Tony Ballantyne

Common Lisp Loops

I can’t think why you wouldn’t use the Common Lisp loop macro in Emacs. Don’t forget to (require ‘cl-lib)

Simple Loops

;; keywords: to, upto, below, downto, above, by
;; keywords: collect, append, nconc, count, sum, maximize, minimize

(cl-loop
    do (princ 1))
=> 111111... infinite loop

(cl-loop repeat 5
     do (princ 1)
     (princ 2))
=> 1212121212

(cl-loop for i from 1 below 10
     maximize i)
=> 9

(cl-loop for x from 9 downto 1
     collect x)
=> (9 8 7 6 5 4 3 2 1)

(cl-loop for i from 10 above 1 by 2
     collect i)
=> (10 8 6 4 2)

Looping over Sets and Arrays

;; keywords: in, on, across, by
;; Remember in for lists
;; across for vectors

(cl-loop for l in '(1 2 3 4 5)
     do (princ l))
=> 12345

(cl-loop for l in '(1 2 3 4 5) by #'cddr
     collect l)
=> (1 3 5)

(cl-loop for l on '(1 2 3 4 5)
     collect l)
=> ((1 2 3 4 5) (2 3 4 5) (3 4 5) (4 5) (5))

(cl-loop for l on '(1 2 3 4 5) by #'cddr
     collect l)
=> ((1 2 3 4 5) (3 4 5) (5))

;; Remember that a string is an array in lisp, so...
;; Add up the digits in 90125
(cl-loop for d across "90125"
    sum (- d 48))
=> 17

Destructuring

(cl-loop for (a nil) in '((1 2) (2 3) (3 4))
     collect a)
=> (1 2 3)

(cl-loop for (a b) in '((1 2) (2 3) (3 4))
     collect (* a b))
=> (2 6 12)

(cl-loop for (a b) on '(1 2 3 4 5) while b
      collect (+ a b))
=> (3 5 7 9)

(cl-loop for (a b) on '(1 2 3 4 5) by #'cddr while b
     collect (+ a b))
=> (3 7)

Hashtables

(cl-loop for key being the hash-keys of myhashtable
        using (hash-value value)
        do (princ value))

Parallel fors

(cl-loop for i from 1 to 5
     for l in '(a b c d)
     collect (list i l))
=> ((1 a) (2 b) (3 c) (4 d))

(cl-loop for i from 1 to 5
     for j from 2 to 10 by 2
     collect (* i j))
=> (2 8 18 32 50)

Nested fors

(cl-loop for i from 1 to 5
     collect (cl-loop for j from 1 to 5
             collect (* i j)))
=> ((1 2 3 4 5) (2 4 6 8 10) (3 6 9 12 15) (4 8 12 16 20) (5 10 15 20 25))

(cl-loop for i from 1 to 5
     append (cl-loop for j from 1 to 5
             collect (* i j)))
=> (1 2 3 4 5 2 4 6 8 10 3 6 ...)

(cl-loop for i from 1 to 5
     collect (cl-loop for j from 1 to 5
             sum (* i j)))
=> (15 30 45 60 75)

(cl-loop for i from 1 to 5
     sum (cl-loop for j from 1 to 5
             sum (* i j)))
=> 225

Selection

;; if, when, unless

(cl-loop for i from 1 to 20
     unless (cl-evenp i) collect i)
=> (1 3 5 7 9 11 13 15 17 19)

(cl-loop for i from 1 to 20
     when (= (% i 3) 0) collect i into fizz
     when (= (% i 5) 0) collect i into buzz
     finally return (list fizz buzz))
=> ((3 6 9 12 15 18) (5 10 15 20))

(cl-loop for i from 1 to 20
     if (and (= (% i 3) 0) (= (% i 5) 0)) collect i into fizzbuzz
     else if (= (% i 3) 0) collect i into fizz
     else if (= (% i 5) 0) collect i into buzz
     finally return (list fizz buzz fizzbuzz))
=> ((3 6 9 12 18) (5 10 20) (15))

(cl-loop for i from 1 to 10
     if (cl-evenp i)
     collect i into evens
     and sum i into evensum
     else
     collect i into odds
     and sum i into oddsum
     finally return (list evens evensum odds oddsum))
=> ((2 4 6 8 10) 30 (1 3 5 7 9) 25)

