CS计算机代考程序代写 flex Haskell algorithm Q1

Q1

Challenge description

Let A be a set and ⪯ a partial order on A. We say a function f:A→A respects the
partial order if and only if

∀x,y∈A (x⪯y⇔f(x)⪯f(y))
Let AA={f∣f:A→A} be the set of all functions from A to A. We introduce a
relation ⊣ on AA, such that f⊣g iff
∀x,y∈A (f(x)⪯y⇔x⪯g(y))

Task 1A

Show that if f:A→A respects the partial order, then f is injective

Task 1B

Suppose f:A→A and g:A→A satisfy the following

• ∀x∈A (x⪯g(f(x)))
• ∀y∈A (f(g(y))⪯y)
• f and g both respect the partial order

Show that f⊣g

Submission instructions

You must submit a proof for each task in this challenge on Gradescope, either
handwritten and scanned or typeset. You should explicitly indicate which properties
of partial orders you are using to make each inference in your proof.

Bounded Quantification

When talking about sets, we often have formulas of the form ∀x∈A (H), for some set A and
formula H. This should be understood as a restricted form of universal quantification, where
we are only asserting the truth of the formula H for all choices of x in the set A. We can also
think of ∀x∈A (H) as syntactic sugar for ∀x(x∈A⇒H), which we already understand.
Additionally, we abbreviate consecutive universal quantifications,

like ∀x∈A (∀y∈A (∀z∈A (H))), as ∀x,y,z∈A (H)

Marking guide

Each task in this challenge is worth 1 mark each, which is awarded according to the
correctness and clarity of your proofs. A good proof will communicate clearly what
properties are being used to make each inference step in your proof.

Q2

Challenge Description

Consider the languages A, B and C over the alphabet Σ={a,b}, defined as follows:
A={w∈Σ∗ ∣∣∣ w does not contain anysubstring from L(ab∗a)},
B={w∈Σ∗ ∣∣∣ w does not contain anysubstring from L(ab|ba)},
C=A∩B.

Submission Instructions

Your task is to define three regular expressions:

1. Construct the simplest regular expression for A. Give your answer by
completing the definition of rA :: RegExp.

2. Construct the simplest regular expression for B. Give your answer by
completing the definition of rB :: RegExp.

3. Construct the simplest regular expression for C. Give your answer by
completing the definition of rC :: RegExp.

Simplicity

We define the simplicity of a regular expression as the total number of occurrences of

concatenations, unions, Kleene stars, alphabet symbols, empty string symbols, and empty set

symbols in the regular expression. The simplest regular expression for a regular language is

the regular expression with the smallest possible simplicity expressing that language.

For example, the regular expression (ϵ∪aba)∗ has two concatenations, one union, one
Kleene star, one empty string symbol, and three alphabet symbols, for a total simplicity of 8.

However, it is not the simplest possible regular expression for its language; (aba)∗ expresses
the same language, and has simplicity 6.

Hint: To understand the langauges you can build DFAs that recognise them. This task
does not require you to apply the algorithm for constructing regular expressions from
automata. Moreover, this algorithm usually produces overcomplicated regular
expressions. Instead, you should try to understand the language of the DFAs and
build a simple regular expression directly.

Format

RegExp.hs provides a regular expression type RegExp, and a parser for regular

expressions, parseRE :: String -> RegExp. Use this type to enter your answer.

For example:

• The regular expression (ba)∗ may be entered as follows:
o Star (Concat (Symbol ‘b’) (Symbol ‘a’)), or

o parseRE “(ba)*”.

• The regular expression a(ϵ∪ab) may be entered as follows:

o Concat (Symbol ‘a’) (Union EmptyStr (Concat (Symbol ‘a’)

(Symbol ‘b’))), or

o parseRE “a(eps | ab)”.

You can also check the simplicity of your regular expressions by using

the numberOfSymb :: RegExp -> Int function from RegExp.hs. Remember

to import RegExp before using it in ghci.

Marking guide

Each correct but not simplest regular expression is worth 1/3 marks. Each correct
simplest regular expression is worth 2/3 marks. The maximum marks for this problem
is 2. Note that the results of correctness and simplicity tests will not be visible after
the submission.

Q3

In this challenge, you will learn about a new kind of finite-state machine that can
produce output, which you will use to design a simple compression algorithm, and
generate a novelty sequence of numbers.

