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How can I iterate a for loop in Python in the increments of sum of another variable and iterator

I am trying to make a python program for "The Sieve Of Eratosthenes"
but I am stuck at one place.
The c program I wrote looks something like this :
for (i = 2; i <= n; i++)
{
if (prime[i] == 0)
{
for (j = i*i; j <= n; j += i)
{
prime[j] = 1;
}
}
}
How will I write the code for for(j=i*i;j<=n;j+=i) in Python?
As by default python increment the iterator by 1

I think you can do something like:
# rest of the code
# .
# .
j = i*i
while j <=n:
# resut of the code
j += i

You code could be rewritten like this
for i in range(2,n+1, 1):
if prime[i]==0:
for j in range(i*i, n+1, i):
prime[j]=1
The range function takes upto 3 values, the first is the starting value, the second is the ending value (exclusive) and the third is the amount to step each time.
The full semantics of range can be found here

Im a little confused because I have no idea what the syntax of C is, but I assume you want to do something along the lines of
for i in range(n): # run from 0 to the value of n (int)
if prime[i] == 0: # assuming prime is a list
for j in range(i**2): # runs from 0 to the value of i squared
prime[j] = 1
This code should work in terms of syntax and actually running (assuming you have defined prime and n somewhere), but you may have to do some tweaking to the for loops to make it acheive the desired effect. Remember, a for loop in that way of construction will have a temporary variable called i that is 0, and as long as the value of i is less than the value of n, the code will run and each time i will increment itself by one. If you want to change it a bit, you could use range(start, stop, step), in which the start is the starting value of i (not 0), and stop is the stopping value (n). step is also the amount in which the variable of i will increment itself by each time the code loops.
I hope this helps. please ask any questions you have.

Filtering a numerical vector in R (vs Python as a list)

I'm trying to learn about R; I was writing a function to return the (proper) divisors of a number.
In Python I can use the filter method, like this:
def divisors(n):
numbers = list(range(1,n))
an_iterator = filter(lambda x: n % x ==0, numbers)
return(list(an_iterator))
What is a good way to write that sort of function, this time working with a vector in R? So far I have this:
divisors <- function(n){
vec <- 1:(n-1)
vec <- ifelse(n %% vec == 0, vec, FALSE)
vec <- vec[vec!=FALSE]
return(vec)
}
EDIT: I also found from below the "which()" function:
divisors <- function(n) {
vec <- 1:(n-1)
vec <- vec[which(n %% vec == 0)]
return(vec)
}

i think the best way is to take advantage of R's vector approach, that can be thought of as a very efficient for loop. For example:
divisors = function (n) {
div = 1:(n - 1)
div[n %% div == 0]
}
(note: i use the = operator to assign, that is not a common practice in R where the <- is by far more common. I use it here because is resembles the way assignments are made in other languages, including Python)
Here div = 1:(1 - n) will create a vector of possible divisors. n %% div will return the remainder for n / div in the form of a vector of the same length as div. This is as if we had made a for loop ranging through each element of div and concatenating every result. This is what people refer to as vectorization in R, this process is written in C and really efficient. The last bit is subsetting (filter) div to get only the proper divisors. Again i use here the vector approach, an generate a logical vector with n %% div == 0. Subsetting with this will return only those values from div for which the condition holds, so
> divisors(26)
[1] 1 2 13
Best,

Many functions in R work very well by default with vectors, so for example if you evaluate 6 %% 1:6 you will get 0 0 0 2 1 0. All of the 0s represent where 6 was divided into evenly. Since you want the actual numbers that divided evenly, not the result, you want to compare the results to 0, or ask 0 0 0 2 1 0 == 0 which will return TRUE TRUE TRUE FALSE FALSE TRUE. With this result you can index your original vector to return the correct subset (1:6)[(6 %% 1:6) == 0] however since the indices themselves indicate the correct subset in our case I used the R function which that returns the indices for TRUE values in a vector.
Also to find the proper divisors you only need to go up to half of the value you are investigating. The floor function ensures the result is an integer for odd numbers.
divisors <- function(n) {
which((n %% 1:floor(n / 2)) == 0)
}