Find c from comp where diff is never a member of squares
(cl-loop for c in comp
     if  (cl-loop for p in pri
              for diff = (/ (- c p) 2)
              never (member diff squares))
     collect c)

Then Iteration

(cl-loop for i from 1 to 5
     for square = (* i i)
     collect square)

;; Though you'd be better with
(cl-loop for i from 1 to 5
     collect (* i i))

;; However, this leads to Triangle Numbers
(cl-loop for n from 1 to 10
     for triangle = 1 then (+ triangle n)
     collect triangle)
=> (1 3 6 10 15 21 28 36 45 55)

(cl-loop for x = 0 then y
     for y = 1 then (+ x y)
     while (< y 30)
     collect y)
=> (1 2 4 8 16)

;; Fibonacci Sequence (note the and)
(cl-loop for x = 0 then y
     and y = 1 then (+ x y)
     while (< y 30)
     collect y)
=> (1 1 2 3 5 8 13 21)

Termination

;; while, until, always, never, and thereis

while and until are straightforward

(cl-loop for n from 1
for tri = 1 then (+ tri n)
until (> (divs tri) 500)
finally return tri)

never, thereis and always are shorthand for a combination of when-return

(defun isprime(n)
  (cond
   ((< n 2) nil)
   ((= n 2) t)
   ((cl-evenp n) nil)
   (t
    (cl-loop for i from 3 to (sqrt n) by 2
         never (= (% n i) 0)))))

– More Collections: Return to Lisp Problems Solutions

1 Last Item

Find the last item in an List of String objects

1.1 Solution

1: String lastItem(List<String> a)
2: {
3:     return  a.get(a.size()-1);
4: }

2 Last Item Generic

Find the last item in a Generic List

2.1 Solution

1: Object gLastItem(List<?> a)
2: {
3:     return  a.get(a.size()-1);
4: }

3 Last But One

Find the last but one item in a List

3.1 Solution

1: Object lastButOneItem(List<?> a)
2: {
3:     if (a.size() > 1) 
4:  return  a.get(a.size()-2);
5:     else  
6:  return "List too small";
7: }

4 Iterate through the Elements of a List

Iterate through the Elements of a List, printing them out in turn

4.1 Solution

1: void printEach(List<?> a)
2: {
3:     for (Object e: a)
4:     {
5:  System.out.println(e);
6:     }        
7: }

or using a stream

1: void streamPrintEach(List<?> list)
2: {
3:     list
4:  .stream()
5:  .forEach(System.out::println);
6: 
7: }

5 Eliminate consecutive duplicates from a List.

Write a method that removes consecutive duplicates from a List. The order of the elements should not be changed.

5.1 Solution

1: List<?> removeDuplicates(List<?> a)
2: {
3:     for (int i = 0; i<a.size()-1; i++)
4:     {
5:  if(a.get(i) == a.get(i+1)) a.remove(a.get(i));        
6:     }
7:     return a;
8: }

6 Eliminate consecutive duplicates from a List using HashSet

Remembering that a HashSet cannot contain duplicates, repeat the above question using a HashSet

6.1 Solution

1: HashSet<String> hashRemoveDuplicates(List<String> a)
2: {
3:     HashSet<String> hSet = new HashSet<String>(a);
4:     return hSet;
5: }

7 RLE

Run-length encoding of a list. Consecutive duplicates of elements are encoded as pairs where N is the number of duplicates of the element E.

Note this solution uses the Pair class, defined in the question

7.1 Solution

 1: List<Pair> RLE(List<?> list)
 2: {
 3:     List<Pair> encoded = new ArrayList<Pair>();
 4:     for (int i = 0; i < list.size(); i++)
 5:     {
 6:     int runLength = 1;
 7:     while (i < list.size() -1 && list.get(i) == list.get(i+1))
 8:     {
 9:         runLength++;
10:         i++;
11:     }
12: 
13:      Pair<Object> pair = new Pair<Object>(runLength, list.get(i));
14:      encoded.add(pair);
15:     }
16:     return encoded;
17: 
18: }

8 Decode a run-length encoded list

Given a run-length code list generated in a previous problem, construct its uncompressed version.

