Diamond Challenge!

[Tigerbomb8]

Okay guys, I was going through some old python programs of mine and i would like to see some other, better methods to solve this problem.

 

The Challenge!!

You must make a program that outputs a diamond pattern, where 'n' as the maximum width of the diamond.

E.g.

 

n = 5    *   * *  * * * * * * ** * * * * * * * *  * * *   * *    *n = 3  * * ** * * * *  *

width = int(input("Enter width: "))loopPassA = 1loopPassC = 1maxPasses = width * 2 - 1line = ""while loopPassA <= maxPasses:if loopPassA <= width:spacesNeeded = width - loopPassAstarsNeeded = loopPassAline = ""loopPassB = 0while len(line) < spacesNeeded:line = line + " "while loopPassB < starsNeeded:line = line + "* "loopPassB = loopPassB + 1print(line.rstrip())if loopPassA > width:spacesNeeded = loopPassCstarsNeeded = width - loopPassCline = ""loopPassB = 0while len(line) < spacesNeeded:line = line + " "while loopPassB < starsNeeded:line = line + "* "loopPassB = loopPassB + 1loopPassC = loopPassC + 1print(line.rstrip())loopPassA +=1STOP = input("STOP")

Works with python 3.4

 

[Butterfish]

Sorry this is out of my nerd comfort zone

[fizzlesticks]

print('\r\n'.join(['{}{}'.format(' '*(n - (i if i<=n else n-abs(n-i))),'* '*(i if i<=n else n-abs(n-i))) for n in [int(input('Width: '))] for i in range(2*n)]))

[Guest]

(display "n: ")

(let ([width (string->number (read-line))])

(for ([i (in-range 1 (* 2 width))])

(let ([stars (if (> i width) (- width (abs (- width i))) i)])

(display

(string-append

(make-string (- width stars) #\space)

(apply string-append (make-list stars "* "))

"\n")))))

[Tigerbomb8]

Well those answers are ridiculously simple compared to my method.

[Guest]

Well those answers are ridiculously simple compared to my method.

 

Looks like you were just over-thinking the solution. Happens.

[eikast]

lol. This was a question on one of my homework assignments in a C++ class. If I remember correctly, it was a recursive function.