there is an issue that i am trying to solve which requires me to generate the indices for an n - dimensional list. Eg: [5, 4, 3] is a 3 dimensional list so valid indices are [0, 0, 0], [0, 0, 1], [0, 1, 0] ... [2, 2, 1] ..... [4, 3, 2]. The best I could come up with a recursive algorithm but this isn't constant space
def f1(dims):
def recur(res, lst, depth, dims):
if depth == len(dims):
res.append(lst[::])
return
curr = dims[depth]
for i in range(curr):
lst[depth] = i
recur(res, lst, depth + 1, dims)
res = []
lst = [0] * len(dims)
recur(res, lst, 0, dims)
return res
the dimensions can be any number , ie: 4D, 5D, 15D etc. Each time it would be given in the form of a list . Eg: 5D would be [3,2,1,5,2] and I would need to generate all the valid indices for these while using constant space ( just while loops and indices processing ) . How would I go about generating these efficiently without the help of any in built python functions ( just while, for loops etc )
This is a working solution in constant space (a single loop variable i, and a vector idx of which modified copies are being yielded, and a single temporary length variable n to avoid calling len() on every iteration):
def all_indices(dimensions):
n = len(dimensions)
idx = [0] * n
while True:
for i in range(n):
yield tuple(idx)
if idx[i] + 1 < dimensions[i]:
idx[i] += 1
break
else:
idx[i] = 0
if not any(idx):
break
print(list(all_indices([3, 2, 1])))
Result:
[(0, 0, 0), (1, 0, 0), (2, 0, 0), (0, 0, 0), (0, 1, 0), (1, 1, 0), (2, 1, 0), (0, 1, 0), (0, 0, 0)]
As pointed out in the comments, there's duplicates there, a bit sloppy, this is cleaner:
def all_indices(dimensions):
n = len(dimensions)
idx = [0] * n
yield tuple(idx) # yield the initial 'all zeroes' state
while True:
for i in range(n):
if idx[i] + 1 < dimensions[i]:
idx[i] += 1
yield tuple(idx) # yield new states
break
else:
idx[i] = 0 # no yield, repeated state
if not any(idx):
break
print(list(all_indices([3, 2, 1])))
Alternatively, you could yield before the break instead of at the start of the loop, but I feel having the 'all zeroes' at the start looks cleaner.
The break is there to force a depth first on running through the indices, which ensures the loop never reaches 'all zeroes' again before having passed all possibilities. Try removing the break and then passing in something like [2, 1, 2] and you'll find it is missing a result.
I think a break is actually the 'clean' way to do it, since it allows using a simple for instead of using a while with a more complicated condition and a separate increment statement. You could do that though:
def all_indices3(dimensions):
n = len(dimensions)
idx = [1] + [0] * (n - 1)
yield tuple([0] * n)
while any(idx):
yield tuple(idx)
i = 0
while i < n and idx[i] + 1 == dimensions[i]:
idx[i] = 0
i += 1 % n
if i < n:
idx[i] += 1
This has the same result, but only uses while, if and yields the results in the same order.
My problem is: given an array of negative and positive integers. You are given value j to jump and r to rest. After each jump, you need to rest for r steps. Moreover, you are allowed to move 1 more step forward even when you have the ability to jump. The problem is to minimize the sum of the array.
Ex.1 r = 2, j = 2, [5, 3, -4, -2, 1, 2, 3] => -4 + -2 + 3 = -3 (Jump 5, 3, Rest -4,-2, Jump 1,2, Rest 3)
Ex.2 r = 2, j = 3, [90, 91, 92, -2, 3, 55, 3] => -2 + 3 + 55 + 3 = 59 (Jump 90,91,92 Rest -2,3,55,3)
My Idea: I decided to use DP to solve this. This is my pseudocode.
def minimalSum (MIN, array, jump, rest, steps_left_to_jump, i):
if MIN[i] is not empty:
return MIN[i]
if i == len(array) - 1:
MIN[i] = array[i]
else:
if steps_left_to_jump == 0:
if i == 0:
MIN[i] = minimalSum(MIN, array, jump, rest, rest - 1, jump)
else:
if i + jump + 1 < len(array):
MIN[i] = array[i] + minimalSum(MIN, array, jump, rest, rest - 1, i + jump + 1)
o1 = array[i] + minimalSum(MIN, array, jump, rest, 0, i + 1)
if MIN[i] is not None:
if o1 < MIN[i]:
MIN[i] = o1
else:
MIN[i] = o1
else:
MIN[i] = array[i] + minimalSum(MIN, array, jump, rest, steps_left_to_jump - 1, i + 1)
return MIN[i]
MIN is array used to store best sums.
