Kenken solver python

Each cage in a KenKen contains a target number and most contain an operator. If you see a single-cell cage with just a number and no operator, it means that the value in that cell is the target number. Such single-cell cages work like givens in Sudoku puzzles.

KenKen Solver: Online!

In this puzzle, we can immediately place a 4 in the upper right cell:. Whenever we place a number, this narrows down the possibilities for other cells, so we want to look for that.

kenken solver python

Remember that a cage can repeat numbers in an irregularly shaped cage as long as no number is duplicated within a single row or column. For more complex deductions, it can be useful to take notes in the puzzle, which you can do by using the right side of the on-screen keyboard, holding down the Shift key while typing on a physical keyboard, or switching to Pencil input with TouchWrite.

Next, we can look at the 1- cage in the first column. Whichever it is, it means the 1- cage will definitely contain a 3. And that means the bottom left cell cannot be a 3, which means it must be a That also lets us place the 3 in the bottom row and then the 4 in the second row tells us how to place the 1 and the 4 above how to see timestamp on facebook messenger. Next, we finish up the first column, and we can tell the order because the third row already has a 2 in it:.

kenken solver python

Just tap on the icon to fill in a value or remove mistakes. All rights reserved.To those of you unfamilar with a Sudoku puzzle, they are a logic puzzle that has recently swept the UK and US, appearing in most newspapers alongside the crosswords and other diversions. Congratulations to Carol Vorderman of "Countdown" fame for cashing in early with her book. The puzzle was introduced in the USA as Number Placeand later became popular in Japan as Nanpure or Sudoku -- an abbreviation of suji wa dokushin ni kagiru only single numbers allowedwhich is tradedmarked in Japan by Nikoli.

Other mis spellings include So DukoSoduko and Suduko. Paul Stephens has a very nice introduction site, Sudoku Paulspageswhich includes a rather detailed description of some basic tactics for How to solve Sudoku. There are more tactics pages available of course, for example by Simon Armstrong and Angus Johnson.

There is also this Sudoku Dictionaryand Dan Rice's Sudoku Blog which includes puzzles to try and discussions of tactics etc.

Neil reported it took him about three hours to write his solver, mine took a similar time but I was distracted by trying to solve a very nasty puzzle I found online - which I now suspect to be impossible or at least, ambigous.

Of course, any good programmer should test their handy work. So, where can we find lots of hard Sudoku puzzles to test with? It turns out there is an active group on this web forum and they have compiled a whole bunch of test puzzles hosted here including:.

These are simple text files, with one line of 81 characters per puzzle. They use a full stop period for an unknown entry. Most "introductions" to solving Sudoku will generally discuss or use what I call "candidate lists".

In each empty grid square or cellthey write down all nine digits very small! To me, this lead to an obvious way to store the information in python - for each of the 81 cells, I would have a python list of the possible candidates as integers. If you are using a recent version of Python it might be more efficient to use a set object instead of a list I decided to store the 81 cells as a list of lists the closest I could get to a 9x9 array in simple Python.

When any cell is solved, then that number can be removed from the candidate lists for all the other cells in the same row, column or 3x3 square. Note that if you reduce any cell's list of possibilites to none, then you have made a mistake - or possibly been given an impossible puzzle.

Each digit must appear once and only once in each of the nine rows, nine columns, nine 3x3 sub-sqaures. You don't actually need this a few levels of supposition will do, its just a bit slow. Anyway, this works by considering each of the nine numbers in each of the nine 3x3 blocks. If all the candidates fall into a single row or column then that number can be removed from the other two rows or columns.

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Just doing the above simple logical steps above turns out to be enough to solve most "easy" puzzles you will find. In fact, most newspapers etc seem to grade their puzzles using computer programs - if this is enough then its an easy puzzle.GitHub is home to over 40 million developers working together to host and review code, manage projects, and build software together.

If nothing happens, download GitHub Desktop and try again. If nothing happens, download Xcode and try again. If nothing happens, download the GitHub extension for Visual Studio and try again. A program in Python 3. The KenKen board is represented by a square n-by-n grid of cells. The grid may contain between 1 and n boxes cages represented by a heavily outlined perimeter. Each cage will contain in superscript: the target digit value for the cage followed by a mathematical operator.

The below table represents the number of assigments used from each algorithm, to solve different size puzzles. Artificial Intelligence: A Modern Approach.

