Yes, I found it easy too.
Tripple click to read the walkthrough that I wrote:1a. 17/2 @ r34c9 = {89} (NP @ c9, HW @ r234c159)
b. 12/2 @ r34c1 = {57} (NP @ c1, HW @ r234c159)
c. 23/3 @ r678c5 = {689} (NT @ c5, HW @ r678c159)
d. 9/3 @ r234c5 = {234} (NT @ c5, HW @ r234c159)
e. r2c19 = NP {16} @ r2
f. 7/2 @ r12c9 = {16} (NP @ c9, n3)
g. 10/2 @ r67c9 = {37} (NP @ c9, HW @ r678c159)
h. 5/2 @ r67c1 = {14} (NP @ c1, HW @ r678c159)
i. r2c19 = [61], r1c9 = 6
j. 11/2 @ r89c1 = [29], r8c9 = 5 (last digit @ HW @ r678c159)
2a. Innies @ W @ r678c678 -> r8c67 = 5 = {14} (NP @ r8, W @ r678c678)
b. 9/2 @ r89c7 = [18], r8c6 = 4
c. 15/3 @ n9 -> r9c89 = [64]
d. Innies @ W @ r678c234 -> r8c34 = 11 = [83]
e. r5c9 = 2, 11/3 @ n6 -> r5c78 = {45} (NP @ r5, n6, HW @ r159c678)
f. 14/3 @ r678c8 = [329], 10/2 @ r67c9 = [73]
g. r7c7 = 7, r8c25 = [76]
h. 14/3 @ r234c8 = {158} -> r4c8 = 1, r23c8 = NP {58} @ c8, n3, W @ r234c678
i. 17/2 @ r34c9 = [98], 11/3 @ r5c789 = [542]
j. r1c8 = 7, 14/3 @ n3 -> r12c7 = [34]
k. r3c7 = 2, 14/3 -> r3c6 = 3, r4c7 = 9
...
Tripple click to read the solution that I wrote:+-------+-------+-------+
| 8 4 2 | 9 5 1 | 3 7 6 |
| 6 9 3 | 8 2 7 | 4 5 1 |
| 7 5 1 | 6 4 3 | 2 8 9 |
+-------+-------+-------+
| 5 2 7 | 4 3 6 | 9 1 8 |
| 3 8 6 | 7 1 9 | 5 4 2 |
| 4 1 9 | 2 8 5 | 6 3 7 |
+-------+-------+-------+
| 1 6 4 | 5 9 8 | 7 2 3 |
| 2 7 8 | 3 6 4 | 1 9 5 |
| 9 3 5 | 1 7 2 | 8 6 4 |
+-------+-------+-------+
I already prepared a windoku killer a few days ago, but will wait some days again before publishing it. I'm next to publish a windoku killer, OK ?
PS There's nothing wrong with uniqueness techniques, as long as you do not use them to
prove the puzzle has a unique solution, because all uniqueness techniques
assume the puzzle has a unique solution.