Swift equivalent of Python slice assignment

In Python, one can have a list (similar to an array in swift):
>>> li=[0,1,2,3,4,5]
And perform a slice assignment on any / all of the list:
>>> li[2:]=[99] # note then end index is not needed if you mean 'to the end'
>>> li
[0, 1, 99]
Swift has a similar slice assignment (this is in the swift interactive shell):
1> var arr=[0,1,2,3,4,5]
arr: [Int] = 6 values {
[0] = 0
[1] = 1
[2] = 2
[3] = 3
[4] = 4
[5] = 5
}
2> arr[2...arr.endIndex-1]=[99]
3> arr
$R0: [Int] = 3 values {
[0] = 0
[1] = 1
[2] = 99
}
So far, so good. But, there are a couple of issues.
First, swift does not work for an empty list or if the index is after the endIndex. Python appends if the slice index is after then end index:
>>> li=[] # empty
>>> li[2:]=[6,7,8]
>>> li
[6, 7, 8]
>>> li=[0,1,2]
>>> li[999:]=[999]
>>> li
[0, 1, 2, 999]
The equivalent in swift is an error:
4> var arr=[Int]()
arr: [Int] = 0 values
5> arr[2...arr.endIndex-1]=[99]
fatal error: Can't form Range with end < start
That is easy to test and code around.
Second issue is the killer: it is really slow in swift. Consider this Python code to to perform exact summations of a list of floats:
def msum(iterable):
"Full precision summation using multiple floats for intermediate values"
# Rounded x+y stored in hi with the round-off stored in lo. Together
# hi+lo are exactly equal to x+y. The inner loop applies hi/lo summation
# to each partial so that the list of partial sums remains exact.
# Depends on IEEE-754 arithmetic guarantees. See proof of correctness at:
# www-2.cs.cmu.edu/afs/cs/project/quake/public/papers/robust-arithmetic.ps
partials = [] # sorted, non-overlapping partial sums
for x in iterable:
i = 0
for y in partials:
if abs(x) < abs(y):
x, y = y, x
hi = x + y
lo = y - (hi - x)
if lo:
partials[i] = lo
i += 1
x = hi
partials[i:] = [x]
return sum(partials, 0.0)
It works by maintaining a hi/lo partial summations so that msum([.1]*10) produces 1.0 exactly rather than 0.9999999999999999. The C equivalent of msum is part of the math library in Python.
I have attempted to replicate in swift:
func msum(it:[Double])->Double {
// Full precision summation using multiple floats for intermediate values
var partials=[Double]()
for var x in it {
var i=0
for var y in partials{
if abs(x) < abs(y){
(x, y)=(y, x)
}
let hi=x+y
let lo=y-(hi-x)
if abs(lo)>0.0 {
partials[i]=lo
i+=1
}
x=hi
}
// slow part trying to replicate Python's slice assignment partials[i:]=[x]
if partials.endIndex>i {
partials[i...partials.endIndex-1]=[x]
}
else {
partials.append(x)
}
}
return partials.reduce(0.0, combine: +)
}
Test the function and speed:
import Foundation
var arr=[Double]()
for _ in 1...1000000 {
arr+=[10, 1e100, 10, -1e100]
}
print(arr.reduce(0, combine: +)) // will be 0.0
var startTime: CFAbsoluteTime!
startTime = CFAbsoluteTimeGetCurrent()
print(msum(arr), arr.count*5) // should be arr.count * 5
print(CFAbsoluteTimeGetCurrent() - startTime)
On my machine, that takes 7 seconds to complete. Python native msum takes 2.2 seconds (about 4x faster) and the library fsum function takes 0.09 seconds (almost 90x faster)
I have tried to replace partials[i...partials.endIndex-1]=[x] with arr.removeRange(i..<arr.endIndex) and then appending. Little faster but not much.
Question:
Is this idiomatic swift: partials[i...partials.endIndex-1]=[x]
Is there a faster / better way?

