A stem cell holds N candidates. For each of its candidates there is one almost locked set (a petal) that contains that digit, with every copy of it in the petal seeing the stem's copy. Whatever the stem turns out to be, one petal loses that digit and becomes locked. Any digit Z that is in every petal, with no copy in the stem, can be removed from every cell that sees all the petals' copies of Z.
Start with two petals
The smallest death blossom has a stem with two candidates and two petals. Here, row seven, column five holds 2 and 9 and serves as the stem.
The petal for the 2 is row seven, column two, which holds 1 and 2 and sees the stem along row seven. The petal for the 9 is row eight, columns four and five, two cells holding 1, 4 and 9 between them. Its only 9 is in column five, directly below the stem, so it sees the stem.
If the stem is the 2, row seven, column two loses its 2 and must be the 1. If it is the 9, the petal in row eight loses its 9 and its two cells lock on 1 and 4. Only column four holds a 1, so that cell takes it. Either way, a 1 sits in row seven, column two or row eight, column four.
Both petals hold a 1 and the stem has none, so the 1 is Z. Row seven, column six and row eight, column one each see both of those cells, so both lose their 1.
Read it petal, stem, petal and you have an ALS chain of three sets, with the stem as the one-cell set in the middle. Every death blossom with a two-candidate stem works out this way. Here the game's ALS chain check removes the same two 1s, and so does its XY-chain check, because all four cells hold two candidates.
The blossom becomes a move of its own when the stem holds three candidates. A cell with three candidates is not an almost locked set, so an ALS chain can't use it as one of its sets.
Where to look
Start with a cell that holds three candidates. For each of them, look for a petal: a bivalue cell or a small almost locked set where every copy of that digit sees the stem. The petals must not share a cell. Then look for a digit every petal holds and the stem lacks, and a cell outside the pattern that sees every copy of it.
Pick the stem
Row nine, column six holds 1, 2 and 6. It needs three petals, one for each of its candidates.
A petal for the 1
Row eight, columns four and five hold 1, 4 and 9 between them. Two cells with three digits make an almost locked set. Its only 1 is in column four, and that cell shares box eight with the stem, so it sees the stem.
A petal for the 2
Row nine, column three holds 2 and 9, a one-cell almost locked set. Its 2 sees the stem along row nine.
A petal for the 6
Column one has the third petal. Row six, column one holds 1 and 9, and row nine, column one holds 1, 6 and 9: two cells, three digits. Only the lower cell holds a 6, and it sees the stem along row nine. The upper cell has no 6, so it does not need to see the stem.
Find the digit every petal holds
The petals hold 1, 4 and 9, then 2 and 9, then 1, 6 and 9. Only the 9 is in all three, and the stem has no 9, so the 9 is Z. Its copies sit in row eight, column five, in row nine, column three, and in column one at rows six and nine.
Strike the 9
Row eight, column one sees all four 9s: one along row eight, one inside box seven and two down column one. Its 9 goes, and the cell keeps 1 and 6.
Why it works
The stem must take one of its three digits. Each board below follows one choice.
In all three cases a petal locks and places a 9 in a cell that row eight, column one sees. Whatever the stem turns out to be, this cell can't be the 9.
In general, whichever digit the stem takes, that digit's petal loses it and locks, and a locked petal must place every digit it still holds, Z included. So a cell that sees every copy of Z loses it.
The two-petal blossom earlier took this cell's 1. With the 9 gone as well, row eight, column one is the 6.
Rows seven and eight of column nine hold 1, 6 and 9 between them, an almost locked set with a 6. But neither cell sees the stem. A 6 in the stem would not take the 6 out of this set, so it can't be the petal.
Its only 6 sees the stem, so a 6 in the stem removes it.
Bigger versions
A stem can hold four candidates, and then it needs four petals. Petals can be larger sets too. The rule does not change: Z must be in every petal and not in the stem, and the cell you clear must see every copy of Z.
The in-game hint shows stems with two to four candidates and petals of one to four cells, each petal inside one row, column or box. With two candidates, the blossom is the ALS chain from the top of this page.
How it relates to other moves
A Y-wing is a death blossom with one-cell petals. The pivot is a two-candidate stem and each wing is a petal. On the Y-wing article's grid, the game's death blossom check takes the pivot and both wings and removes the same 5.
As shown above, a two-candidate stem makes an ALS chain, and a three-candidate stem can't be a set in one. On this grid, the game's ALS chain, restricted common candidate and Sue de Coq checks all leave the 9 in row eight, column one alone.
Common slips
Taking a set whose copies of the stem digit don't all see the stem. Column nine's set holds two 6s, and neither sees row nine, column six.
Counting a stem digit as Z. The 1 is in the first and third petals, but the stem holds a 1 too, so it can't be Z. Z must be missing from the stem.
Clearing a cell that sees only some copies of Z. Row nine, column nine sees the 9s in row nine but not the other two, so this blossom leaves its 9 alone.
Letting two petals share a cell, or putting the stem inside a petal.
What the in-game hint shows
The in-game hint may show a different or simpler step first. On this grid it opens with a W-wing. Row four, column six and row seven, column two both hold 1 and 2. Row nine's only 2s sit in columns three and six, and each sees one of those cells. So one of them is the 1, and row seven, column six loses its 1. That cell is outside the three-petal blossom, so the pattern stands.
Simple coloring and an X-wing follow, then a coloring step that strikes the stem's 2. Singles come next, and the 9 in row eight, column one goes at the nineteenth hint, when row eight, column nine becomes the 9.
On the starting grid, no step the hint checks before Nishio removes this 9. Nishio does. If row eight, column one is the 9, row six, column one must be the 1 and row nine, column one the 6. Row eight, column four is then the only cell in column four with a 1. That leaves row nine, columns three and six both needing the 2. The game checks Nishio before this move, so the hint seldom shows it. A different route can still be right.
When a cell with three candidates has, for each one, a small almost locked set whose copies of that digit all see the cell, look for a digit every set holds and the cell lacks. A cell that sees every copy of that digit in the sets can lose it.
Where this sits on the path
This is the seventh technique of the last chapter, after Sue de Coq, and the last one on the learning path.
On a different grid, row one, column three holds 3 and 4 and is the stem. Row one, column seven holds 2 and 3 and is the petal for the 3, marked purple. Row seven, columns three and eight hold 2, 4 and 8 between them and form the petal for the 4, marked blue. Which candidate does this blossom remove?
Find the digit both petals hold and the stem lacks, then the cell that sees every copy of it.
الخلاصة
- ابحث عن
- A cell with two to four candidates, and for each candidate a small almost locked set whose copies of that digit all see the cell.
- النطاق
- Cells outside the stem and petals that see every copy of the shared digit.
- النتيجة
- The digit every petal holds and the stem lacks leaves every cell that sees all its copies in the petals.
- العلامة
- A three-candidate cell with bivalue cells and small sets around it that share one extra digit.
- إذا لم تنجح
- If a petal's copy of its digit doesn't see the stem, or no digit is in every petal, try another set or another stem.