Programming Fundamentals 3 rd lecture Szabolcs Papp

Programming Fundamentals

3rd lecture

Szabolcs Papp

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Loops –

specification+algorithm+codingFirst example for arraysThe arrayArray instead of conditional statement

Constant arrays

Content

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Loops

Problem:Let’s calculate the non-1 least divisor of and integer (N>1)!

Specification:Input: N:IntegerOutput: D:IntegerPrecondition: N>1Post condition: 1<DN and D|

N and i (2i<D): i ł N

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Loops

An idea for solution:Let’s try 2; if not good then try 3; if still not good then try 4, …, worst case N will be a divisor!

Algorithm expressing this:i:=2i ł Ni:=i+1

D:=i

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Loops

Problem:Let’s calculate both the non-1 least and the non-N greatest divisor of an integer (N>1)!

Specification:Input: N:IntegerOutput: LD,GD:IntegerPrecondition: N>1Post condition: 1<LDN and

1GD<N andLD|N andi (2i<LD): i ł N andGD|N andi (GD<i<N): i ł N

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Loops

Note:By knowing LD, GD can be definied in post condition in another way: LD*GD=N!

Algorithm expressing this:i:=2

i ł Ni:=i+1

LD:=IGD:=N Div LD

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Loops

Problem: Let’s calculate the non-1 and non-N least divisor of and integer (N>1), if exists!

Specification:Input: N:IntegerOutput: D:Integer,

EXISTS:BooleanPrecondition: N>1Post condition: EXISTS=i (2i<N): i|

N andEXISTS D|N and2D<N andi (2i<D): i ł N

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Loops

Algorithm:

Note:Whenever i is a divisor of N, then (N Div i) is also a divisor, so we are good to go until the square root of N!

i:=2i<N and i ł

Ni:=i+1

EXISTS:=i<N

EXISTSD:=i

Ni

I N 2i N Div i N Div 2

thusi*i N

thusi N

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Loops

Problem:Let’s calculate the sum of all the divisors of an integer (N>1)!

Specification:Input: N:IntegerOutput: S:IntegerPrecondition: N>1Post condition:

N

Ni

1i

iS

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Loops

Algorithm:

S:=0i=1..N

i|NS:=S+i

I N

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Loops

Problem: Let’s calculate the sum of all odd divisors of an integer (N>1)!

Specification:Input: N:IntegerOutput: S:IntegerPrecondition: N>1Post condition: S=

N

i

i

odd(i) and Ni

1

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Algorithm1:

Algorithm2:

Loops

S:=0i=1..N

i|N és odd(i)S:=S+i

S:=0i=1..N; by 2

i|NS:=S+i

I N

I N

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Loops

Problem:Let’s calculate the sum of all prime divisors of an integer (N>1)!


isprime(i)???

N

isprime(i)andNi

2i

i

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Loops

Algorithm:The least divisor is prime for sure; let’s divide N by it as many times as we can; the next divisor of the reduced N will be prime again.S:=0

i:=2iN

i|NS:=S+i

i|NN:=N Div i

i:=i+1

I N

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Loops

Problem:Let’s determine the count of i,j (i,j>1) integer pairs, where i*j<N!


2 Div N

2i

2 Div N

Nji2j

1

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Loops

S:=0i=2..N Div 2

j=2..N Div 2i*j<N

S:=S+1

Algorithm1:

I N

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Loops

S:=0i=2..N Div 2

j:=2i*j≤N

S:=S+1j:=j+1

Algorithm2:

Algorithm3:S:=0

i=2..N Div 2S:=S+(N Div i)-1

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Loops

Let’s observe that: If there is , , or sign in post condition,

then the solution always contains loop! If there is or sign in post condition,

then the solution is mostly a conditional loop!

If there is sign in post condition, then the solution is mostly a for loop! ( also)

In case of two embedded signs, we will have two for loops embedded in each other.

In case of conditional , there will be a conditional statement within the loop.