First (as already said in the comments), there is a huge
difference between non-optimized and optimised code in Swift
("-Onone" vs "-O" compiler option, or Debug vs. Release configuration), so for performance test make sure that the "Release" configuration
is selected. ("Release" is also the default configuration if you
profile the code with Instruments).
It has some advantages to use half-open ranges:
var arr = [0,1,2,3,4,5]
arr[2 ..< arr.endIndex] = [99]
print(arr) // [0, 1, 99]
In fact, that's how a range is stored internally, and it allows you
to insert a slice at the end of the array (but not beyond that as in Python):
var arr = [Int]()
arr[0 ..< arr.endIndex] = [99]
print(arr) // [99]
So
if partials.endIndex > i {
partials[i...partials.endIndex-1]=[x]
}
else {
partials.append(x)
}
is equivalent to
partials[i ..< partials.endIndex] = [x]
// Or: partials.replaceRange(i ..< partials.endIndex, with: [x])
However, that is not a performance improvement. It seems that
replacing a slice is slow in Swift. Truncating the array and
appending the new element with
partials.replaceRange(i ..< partials.endIndex, with: [])
partials.append(x)
reduced the time for your test code from about 1.25 to 0.75 seconds on my
computer.

As #MartinR points out, replaceRange is faster than slice assignment.
If you want maximum speed (based on my tests), your best bet is probably:
partials.replaceRange(i..<partials.endIndex, with: CollectionOfOne(x))
CollectionOfOne is faster than [x] because it just stores the element inline within the struct, rather than allocating memory like an array.

Find lists items closest to certain values in Python

I have a list of sorted floats y, as well as a list of unsorted floats x.
Now, I need to find out for every element in x between which values of y it lies, preferably by index of y. So for example, if
y=[1,2,3,4,5]
x[0]=3.5
I would need the output for index 0 of x to be (2,3), because 3.5 is between y[2] and y[3].
Basically, it is the same as seeing y as bin edges and sorting x to those bins, I guess.
What would be the easiest way yo accomplish that?

I would use zip (itertools.izip in Python 2.x) to accomplish this:
from itertools import islice#, izip as zip # if Python 2.x
def nearest_neighbours(x, lst):
for l1, l2 in zip(lst, islice(lst, 1, None)):
if l1 <= x <= l2:
return l1, l2
else:
# ?
Example usage:
>>> nearest_neighbours(3.5, range(1, 6))
(3, 4)
You will have to decide what you want to happen if x isn't between any pair in lst (i.e. replace # ?!) If you want indices (although your example isn't using them), have a play with enumerate.

Thanks - I'm aware of how to code that step-by-step. However, I was looking for a pretty/easy/elegant solution and now I am using numpy.digitize(), wich looks pretty to me and works nicely.

Q: What would be the easiest way yo accomplish that?
Instead of giving you the code, I think you should see this pseudo-code and try to write your own code! Don't just copy paste code from the internet, if you want to educate yourself!
Pseudocode:
// Assume that when you have a tie,
// you put the number in the smallest range
// Here, b is between 2.1 and 3.5, instead of
// 3.5 and 4.1
float a[5] = {0.1, 1.1, 2.1, 3.5, 4.1}; // your y
float b = 3.5; // your x
// counter for the loop and indexes. Init i to second element
integer i = 1, prev = -1, next;
// while we are not in the end of the array
while(i < 5) {
// if b is in the range of ( a(i-1), a(i) ]
if(b <= a[i] && b > a[i - 1]) {
// mark the indexes
prev = i - 1;
next = i;
}
// go to next element
i++;
}
if(prev = -1)
print "Number is not between some numbers"
else
print "prev, next"
I think that this can make you understand the point and then be able to select the easiest way for you.

