An implementation of a brute-force algorithm to solve Sudoku puzzles fails if a cell is discovered in which placing any of the digits 1-9 would be an illegal move.
The implementation is written in C, with the board represented by a 9x9 array. The solver counts down from 9 until a legal number's reached, and if none can be reached, it outputs a zero in its place.
A zero also represents a cell to be filled in. Here's the output (truncated) if a string of zeros (an empty board) is the input:
9 8 7 6 5 4 3 2 1
6 5 4 9 8 7 0 0 0
Those last three zeros are there because the values filled in previously aren't changing. How can I stop the solver from failing like this?
If you would currently put a zero in a spot, instead go back to the previous spot you put a number in and continue to count down till you find another value number for that spot.
For instance, in your example:
9 8 7 6 5 4 3 2 1
6 5 4 9 8 7 0 0 0
Instead of putting the zero in below the three, you would instead go back and try putting a 6 in below the 4.
don't treat every "move" like the right move. E.g. placing the last 7 seemed ok but makes it so that in the next cell no valid moves are left. So upon hitting the "no move possible" situation, go back, and try the next option. Iterate and you will have your solution.
A better way of course would be to start brute forcing for places with a small set of options left; run through all cells and start brute forcing with the cell with the least number of options left. When starting out with all-zero, you would then end up with
9 8 7 6 5 4 3 2 1
6 5 4 0 0 0 0 0 0
3 2 1 0 0 0 0 0 0
which is legal, without backtracking once.
You can do this by pushing your guesses onto a stack. Every time you end up wanting to output a zero, instead pop your last answer off the board and continue counting from it.
So if you guess 3 in (2,3) and next you're looking at (3,3) and get to zero, go back to (2,3) and try 2, then 1, then pop to before your (2,3) guess, etc.
Related
There is an existing array of size 64 that has values 6 values distributed as 0 1 2 3 4 5 0 1 2 3 4 5 0 1 2 3 4 5 ...
Please see the image for complete data.
The number of occurrence of 0 in the array is 11 times (at every 6th index), 1 is 11 times ... where as 4 and 5 occurs 10 times each.
There is a necessity to reduce the occurrence of any of these numbers [0 to 5] to a lesser number that could be any number from 0 to 10.
For example, it could be to reduce occurrence of 0 to 6 and 1 to 9.
I am looking for a solid idea to do this. Certainly all the numbers are to be evenly distributed and not something like 0 0 0 0 0 2 2 2 2 2 2 ...
I tried to find the index/position where the reduced value has to filled (64/occurrence of 0 or 2). But at times the index collide with each other and thus is not robust one.
From the example I quoted above, number of occurrence of 0 must be changed to 6 and occurrence of 1 to 9, the result after my algorithm is below -
New location to fill 0 = (Array size)/(new occurrence of 0) = 64/6 = ~10th index
New location to fill 1 = (Array size)/(new occurrence of 1) = 64/9 = ~7 index
For filling 6 0's and 9 1's, first the array is reset after which each of the values are filled to maintain balanced distribution.
After filling 6 0's, the array would be come like this:
Then, after filling 9 1's, the array would be come like this:
The index at 55 already has value 0 and apparently 8th 1 also index to 55 that creates a collision. So I believe, this algorithm to balance the distribution does not work.
How do I populate 6 's, 9 1's and rest of the numbers {2, 3, 4, 5} in the array in a balanced way?
I am new to python.
May I ask why the sixth outcome is 8 instead of 5? As I learnt from "scope" that the later statement should not be affected by whatever happened in another inner scope, so i+=3 should have no effect on what "i" is going to be printed? Thank you for the help.
for i in range (0,10):
if i==5:
i+=3
print i
outcome:
0
1
2
3
4
8
6
7
8
9
In the code you created a condition that if the i reaches number 5 it will add +3 giving an 8.
the += adds do not replace.
if you expect that change the number 5 with a 3 try:
for i in range (0,10):
if i==5:
i = 3
print i
Problem 9-3 of the textbook Intro to Algorithms (CLRS) describes a fast O(n) algorithm for finding the k-th order statistic (k-th element in the array when sorted) of a length-n array, for the particular case that k is much smaller than n. I am not certain about the correctness of this algorithm when n is odd, and want to see a way to prove that it is correct.
