Hard Picross puzzle featuring a lion design to be solved using number clues.

Let’s solve this Picross puzzle step by step. The goal is to fill in the grid so that each row and column matches the numbers on the side — those numbers tell you how many consecutive black squares there are, with at least one white square between groups.

We’ll start with rows or columns that have big numbers or only one possibility — these are easiest to figure out first.

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Step 1: Look for rows or columns with a single large number

Look at Row 8 (counting from top): it says “12 1”. That means 12 black squares, then at least one white, then 1 black. But the whole row is only 20 squares long! Let’s check:

Total needed = 12 + 1 (minimum gap) + 1 = 14 → fits in 20.

But actually, let’s look at Row 7: “11 1” → 11 + 1 + 1 = 13 → also fits.

Wait — better idea: look at Column 10 (from left). It has “18” as its clue. That’s huge! The grid is 20 columns wide? Wait — let’s count the columns.

Actually, looking at the top clues: the last column has “0”, meaning no black squares at all. And the first column has “0” too? Wait — let’s list the column clues properly.

From left to right, column clues (top of grid):

Col 1: 0
Col 2: 1
Col 3: 1
Col 4: 2
Col 5: 3
Col 6: 2
Col 7: 7
Col 8: 6
Col 9: 5
Col 10: 11
Col 11: 16
Col 12: 1
Col 13: 2
Col 14: 9
Col 15: 4
Col 16: 2
Col 17: 2
Col 18: 1
Col 19: 2
Col 20: 0

Wait — Col 11 has “16” — that’s very big. The grid height is 20 rows? Let’s count the rows.

Rows (left side clues):

Row 1: 4 1
Row 2: 2 4 2
Row 3: 2 7 2
Row 4: 3 6 1
Row 5: 3 9 1
Row 6: 3 7 1
Row 7: 11 1
Row 8: 8 1
Row 9: 12 1
Row 10: 11 1
Row 11: 4 1
Row 12: 7 2
Row 13: 9 1
Row 14: 11 1
Row 15: 9 3
Row 16: 4 5
Row 17: 2 6
Row 18: 3 2 3
Row 19: 3 3
Row 20: 4 3

So 20 rows × 20 columns.

Back to Column 11: clue is “16”. That means 16 consecutive black squares in that column. Since the column is 20 cells tall, we can place 16 blacks starting from row 1? Or row 5? Let’s see where they can fit.

The only way to fit 16 in 20 is if they start at row 1, 2, 3, 4, or 5.

Start at row 1: ends at row 16 → leaves rows 17-20 empty → possible.
Start at row 2: ends at row 17 → leaves row 1 and 18-20 → but row 1 must be empty? Not necessarily.
But wait — we need to check which rows allow 16 consecutive blacks without conflicting with their own row clues.

This might be messy. Let’s try a different approach.

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Better Strategy: Start with rows that have only two numbers and one is very large.

Look at Row 9: “12 1”

That’s 12 blacks, then at least one white, then 1 black. Total minimum space: 12 + 1 + 1 = 14. Grid width is 20, so plenty of room.

But where can it go?

The 12-block must be placed such that after it, there’s at least one white, then the 1-black.

Also, the 1-black at the end could be anywhere after the gap.

But here’s a trick: if a row has a big number like 12, and the row is 20 long, the 12-block can’t start too late.

Maximum start position for 12-block: if it starts at column 9, it goes to 20 → that’s 12 cells (9 to 20 inclusive is 12 cells? 20-9+1=12 → yes). Then we need a white after? No — the next group is “1”, so after the 12-block, we need at least one white, then the 1-black. So if 12-block ends at 20, there’s no room for the white and the 1-black. So 12-block cannot end at 20.

Similarly, if 12-block ends at 19, then column 20 must be the 1-black? But then no white between them — not allowed. So 12-block must end by column 18 at latest, so that column 19 is white, column 20 is the 1-black.

So 12-block ends at 18 → starts at 7 (since 18-12+1=7).

Or ends at 17 → starts at 6, etc.

But let’s think differently. What if we look at columns with “0”? Those are easy.

Column 1: clue “0” → entire column is white. Mark all cells in column 1 as white (or X).

Column 20: clue “0” → entire column is white. Mark all cells in column 20 as white.

Now, any row that has a clue including a number that would require a black in column 1 or 20 must adjust.

For example, Row 1: “4 1” — total 5 blacks plus gaps. If column 1 and 20 are white, then the blacks must be in columns 2-19.

Similarly, Row 20: “4 3” — same thing.

Now, let’s look at Row 7: “11 1”

11 blacks, then gap, then 1 black.

Minimum space: 11 + 1 + 1 = 13.

Since col 1 and 20 are white, available columns: 2 to 19 → 18 columns.

Where can 11-block go?

If it starts at col 2, ends at 12 → then col 13 must be white (gap), then col 14 could be the 1-black? But then cols 15-19 are free — but the row clue doesn’t say anything else, so they must be white. Is that okay? Yes.

But what if the 1-black is at col 19? Then the 11-block must end by col 17 (so col 18 is white, col 19 is black).

So 11-block can start as early as col 2, or as late as col 7 (because 7+11-1=17, then col 18 white, col 19 black).

Still multiple options.

Let’s try a column with a high number.

Column 11: clue “16”

16 consecutive blacks in a 20-row column.

Possible placements:

- Rows 1-16
- Rows 2-17
- Rows 3-18
- Rows 4-19
- Rows 5-20

Now, check which of these are compatible with the row clues.

For example, if we put blacks in rows 1-16 for column 11, then for Row 1, which has clue “4 1”, having a black in col 11 might be part of the “4” or the “1”.

But Row 1’s “4 1” — if col 11 is black, it could be in the first group or second.

This is getting complicated. Maybe we should look for rows that are forced.

Another idea: look at Row 10: “11 1” — same as Row 7.

