Example: Let to suppose that there are two private keys A and B and A > B
If P(A) > P(B) then P(A) - P(B) = P(C) Then P(C) is a positive Value if the value of C is under some low bit range lets to say less than 2^70 (70 bits) you can determine the value of P(C) with some tool like kangaroo of BSGS easily with no effort
So ONLY if the Difference P(A) - P(B) or P(B) - P(A) is under some value easy to find you can know which one of those publickeys are bigger.
If the difference is bigger than 90 bits or something other high value then you CAN NOT know it. Unless you have some GPU farm to calculate those values.
Example:
P(B) = 02fde2347f83e21198fc48b918f5657c188ffcdd8611b39b987230addb91d05d80
Differences:
P(A) - P(B) = 0379be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
P(B) - P(A) = 0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
Solve it with keyhunt:
0379be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
$ ./keyhunt -m bsgs -f input.txt -k 128 -t 4 -S -r 1:100000000000000
[+] Version 0.2.211117 SSE Trick or treat ¡Beta!, developed by AlbertoBSD
[+] K factor 128
[+] Threads : 4
[+] Mode BSGS secuential
[+] Opening file input.txt
[+] Added 2 points from file
[+] Range
[+] -- from : 0x1
[+] -- to : 0x100000000000000
[+] N = 0x100000000000
[+] Bloom filter for 536870912 elements : 1840.33 MB
[+] Bloom filter for 16777216 elements : 57.51 MB
[+] Bloom filter for 524288 elements : 1.80 MB
[+] Allocating 8.00 MB for 524288 bP Points
[+] Reading bloom filter from file keyhunt_bsgs_4_536870912.blm .... Done!
[+] Reading bloom filter from file keyhunt_bsgs_6_16777216.blm .... Done!
[+] Reading bP Table from file keyhunt_bsgs_2_524288.tbl .... Done!
[+] Reading bloom filter from file keyhunt_bsgs_7_524288.blm .... Done!
[+] Thread Key found privkey 1
[+] Publickey 0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
[+] Thread 0xfff00000000001
End
The value of 0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798 is 1 Positive then if the result of the subtraction P(B) - P(A) is positive, hence P(B) > P(A) for this example.
@PawGo sorry to divert your topic, but i want to reply that question.