Andzhig
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March 09, 2021, 01:15:55 AM Last edit: March 09, 2021, 09:40:58 AM by Andzhig |
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hi there andzhig,
could you make it so it searches the address starting with 16 only or show and search only the 64th range with the same way you already beautifully described and explained thanks a lot sir.
16 in the sense of a number or name of an address... to make the number start a1= [(1, 0), (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7)] a1= [(1, 6)] all the first similarly (9, 2), (9, 3), (9, 4), (9, 5)... especially since we do not need more than (9, 9) in sets (11, 1), (14, 0) etc... a1= [(0, 3), (3, 0), (1, 2), (2, 1)] a2= [(0, 11), (11, 0), (1, 10), (2, 9), (3, Cool, (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), (9, 2), (10, 1)] a3= [(0, 11), (11, 0), (1, 10), (2, 9), (3, Cool, (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), (9, 2), (10, 1)] a4= [(0, 14), (14, 0), (1, 13), (2, 12), (3, 11), (4, 10), (5, 9), (6, Cool, (7, 7), (8, 6), (9, 5), (10, 4), (11, 3), (12, 2), (13, 1)] a5= [(0, 4), (4, 0), (1, 3), (2, 2), (3, 1)] a6= [(0, 2), (2, 0), (1, 1)] a7= [(0, 10), (10, 0), (1, 9), (2, Cool, (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)] a8= [(0, 2), (2, 0), (1, 1)] a9= [(0, 10), (10, 0), (1, 9), (2, Cool, (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)] a10=[(0, 10), (10, 0), (1, 9), (2, Cool, (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)] - a1= [(0, 3), (3, 0), (1, 2), (2, 1)] a2= [(2, 9), (3, , (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), (9, 2)] a3= [(2, 9), (3, , (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), (9, 2)] a4= [(5, 9), (6, , (7, 7), (8, 6), (9, 5)] a5= [(0, 4), (4, 0), (1, 3), (2, 2), (3, 1)] a6= [(0, 2), (2, 0), (1, 1)] a7= [(1, 9), (2, , (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)] a8= [(0, 2), (2, 0), (1, 1)] a9= [(1, 9), (2, , (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)] a10=[(1, 9), (2, , (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)] instead of all brute force after discarding, 41990400 whole run, finding at 15513701 steps. import time a1= [(0, 3), (3, 0), (1, 2), (2, 1)] a2= [(2, 9), (3, , (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), (9, 2)] a3= [(2, 9), (3, , (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), (9, 2)] a4= [(5, 9), (6, , (7, 7), (8, 6), (9, 5)] a5= [(0, 4), (4, 0), (1, 3), (2, 2), (3, 1)] a6= [(0, 2), (2, 0), (1, 1)] a7= [(1, 9), (2, , (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)] a8= [(0, 2), (2, 0), (1, 1)] a9= [(1, 9), (2, , (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)] a10=[(1, 9), (2, , (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)] count = 0 for b1 in a1: for b2 in a2: for b3 in a3: for b4 in a4: for b5 in a5: for b6 in a6: for b7 in a7: for b8 in a8: for b9 in a9: for b10 in a10: count += 1 d = b1+b2+b3+b4+b5+b6+b7+b8+b9+b10 if d == (3, 0, 5, 6, 8, 3, 7, 7, 3, 1, 2, 0, 6, 4, 2, 0, 2, 8, 5, 5): print(d) print(count) print(count) time.sleep(260.0) in any case, even discarding unnecessary (not need more than (9, 9) in sets (11, 1), (14, 0) etc...) 18×18×18×18×18×18×18×18×18×18 = 3570467226624 / 2 = 1785233613312 fixed combinations in which only 3 to 4 may be needed. again a performance question. If could it run 10-50 million at least in a second 1×18×18×18×18×18×18×18×18×18 1 sec 2×18×18×18×18×18×18×18×18×18 2 sec 3×18×18×18×18×18×18×18×18×18 3 sec 4×18×18×18×18×18×18×18×18×18 4 sec 18×18×18×18×18×18×18×18×18×1 blablabla sec 18×18×18×18×18×18×18×18×18×2 blablabla sec 18×18×18×18×18×1×18×18×18×18 blablabla sec 18×18×18×18×18×2×18×18×18×18 blablabla sec ... try it on gpu (who can write the code) . import random from bit import Key import time import itertools
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def funcc(u): number = int(u) lst = [] tmp = sorted((0, number)) lst.append(tmp)
for j in range(number): for i in range(number): if j + i == number: tmp = (j, i) if tmp not in lst: lst.append(tmp) #print(*tmp)
tmp = sorted((number, 0)) lst.append(tmp) return lst
def func(): gennum1 = random.randint(0,18) gennum = gennum1 print(gennum,funcc(gennum)) number = int(gennum) lst = [] tmp = (0, number) if tmp[0] <=9: if tmp[1] <=9: lst.append(tmp)
for j in range(number): for i in range(number): if j + i == number: tmp = (j, i) if tmp[0] <=9: if tmp[1] <=9: if tmp not in lst: lst.append(tmp) #print(*tmp) tmp = (number, 0) if tmp[0] <=9: if tmp[1] <=9: lst.append(tmp) return lst
print("18×18×18×18×18×18×18×18×18×18 = 3570467226624") print(" ")
a1= [(1, 0), (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7)] a2= func() a3= func() a4= func() a5= func() a6= func() a7= func() a8= func() a9= func() a10=func()
print(" ") print("***") print("cut off set (18, 0), (0, 18), (17, 1), (1, 17)... = 1785233613312") print(" ")
print(a1) print(a2) print(a3) print(a4) print(a5) print(a6) print(a7) print(a8) print(a9) print(a10)
print(" ") print("***")
time.sleep(3.0) count = 0
for x in itertools.product(a1, a2, a3, a4, a5, a6, a7, a8, a9, a10): count += 1
d = ("".join(map(str, (item for sublist in x for item in sublist)))) if len(d) <= 50: ran = int(d) key1 = Key.from_int(ran) addr1 = key1.address if addr1 in list:
print (ran,"found!!!")
s5 = str(ran) f=open(u"C:/a.txt","a") f.write(s5 + '\n') f.close()
break
else: #pass print(count,ran,len(d),addr1)
[/size] *** don’t understand how to count, we removed unnecessary and we get options 10×10×10×10×10×10×10×10×10×10 = 10000000000 possible, among which there are 400000000 (exl 8 8 6 8 10 9 7 7 3 10 = 406425600) and several thousand (8 1 3 1 3 2 3 2 7 10 = 60480) transformation 18×18×18×18×18×18×18×18×18×18 10×10×10×10×10×10×10×10×10×10 2×8×4×4×5×3×13×15×11×2 16473600 1135041350219496382 3×9×5×5×6×4×6×4×8×3 9331200 5×7×15×12×6×7×14×9×17×2 1133546400 1425787542618654982 6×8×4×7×7×8×5×10×2×3 22579200 12×8×10×7×6×5×15×4×6×2 145152000 3908372542507822062 7×9×9×8×7×6×4×5×7×3 80015040 17×12×4×18×13×7×8×15×13×8 16680867840 8993229949524469768 2×7×5×1×6×8×9×4×6×9 6531840 what can factorization give to produce exactly such in the set 3×9×5×5×6×4×6×4×8×3 9331200 how will it look to 9331201 9331202 9331203 ...
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