I use Maxima CAS to create the list:
a:makelist(i,i,1,20);
result:
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]
I want to slim the list and leave only every third element. To find it I check index i of the list a :
mod(i,3)>0
to find elements.
My code :
l:length(a);
for i:1 thru l step 1 do if (mod(i,3)>0) then a:delete(a[i],a);
Of course it does not work because length of a is changing.
I can do it using second list:
b:[];
for i:1 thru l step 1 do if (mod(i,3)=0) then b:cons(a[i],b);
Is it the best method ?
There are different ways to solve this, as know already. My advice is to construct a list of the indices you want to keep, and then construct the list of elements from that. E.g.:
(%i1) a:makelist(i,i,1,20);
(%o1) [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]
(%i2) ii : sublist (a, lambda ([a1], mod(a1, 3) = 0));
(%o2) [3, 6, 9, 12, 15, 18]
(%i3) makelist (a[i], i, ii);
(%o3) [3, 6, 9, 12, 15, 18]
The key part is the last step, makelist(a[i], i, ii), where ii is the list of indices you want to select. ii might be constructed in various ways. Here is a different way to construct the list of indices:
(%i4) ii : makelist (3*i, i, 1, 6);
(%o4) [3, 6, 9, 12, 15, 18]
One simple way (I do not know which one is best or faster) with compact code: makelist(a[3*i],i,1,length(a)/3)
Test example:
l1:makelist(i,i,1,12)$
l2:makelist(i,i,1,14)$
l3:[2,3,5,7,11,13,17,19,23,29]$
for a in [l1,l2,l3] do (
b:makelist(a[3*i],i,1,length(a)/3),
print(a,"=>",b)
)$
Result:
[1,2,3,4,5,6,7,8,9,10,11,12] => [3,6,9,12]
[1,2,3,4,5,6,7,8,9,10,11,12,13,14] => [3,6,9,12]
[2,3,5,7,11,13,17,19,23,29] => [5,13,23]
Related
The question I am given is
We are given an array.
In one operation we can replace any element of the array with any two elements that sum to that element.
For example: array = {4, 11, 7}. In one operation you can replace array[1] with 5 and 6 which sums to 11. So the array becomes array = {4, 5, 6, 7}
Return the minimum number of steps in which the whole array can be sorted in non-decreasing order. Along with array in sorted order.
For example: array = {3,9,3}
I think the answer will be 9 will be converted to 3,3,3
But I cannot think of a general formula of doing it.
My thoughts on the solution are
Suppose we want to convert number 6 and 9
We use if and else
IF
we see that we divide a number by 2 and take ceiling but it is greater than the number on it's right side(last example in the question) then we keep subtracting that number(3) until we get integer 0.
That is 9 = 3(number on right of 9 in array in last example) - 3 - 3
ELSE
simply do ceiling(num / 2) to get first number and then num - ceil(num / 2) to ger second. 7 will be 4 and 3.
Please can someone think of a general formula for doing it?
Edy's way (as I interpret it) in Python:
def solve(xs):
limit = 10**100
out = []
for x in reversed(xs):
parts = (x - 1) // limit + 1
limit, extra = divmod(x, parts)
out += extra * [limit+1] + (parts - extra) * [limit]
print(len(out) - len(xs), out[::-1])
solve([4, 11, 7])
solve([3, 9, 3])
solve([9, 4, 15, 15, 28, 23, 13])
Output showing steps and result array for the three test cases (Try it online!):
1 [4, 5, 6, 7]
2 [3, 3, 3, 3, 3]
8 [3, 3, 3, 4, 5, 5, 5, 7, 8, 9, 9, 10, 11, 12, 13]
An output illustrating the progress:
[4, 11, 7] = (input)
[4, 11, [7]]
[4, [5, 6], [7]]
[[4], [5, 6], [7]]
[3, 9, 3] = (input)
[3, 9, [3]]
[3, [3, 3, 3], [3]]
[[3], [3, 3, 3], [3]]
[9, 4, 15, 15, 28, 23, 13] = (input)
[9, 4, 15, 15, 28, 23, [13]]
[9, 4, 15, 15, 28, [11, 12], [13]]
[9, 4, 15, 15, [9, 9, 10], [11, 12], [13]]
[9, 4, 15, [7, 8], [9, 9, 10], [11, 12], [13]]
[9, 4, [5, 5, 5], [7, 8], [9, 9, 10], [11, 12], [13]]
[9, [4], [5, 5, 5], [7, 8], [9, 9, 10], [11, 12], [13]]
[[3, 3, 3], [4], [5, 5, 5], [7, 8], [9, 9, 10], [11, 12], [13]]
Code for that (Try it online!):
def solve(xs):
print(xs, '= (input)')
limit = 10**100
for i, x in enumerate(reversed(xs)):
parts = (x - 1) // limit + 1
limit, extra = divmod(x, parts)
xs[~i] = (parts - extra) * [limit] + extra * [limit+1]
print(xs)
print()
You would want to scan from the right to the left. For convenient explanation, let's mark the right-most element x_0, and the left-most x_{n-1} (n can increase as you split a number into two).