Note this solution uses the Pair class, defined in the question

8.1 Solution

 1: List<Object> decodeRLE(List<Pair> list)
 2: {
 3: 
 4:     List<Object> decoded = new ArrayList<Object>();
 5:     for (Pair pair: list)
 6:     {
 7:     for (int i =0; i<(int)pair.count; i++)
 8:     {
 9:         decoded.add(pair.value);
10:     }
11:     }
12: 
13:     return decoded;
14: }

9 Duplicate the elements of a list

9.1 Solution

 1: List<String> dupli(List<String> list)
 2: {
 3:     List<String> dup = new ArrayList<String>();
 4: 
 5:     for(String s: list)
 6:     {
 7:     dup.add(s);
 8:     dup.add(s);
 9:     }
10: 
11:     return dup;
12: }

or, using a stream

1: List<String> streamDupli(List<String> list)
2: {
3:     List<String> dupli = list
4:              .stream()
5:              .collect(Collectors.toList());
6: 
7:     return dupli;        
8: 
9: }

10 Replicate the elements of a list a given number of times

10.1 Solution

 1: List<String> repli(List<String> list, int num)
 2: {
 3:     List<String> rep = new ArrayList<String>();
 4: 
 5:     for(String s: list)
 6:     {
 7:     for (int i=0; i<num; i++)
 8:     {
 9:         rep.add(s);
10:     }
11:     }
12: 
13:     return rep;
14: }

7: More Collections: Return to Lisp Problems

These problems were inspired by the 99 Lisp problems. Lisp is a very different language to Java; its use of lists means it’s not appropriate to emulate all the original problems in Java, however, there is some benefit in practising the use of Java collections.

The introduction of lambda functions in Java8 made me rethink how best to approach these problems. There are three things being practised here: collections, generics and lambdas. I’ve split up the questions so that they give a chance to practice these different skills.

1 Last Item

Find the last item in an List of String objects

1.1 Example

1: List<String> myList = new ArrayList<>(asList("knife", "fork", "spoon"));
2: System.out.println(lastItem(myList));
3:  *** Output ***
4: spoon

2 Last Item Generic

Find the last item in a Generic List

2.1 Example

1: List<String> myList = new ArrayList<>(asList("knife", "fork", "spoon"));
2: List<Integer> anotherList = new ArrayList<>(asList(1,2,3));
3: System.out.println(gLastItem(myList));
4: System.out.println(gLastItem(anotherList));
5:  *** Output ***
6: spoon
7: 3

3 Last But One

Find the last but one item in a List

3.1 Example

1: List<String> myList = new ArrayList<>(asList("knife", "fork", "spoon"));
2: System.out.println(lastButOneItem(myList));
3:  *** Output ***
4: fork

4 Iterate through the Elements of a List

Iterate through the Elements of a List, printing them out in turn

4.1 Example

1: List<String> myList = new ArrayList<>(asList("knife", "fork", "spoon"));
2: System.out.println(printEach(myList));
3:  *** Output ***
4: knife
5: fork
6: spoon

5 Eliminate consecutive duplicates from a List

Write a method that removes consecutive duplicates from a List. The order of the elements should not be changed.

5.1 Example

1: List<String> myList = new ArrayList<>(asList("knife", "knife", "fork", "spoon","spoon"));
2: printEach(removeDuplicates(myList));
3:  *** Output ***
4: knife
5: fork
6: spoon

6 Eliminate consecutive duplicates from a List using HashSet

Remembering that a HashSet cannot contain duplicates, repeat the above question using a HashSet

6.1 Example

1: List<String> myList = new ArrayList<>(asList("knife", "knife", "fork", "spoon","spoon"));
2: printEach(removeDuplicates(myList));
3:  *** Output ***
4: knife
5: fork
6: spoon

Remembering that a HashSet cannot contain duplicates

7 RLE

Run-length encoding of a list. Consecutive duplicates of elements are encoded as pairs where N is the number of duplicates of the element E.

Use this Pair class

 1: class Pair<T>
 2:    {
 3:        int count;
 4:        T value;
 5: 
 6:        Pair(int count, T value) 
 7:        {
 8:        this.count = count;
 9:        this.value = value;
10:        }
11: 
12:        void print()
13:        {
14:        System.out.print("("+ count + ", " + value +") ");
15:        }
16:    }

7.1 Example

1: List<String> myList = new ArrayList<>(asList("knife", "knife", "fork", "spoon","spoon"));
2:  List<Pair> rle = RLE(myList);
3:  for (Pair r: rle)
4:  {
5:      r.print();
6:  }
7:  System.out.println("");
8:  *** Output ***
9: (2, knife) (1, fork) (2, spoon)

8 Decode a run-length encoded list

Given a run-length code list generated in a previous problem, construct its uncompressed version.