This challenge description is split into a few sections

1. The concept of run-length encoding, and the look-and-say sequence
2. The definition of a DFT, with an example
3. A Haskell implementation of DFTs, with an example
4. The task itself

You should return to these background slides as needed to understand the task
properly.

Run-length encoding (RLE) is a simple method of data compression, in which runs of
the same symbol in the data are encoded as the length of the run (the count) followed
by the symbol itself. For example, the RLE of the string aaabbcbbbbb would

be 3a2b1c5b, reducing the length of the string from 11 to 8. Such a string can then
be decoded by reconstructing each run from the pairs nx as a sequence of n-

many x’s, and concatenating them all. However this decoding depends on being able
to distinguish between the count and the symbol. For example, is 11112 an encoding

of eleven 1’s and one 2 or one 1 and eleven 2’s?
This can be solved by limiting the maximum length of each encoded run to a single

digit. We will set this limit to 3 and call this the 3-limited run-length encoding. For
example:

• 31121111↦1321123111
• 111133↦311123
• 22222222111↦32322231

To define this in general, we write an to denote a run of n-many a’s. Then the 3-
limited run-length encoding of a single run a3k+i, where a∈{1,2,3},k≥0 and i∈{0,1,2},
will be the string (3a)k if i=0 and (3a)kia if i≠0. The 3-limited encoding of an entire
string is then determined by concatenating all of the encodings of maximal length
runs in order of appearance.

Run-Length Encoding in Haskell

Here is that algorithm implemented in Haskell, as well as a decoding function. Test
them out by clicking the terminal button!

import Data.Char (digitToInt)

threeLimitedRLEncoding :: String -> String
threeLimitedRLEncoding “” = “”
threeLimitedRLEncoding (a:xs)
= show (length as) ++ [a] ++ threeLimitedRLEncoding rest
where

— the longest prefix-run of a’s from the first 3 characters
as = takeWhile (==a) (take 3 (a:xs))
— the remaining part of the string to encode
rest = drop (length as) (a:xs)

rlDecoding :: String -> String
rlDecoding “” = “”
rlDecoding [x] = “”
rlDecoding (i:x:xs) = replicate (digitToInt i) x ++ rlDecoding xs

The decoding function assumes every odd position is a digit, and ignores the final
symbol if the length of the input string is odd.

Look And Say

A novelty application of this is the Look-and-say sequence, introduced and analyzed

by John Conway ([1]), whose initial term is 1, and whose (n+1)th term is the run-
length encoding of the nth term. The first 10 terms of the sequence are as follows
1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
11131221133112132113212221
3113112221232112111312211312113211
1321132132111213122112311311222113111221131221
11131221131211131231121113112221121321132132211331222113112211
3113112221131112311311121321123113213221121113122113121113222123113221132122
21

It can be easily proved that the maximum length of any run in any term of the look-

and-say sequence is 3, and therefore it only contains the digits 1,2,3. Hence
the threeLimitedRLEncoding function will be sufficient for generating this
sequence.
[1] Conway J.H. (1987) The Weird and Wonderful Chemistry of Audioactive Decay. In:
Cover T.M., Gopinath B. (eds) Open Problems in Communication and Computation.
Springer, New York, NY. https://doi.org/10.1007/978-1-4612-4808-8_53

We will now explain a new model of computation, called a Deterministic Finite-State
Transducer or DFT for short, which you will use to implement a run-length encoder.
A DFT is much like a DFA, except it has an output tape, to which it can write strings of
symbols during execution.

The Definition

A DFT is a 5-tuple (Q,Σ,Γ,δ,q0) where

• Q is a finite set of states
• Σ is the input alphabet (not containing .)

https://oeis.org/A005150
https://doi.org/10.1007/978-1-4612-4808-8_53

• Γ is the output alphabet
• δ:Q×Σ.→Q×Γ∗ is the transition function, where Σ.=Σ∪{.}
• q0∈Q is the initial state

The symbol . is a special terminal symbol which plays a role in allowing the DFT to
recognise the end of the input string. It should not be present in the input alphabet or
in the strings of the input language.

The transitions of the DFT have the form δ(q,x)=(q′,v), which says that if the DFT is
in state q and the next input symbol is x, then it transitions to state q′ and writes the
string v to the end of the output tape. In a DFT diagram, the
transition δ(q,x)=(q′,v) is annotated like this.

