I am solving a puzzle which is as follows.
There is a 5x5 matrix with one element as "-" and all others as whole numbers.
I can swap any element with the "-" straight(not diagonally).
Finally, I have to sort the matrix.
These are the steps I follow:
1) Receive user input for 5x5 matrix
2) Locate the position of "-"
3) Find the eligible candidates to be swapped with "-"
4) Apply some algorithm and find the most eligible candidate
5) Swap the element with "-"
6) Repeat the steps 3-5 until matrix is sorted
I have completed till step 3. However I have no idea what logic to be applied for step 4.
Can someone give some thoughts, how to find the most eligible candidate?
Examples
Input Matrix
17 7 9 18 3
15 11 1 12 14
2 - 4 21 24
5 19 6 18 8
10 13 16 19 20
Eligible candidates to swap with "-" are 11,2,4,19
Sorted matrix
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 -
this isn't the most easiest task. Here are two links:
http://en.wikipedia.org/wiki/N-puzzle#Solvability
http://cseweb.ucsd.edu/~ccalabro/essays/15_puzzle.pdf
It is better to use some AI algorithm like A* with the Manhattan distance heuristic.
For A* algorithm read here
For the Manhattan distance heuristic read here
Related
I did some research but i have difficulties finding an answer.
I am using python 2.7 and pandas so far but i am still learning.
I have two CSVs, let say it's the alphabet A-Z in one and digits in the second one, 0-100.
I want to merge the two files to have A0 to A100 up through Z.
For information the two files have DNA sequence so i believe they are strings.
I tried to create arrays with numpy and create a matrix but to no available..
here is a preview of the files:
barcode
0 GGAAGAA
1 CCAAGAA
2 GAGAGAA
3 AGGAGAA
4 TCGAGAA
5 CTGAGAA
6 CACAGAA
7 TGCAGAA
8 ACCAGAA
9 GTCAGAA
10 CGTAGAA
11 GCTAGAA
12 GAAGGAA
13 AGAGGAA
14 TCAGGAA
659
barcode
0 CGGAAGAA
1 GCGAAGAA
2 GGCAAGAA
3 GGAGAGAA
4 CCAGAGAA
5 GAGGAGAA
6 ACGGAGAA
7 CTGGAGAA
8 CACGAGAA
9 AGCGAGAA
10 TCCGAGAA
11 GTCGAGAA
12 CGTGAGAA
13 GCTGAGAA
14 CGACAGAA
1995
I am putting here the way i found to do it, there might be a sexier way:
index = pd.MultiIndex.from_product([df8.barcode, df7.barcode], names = ["df8", "df7"])
df = pd.DataFrame(index = index).reset_index()
def concat_BC(x):#concatenate the two sequences into one new column
return str(x["df8"]) + str(x["df7"])
df["BC"] = df.apply(concat_BC, axis=1)
– Stephane Chiron
I've encountered an interesting situation while calculating the inter-quartile range. Assuming we have a dataframe such as:
import pandas as pd
index=pd.date_range('2014 01 01',periods=10,freq='D')
data=pd.np.random.randint(0,100,(10,5))
data = pd.DataFrame(index=index,data=data)
data
Out[90]:
0 1 2 3 4
2014-01-01 33 31 82 3 26
2014-01-02 46 59 0 34 48
2014-01-03 71 2 56 67 54
2014-01-04 90 18 71 12 2
2014-01-05 71 53 5 56 65
2014-01-06 42 78 34 54 40
2014-01-07 80 5 76 12 90
2014-01-08 60 90 84 55 78
2014-01-09 33 11 66 90 8
2014-01-10 40 8 35 36 98
# test for q1 values (this works)
data.quantile(0.25)
Out[111]:
0 40.50
1 8.75
2 34.25
3 17.50
4 29.50
# break it by inserting row of nans
data.iloc[-1] = pd.np.NaN
data.quantile(0.25)
Out[115]:
0 42
1 11
2 34
3 12
4 26
The first quartile can be calculated by taking the median of values in the dataframe that fall below the overall median, so we can see what data.quantile(0.25) should have yielded. e.g.
med = data.median()
q1 = data[data<med].median()
q1
Out[119]:
0 37.5
1 8.0
2 19.5
3 12.0
4 17.0
It seems that quantile is failing to provide an appropriate representation of q1 etc. since it is not doing a good job of handling the NaN values (i.e. it works without NaNs, but not with NaNs).
I thought this may not be a "NaN" issue, rather it might be quantile failing to handle even-numbered data sets (i.e. where the median must be calculated as the mean of the two central numbers). However, after testing with dataframes with both even and odd-numbers of rows I saw that quantile handled these situations properly. The problem seems to arise only when NaN values are present in the dataframe.
I would like to use quntile to calculate the rolling q1/q3 values in my dataframe, however, this will not work with NaN's present. Can anyone provide a solution to this issue?
Internally, quantile uses numpy.percentile over the non-null values. When you change the last row of data to NaNs you're essentially left with an array array([ 33., 46., 71., 90., 71., 42., 80., 60., 33.]) in the first column
Calculating np.percentile(array([ 33., 46., 71., 90., 71., 42., 80., 60., 33.]) gives 42.
From the docstring:
Given a vector V of length N, the qth percentile of V is the qth ranked
value in a sorted copy of V. A weighted average of the two nearest
neighbors is used if the normalized ranking does not match q exactly.
The same as the median if q=50, the same as the minimum if q=0
and the same as the maximum if q=100.