8.1 Example

 1: List<String> myList = new ArrayList<>(asList("knife", "knife", "fork", "spoon","spoon"));
 2:     List<Pair> rle = RLE(myList);
 3:     for (Pair r: rle)
 4:     {
 5:         r.print();
 6:     }
 7:     System.out.println("");
 8: 
 9:     List<Object> decoded = decodeRLE(rle);
10: 
11:     for (Object o: decoded)
12:     {
13:         System.out.println(o);
14:     }
15:  *** Output ***
16: (2, knife) (1, fork) (2, spoon) 
17: knife
18: knife
19: fork
20: spoon
21: spoon

9 Duplicate the elements of a list

9.1 Example

 1: List<String> myList = new ArrayList<>(asList("knife", "knife", "fork", "spoon","spoon"));
 2: printEach(dupli(myList));
 3:  *** Output ***
 4: knife
 5: knife
 6: knife
 7: knife
 8: fork
 9: fork
10: spoon
11: spoon
12: spoon
13: spoon

10 Replicate the elements of a list a given number of times

10.1 Example

 1: List<String> myList = new ArrayList<>(asList("knife", "knife", "fork", "spoon","spoon"));
 2: printEach(repli(myList,3));
 3:  *** Output ***
 4: knife
 5: knife
 6: knife
 7: knife
 8: knife
 9: knife
10: fork
11: fork
12: fork
13: spoon
14: spoon
15: spoon
16: spoon
17: spoon
18: spoon

Welcome

This blog’s tagline is adapted from the Emacs Org-Mode motto. It seemed appropriate, as I seem to have spent most of my life writing novels and short stories (of which you can find out more at tonyballantyne.com) or teaching computer coding.

I’ve amassed a lot of material over the years, and I wanted to share it with people who may not have had the same access to education as people living in my country are lucky enough to have. If you want to change the world, become a teacher.

As the the teaching of coding seems to be coming back into fashion, I’ve also included my thoughts on the pedagogy of this subject.

All comments are gratefully received.

No Charge, No Adverts – however be aware that as an Amazon Associate I earn from qualifying purchases. If you’d like to show your appreciation, please follow me on Twitter @TonyBallantyne

All Materials Copyright (c) Tony Ballantyne 2022

8 – Strings and Characters

Sample Code

String, char, int

public class StrAsc
{
    public static void main (String args [])
    {
        String s = "A String";
        System.out.println("The character at index 2 is " + s.charAt(2));
        System.out.println("The ASCII equivalent is " + (int)s.charAt(2));
    }
}

Exercises

Convert the following string to its ASCII values: “I never saw a purple cow”
If a = 1, b=2, c=3… convert the following String to its equivalent character codes: “DailyJava”
ROT13 (“rotate by 13 places”, sometimes hyphenated ROT-13) is a simple letter substitution cipher that replaces a letter with the letter 13 letters after it in the alphabet. ROT13 is an example of the Caesar cipher, developed in ancient Rome. Write a program that will accept a String as input then output that string under a ROT13 transformation, so input of HELLO will result in output of URYYB
Write a ROT-N cipher, similar to a ROT13 cipher, where a string and a shift are input, and a string is outputted with the characters shifted by N, so if the input is “DAD” and 1, the output is “EBE”
Write a program that uses ASCII values to convert lowercase characters to uppercase, so input of “this” will result in output of “THIS”. DO NOT use library methods such as toUpperCase()
There are 62 Alphanumeric Characters: [A-Za-z0-9]. Any other character, such as %,(): is non-alphanumeric. There are also a number of control or non-printing characters. These include Line Feed, Carriage Return and Tab. Write a program that imports a text file and prints the number of alphanumeric characters it contains.
Write a program that accepts a string cipher as an input and ouputs a string plaintext containing every second letter from input. Test your program using the input “Knives” and “Forks”. You should get the output “nvs” and “ok” respectively

7 – Methods Answers

1) Write a method that accepts the length and width of a rectangle and returns the perimeter of the rectangle

double Perimeter(double length, double width)
{
    return length*2 + width*2;
}

2) Write a method that accepts the base and height of a triangle and returns the area of the triangle

double areaTriangle(double base, double height)
{
   return 0.5*base*height;
}

3) Write a method that accepts three integers as paramaters and returns the average of the integers.