Skip to content. Dismiss Join GitHub today GitHub is home to over 40 million developers working together to host and review code, manage projects, and build software together. Sign up. Python Branch: master. Find file. Sign in Sign up. Go back. Launching Xcode If nothing happens, download Xcode and try again. Latest commit Fetching latest commit…. Overview A program in Python 3. Puzzle representation Board The KenKen board is represented by a square n-by-n grid of cells.

Constraints Each valid solution must follow the below rules: The only numbers you may write are 1 to N for a NxN size puzzle.

A number cannot be reapeated within the same row. A number cannot be reapeated within the same solumn. In a one-cell cage, just write the target number in that cell.

Each "cage" region bounded by a heavy border contains a "target number" and an arithmetic operation. You must fill that cage with numbers that produce the target number, using only the specified arithmetic operation. Numbers may be repeated within a cage, if necessary, as long as they do not repeat within a single row or column.Begin by selecting a template either a blank NxN puzzle, or one of the demos from the drop-down list over the playfield. Use the inputs below the playfield to change the rules governing the highlighted cage.

Please note two points:. Click the "Solve" button to search for and display a solution to the puzzle. You may also click the "Make Puzzle" button; this will search for a solution, but it will not show it to you immediately. Instead, it will present the puzzle you just created in our playfield interface. This interface allows you to get cell-by-cell hints, and to continue to work on the puzzle, as opposed to simply seeing the entire answer revealed at once.

You might be interested in some blog articles describing the development of the code behind this solver:. Highlighted cell s must equal produce a sum of produce a difference of produce a product of produce a quotient of produce a result of. Want to play over a thousand free puzzles? Visit our puzzle archive! Introduction This is a solver for KenKen puzzles. Lay out the cages: Click next to a highlighted cell to add the clicked cell to the highlighted cage.

Click on a highlighted cell to remove the clicked cell from the highlighted cage. Click on any other cell to highlight the cage containing the clicked cell. Please note two points: The final choice in the dropdown "produce a result of" assigns an "unknown" operator to the cage; this is how you lay out "no-op" puzzles. If you wish to leave a cage completely unconstrained, assign it a blank value.

Further Reading You might be interested in some blog articles describing the development of the code behind this solver: KenKen vs.By using our site, you acknowledge that you have read and understand our Cookie PolicyPrivacy Policyand our Terms of Service. The dark mode beta is finally here. Change your preferences any time. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information.

kenken solver python

A KenKen puzzle is a Latin square divided into edge-connected domains: a single cell, two adjacent cells within the same row or column, three cells arranged in a row or in an ell, etc. If the domain has just one cell, there is no operator given, just a target the square is solved for you.

The aim of the puzzle is to re construct the Latin square consistent with the domains' boundaries and labels. I think that I have seen a puzzle with a non-unique solution just once. The number in a cell can range from 1 to the width height of the puzzle; commonly, puzzles are 4 or 6 cells on a side, but consider puzzles of any size.

Domains in published puzzles 4x4 or 6x6 typically have no more than 5 cells, but, again, this does not seem to be a hard limit. However, if the puzzle had just one domain, there would be as many solutions as there are Latin squares of that dimension A first step to writing a KenKen solver would be to have routines that can produce the possible combinations of numbers in any domain, at first neglecting the domain's geometry.

A linear domain, like a line of three cells, cannot have duplicate numbers in the solved puzzle, but we ignore this for the moment. I've been able to write a Python function which handles addition labels case by case: give it the width of the puzzle, the number of cells in the domain, and the target sum, and it returns a list of tuples of valid numbers adding up to the target. The multiplication case eludes me. I can get a dictionary with keys equal to the products attainable in a domain of a given size in a puzzle of a given size, with the values being lists of tuples containing the factors giving the product, but I can't work out a case-by-case routine, not even a bad one.

Factoring a given product into primes seems easy, but then partitioning the list of primes into the desired number of factors stumps me. I have meditated on Fascicle 3 of Volume 4 of Knuth's TAOCP, but I have not learned how to 'grok' his algorithm descriptions, so I do not know whether his algorithms for set partitioning would be a starting point. Understanding Knuth's descriptions could be another question! I'm quite happy to precompute the 'multiply' dictionaries for common domain and puzzle sizes and just chalk the loading time up to overhead, but that approach would not seem an efficient way to deal with, say, puzzles cells on a side and domains from 2 to 50 cells in size.

Simplified goal: you need to enumerate all integer combinations that multiply together to form a certain product, where the number of integers is fixed. To solve this, all you need is a prime factorization of your target number, and then use a combinatorial approach to form all possible sub-products from these factors. The trick to this is to form all possible sub-products, and the code below does this. It works by looping through the factors forming all possible single pairs, and then doing this recursively, to give all possible sets of all single or multiple pairings.

