I have a triangle with two-hundred rows, where I have to find the maximum distance to get from the top to the bottom of the triangle.
5
9 8
5 4 6
9 7 3 4
Here, the shortest distance would be 5+8+4+3=20. The maximum distance would be 5+9+5+9=28.
I have a good idea of the algorithm I want to implement but I am struggling to turn it into code.
My plan is: start at the 2nd to last row, add the maximum of the possible paths from the bottom row, and iterate to the top.
For instance, the above triangle would turn into:
28
23 19
14 11 10
9 7 3 4
This is vastly more efficient than brute-forcing, but I have two general questions:
Using brute-force, how do I list all the possible paths from top to
bottom (can only move to adjacent points)? I tried using this
(triangle is the list of lists containing the triangle):
points=list(itertools.product(*triangle))
but this contains all possible combinations from each row, not just
adjacent members.
Project Euler #18 - how to brute force all possible paths in tree-like structure using Python?
This somewhat explains a possible approach, but I'd like to use
itertools and any other modules (as pythonic as possible)
How would I go about iterating the strategy of adding each maximum
from the previous row and iterating to the top? I know I have to
implement a nested loop:
for x in triangle:
for i in x:
i+=? #<-Not sure if this would even increment it
edit:
what I was thinking was:
triangle[y][x] = max([triangle[y+1][x],triangle[y+1][x+1]])
It does not use itertools, it is recursive, but I memoize the results, so its still fast...
def memoize(function):
memo = {}
def wrapper(*args):
if args in memo:
return memo[args]
else:
rv = function(*args)
memo[args] = rv
return rv
return wrapper
#memoize
def getmaxofsub(x, y):
if y == len(triangle) or x>y: return 0
#print x, y
return triangle[y][x] + max(getmaxofsub(x, y+1), getmaxofsub(x+1, y+1))
getmaxofsub(0,0)
I read your algorithm suggestion some more times and your "cumulative triangle" is stored in memoof the memoized decorator, so in the end it is very similar. if you want to prevent that there is big stack during recursive "down calling" through the triangle, you can fill the cache of memoize by calling getmaxofsub() bottom -> up.
for i in reversed(range(len(triangle))):
getmaxofsub(0, i), getmaxofsub(i//2, i), getmaxofsub(i, i)
print getmaxofsub(0,0)
Edit
getmaxofsub: How does this function work? First you have to know, that you can't divide your triangle in sub triangles. I take your triangle as an example:
5
9 8
5 4 6
9 7 3 4
That's the complete one. The "coordinates" of the peak are x=0, y=0.
Now I extract the sub triangle of the peak x=0, y=1:
9
5 4
9 7 3
or x=1, y=2
4
7 3
So this is how my algorithm works: The peak of the whole triangle (x=0, y=0) asks its sub triangles (x=0, y=1) and (x=1, y=1), "What is your maximum distance to the ground?" And each of them will ask their sub-triangles and so on…
this will go on until the function reaches the ground/y==len(triangle): The ground-entries want to ask their sub triangles, but since their is none of those, they get the answer 0.
After each triangle has called their sub triangles, it decides, which one is the greater one, add their own value and return this sum.
So now you see, what is the principle of this algorithm. Those algorithms are called recursive algorithms. You see, a function calling itself is pretty standard… and it works…
So, if you think about this whole algorithm, you would see that a lot of sub-triangles are called several times and they would ask their sub-triangles and so on… But each time they return the same value. That is why I used the memorize-decorator: If a function is called with the same arguments x and y, the decorator returns the last calculated value for those arguments and prevents the time-consuming calculation… It is a simple cache…
That is why this function is as easy to implement as a recursive algorithm and as fast as a iteration...
To answer your first question (how to brute-force iterate over all paths): If you start at the top of the triangle and move down along some random path, you have to make a decision to go left or right for every level that you go down. The number of different paths is thus 2^(nrows-1). For your problem with 200 rows, there are thus 8e59 different paths, way to much to check in a brute-force way.