double average(int a, int b, int c)
{
    return((a+b+c)/3.0);
}

4) Write a method that accepts an integer array as a parameter and returns the average of the values of that array.

double average(int [] numbers)
{
    double av = 0;
    double total = 0;
    for(int n: numbers)
    {
    total += n;
    }
    return total/numbers.length;
}

5) Write a method that accepts an integer array as a parameter and returns the minium value in that array

double min(int [] numbers)
{
    int minVal = numbers[0];
    for (int n : numbers)
    {
    if (n < minVal) minVal = n;
    }
    return minVal;
}

double min(int [] numbers)
{
    Arrays.sort(numbers);
    return numbers[0];
}

6) Write a method that returns the hypotenuse of a triangle when the other two sides are int a and int b. (Remember: hypotenuse squared equals a squared plus b squared)

double hypotenuse(double a, double b)
{
    return Math.sqrt(a*a + b*b);
}

7) The scalar product of u=(u1,u2,u3) and v=(v1,v2,v3) is defined to be u1v1+u2v2+u3v3. Write a method that accepts two int arrays as parameters and returns an int representing the scalar product of those two arrays

double scalarProduct(int [] u, int [] v) throws Exception
{
    if (u.length != v.length) throw new Exception();
    double product = 0;
    for(int i = 0; i<u.length; i++)
    {
    product += u[i]*v[i];
    }
    return product;
}

8) If A = (a1,a2, …an) and B = (b1,b2, …bn) then the vector sum of the two arrays A + B = (a1+b1, a2+b2, … , an+bn). Write a method that accepts two arrays as parameters and returns an array representing the vector sum of those two arrays.

double [] vectorSum(int [] u, int [] v) throws Exception
{
    if (u.length != v.length) throw new Exception();
    double [] sum = new double[u.length];
    for(int i = 0; i<u.length; i++)
    {
    sum[i] = u[i] + v[i];
    }
    return sum;
}

9) The Euclidean distance between two points A = (a1,a2, …an) and B = (b1,b2, …bn) is defined as sqrt((a1-b1)² + (a2-b2)² +… + (an-bn)²). Write a method that accepts two int arrays representing A and B as parameters and returns a double representing the Euclidean distance between them.

double [] eDistance(int [] u, int [] v) throws Exception
{
  if (u.length != v.length) throw new Exception();
  double [] sum = new double[u.length];
  double dist = 0;
  for(int i = 0; i<u.length; i++)
  {
       dist += Math.pow((u[i]-v[i]),2);
  }
  return sum;
}

Emacs Characters 2

I wrote about inserting characters in Emacs in this post.

There I pointed out that it’s easy to insert characters such as è and ä by using the C-x 8 key combination. So, for example:

C-x 8 ' e prints é
C-x 8 `e prints è
C-x 8 ^ e prints ê
C-x 8 " u prints ü
C-x 8 / / prints ÷
C-x 8 C prints © copyright

What I didn’t realise at the time is there was an easier combination formed by simply entering the Unicode name of a character.

For example, to insert é, use the combination

C-x 8 [return] LATIN SMALL LETTER E ACUTE

Capital letters begin, unsurprisingly, with LATIN CAPITAL.

At first glance the above doesn’t look easier, even allowing for the fact that Emacs allows you to use a few shortcuts. With tab completion, I got the key sequence down to…

C-x 8 [return] lat [tab] sm [tab] e [space] a [tab]

… but that’s still not as compact as the original examples.

So why is that an easier combination?

Well, it’s easier in the sense that it’s easier to remember, and therefore it can be quicker to use for obscure characters than having to look up a character code.

Just as an experiment, I tried to put in a British pound sign without using the appropriate key on my UK keyboard.

I used C-x 8 [return] and typed po [tab], and there was pound sign (along with

POODLE, POULTRY LEG and POUTING CAT FACE).