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Loopsalgorithm – code

Conditional loop:

For loop:

conditionstatement

while (condition){ statements}

i=1..Nstateme

nt

for (int i=1;i<=N;i++){ statements}

i=1..N; by xstatement

for (int i=1;i<=N;i+=x){ statements}

statement

condition

do{ statement}while (condition)

Typical usage at Typical usage at console inputsconsole inputs

((seesee 2 2ndnd lecture lecture))

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Sample exercise for conditional

statement, or do we need something

else?Problem:The Japanese calendar contains 60 years cycles. The years are paired, and a color is assigned to each pair (green, red, yellow, white and black).

o 1,2,11,12, …,51,52: green yearso 3,4,13,14,…,53,54: red yearso 5,6,15,16,…55,56: yellow yearso 7,8,17,18,…57,58: white yearso 9,10,19,20,…,59,60: black years

We know that the last cycle has started in 1984 that will end in 2043. Let’s write a computer program which determines the color assigned to a given M year (1984≤M≤2043)!

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Conditional statement

or do we need something

else?Specification1: Input year:Integer Output: c:Color Precondition: 1984≤year and year≤2043 Post cond. ((year-1984) Mod 10) Div 2=0

and c=”green” or((year-1984) Mod 10) Div 2=1

and c=”red” or …

Definition: Color:=(”green”,”red”, ”yellow”,”white”,

”black”) Text

This is still an

“unknown”set.

We define the set Color,

which is derived

from Text

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else?Specification2: Input: year:Integer Output: c:Color

Color=(”green”,”red”,”yellow”, ”white”,”black”) Text

Precondition: 1984≤year and year≤2043

Post cond: ((year-1984) Mod 10) Div 2=0 and

c=”green” or((year-1984) Mod 10) Div 2=1

and c=”red” or …

We can define

Color set here, too

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else?

y:=((year-1984) Mod 10) Div 2y=0 y=1 y=2 y=3 y=4

sc:=”green”

s:= ”red”

s:= ”yellow”

s:= ”white”

s:= ”black”

Question:Would we do the same if we had 90 ways instead of 5?

Before answering this, a new structure: array

Algorithm:

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Arrays

Related concepts: Sequence: a countable sequence of

same typed data elements Array: finite length sequence, we can

define operations for ith element (we know the least and the greatest index, or the count).

Index: sometimes 1..N, sometimes 0..N–1.

Operations on elements of an array: reference to an element, modification of an element (the element is selected by a given index).

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Arrays(in algorithm)

Examples for declaration:X:Array[1..N:Integer]Y:Array[0..4:Text]We can represent the Color set from the previous example with a constant array:Constant Color:Array[0..4:Text]= (”green”,”red”,”yellow”,”white”,”black”)

Examples for referencing:X[1]:=X[N]; Y[i]:=Color[0]

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Declaration:X:Array[1..N:Integer]Y:Array[0..4:Text]We can represent the Color set from the previous example with a constant array:Constant Color:Array[0..4:Text]= (”green”,”red”,”yellow”,”white”,”black”)

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Arrays(in algorithm and code)

string Y[5]

const string Color[5]={"green","red", "yellow","white","black"};

int X[N+1]

Tömb-Tömb-elemszámelemszám

indexelésindexelés 0 0..N..N

indexindex 0 0..??? ..??? count of count of elementselements

in C++:

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Arrays(C++ code – overview)

Static arrays: Declaration:

References:

const int MaxN=???;//maxcount type array[MaxN]; //array declaration //with indices 0..MaxN-1 …

… array[ind] //reference to an element … array[ind]=expr;//modification of elem. …

This is a This is a difference difference between usual between usual

algorithm and code!algorithm and code!

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Arrays (C++ code – overview)

Static array constants: Declaration:

or

const int N=???;//number of elements const type array[N]={t1,t2,…,tN}; //declaration of constant array, //with indices 0..N-1

const type array[]={t1,t2,…,tN}; //declaration of const. array const int N=sizeof array/sizeof(type); //number of elements //indices: 0..N-1

Note that syntax is Note that syntax is different in case of different in case of

identifiers and types!identifiers and types!

