# Puzzle 22: Cross Sums - The Zotmeister

#### solving the puzzle of life one entry at a time

## Dec. 20th, 2005

### 05:40 pm - *Puzzle 22: **Cross Sums*

*Cross Sums*

I couldn't do it. I just couldn't bring myself to rename it. Nothing else I'd come up with was even remotely clever, anyway. I'm guessing Dell had this title trademarked at some point, but they don't now, or at least aren't enforcing it (Kappa publications use the *Cross Sums* name now as well). Nikoli calls it *Kakro*; the multitude of newspapers trying to expand out a bit from *Sudoku* are calling it "Kakuro" (I *REFUSE* to italicize that atrocious re-spelling). To me, however, it is, and always shall be, *Cross Sums*.

You probably already know how to solve these if you're even remotely familiar with English or Japanese puzzle magazines. Note that like Nikoli, I construct my puzzles by hand, so try not to solve these with guesses; you should be able to simply *see* the next move if you look - and think - carefully enough.

I like the way the graphics came out, but was it ever a timesink to create:

The "fireball" on the left is a cute little

The obligatory minimalist-but-rigorous instructions are a single sentence: enter a non-zero single-decimal-digit number into each cell such that each "entry" (orthogonal line of contiguous cells) contains no duplicate numbers and that the sum of the numbers in each entry matches the number in the "flag" (black triangle) incident upon and collinear with that entry.

Simple, really. ...Well, have

1) See all the little black triangles sticking out of the grid? I call those "flags". You may well call them "clues", just like a crossword. Each one has a number in it. Now those flags are positioned at the top of every column and to the left of every row, and also in front of parts of rows and columns separated by gaps in the grid. Each line of cells - across or down - that has its own flag in front of it is called an "entry", and what you have to do is put numbers into the cells of each entry that add up to the number in its flag.

2) Not just any numbers will do. You have to use only whole numbers from 1 to 9. No zeroes, no fractions.

3) Here's the kicker: there can't be any

Consider the top row of the sample puzzle grid: it has two cells, and the flag to their left reads "5". (The "16" refers to the column it's sitting on top of.) This means the two numbers put in those cells must add up to five; 1 + 4 could do it, as could 2 + 3. You also need to figure out what order to put those numbers in, so there would be four permutations to consider: 14, 23, 32, and 41. By looking at an entry that crosses it, it can be deduced which is correct.

For those who are still lost, perhaps you can understand by doing:

(Start of sample puzzle solution)

I built the sample puzzle to be particularly instructive. There are four places in the grid where two-cell entries intersect, and I varied those to show what to look for.

I'll start on the right-center. There's a flag reading "4" in front of a two-cell entry running across - a "4-in-two" - and this intersects a two-cell entry running down with a '3' in the flag above it - a "3-in-two". Three is easy: 1 + 2 is the only way to make it. Four, as noted in the instructions, can't be 2 + 2 here, so it must be 1 + 3. These two combinations only have one number

Now I look at the left-center, where a "14-in-two" crosses a "16-in-two". A 16-in-two must be 9 + 7, since two eights would be a duplication. Now "14-in-two" has two combinations that work, but we don't even need to look at them: if the 16-in-two's '7' were in the 14-in-two, we'd need another '7' to complete it - a repeat. So the '9' goes in the cell where they cross; a '7' goes above it; 14 - 9 = 5, so '5' goes to its right.

The top-center is also defined. There's a 5-in-two and a 13-in-two crossing there. Both have multiple combinations, but forget combinations for this one - the difference is so large that there's only one number they can share. The

Where should we look next? I'll give you a hint: it's NOT the bottom-center. Five and thirteen were

...Found it yet?...

...Congratulations if you saw where the 11-in-four crosses the 29-in-four! If not, don't sweat it. It's a good idea to remember at least this much:

1 + 2 + 3 + 4 = 10

9 + 8 + 7 + 6 = 30

Those are the largest and smallest totals you can get for a four-cell entry, and they're nice round numbers. Let's look at the top one: that makes ten, but we need eleven, so we need to add one to a summand. But we can't have any duplicates, so we need to add one to the largest of them. That's right: 11-in-four can only be 1 + 2 + 3 + 5. Similarly, 29-in-four can only be 9 + 8 + 7 + 5. Sure enough, they only have the '5' in common, so in it goes. From there, the only cell in the 29-in-four not yet filled in gets the '8'; some quick arithmetic yields '4' to finish the 18-in-four.

No, I'm

(End of sample puzzle solution)

The better one gets at these, the less arithmetic one needs to solve them and the more instinctual they become. Learn to see the combinations in your mind and try to position them in the grid without guesswork, or complex formulae, or writing several little numbers in a cell until you figure out which one is right. Those should all be last resorts.

This one certainly has a Nikoli bent to it. It's NOT easy for someone's first puzzle, but technically it's your second if you've tried the sample puzzle. Mind you, it's not easy as a second puzzle, either, but hopefully it isn't insurmountable. If I were to guess at par times for solving, I'd say complete beginners will need at most three hours, experienced solvers at most one hour, someone like myself at most twenty minutes, and a world-class competitor ten minutes on a VERY bad day. But those are guesses. Feel free to email me your solution and tell me how long it took you. There might even be a prize in it for you, and it might not necessarily go to the first submission or fastest solver. Then again, maybe it will. You just can't know, can you? - ZM

The "fireball" on the left is a cute little

*Cross Sums*puzzle; the unique solution to same is on the right, scribed in what my graphics program referred to as simply "Brown".The obligatory minimalist-but-rigorous instructions are a single sentence: enter a non-zero single-decimal-digit number into each cell such that each "entry" (orthogonal line of contiguous cells) contains no duplicate numbers and that the sum of the numbers in each entry matches the number in the "flag" (black triangle) incident upon and collinear with that entry.