I'm trying to code a model that can solve the Multiple Choice Knapsack Problem (MCKP) as described in Knapsack Problems involving dimensions, demands and multiple
choice constraints: generalization and transformations between
formulations (Found here, see figures 8 an 9). You can find an example GMPL model of the basic knapsack problem here. For anyone looking for a quick explanation of the knapsack problem read the following illustration:
You are an adventurer and have stumbled upon a treasure trove. There are hundreds of wonderful items 'i' that each have a weight 'w' and a profit 'p'. Say you have a knapsack with weight capacity as 'c' and you want to make the most profit without overfilling your knapsack. What is the best combination of items such that you make the most profit?
In code:
maximize obj :
sum{(i,w,p) in I} p*x[i];
Where 'I' is the basket of items, and x[i] is the binary variable (0 = not chosen, 1 = chosen)
The problem that I am having trouble with is the addition of multiple groups. MCKP requires exactly one item to be selected from each group. So, for example, lets say we have three groups from which to choose. They could be represented as follows (ignore actual values):
# Items: index, weight, profit
set ONE :=
1 10 10
2 10 10
3 15 15
4 20 20
5 20 20
6 24 24
7 24 24
8 50 50;
# Items: index, weight, profit
set TWO :=
1 10 10
2 10 10
3 15 15
4 20 20
5 20 20
6 24 24
7 24 24
8 50 50;
# Items: index, weight, profit
set THREE :=
1 10 10
2 10 10
3 15 15
4 20 20
5 20 20
6 24 24
7 24 24
8 50 50;
I am confused on how I can iterate over each group and how I would define the variable x. I assume it would look something like:
var x{i,j} binary;
Where i is the index of items in j of groups. This assumes I define a set of sets:
set Groups{ONE,TWO,THREE}
Then I'd iterate over the groups of items:
sum{j in Groups, (i,w,p) in Groups[j]} p*x[i,j];
But I am concerned because I believe GMPL does not support ordered sets. I have seen this related question where the answer suggests defining a set within a set. However, I am not sure how it would apply in this particular scenario.
My main question, to be clear: In GMPL, how can I iterate over sets of sets (in this case a set of groups where each group has a set of items)?
Unlike AMPL, GMPL doesn't support sets of sets. Here's how to do it in AMPL:
set Groups;
set Items{Groups} dimen 3;
# define x and additional constraints
# ...
maximize obj: sum{g in Groups, (i,w,p) in Items[g]} p*x[i];
data;
set Groups := ONE TWO THREE;
# Items: index, weight, profit
set Items[ONE] :=
1 10 10
2 10 10
3 15 15
4 20 20
5 20 20
6 24 24
7 24 24
8 50 50;
# Items: index, weight, profit
set Items[TWO] :=
1 10 10
2 10 10
3 15 15
4 20 20
5 20 20
6 24 24
7 24 24
8 50 50;
# Items: index, weight, profit
set Items[THREE] :=
1 10 10
2 10 10
3 15 15
4 20 20
5 20 20
6 24 24
7 24 24
8 50 50;
If you have no more than 300 variables, you can use a free student version of AMPL and solvers (e.g. CPLEX or Gurobi).
Based on this gnu mailing list thread, I believe GMPL/MathProg has support for what you want to do. Here's their example:
set WORKERS;
param number_of_shifts, integer, >= 1;
set WORKER_CLIQUE{1..number_of_shifts}, within WORKERS;
data;
set WORKERS := Jack Kate Sawyer Sun Juliet Richard Desmond Hugo;
param number_of_shifts := 2;
set WORKER_CLIQUE[1] := Sawyer, Juliet;
set WORKER_CLIQUE[2] := Jack, Kate, Hugo;
In your example, I assume you'd use something like, set Items{1..3}, within Groups; with the data block from #vitaut's answer.
Can anyone help me with this pratice question: 33 31 11 47 2 20 24 12 2 43. I am trying to figure out what the contents of the two output lists would be after the first pass of the Merge Sort.
The answer is supposedly:
List 1: 33 11 47 12
List 2: 31 2 20 24 2 43
Not making any sense to me sense I was under the impression that the first pass was where it divided it into two lists at the middle....
33 31 11 47 2 20 24 12
At first the list divides into individual elements such that when a single ton list is formed each element is compared with the one next to it. So after first pass we have
31 33 11 47 2 20 12 24
After that
11 31 33 47 2 12 20 24
and then
2 11 12 20 24 3133 37
I just need some verification on whether or not I'm doing this correctly. I checked the wiki for a heapsort, but it seems in the animation to build the heap it inserts the numbers into the nodes and orders it as it goes.
The question asks to "Draw a valid heap with these elements.. {7, 12, 1, 3, 22, 5, 11} as a tree"
I've tried it on a few examples, and it seems I should layout the nodes first then reorder the nodes instead of ordering it as I go. Is the way I'm doing it correct?
For example, putting elements into the nodes
7
12 1
3 22 5 11
Ordering begins here: swap 1 and 7
7
12 1
3 22 5 11
Swap 3 and 12
1
12 7
3 22 5 11
Swap 5 and 7
1
3 7
12 22 5 11
done.
1
3 5
12 22 7 11
Actually that's not right.
The answer given is
1
7 3
12 22 5 11
If I begin to re-order the heap from the left side first (beginning with 3) then I get the right answer.
Actually, the heap data structure as only one property, which can be defined as follows: "In a heap T, for every node v other than the root, the key stored at v is greater than or equal to the key stored at v's parent."
So there are many right representations of the heap based on elements (7, 12, 1, 3, 22, 5, 11). With these elements I applied an "insert-and-sort" algorithm and the result gives yet another version:
1
3 5
12 22 7 11