The basic idea is that we first split the array into two halves, the first with floor(n/2) elements, and the second with ceil(n/2) elements. Then, we "partner" each element in the first half with the corresponding element in the second half. When n is odd this leaves a remaining unpartnered element.
For each pair of partners, we make sure that the left partner is >= the right partner, swapping the two if not. Then, recursively find the k-th order statistic of the second half, mirroring any swaps made in the second half with corresponding swaps in the first half. After this, the k-th order statistic of the entire array must be either in the first k elements in the first half, or the first k elements in the second half.
My confusion comes from the case when the array length n is odd, and there is a lone element in the second half that has no partner. Since the recursion is performed on the second half, consisting of the last ceil(n/2) elements of the array, including the lone partnerless last element, and we are supposed to mirror all swaps made in second half with swaps made within the corresponding partners in the first half, it is unclear what to do when one of the swaps involves the final element, since it has no partner.
The textbook doesn't seem to take particular care on this issue, so I'm assuming that when a swap involves the final element, then just don't make any mirror moves of the partner in the first half at all. As a result, the final element simply "steals" the partner of whoever it got swapped with. However, in this case, is there an easy way to see if the algorithm is still correct? What if when the last element steals someone else's partner, the partner is actually the k-th order statistic, and gets swapped later on to an inaccessible location? The mechanics of the recursion and partitioning involving in order-statistic selection are sufficiently opaque to me such that I cannot confidently rule out that scenario.
I don't think your description of the algorithm is entirely accurate (but then the explanation you linked to is far from clear). As I understand it, the reason why the algorithm is correct for an odd-length array is as follows:
Let's first look at a few examples of even-length arrays, with n=10 and k=3 (i.e. we're looking for the third-smallest element, which is 2):
a. 5 2 7 6 1 9 3 8 4 0
b. 5 1 7 6 2 9 3 8 4 0
c. 5 0 7 6 2 9 3 8 4 1
d. 5 0 7 6 2 9 3 8 1 4
If we split the arrays into two parts, we get:
a. 5 2 7 6 1 9 3 8 4 0
b. 5 1 7 6 2 9 3 8 4 0
c. 5 0 7 6 2 9 3 8 4 1
d. 5 0 7 6 2 9 3 8 1 4
and these couples:
a. (5,9) (2,3) (7,8) (6,4) (1,0) <- 0 coupled with 1
b. (5,9) (1,3) (7,8) (6,4) (2,0) <- 0 coupled with 2
c. (5,9) (0,3) (7,8) (6,4) (2,1) <- 1 coupled with 2
d. (5,9) (0,3) (7,8) (6,1) (2,4) <- 0, 1 and 2 not coupled with each other
After comparing and swapping the couples so that their smallest element is in the first group, we get:
a. 5 2 7 4 0 9 3 8 6 1
b. 5 1 7 4 0 9 3 8 6 2
c. 5 0 7 4 1 9 3 8 6 2
d. 5 0 7 1 2 9 3 8 6 4
You'll see that the smallest element 0 will always be in the first group. The second-smallest element 1 will be either in the first group, or in the second group if it was coupled with the smallest element 0. The third-smallest element 2 will be either in the first group, or in the second group if it was coupled with either the smallest element 0 or the second-smallest element 1.
So the smallest element is in the first group, and the second- and third-smallest elements can be in either group. That means that the third-smallest element is either one of the 3 smallest elements in the first group, or one of the 2 (!) smallest elements in the second group.
a. 5 2 7 4 0 9 3 8 6 1 -> 0 2 4 + 1 3
b. 5 1 7 4 0 9 3 8 6 2 -> 0 1 4 + 2 3
c. 5 0 7 4 1 9 3 8 6 2 -> 0 1 4 + 2 3
d. 5 0 7 1 2 9 3 8 6 4 -> 0 1 2 + 3 4
So if we say that the k-th smallest element of the whole array is now one of the k-th smallest elements in either of the groups, there is an available spot in the the second group, and that's why, in an odd-length array, we'd add the uncoupled element to the second group. Whether or not the uncoupled element is the element we're looking for, it will certainly be one of the k-th smallest elements in either of the groups.
