Exercise 3: Chess Board
Write a program that creates a string that represents an 8 x 8 grid, using newline characters to separate lines.
At each position of the grid there is either a space or a “#” character.
The characters should form a chess board. Passing this string to console.log should show something like this:
When you have a program that generates this pattern, define a variable size = 8 and change the program so that it works for any size, outputting a grid of the given width and height.
Approach
Loop – “Vertical”:
- This will be the outermost for loop
- This loop adds height to the output
- Let i represent the number of iterations
- i is of type number
- Let size represent the width and height of the square
- size is of type number
- By the end of the loop, i == size
Loop – “Horizontal”:
-Increases characters horizontally per line
-Each new line is a new iteration
-Repeating characters: # and a space
-When i is an even number, the line starts with a # character
– Repeated characters when i is an even number: #_
-When i is an odd number, the line starts with a space (will rep. with _ character in notes) -Repeated characters when i is an odd number: #_ , but one _ exists before any repeats
- This will be an inner for loop
- This loop adds width to the output
- Let char represent the characters on the line
- char is of type string
- char.length returns the length (number of characters) of char
- By the end of the loop, i == size == char.length
Questions:
1. How do you repeat string characters horizontally when the number of repeats required is unknown?
You can concatenate strings i.e.
console.log("# " + "# ")
//-> # #
EDIT (after writing Solution Attempt 4 – Final Submission) : You CAN repeat strings horizontally when the number of repeats is unknown.
What is needed: Put console.log outside the for loop (so whatever is printed is the final result of repeated string concatenation). See the next blog post “Exercise 2.3 Part 2” for this solution.
But recall that no arithmetic operation can be applied to strings i.e.
var test = "Hi"; test *= 2; console.log(test); //-> NaN
I will not be able to repeat characters horizontally using arithmetic multiplication i.e.
var char = "# " * 3; console.log(char); //-> NaN
or
console.log("10" * 2);
//-> 20
console.log(typeof("10" * 2));
//-> number
Observe string concatenation between string values and number values:
//Example 1
console.log("150" + 1010);
//-> 1501010
console.log(typeof("150" + 1010));
//-> string
//Example 2
console.log(150 + "1010");
//-> 1501010
console.log(typeof(150 + "1010"));
//-> string
//Example 3
console.log("#" + 10);
//-> #10
console.log(typeof("#" + 10));
//-> string
Keep in mind that the Number and String functions convert a value to type number and type string respectively.
Though that still isn’t helpful in repeating string characters horizontally when the number of repeats required is unknown.
It passed the mind to replace # and space with 1 and 0, and at the end, replace 1 and 0 with # and space again. Using numbers would allow for repetition via arithmetic operations. However, this idea died a swift death upon the realisation that when a string value’s type is converted to type number, (“10” with 10), any arithmetic operation on this value will be governed by mathematical laws i.e. 10 * 2 = 20, not 1010 i.e.
And why it wouldn’t work with # :
.
.
.
2. What is the update/final-expression in the inner for loop?
Can it be empty?
.
.
.
Outside the scope of Chapter 2, there is a JavaScript String repeat() method. Let’s try this out for Solution 1.
// Allow any size to be inputted
var size = prompt("Enter a positive integer.");
// Convert string value inputted in prompt to type number
Number(size);
// Outer loop to track vertical iteration
for (var i = 0; i <= size; i++) {
// Inner loop to add characters horizontally per line
for(var char = ""; char.length <= size; ) {
//If the line number is odd
if (size % 2 !== 0) {
// Begin the line with a space
char = " ";
// How many times does the pattern have to repeat?
var numRepeats1 = (((size + 1)/2) - 1);
var pattern1 = "# ";
// Concatenate the first space with the pattern
// Use string repeat() method to repeat the pattern the appropriate number of times
char += pattern1.repeat(numRepeats1);
}
//If the line number is even
if (size % 2 == 0) {
//Pattern to be repeated
var pattern2 = "# ";
// Repeat the pattern this amount of times
var numRepeats2 = size/2;
//Update char value
char += pattern2.repeat(numRepeats2);
}
.
.
.