For a small triangle, you can still iterate over all possible paths in a brute-force way, for example like this:
In [10]: from itertools import product
In [11]: triangle = [[5], [9,8], [5,4,6], [9,7,3,4]]
In [12]: for decisions in product((0,1), repeat = len(triangle)-1):
...: pos = 0
...: path = [triangle[0][0]]
...: for lr, row in zip(decisions, triangle[1:]):
...: pos += lr # cumulative sum of left-right decisions
...: path.append(row[pos])
...: print path
[5, 9, 5, 9]
[5, 9, 5, 7]
[5, 9, 4, 7]
[5, 9, 4, 3]
[5, 8, 4, 7]
[5, 8, 4, 3]
[5, 8, 6, 3]
[5, 8, 6, 4]
The way this works is to use itertools.product to iterate over all possible combinations of nrows-1 left/right decisisions, where a 0 means go left and a 1 means go right (so you are more or less generating the bits of all binary numbers up to 2^(nrows-1)). If you store the triangle as a list of lists, going left means staying at the same index in the next row, while going right means adding 1. To keep track of the position in the row, you thus simply calculate the cumulative sum of all left/right decisions.
To answer your second question: First of all, your algorithm seems pretty good, you only need to iterate once backwards over all rows and you do not have the exponential number of cases to check as in the brute-force solution. The only thing I would add to that is to build a new triangle, which indicates at every step whether the maximum was found to the left or to the right. This is useful to reconstruct the optimal path afterwards. All this can be implemented like this:
mx = triangle[-1] # maximum distances so far, start with last row
directions = [] # upside down triangle with left/right direction towards max
for row in reversed(triangle[:-1]): # iterate from penultimate row backwards
directions.append([l < r for l, r in zip(mx[:-1], mx[1:])])
mx = [x + max(l, r) for x, l, r in zip(row, mx[:-1], mx[1:])]
print 'Maximum so far:', mx
print 'The maximum distance is', mx[0]
directions.reverse()
pos = 0
path = [triangle[0][0]]
for direction, row in zip(directions, triangle[1:]):
pos += direction[pos]
path.append(row[pos])
print 'The optimal path is', path
As before, I used the trick that False = 0 and True = 1 to indicate going left and right. Using the same triangle as before, the result:
Maximum so far: [14, 11, 10]
Maximum so far: [23, 19]
Maximum so far: [28]
The maximum distance is 28
The optimal path is [5, 9, 5, 9]
Related
I have a list of numbers. I also have a certain sum. The sum is made from a few numbers from my list (I may/may not know how many numbers it's made from). Is there a fast algorithm to get a list of possible numbers? Written in Python would be great, but pseudo-code's good too. (I can't yet read anything other than Python :P )
Example
list = [1,2,3,10]
sum = 12
result = [2,10]
NOTE: I do know of Algorithm to find which numbers from a list of size n sum to another number (but I cannot read C# and I'm unable to check if it works for my needs. I'm on Linux and I tried using Mono but I get errors and I can't figure out how to work C# :(
AND I do know of algorithm to sum up a list of numbers for all combinations (but it seems to be fairly inefficient. I don't need all combinations.)
This problem reduces to the 0-1 Knapsack Problem, where you are trying to find a set with an exact sum. The solution depends on the constraints, in the general case this problem is NP-Complete.
However, if the maximum search sum (let's call it S) is not too high, then you can solve the problem using dynamic programming. I will explain it using a recursive function and memoization, which is easier to understand than a bottom-up approach.
Let's code a function f(v, i, S), such that it returns the number of subsets in v[i:] that sums exactly to S. To solve it recursively, first we have to analyze the base (i.e.: v[i:] is empty):
S == 0: The only subset of [] has sum 0, so it is a valid subset. Because of this, the function should return 1.
S != 0: As the only subset of [] has sum 0, there is not a valid subset. Because of this, the function should return 0.