If you’re interested what POUTING CAT FACE looks like (I certainly was) here’s a link: http://www.fileformat.info/info/unicode/char/1f63e/index.htm

7 – Methods (Level 5)

Sample Code

public class  Meth
{
    final double PI = 3.1415;
    Meth()
    {
    int r = 4;
    System.out.println("The Area of a circle radius " + r + " is " + area(r));
    System.out.println("The Circumference of a circle radius " + r + " is " + circumference(r));
    }
    double area(int r)
    {
    return PI*r*r;
    }
    double circumference(int r)
    {
    return 2*PI*r;
    }
    public static void main (String args [])
    {
    new Meth();
    }
}

Exercises

Write a method that accepts the length and width of a rectangle and returns the perimeter of the rectangle
Write a method that accepts the base and height of a triangle and returns the area of the triangle
Write a method that accepts three integers as paramaters and returns the average of the integers.
Write a method that accepts an integer array as a parameter and returns the average of the values of that array.
Write a method that accepts an integer array as a parameter and returns the minium value in that array
Write a method that returns the hypotenuse of a triangle when the other two sides are int a and int b. (Remember: hypotenuse squared equals a squared plus b squared)
The scalar product of u=(u1,u2,u3) and v=(v1,v2,v3) is defined to be u1v1+u2v2+u3v3. Write a method that accepts two int arrays as parameters and returns an int representing the scalar product of those two arrays
If A = (a1,a2, …an) and B = (b1,b2, …bn) then the vector sum of the two arrays A + B = (a1+b1, a2+b2, … , an+bn). Write a method that accepts two arrays as parameters and returns an array representing the vector sum of those two arrays.
The Euclidean distance between two points A = (a1,a2, …an) and B = (b1,b2, …bn) is defined as sqrt((a1-b1)² + (a2-b2)² +… + (an-bn)²). Write a method that accepts two int arrays representing A and B as parameters and returns a double representing the Euclidean distance between them.

6- Nesting Answers

1) Print out the following shapes: \/ \/\/ \/\/\/ \/\/\/\/ \/\/\/\/\/

for (int i = 1; i<6; i++)
{
    for(int j = 0; j<i; j++)
    {
    System.out.print("\\/");
    }
    System.out.print(" ");
}

2) Print out the following 54321,4321,321,21,1

for (int i = 5; i>0; i--)
{
    for(int j =i; j>0; j--)
    {
    System.out.print(j);
    }
    System.out.print(" ");
}

3) Print out the following shapes */ */ */ */ *****/

for (int i = 1; i<6; i++)
{
    System.out.print("\\");
    for(int j = 0; j<i; j++)
    {
    System.out.print("*");
    }
    System.out.println("/");
}

4) Print out a 10 x 10 table square

for (int i = 1; i<11; i++)
{
    for(int j = 1; j<11; j++)
    {
    System.out.print("\t"+ i*j + "\t |");
    }
    System.out.println();
}

5) Print out the following shapes \/ \// \\/// \\//// \\\/////

String seed = "";
for (int i = 1; i<6; i++)
{
    seed = "\\" + seed + "/";
    System.out.print(seed + " ");
}

6) Print out an 8 x 8 chessboard. Use * for black and – for white

boolean isBlack = true;
for (int i = 1; i<9; i++)
{
    for(int j = 1; j<9; j++)
    {
    System.out.print(isBlack ? "*" : "-");
    isBlack = !isBlack;
    }
    isBlack=!isBlack;
    System.out.println();
}

7) Print out the following shapes:

*
**
**
***
***
***
****
****
****
****
*****
*****
*****
*****
*****

String stars = "";
for (int i = 1; i<6; i++)
{
    stars = stars + "*";
    for(int j = 0; j<i; j++)
    {
    System.out.println(stars);
    }
    System.out.println();
}

6 – Nesting (Level 5)

Sample Code

Squares

public class Triangle
{
    public static void main (String args [])
    {
        for(int i=0;i<5;i++)
        {
            for (int j = 0; j<i; j++)
                   {
                   System.out.print("*");
                   }
            System.out.println("");
        }
    }
}

Table Square

public class nest1
{
    public static void main (String args [])
    {
        for(int i = 1; i<4; i++)
        {
        for(int j = 1; j<4; j++)
        {
            System.out.println(i + " x " + j + " = " + i*j );
        }
        }
    }
}

Exercise

Print out the following shapes: \/ \/\/ \/\/\/ \/\/\/\/ \/\/\/\/\/
Print out the following 54321 4321 321 21 1
Print out the following shapes \*/ \**/ \***/ \****/ \*****/
Print out a 10 x 10 table square
Print out the following shapes \/ \\// \\\/// \\\\//// \\\\\/////
Print out an 8 x 8 chessboard. Use * for black and – for white
Print out the following shapes:

*
**
**
***
***
***
****
****
****
****
*****
*****
*****
*****
*****