It's inefficiently, but even large numbers have a small prime factorization :. Learn more. Find all possible factors in KenKen puzzle 'multiply' domain Ask Question.By using our site, you acknowledge that you have read and understand our Cookie PolicyPrivacy Policyand our Terms of Service. Code Review Stack Exchange is a question and answer site for peer programmer code reviews.

It only takes a minute to sign up. The user must fill in each embolded regions with numbers in the range 1-n where n is the board dimensions e. So the top square must use the numbers to create 36 using the multiplication operand e. This is valid if the two threes are placed in a diagonal relationship to each other.

When filling in a region, the order is arbitrary. So for example if considering the bottom left corner with a target of 2 and operand of "minus" then placing 2 above 1 or 1 above 2 would not be relevant. If you don't understand, please see here. You use numpy to represent your board but don't take advantage of some of its capabilities to simplify your code.

Some loops are unnecessary as they can be performed by a numpy operation. You also use itertools. This will also allows you to check for completion of the board, if the generator is exhausted and raise StopIteration. I'd put the translation dictionary outside of the function.

I think it makes more sense as a "global constant" than it does as something local to that function. That also gives you a way to check for the validity of an operator string for free if you ever need that:. Your accumulation loop that splits the head off then iterates the rest could be made much neater by using a reduction. If you don't supply an initial value for a reduction, it automatically uses the first element as the starting accumulator.

Your return could be simplified down to just returning the condition. I also wouldn't put the return True inside the if body, then not put the return False in an else. If board were immutable, ya, that would make sense. You're directly mutating the board that's passed in though.

kenken solver python

I think it creates necessary bulk, especially in lines like:. With how you have it, I would just return the boolean status. It may make sense however to make a copy of the input board, then mutate and return the copy.

This prevents the original board from getting mutated by accident.By using our site, you acknowledge that you have read and understand our Cookie PolicyPrivacy Policyand our Terms of Service.

Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. Having tested several of the more promising answers, I'm going to award the answer-prize to Lennart Regebro, because:. Thanks, Lennart. A remark: Alex Martelli's bare-bones recursive algorithm is an example of making every possible combination and throwing them all at a sieve and seeing which go through the holes.

Moral: Not generating every possible combination is good. Another remark: Lennart's and my routines generate the same answers in the same order. Are they in fact the same algorithm seen from different angles? I don't know. Something occurs to me. Is this related to "informational entropy"? What's the proper metric for looking at it? Is there an algorithm that producs the combinations in descending or ascending order of heat?

This one doesn't, as far as I can see, although it's close over short stretches, looking at normalized std. First of all, I'd use variable names that mean something, so that the code gets comprehensible. Then, after I understood the problem, it's clearly a recursive problem, as once you have chosen one number, the question of finding the possible values for the rest of the squares are exactly the same problem, but with different values in.

I also notice that your solution still seems buggy. I'm not sure if that means I've missed a rule in this or not, but as I understand the rules that should be a valid option.

Python Sudoku Solver - Computerphile

Here's a version using generators, It saves a couple of lines, and memory if the values are really big, but as recursion, generators can be tricky to "get". Your algorithm seems pretty good at first blush, and I don't think OO or another language would improve the code.

I can't say if recursion would have helped but I admire the non-recursive approach. I bet it was harder to get working and it's harder to read but it likely is more efficient and it's definitely quite clever.

To be honest I didn't analyze the algorithm in detail but it certainly looks like something that took a long while to get working correctly.

I bet there were lots of off-by-1 errors and weird edge cases you had to think through, eh? Given all that, basically all I tried to do was pretty up your code as best I could by replacing the numerous C-isms with more idiomatic Python-isms. Often times what requires a loop in C can be done in one line in Python. Also I tried to rename things to follow Python naming conventions better and cleaned up the comments a bit.

Hope I don't offend you with any of my changes. You can take what you want and leave the rest. I'm developing it from scratch. Here's a version without any optimization at all, just for simplicity:. The for tries all possibilities for "this" item and loops over whatever the next-lower-down level of recursion is still able to yield. I corresponded with Kent Beck to properly attribute this quote in "Python in a Nutshell", and he tells me he got it from his dad, whose job was actually unrelated to programming.

In this case, it seems to me that the key issue is understanding what's going on, and any optimization might interfere, so I'm going all out for "simple and understandable"; we can, if need be! Sorry to say, your code is kind of long and not particularly readable. If you can try to summarize it somehow, maybe someone can help you write it more clearly. As for the problem itself, my first thought would be to use recursion.

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