KenKen puzzle addends: REDUX A (corrected) non-recursive algorithm

This question relates to those parts of the KenKen Latin Square puzzles which ask you to find all possible combinations of ncells numbers with values x such that 1 <= x <= maxval and x(1) + ... + x(ncells) = targetsum. Having tested several of the more promising answers, I'm going to award the answer-prize to Lennart Regebro, because:
his routine is as fast as mine (+-5%), and
he pointed out that my original routine had a bug somewhere, which led me to see what it was really trying to do. Thanks, Lennart.
chrispy contributed an algorithm that seems equivalent to Lennart's, but 5 hrs later, sooo, first to the wire gets it.
A remark: Alex Martelli's bare-bones recursive algorithm is an example of making every possible combination and throwing them all at a sieve and seeing which go through the holes. This approach takes 20+ times longer than Lennart's or mine. (Jack up the input to max_val = 100, n_cells = 5, target_sum = 250 and on my box it's 18 secs vs. 8+ mins.) Moral: Not generating every possible combination is good.
Another remark: Lennart's and my routines generate the same answers in the same order. Are they in fact the same algorithm seen from different angles? I don't know.
Something occurs to me. If you sort the answers, starting, say, with (8,8,2,1,1) and ending with (4,4,4,4,4) (what you get with max_val=8, n_cells=5, target_sum=20), the series forms kind of a "slowest descent", with the first ones being "hot" and the last one being "cold" and the greatest possible number of stages in between. Is this related to "informational entropy"? What's the proper metric for looking at it? Is there an algorithm that producs the combinations in descending (or ascending) order of heat? (This one doesn't, as far as I can see, although it's close over short stretches, looking at normalized std. dev.)
Here's the Python routine:
#!/usr/bin/env python
#filename: makeAddCombos.07.py -- stripped for StackOverflow
def initialize_combo( max_val, n_cells, target_sum):
"""returns combo
Starting from left, fills combo to max_val or an intermediate value from 1 up.
E.g.: Given max_val = 5, n_cells=4, target_sum = 11, creates [5,4,1,1].
"""
combo = []
#Put 1 in each cell.
combo += [1] * n_cells
need = target_sum - sum(combo)
#Fill as many cells as possible to max_val.
n_full_cells = need //(max_val - 1)
top_up = max_val - 1
for i in range( n_full_cells): combo[i] += top_up
need = target_sum - sum(combo)
# Then add the rest to next item.
if need > 0:
combo[n_full_cells] += need
return combo
#def initialize_combo()
def scrunch_left( combo):
"""returns (new_combo,done)
done Boolean; if True, ignore new_combo, all done;
if Falso, new_combo is valid.
Starts a new combo list. Scanning from right to left, looks for first
element at least 2 greater than right-end element.
If one is found, decrements it, then scrunches all available counts on its
right up against its right-hand side. Returns the modified combo.
If none found, (that is, either no step or single step of 1), process
done.
"""
new_combo = []
right_end = combo[-1]
length = len(combo)
c_range = range(length-1, -1, -1)
found_step_gt_1 = False
for index in c_range:
value = combo[index]
if (value - right_end) > 1:
found_step_gt_1 = True
break
if not found_step_gt_1:
return ( new_combo,True)
if index > 0:
new_combo += combo[:index]
ceil = combo[index] - 1
new_combo += [ceil]
new_combo += [1] * ((length - 1) - index)
need = sum(combo[index:]) - sum(new_combo[index:])
fill_height = ceil - 1
ndivf = need // fill_height
nmodf = need % fill_height
if ndivf > 0:
for j in range(index + 1, index + ndivf + 1):
new_combo[j] += fill_height
if nmodf > 0:
new_combo[index + ndivf + 1] += nmodf
return (new_combo, False)
#def scrunch_left()
def make_combos_n_cells_ge_two( combos, max_val, n_cells, target_sum):
"""
Build combos, list of tuples of 2 or more addends.
"""
combo = initialize_combo( max_val, n_cells, target_sum)
combos.append( tuple( combo))
while True:
(combo, done) = scrunch_left( combo)
if done:
break
else:
combos.append( tuple( combo))
return combos
#def make_combos_n_cells_ge_two()
if __name__ == '__main__':
combos = []
max_val = 8
n_cells = 5
target_sum = 20
if n_cells == 1: combos.append( (target_sum,))
else:
combos = make_combos_n_cells_ge_two( combos, max_val, n_cells, target_sum)
import pprint
pprint.pprint( combos)