If x_{i} > x_{i-1}, you would want to divide x_{i} into ((x_{i} - 1) / x_{i-1}) + 1 parts, where / is integer division, as evenly as possible.
So for example:
If x_{i} = 15, x_{i-1] = 5, divide x_{i} into (15-1)/5 + 1 = 3 parts: (5, 5, 5).
If x_{i} = 19, x_{i-1] = 5, divide x_{i} into (19-1)/5 + 1 = 4 parts: (4, 5, 5, 5).
(To divide a number equally into a non-decreasing sequence would require a bit of calculation, which shouldn't be too difficult.)
Once you know the sequence, it would be straightforward to repeatedly split a number into 2 to produce that sequence.
I'm trying to make a program that for a sublist of numbers, uses index as a variable and selects each number from the list of lists
so if my numbest = [[1, 2, 3, 4, 5], [2, 4, 6, 8, 10], [3, 5, 7, 9, 11]]
I want to be able to call the function like this
column_sum(2, [[1, 2, 3, 4, 5], [2, 4, 6, 8, 10], [3, 5, 7, 9, 11]]) will add the numbers at index 2 in each sublist (3, 6, and 7) and will return the number 16."
I can't for the life of me figure out how to print
for i in numlist:
print numbest[index]
Looks like Python, so imma say that all you need to do is have a variable that is a running total, add up all the numbers that are the values at the index you specify, and then return that value.
Alexander is also right and if his way is easier for you, you can find resources https://www.w3schools.com/python/ref_func_sum.asp and https://www.w3schools.com/python/python_lists_comprehension.asp
Implement the next permutation, which rearranges numbers into the numerically next greater permutation of numbers for a given array A of size N.
If such arrangement is not possible, it must be rearranged as the lowest possible order i.e., sorted in an ascending order.
The test cases of this problem include :
Input :
A = [20, 50, 113]
Output :
[20, 113, 50]
How is [20, 113, 50] greater than [20, 50, 113]?
Similarly ,
Input :
A = [5, 18, 9]
Output :
[9, 5, 18]
How is this next permutation rather than [5,9,18]?
How is [20, 113, 50] greater than [20, 50, 113]?
Because it's lexicographically greater than, this works as follows:
If the first item is greater than or less than then this is the result.
Otherwise if they're equal then if the second item is greater than or less than then this is the result.
Otherwise if they're equal then if the third item is greater than or less than then this is the result.
Otherwise they're equal.
So [20, 113, 50] > [20, 50, 113] because in step 1: 20 == 20 and in step 2: 113 > 50.
Likewise: [5, 9, 18] < [5, 18, 9] because 9 < 18 and [9, 5, 18] > [5, 18, 9] because 9 > 5.
How is [20, 113, 50] greater than [20, 50, 113]?
In [20, 50, 113], the last increasing length-2 interval is [50, 113], then we pick the first element in [113, 113](eq. to [113]) from right to left so that it is greater than 50. Swap it with 50, then reverse the interval (50, 113], This will give you [20, 113, 50].
Output : [9, 5, 18]\n How is this next permutation rather than [5,9,18]?
From A = [5, 18, 9] Found [5, 18],
From [18, 9] Found 9,
Swap(5,9) Get [9, 18, 5],
Reverse([18, 5]) Get [5, 18],
So you get: [9, 5, 18].
the next permutation means the next largest number
ex-- like if an array is arr = {1, 2, 3, 6, 5, 4}
then the number is 123654
its next greater element is 124356;
so return the array {1, 2, 4, 3, 5, 6}.
possible_list = []
bigger_list = []
new_list= [0, 25, 2, 1, 14, 1, 14, 1, 4, 6, 6, 7, 0, 10, 11]
for i in range(0,len(new_list)):
# if the next index is not greater than the length of the list
if (i + 1) < (len(new_list)):
#if the current value is less than the next value
if new_list[i] <= new_list[i+1]:
# add the current value to this sublist
possible_list.append(new_list[i])
# if the current value is greater than the next, close the list and append it to the lager list
bigger_list.append(possible_list)
print bigger_list
How do I find the longest consistent increment in the list called new_list?