Row 14: “11 1” — again.

There are three rows with “11 1”: rows 7, 10, 14.

And Column 11 has “16” — which is close to 20, so likely spans most of the column.

Perhaps the 16 in column 11 corresponds to the 11s in those rows being aligned.

Let’s assume that in column 11, the 16 blacks are from row 3 to row 18. Why? Because rows 7,10,14 are within that range, and they have 11s, which might include column 11.

But let’s calculate: if column 11 has blacks from row 3 to 18, that’s 16 rows (18-3+1=16) — perfect.

So mark rows 3 to 18, column 11 as black.

Now, for each of those rows, we know that column 11 is black, so it must be part of one of their groups.

Take Row 7: “11 1” — if col 11 is black, and it's part of the 11-group, then the 11-group must include col 11.

Similarly for Row 10 and 14.

Now, for Row 7: “11 1” — suppose the 11-group includes col 11. Where can it start and end?

If the 11-group ends at col 11, it starts at col 1 (11-1+1=1), but col 1 is white (clue 0), so impossible.

If it ends at col 12, starts at col 2.

Ends at col 13, starts at col 3, etc.

But col 11 is black, so the 11-group must cover col 11.

The earliest it can start is col 2 (ending at 12), latest is col 11 (ending at 21 — too far, max col 20).

If it starts at col 11, ends at 21 — invalid.

So maximum end is col 20, so start at col 10 (20-11+1=10).

So for Row 7, the 11-group can start from col 2 to col 10.

But we have col 11 black, so if the 11-group includes col 11, it must start by col 1 at latest? No.

If the 11-group starts at col k, it occupies k to k+10.

Set k+10 >= 11 => k >=1, and k <=11.

But col 1 is white, so k>=2.

Also, after the 11-group, there must be at least one white, then the 1-black.

So if the 11-group ends at e, then e+1 is white, e+2 is the 1-black (if e+2<=20).

Or the 1-black could be later, but since there's only one more group, and it's size 1, it must be immediately after the gap.

The clue is "11 1", so exactly two groups: 11 blacks, then at least one white, then 1 black.

So the 1-black is right after the gap.

So positions: 11-group from s to s+10, then white at s+11, then black at s+12.

s+12 <=20, so s<=8.

Also s>=2 (since col 1 white).

So s can be 2 to 8.

Now, col 11 is black, so s+10 >=11 => s>=1, which is already satisfied.

s+10 >=11 => s>=1, but s>=2.

If s=2, ends at 12, so col 11 is included (yes, 2 to 12 includes 11).

If s=3, ends at 13, includes 11.

...

If s=8, ends at 18, includes 11.

All good.

But we also have the constraint that after the 11-group, there's a white at s+11, and black at s+12.

For example, if s=2, then white at 13, black at 14.

If s=8, white at 19, black at 20 — but col 20 is white (clue 0), so black at 20 is impossible. So s cannot be 8.

Similarly, s=7: white at 18, black at 19 — col 19 is not necessarily white, so possible.

s=6: white at 17, black at 18.

etc.

Now, recall that in column 11, we have black from row 3 to 18, so for Row 7, col 11 is black, which is consistent.

But we need to find which s works.

Perhaps look at other constraints.

Let's consider Row 9: "12 1"

12 blacks, then gap, then 1 black.

Minimum space: 12+1+1=14.

Col 1 and 20 white, so available 2-19.

12-group from s to s+11.

Then white at s+12, black at s+13.

s+13 <=20, so s<=7.

s>=2.

Also, col 11 is black for rows 3-18, so for Row 9, col 11 is black, so s+11 >=11 => s>=0, ok.

s+11 >=11 always true for s>=2.

But s+11 must be at least 11, which it is.

Now, if s=2, ends at 13, so col 11 included.

s=7, ends at 18, includes 11.

Black at s+13: if s=7, black at 20 — but col 20 is white, impossible.

s=6, black at 19 — possible.

s=5, black at 18.

etc.

Still many options.

Perhaps we can look at the intersection.

Another idea: look at Column 10: clue "11"

11 consecutive blacks.

Column 10, rows ?

Clue is "11", so 11 blacks in a row in that column.

Grid has 20 rows, so possible.

But we have column 11 with 16 blacks from row 3 to 18.

For Column 10, if it has 11 blacks, where?

Perhaps aligned with the rows.

Let's list the row clues again and see which rows have large numbers that might correspond.

Notice that Rows 7,9,10,14 have large first numbers: 11,12,11,11.

And Column 11 has 16, which might be covering those.

Perhaps the lion's body is in the middle.

Let's try to fill in what we know.

We know:

- Col 1: all white
- Col 20: all white
- Col 11: rows 3 to 18 are black (we assumed, but is it correct?)

Is there a better way? Let's look for rows that are completely determined.

For example, Row 1: "4 1"

With col 1 and 20 white, the 4 and 1 must be in 2-19.

The 4-group and 1-group with at least one white between.

Possible configurations:

- 4-group from 2-5, then white at 6, then 1 at 7
- 2-5, white 6, 1 at 8
- ...
- up to 4-group from 15-18, white 19, 1 at 20 — but col 20 white, so 1 at 20 impossible.

So 1-group cannot be at 20.

Similarly, if 1-group is at 19, then white at 18, 4-group ends at 17, so starts at 14 (17-4+1=14).

So possible.

But still many.

Perhaps start with the bottom or top.

Another thought: the title is "Lion", so probably symmetric or has a shape.

Perhaps the mane is at the top, body in middle, legs at bottom.

Look at Row 1: "4 1" — small, perhaps head.

Row 2: "2 4 2" — wider.

Row 3: "2 7 2" — even wider.

Row 4: "3 6 1"

Row 5: "3 9 1"

Row 6: "3 7 1"

Row 7: "11 1" — very wide

Row 8: "8 1"

Row 9: "12 1" — widest

Row 10: "11 1"

Row 11: "4 1" — narrow again

So likely, the lion's body is from row 7 to 10 or so, with row 9 being the widest.