Your algorithm seems pretty good at first blush, and I don't think OO or another language would improve the code. I can't say if recursion would have helped but I admire the non-recursive approach. I bet it was harder to get working and it's harder to read but it likely is more efficient and it's definitely quite clever. To be honest I didn't analyze the algorithm in detail but it certainly looks like something that took a long while to get working correctly. I bet there were lots of off-by-1 errors and weird edge cases you had to think through, eh?
Given all that, basically all I tried to do was pretty up your code as best I could by replacing the numerous C-isms with more idiomatic Python-isms. Often times what requires a loop in C can be done in one line in Python. Also I tried to rename things to follow Python naming conventions better and cleaned up the comments a bit. Hope I don't offend you with any of my changes. You can take what you want and leave the rest. :-)
Here are the notes I took as I worked:
Changed the code that initializes tmp to a bunch of 1's to the more idiomatic tmp = [1] * n_cells.
Changed for loop that sums up tmp_sum to idiomatic sum(tmp).
Then replaced all the loops with a tmp = <list> + <list> one-liner.
Moved raise doneException to init_tmp_new_ceiling and got rid of the succeeded flag.
The check in init_tmp_new_ceiling actually seems unnecessary. Removing it, the only raises left were in make_combos_n_cells, so I just changed those to regular returns and dropped doneException entirely.
Normalized mix of 4 spaces and 8 spaces for indentation.
Removed unnecessary parentheses around your if conditions.
tmp[p2] - tmp[p1] == 0 is the same thing as tmp[p2] == tmp[p1].
Changed while True: if new_ceiling_flag: break to while not new_ceiling_flag.
You don't need to initialize variables to 0 at the top of your functions.
Removed combos list and changed function to yield its tuples as they are generated.
Renamed tmp to combo.
Renamed new_ceiling_flag to ceiling_changed.
And here's the code for your perusal:
def initial_combo(ceiling=5, target_sum=13, num_cells=4):
"""
Returns a list of possible addends, probably to be modified further.
Starts a new combo list, then, starting from left, fills items to ceiling
or intermediate between 1 and ceiling or just 1. E.g.:
Given ceiling = 5, target_sum = 13, num_cells = 4: creates [5,5,2,1].
"""
num_full_cells = (target_sum - num_cells) // (ceiling - 1)
combo = [ceiling] * num_full_cells \
+ [1] * (num_cells - num_full_cells)
if num_cells > num_full_cells:
combo[num_full_cells] += target_sum - sum(combo)
return combo
def all_combos(ceiling, target_sum, num_cells):
# p0 points at the rightmost item and moves left under some conditions
# p1 starts out at rightmost items and steps left
# p2 starts out immediately to the left of p1 and steps left as p1 does
# So, combo[p2] and combo[p1] always point at a pair of adjacent items.
# d combo[p2] - combo[p1]; immediate difference
# cd combo[p2] - combo[p0]; cumulative difference
# The ceiling decreases by 1 each iteration.
while True:
combo = initial_combo(ceiling, target_sum, num_cells)
yield tuple(combo)
ceiling_changed = False
# Generate all of the remaining combos with this ceiling.
while not ceiling_changed:
p2, p1, p0 = -2, -1, -1
while combo[p2] == combo[p1] and abs(p2) <= num_cells:
# 3,3,3,3
if abs(p2) == num_cells:
return
p2 -= 1
p1 -= 1
p0 -= 1
cd = 0
# slide_ptrs_left loop
while abs(p2) <= num_cells:
d = combo[p2] - combo[p1]
cd += d
# 5,5,3,3 or 5,5,4,3
if cd > 1:
if abs(p2) < num_cells:
# 5,5,3,3 --> 5,4,4,3
if d > 1:
combo[p2] -= 1
combo[p1] += 1
# d == 1; 5,5,4,3 --> 5,4,4,4
else:
combo[p2] -= 1
combo[p0] += 1
yield tuple(combo)
# abs(p2) == num_cells; 5,4,4,3
else:
ceiling -= 1
ceiling_changed = True
# Resume at make_combo_same_ceiling while
# and follow branch.
break
# 4,3,3,3 or 4,4,3,3
elif cd == 1:
if abs(p2) == num_cells:
return
p1 -= 1
p2 -= 1
if __name__ == '__main__':
print list(all_combos(ceiling=6, target_sum=12, num_cells=4))

