to
V
in all solutions to ctr and if
v\in \overline{V}\setminus \underline{V}
v
V
in at least one solution and is excluded from this set in at least one solution.
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