So for Row 9: "12 1", and it's the widest, so probably the 12-group is centered or something.

Assume that for Row 9, the 12-group is from col 4 to 15? 15-4+1=12, then white at 16, black at 17.

Col 17 is not white, so possible.

Or from 5 to 16, white 17, black 18.

etc.

But earlier we have col 11 black for rows 3-18, so for Row 9, col 11 is black, which is within 4-15 or 5-16, etc.

Now, let's consider Column 14: clue "9"

9 consecutive blacks.

Where can it be?

Rows 1 to 20.

If we assume the lion's body is around rows 7-10, perhaps col 14 has blacks there.

But let's calculate possible positions.

Perhaps use the fact that some rows have only two groups, and the sum of the numbers plus minimum gaps gives the minimum length, and if it equals the row length minus the white columns, it might be forced.

For example, Row 9: "12 1" , min length 12+1+1=14, but row has 20 cells, minus col 1 and 20 white, so 18 cells available, so not forced.

Row 7: "11 1" , min 13, available 18, not forced.

Let's look at Row 15: "9 3" — 9+1+3=13, available 18, not forced.

Row 16: "4 5" — 4+1+5=10, not forced.

Perhaps start with columns that have small numbers or 0.

We have col 1 and 20 as 0, so all white.

Col 2: clue "1" — so only one black in the entire column.

Similarly, col 3: "1" — one black.

Col 4: "2" — two blacks, possibly consecutive or not, but since it's a single number, it means one group of 2, so consecutive.

In Picross, if a column has a single number, it means one group of that size, so consecutive blacks.

If multiple numbers, multiple groups with gaps.

So for col 2: "1" — exactly one black cell in the column.

Similarly, col 3: "1" — one black cell.

Col 4: "2" — two consecutive black cells.

Col 5: "3" — three consecutive.

Col 6: "2" — two consecutive.

Col 7: "7" — seven consecutive.

Col 8: "6" — six consecutive.

Col 9: "5" — five consecutive.

Col 10: "11" — eleven consecutive.

Col 11: "16" — sixteen consecutive.

Col 12: "1" — one black.

Col 13: "2" — two consecutive.

Col 14: "9" — nine consecutive.

Col 15: "4" — four consecutive.

Col 16: "2" — two consecutive.

Col 17: "2" — two consecutive.

Col 18: "1" — one black.

Col 19: "2" — two consecutive.

Col 20: "0" — none.

Now, for col 11: "16" — 16 consecutive blacks. As before, possible from row 1-16, 2-17, 3-18, 4-19, 5-20.

Now, let's see which one is compatible with the row clues.

Consider Row 1: "4 1" — if col 11 is black, it must be part of the 4 or the 1.

If it's part of the 4, then the 4-group includes col 11.

If part of the 1, then the 1 is at col 11.

But the 1 is a single black, so if col 11 is the 1 for Row 1, then the 4-group is elsewhere.

But let's see the position.

Suppose for col 11, we choose rows 3-18 as black.

Then for Row 1, col 11 is white (since row 1 < 3), so not black.

For Row 2, col 11 is white (row 2<3).

Row 3: col 11 is black.

Row 3 clue: "2 7 2" — so three groups: 2, then 7, then 2.

Col 11 is black, so it must be in one of the groups.

Likely in the 7-group, since 7 is large.

Similarly for other rows.

Now, for Row 3: "2 7 2" — with col 11 black, and assuming it's in the 7-group.

The 7-group must include col 11.

So the 7-group from s to s+6, with s+6 >=11, s<=11.

s>=1, but col 1 white, so s>=2.

s+6 >=11 => s>=5.

s<=11, but s+6<=20, s<=14, but also after the 7-group, there is a gap, then the 2-group.

So if 7-group ends at e, then white at e+1, then 2-group at e+2 to e+3.

e+3 <=20, so e<=17.

s+6 = e, so s<=11.

Also s>=5.

So s from 5 to 11.

If s=5, 7-group 5-11, then white 12, 2-group 13-14.

If s=6, 6-12, white 13, 2-group 14-15.

...
s=11, 11-17, white 18, 2-group 19-20 — but col 20 white, so 2-group at 19-20 impossible. So s<=10.

s=10, 10-16, white 17, 2-group 18-19 — col 19 is not white, so possible.

s=5 to 10.

Col 11 is in 5-11, 6-12, ..., 10-16, all include 11, good.

Now, similarly for other rows.

But let's look at Row 9: "12 1" — with col 11 black, and we want to place the 12-group.

Suppose the 12-group includes col 11.

s to s+11, s+11 >=11, s>=0, s>=2.

s+11 <=20, s<=9.

After 12-group, white at s+12, black at s+13.

s+13 <=20, s<=7.

So s from 2 to 7.

If s=2, 2-13, white 14, black 15.

s=3, 3-14, white 15, black 16.

s=4, 4-15, white 16, black 17.

s=5, 5-16, white 17, black 18.

s=6, 6-17, white 18, black 19.

s=7, 7-18, white 19, black 20 — but col 20 white, impossible. So s<=6.

So s=2 to 6.

Col 11 is in all these ranges.

Now, to narrow down, perhaps look at the column clues for the sides.

For example, col 2: "1" — only one black in the column.

Which row can have a black in col 2?

Look at row clues that have a group that could include col 2.

For example, Row 1: "4 1" — could have 4-group starting at 2, so col 2,3,4,5 black.

Then col 2 would be black for Row 1.

But col 2 has only one black in the entire column, so if Row 1 has black in col 2, then no other row can have black in col 2.

Similarly, if Row 2 has black in col 2, etc.

Row 2: "2 4 2" — could have 2-group at 2-3, so col 2 black.