First of all, I'd use variable names that mean something, so that the code gets comprehensible. Then, after I understood the problem, it's clearly a recursive problem, as once you have chosen one number, the question of finding the possible values for the rest of the squares are exactly the same problem, but with different values in.
So I would do it like this:
from __future__ import division
from math import ceil
def make_combos(max_val,target_sum,n_cells):
combos = []
# The highest possible value of the next cell is whatever is
# largest of the max_val, or the target_sum minus the number
# of remaining cells (as you can't enter 0).
highest = min(max_val, target_sum - n_cells + 1)
# The lowest is the lowest number you can have that will add upp to
# target_sum if you multiply it with n_cells.
lowest = int(ceil(target_sum/n_cells))
for x in range(highest, lowest-1, -1):
if n_cells == 1: # This is the last cell, no more recursion.
combos.append((x,))
break
# Recurse to get the next cell:
# Set the max to x (or we'll get duplicates like
# (6,3,2,1) and (6,2,3,1), which is pointless.
# Reduce the target_sum with x to keep the sum correct.
# Reduce the number of cells with 1.
for combo in make_combos(x, target_sum-x, n_cells-1):
combos.append((x,)+combo)
return combos
if __name__ == '__main__':
import pprint
# And by using pprint the output gets easier to read
pprint.pprint(make_combos( 6,12,4))
I also notice that your solution still seems buggy. For the values max_val=8, target_sum=20 and n_cells=5 your code doesn't find the solution (8,6,4,1,1,), as an example. I'm not sure if that means I've missed a rule in this or not, but as I understand the rules that should be a valid option.
Here's a version using generators, It saves a couple of lines, and memory if the values are really big, but as recursion, generators can be tricky to "get".
from __future__ import division
from math import ceil
def make_combos(max_val,target_sum,n_cells):
highest = min(max_val, target_sum - n_cells + 1)
lowest = int(ceil(target_sum/n_cells))
for x in xrange(highest, lowest-1, -1):
if n_cells == 1:
yield (x,)
break
for combo in make_combos(x, target_sum-x, n_cells-1):
yield (x,)+combo
if __name__ == '__main__':
import pprint
pprint.pprint(list(make_combos( 6,12,4)))

Here's the simplest recursive solution that I can think of to "find all possible combinations of n numbers with values x such that 1 <= x <= max_val and x(1) + ... + x(n) = target". I'm developing it from scratch. Here's a version without any optimization at all, just for simplicity:
def apcnx(n, max_val, target, xsofar=(), sumsofar=0):
if n==0:
if sumsofar==target:
yield xsofar
return
if xsofar:
minx = xsofar[-1] - 1
else:
minx = 0
for x in xrange(minx, max_val):
for xposs in apcnx(n-1, max_val, target, xsofar + (x+1,), sumsofar+x+1):
yield xposs
for xs in apcnx(4, 6, 12):
print xs
The base case n==0 (where we can't yield any more numbers) either yield the tuple so far if it satisfies the condition, or nothing, then finishes (returns).
If we're supposed to yield longer tuples than we've built so far, the if/else makes sure we only yield non-decreasing tuples, to avoid repetition (you did say "combination" rather than "permutation").
The for tries all possibilities for "this" item and loops over whatever the next-lower-down level of recursion is still able to yield.
The output I see is:
(1, 1, 4, 6)
(1, 1, 5, 5)
(1, 2, 3, 6)
(1, 2, 4, 5)
(1, 3, 3, 5)
(1, 3, 4, 4)
(2, 2, 2, 6)
(2, 2, 3, 5)
(2, 2, 4, 4)
(2, 3, 3, 4)
(3, 3, 3, 3)
which seems correct.
There are a bazillion possible optimizations, but, remember:
First make it work, then make it fast
I corresponded with Kent Beck to properly attribute this quote in "Python in a Nutshell", and he tells me he got it from his dad, whose job was actually unrelated to programming;-).
In this case, it seems to me that the key issue is understanding what's going on, and any optimization might interfere, so I'm going all out for "simple and understandable"; we can, if need be!, optimize the socks off it once the OP confirms they can understand what's going on in this sheer, unoptimized version!

Sorry to say, your code is kind of long and not particularly readable. If you can try to summarize it somehow, maybe someone can help you write it more clearly.
As for the problem itself, my first thought would be to use recursion. (For all I know, you're already doing that. Sorry again for my inability to read your code.) Think of a way that you can reduce the problem to a smaller easier version of the same problem, repeatedly, until you have a trivial case with a very simple answer.
To be a bit more concrete, you have these three parameters, max_val, target_sum, and n_cells. Can you set one of those numbers to some particular value, in order to give you an extremely simple problem requiring no thought at all? Once you have that, can you reduce the slightly harder version of the problem to the already solved one?
EDIT: Here is my code. I don't like the way it does de-duplication. I'm sure there's a more Pythonic way. Also, it disallows using the same number twice in one combination. To undo this behavior, just take out the line if n not in numlist:. I'm not sure if this is completely correct, but it seems to work and is (IMHO) more readable. You could easily add memoization and that would probably speed it up quite a bit.
def get_combos(max_val, target, n_cells):
if target <= 0:
return []
if n_cells is 1:
if target > max_val:
return []
else:
return [[target]]
else:
combos = []
for n in range(1, max_val+1, 1):
for numlist in get_combos(max_val, target-n, n_cells-1):
if n not in numlist:
combos.append(numlist + [n])
return combos
def deduplicate(combos):
for numlist in combos:
numlist.sort()
answer = [tuple(numlist) for numlist in combos]
return set(answer)
def kenken(max_val, target, n_cells):
return deduplicate(get_combos(max_val, target, n_cells))

