How to shorten the Python code? Part II

Question

Code Python 3, 245 bytes

E=print
def F():
	B,C,A=1,1,1;yield 0
	while True:yield B;B,C,A=B+C+A+1,C+A+1,A+1
C=int(input())
D=0
for A in range(C):
	E(end='*'*(C-(C-A)+1));B=A
	while 0<B:E(end=' '+chr(42+B)*B);B-=1
	E();D+=1
for A in F():
	if D==0:E(end=str(A));break
	D-=1

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What it does?

Takes a number from input, loops from 0 to \$x - 1\$ (let's call iterator \$y\$), prints ASCII 41 ("*"), \$y\$ times with a space and prints ASCII 42, \$y - 1\$ times with a space and repeat until the \$y - x\$ is \$0\$, print a newline and go to next iteration.

If the loop is ended, print the total of the output (not including the output of spaces and newlines).

Answer 1

Python 3, 106 bytes

C=int(input())
i=0
while i<C:i+=1;print(*[j*chr(42+j%i)for j in range(i,0,-1)])
print(C*~C*~-~C//6,end='')

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The grand total can be computed as C*(C+1)*(C+2)//6 (see tetrahedral numbers). So you don't need F at all! This saves a lot of bytes.

This formula can be golfed down to C*-~C*-~-~C//6, and then further to C*~C*~-~C//6 because two of the -s cancel out in the product.

Then, I just wrote some simpler code to print the ASCII characters. Usually when you want output separated by spaces, print(*[list comprehension]) is pretty good, so I tried that.

The clever thing here is using j%i to count like 0, i-1, i-2, ... 1 on each row. Try replacing it with just j and compare the output.

There are only two print calls now, so E=print doesn't save bytes anymore.

print(end=str(num)) is longer than print(num,end='').

Tiny golfs in your code

• B,C,A=1,1,1 can be B=C=A=1.

• while True can be while 1.

• (C-(C-A)+1) equals (A+1) equals -~A.

• while 0<B can be while B here.

Making a generator and extracting the D-th value seems painful. You can "inline" the iterator: delete the definition of F and just perform the logic inside F, D times at the end of your code:

E=print
C=int(input())
D=0
for A in range(C):
    E(end='*'*-~A);B=A
    while B:E(end=' '+chr(42+B)*B);B-=1
    E();D+=1
b=c=a=0
while D:b,c,a=b+c+a+1,c+a+1,a+1;D-=1
print(b,end='')

If we stare at the assignments long enough, we can figure out that b,c,a=b+c+a+1,c+a+1,a+1 can be a+=1;c+=a;b+=c.

Also, we count up from D=0 to D=C, so we can just use C instead of D:

E=print
C=int(input())
for A in range(C):
    E(end='*'*-~A);B=A
    while B:E(end=' '+chr(42+B)*B);B-=1
    E()
b=c=a=0
while C:a+=1;c+=a;b+=c;C-=1
print(b,end='')

But an even shorter way to just repeat some code C times is this:

exec("a+=1;c+=a;b+=c;"*C)

Of course, the tetrahedral number formula is even shorter than any of this.

Answer 2

Make sure to check out Lynn's answer. It contains some great advice too.

Python 3, 237 bytes

E=print
def F():
 B=C=A=1;yield 0
 while 1:yield B;B,C,A=B+C+A+1,C+A+1,A+1
C=int(input())
D=0
for A in range(C):
 E(end='*'*(C-(C-A)+1));B=A
 while 0<B:E(end=' '+chr(42+B)*B);B-=1
 E();D+=1
for A in F():
 if D<1:E(end=str(A));break
 D-=1

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My go at this. Will keep golfing later.

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Answer 3

Python 3, 132 bytes

Here is my approach to this problem.

C=int(input())
print(*(' '.join(['*'*(A+1)]+[B*chr(42+B)for B in range(A,0,-1)])for A in range(C)),C*(C+1)*(C+2)//6,sep='\n',end='')

Please refer to Lynn's answer for more details and for a much shorter code. For example, my code could be further optimized by replacing C*(C+1)*(C+2)//6 with C*~C*~-~C//6.

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