Row 3: "2 7 2" — 2-group at 2-3, col 2 black.

Row 4: "3 6 1" — 3-group at 2-4, col 2 black.

And so on.

So many rows could potentially have black in col 2, but only one can, because col 2 has only one black.

So we need to find which row has the black in col 2.

Similarly for col 3: "1" — only one black.

Col 4: "2" — two consecutive, so one pair of rows have black in col 4 and 5? No, col 4 has "2", so two consecutive rows have black in col 4? No.

Clarify: for a column, the clue tells you the groups of black cells in that column, vertically.

So for col 4: "2" means there is one group of 2 consecutive black cells in that column, so two adjacent rows have black in col 4.

Similarly, for col 2: "1" means exactly one row has black in col 2, and it's isolated (but since it's size 1, no issue).

So for col 2, only one cell is black in the entire column.

For col 3, only one cell is black.

For col 4, two consecutive cells are black.

Etc.

Now, let's consider the top rows.

Row 1: "4 1" — likely the head, so perhaps the 4-group is on the left or right.

Suppose the 4-group is on the left: cols 2-5 black, then white, then 1 at say col 7.

Then col 2,3,4,5 are black for Row 1.

But col 2 has only one black in the column, so if Row 1 has black in col 2, then no other row can have black in col 2.

Similarly, col 3 has only one black, so if Row 1 has black in col 3, then no other row can.

But in this case, for Row 1, if 4-group is 2-5, then col 2,3,4,5 are all black for Row 1.

Then for col 2, since only one black allowed, and it's in Row 1, so for all other rows, col 2 is white.

Similarly for col 3, if Row 1 has black in col 3, then no other row can have black in col 3.

But col 3 has clue "1", so only one black, so if Row 1 has it, good.

Col 4 has clue "2", so two consecutive blacks in col 4.

If Row 1 has black in col 4, then since col 4 needs two consecutive, another row must have black in col 4, and it must be adjacent to Row 1, so Row 2 must have black in col 4.

Similarly, col 5 has clue "3", so three consecutive blacks in col 5.

If Row 1 has black in col 5, then Rows 2 and 3 must also have black in col 5 to make three consecutive.

So let's assume that for Row 1, the 4-group is at cols 2-5.

Then:

- Col 2: black in Row 1, and since only one black allowed, all other rows have white in col 2.

- Col 3: black in Row 1, and only one black allowed, so all other rows have white in col 3.

- Col 4: black in Row 1, and needs two consecutive, so Row 2 must also have black in col 4.

- Col 5: black in Row 1, and needs three consecutive, so Rows 2 and 3 must also have black in col 5.

Now, check Row 2: "2 4 2"

If col 4 and 5 are black for Row 2, and col 2 and 3 are white (because col 2 and 3 have only one black each, already used in Row 1), so for Row 2, col 2 and 3 are white.

So the first group "2" cannot be at col 2-3, since they are white.

Could be at col 4-5? But col 4 and 5 are black, and "2" could be there, but then the next group is "4", which would start at col 6 or later.

But col 4-5 are two blacks, so if the first group is 2 at 4-5, then after that, gap, then 4-group.

But col 6 may be available.

Row 2 clue: "2 4 2" — so after the first 2-group, gap, then 4-group, then gap, then 2-group.

If first 2-group is at 4-5, then gap at 6, then 4-group at 7-10, then gap at 11, then 2-group at 12-13.

But we have col 11 is black for rows 3-18, so for Row 2, col 11 is white (since row 2<3), so if 2-group at 12-13, col 11 is white, good.

But is col 6 white? We don't know yet.

Also, for col 5, we need three consecutive blacks: Rows 1,2,3.

So Row 3 must have black in col 5.

Row 3: "2 7 2"

Col 2 and 3 are white (because col 2 and 3 have only one black each, in Row 1), so for Row 3, col 2 and 3 are white.

So the first "2" group cannot be at 2-3.

Could be at 4-5? But col 4 and 5: col 4 has clue "2", which requires two consecutive blacks in the column.

We have Row 1 and Row 2 have black in col 4, so that's two, so for col 4, the two consecutive are Rows 1-2, so Row 3 must have white in col 4.

Similarly, col 5 has clue "3", so three consecutive: Rows 1,2,3 must have black in col 5.

So for Row 3, col 5 is black.

Col 4 is white (as above).

Col 2 and 3 are white.

So for Row 3, col 2,3,4 are white, col 5 is black.

The first group "2" must be somewhere. Could be at col 5-6? But col 5 is black, so if 2-group at 5-6, then col 6 must be black.

Then after that, gap, then 7-group, then gap, then 2-group.

But col 5 is black, and if 2-group at 5-6, then it's size 2, good.

Then gap at 7, then 7-group at 8-14, then gap at 15, then 2-group at 16-17.

Now, check if this works with col clues.

Col 6: clue "2" — two consecutive blacks in the column.

If Row 3 has black in col 6 (as part of the 2-group), then since col 6 needs two consecutive, another row must have black in col 6, adjacent to Row 3.

So either Row 2 or Row 4.

Row 2: if we have for Row 2, first 2-group at 4-5, then gap at 6, so col 6 is white for Row 2.

So not Row 2.

Row 4: may have black in col 6.

So possible.

Also, for col 5, we have Rows 1,2,3 black, good for "3".

Col 4: Rows 1,2 black, good for "2".

Col 2: only Row 1 black, good for "1".

Col 3: only Row 1 black, good for "1".

Now, for Row 2: we have col 4 and 5 black, and we said first 2-group at 4-5.

Then gap at 6, so col 6 white for Row 2.

Then 4-group at 7-10, so col 7,8,9,10 black for Row 2.

Then gap at 11, so col 11 white for Row 2 (which is good, since col 11 starts at row 3).

Then 2-group at 12-13, so col 12,13 black for Row 2.

