50 pages and 2.25BTC unlocked in less than 2 months ! Rather a good score, no ?

yes this is good score.

Not listen all not health critics.

No!

@arulbero test with 1000 pub keys fulfilled.
Workfile taken from puzzle 49 bit and prepared to work in the range 54 bit.
Each wild DPs tamed. All DPs multipled by 32 and checked (x-coordinate was correct when distance*G' )
-xbit was 5 so G'=G*inv(32) = (bb2228d3ea32cb3c1eb160cc824a4ba8115f9a7f415d18ddcaac8193defc2c47,71f8c9c7cc35f99b8b2abcdcab86d12cb7539b3fbb45bc433b3bd9421b35be53)
1000 keys was generated in range 40000000000000:7fffffffffffff
Each pubkeys was multipled by inv(32) and solved with G'and Kangaroo v1.4
Expected operations 2^28.06
Total op was 261530636052 = 2^37.93
In average 2^27.98

Many thanks.

The gain (- 5,4%) is little, how many DPs have you generated for the key in 49 bit?

If you have generated 2^25.08 points, then:

(2^x  +  2^25.08) tame * (2^x) wild = 2^54 couples

2^(2x) + 2^(25.08)*2^x - 2^54 = 0

--> x = 2^26.81

Then you should have computed on average 2^26.81 tame-steps + 2^26.81 wild-steps = 2^27.81 steps

DP Count  : 240568 2^17.876 in 49bit workfile. i have source file and converted file. I can share if you want to verify DPs. But each DP after multiplication was verifed with G' and x-coordinate correct.
For 1000 pubkeys

Test done on CPU

expected op for 1 pub = 2^28.06
1000 pubkeys ~2^ 9.965
2^28.06*2^9.965=(28.06+9.965)=2^38.025
So 2^37.93 is little-little bit faster then expected

I think that for birthday paradox is a big difference between have 16Tame DP and 2 wild DP or 9tame and 9wild.
Maybe if Kangaroo app first launch wild (as you say somewhere above) gain the same amount as tame DPs maybe in this case we will have faster result.

The best time complexity with the birthday paradox and kangaroo having a starting position uniformly distributed in [0..N] is obtained when drawing alternatively one tame and one wild. When using DP0, you get ~2.sqrt(N). When using DPx, it is like drawing alternatively 2^x TAME and then 2^x WILD, you have then a DP overhead.
I managed to get the formula for parallel version which is: ~2.CubicRoot( N (k.theta + sqrt(N)) ) where theta=2^dpbit and k the number of kangaroo-2 running in parallel.
For theta=1 (DP0) and k=0 (or when k.theta << sqrt(N)) we well get ~2.sqrt(N), when k.theta >> sqrt(N), the time complexity tend to ~2.CubicRoot(k.N.theta).
So it is important to choose a DP such as k.theta << sqrt(N).

Etar
if i am not wrong
you have 49 bit 1000 pubkeys
you shift dp file to 54 bit
and you have this info
" Expected operations 2^28.06, Total op was 261530636052 = 2^37.93 "
if i am not wrong you total op 2^38 work = to search 1 pubkey in 2^76 bitrange
76 bit = 75557863725914323419135
54 bit = 18014398509481983
distance = 4194303 for 1000 pub key
and for 1 pubkey = 419.4303
and for 32*g = 134217.72
in short you used tooo much time as compare to equal 76 bit work
hope u understand