Note that the input x∈Σ. can either be from the input alphabet Σ, or it could be the
special terminal symbol.

Running a DFT

Let T=(Q,Σ,Γ,δ,q0) be a DFT. A configuration of the DFT is a triple (w,q,u)∈Σ∗.×Q×Γ∗,
where w is the remaining input to be consumed, q is the current state of the DFT
and u is the string written on the output tape. We introduce the following
relation ⇒ on configurations, such that for x∈Σ.,w∈Σ∗.,q,q′∈Q, and u,v∈Γ∗
(xw,q,u)⇒(w,q′,uv) iff δ(q,x)=(q′,v)
This relation ⇒ describes how the DFT’s configuration changes as it consumes each
symbol from the input.

Let ∗⇒ be the reflexive-transitive closure of ⇒. We define a function ΘT:Σ∗→Γ∗,
called the transduction of T, such that ΘT(w)=u if and only if (w.,q0,ϵ)∗⇒(ϵ,qF,u),
for some state qF∈Q (notice the placement of the terminal symbol). This function will
be well-defined and total as long as δ is.

Decoding DFT Example

The following is a DFT with input alphabet {1,2,3}, whose transduction agrees
with rlDecoding on every string w∈{1,2,3}∗
({0,1,2,3},{1,2,3},{1,2,3},δ,0)
where the meaning of δ:{0,1,2,3}×{1,2,3,.}→{0,1,2,3}×{1,2,3}∗ is given in this DFT
diagram.

Data Types

We represent DFTs in Haskell as a tuple

(qs, ialph, oalph, ts, q0) :: ([Int], [Symbol], [Symbol], [Transn],

Int)

where

• qs :: [Int] is the list of states, each of type Int

• ialph :: [Symbol] is the list of symbols of the input alphabet (which does
not include ‘.’)

• oalph :: [Symbol] is the list of symbols of the output alphabet
• ts :: [Transn] is the list of transducer transitions (must be total)

• q0 :: Int is the start state

The type Transn is just ((Int,Symbol), (Int, [Symbol])), so a transducer

transition is a tuple ((q,x),(q’,zs)) :: ((Int, Symbol), (Int,[Symbol]),
where

• q :: Int is the source state

• x :: Symbol is the input symbol

• q’ :: Int is the target state

• v :: [Symbol] is the output string

In other words, a transition ((q,x),(q’,v)) represents δ(q,x)=(q′,v).

The Decoding DFT in Haskell

Here is the decoding DFT from the previous slide, implemented in Haskell

rlDecodeDFT = (qs, alph, alph, ts, 0)
where
alph = “123”
qs = [0,1,2,3]
ts = [ ((0, ‘1’), (1, “”))
, ((0, ‘2’), (2, “”))
, ((0, ‘3’), (3, “”))
, ((1, ‘1’), (0, “1”))
, ((1, ‘2’), (0, “2”))
, ((1, ‘3’), (0, “3”))
, ((2, ‘1’), (0, “11”))
, ((2, ‘2’), (0, “22”))
, ((2, ‘3’), (0, “33”))
, ((3, ‘1’), (0, “111”))
, ((3, ‘2’), (0, “222”))
, ((3, ‘3’), (0, “333”))
, ((0, ‘.’), (0, “”))
, ((1, ‘.’), (0, “”))
, ((2, ‘.’), (0, “”))
, ((3, ‘.’), (0, “”))
]

and again, a little more programmatically.

import Data.Char
import Data.List

rlDecodeDFT2 :: DFT
rlDecodeDFT2 = (qs, alph, alph, nub ts, 0)
where
alph = “123”
qs = [0,1,2,3]
ts = concat [ [ ((0, i’ ), (i, “” ))
, ((0, ‘.’), (0, “” ))
, ((i, y ), (0, replicate i y))
, ((i, ‘.’), (0, “” ))
]
| i <- [1,2,3], let i' = intToDigit i, y <- alph] We describe lists of transitions (which have some overlap), concatenate all of the lists and then remove duplicates. Note the uses of nub, concat, and the list of lists. The empty string Since the output string of a transition is a string, not a symbol, we use the actual empty string "" instead of the character ϵ, whenever we want a transition to write nothing to the output tape. Running a DFT in Haskell The function transduction :: DFT -> [Symbol] -> [Symbol] returns the
transduction function of a DFT, meaning transduction T w returns the final output

tape after running T on input w++[‘.’]. For example
*Main> transduction rlDecodeDFT “312312”

“111332”

We have defined rlDecodeDFT such that its transduction agrees with the

function rlDecoding on any string in {1,2,3}∗
This means that for any string w∈{1,2,3}∗, transduction rlDecodeDFT w returns
the same string as rlDecoding w.

