Starry Metagolf
Python 3, 17071 11845
from functools import lru_cache
import heapq
import time
cases = r"""
Hello, World!
pneumonoultramicroscopicsilicovolcanoconiosis
.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.
Hickory, dickory, dock. The mouse ran up the clock. The clock struck 1. The mouse ran down. Hickory, dickory, dock.
36912059868043514648560046917066768694455682545071266675083273015450033938555319356951628735735013250100789433961153496780296165
bVZ48121347GLtpYnt76CZSxTpMDs6791EJE808077eySXldY162424ddTB90707UupwlWGb63618542VhA252989453TXrWgqGm85899uHOAY2oAKE198GOVUttvW63
7MYxoWBNt180CDHS5xBGvU70HHVB17bh8jYzIIiU6n6g98Rose1nOe8Svcg56nax20q30kT3Ttb2jHl5q2Iuf1vPbjPxm9cyKXwxc0OUK8pr13b2n7U9Y7RwQTc26A1I
n9}unwxVa}[rj+5em6K#-H@= p^X/:DS]b*Jv/_x4.a5vT/So2R`yKy=in7-15B=g _BD`Bw=Z`Br;UwwF[{q]cS|&i;Gn4)q=`!G]8"eFP`Mn:zt-#mfCV2AL2^fL"A
""".strip().splitlines()
@lru_cache(maxsize=128)
def shortest_m_to_n(m, n):
if m is None:
L = []
else:
L = [m]
to_search = [[0, "", L]]
seen = set()
while True:
length, code, stack = heapq.heappop(to_search)
if len(stack) == 1 and stack[-1] == n:
return code
seen.add(tuple(stack))
options = []
for i in range(1, 11):
new_stack = stack[:] + [i]
new_code = code + ' '*(i+5) + '+'
options.append([len(new_code), new_code, new_stack])
if stack:
new_stack = stack[:] + [stack[-1]]
new_code = code + " +"
options.append([len(new_code), new_code, new_stack])
if len(stack) >= 2:
x, y = stack[-2:]
for i, op in enumerate(['+', '-', '*', '//', '%']):
try:
new_elem = eval("{}{}{}".format(x, op, y))
new_stack = stack[:-2] + [new_elem]
new_code = code + ' '*i + '*'
options.append([len(new_code), new_code, new_stack])
except ZeroDivisionError:
pass
for op in options:
if tuple(op[2]) in seen or len(op[2]) > 4 or op[2][-1] > 200:
continue
heapq.heappush(to_search, op)
def lcs(s1, s2):
dp_row = [""]*(len(s2)+1)
for i, c1 in enumerate(s1):
new_dp_row = [""]
for j, c2 in enumerate(s2):
if c1 == c2 and not c1.isdigit():
new_dp_row.append(dp_row[j] + c1)
else:
new_dp_row.append(max(dp_row[j+1], new_dp_row[-1], key=len))
dp_row = new_dp_row
return dp_row[-1]
def metagolf(s):
keep = ""
split_index = 0
for i in range(1, len(s)):
l = lcs(s[:i], s[i:][::-1])
if len(l) > len(keep):
keep = l
split_index = i
code = []
stack = []
keep_ptr = 0
i = 0
while i < len(s):
c = s[i]
n = ord(c)
if c in "0123456789":
code += [" "*(int(c)+5) + "+."]
i += 1
continue
if stack:
if stack[-1] == n:
code += [" +", " ."]
elif len(stack) >= 2 and stack[-2] == n:
for j in range(len(code)):
if code[~j] == " +":
code[~j] = ""
break
code += [" +", " ."]
stack.pop()
else:
code += [shortest_m_to_n(stack[-1], n), " +", " ."]
stack[-1] = n
else:
code += [shortest_m_to_n(None, n), " +", " ."]
stack.append(n)
while i < split_index and keep[keep_ptr:][:1] == c:
code += [" +"]
keep_ptr += 1
stack.append(n)
i += 1
code = "".join(code)
if code[-4:] == " + .":
code = code[:-4] + " ."
return code
total = 0
for case in cases:
start_time = time.time()
s = metagolf(case)
print(len(s), time.time() - start_time)
total += len(s)
print(s)
print('='*50)
print(total)
The relevant function is the aptly named metagolf
.
The results are:
210
676
684
2007
1463
2071
2204
2530
Total: 11845
You can find the full output here.
Brief explanation
I'm going to keep the explanation brief since there's many things still to be improved.
The basic algorithm just looks at pairs of chars, and finds the optimal way to transition from one char to another via BFS. Digits are currently pushed and printed immediately, although this will be changed later.
There's also a little longest-common-subsequence going on, to leave a few elements on the stack for reuse later. It's not as good as repetition, but provides decent savings.
