Possibly a breakthrough programming language?

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HS

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Re: Possibly a breakthrough programming language?
« Reply #45 on: February 11, 2020, 01:24:01 am »
Wow that's a lot of progress. I'm still stuck on the first question.

Screenshot-772" border="0

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« Last Edit: February 11, 2020, 03:24:49 am by LOCKSUIT »
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Re: Possibly a breakthrough programming language?
« Reply #47 on: February 11, 2020, 06:00:56 am »
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Re: Possibly a breakthrough programming language?
« Reply #48 on: February 11, 2020, 06:08:26 am »
Hmm, it has less bvlocks but longer code :)

above version:

from numbers import Number
tree = ['']
window_start = 1
window_end = 4
for count2 in range(25):
  window = 'abbacbccacbbbababcabccacabcbcb'[window_start - 1 : window_end]
  window_start = (window_start if isinstance(window_start, Number) else 0) + 1
  window_end = (window_end if isinstance(window_end, Number) else 0) + 1
  char_location = 1
  node = 1
  for count in range(4):
    char_in_window = window[char_location - 1]
    char_location = (char_location if isinstance(char_location, Number) else 0) + 1
    char_index = tree[node - 1].find(char_in_window) + 1
    if char_index == 0:
      tree[node - 1] = str(tree[node - 1]) + str(char_in_window)
      if char_location != 5:
        if node == len(tree):
          tree.append([])
        tree[(node + 1 - 1)].append(len(tree) + 1)
        tree.append('')
      node = len(tree)
    else:
      if char_location != 5:
        goto = node + 1
      else:
        goto = node
      node = tree[goto - 1][char_index - 1]
print(tree)

previous version:

tree = ['']
window_start = 1
window_end = 4
for count2 in range(25):
  window = 'abbacbccacbbbababcabccacabcbcb'[window_start - 1 : window_end]
  window_start = window_start + 1
  window_end = window_end + 1
  char_location = 1
  node = 1
  for count in range(4):
    char_in_window = window[char_location - 1]
    char_location = char_location + 1
    char_index = tree[node - 1].find(char_in_window) + 1
    if char_index == 0:
      tree[node - 1] = str(tree[node - 1]) + str(char_in_window)
      if char_location != 5:
        if node == len(tree):
          tree.append([])
        tree[(node + 1 - 1)].append(len(tree) + 1)
        tree.append('')
      node = len(tree)
    else:
      if char_location != 5:
        goto = node + 1
      else:
        goto = node
      node = tree[goto - 1][char_index - 1]
print(tree)
Emergent          https://openai.com/blog/

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Re: Possibly a breakthrough programming language?
« Reply #49 on: February 11, 2020, 07:00:54 am »
now this is getting interesting, conversion type blocks and more? Code2Blocks? Hmm....sweet!
My code makes it generate the blocks, paste my code from above :)
Seems buggy though, I like Google's.
https://think.cs.vt.edu/blockpy/blockpy/load?settings-display.instructor=true
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Re: Possibly a breakthrough programming language?
« Reply #50 on: February 11, 2020, 07:26:21 am »
Wow, 45 seconds for the bigger code (post # 48), 30 secs for the smaller code. So it's a keeper hmm....
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Re: Possibly a breakthrough programming language?
« Reply #51 on: February 11, 2020, 07:52:42 pm »
refactor!
https://blockly-demo.appspot.com/static/demos/code/index.html#xhyzuf

tree = ['']
window_start = 1
window_end = 4
for count2 in range(8000000):
  window = text[window_start - 1 : window_end]
  window_start = window_start + 1
  window_end = window_end + 1
  char_location = 1
  node = 1
  for count in range(4):
    char_in_window = window[char_location - 1]
    char_location = char_location + 1
    char_index = tree[node - 1].find(char_in_window) + 1
    if char_index == 0:
      tree[node - 1] = str(tree[node - 1]) + str(char_in_window)
      if node == len(tree):
        tree.append([])
      tree[(node + 1 - 1)].append(len(tree) + 1)
      tree.append('')
      node = len(tree)
    else:
      if char_location != 5:
        goto = node + 1
      else:
        goto = node
      node = tree[goto - 1][char_index - 1]
print(tree)
Emergent          https://openai.com/blog/