EPIPHANY: I don’t need the second for loop. With the string repeat method, there is no need for a loop to repeat the characters horizontally.
EPIPHANY 2 (after Solution Attempt 1 Test 1): Nevermind, I do need the second loop. Without this loop, a conditional branch would be left out depending on what the value of size is i.e. if size = 2, the odd number if-statement branch is not considered at all, so instead of printing out line 1, line 2, only line 2 with weird stuff is printed.
Solution Attempt 1
var char = ""
var size = prompt("Enter a positive integer.");
Number(size);
for (var i = 0; i <= size; i++) {
if (size % 2 !== 0) {
char = " ";
var pattern1 = "# ";
var numRepeats1 = (((size + 1)/2) - 1);
char += pattern1.repeat(numRepeats1);
console.log(char);
}
if (size % 2 == 0) {
var pattern2 = "# ";
var numRepeats2 = size/2;
char += pattern2.repeat(numRepeats2);
console.log(char);
}}
No plan survives first contact with the enemy. I think I’ll become quite fond of that saying.
What I think went wrong:
Firstly, the initial value of i should be set to 1, not zero. This way, the iteration number is equal to the row number (of the output).
Secondly, the if-statement conditions are wrong. Variable i instead of variable size should be used. That is:
// for odd-number lines
if (i % 2 !== 0) {
}
// for even-number lines
if (i % 2 == 0) {
}
Why? Because it is the iteration value that tracks the row that we are on. The value of i increases with each iteration, while the value of size stays the same (it is whatever value was inputted in prompt). Size is the maximum width and height of the grid. We just need the current row number of the grid (i).
Let’s say size = 2. The odd-numbered line if-statement branch {check wording} is evaluated to be false since size is an even-numbered value. That’s why char = ” ” didn’t manifest. Execution moves to the even-numbered line if-statement branch, and its statement body is executed.
NB: The value of i is compared to the value of size at the beginning of each iteration to see if another iteration is required. {check wording and understanding: is the updated value compared with the value in the condition at the END or the BEGINNING of each ITERATION (or do we say LOOP)?}
What happened?
Look at numRepeats1 and numRepeats2. Are their formulas correct? Seems to be so, yes. Next, look at char +=.
QUESTION 1: Why did a change from Line 17 char += pattern2.repeat(numRepeats2); to char = pattern2.repeat(numRepeats2); give the correct output for the 2nd line?
Testing
1) Line 17 –> char += re-added
Line 7 –> char = “1”
Changes are to see whether the new value of char bleeds out of its if-statement into the rest of the program. Apparently yes. char = “1” should only affect odd number iterations.
ANSWER to QUESTION 1: For some reason, the value of char remains “1# # # … #” even when the iteration is a new, even-numbered iteration. That the iteration is even-numbered means that execution should bypass the first if statement (Line 6) and go on to the second if statement (Line 14). char’s value should be an empty string when execution starts at Line 14, not holding on to whatever value it gained from Line 10.
We must also consider the += operator on Line 17. What happens when it is removed?
2) Line 17 –> = operator replaces += operator
ANSWER to QUESTION 1: When char (whose value seems to cling to the result from Line 10) is not concatenated to pattern2 on Line 17, the output for iteration number 2 looks proper.
if-statement for even-number lines seem to be working properly. Something’s wrong with the if-statement for odd-number lines branch. Suspect are Lines 9 and/or 10.
Looking at Line 9 — changed (((size + 1)/2) – 1); to (size):
Testing the repeat formula:
That means something is wrong with char += or size input, not the formula.
EDIT: At least, for when size holds ODD NUMBERS. How about EVEN NUMBERS?
EPIPHANY: Whenever size is an odd number, the repeating formula in the first if-statement works fine. But the repeating formula in the second if-statement doesn’t work properly. AND VICE VERSA. All because of variable size, which is used in the repeating formula. FIX THE FORMULAS.
Repeating a string 2.5 times doesn’t happen. It repeats 2 times.
But the if-statement for odd-number and the if-statement for even numbers should filter incoming numbers, and slide them into their respective formulas.
The String repeat() method and char += … seems to be working fine.