Now, check col 7: clue "7" — seven consecutive blacks in the column.

If Row 2 has black in col 7, then it could be part of the 7-group.

Similarly, col 8: "6", col 9: "5", etc.

But let's see Row 3: we have 2-group at 5-6, so col 5,6 black.

Then gap at 7, so col 7 white for Row 3.

Then 7-group at 8-14, so col 8 to 14 black for Row 3.

Then gap at 15, so col 15 white.

Then 2-group at 16-17, so col 16,17 black.

Now, col 7: for Row 2, black; Row 3, white; so not consecutive yet.

Col 7 clue "7" — needs seven consecutive blacks.

So if Row 2 has black, and Row 3 has white, then the 7-group cannot include both, so perhaps starts at Row 4 or later.

But let's continue.

For Row 4: "3 6 1"

Col 2,3,4 are white (col 2,3 only one black in Row 1, col 4 has two blacks in Rows 1-2, so Row 4 must have white in col 4).

Col 5: for col 5, we have three consecutive in Rows 1-2-3, so Row 4 must have white in col 5 (since "3" is done).

Col 6: clue "2" — two consecutive blacks. We have Row 3 has black in col 6 (as part of 2-group at 5-6), so if col 6 needs two consecutive, then Row 4 must have black in col 6, and it must be adjacent, so Row 4 col 6 black.

Similarly, col 7: clue "7", not yet filled.

So for Row 4, col 6 is black.

Col 2,3,4,5 are white.

So the first group "3" could be at col 6-8? But col 6 is black, so if 3-group at 6-8, then col 7,8 also black.

Then after that, gap, then 6-group, then gap, then 1-group.

But let's see.

Perhaps the 3-group is at 6-8.

Then col 6,7,8 black for Row 4.

Then gap at 9, so col 9 white.

Then 6-group at 10-15, so col 10 to 15 black.

Then gap at 16, so col 16 white.

Then 1-group at 17, so col 17 black.

Now, check col 6: we have Row 3 and Row 4 have black in col 6, so that's two consecutive, good for "2".

Col 7: Row 4 has black, Row 3 has white (from earlier), Row 2 has black (from Row 2's 4-group at 7-10), so Row 2 and Row 4 have black, but Row 3 has white, so not consecutive. For col 7 to have seven consecutive, it needs a block of seven rows with black in col 7.

Currently, Row 2: black, Row 3: white, Row 4: black, so interrupted.

So perhaps our assumption is wrong.

Maybe for Row 1, the 4-group is not at 2-5.

Perhaps it's at the right side.

For example, Row 1: "4 1" — suppose the 4-group is at cols 15-18, then white at 19, but col 20 white, so 1-group cannot be at 20, so perhaps 1-group at 14 or something.

Let's try a different strategy.

Let's look at the answer or think about the lion shape.

Perhaps the lion has a mane at the top, so rows 1-6 have wider patterns, then body, then legs.

Row 9 is "12 1", which is very wide, so likely the chest or something.

Another idea: use the fact that for a row with "A B", the possible positions can be limited by the column clues.

Perhaps start with the column that has "0" — we did, col 1 and 20 are all white.

Then col 2: "1" — only one black.

Which row is most likely to have the black in col 2? Probably a row that has a group on the left.

For example, Row 3: "2 7 2" — the first "2" could be at 2-3.

Similarly, Row 4: "3 6 1" — first "3" at 2-4.

But if we put the black in col 2 for Row 3, then for col 2, only Row 3 has black, so other rows have white in col 2.

Then for Row 3, if first 2-group at 2-3, then col 2,3 black.

Then for col 3, if Row 3 has black, and col 3 has clue "1", so only one black, so no other row can have black in col 3.

Then for Row 4, if it wants to have 3-group at 2-4, but col 2 and 3 are white for Row 4 (since only Row 3 has black in col 2 and 3), so cannot.

So perhaps for Row 4, the 3-group is not at 2-4.

This is taking too long. Perhaps I can search for a standard solution or think logically.

Let's consider Row 9: "12 1" — and it's likely that the 12-group is from col 4 to 15, for example, then white at 16, black at 17.

Then col 17 is black for Row 9.

Col 17 has clue "2" — two consecutive blacks in the column.

So if Row 9 has black in col 17, then either Row 8 or Row 10 must also have black in col 17.

Row 8: "8 1" — could have the 1 at col 17.

Row 10: "11 1" — could have the 1 at col 17.

So possible.

Similarly, for col 11, we have 16 blacks from row 3 to 18.

For Row 9, col 11 is black, and if 12-group is 4-15, then col 11 is included, good.

Now, let's assume that for Row 9, 12-group is from col 4 to 15.

Then white at 16, black at 17.

So col 16 white, col 17 black for Row 9.

Now, for col 16: clue "2" — two consecutive blacks.

If Row 9 has white in col 16, then the two consecutive must be in other rows.

Similarly, col 17: "2" — two consecutive, and Row 9 has black, so need one more adjacent row.

Say Row 8 or 10.

Now, for Row 8: "8 1" — if the 1 is at col 17, then the 8-group is elsewhere.

Similarly for Row 10.

Perhaps the 1 for Row 8 is at col 17, so col 17 black for Row 8.

Then for col 17, Rows 8 and 9 have black, so that's two consecutive, good for "2".

Then for Row 8, if 1-group at col 17, then the 8-group is before, with gap.

So 8-group from s to s+7, then white at s+8, then black at s+9 =17, so s+9=17, s=8.

So 8-group from 8 to 15, white at 16, black at 17.

Perfect.

So for Row 8: 8-group cols 8-15, white 16, black 17.

Similarly, for Row 10: "11 1" — if the 1 is at col 17, then 11-group from s to s+10, white at s+11, black at s+12=17, so s+12=17, s=5.

So 11-group from 5 to 15, white at 16, black at 17.

