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By : user2951577
Date : November 18 2020, 11:13 AM
Hope this helps The following algorithm can be used provided that every cost is >= 0 code :
``````create an 'output' array the same size as the 'cost' array
set all entries in the 'output' array to 'unknown'
the first entry in the 'output' array equals the first entry of the 'cost' array
write the coordinates of the first entry {0,0} to a queue
while the queue is not empty
{
read a set of coordinates from the queue
for each neighboring square
{
compute the cost to move from the current square to the neighbor
if ( the current cost of the neighbor is unknown or higher )
{
set the cost of the neighboring square to the new lower cost
add the coordinates of the neighboring square to the queue
}
}
}
`````` ## Cheapest path on a checkerboard from A to B. Turning and moving cost differently

By : Charne van der Merwe
Date : March 29 2020, 07:55 AM
it helps some times The minimum number of moves is always |XB-XA|+|YB-YA| and we can do it always with only one turn.
Example: move from (1,2) to (4,6):
code :
``````3 moves right: 1500
1 turn:         100
4 moves down:  2000
----
3600
`````` ## Finding a path through a checkerboard which is closest to a given cost

By : Anisha K.M.
Date : March 29 2020, 07:55 AM
help you fix your problem No. If all the costs are integers, at each cell you need to store at most O(M) elements. So you need O(MN^2) memory. If the sum is >M you just ignore it.
In this paper there is a mention of a pseudo polynomial algorithm to solve similar problem (exact cost). You can either use same algorithm multiple time with exact cost = M..1, or maybe read the algorithm and find a variation that solves your problem directly. ## Finding a path travel cost closest to a given value

By : Nokardron
Date : March 29 2020, 07:55 AM
will help you I have been stuck on a problem recently. I have to find the cost of a path from the top left corner to the bottom right corner of a multidimensional array of integers. This path's cost has to be the greatest cost that does not go over a certain provided value. , You can update your memoize function to something like this:
code :
``````def memoize (f):
memo = {}
def helper(*args):
if args not in memo:
memo[args] = f(*args)
return memo[args]
return helper
``````
``````grid = [[0,2,5],[1,1,3],[2,1,1]]
# 11
``````
``````def fn(*args):
print(args)

fn(1,2,3)
(1, 2, 3)
`````` ## Error using Max and Min based on Avg to determine lowest average cost and highest average cost (MySQL)

By : tapavko
Date : March 29 2020, 07:55 AM
I wish this help you I have determined the average subject cost as per the following: , I think that what you are trying to achieve is
code :
``````SELECT *
FROM BOOK
JOIN
(
SELECT BOOK_SUBJECT,
AVG(BOOK_COST) AS AVGCOST,
MIN(BOOK_COST) AS MINCOST,
MAX(BOOK_COST) AS MAXCOST
FROM BOOK
GROUP BY BOOK_SUBJECT
) AS SUB USING (BOOK_SUBJECT)
ORDER BY BOOK_TITLE;
``````
``````BOOK_SUBJECT BOOK_TITLE BOOK_COST
------------ ---------- ---------------------
MATHS        Book 1     15.99
MATHS        Book 2     14.99
MATHS        Book 3     13.99
ENGLISH      Book A     10.99
ENGLISH      Book B     9.99
``````
``````BOOK_SUBJECT BOOK_TITLE BOOK_COST AVGCOST MINCOST MAXCOST
------------ ---------- --------- ------- ------- -------
MATHS        Book 1     15.99     14.99   13.99   15.99
MATHS        Book 2     14.99     14.99   13.99   15.99
MATHS        Book 3     13.99     14.99   13.99   15.99
ENGLISH      Book A     10.99     10.49   9.99    10.99
ENGLISH      Book B     9.99      10.49   9.99    10.99
``````
``````SELECT
ROUND(MIN(MM.AVGCOST), 2) AS MINAVG,
ROUND(MAX(MM.AVGCOST), 2) AS MAXAVG
FROM
(
SELECT BOOK_SUBJECT,
AVG(BOOK_COST) AS AVGCOST
FROM BOOK
GROUP BY BOOK_SUBJECT
) MM
``````
``````MINAVG    MAXAVG
------    ------
10.49     14.99
`````` ## Split a group in two based on lowest possible travel distance

By : Sissitest Giannitest
Date : March 29 2020, 07:55 AM
With these it helps You can model the math problem like this.
http://mathb.in/31885?key=216f0d8271c8a65ecbce2faff12735042a4b7684 