Then, let's analyze the recursive case (i.e.: v[i:] is not empty). There are two choices: include the number v[i] in the current subset, or not include it. If we include v[i], then we are looking subsets that have sum S - v[i], otherwise, we are still looking for subsets with sum S. The function f might be implemented in the following way:
def f(v, i, S):
if i >= len(v): return 1 if S == 0 else 0
count = f(v, i + 1, S)
count += f(v, i + 1, S - v[i])
return count
v = [1, 2, 3, 10]
sum = 12
print(f(v, 0, sum))
By checking f(v, 0, S) > 0, you can know if there is a solution to your problem. However, this code is too slow, each recursive call spawns two new calls, which leads to an O(2^n) algorithm. Now, we can apply memoization to make it run in time O(n*S), which is faster if S is not too big:
def f(v, i, S, memo):
if i >= len(v): return 1 if S == 0 else 0
if (i, S) not in memo: # <-- Check if value has not been calculated.
count = f(v, i + 1, S, memo)
count += f(v, i + 1, S - v[i], memo)
memo[(i, S)] = count # <-- Memoize calculated result.
return memo[(i, S)] # <-- Return memoized value.
v = [1, 2, 3, 10]
sum = 12
memo = dict()
print(f(v, 0, sum, memo))
Now, it is possible to code a function g that returns one subset that sums S. To do this, it is enough to add elements only if there is at least one solution including them:
def f(v, i, S, memo):
# ... same as before ...
def g(v, S, memo):
subset = []
for i, x in enumerate(v):
# Check if there is still a solution if we include v[i]
if f(v, i + 1, S - x, memo) > 0:
subset.append(x)
S -= x
return subset
v = [1, 2, 3, 10]
sum = 12
memo = dict()
if f(v, 0, sum, memo) == 0: print("There are no valid subsets.")
else: print(g(v, sum, memo))
Disclaimer: This solution says there are two subsets of [10, 10] that sums 10. This is because it assumes that the first ten is different to the second ten. The algorithm can be fixed to assume that both tens are equal (and thus answer one), but that is a bit more complicated.
I know I'm giving an answer 10 years later since you asked this, but i really needed to know how to do this an the way jbernadas did it was too hard for me, so i googled it for an hour and I found a python library itertools that gets the job done!
I hope this help to future newbie programmers.
You just have to import the library and use the .combinations() method, it is that simple, it returns all the subsets in a set with order, I mean:
For the set [1, 2, 3, 4] and a subset with length 3 it will not return [1, 2, 3][1, 3, 2][2, 3, 1] it will return just [1, 2, 3]
As you want ALL the subsets of a set you can iterate it:
import itertools
sequence = [1, 2, 3, 4]
for i in range(len(sequence)):
for j in itertools.combinations(sequence, i):
print(j)
The output will be
()
(1,)
(2,)
(3,)
(4,)
(1, 2)
(1, 3)
(1, 4)
(2, 3)
(2, 4)
(3, 4)
(1, 2, 3)
(1, 2, 4)
(1, 3, 4)
(2, 3, 4)
Hope this help!
So, the logic is to reverse sort the numbers,and suppose the list of numbers is l and sum to be formed is s.
for i in b:
if(a(round(n-i,2),b[b.index(i)+1:])):
r.append(i)
return True
return False
then, we go through this loop and a number is selected from l in order and let say it is i .
there are 2 possible cases either i is the part of sum or not.
So, we assume that i is part of solution and then the problem reduces to l being l[l.index(i+1):] and s being s-i so, if our function is a(l,s) then we call a(l[l.index(i+1):] ,s-i). and if i is not a part of s then we have to form s from l[l.index(i+1):] list.
So it is similar in both the cases , only change is if i is part of s, then s=s-i and otherwise s=s only.
now to reduce the problem such that in case numbers in l are greater than s we remove them to reduce the complexity until l is empty and in that case the numbers which are selected are not a part of our solution and we return false.
if(len(b)==0):
return False
while(b[0]>n):
b.remove(b[0])
if(len(b)==0):
return False
and in case l has only 1 element left then either it can be part of s then we return true or it is not then we return false and loop will go through other number.
if(b[0]==n):
r.append(b[0])
return True
if(len(b)==1):
return False
note in the loop if have used b..but b is our list only.and i have rounded wherever it is possible, so that we should not get wrong answer due to floating point calculations in python.
r=[]
list_of_numbers=[61.12,13.11,100.12,12.32,200,60.00,145.34,14.22,100.21,14.77,214.35,200.32,65.43,0.49,132.13,143.21,156.34,11.32,12.34,15.67,17.89,21.23,14.21,12,122,134]
list_of_numbers=sorted(list_of_numbers)
list_of_numbers.reverse()
sum_to_be_formed=401.54
def a(n,b):
global r
if(len(b)==0):
return False
while(b[0]>n):
b.remove(b[0])
if(len(b)==0):
return False
if(b[0]==n):
r.append(b[0])
return True
if(len(b)==1):
return False
for i in b:
if(a(round(n-i,2),b[b.index(i)+1:])):
r.append(i)
return True
return False
if(a(sum_to_be_formed,list_of_numbers)):
print(r)
this solution works fast.more fast than one explained above.
