another search option, but not fast
each of the 3 according to their position...
i.e 2^... trimming and structuring from smallest to largest, so that all sets are completely filled from 1 to 1000, from 1 to 1000000 etc.
2 numbers
970
436
974
005
6 05 0 0 1
51 50 1 1 0
023
690
481
3 numbers
970
80 079 2 1 0
98 097 1 2 0
710 709 2 0 1
791 790 1 0 2
908 907 0 2 1
971 970 0 1 2
436
347 346 1 0 2
365 364 2 0 1
437 436 0 1 2
464 463 0 2 1
635 634 2 1 0
644 643 1 2 0
974
480 479 2 1 0
498 497 1 2 0
750 749 2 0 1
795 794 1 0 2
948 947 0 2 1
975 974 0 1 2
005
6 005 0 0 2
16 015 0 0 2
26 025 0 0 2
36 035 0 0 2
46 045 0 0 2
51 050 0 0 1
52 051 0 0 1
53 052 0 0 1
54 053 0 0 1
55 054 0 0 1
56 055 0 0 1
57 056 0 0 1
58 057 0 0 1
59 058 0 0 1
60 059 0 0 1
66 065 0 0 2
76 075 0 0 2
86 085 0 0 2
96 095 0 0 2
106 105 1 1 2
151 150 2 2 1
206 205 1 1 2
251 250 2 2 1
306 305 1 1 2
351 350 2 2 1
406 405 1 1 2
451 450 2 2 1
501 500 1 1 0
502 501 1 1 0
503 502 1 1 0
504 503 1 1 0
505 504 1 1 0
506 505 1 1 0
507 506 1 1 0
508 507 1 1 0
509 508 1 1 0
510 509 1 1 0
511 510 2 2 0
521 520 2 2 0
531 530 2 2 0
541 540 2 2 0
551 550 2 2 0
561 560 2 2 0
571 570 2 2 0
581 580 2 2 0
591 590 2 2 0
606 605 1 1 2
651 650 2 2 1
706 705 1 1 2
751 750 2 2 1
806 805 1 1 2
851 850 2 2 1
906 905 1 1 2
951 950 2 2 1
023
24 023 0 1 2
33 032 0 2 1
204 203 1 0 2
231 230 2 0 1
303 302 1 2 0
321 320 2 1 0
690
70 069 1 2 0
97 096 2 1 0
610 609 0 2 1
691 690 0 1 2
907 906 2 0 1
961 960 1 0 2
481
149 148 1 2 0
185 184 2 1 0
419 418 0 2 1
482 481 0 1 2
815 814 2 0 1
842 841 1 0 2
or 6 numbers
count = 0
v = []
S=[]
Nn =['970', '436', '974', '005', '023', '690', '481'] #970 436 974 005 023 690 481
for elem in Nn:
h1 = elem #970436974005023690481 108717677802902655490863978809014617508706260176761067916188909553243434167589
print(" ")
print(elem)
g1 = ([h1[i:i + 1] for i in range(0, len(h1), 1)])
g = g1
i = 256
while i <= 10000:
a=pow(2,i)
b = str(a)[1:6]
v.append(b)
i=i+1
j = set(v)
jj = sorted(j)
for elem in jj:
count += 1
if g[0] in elem:
if g[1] in elem:
if g[2] in elem:
print(count,elem,elem.index(g[0]),elem.index(g[1]),elem.index(g[2]))
j = (elem,elem.index(g[0]),elem.index(g[1]),elem.index(g[2]))
S.append(j)
#print(sorted(S))
#print(elem)
count = 0
well, we get at 3 numbers combinations 0 0 0 2 2 2, at 6 nembers 0 0 0 4 4 4. 2x2x2=8 8x8x8x8x8x8x8 = 2097152, 4x4x4 = 64 64x64x64x64x64x64x64 = 44392781971456
at 3 numbers the spread is large for step by step
at 6 nembers less spread but more combinations 0 0 0 4 4 4.
that is, the search looks like this
970
530 04739 4 2 0
436
516 04639 1 3 2
974
530 04739 4 2 1
005
500 04507 0 0 2
023
584 05236 0 2 3
690
516 04639 2 4 0
481
511 04581 1 3 4
from 500 to 600 step by step 100×100×100×100×100×100×100 = 100000000000000
maybe can get something out of this...
or for 22 pz. lenght 0 1, 10×10×10×10×10×10×10×10×10×10×10= 100000000000
count = 0
v = []
S=[]
h2 = ("9704369740050236904811") #970436974005023690481 108717677802902655490863978809014617508706260176761067916188909553243434167589
g2 = ([h2[i:i + 1] for i in range(0, len(h2), 1)])
Nn =g2 #970 436 974 005 023 690 481
for elem in Nn:
h1 = elem #970436974005023690481 108717677802902655490863978809014617508706260176761067916188909553243434167589
print(" ")
print(elem)
g1 = ([h1[i:i + 1] for i in range(0, len(h1), 1)])
g = g1
i = 256
while i <= 10000:
a=pow(2,i)
b = str(a)[1:3]
v.append(b)
i=i+1
j = set(v)
jj = sorted(j)
for elem in jj:
count += 1
if g[0] in elem:
if count >= 50:
if count <= 60:
print(count,elem,elem.index(g[0]))
j = (elem,elem.index(g[0]))
S.append(j)
#print(sorted(S))
#print(elem)
count = 0
9
50 49 1
60 59 1
7
58 57 1
0
51 50 1
4
50 49 0
55 54 1
3
54 53 1
6
57 56 1
9
50 49 1
60 59 1
7
58 57 1
4
50 49 0
55 54 1
0
51 50 1
0
51 50 1
5
51 50 0
52 51 0
53 52 0
54 53 0
55 54 0
56 55 0
57 56 0
58 57 0
59 58 0
60 59 0
0
51 50 1
2
53 52 1
3
54 53 1
6
57 56 1
9
50 49 1
60 59 1
0
51 50 1
4
50 49 0
55 54 1
8
59 58 1
1
52 51 1
1
52 51 1
if I think correctly 2 positions 0 and 1 for 11 seats calculated as 2×2×2×2×2×2×2×2×2×2×2=2048, by 10 step by step, 10×10×10×10×10×10×10×10×10×10×10= 100000000000, 2048×100000000000 = 204800000000000.
204800000000000
vs
9999999999999999999999
although need 10 hike 22 multiply(( but anyway an interesting result from 50 to 60, from 0 to 10 another result.
111111101111111 119666659114170 32639
111111111111111 191206974700443 32767
111111110111010 409118905032525 32698
111111111111111 611140496167764 32767
1101111010101111 2058769515153876 57007
1111110111111111 4216495639600700 65023
1111111111111111 6763683971478124 65535
1111100111111111 9974455244496707 63999
11110111111111111 30045390491869460 126975
11111111111111010 44218742292676575 131066
111110101111111111 138245758910846492 257023
111111111111111111 199976667976342049 262143
010111111101111111 525070384258266191 98175
1110111101111111111 1135041350219496382 490495
1110111011111101111 1425787542618654982 489455
1111111011011111111 3908372542507822062 521983
1111111111011111111 8993229949524469768 524031
11011111111111111100 30568377312064202855 917500
111111111110111111111 970436974005023690481 2096639
11011111111111111111111 22538323240989823823367 7340031
...
4
50 49 0
55 54 1
...
such can be considered by 1