Testing prompt input:
Something’s wrong with the prompt input.
More proof:
By process of elimination, these are the remaining potential errors:
- Repeating formulas — fix both the odd and even one; use something other than size or adjust the numbers to accommodate variable size
- REPEATING PATTERN ISSUE and discrepancy between value of i and value of size
- Don’t cheat with using blanks as noncharacters. Spaces count as characters and you want to print the specified character amount, no more no less.
- Prompt input
- for loop? lack of a secondary loop?
THE BIG EPIPHANY
When size holds an odd number value, the numRepeats1 formula works properly. But the numRepeats2 formula does not. When size holds an even number value, the numRepeats2 formula works properly, but the numRepeats1 formula does not.
The solution:
if size = odd number {
if i = odd number
Adjust repeating formula accordingly
if i = even number
Adjust repeating formula accordingly
}
else if size = even number {
if i = odd number
Adjust repeating formula accordingly
if i = even number
Adjust repeating formula accordingly
}
Place these if statements after the for loop. Each if statement will have a nested if statement (for i = odd number) and else-if statement (for i = even number).
Solution Attempt 2
var char = ""; var size = 2; Number(size); for (var i = 1; i <= size; i++) { // if size is an odd number if (size % 2 !== 0) { //if the row number (iteration) is an odd number if (i % 2 !== 0) { char = " "; var pattern1 = "# "; var numRepeats1 = (((size + 1)/2) - 1); char += pattern1.repeat(numRepeats1); console.log(char); } //if the row number is an even number if (i % 2 == 0) { char = "#" var pattern2 = " #"; var numRepeats2 = ((size -1)/2); char += pattern2.repeat(numRepeats2); console.log(char); }} //If size is an even number else if (size % 2 == 0) { //if the row number is odd if (i % 2 !== 0) { var pattern1 = " #"; var numRepeats1 = (size/2); char += pattern1.repeat(numRepeats1); console.log(char); } //if the row number is even if (i % 2 == 0) { var pattern2 = "# "; var numRepeats2 = (size/2); char += pattern2.repeat(numRepeats2); console.log(char); }}}
^Makes more structural sense for the currently outermost for loop to be inside “if size is odd” and “if size is even” statements. Is the structure causing the following problem though? This structure, while disorganized, should still work properly.
** “if size is odd” if statement branch is working properly. Something’s wrong with the “if size is even” statement
Get rid of += on Line 33:
Highlighted yellow is the change:
Why this change when += was replaced by = , on Line 33?
Now there are issues with the += on Line 26:
The pattern marked “2” is the pattern on Line 24, repeated twice. But where did the pattern marked “1” come from?
char += pattern1.repeat(numRepeats1); //equivalent to: char = char + pattern1.repeat(numRepeats1);
char should be an empty string, until it is concatenated with pattern1.
Highlighted yellow is the change from += to + on Line 26:
So char = “# # ” when execution hit upon Line 26? QUESTION 2: How?
Using other values for size i.e. 3, 5, 6, 8, 10, the program now displays the proper output. On Line 2, I re-added prompt(), which also works fine now.
EDIT (during Solution Attempt 3): Oh no, prompt() doesn’t work fine for odd numbers.
Solution Attempt 3
1 var char = ""; 2 var size = prompt("Enter a positive integer."); 3 Number(size); 4 5 for (var i = 1; i <= size; i++) { 6 if (size % 2 !== 0) { 7 if (i % 2 !== 0) { 8 char = " "; 9 var pattern1 = "# "; 10 var numRepeats1 = (((size + 1)/2) - 1); 11 char += pattern1.repeat(numRepeats1); 12 console.log(char); 13 } 14 15 if (i % 2 == 0) { 16 char = "#"; 17 var pattern2 = " #"; 18 var numRepeats2 = ((size -1)/2); 19 char += pattern2.repeat(numRepeats2); 20 console.log(char); 21 }} 22 else if (size % 2 == 0) { 23 if (i % 2 !== 0) { 24 var pattern1 = " #"; 25 var numRepeats1 = (size/2); 26 char = pattern1.repeat(numRepeats1); 27 console.log(char); 28 } 29 30 if (i % 2 == 0) { 31 var pattern2 = "# "; 32 var numRepeats2 = (size/2); 33 char = pattern2.repeat(numRepeats2); 34 console.log(char); 35 }}}
…
I celebrated too soon.