However this works for positive numbers only.
However also it works good if there is a solution only otherwise it takes to much time to get out of loops.
an example run is like this lets say
l=[1,6,7,8,10]
and s=22 i.e. s=1+6+7+8
so it goes through like this
1.) [10, 8, 7, 6, 1] 22
i.e. 10 is selected to be part of 22..so s=22-10=12 and l=l.remove(10)
2.) [8, 7, 6, 1] 12
i.e. 8 is selected to be part of 12..so s=12-8=4 and l=l.remove(8)
3.) [7, 6, 1] 4
now 7,6 are removed and 1!=4 so it will return false for this execution where 8 is selected.
4.)[6, 1] 5
i.e. 7 is selected to be part of 12..so s=12-7=5 and l=l.remove(7)
now 6 are removed and 1!=5 so it will return false for this execution where 7 is selected.
5.)[1] 6
i.e. 6 is selected to be part of 12..so s=12-6=6 and l=l.remove(6)
now 1!=6 so it will return false for this execution where 6 is selected.
6.)[] 11
i.e. 1 is selected to be part of 12..so s=12-1=1 and l=l.remove(1)
now l is empty so all the cases for which 10 was a part of s are false and so 10 is not a part of s and we now start with 8 and same cases follow.
7.)[7, 6, 1] 14
8.)[6, 1] 7
9.)[1] 1
just to give a comparison which i ran on my computer which is not so good.
using
l=[61.12,13.11,100.12,12.32,200,60.00,145.34,14.22,100.21,14.77,214.35,145.21,123.56,11.90,200.32,65.43,0.49,132.13,143.21,156.34,11.32,12.34,15.67,17.89,21.23,14.21,12,122,134]
and
s=2000
my loop ran 1018 times and 31 ms.
and previous code loop ran 3415587 times and took somewhere near 16 seconds.
however in case a solution does not exist my code ran more than few minutes so i stopped it and previous code ran near around 17 ms only and previous code works with negative numbers also.
so i thing some improvements can be done.
#!/usr/bin/python2
ylist = [1, 2, 3, 4, 5, 6, 7, 9, 2, 5, 3, -1]
print ylist
target = int(raw_input("enter the target number"))
for i in xrange(len(ylist)):
sno = target-ylist[i]
for j in xrange(i+1, len(ylist)):
if ylist[j] == sno:
print ylist[i], ylist[j]
This python code do what you asked, it will print the unique pair of numbers whose sum is equal to the target variable.
if target number is 8, it will print:
1 7
2 6
3 5
3 5
5 3
6 2
9 -1
5 3
I have found an answer which has run-time complexity O(n) and space complexity about O(2n), where n is the length of the list.
The answer satisfies the following constraints:
List can contain duplicates, e.g. [1,1,1,2,3] and you want to find pairs sum to 2
List can contain both positive and negative integers
The code is as below, and followed by the explanation:
def countPairs(k, a):
# List a, sum is k
temp = dict()
count = 0
for iter1 in a:
temp[iter1] = 0
temp[k-iter1] = 0
for iter2 in a:
temp[iter2] += 1
for iter3 in list(temp.keys()):
if iter3 == k / 2 and temp[iter3] > 1:
count += temp[iter3] * (temp[k-iter3] - 1) / 2
elif iter3 == k / 2 and temp[iter3] <= 1:
continue
else:
count += temp[iter3] * temp[k-iter3] / 2
return int(count)
Create an empty dictionary, iterate through the list and put all the possible keys in the dict with initial value 0.
Note that the key (k-iter1) is necessary to specify, e.g. if the list contains 1 but not contains 4, and the sum is 5. Then when we look at 1, we would like to find how many 4 do we have, but if 4 is not in the dict, then it will raise an error.
