The Nurikabe of Peace

Today is the 75th anniversary of Sir Winston Churchill's speech "The Sinews of Peace" which introduced to the world the phrase "The Iron Curtain". This puzzle is inspired by that phrase: in addition to normal Nurikabe rules, the completed grid must also have an "iron curtain", a full row of cells that are all shaded. I hope you enjoy!

Solver Notes

This is not a particularly hard puzzle; I was noodling with the "iron curtain" idea, and when I looked up Churchill's birth and death dates, I saw that today was this significant anniversary, so I had to rush a little bit. But that said, while finding the solution is not hard, doing so logically in a fashion that shows the solution is unique is more of a challenge, so please provide logical steps as part of your answer.

I believe this is the solution:

Starting off:

We can immediately shade in some of our cells to separate the 1 cell regions and avoid two regions with different numbers from joining.

Now, we note that

the 5 cell region on the lower right must continue upward, because if it went to the left then it would produce a shaded cell island (R8C8). This gives us the following:

Next,

we will use the "iron curtain" stipulation. Clearly, rows 1-3 and 6-9 cannot have all of their cells shaded. If row 4 were completely shaded, then that would leave us with 20 white cells in the upper area to divide among the 7, 8, and 4 unshaded regions which take up 19 cells in total. However, we need at least two shaded cells to separate these regions, so we would need at least 21 cells in order to complete the upper region compared to the 20 available cells we are given. Thus, row 4 cannot be completely shaded, leaving us with row 5 as our "iron curtain":

Things get a bit trickier from here on out - if any step doesn't seem well-substantiated, let me know, and I'll try to flesh out my explanation.

We start with the lower region by focusing on R8C3. If that cell were not shaded, then the three shaded cell block in the lower left (R8C1-2, R9C1) would necessarily have to connect with row 5 via R6-7C1, and the one shaded cell block (R9C3) would have to connect with R9C6 via R9C4-5. This leaves 17 available cells to divide into two regions of 9 and 6, as shown below. Clearly, this is not possible, as we can only use 2 shaded cells to separate the two regions, which is not enough. Thus, R8C3 must be shaded.

Now, if we

try to connect the lower left shaded region to row 5 via R6-7C1 again, we will still find that it is impossible to create the 9 and 6 regions: we cannot connect to the R9C6-7 shaded block because that would lead to too few remaining cells to create our unshaded regions, and if we try to connect the R9C6-7 block to row 5 via R6-9C5, then the 9 region will only have 8 unshaded cells. Thus, we must connect our lower left region with R9C6-7 via R9C4-5. At this point, we are left with 18 cells to create our 9 and 6 regions, which we can do easily:

We move on to

completing the upper region. Again, we will consider how a particular shaded block can connect to other shaded blocks. Specifically, we look at R3C4: it can connect to the upper left hand blocks via R3C3 or to row 5 via R4C4. If R3C3 were shaded, then the options for forming the 8 region are severely reduced, as it will have to go toward the right while avoiding including any of the numbered cells. The following shows the only way we can create the 8 unshaded cell region, and we can clearly see that R2C5 cannot be connected to other shaded cells:

Thus,

R3C3 cannot be shaded, meaning that R4C3 must connect to row 5 via R4C4. This further allows us to deduce that R4C1 and R3C2 must be shaded in order to connect it to row 5, since there are too few unshaded cells on the left side to form the 8 region. Instead, we form the 8 region by shading R1-2C6 after that, since we only need one more unshaded cell in order to complete that region:

Finally,

we need to connect the three shaded cell block (R1C6, R2C5-6) to row 5. The only way to do so is through C9, since any closer column would result in the 7 region having too few cells. The rest can be deduced fairly straightforwardly, yielding the final solution:

• You got it. Nice explanation, thank you...I hope you enjoyed. Commented Mar 6, 2021 at 0:34
• @JeremyDover It was a neat idea, I liked it a lot :)
– HTM
Commented Mar 6, 2021 at 0:50