Simple, really. ...Well, have

*you*ever tried to provide rigorous instructions for, say, a crossword? No? Didn't think so! If you think of this puzzle as a crossword that has numbers in it, you're on the right track:1) See all the little black triangles sticking out of the grid? I call those "flags". You may well call them "clues", just like a crossword. Each one has a number in it. Now those flags are positioned at the top of every column and to the left of every row, and also in front of parts of rows and columns separated by gaps in the grid. Each line of cells - across or down - that has its own flag in front of it is called an "entry", and what you have to do is put numbers into the cells of each entry that add up to the number in its flag.

2) Not just any numbers will do. You have to use only whole numbers from 1 to 9. No zeroes, no fractions.

3) Here's the kicker: there can't be any

*repeated*numbers in an entry. For example, if there's a two-cell entry whose flag reads "4", you can't put '2's in both cells. You'll have to use a '1' and a '3' for that (that is, once you figure out which order to put them in!).Consider the top row of the sample puzzle grid: it has two cells, and the flag to their left reads "5". (The "16" refers to the column it's sitting on top of.) This means the two numbers put in those cells must add up to five; 1 + 4 could do it, as could 2 + 3. You also need to figure out what order to put those numbers in, so there would be four permutations to consider: 14, 23, 32, and 41. By looking at an entry that crosses it, it can be deduced which is correct.

For those who are still lost, perhaps you can understand by doing:

(Start of sample puzzle solution)

I built the sample puzzle to be particularly instructive. There are four places in the grid where two-cell entries intersect, and I varied those to show what to look for.

I'll start on the right-center. There's a flag reading "4" in front of a two-cell entry running across - a "4-in-two" - and this intersects a two-cell entry running down with a '3' in the flag above it - a "3-in-two". Three is easy: 1 + 2 is the only way to make it. Four, as noted in the instructions, can't be 2 + 2 here, so it must be 1 + 3. These two combinations only have one number

*in common*: the '1'. So the '1' goes in the cell where they intersect; the '3' goes to its left to make four; the '2' goes below it to make three. Not so bad, right?Now I look at the left-center, where a "14-in-two" crosses a "16-in-two". A 16-in-two must be 9 + 7, since two eights would be a duplication. Now "14-in-two" has two combinations that work, but we don't even need to look at them: if the 16-in-two's '7' were in the 14-in-two, we'd need another '7' to complete it - a repeat. So the '9' goes in the cell where they cross; a '7' goes above it; 14 - 9 = 5, so '5' goes to its right.

The top-center is also defined. There's a 5-in-two and a 13-in-two crossing there. Both have multiple combinations, but forget combinations for this one - the difference is so large that there's only one number they can share. The

*largest*number a 5-in-two can hold is a four (which is when paired with a one); the*smallest*number a 13-in-two can hold is a four (which is when paired with a nine). Therefore, four is both the*largest*and*smallest*number their intersection can hold. This of course means it*is*a four. A '1' goes to its left and a '9' goes below it.Where should we look next? I'll give you a hint: it's NOT the bottom-center. Five and thirteen were

*just enough*to make their intersection unique; seven and fourteen are closer to each other. I put this one in as an example of what NOT to waste your time on; in fact, I*end*the puzzle with those. Don't scribble in possible combinations to sort out later; find something with only one possibility....Found it yet?...

...Congratulations if you saw where the 11-in-four crosses the 29-in-four! If not, don't sweat it. It's a good idea to remember at least this much:

1 + 2 + 3 + 4 = 10

9 + 8 + 7 + 6 = 30

Those are the largest and smallest totals you can get for a four-cell entry, and they're nice round numbers. Let's look at the top one: that makes ten, but we need eleven, so we need to add one to a summand. But we can't have any duplicates, so we need to add one to the largest of them. That's right: 11-in-four can only be 1 + 2 + 3 + 5. Similarly, 29-in-four can only be 9 + 8 + 7 + 5. Sure enough, they only have the '5' in common, so in it goes. From there, the only cell in the 29-in-four not yet filled in gets the '8'; some quick arithmetic yields '4' to finish the 18-in-four.

No, I'm

*still*not going to work out combinations for the 7-in-two and 14-in-two - not when the 11-in-four is simpler! It still needs a '1' and a '2', and if you look alongside it, the '2' in the 3-in-two forces the order. The 16-in-four can't have duplicate '2's, or any other duplicate for that matter; I don't even need to read the "16". Putting the '1' above the '2', two simple subtractions finish the puzzle. I'll test my answer by adding up that last entry: 4 + 9 + 1 + 2 = 16. All done.(End of sample puzzle solution)

The better one gets at these, the less arithmetic one needs to solve them and the more instinctual they become. Learn to see the combinations in your mind and try to position them in the grid without guesswork, or complex formulae, or writing several little numbers in a cell until you figure out which one is right. Those should all be last resorts.

This one certainly has a Nikoli bent to it. It's NOT easy for someone's first puzzle, but technically it's your second if you've tried the sample puzzle. Mind you, it's not easy as a second puzzle, either, but hopefully it isn't insurmountable. If I were to guess at par times for solving, I'd say complete beginners will need at most three hours, experienced solvers at most one hour, someone like myself at most twenty minutes, and a world-class competitor ten minutes on a VERY bad day. But those are guesses. Feel free to email me your solution and tell me how long it took you. There might even be a prize in it for you, and it might not necessarily go to the first submission or fastest solver. Then again, maybe it will. You just can't know, can you? - ZM