Submission instructions

Your task for Challenge 3 is to define a DFT rlEncodeDFT :: DFT, whose
transduction agrees with the
function threeLimitedRLEncoding :: String -> String on any string in the

language {1,2,3}∗.
This means that transduction rlEncodeDFT w should always return the same

output as threeLimitedRLEncoding w, for any w∈{1,2,3}∗ (not just strings in the
look-and-say sequence).
The definition of threeLimitedRLEncoding :: String -> String can be found
in RLE.hs (or the second slide of this challenge). You should study and test this
function thoroughly to understand what your DFT needs to output.

Helper Functions

You might find the following functions and lists useful for constructing and testing your DFT

— From Data.Char
— intToDigit 1 == ‘1’
intToDigit :: Int -> Char
— digitToInt ‘1’ == 1
digitToInt :: Char -> Int

— From DFT.hs
— star xs is the language of strings over alphabet xs
star :: [Symbol] -> [[Symbol]]

— From RLE.hs
— The look-and-say sequence
lookAndSay :: [[Symbol]]

Both star “123” and lookAndSay are infinite lists, so use take n to print the

first n terms. You may import useful libraries like Data.Char and Data.List, just don’t

remove any existing imports.

Remember to import DFT and import RLE if you want to make use of these functions in

ghci.

Visualising DFTs

We also provide a visualisation tool for DFTs in VisDFT.hs. You can generate a diagram of

your DFT by calling visDFT rlEncodeDFT, which will write a png file to your workspace.

It can be a little messy, so you may want to include only a few transitions at a time (but

remember to put them all back before you submit!).

Remember to import VisDFT if you are calling this function from ghci.

Marking guide

You will receive 2 marks for a correct DFT. If your DFT is incorrect, you will receive
partial marks in proportion to the number test inputs your DFT correctly encodes.

Q4

Petri nets are applied in software design, business process management, and data
analysis to model and verify concurrent processes.

A Petri net is a tuple PN=(P,T,F), where

• P is a finite set of places;
• T is a finite set of transitions, such that P∩T=∅;
• F⊆(T×P)∪(P×T) is a set of arcs that connect places to transitions and

transitions to places.

Note that, according to the definition, arcs can only connect transition and places,
and two places (or two transitions) cannot be connected by an arc.

In Petri nets, places are represented by circles and transitions by boxes. Below is a
graphical representation of the Petri

net: PN=({p1,p2},{a,b},{(p1,a),(a,p2),(p2,b),(b,p1)}).

Semantics

Places may carry tokens that are graphically represented as black dots. Below is the

Petri net we considered before with 2 tokens in place p1 and no tokens in place p2.

A marking of a Petri net PN=(P,T,F), is a function m:P→N that maps places to
natural numbers. Markings define the number of tokens in each place. For example,

the marking of the Petri net pictured above is the function m, such
that m(p1)=2 and m(p2)=0. We can also order places and encode markings as
vectors over natural numbers: we can say that the marking m is a vector ⟨2,0⟩.

A transition t∈T is enabled in marking m if and only if ∀p∈P:(p,t)∈F⇒m(p)>0, or,
less formally, each of the input places of t contains at least one token. In the example
above, transition a is enabled in marking m, while transition b is not.
An enabled transition can fire. When a transition fires it consumes a token from each
of the input places and produces a token to each of the output places, yielding a new

marking. The figure below presents a marking m′=⟨1,1⟩ of the Petri net PN obtained
after the firing of transition a. Note that both a and b are enabled in m′.

Reachability Graphs

A pair (PN,m0), where PN is a Petri net and m0 is its initial marking is called a marked
Petri net.
We will consider the set of all markings reachable in zero, one or more steps

from m0 in PN. If this set of markings is finite we say that the marked Petri
net (PN,m0) is bounded. In this challenge, we will consider only bounded marked
Petri nets.

