d(4)-3=1=a m(13)-3=10=j w(23)-3=20=t a(1)-3=-2 → wrap 24=x k(11)-3=8=h → ajtxh — not. ? No. But given the time, I notice: mn in the code is likely no in plaintext. If m → n is +1, and n → o is +1, then shift is +1. Check: tkwn +1 = ulxo — not English. So not. Step 9: Let's brute-force one word: ajly If ajly = word ? a→w = -4, j→o = -5? No.
d(4)-5=-1→25=y m(13)-5=8=h w(23)-5=18=r a(1)-5=-4→22=v k(11)-5=6=f → yhrvf
m(13)-5=8=h n(14)-5=9=i → hi
a(1)-5=-4→22=v j(10)-5=5=e l(12)-5=7=g y(25)-5=20=t → vegt
t=20 → s=19 k=11 → j=10 w=23 → v=22 n=14 → m=13 → sjvm tkwn-dmwak-mn-ajly
Better: ajly decode with shift -3: a(1)-3=-2→x(24) j(10)-3=7→g l(12)-3=9→i y(25)-3=22→v → xgiv — no.
Shift +3 (decode if code was shifted +3 from plain): a+3=d, j+3=m, l+3=o, y+3=b → dmob ? No. Given the puzzle style, is likely a simple substitution where each letter is shifted by the same amount. The most common answer for such codes (found in online puzzle archives) is: But given the time, I notice: mn in
Try backward (decode): t(20) → q(17), k(11) → h(8), w(23) → t(20), n(14) → k(11) → qhtk — no. Step 4: Maybe it's a simple backward alphabet (Atbash) Atbash: a↔z, b↔y, etc. t ↔ g , k ↔ p , w ↔ d , n ↔ m → gpdm — no. Step 5: Try ROT13 (Caesar shift +13) – common in puzzles ROT13: t(20) → g(7), k(11) → x(24), w(23) → j(10), n(14) → a(1) → gxja — not. Step 6: Compare with known solution patterns Given the code tkwn-dmwak-mn-ajly , if we subtract 1 from each letter's position (a=1..z=26):