But col 16 is white for both, good.

Now, for Row 9: we have 12-group from 4 to 15, white 16, black 17.

So all three rows have black in col 17, but for col 17, clue is "2", so only two consecutive blacks allowed, but here Rows 8,9,10 all have black in col 17, which is three, not allowed.

Problem.

So cannot have all three with black in col 17.

Perhaps for Row 10, the 1 is not at 17.

Or for Row 8, not at 17.

Perhaps for Row 9, the 1 is at a different position.

Earlier for Row 9, if 12-group from 4 to 15, then 1 at 17, but perhaps 1 at 18 or 19.

If 1 at 18, then white at 16, black at 18, so col 17 white.

Then for col 17, clue "2", so two consecutive elsewhere.

For Row 8: "8 1" — if 1 at 18, then 8-group from s to s+7, white at s+8, black at s+9=18, so s=9, 8-group 9-16, white 17, black 18.

Then col 17 white for Row 8.

For Row 10: "11 1" — if 1 at 18, then 11-group from s to s+10, white at s+11, black at s+12=18, s=6, 11-group 6-16, white 17, black 18.

Then for Row 9: 12-group 4-15, white 16, black 18 — but then col 17 is white for all, good.

But for col 18: clue "1" — only one black in the column.

Here, Rows 8,9,10 all have black in col 18, which is three, but clue is "1", so only one allowed. Impossible.

So cannot have the 1 at 18 for all.

Perhaps for some rows, the 1 is at different positions.

For example, for Row 9, 1 at 17, for Row 8, 1 at 16, etc.

Let's try for Row 9: 12-group from 5 to 16, then white at 17, black at 18.

Then col 17 white, col 18 black for Row 9.

For col 18: "1" — only one black, so if Row 9 has it, then no other row can have black in col 18.

For Row 8: "8 1" — if 1 at 18, but then conflict, so perhaps 1 at other position.

Suppose for Row 8, 1 at 17, then 8-group from 8 to 15, white 16, black 17.

Then col 17 black for Row 8.

For Row 9: if 12-group from 5 to 16, then white at 17, black at 18, so col 17 white for Row 9.

Good.

For Row 10: "11 1" — suppose 1 at 17, then 11-group from 5 to 15, white 16, black 17.

Then col 17 black for Row 10.

But for col 17, clue "2" — two consecutive blacks.

Here, Row 8 and Row 10 have black in col 17, but Row 9 has white, so not consecutive. So not good for "2" which requires consecutive.

So must be adjacent rows.

So perhaps Row 8 and Row 9 have black in col 17, or Row 9 and Row 10.

Suppose Row 9 and Row 10 have black in col 17.

For Row 9: if 1 at 17, then 12-group from s to s+11, white at s+12, black at s+13=17, so s=4, 12-group 4-15, white 16, black 17.

For Row 10: if 1 at 17, then 11-group from s to s+10, white at s+11, black at s+12=17, s=5, 11-group 5-15, white 16, black 17.

Then for col 17, Rows 9 and 10 have black, so two consecutive, good for "2".

For Row 8: "8 1" — if 1 at 16, then 8-group from 8 to 15, white 16, black 16? No, white at s+8, black at s+9.

If 1 at 16, then s+9=16, s=7, 8-group 7-14, white 15, black 16.

Then col 16 black for Row 8.

For Row 9: white at 16, so col 16 white for Row 9.

For Row 10: white at 16, so col 16 white for Row 10.

So for col 16: clue "2" — two consecutive blacks.

Here, only Row 8 has black in col 16, so need another row.

Row 7 or Row 9, but Row 9 has white, so perhaps Row 7 has black in col 16.

Row 7: "11 1" — if 1 at 16, then 11-group from s to s+10, white at s+11, black at s+12=16, s=4, 11-group 4-14, white 15, black 16.

Then col 16 black for Row 7.

Then for col 16, Rows 7 and 8 have black, so two consecutive, good for "2".

Perfect.

So let's summarize what we have so far:

- Col 1: all white
- Col 20: all white
- Col 11: rows 3 to 18 black (assumed, but let's verify later)
- For Row 7: 11-group cols 4-14, white 15, black 16
- For Row 8: 8-group cols 7-14, white 15, black 16? Earlier I said for Row 8: 8-group 7-14, white 15, black 16 — but then col 16 black, and for Row 7 also col 16 black, good.

But for Row 8, if 8-group 7-14, then col 7 to 14 black, white 15, black 16.

For Row 7: 11-group 4-14, so col 4 to 14 black, white 15, black 16.

So both have col 4 to 14 black for their groups, but for Row 7, 11-group 4-14, for Row 8, 8-group 7-14, so col 4-6 are black for Row 7 but not for Row 8? For Row 8, col 4-6 are not in the 8-group, and since the 8-group starts at 7, col 4-6 are white for Row 8.

But for col 4: clue "2" — two consecutive blacks in the column.

If Row 7 has black in col 4, and Row 8 has white, then need another row.

Row 6 or Row 9.

Row 6: "3 7 1" — may have black in col 4.

Row 9: "12 1" — we have for Row 9: 12-group 4-15, so col 4 black.

So if Row 7,9 have black in col 4, but not consecutive, since Row 8 has white.

So not good for "2" which requires consecutive.

So problem.

Perhaps for Row 8, the 8-group is not at 7-14.

Earlier for Row 8, if 1 at 16, then 8-group from s to s+7, white at s+8, black at s+9=16, so s=7, 8-group 7-14.

Must be.

Unless the 1 is not at 16.

For Row 8, if 1 at 17, then 8-group from 8 to 15, white 16, black 17.

Then for col 17, if Row 9 and 10 have black, as before.

For Row 9: 12-group 4-15, white 16, black 17.

For Row 10: 11-group 5-15, white 16, black 17.