The problem that it does not work for all inputs, can you help me spot where I am going wrong. Consider the example
r = 2, j = 2 , [2 ,-2 ,-3,1 ,3 ,4]. The answer should be 1 (Visit 2, -2, Jump -3, Rest 4) 2-2-3+4 = 1, but my program outputs 5
Your problem seems to be in this line:
if i == 0:
MIN[i] = minimalSum(MIN, array, jump, rest, rest - 1, jump)
This prevents you from ever choosing Visit whenever i is 0, since you ALWAYS jump in your first step. I don't know about your full code, but this part should be:
if i == 0:
MIN[i] = min(minimalSum(MIN, array, jump, rest, rest - 1, jump) , # case where you jump at 0
array[0] + minimalSum(MIN, array, jump, rest, 0, 1) # case where you visit index 0
)
Also, your code give you an out-of-bounds error if jump>len(MIN)-1. If this condition is true, you know you should ALWAYS visit.
Given all of this, I'm going to write the recursive formula, you can then memoize it:
def opt_sum(array, r, j):
if j > len(array)-1:
return sum(array) # cannot jump, return sum of full array
else:
visit_val = array[0] + opt_sum(array[1:], r, j) # best sum if you choose to visit index 0
jump_val = (
sum(array[j:j+r]) # total mandatory resting visits
+ opt_sum(array[j+r:], r, j) # the optimal sum of the subarray
) # starting after the rest
return min(visit_val, jump_val) # return best of both policies
For instance, given list [1, 0, 1] the code would return [1,1,0]. Other examples:
[1,1,1] -- > [1,0,0,0]
[1,0,0,1] --> [1,0,1,0]
I'm having most trouble understanding what my base case for recursion would be and then how to implement for the (n-1) case.
def increment_helper(number):
newNum = []
if len(number) ==1:
if number[0] == 1:
carry = 1
newNum.append(0)
else:
carry = 0
newNum.append(1)
else:
return increment_helper(number-1)
return newNum
So I'm sure that there are a lot of errors in here specifically how I am calling my recursion because I am not sure how to recurse on the list while storing the number that is removed somehow. The else return statement is obviously incorrect but I am using that as a placeholder. I am unsure of what condition to use as my base case for incrementation. I think I should be using a carry variable that keeps track of whether I am carrying a one over but other than that I am stuck on how to proceed.
Ah, ha! Okay, you have some idea of what you're doing. The basic outline is
Base case: how do I know when I'm done?
You're done when you run out of digits. number should be a list of individual digits; check its length to figure out when not to recur.
Recursion case: what next?
The general concept of recursion is "do something simple, reduce the problem by a small amount, and recur with the smaller problem." Your job in this part is to do the addition for one digit. If you need to keep going (is there a carry from that digit?), then recur. Otherwise, you have all the info you need to finish.
Specific application
Your recursion step will involve calling increment_helper with one digit less: not number - 1, but number[:-1].
After you return from each recursion, you'lll then want to append the digit you just finished. For instance, if you're incrementing 1101, your first call will see that the right-hand one, incremented, has a carry. The new digit is 0, and you have to recur. Hold onto the 0 for a moment, call yourself with 110, and get the result of that call. Append your saved 0 to that, and return to your main program.
Does that get you moving?
This one is "tricky" because you have two things going at each step:
what is the current number I'm looking at? (binary question 0/1)
should it be incremented? (do we have carry?) (binary question yes/no)
This leads to 4 cases.
There is the extra "case" of did we even get a list but it isn't that interesting.
So I would describe it as follows:
if not carry:
# technically meaningless so just return the entire list immediately
return list
# we have "carry", things will change
if not list:
# assumes [] == [0]
return [1]
if list[0]:
# step on `1` (make it 0 and carry)
return [0] + increment(rest_of_list)
# step on `0` (make it 1 and no more carry)
return [1] + rest_of_list
I strongly advise to change lists to tuples and work with these.
Also note that the recursion is on the reversed list, so apply as follows:
def increment_helper(lst, carry):
if not carry:
return lst
if not lst:
return [1]
if lst[0]:
return [0] + increment_helper(lst[1:], True)
return [1] + lst[1:]
def increment(lst):
# long but really just reverse input and output
return increment_helper(lst[::-1], True)[::-1]
I used some shortcuts by returning lists immediately (short-circuiting), but you can make it more "pure" by carrying on the recursion even without carry.
Another recursive approach.
Is the current input an empty list?
If yes return [1]
If no, continue
Is the sum (value) of the last element in the list and 1 greater than 1?
If so recursively call your function on the list without the last element (number_list[:-1]) and append [0] to the result.
If no, set the last element of the list to the sum.
