Puzzle Linking Pieces
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In this page:divide and conquer,unique partitions,unique intersections,intersections,elimination,summary,A simple puzzle
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Here is a Kakuro puzzle which will turn out to be very simple to solve. Beforewe start on it, let's give names to each square. We give every column aname: the first one is A, the second, B, and so on to the tenth, J. To eachrow we also give a name: the top row is a, the next, b, and so on to thebottom row, which is called j. Each square lies in the intersection of a rowand a column, so we can name it uniquely by giving both the row and columnnames. For example, the bottom right square is called Jj.
In trying to solve a puzzle we will need to have a name for the unknownnumber that goes into a square. So, for example, when I need to talkabout the number that goes into the square Jj I will write [Jj]. Similarlyfor any square, when you see square brackets around the name of a square,it means the number that goes into that square.
A final convention before we move on to solving the puzzle above. Theleft-most clue in row c involves breaking 7 into 3 parts. The left-mostclue in row j involves breaking 7 into 2 parts. We will distinguish theseby a subscript on the clue: 73 for the clue inrow c and 72 for the other. The number is theclue, and the subscript is the number of pieces it should be broken into.
Divide and conquer
Now let us work up a strategy for splitting a puzzle into smaller pieces,because smaller puzzles are easier to solve. Take the squares Bb, Bc, Cband Cc. The cluestell us that the sum of these four numbers taken column-wise must equal thesum of the column clues, ie, 10+13=23. But when the same fournumbers are summed row-wise, then the result has to be the same. The sum inthe row b has to be 17, because it must match the clue. So the remaining twonumbers, [Bc] and [Cc], must add up to 23-17=6. This forces thenumber in [Dc] to be 1. Magic? We have converted partial knowledge about foursquares into full knowledge of one square. The small 2×2 square isnow isolated from the rest of the puzzle, and can be solved by itself. Wehave divided. Next we will conquer. But first we will state the rule.
The rule of divide and conquer
When a block of squares isconnected to the rest of the puzzle through exactly one square, thencall this the linking square. The value in the linking square is foundas follows:
- If the linking square is part of a row clue
- Find the sum of all the column clues (this was 23 in the example above).Next find the sum of all the row clues except the one that involves thelinking square (this was 17 in the example). Subtract this partial sum fromthe sum of the column clues (in the example this was 6). Subtract this numberfrom the remaining clue to get the value in the linking square (in theexample we got 7-6=1).
- If the linking square is part of a column clue
- Interchange column and row in the procedure given above.
Applying the rule
The linking squares can be identified without looking at the clues. Theydepend only on the placement of the red and white squares. We can say thatthe linking squares depend on the shape of the puzzle. We have circled thelinking squares in the figure alongside. Now let's fill in the values usingthe rule of divide and conquer.
Start with the square Fd. This is part of a column clue. The sum of the rowclues behind it is 3+11=14. The sum of the column clues exceptthe one it is part of is 11. The difference is 3. Therefore[Fd]=6-3=3. It is a straightforward matter to work out[If]=9 and [Ih]=4.
It might seem that finding [Eh] is more complicated, but actually it is not.The square Eh is part of a row clue. So we first find the sum of the columnclues behind it, 14+11+4=29. Next the sum of the row cluesexcept the one to which it belongs is 6+13=19. The difference is29-19=10. So [Eh]=19-10=9.
This tool cuts finer
The trick we have just developed actually cuts the puzzle into even smallerpieces. But before we do that, let's polish off the one square whose valuewe can write down without tricks. We see that [Hf]=2.
Now let's divide a bit more so that we can conquer more easily. You caneasily convince yourself that you can figure out the values to be filledinto the two squares which are circled in the picture alongside. Lookat the square Gi. Since we have already filled in [Eh]=9,we know that [Ei]+[Ej]=13-9=4. So the 2×2 block behindGi looks like a closed small puzzle. So Gi is a linking square. Findingthe number to put down there is also strightforward. Since it is part ofa row clue, take the sum of the column clues behind it:15+(13-9)=15+4=19. From this we subtract the row clues whichdo not involve the linking square: 19-7=12. So we get[Gi]=17-12=5.You can go through similar reasoning to convince yourself that Ef is alsoa linking square which should contain the value 9.
Once Ef is filled up, it is clear that [Ff]=2. From there itis another simple step to write down [Fg]=1. With this thestate of the puzzle is as shown above. Now it is time to solve the eightsimpler and smaller puzzles to which we have reduced this.
Only eliminate
Consider the simple sub puzzle at the top left corner, which isextracted here. Now we use a simple fact: if you want to write 17 asthe sum of two numbers, then there is only one pair which is possible:172=9+8 (the subscript two means that you want tobreak up 17 into two numbers). So [Bb] can be one of 9 and 8,as shown in the figure alongside. If it is 9, then [Bc]=1,and if it is 8 then [Bc]=2. Since the row already containsa 1, the first possibility is eliminated, and we have the solution[Ba]=9. The other three squares are child's play.