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Array variables: Declaration:

Creation:

References (the same):

int N; //actual number of elements …

N=???;//N to be defined, e.g. input type array[N];//an array containing N //elements with type “type”

… array[ind] //reference to an element … array[ind]=expr;//modification of elem. …

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Array input (with check):

bool err; //is there an error? … do{ cout << "Count:"; cin >> N; err=!!IsCorrectN(N); if (err) { cout << „error message" << endl; } }while (err);


IsCorrectN does both IsCorrectN does both syntax and semantic syntax and semantic

checkcheck


In algorithmIn algorithm::InputInput:: N N [[IsCorrectIsCorrectNN(N))(N))]]

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Array input (with check):

Let’s get back to our question…



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for (intfor (int i i==00;;ii<N<N;;++i++i)){{ do{ do{ cout << cout << ii+1+1 << << ""..:":"; cin >> ; cin >> v[i]v[i];; errerr=!=!CorrectCorrect(v[i])(v[i]);; if ( if (errerr)) { { cout << " cout << "hibaüzenethibaüzenet" << endl;" << endl; }; }; }while ( }while (hibahiba););}}

for (intfor (int i i==00;;ii<N<N;;++i++i)){{ do{ do{ cout << cout << ii+1+1 << << ""..:":"; cin >> ; cin >> v[i]v[i];; errerr=!=!CorrectCorrect(v[i])(v[i]);; if ( if (errerr)) { { cout << " cout << "hibaüzenethibaüzenet" << endl;" << endl; }; }; }while ( }while (hibahiba););}}

In algorithmIn algorithm::InputInput:: v(1..N) v(1..N) [[HelyesHelyes(v(1..N))(v(1..N))]]

Correct does both Correct does both syntax and semantic syntax and semantic

checkcheck


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Array instead of conditional statement

Specification: Input: year:Integer Output: c:Color Color=(”green”,”red”,”yellow”,

”white”,”black”) Text Constant Colors:Array[0..4:Color]= (”green”,”red”,”yellow”,”white”,”black”)

Precondition: 1984≤year and year≤2043 Post condition:

c=Colors[((year-1984) Mod 10) Div 2]

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Array instead of conditional statement

Algorithm:

y:=((year-1984) Mod 10) Div 2c:=Colors[y]

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Applying constant arrays

Problem:Let’s create a computer program, which writes an integer between 1 and 99 with letters!

Specification: Input: N:Integer

Constant ones:Arrays[0..19:Text]=

(””, ”one”,…,”nineteen”)Constant tens: Array[2..9:Text]=

(”twenty”,…,”ninety”) Output: S:Text Preconditon: 1N99

This is a This is a better place better place

here.here.Constant Constant

array is an array is an output from output from the point of the point of view of the view of the algorithm.algorithm.

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Post condition: N<20 S=ones[N] and

N≥20 S=tens[N Div 10]+ ones[N Mod 10]

Algorithm:

N<20 otherwhiseS:=ones[N]

S:=tens[N Div 10]+ ones[N Mod 10]

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Problem:Let’s create a computer program that writes the serial number of a month based on the name of the month.

Specification: Input: H:Text Constant

MonName:Array[1..12:Text]= (”January”,…,”December”)

Output: S:Integer Precondition: MMonName Post cond.: 1S12 and

MonName[S]=M

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Algorithm:

Question: What would happen if we didn’t meet the precondition? Runtime error? Infinite loop?


i:=1

MonName[i]M

i:=i+1

S:=i

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Constant arrays – What do we store?

Problem:We have a non leap year. What is the serial number of a given day (month, day)?

Specification1: Input: M,D:Integer

Constant mon:Array[1..12:Integer]= (31,28,31,…,31)

Output: S:Integer Precondition: 1≤M≤12 és 1≤D≤mon[M] Post condition:

1M

1i

mon[i]DS

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Algorithm:

Homework: How to modify this if we allow leap year?

S:=D

i=1..M–1

S:=S+mon[i]

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Another solution:For each months, let’s store the total number of days preceding the given month!

Specification2: Input: …

Constant mon:Array[1..12:Integer]=

(0,31,59,90,…,334) Post condition: S=D+mon[M]

Question: Is it a better solution? How do we express the precondition?

Programming Fundamentals

End of 3rd lecture

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Programming Fundamentals 3 rd lecture Szabolcs Papp