Two Diagonals (A..E)
⭐ Pattern Output
Output
A A
B B
C C
D D
E1
Node.js / console version
JavaScript
const A = "A".charCodeAt(0);
const top = "E".charCodeAt(0);
for (let i = A; i <= top; i++) {
let line = "";
for (let j = A; j <= top; j++) line += (i === j ? String.fromCharCode(j) : " ");
for (let k = top - 1; k >= A; k--) line += (i === k ? String.fromCharCode(k) : " ");
console.log(line);
}2Browser version (
Browser version (document.write)
HTML
<!DOCTYPE html>
<html>
<body>
<script>
const A = 65;
const top = 69;
for (let i = A; i <= top; i++) {
for (let j = A; j <= top; j++) {
if (i === j) document.write(String.fromCharCode(j));
else document.write(" ");
}
for (let k = top - 1; k >= A; k--) {
if (i === k) document.write(String.fromCharCode(k));
else document.write(" ");
}
document.write("<br>");
}
</script>
</body>
</html>🧠 How It Works
1
Core idea
Outer i runs from A to E. First inner loop scans j = A..E and prints a letter only when i === j. Second inner loop scans k = D..A and prints a letter only when i === k. Other positions are blanks, creating two diagonals that meet in the middle.
Logic
2
Loops + condition
An outer loop picks the row, inner loop(s) decide what to print (letter / star / blank) based on comparisons.
Nested loops
=
Connect it to the output
Compare each loop boundary with the pattern output above — each row corresponds to one outer iteration.
❓ Frequently Asked Questions
On the last row, the two diagonals meet at the center. Both inner checks point to the same position, so you only see a single
E.HTML collapses normal spaces, so the diagonal alignment breaks. Using
(or fixed-width cells) preserves the grid spacing.Change
top from E to a later letter (for example "G".charCodeAt(0)). Keep the left scan as A..top and the right scan as top - 1..A.More JavaScript alphabet patterns
Browse the full list for more variations.
About the author
10 people found this page helpful