Ruby, 13461 10997
$s = {};
def shortest a,b=nil
return $s[[a,b]] if $s[[a,b]]
l = []
if b
if a == b
return $s[[a,b]] = ""
elsif a > b
l.push shortest(a-b)+" *"
l.push " + *"+shortest(1,b) if a > 1
l.push " + *"+shortest(0,b) if a > 0
l.push " +"+shortest(b)
elsif a < b
l.push " + *"+shortest(a*a,b) if a*a>a && a*a<=b
l.push " +*"+shortest(a+a,b) if a+a<=b && a+a>a
l.push shortest(b-a)+"*"
l.push " +"+shortest(a,b/a)+" *" if a>2 && b%a == 0
l.push " +"+shortest(a,b-a)+"*" if a>1 && b>a*2
end
else
l.push ' '*(a+5)+'+' #if a < 6
(1..a/2).each {|n|
l.push shortest(n)+shortest(n,a)
}
end
return $s[[a,b]] = l.min_by{|x|x.length}
end
def starry(str)
arr = str.bytes.map{|b|
if b>47 && b<58
b-48# change digets to numbers
else
b
end
}
startNum = (1..128).min_by{|x|arr.inject{|s,y|s + [shortest(x,y).length+2,shortest(y).length].min}+shortest(x).length}
#one number to be put on the stack at the start.
code = shortest(startNum)
code += [
shortest(arr[0]),
" +"+shortest(startNum, arr[0])
].min_by{|x|x.length}
arr.each_cons(2) do |a|
pr = a[0]<10?'.':' .'
code += [
pr+shortest(a[1]),
" +"+pr+shortest(a[0], a[1]),
pr+" +"+shortest(startNum, a[1])
].min_by{|x|x.length}
end
code += arr[-1]<10?'.':' .'
end
a = ["Hello, World!",
"pneumonoultramicroscopicsilicovolcanoconiosis",
".oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.oOo.",
"Hickory, dickory, dock. The mouse ran up the clock. The clock struck 1. The mouse ran down. Hickory, dickory, dock.",
"36912059868043514648560046917066768694455682545071266675083273015450033938555319356951628735735013250100789433961153496780296165",
"bVZ48121347GLtpYnt76CZSxTpMDs6791EJE808077eySXldY162424ddTB90707UupwlWGb63618542VhA252989453TXrWgqGm85899uHOAY2oAKE198GOVUttvW63",
"7MYxoWBNt180CDHS5xBGvU70HHVB17bh8jYzIIiU6n6g98Rose1nOe8Svcg56nax20q30kT3Ttb2jHl5q2Iuf1vPbjPxm9cyKXwxc0OUK8pr13b2n7U9Y7RwQTc26A1I",
"n9}unwxVa}[rj+5em6K#-H@= p^X/:DS]b*Jv/_x4.a5vT/So2R`yKy=in7-15B=g _BD`Bw=Z`Br;UwwF[{q]cS|&i;Gn4)q=`!G]8\"eFP`Mn:zt-#mfCV2AL2^fL\"A"]
c = a.map{
|s|
starry(s).length
}
p c.inject(0){|a,b|a+b}
The method starry
answers the given question.
Results:
230
639
682
1974
1024
1897
2115
2436
Total: 10997
How it works
shortest
is the main algorithm.
It takes one number and finds the shortest way to place it on the stack, or it takes two numbers, and returns code to put the second on the stack assuming the first one is already on. $s
is a Hash to hold the results of these operations for further use.
starry
takes a string and splits it into an array of character codes (or numbers for digests). It starts the code with one number on the bottom of the stack. Next it calculates the shortest way it can generate each successive number, possibly copying the last one or using the number put on the stack at the beginning.
JavaScript, 25158 23778
Now ES5-compatible!
String.prototype.repeat = String.prototype.repeat || function (n) { return Array(n+1).join(this); }
function starrify(x) {
function c(x){return x.charCodeAt()}
var char = x[0], result = ' '.repeat(c(char)+5)+'+ + .';
x=x.slice(1);
for(var i in x) {
if (char < x[i]) {
result += ' '.repeat(c(x[i])-c(char)+5)+'+* + .';
} else if (char > x[i]) {
if(c(char)-c(x[i]) < c(x[i])) {
result += ' '.repeat(c(char)-c(x[i])+5)+'+ * + .';
} else {
result += ' '.repeat(c(x[i])+5)+'+ + .';
}
} else {
result += ' + .';
}
char = x[i];
}
return result;
}
Results:
432
949
2465
3996
1805
3551
5205
5375
Total: 23778
A good start in my opinion, but obviously not finished. Instead of creating each char separately, it adds or subtracts from the previous char code. I'll add an full explanation when I'm done meta-golfing.