Pipsum rules and solving techniques
Pipsum is a logic puzzle made of sums and dice pips. It follows the rules of Kakuro, the Japanese cross-sums puzzle - with one difference that makes everything easier: there are only the pip counts 1 to 6, not 1 to 9.
There is little arithmetic and a lot of inference: every cell follows by deduction from the runs that cross on it. On striduel there are new tasks every day, free in the browser and playable without an account.
The rules
Some cells are blocked and set apart in colour. Some of those carry sums: the number at the top right applies to the cells to its right, the number at the bottom left to the cells below it. One such run of adjacent free cells is called a block.
- Every free cell takes a pip count from 1 to 6. Tap the cell, then the number below; a tap on a filled cell clears it again.
- Every block adds up to exactly its sum. No more, no less.
- Within a block no pip count appears twice. Across block boundaries it may very well - that is the most common mistake at the start.
From the second and third rule follows one that saves a lot of work: a block over five cells holds five different pip counts out of six. Exactly one is missing, and the sum gives it away - 21 minus the sum of the block. A five-cell block is always a gift, never a hurdle.
Every task is solvable without guessing and has exactly one solution.
Strategy: three inferences that always hold
The work is not in the arithmetic - no run has more than three decompositions, at any length. It is in which runs cross where.
- Decomposition. A run of two cells summing to 11: with pip counts from 1 to 6 that has to be 5 and 6, and there is no other split. Which value goes where is still open - which two they are is settled.
- Crossing. Every free cell belongs to one run across and one run down. If one of them allows only 5 or 6 there and the other only 2 or 6, the cell is a 6 by force - there is no other agreement. This is the step Pipsum is made of.
- Remainder. Once every cell of a run but one is filled, the last one is the difference to the run's sum, by force. No search, a subtraction.
An example, step by step
A real 5×5 grid on the medium tier, solved with the same rules that apply in the game. A few pip counts are given from the start; they stand out less than the ones that arrive during the walkthrough. Each step names the technique, the reason, and the cell that follows from it.
Step 1Remainder - The across block starting at row 5, column 2 has the sum 7, and only one cell in it is still empty. The other cell holds 3, so 4 is what is left for the last one. That settles: row 5, column 3 = 4.
Step 2Remainder - The down block starting at row 4, column 2 has the sum 9, and only one cell in it is still empty. The other cell holds 3, so 6 is what is left for the last one. That settles: row 4, column 2 = 6.
Step 3Crossing - Two blocks meet at row 2, column 4: the across block with the sum 10 and the down block with the sum 13. Across, 4 or 6 would still fit there; down, 2 or 6 - together only one pip count is left. That settles: row 2, column 4 = 6.
Step 4Decomposition - The across block starting at row 2, column 4 has the sum 10 over 2 cells, and there is only one way to make it: 4 + 6. The 6 is already there - so the pip count that is missing can be read off rather than worked out. That settles: row 2, column 5 = 4.
Step 5Remainder - The down block starting at row 2, column 4 has the sum 13, and only one cell in it is still empty. The rest hold 11 between them, so 2 is what is left for the last one. That settles: row 4, column 4 = 2.
Step 6Remainder - The across block starting at row 4, column 2 has the sum 13, and only one cell in it is still empty. The rest hold 8 between them, so 5 is what is left for the last one. That settles: row 4, column 3 = 5.
Step 7Remainder - The down block starting at row 3, column 3 has the sum 11, and only one cell in it is still empty. The rest hold 9 between them, so 2 is what is left for the last one. That settles: row 3, column 3 = 2.
Step 8Remainder - The across block starting at row 3, column 3 has the sum 10, and only one cell in it is still empty. The rest hold 7 between them, so 3 is what is left for the last one. That settles: row 3, column 5 = 3.
The techniques by name
These are the same names the game uses when it names a technique - in Deduction Rush it stands there after every answer. Read them here, play there, and you meet the same word again.
- Decomposition. Some sums can be formed over their length in only one way - 3 over two cells is always 1+2, 11 over two cells always 5+6. You write blocks like that down without doing arithmetic. No block has more than three decompositions, at any length; the difficulty of Pipsum never lies in the arithmetic.
- Remainder. Once only one cell in a block is empty, its value comes out as the sum minus what is already there. The most common step there is - and the one that turns one filled cell into the next two.
- Crossing. Every free cell belongs to one across block and one down block. Only what is possible in both stays possible. This is the technique that makes Pipsum something other than an arithmetic exercise.
- Block exclusion. A pip count that cannot be placed in any of a block's remaining decompositions drops out for that block's cells - even when no single cell forbids it directly.
- Trial. On the hard tier only: you place a pip count on trial. If that leads to a contradiction, it was the wrong one.
Practise it yourself
In practice mode you can play Pipsum on its own, as often as you like and on the tier you choose. None of it counts towards a leaderboard - it is the place to try things out.