When size = 11 without prompt(), we get this:
When size = 11 via prompt(), we get this mess:
Inputting size values into prompt() that are even and greater than 10 i.e. 12, 14, 16, etc., the output is proper as expected. However, when inputting size values into prompt() that are ODD and greater than 10 i.e. 11, 13, 15, 17, etc., the output is all wrong.
EDIT: Nope, odd-numbers below 10, when inputted via prompt() for size, also do not work properly.
Therefore, we’re dealing with either an error(s) with 1) Lines 5 – Line 18 (when size is an odd number) or 2) the prompt() method or 3) both or 4) something else.
Note that the even-numbered rows do show the expected output. The odd-numbered rows show more than the expected characters. So the error is somewhere in Lines 6-11 if we’re looking at option 1).
Suspects:
Indentation? Line 13 if statement is indented more than it should be? This is unlikely to be the problem.(Check: Yeah, it isn’t the issue.)- prompt() The main suspect
The += operator on Line 10(Check: Not the issue.)usage of same variable for different values i.e. char (Line 7, Line 10) ??(Check:Not an issue.)
Tentative conclusion: It looks like a prompt() issue, because inputting any value manually gives the correct output. RETURN TO THIS.
Cleaning up the code:
see Solution Attempt 4
- defined the variable char, but did not initialize it –> its value is undefined; will give char a value depending on the nested if statement
- declared and initialized variables patternsharp and patternspace outside of the for loop –> prevents re-writing these four times inside the loop’s nested if and else if statement bodies
- renamed numRepeats1 and numRepeats2 –> under one variable name: numRepeats (see Line 9), because they are contained within their nested if statements
- under the else-if statement on Line 19, pulled two numRepeats out of their nested if-statements, and placed it as one numRepeats statement (Line 20) –> this applied to both nest if-statements and saves space (no need to rewrite numRepeats twice since they have the same formula of size/2)
Solution Attempt 4: Final Submission
using String repeat() method
still having issues with prompt()
Solution Attempt 4 with Comments
var char; var patternsharp = "# "; var patternspace = " #"; var size = 3; Number(size); for (var i = 1; i <= size; i++) { // size is an odd number if (size % 2 !== 0) { // row number is an odd number if (i % 2 !== 0) { //odd rows start with a space char = " "; // the number of times patternsharp shoud be repeated var numRepeats = (((size + 1)/2) - 1); /*concatenating " " and patternsharp however many times it's repeated*/ char += patternsharp.repeat(numRepeats); console.log(char); } //row number is an even number if (i % 2 == 0) { //even rows start with a sharp char = "#"; //the number of times patternspace should be repeated numRepeats = ((size-1)/2); char += patternspace.repeat(numRepeats); console.log(char); }} //size is an even number else if (size % 2 == 0) { /*the number of times patternsharp or patternspace should be repeated */ numRepeats = (size/2); //row number is odd if (i % 2 !== 0) { char = patternspace.repeat(numRepeats); console.log(char); } //row number is even if (i % 2 == 0) { char = patternsharp.repeat(numRepeats); console.log(char); }}}
QUESTION 3:
I defined and initialized the variable numRepeats on Line 9. The later usage of numRepeats on Lines 15, 21, and 27 do no require being defined again, even when they are inside different if/else-if statements. QUESTION: Was that defining a variable ‘locally’? Versus ‘globally’ before the for loop?
QUESTION 4:
What happens when you take out the console.log(char); from the if statement body, and place it outside of it? See Line 13.
For the Future
- I didn’t try a solution without String repeat() method. Try that.
- Answer Questions 1-4 properly.
- Figure out what is wrong with prompt() or Lines 6-11.
- Consider other methods to solve Chess Board — replace(), a second for loop, newline character, etc.
- Then check your solutions with the Book Solutions.
Ode to STEM 