It is in fact more correct to say that the k-th smallest element is either one of the k smallest elements in the first group, or one of the k/2+1 smallest elements in the second group. I'm actually not sure that the algorithm is optimal, or even correct. There's a lot of repeated comparing and swapping going on, and the idea of keeping track of the couples and swapping elements in one group when their corresponding elements in the other group are swapped doesn't seem to make sense.
On the internet I only find code for the algorithm but I need understand in form of text first because I have trouble understand things from code only. And other description of the algorithm are very complicated for me (on Wikipedia and other sites).
Here is what I understand for far:
Let say we want search in array the element 10:
Index i 0 1 2 3 4
2 3 4 10 40
Some fibonacci number here:
Index j 0 1 2 3 4 5 6 7 8 9
0 1 1 2 3 5 8 13 21 34
First thing we do is find fibonacci number that is greater-equal to array length. Array length is 4 so we need take fibonacci number 5 that is in index position j=5.
But where we divide the array now and how continue? I really don't understand it.. Please help understand for exam...
The algorithm goes in the following way:
The length of the array is 5, so the fibonacci number which is greater than or equal to 5 is 5. The two numbers which are preceding in the Fibonacci sequence are 2 [n-2] and 3 [n-1] - (2, 3, 5).
So, arr[n-2] i.e. arr[2] is compared with the number to be searched which is 10.
If the element is smaller than the number, then the sequence is shifted 1 time to the left. Also, the previous index is saved for next iteration to give an offset for the index. In this case, since 4 is smaller, n-2 becomes 1 (1, 2, 3). arr[1 + 2(prev)] = arr[3] = 10. So, the index of the number is 3.
If the element is larger, the sequence is shifted 2 times to the left.
Always the min(n-2+offset,n)th element is compared with number to get the matching result.
The problem is asking to take any amount of numbers, and find the highest possible sum of difference(using absolute value) between consecutive numbers. For example numbers 1 2 and 3 would be arranged 3 1 2 to get a sum of 3 (3-1 = 2, and 1-2 = 1).
Now my first thoughts were to take the highest number in the list followed by the lowest number and arrange in that way through the end, but that doesnt work out as the end of the list will end up having all of the numbers in the middle accumulating almost no differences. The only other thing I have thought of is to find every single possible order and return the highest sum, but with a longer list this will take way too long and I assume there might be a better way.
For reference here are some sample input and output numbers
9 2 5 3 1 -> 21
7 3 4 5 5 7 6 8 5 4 -> 24
Any help at all would be much appreciated, even if its just pointing me in the right direction.
There are 2 approaches to this problem.
Approach 1:
Brute force.
Approach 2:
Figure out an algorithm for how to arrange the numbers.
I always like approach 2 better if it is feasible.
It seems reasonable that you would get a high sum if you order the numbers high-low-high-low-high...
So start by sorting the numbers and then divide them into two equally large groups of low and high numbers. If there is an odd number of numbers the middle number will be left over.
Then you just pick numbers alternately from the two groups.
It is easy to prove that the order of the interior numbers doesn't matter as long as you stick with the high-low-high-low ordering.
However, since the start and end number only has one neighbour, the first and last number should be the middle numbers.
Finally, if you have an odd number of numbers, place the last number at the start or end, whatever gives the biggest difference.
Example:
7 3 4 5 5 7 6 8 5 4 -> [sort] -> 3 4 4 5 5 5 6 7 7 8
high numbers: 5 6 7 7 8
low numbers: 3 4 4 5 5
Arranged:
5 3 6 4 7 4 7 5 8 5 = 24
Example:
9 2 5 3 1 -> [sort] -> 1 2 3 5 9
high numbers: 5 9
low numbers: 1 2
left over: 3
Arranged:
3 5 1 9 2 = 21 (3 goes at the start, because |3-5| > |3-2|)