Boustrophedon transform: Difference between revisions
Added Python implementation for Boustrophedon transform task
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<span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
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=={{header|Python}}==
<syntaxhighlight lang="python">
# bt_transform.py by Xing216
from math import factorial
from itertools import islice
from sys import setrecursionlimit
setrecursionlimit(1000000)
def gen_one_followed_by_zeros():
yield 1
while True:
yield 0
def gen_ones():
while True:
yield 1
def gen_alternating_signs():
s = 1
while True:
yield s
s *= -1
def gen_primes():
"""Code by David Eppstein, UC Irvine, 28 Feb 2002 http://code.activestate.com/recipes/117119/"""
D = {}
q = 2
while True:
if q not in D:
yield q
D[q * q] = [q]
else:
for p in D[q]:
D.setdefault(p + q, []).append(p)
del D[q]
q += 1
def gen_fibonacci():
a=0
b=1
while True:
yield b
a,b= b,a+b
def gen_factorials():
f = 0
while True:
yield factorial(f)
f += 1
def compressed(n):
strn = str(n)
return f"{strn[:20]}...{strn[-20:]} ({len(strn)} digits)"
def boustrophedon(a):
# Copied from the Wren Solution
k = len(a)
cache = [None] * k
for i in range(k): cache[i] = [0] * k
b = [0] * k
b[0] = a[0]
def T(k,n):
if n == 0: return a[k]
if cache[k][n] > 0: return cache[k][n]
cache[k][n] = T(k,n-1) + T(k-1,k-n)
return cache[k][n]
for n in range(1,k):
b[n] = T(n,n)
return b
funcs = {"One followed by an infinite series of zeros:":gen_one_followed_by_zeros,
"Infinite sequence of ones:": gen_ones,
"Alternating 1, -1, 1, -1:": gen_alternating_signs,
"Sequence of prime numbers:": gen_primes,
"Sequence of Fibonacci numbers:": gen_fibonacci,
"Sequence of factorial numbers:": gen_factorials}
for title, func in funcs.items():
x = list(islice(func(), 1000))
y = boustrophedon(x)
print(title)
print(y[:15])
print(f"1000th element: {compressed(y[-1])}\n")
</syntaxhighlight>
{{out}}
<pre style="height:15em">
[1, 2, 4, 9, 24, 77, 294, 1309, 6664, 38177, 243034, 1701909, 13001604, 107601977, 959021574]
1000th element: 29375506567920455903...86575529609495110509 (2370 digits)
Alternating 1, -1, 1, -1:
[1, 0, 0, 1, 0, 5, 10, 61, 280, 1665, 10470, 73621, 561660, 4650425, 41441530]
1000th element: 12694307397830194676...15354198638855512941 (2369 digits)
Sequence of prime numbers:
[2, 5, 13, 35, 103, 345, 1325, 5911, 30067, 172237, 1096319, 7677155, 58648421, 485377457, 4326008691]
1000th element: 13250869953362054385...82450325540640498987 (2371 digits)
Sequence of Fibonacci numbers:
[1, 2, 5, 14, 42, 144, 563, 2526, 12877, 73778, 469616, 3288428, 25121097, 207902202, 1852961189]
1000th element: 56757474139659741321...66135597559209657242 (2370 digits)
Sequence of factorial numbers:
[1, 2, 5, 17, 73, 381, 2347, 16701, 134993, 1222873, 12279251, 135425553, 1627809401, 21183890469, 296773827547]
1000th element: 13714256926920345740...19230014799151339821 (2566 digits)
</pre>
=={{header|Raku}}==
<syntaxhighlight lang="raku" line>sub boustrophedon-transform (@seq) { map *.tail, (@seq[0], {[[\+] flat @seq[++$ ], .reverse]}…*) }
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