Then col 17 black for Rows 9,10, and if Row 8 has black at 17, then three, but col 17 clue "2", so only two allowed, so cannot have Row 8 also.

So for Row 8, if 1 at 17, then col 17 black, but then with Rows 9,10, it's three, not good.

So perhaps for Row 10, the 1 is not at 17.

Let's set for Row 9: 12-group 4-15, white 16, black 17.

For Row 10: "11 1" — suppose the 1 is at 18, then 11-group from 6 to 16, white 17, black 18.

Then col 17 white for Row 10, col 18 black.

For col 18: "1" — only one black, so if Row 10 has it, good.

For Row 8: "8 1" — suppose 1 at 16, then 8-group 8-15, white 16, black 16? No, white at s+8, black at s+9.

If 1 at 16, s+9=16, s=7, 8-group 7-14, white 15, black 16.

Then col 16 black for Row 8.

For Row 9: white at 16, so col 16 white.

For Row 10: white at 17, so col 16 may be black or white; in this case, 11-group 6-16, so col 16 black for Row 10.

So for col 16: Row 8 and Row 10 have black, but Row 9 has white, so not consecutive. For "2" in col 16, need two consecutive, so not good.

If for Row 10, the 11-group is 5-15, then col 16 white, but then 1 at 18, so white at 16, black at 18, so col 17 white.

Then for col 16, only Row 8 has black, need another.

Row 7 or 9.

Row 7: if 1 at 16, then 11-group 4-14, white 15, black 16, so col 16 black.

Then Rows 7 and 8 have black in col 16, good for "2".

For Row 9: white at 16, so col 16 white.

For Row 10: if 11-group 5-15, then col 16 white.

So col 16: Rows 7 and 8 black, good.

Col 17: for Row 9: black (since 1 at 17), for Row 8: if 1 at 16, then col 17 may be white or black; in 8-group 7-14, so col 17 not in group, and after white at 15, black at 16, so col 17 is after, so white for Row 8.

For Row 10: if 11-group 5-15, white 16, black 18, so col 17 white.

So col 17: only Row 9 has black, but clue is "2", so need two consecutive, so not good.

So still problem.

Perhaps for Row 9, the 1 is at 18.

Let's try:

Row 9: 12-group 4-15, white 16, black 18. So col 17 white, col 18 black.

Then for col 18: "1" — only one black, so if Row 9 has it, good.

For Row 8: "8 1" — suppose 1 at 17, then 8-group 8-15, white 16, black 17.

Then col 17 black for Row 8.

For Row 10: "11 1" — suppose 1 at 17, then 11-group 5-15, white 16, black 17.

Then col 17 black for Row 10.

But then for col 17, Rows 8 and 10 have black, but Row 9 has white, so not consecutive. For "2" in col 17, need consecutive, so not good.

If for Row 10, 1 at 16, then 11-group 5-15, white 16, black 16? No, white at s+11, black at s+12.

If 1 at 16, s+12=16, s=4, 11-group 4-14, white 15, black 16.

Then col 16 black for Row 10.

For Row 8: if 1 at 17, col 16 may be white; in 8-group 8-15, so col 16 not in group, and after white at 16? White at s+8=8+8=16, so col 16 white for Row 8.

For Row 9: white at 16, so col 16 white.

So col 16: only Row 10 has black, need another for "2".

Row 7 or 9, but Row 9 white, so Row 7.

Row 7: "11 1" — if 1 at 16, then 11-group 4-14, white 15, black 16, so col 16 black.

Then Rows 7 and 10 have black in col 16, but not consecutive, since Row 8 and 9 have white.

So not good for "2".

This is frustrating.

Perhaps the 16 in col 11 is not from 3 to 18.

Let's try from 4 to 19.

Then for Row 4 to 19, col 11 black.

Then for Row 3, col 11 white.

Row 3: "2 7 2" — if col 11 white, then the 7-group may not include it.

But let's give up and look for a different approach or accept that we need to solve it systematically.

Perhaps the answer is to fill the grid as per the lion shape, and since it's "Hard", it might be symmetric or have a specific pattern.

Another idea: the sum of the row clues should equal the sum of the column clues, as both represent the total number of black cells.

Let's calculate the sum of row clues.

Row 1: 4+1=5
Row 2: 2+4+2=8
Row 3: 2+7+2=11
Row 4: 3+6+1=10
Row 5: 3+9+1=13
Row 6: 3+7+1=11
Row 7: 11+1=12
Row 8: 8+1=9
Row 9: 12+1=13
Row 10: 11+1=12
Row 11: 4+1=5
Row 12: 7+2=9
Row 13: 9+1=10
Row 14: 11+1=12
Row 15: 9+3=12
Row 16: 4+5=9
Row 17: 2+6=8
Row 18: 3+2+3=8
Row 19: 3+3=6
Row 20: 4+3=7

Sum: let's add:

5+8=13
13+11=24
24+10=34
34+13=47
47+11=58
58+12=70
70+9=79
79+13=92
92+12=104
104+5=109
109+9=118
118+10=128
128+12=140
140+12=152
152+9=161
161+8=169
169+8=177
177+6=183
183+7=190

So 190 black cells.

Now sum of column clues:

Col 1: 0
Col 2: 1
Col 3: 1
Col 4: 2
Col 5: 3
Col 6: 2
Col 7: 7
Col 8: 6
Col 9: 5
Col 10: 11
Col 11: 16
Col 12: 1
Col 13: 2
Col 14: 9
Col 15: 4
Col 16: 2
Col 17: 2
Col 18: 1
Col 19: 2
Col 20: 0

Sum: 0+1+1+2+3+2+7+6+5+11+16+1+2+9+4+2+2+1+2+0 = let's calculate:

1+1=2
2+2=4
4+3=7
7+2=9
9+7=16
16+6=22
22+5=27
27+11=38
38+16=54
54+1=55
55+2=57
57+9=66
66+4=70
70+2=72
72+2=74
74+1=75
75+2=77
77+0=77

77? But earlier row sum is 190, inconsistency.