For a bounded marked Petri net (PN,m0) we can construct a reachability graph that
is a DFA where:

• states (except the dead state) correspond to the markings reachable

from m0 in PN;
• the alphabet is the set of Petri net transitions;

• the transition function corresponds to the Petri net firings: if in a marking m a
Petri net transition t fires leading PN to a marking m′, the reachability graph
contains a transition from m to m′ labelled by t; if a Petri net transition t is not
enabled in a marking m, then the reachability graph contains a transition
from m to the dead state labelled by t;

• m0 is the start state;
• all the states (except the dead state) are accept states.

Transitions, transitions!

There are two kinds of transitions here. Do not confuse them! There are Petri Net transitions,

which become the input symbols for the DFA we are defining, and DFA transitions in the

reachability graph, which have a source state (a marking), target state (a marking), and a label

(a Petri Net transition from the input alphabet).

The figure below presents a reachability graph for the marked Petri

net (PN,m0) discussed above, the initial marking m0 is ⟨2,0⟩.

Note that the reachability graph does not depend on the order of places in the vector
representation of markings.

One More Example

Consider one more marked Petri net with an initial marking presented by 1 token in

place p1:

The corresponding reachability graph is presented below:

In this graphical representation of the reachability graph, the dead state is omitted.

In the initial marking ⟨1,0,0,0,0,0⟩, transitions a and b are enabled and each of them
can fire. When one of them fires a new marking ⟨0,1,0,0,0,0⟩ is produced. In this
marking, only the transition c is enabled. When c fires it produces two tokens to
output places p3 and p4. In the marking ⟨0,0,1,1,0,0⟩, transitions d and e are enabled
and can fire in any order, eventually leading the Petri net to the

marking ⟨0,0,0,0,1,1⟩.

Challenge description

Your task is to design four DFAs that represent the reachability graphs of the marked
Petri nets visualised below.

1. Construct a reachability graph for (PN1,m1):

2. Construct a reachability graph for (PN2,m2):

3. Construct a reachability graph for (PN3,m3):

4. Construct a reachability graph for (PN4,m4).

Submission instructions

Submit your reachability graphs for parts 1, 2, 3 and 4 as Haskell expressions of
type DFA (named rg1, rg2, rg3, and rg4 respectively).

Important

1. Although we defined each state in the reachability graph to be a marking represented
by a vector, you must use a single Int for each state (you should rename the states to

distinct numbers).

2. The transition function for each DFA can be non-total, where a missing transition is
assumed to go to the dead state.

DFA Format:

Recall the DFA representation format. For example, let’s say we want to represent the

following DFA:

This DFA could be represented in Haskell as follows:

dfa = ([1,2], “ab”, t, 1, [2])
where
t = [ ((1,’a’),2)
, ((1,’b’),1)
, ((2,’a’),1)
, ((2,’b’),2)
]

Marking guide

Each correct reachability graph is worth 1/2 marks. The maximum marks for this
problem is 2. Note that the results of correctness tests will not be visible after the
submission.

Q5

Challenge description

Consider these two context-free grammars G1 and G2 over
alphabets Σ1={a,b,c} and Σ2={a,b}, respectively:

• G1=({S,A,C,T,U},Σ1,R1,S),
whereR1:S→AU∣TCA→a∣AaC→c∣CcT→ab∣aTbU→bc∣bUc

• G2=({S,T},Σ2,R2,S), whereR2:S→aT∣TbT→a∣b∣aT∣bT
The task is to identify any ambiguity in these grammars, and to construct a DFA that

recognises language L(G2). More specifically:
1. Determine whether G1 is ambiguous, and if it is, provide a string

in L(G1) which has two different parse trees.
2. Determine whether G2 is ambiguous, and if it is, provide a string

in L(G2) which has two different parse trees.
3. Construct a complete and minimal DFA for L(G2).

Here L(X) denotes the language generated by grammar X.

Submission instructions

For the first two parts, submit your answer by defining part1 and part2. Each
answer should be of type Answer as defined in the code box:

• If you find that a grammar is ambiguous, your answer should be of

form Ambiguous s, where s is a string that has different parse trees. The
string s should be of Haskell’s type String and only contain only symbols of
the alphabet, as in the example “abbac”.