Iterate through the list again, and count how many times that each integer occurs and store the results to the dict.
Iterate through through the dict, this time is to find how many pairs do we have. We need to consider 3 conditions:
3.1 The key is just half of the sum and this key occurs more than once in the list, e.g. list is [1,1,1], sum is 2. We treat this special condition as what the code does.
3.2 The key is just half of the sum and this key occurs only once in the list, we skip this condition.
3.3 For other cases that key is not half of the sum, just multiply the its value with another key's value where these two keys sum to the given value. E.g. If sum is 6, we multiply temp[1] and temp[5], temp[2] and temp[4], etc... (I didn't list cases where numbers are negative, but idea is the same.)
The most complex step is step 3, which involves searching the dictionary, but as searching the dictionary is usually fast, nearly constant complexity. (Although worst case is O(n), but should not happen for integer keys.) Thus, with assuming the searching is constant complexity, the total complexity is O(n) as we only iterate the list many times separately.
Advice for a better solution is welcomed :)
Hi I have a problem as below:
Given a set of n segments {[a0, b0], [a1, b1], . . . , [an-1, bn-1]} with integer coordinates on a line, find the minimum number m of points such that each segment contains at least one point. That is, find a set of integers X of the minimum size such that for any segment [ai,bi] there is a point x ∈ X such that ai ≤ x ≤ bi.
Input Format: The first line of the input contains the number n of segments. Each of the following n lines contains two integers ai and bi (separated by a space) defining the coordinates of endpoints of the i-th segment.
Output Format: Output the minimum number m of points on the first line and the integer coordinates of m points (separated by spaces) on the second line. You can output the points in any order. If there are many such sets of points, you can output any set. (It is not difficult to see that there always exist a set of points of the minimum size such that all the coordinates of the points are integers.)
Sample 1:
Input: 3
1 3
2 5
3 6
Output: 1 3
Explanation:
In this sample, we have three segments: [1,3],[2,5],[3,6] (of length 2,3,3 respectively). All of them contain the point with coordinate 3: 1 ≤3 ≤3, 2 ≤3 ≤5, 3 ≤ 3 ≤ 6.
Sample 2:
Input: 4
4 7
1 3
2 5
5 6
Output: 2
3 6
Explanation:
The second and the third segments contain the point with coordinate 3 while the first and the fourth segments contain the point with coordinate 6. All the four segments cannot be covered by a single point, since the segments [1, 3] and [5, 6] are disjoint.
Solution:
The greedy choice is selecting the minimum right endpoint. Then remove all segments that contains that endpoint. Keep choosing minimum right endpoint and removing segments.
I followed the solution. I found the minimum right endpoint, removed all segments that contain that endpoint in my code. Then execute the function again with the new segments list (Keep choosing minimum right endpoint and removing segments - Recursive) but I'm stuck with the order of my code and can't make it works.
list_time = [[4,7],[1,3],[2,5],[5,6]]
def check_inside_range(n, lst): #Function to check if a number is inside the range of start and end of a list
#for example 3 is in [3,5], 4 is not in [5,6], return False if in
if lst[1]-n>=0 and n-lst[0]>=0:
return False
else:
return True
def lay_chu_ki(list_time):
list_time.sort(key = lambda x: x[1]) #Sort according to the end of each segments [1,3],[2,5],[5,6],[4,7]
first_end = list_time[0][1] #Selecting the minimum right endpoint
list_after_remove = list(filter(lambda x: check_inside_range(first_end, x),list_time))
#Remove all segments that contains that endpoint
lay_chu_ki(list_after_remove) #Keep doing the function again with new segments list
#(Keep choosing minimum right endpoint and removing segments)
return first_end #I don't know where to put this line.
print(lay_chu_ki(list_time))
As you can see, I've already done 3 steps: Selecting the minimum right endpoint; Remove all segments that contains that endpoint; Keep choosing minimum right endpoint and removing segments but it won't work somehow. I tried to print two numbers 3 and 6 first (the return result of each recursive call). I also tried to create a count variable to count each recursive call (count +=1) but it didn't work too since it reset count = 0 for each call.