The sub-puzzle shown alongside is solved in a similar way. Thereis an unique partition 32=1+2, that is to say,there is only one way to write 3 as the sum of two numbers. So [Gb]must be either 1 or 2. If it is 1 then [Gc]=10, and thatis not allowed by the rules of Kakuro. Therefore, we must have[Gb]=2. Once you are able to fill in a single number intoany square in a self-contained 2×2 block of squares, the remainderare completely fixed by the clues. So this piece of the puzzle is alsodone.
With a little thought, you can use the unique partition32=1+2 to complete the 2×2 block at thebottom right hand corner of the puzzle. The block involving Dd, De, Ed, Ee is yours once you know theunique partition 42=1+3.I leave this in your, by now, capablehands.
Unique partitions
The solution of Kakuro puzzles is helped along by several unique partitions.We have used
32=1+2 and 172=9+8.Closely related are two more 42=1+3 and162=9+7. I put down a complete list of unique partitionsof length upto 4 here:| 32=1+2 | 42=1+3 | 172=9+8 | 162=9+7 |
| 63=1+2+3 | 73=1+2+4 | 243=9+8+7 | 233=9+8+6 |
| 104=1+2+3+4 | 114=1+2+3+5 | 304=9+8+7+6 | 294=9+8+7+5 |
Two unique partitions with anunique intersection
Look at the 2×2 sub-puzzle involving the squares Gg, Gh, Hg andHh. The clue in column H is 18 and has a 2 filled in already so theremainder involves the unique partition 162.The clue in row h is 21 and has a 4 filled in already, so the remainderinvolves the unique partition 172. The twoclues intersect at the square Hh, so [Hh] must have a value which iscommon to these two unique partitions. That is the number 9. This issituation that may arise often, so it is a good idea to keep a watchfor unique intersections. It might be useful to keep in mind a fewunique intersections; the most common ones are32&42=1,162&172=9,42&73=1,162&233=9.
Logic rules
We have now solved a large part of the puzzle. Three sub-puzzles remain.We will now solve the two remaining 2×2 pieces first. Start with thepiece containing the squares Ei, Ej, Fi and Fj. Since the column E containsas clue 133, and a 9 is known, the remainder of theclue is 42. Since this is an unique decomposition,one can check each of the two possibilities for [Ei] to find the solution.Instead, I will give you a logic without using unique partitions.
The clue in row j is 72 which can be writtenin three ways: 72 = 1+6 = 2+5 = 3+4. Nowlook at which of these six numbers can go into the square Fj. Sincethe column clue is 152, the only possibilityis [Fj]=6. Nothing else gives a single digit in Fi. Itis useful to give a name to this kind of logic, because that helpsus remember it. I call it 'unique intersection without uniquepartition'.
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Next we take up the 2×2 block containing the squares Id, Ie, Jd and Je.This block contains the clue 62 twice. Since62 = 1+5 = 2+4, there are 4 possibilities for[Jd]. The values 1 and 2 are ruled out because that can't give a singledigit value for [Id]. Also, we know that 9 cannot appear in Id, since itappears again in the solution of the column clue. So we rule out 4 in thesquare Jd, leaving the value [Jd]=5. Such a process of eliminationgets better the more circumstantial evidence you have.
Larger pieces need not be much tougher
One last piece of the puzzle remains. The seven squares in this piece cannotbe decomposed into simpler puzzles. The column clue 42in column D gives two possibilities in the square Dh. The possibility that[Dh]=1 is eliminated because one would then have [Ch]=9, leading to a repeated digit in the solution of theclue 193 in row h. So the solution is[Dh]=3, [Dg]=1 and [Ch]=7. This isa seconf example of elimination using circumstantial evidence.
Now there remains only a 2×2 puzzle block. The clue in column Creduces (becuase of the value already filled in Ch) to 42. Then, we get[Cg]=3, because the value 1 would lead to a repeated digit in thesolution of the row clue. Elimination using circumstantial evidence again.The remaining three squares can now be filled inby a simple process of subtraction. And that's the whole big puzzle completelysolved!
Tips to remember
- Divide and conquer
- Divide the puzzle into smaller pieces, if possible, by finding linking squares. The values in the linking squares can be easily filled in by the rule of divide and conquer. Divide and conquer sometimes can be used several times over. After one set of linking squares have been filled in, other squares may become linking squares.
- Simplify using unique partitions
- Some clues have an unique partition. Every unique partition reduced the number of possibilities you have to check. Some unique partitions are listed earlier in this page.
- Unique intersections of unique partitions
- When clues with unique partitions intersect, then sometimes we have unique intersections. Some unique intersections are listed earlier in this page.
- Unique intersections without unique partitions
- Sometimes you get unique intersections without unique partitions. An example was given above.
- Elimination using circumstantial evidence
- Elimination works better and better as you fill in more and more squares. This is the principle of elimination using circumstantial evidence.