for example:
in subgroup generated by
0447316cb65cc8f20d539616cf65bc78479c686c3f70454cf5aab84c579b57efcd18a6a630cef25 44625d80b0297017dc9fef77712bd494fb3374974096e4dc278
that has scalar(61168582499785340698020811768434254152333414806039741990912550463524917977698) in secp256k1 subgroup
0479be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798483ada7726a3c 4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8 G of secp256k1
will be at position(have scalar) 54229698599845083480280347574976582697435195709826937379446800456474139525780
0400000000000000000000003b78ce563f89a0ed9414f5aa28ad0d96d6795f9c63c0c686408d517 dfd67c2367651380d00d126e4229631fd03f8ff35eef1a61e3c
will be at position(have scalar) 27114849299922541740140173787488291348717597854913468689723400228237069762890
we can find generator point so that some point be at certain position in the subgroup.
or we can retrieve point by generator and position.
and we can do so with any point from secp256k1 if we take point and its scalar. good for research only.
will not be able to break secp256k1 curve with that.
in your
G = 0447316cb65cc8f20d539616cf65bc78479c686c3f70454cf5aab84c579b57efcd18a6a630cef2544625d80b0297017dc9fef77712bd494fb3374974096e4dc278
k=27114849299922541740140173787488291348717597854913468689723400228237069762890
k.G = 0400000000000000000000003b78ce563f89a0ed9414f5aa28ad0d96d6795f9c63c0c686408d517dfd67c2367651380d00d126e4229631fd03f8ff35eef1a61e3c
Nice generation of points!
how do you find such point with x having many leading zeros and the corresponding scalar?
I name it
pt0 for convenient
do you start from it and randomly pick a scalar that point pt0/scalar = G
of inversly fix a random G and randomly generate a scalar k unless you find a point with x having sufficient leading zero?