I think recursion overcomplicates the implementation. While it's still feasible, you have to pass in a bunch of extra parameters, which could be difficult to track. In my opinion, it's much simpler to implement this approach iteratively.
Also, your approach repeatedly uses filter() and list(), which takes linear time every time you do it (to clarify, "linear" means linear in the size of the input list). In the worst case, you would perform that operation for every element in the list, which means that the runtime of your original implementation is quadratic (assuming you fix the existing issues with your code). This approach avoids that by making a single pass through the list:
def lay_chu_ki(list_time):
list_time.sort(key=lambda x: x[1])
idx = 0
selected_points = []
while idx != len(list_time):
selected_point = list_time[idx][1]
while idx != len(list_time) and list_time[idx][0] <= selected_point:
idx += 1
selected_points.append(selected_point)
return selected_points
result = lay_chu_ki(list_time)
print(len(result))
print(' '.join(map(str, result)))
With the given list, this outputs:
2
3 6
I am new to Genetic Algorithms and am working on a python implementation. I am up to the crossover step and am attempting a Partially Matched Crossover. For my final output I am hoping for a list that contains no duplicated numbers. However, in some cases, I am introducing duplicates.
For example, Take the lists
Mate 1 [1,2,3,5,4,6]
Mate 2 [6,5,4,3,2,1]
If the crossover portion is [3,5,4] -> [4,3,2]
Then the offspring before mapping becomes [1,2,4,3,2,6]. My understanding of the algorithm is the mapping outside the crossover is 4 -> 3, 5 -> 3 and 2 -> 4. However, this results in an output of [1,4,4,3,2,6] which has duplicates and is missing a 5.
How do I work around this problem? Does the first 4 just become a 5? And how would this scale to larger lists that might introduce multiple duplicates?
I am not sure you have implemented it right:
for Partially Matched Crossover (see explanation), if your crossover points are 2 and 5 as suggested in the example then you can only obtain
offspring1 = [6, 2, 3, 5, 4, 1]
offspring2 = [1, 5, 4, 3, 2, 6]
if you select 3,5,4 from mate1 and fill the rest in the order of mate2 you will get offspring 1 but if you select 4,3,2 from mate2 and fill the rest in the order of mate 1 you will get offspring 2
See implementation below:
mate1 = [1,2,3,5,4,6]
mate2 = [6,5,4,3,2,1]
crossoverpoint1 = 2
crossoverpoint2=5
child = []
#fill in the initial genes in order of mate1
count = 0
for i in mate1:
if(count == crossoverpoint1):
break
if(i not in mate2[crossoverpoint1:crossoverpoint2]):
child.append(i)
count= count+1
#select the genes within the crossover points from mate2
child.extend(mate2[crossoverpoint1:crossoverpoint2])
#fill in the remaining genes in order of mate1
child.extend([x for x in mate1 if x not in child])
print(child)
output:
[1, 5, 4, 3, 2, 6]
to obtain offspring1 swap mate1 for mate2.
you can also try different crossover points, let me know if this helps
I need help with my code when answering the following question.
An arithmetic progression is a sequence of numbers in which the distance (or difference) between any two successive numbers is the same. This in the sequence 1, 3, 5, 7, ..., the distance is 2 while in the sequence 6, 12, 18, 24, ..., the distance is 6.
Given the positive integer distance and the non-negative integer n, create a list consisting of the arithmetic progression between (and including) 1 and n with a distance of distance. For example, if distance is 2 and n is 8, the list would be [1, 3, 5, 7].
Associate the list with the variable arith_prog.
I updated my progress:
arith_prog = []
for i in range(1, n, distance):
arith_prog.append(n)
total = n + distance
While the suggestions made so far were helpful, I still haven't arrived at the correct solution turingscraft codelab is looking for.
The range function takes up to three arguments; start, stop and step. You want
list(range(1, n, distance))
I'm responding to this as a homework question, since you seem to be indicating that's what it is:
First of all, you never initialize n. What starting value should it
have?
Second, you don't need two loops here - all you need is one.
Third, why are you passing distance to range()? If you pass two
arguments to range() they're treated as a lower and upper bound,
respectively - and distance is probably not a bound.
