Programming Fundamentals 9 th lecture Szabolcs Papp
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Programming Fundamentals
9th lecture
Szabolcs Papp
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More Patterns of Algorithms (PoA)
Copy – calculation with function Multiple item selection Partitioning Intersection Union
PoAs – summary
Content
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Further PoAs
What is Pattern of Algorithm (PoA)?It is a general solution for a typical programming task.
Array single valueArray ArrayArray ArraysArrays Array
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7. Copy calculation with
function
Example tasks: Let's define the absolute values of
all elements in an array! Let's convert all letters of a text
to lowercase! Let's calculate the sum of two
vectors! Let's calculate sin(x) values for a
given x series! We know the ordinal number of N
months, let's give their names!
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7. Copy calculation with
function
What is in common? We have N something, and we have to assign N other things to these. The type of assigned values can differ from the type of original values, but the count (N) remains the same, as well as the order.
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7. Copy calculation with
function
Specification: Input: N:Integer,
X:Array[1..N:Type1]
Output: Y:Array[1..N:Type2] Precondition: N0 Post condition: i (1≤i≤N):
Y[i]=f(X[i])
Algorithm:i=1..N
Y[i]:=f(X[i])
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7. Copy calculation with
function
Specification (a frequent special case): Input: N:Integer,
X:Array[1..N:Type1]
Output: Y:Array[1..N:Type2] Precondition: N0 Post cond.: i (1≤i≤N):
Algorithm:i=1..N
T(X[i])
Y[i]:=f(X[i])
Y[i]:=X[i]
otherwise, iX
iX Tif, iXfY[i]
Y N
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7. Copy calculation with
function
Specification (other special case): Input: N:Integer,
X,Z:Array[1..N:Real] Output: Y:Array[1..N:Real] Precondition: N0 Post condition: i(1≤i≤N): Y[i]=X[i]
+Z[i]Algorithm:
i=1..N
Y[i]:=X[i]+Z[i]
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8. Multiple item selection
Example tasks: Let's define all excellent students
in a class! Let's calculate the divisors of an
integer! Let's list all the vowels from an
English word! Let's garble all the people who are
taller than 180cm, from a set of people!
Let's list those days of the year when it didn't freeze!
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8. Multiple item selection
What is in common?We have to list all elements from N "something", which have a common attribute T.
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8. Multiple item selection
Specification: Input: N:Integer,
X:Array[1..N:Type] Output: Cnt:Integer,
Y:Array[1..N:Integer] Precondition: N0 Post cond.: Cnt= and
i(1≤i≤Cnt): T(X[Y[i]])
and Y(1,2,…,N)
N
])i[X(T1i
1See PoA "Count"
In case of static In case of static array array
declarationdeclaration
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8. Multiple item selection
Algorithm:
Comment:Collecting ordinal numbers is more general than collecting elements. If we needed elements then we would have: Y[Cnt]:=X[i]. (And accordingly modified specification, too.)
Cnt:=0
i=1..N
T(X[i])
Cnt:=Cnt+1 Y[Cnt]:=i
Y N
See PoA "Count"
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8. Multiple item selection (not freezing days)
Specification: Input: N:Integer,
H:Array[1..N:Real] Output:
Cnt:Integer,NF:Array[1..N:Integer] Precondition: N0 Post cond.: Cnt= and
i(1≤i≤Cnt): H[NF[i]]>0 and
NF(1,2,…,N)
N
0H[i]1i
1
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8. Multiple item selection (not freezing days)
Algorithm:
Cnt:=0
i=1..N
H[i]>0
Cnt:=Cnt+1
NF[Cnt]:=i
Y N
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10. Partitioning
Example tasks: Let's define all excellent and non-
excellent students in a class! Let's garble all the people who are
taller than 180cm, and others who are not taller than 180cm, from a set of people!
Let's list all even and all odd numbers from a series!
Let's list those days of the year when it did freeze and didn't freeze!
Let's list all the vowels and consonants from an English word!
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10. Partitioning
What is in common? We have to list all elements from N "something", which have a common attribute T, and then also list those ones not having attribute T.
