2. Conceptualize
Think of your printing space as a grid with
rows and columns
0
1
2
0
*
*
*
*
1
*
*
*
*
2
*
*
*
1
2
3
1
*
*
*
2
*
*
3
*
*
3. Simple right triangle
Where
does a star appear?
1,1; 2,1; 2,2; 3,1; 3,2; 3,3
In each row, the col # with
a star starts at 1 and goes
until the row #.
The same logic applies for a
larger grid.
1
2
1
*
2
*
*
3
*
3
*
1
2
*
3
1
*
2
*
*
3
*
*
*
4
*
*
*
4
*
4. Simple right triangle
Rows
go from 1 to n.
On each row we output a star in each
column until column equals row.
for(int r=1; r<=n; r++){
for(int c=1;c<=r;c++){
cout<<“*”;
}
cout<<endl;
}
5. Finding the middle
You
don’t know what input size the user
will give, so how will you know where the
middle is?
If n=3, then the middle is 2
If n=5, then the middle is 3
1 2 3
middle = n/2+1
1
*
1
2
3
2
*
1
*
3
*
2
*
4
*
3
*
5
*
4
5
6. Isosceles Triangle
The
first thing you may notice
1
2
about this “triangle” is that it
1
doesn’t start at the top of the
2
*
grid. For the sake of space
3
*
*
(and other considerations)
let’s try to figure out (given the number of
columns) how many rows we need.
3
*
8. Starting at Zero
We
often start our rows and
columns at zero instead of
one.
In that case, the formula for
the middle would be:
mid = n/2 instead of n/2+1
At right: n = 3. 3/2=1 (int div)
and 1 is the middle column.
1
2
1
*
2
*
3
3
*
0
1
0
*
1
*
2
*
2
9. Center+diagonals
for (int r = 0; r <= rows; r++) {
for (int c = 0; c <= cols; c++) {
if (c==cntr+r || c==cntr-r || c==cntr)
cout << "*";
else cout << " ";
}
0
1
0
3
4
*
1
2
2
*
*
*
*
*
*
10. Center+diagonals
for (int r = 0; r <= rows; r++) {
for (int c = 0; c <= cols; c++) {
if (c>=cntr-r && c<=cntr+r) cout << "*";
else cout << " ";
}
cout<<endl;
}
0
1
0
3
4
*
1
2
2
*
*
*
*
*
*
*
*