First of all, I am learning Python myself so this solution won't be great but this is just an attempt at solving this. I have tried to solve it recursively and I think a recursive solution would be ideal for this kind of problem although THAT recursive solution might not be this one:
def GetFactors(maxVal, noOfCells, targetSum):
l = []
while(maxVal != 0):
remCells = noOfCells - 1
if(remCells > 2):
retList = GetFactors(maxVal, remCells, targetSum - maxVal)
#Append the returned List to the original List
#But first, add the maxVal to the start of every elem of returned list.
for i in retList:
i.insert(0, maxVal)
l.extend(retList)
else:
remTotal = targetSum - maxVal
for i in range(1, remTotal/2 + 1):
itemToInsert = remTotal - i;
if (i > maxVal or itemToInsert > maxVal):
continue
l.append([maxVal, i, remTotal - i])
maxVal -= 1
return l
if __name__ == "__main__":
l = GetFactors(5, 5, 15)
print l

Here a simple solution in C/C++:
const int max = 6;
int sol[N_CELLS];
void enum_solutions(int target, int n, int min) {
if (target == 0 && n == 0)
report_solution(); /* sol[0]..sol[N_CELLS-1] is a solution */
if (target <= 0 || n == 0) return; /* nothing further to explore */
sol[n - 1] = min; /* remember */
for (int i = min; i <= max; i++)
enum_solutions(target - i, n - 1, i);
}
enum_solutions(12, 4, 1);

Little bit offtopic, but still might help at programming kenken.
I got good results using DLX algorhitm for solving Killer Sudoku (very simmilar as KenKen it has cages, but only sums). It took less than second for most of problems and it was implemented in MATLAB language.
reference this forum
http://www.setbb.com/phpbb/viewtopic.php?t=1274&highlight=&mforum=sudoku
killer sudoku
"look at wikipedia, cant post hyper link" damt spammers

Here is a naive, but succinct, solution using generators:
def descending(v):
"""Decide if a square contains values in descending order"""
return list(reversed(v)) == sorted(v)
def latinSquares(max_val, target_sum, n_cells):
"""Return all descending n_cells-dimensional squares,
no cell larger than max_val, sum equal to target_sum."""
possibilities = itertools.product(range(1,max_val+1),repeat=n_cells)
for square in possibilities:
if descending(square) and sum(square) == target_sum:
yield square
I could have optimized this code by directly enumerating the list of descending grids, but I find itertools.product much clearer for a first-pass solution. Finally, calling the function:
for m in latinSquares(6, 12, 4):
print m

And here is another recursive, generator-based solution, but this time using some simple math to calculate ranges at each step, avoiding needless recursion:
def latinSquares(max_val, target_sum, n_cells):
if n_cells == 1:
assert(max_val >= target_sum >= 1)
return ((target_sum,),)
else:
lower_bound = max(-(-target_sum / n_cells), 1)
upper_bound = min(max_val, target_sum - n_cells + 1)
assert(lower_bound <= upper_bound)
return ((v,) + w for v in xrange(upper_bound, lower_bound - 1, -1)
for w in latinSquares(v, target_sum - v, n_cells - 1))
This code will fail with an AssertionError if you supply parameters that are impossible to satisfy; this is a side-effect of my "correctness criterion" that we never do an unnecessary recursion. If you don't want that side-effect, remove the assertions.
Note the use of -(-x/y) to round up after division. There may be a more pythonic way to write that. Note also I'm using generator expressions instead of yield.
for m in latinSquares(6,12,4):
print m