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Re: Possibly a breakthrough programming language?
« Reply #52 on: February 11, 2020, 08:30:27 pm »
REFACTORRRRRRRRRR!!!!!!!!!!!!!!!!!!
https://blockly-demo.appspot.com/static/demos/code/index.html#95fn2y

tree = ['']
window_start = 1
window_end = 4
for count2 in range(2):
  window = 'ababa'[window_start - 1 : window_end]
  window_start = window_start + 1
  window_end = window_end + 1
  char_location = 1
  node = 1
  for count in range(4):
    char_in_window = window[char_location - 1]
    char_location = char_location + 1
    char_index = tree[node - 1].find(char_in_window) + 1
    if char_index == 0:
      tree[node - 1] = str(tree[node - 1]) + str(char_in_window)
      if node == len(tree):
        tree.append([])
      tree[(node + 1 - 1)].append(len(tree) + 1)
      tree.append('')
      node = len(tree)
    else:
      node = tree[(node + 1 - 1)][char_index - 1]
print(tree)
Emergent          https://openai.com/blog/

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Re: Possibly a breakthrough programming language?
« Reply #53 on: February 11, 2020, 09:26:56 pm »
RE-FACT-OR

tree = ['']
window_start = 1
window_end = 4
for count2 in range(2):
  window = 'ababa'[window_start - 1 : window_end]
  window_start = window_start + 1
  window_end = window_end + 1
  char_location = 1
  node = 1
  for count in range(4):
    char_in_window = window[char_location - 1]
    char_location = char_location + 1
    char_index = tree[node - 1].find(char_in_window) + 1
    if char_index == 0:
      tree[node - 1] = str(tree[node - 1]) + str(char_in_window)
      if node == len(tree):
        tree.append([])
      tree[node].append(len(tree) + 1)
      tree.append('')
      node = len(tree)
    else:
      node = tree[node][char_index - 1]
print(tree)
Emergent          https://openai.com/blog/

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Re: Possibly a breakthrough programming language?
« Reply #54 on: February 11, 2020, 09:42:24 pm »
RE FACT OR

e = ['']
h = 1
g = 4
for count2 in range(2):
  f = 'ababa'[h - 1 : g]
  h = h + 1
  g = g + 1
  c = 1
  d = 1
  for count in range(4):
    a = f[c - 1]
    c = c + 1
    b = e[d - 1].find(a) + 1
    if b == 0:
      e[d - 1] = str(e[d - 1]) + str(a)
      if d == len(e):
        e.append([])
      e[d].append(len(e) + 1)
      e.append('')
      d = len(e)
    else:
      d = e[d][b - 1]
print(e)
Emergent          https://openai.com/blog/

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Re: Possibly a breakthrough programming language?
« Reply #55 on: February 11, 2020, 09:45:22 pm »
R E F A C T O R

e = ['']
g = 4
for count2 in range(2):
  f = 'ababa'[count2 : g]
  g = g + 1
  c = 1
  d = 1
  for count in range(4):
    a = f[count - 1]
    b = e[d - 1].find(a) + 1
    if b == 0:
      e[d - 1] = str(e[d - 1]) + str(a)
      if d == len(e):
        e.append([])
      e[d].append(len(e) + 1)
      e.append('')
      d = len(e)
    else:
      d = e[d][b - 1]
print(e)
Emergent          https://openai.com/blog/

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Re: Possibly a breakthrough programming language?
« Reply #56 on: February 11, 2020, 09:49:36 pm »
ROTCAFER

e=['']
g=4
for count2 in range(2):
  f='ababa'[count2:g]
  g=g+1
  c=1
  d=1
  for count in range(4):
    a=f[count-1]
    b=e[d-1].find(a)+1
    if b==0:
      e[d-1]=str(e[d-1])+str(a)
      if d==len(e):
        e.append([])
      e[d].append(len(e)+1)
      e.append('')
      d=len(e)
    else:
      d=e[d][b-1]
print(e)
Emergent          https://openai.com/blog/

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Re: Possibly a breakthrough programming language?
« Reply #57 on: February 21, 2020, 11:29:09 pm »
By today I SHOULD be done the AI compressor. Just fine tuning it now...
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Re: Possibly a breakthrough programming language?
« Reply #58 on: February 23, 2020, 09:16:37 am »
it seems my alg without its enhancing function already compresses

"thezcatzatzthezczazcatzthatzazcatzthezcatzthezcat"

into

0636763912451316

which in bits is is 10010000110010001000100110110011101011010011110100

so 50 bits to store 392 bits
« Last Edit: February 23, 2020, 09:44:36 am by LOCKSUIT »
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Re: Possibly a breakthrough programming language?
« Reply #59 on: February 23, 2020, 09:25:28 am »
see it wants patterns :)

for "azctheztaehheczathaeezzccaahazcezeccaaathtaczaeta"
959365642149805036228786047993219

for "thethethecatthecatthththaccattheacatthethethethet"
9611206569478968

for "bbbaaabbbaaabbbaaabbbaaabbbaaabbbaaabbbaaabbbaaab"
199905
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