I think I mistake.

For columns, the clue is for the column, so for col 7: "7" means 7 black cells in that column, not 7 consecutive in the sense of the number, but the number is the size of the group, but if it's a single number, it means one group of that size, so 7 consecutive, but the number of black cells is 7.

In Picross, the number indicates the size of the group, so for a column with "7", it means there is one group of 7 consecutive black cells, so 7 black cells in that column.

Similarly, for "16", 16 black cells.

So sum of column clues should equal sum of row clues, both should be the total number of black cells.

But I got 190 for rows, 77 for columns, which is impossible.

I must have misread the column clues.

Let's look back at the image description.

In the user's message, the column clues are given as:

From the text: "0 1 1 2 3 2 7 6 5 11 16 1 2 9 4 2 2 1 2 0" for the columns, but that's 20 numbers, good.

But sum is 0+1+1+2+3+2+7+6+5+11+16+1+2+9+4+2+2+1+2+0 = let's calculate again:

List: 0,1,1,2,3,2,7,6,5,11,16,1,2,9,4,2,2,1,2,0

Add step by step:

Start with 0

+1 =1

+1=2

+2=4

+3=7

+2=9

+7=16

+6=22

+5=27

+11=38

+16=54

+1=55

+2=57

+9=66

+4=70

+2=72

+2=74

+1=75

+2=77

+0=77

Yes, 77.

But row sum is 190, which is much larger, so error in row sum or understanding.

For rows, the clue is for the row, so for Row 1: "4 1" means two groups: one of 4, one of 1, so 5 black cells.

Similarly, all seem correct.

Perhaps the grid is not 20x20? Let's count the number of columns from the clues.

In the text, for columns, there are 20 clues: from "0" to "0", 20 items.

For rows, 20 clues.

But 190 vs 77, impossible.

Unless I miscalculated row sum.

Let me recalculate row sum.

Row 1: 4+1=5

Row 2: 2+4+2=8

Row 3: 2+7+2=11

Row 4: 3+6+1=10

Row 5: 3+9+1=13

Row 6: 3+7+1=11

Row 7: 11+1=12

Row 8: 8+1=9

Row 9: 12+1=13

Row 10: 11+1=12

Row 11: 4+1=5

Row 12: 7+2=9

Row 13: 9+1=10

Row 14: 11+1=12

Row 15: 9+3=12

Row 16: 4+5=9

Row 17: 2+6=8

Row 18: 3+2+3=8

Row 19: 3+3=6

Row 20: 4+3=7

Now sum:

Let's group:

Rows 1-5: 5+8+11+10+13 = 5+8=13, +11=24, +10=34, +13=47

Rows 6-10: 11+12+9+13+12 = 11+12=23, +9=32, +13=45, +12=57

Rows 11-15: 5+9+10+12+12 = 5+9=14, +10=24, +12=36, +12=48

Rows 16-20: 9+8+8+6+7 = 9+8=17, +8=25, +6=31, +7=38

Now total: 47 + 57 = 104

104 + 48 = 152

152 + 38 = 190

Yes.

But column sum 77, so contradiction.

Unless the column clues are not the number of black cells, but in Picross, the number is the size of the group, so for a column with "7", it means 7 black cells in that column (since one group of 7).

Similarly for "16", 16 black cells.

So sum should be equal.

Perhaps for columns with multiple numbers, but in this case, all column clues are single numbers except possibly, let's see the text: "0 1 1 2 3 2 7 6 5 11 16 1 2 9 4 2 2 1 2 0" — all single numbers, so each is the size of the single group in that column, so number of black cells in the column is that number.

So sum should be 77, but row sum 190, impossible.

Unless I misread the row clues.

In the user's message, for rows, it's given as:

"4 1"
"2 4 2"
"2 7 2"
"3 6 1"
"3 9 1"
"3 7 1"
"11 1"
"8 1"
"12 1"
"11 1"
"4 1"
"7 2"
"9 1"
"11 1"
"9 3"
"4 5"
"2 6"
"3 2 3"
"3 3"
"4 3"

All seem correct.

Perhaps the "0" for col 1 and 20 means no black, good.

But 190 > 77, so error.

Unless the grid is not 20x20, but let's count the number of cells.

From the column clues, 20 columns.

From row clues, 20 rows.

Perhaps for the rows, the clue "11 1" means 11 and 1, so 12 black cells, but in the column, "16" means 16, etc.

But 190 vs 77, ratio about 2.47, not integer.

Perhaps I have a mistake in the column sum.

Let's list the column clues again from the text:

"0 1 1 2 3 2 7 6 5 11 16 1 2 9 4 2 2 1 2 0"

Positions: 1:0, 2:1, 3:1, 4:2, 5:3, 6:2, 7:7, 8:6, 9:5, 10:11, 11:16, 12:1, 13:2, 14:9, 15:4, 16:2, 17:2, 18:1, 19:2, 20:0

Sum: let's use calculator in mind:

0+1=1

1+1=2

2+2=4

4+3=7

7+2=9

9+7=16

16+6=22

22+5=27

27+11=38

38+16=54

54+1=55

55+2=57

57+9=66

66+4=70

70+2=72

72+2=74

74+1=75

75+2=77

77+0=77

Yes.

Perhaps for the rows, the clue is not the sum, but in Picross, it is the sum of the group sizes.

Unless the "0" for col 1 and 20 is correct, but perhaps the grid has less than 20 columns, but the clues suggest 20.

Another possibility: in the column clues, "18 17 10" etc, but in the text, it's "0 1 1 2 3 2 7 6 5 11 16 1 2 9 4 2 2 1 2 0" for the bottom row of column clues, but in the image, there might be multiple lines.

Looking back at the user's message:

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