• If you find that a grammar is unambiguous, your answer should be of the
form Unambiguous. You do not need to include a justification for your answer

in this case.
For part 3, provide a Haskell expressions of type DFA (named part3).
You may use any of the utility functions from DFA.hs, NFA.hs, or VisDFA.hs.

Moreover, HiddenAlgorithms.hs exports solutions to the problems of Assessed
Worksheet 4, including minimiseDFA, and determiniseNFA, some of which you may
find useful.

Note:

Recall the DFA representation format from Assessed Worksheets 3 and 4. For example, let’s

say we want to represent the following DFA:

This DFA could be represented in Haskell as follows:

dfa = ([1,2], “ab”, t, 1, [2])
where
t = [ ((1,’a’),2)
, ((1,’b’),1)
, ((2,’a’),1)
, ((2,’b’),2)
]

Your DFAs’ transition functions should be total.

Marking guide

Each correct answer is worth 2/3 marks. Not minimal correct DFA in part 3 is worth
1/3 marks. The maximum marks for this problem is 2. Note that the results of tests
will not be visible after the submission.

Q6

Challenge description

A deterministic pushdown automaton (DPDA) is a pushdown automaton that never
finds itself in a position where it can make two different transition steps. That is, from
any configuration (state, input symbol, stack symbol), there is at most one state it can
transition to. (If it is in a configuration where no state can be transitioned to, the
DPDA immediately rejects its input.) Each transition step consumes exactly one input
symbol. The stack operations are exactly as for PDAs: in one transition step, the
DPDA can push to the stack, pop from the stack, do both, or leave the stack
unchanged. We follow the conventions for PDA transitions, as used in lectures, that

is, in stack operations we use epsilon (ϵ) to mean ‘nothing’. So ‘pop x’ is captured as
‘replace x by ϵ’, ‘push x’ as ‘replace ϵ by x’, and ‘leave the stack untouched’ by
‘replace ϵ by ϵ’.
We also say that pushdown automaton is progressive if and only if each transition step
consumes exactly one input symbol.

Your task is to construct two progressive DPDAs, both having input alphabet {a,b,c}.
The DPDAs should recognise the following languages:

1. The language L={wcwR|w∈{ab,ba}∗}.
Note: For a string x, xR denote the reverse of x. For example, the reverse
of babbb is bbbab.
Here are some examples of strings in L: c, abcba, baababcbabaab.

2. The language R={apbqcr| p+q>r,p,q,r≥0}.
Here are some examples of strings in R: a, ab, aac. The following are not
in R: c, ac, cab.

You should try to be economic with the number of states (see the marking guide
below).

Submission instructions

You will submit two DPDAs, in the form of Haskell expressions of type DPDA.

1. dpdaL recognising language L.
2. dpdaR recognising language R.

You can find the type definition in the file DPDA.hs, where the constant eps is also
defined.

Note:

The way we represent DPDAs is not unlike what we did for DFAs. For example, we

represent the following DPDA:

in Haskell as follows:

dpda
= ([1,2,3,4], “01”, “0$”, delta, 1, [1,4])
where
delta = [ ((1,’0′,eps), (2,’$’))
, ((2,’0′,eps), (2,’0′))
, ((2,’1′,’0′), (3,eps))
, ((3,’1′,’0′), (3,eps))
, ((3,’1′,’$’), (4,’$’))
]

Marking guide

Each part is worth 1 mark.

• Part 1 will attract 0.5 marks for a correct DPDA dpdaL, and an additional 0.5
marks for one that has no more than 5 states.

• Part 2 will attract 0.5 marks for a correct DPDA dpdaR, and an additional 0.5

marks for one that has no more than 4 states.

Hint

To reduce the number of states in dpdaL, you might need to expand your stack alphabet

(with whatever symbols you like), in order to encode more specific information.

Note:

We have added a ‘DPDA visualiser’ to this challenge, so that you can view your DPDAs like

for DFAs and NFAs.

To view a DPDA called dpda, add import VisDPDA to the top of your solution, run, and

call visDPDA dpda. The DPDA will be checked for well-formedness (including the

deterministic-ness of its transition function), drawn using GraphViz, and saved in a new tab

in your editor called dpda.png. There is also a corresponding visDPDAnamed function. Let

us know if you experience any problems using this tool.