The problem is where you have arith_prog.append(n). You need to replace the .append(n) with an .append(i) because we are adding the value in that range to list. I just did this homework for 15 minutes ago and that was one of the correct solutions. I made the same error you did.
do something like this
arith_prog = []
n = 5 #this is just for example, you can use anything you like or do an input
distance = 2 #this is also for example, change it to what ever you like
for i in range(1,n,distance):
arith_prog.append(i)
print(arith_prog) #for example this prints out [1,3]
I also encountered this exercise on myprogramminglab. You were very close. Try this:
arith_prog = []
for i in range(1, n + 1, distance):
arith_prog.append(i)
total = n + distance
Hope this helps.
Working through MPL and came across this problem, accepted answer below:
arith_prog=[]
for i in range(1,n+1,distance):
arith_prog.append(i)
I'm trying to create a sorting technique that sorts a list of numbers. But what it does is that it compares two numbers, the first being the first number in the list, and the other number would be the index of 2k - 1.
2^k - 1 = [1,3,7, 15, 31, 63...]
For example, if I had a list [1, 4, 3, 6, 2, 10, 8, 19]
The length of this list is 8. So the program should find a number in the 2k - 1 list that is less than 8, in this case it will be 7.
So now it will compare the first number in the random list (1) with the 7th number in the same list (19). if it is greater than the second number, it will swap positions.
After this step, it will continue on to 4 and the 7th number after that, but that doesn't exist, so now it should compare with the 3rd number after 4 because 3 is the next number in 2k - 1.
So it should compare 4 with 2 and swap if they are not in the right place. So this should go on and on until I reach 1 in 2k - 1 in which the list will finally be sorted.
I need help getting started on this code.
So far, I've written a small code that makes the 2k - 1 list but thats as far as I've gotten.
a = []
for i in range(10):
a.append(2**(i+1) -1)
print(a)
EXAMPLE:
Consider sorting the sequence V = 17,4,8,2,11,5,14,9,18,12,7,1. The skipping
sequence 1, 3, 7, 15, … yields r=7 as the biggest value which fits, so looking at V, the first sparse subsequence =
17,9, so as we pass along V we produce 9,4,8,2,11,5,14,17,18,12,7,1 after the first swap, and
9,4,8,2,1,5,14,17,18,12,7,11 after using r=7 completely. Using a=3 (the next smaller term in the skipping
sequence), the first sparse subsequence = 9,2,14,12, which when applied to V gives 2,4,8,9,1,5,12,17,18,14,7,11, and the remaining a = 3 sorts give 2,1,8,9,4,5,12,7,18,14,17,11, and then 2,1,5,9,4,8,12,7,11,14,17,18. Finally, with a = 1, we get 1,2,4,5,7,8,9,11,12,14,17,18.
You might wonder, given that at the end we do a sort with no skips, why
this might be any faster than simply doing that final step as the only step at the beginning. Think of it as a comb
going through the sequence -- notice that in the earlier steps we’re using course combs to get distant things in the
right order, using progressively finer combs until at the end our fine-tuning is dealing with a nearly-sorted sequence
needing little adjustment.
p = 0
x = len(V) #finding out the length of V to find indexer in a
for j in a: #for every element in a (1,3,7....)
if x >= j: #if the length is greater than or equal to current checking value
p = j #sets j as p
So that finds what distance it should compare the first number in the list with but now i need to write something that keeps doing that until the distance is out of range so it switches from 3 to 1 and then just checks the smaller distances until the list is sorted.
The sorting algorithm you're describing actually is called Combsort. In fact, the simpler bubblesort is a special case of combsort where the gap is always 1 and doesn't change.
Since you're stuck on how to start this, here's what I recommend:
Implement the bubblesort algorithm first. The logic is simpler and makes it much easier to reason about as you write it.
Once you've done that you have the important algorithmic structure in place and from there it's just a matter of adding gap length calculation into the mix. This means, computing the gap length with your particular formula. You'll then modifying the loop control index and the inner comparison index to use the calculated gap length.
After each iteration of the loop you decrease the gap length(in effect making the comb shorter) by some scaling amount.
The last step would be to experiment with different gap lengths and formulas to see how it affects algorithm efficiency.