I expect the result to be
[[0,2], [2], [1,14], [1,14], [1,4,6,6,7], [0,10,11]]
I can find the remaining solution from there myself.
One problem (but not the only one) with your code is that you are always adding the elements to the same possible_list, thus the lists in bigger_list are in fact all the same list!
Instead, I suggest using [-1] to access the last element of the list of subsequences (i.e. the one to append to) and [-1][-1] to access the last element of that subsequence (for comparing the current element to).
new_list= [0, 25, 2, 1, 14, 1, 14, 1, 4, 6, 6, 7, 0, 10, 11]
subseq = [[]]
for e in new_list:
if not subseq[-1] or subseq[-1][-1] <= e:
subseq[-1].append(e)
else:
subseq.append([e])
This way, subseq ends up the way you want it, and you can use max to get the longest one.
>>> subseq
[[0, 25], [2], [1, 14], [1, 14], [1, 4, 6, 6, 7], [0, 10, 11]]
>>> max(subseq, key=len)
[1, 4, 6, 6, 7]
I need to port a snippet written in Python to C++
but that snippet is using combinations from itertools in python.
The line that I'm really interested to porting over to C++ is this one:
for k in combinations(range(n-i),2*i):
range(n-i) in Python will generate a list from 0 to (n-i) - 1
Let n = 16, i = 5
print range(n-i)
outputs:
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
and python combinations will generate all possible combinations in that list.
e.g.
print list(combinations(range(n-i),2*i))
outputs:
[(0, 1, 2, 3, 4, 5, 6, 7, 8, 9),
(0, 1, 2, 3, 4, 5, 6, 7, 8, 10),
(0, 1, 2, 3, 4, 5, 6, 7, 9, 10),
(0, 1, 2, 3, 4, 5, 6, 8, 9, 10),
(0, 1, 2, 3, 4, 5, 7, 8, 9, 10),
(0, 1, 2, 3, 4, 6, 7, 8, 9, 10),
(0, 1, 2, 3, 5, 6, 7, 8, 9, 10),
(0, 1, 2, 4, 5, 6, 7, 8, 9, 10),
(0, 1, 3, 4, 5, 6, 7, 8, 9, 10),
(0, 2, 3, 4, 5, 6, 7, 8, 9, 10),
(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)]
I want to generate similar output using std::vector and next_permutation from C++ but I'm still getting erroneous results. This is my current approach:
for(int j = 0; j < n-i; j++) {
temp_vector.push_back(j);
}
That snippet is equivalent to range(n-i) in Python.
But the following snippet:
do {
myvector.push_back(temp_vector);
} while(next_permutation(temp_vector.begin(),temp_vector.begin()+2*i));
cout<<myvector.size()<<endl;
Is not equivalent to combinations(range(n-i),2*i)) in Python, and I've tried many variations and still haven't been able to come up with the results I'm expecting.
For example:
Let n = 16
i = 5
Python
>>> print len(list(combinations(range(n-i),2*i)))
11
C++
#include <vector>
#include <iostream>
using namespace std;
int main() {
vector<int> temp_vector;
vector< vector<int> > myvector;
int n = 16, i = 5;
for(int j = 0; j < n - i; j++) {
temp_vector.push_back(j);
}
do {
myvector.push_back(temp_vector);
} while(next_permutation(temp_vector.begin(), temp_vector.begin()+2*i));
cout<<myvector.size()<<endl;
return 0;
}
g++ combinations.cpp
./a.out
3628800
Any guidance will be greatly appreciated! Thanks a lot!
combinations and permutations are not the same thing.
A combination is an unordered list of a subset of the items from another set. A permutation is a unique order of the items in the list.
You're generating all combinations of 10 things from a list of 11 things, so you'll get 11 results, each one missing a different one of the original 11 items.
Generating every permutation will generate every unique order of the original 11 items. Since the items in this case are all unique that means the result would be 11! lists where each contains all 11 items. You're only generating permutations from the first 10 items however, so you're getting 10! lists, none of which contain the 11th item.
You need to find an algorithm for generating combinations instead of permutations.
There's no built-in algorithm for combinations. std::next_permutation can be used as part of an algorithm to generate combinations: See Generating combinations in c++.
Here's an old draft proposal for algorithms for combinations, including code.