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10. Partitioning
Specification: Input: N:Integer,
X:Array[1..N:Type] Output: Cnt:Integer,
Y,Z:Array[1..N:Integer] Precondition: N0 Post cond.: Cnt= and
i(1≤i≤Cnt): T(X[Y[i]]) and i(1≤i≤N–Cnt): not T(X[Z[i]])
and Y(1,2,…,N) and Z(1,2,…,N)
N
])i[X(T1i
1
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10. Partitioning
Algorithm:
Comment:If we needed the partitioned elements, than here we could also have:=X[i] instead of :=i (And accordingly modified specification!)CntZ can be eliminated. How?
Cnt:=0
CntZ:=0
i=1..N
T(X[i])
Cnt:=Cnt+1
CntZ:=CntZ+1
Y[Cnt]:=I Z[CntZ]:=i
Y N
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10. Partitioning
Permutation(1,2,...,N):=set of all permutations of
1..N (set of sets)
Modification:There are exactly N number of elements in Y and Z together, so it's sufficient to have only one array. We will collect the T attributed elements at the beginning of the resulting array Y.
Solution: Input: N:Integer, X:Array[1..N:Type] Output: Cnt:Integer, Y:Array[1..N:Integer] Precondition: N0 Post cond.: Cnt= and
i(1≤i≤Cnt): T(X[Y[i]]) and i(Cnt+1≤i≤N): not T(X[Y[i]])
and YPermutation(1,2,…,N)
N
])i[X(T1i
1
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10. Partitioning
Algorithm:Cnt:=0
CntZ:=N+1
i=1..N
T(X[i])
Cnt:=Cnt+1
CntZ:=CntZ–1
Y[Cnt]:=I Y[CntZ]:=i
Y N
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11. Intersection
Example tasks: Let's list all common divisors of
two integers! We investigate birds in winter
and summer. Let's collect the birds which are not migratory!
Based on the free hours from the calendars of two persons, let's determine when they can meet.
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11. Intersection
What is in common?We have two sets as input (with same typed items), and we have to list all elements that are part of both sets!
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11. Intersection
One item included One item included only onceonly once
Specification: Input: N,M:Integer,
X:Array[1..N:Type] Y:Array[1..M:Type]
Output: Cnt:Integer, Z:Array[1..Cnt:Type]
Precondition: N0 and M0 andIsSet(X) and IsSet(Y)
Post cond.: Cnt= and
i(1≤i≤Cnt): (Z[i]X and Z[i]Y) and
IsSet(Z)
N
YX[i]1i
1
Could be alsoCould be also: : min(N,M)min(N,M)
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11. Intersection
Algorithm:
Comment:The solution is a multiple item selection and a decision.
Cnt:=0i=1..N
j:=1
j≤M and X[i]≠Y[j]
j:=j+1
j≤M
Cnt:=Cnt+1
Z[Cnt]:=X[i]
Y N
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11. Intersection
Variations for intersection:
We have two sets, we have to determine the number of common elements.
We have two sets, we have to determine if they have at least one common element.
We have two sets, we have to provide one common element.
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12. Union
Example tasks: We have two courses, let's list the
students visiting both courses! Let's collect all the birds we can
investigate in winter as well as in summer!
Based on the free hours from the calendars of two persons, let's determine when we can reach at least one of them!
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12. Union
What is in common?We have two sets as input (with same typed items), and we have to list all elements that are included at least in one of the sets.
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12. Union
Specification: Input: N,M:Integer,
X:Array[1..N:Type] Y:Array[1..M:Type]
Output: Cnt:Integer, Z:Array[1..Cnt:Type]
Precondition: N0 and M0 andIsSet(X) and IsSet(Y)
Post cond.: Cnt=N+ and
i(1≤i≤Cnt): (Z[i]X or Z[i]Y) and IsSet(Z)
M
XY[j]
1j1
Could be alsoCould be also: : N+MN+M
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12. Union
Algorithm:Z:=XCnt:=N
j=1..Mi:=1
i≤N and X[i]≠Y[j]
i:=i+1
i>N
Cnt:=Cnt+1
Z[Cnt]:=Y[j]
Y N
CopyCopy
DecisionDecision
Multiple item Multiple item selectionselection
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Patterns of Algorithms (PoAs)
Sequence Sequence7. Copy – calculation with function
8. More item selection9. Sort (will have later) Sequence Sequences10.Partitioning Sequences Sequence11.Intersection12.Union
Programming Fundamentals
End of 9th lecture
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