Return the number_list
Code:
def increment_helper(number_list):
if not number_list:
return [1]
value = number_list[-1] + 1
if value > 1:
number_list = increment_helper(number_list[:-1]) + [0]
else:
number_list[-1] = value
return number_list
Example output:
numbers = [[1, 0, 1], [1,1,1], [1,0,0,1]]
for n in numbers:
print("%r ---> %r" % (n, increment_helper(n)))
#[1, 0, 1] ---> [1, 1, 0]
#[1, 1, 1] ---> [1, 0, 0, 0]
#[1, 0, 0, 1] ---> [1, 0, 1, 0]
Try this:
def add1(listNum):
if listNum.count(0):
oneArr = [[0] * (len(listNum) - 1)] + [1]
sumArr = []
for i in range(len(listNum)):
sumArr.append(sum(listNum[i], oneArr[i]))
newArr = []
for j in range(len(sumArr) - 1):
if sumArr[len(sumArr) - 1 - j] < 2:
newArr.insert(0, sumArr[len(sumArr) - 1 - j])
else:
newArr.insert(0, 1)
sumArr[len(sumArr) - 1 - j] += 1
return sumArr
else:
return [1] + [[0] * len(listNum)]
There aren't many reasons for using recursion for a program as simple as this, which is why I have chosen to provide a non-recursive solution.
In case it interests you, I've calculated the time complexity of this function and it is O(n).
It may be best to use two functions: one to check if a simple increment of the last position would suffice and another to perform the recursion should the previous attempt fail:
vals = [[1, 0, 1], [1,1,1], [1,0,0,1]]
def update_full(d):
if all(i in [1, 0] for i in d):
return d
start = [i for i, a in enumerate(d) if a > 1]
return update_full([1]+[0 if i > 1 else i for i in d] if not start[0] else [a+1 if i == start[0] -1 else 0 if i == start[0] else a for i, a in enumerate(d)])
def increment(d):
if not d[-1]:
return d[:-1]+[1]
return update_full(d[:-1]+[2])
print(list(map(increment, vals)))
Output:
[[1, 1, 0], [1, 0, 0, 0], [1, 0, 1, 0]]
You can treat the (binary) digits recursively by traversing the number from tail to head (just like addition works).
Before performing the recursion you have to check for two special cases:
No increment at the current digit is to be performed. Then just return the unmodified digits.
A single digit remains (you're at the head of the number). Then you possibly need to append the overflow to the head.
For the remaining cases you can increment the current digit and treat all digits before the current one in a recursive manner.
def bin_incr(n, incr=True):
if not incr:
return n
if len(n) == 1:
return [1, 0] if n[0] == 1 else [1]
return (
# `n[-1] == 1` denotes an overflow to the next digit.
bin_incr(n[:-1], n[-1] == 1)
+ [(n[-1] + 1) % 2]
)
I understand you don't want to use decimal addition, but if you must use big endian bit order, converting back to decimal first is probably the most practical. Otherwise, you end up with unnecessary reverse calls or awkward negative array indices
def binary (dec): # big endian bit order
if dec < 2:
return [ dec ]
else:
return binary (dec >> 1) + [ dec & 1 ]
def decimal (bin, acc = 0):
if not bin:
return acc
else:
return decimal (bin[1:], (acc << 1) + bin[0])
def increment (bin):
# sneaky cheat
return binary (decimal (bin) + 1)
for x in range (10):
print (x, binary (x), increment (binary (x)))
# 0 [0] [1]
# 1 [1] [1, 0]
# 2 [1, 0] [1, 1]
# 3 [1, 1] [1, 0, 0]
# 4 [1, 0, 0] [1, 0, 1]
# 5 [1, 0, 1] [1, 1, 0]
# 6 [1, 1, 0] [1, 1, 1]
# 7 [1, 1, 1] [1, 0, 0, 0]
# 8 [1, 0, 0, 0] [1, 0, 0, 1]
# 9 [1, 0, 0, 1] [1, 0, 1, 0]
If however you can represent your binary numbers in little endian bit order, things change. Instead of converting back to decimal, increment can be defined directly as a beautiful recursive function
def binary (dec): # little endian bit order
if dec < 2:
return [ dec ]
else:
return [ dec & 1 ] + binary (dec >> 1)
def increment (bin):
if not bin:
return [1]
elif bin[0] == 0:
return [1] + bin[1:]
else:
return [0] + increment(bin[1:])
for x in range (10):
print (x, binary (x), increment (binary (x)))
# 0 [0] [1]
# 1 [1] [0, 1]
# 2 [0, 1] [1, 1]
# 3 [1, 1] [0, 0, 1]
# 4 [0, 0, 1] [1, 0, 1]
# 5 [1, 0, 1] [0, 1, 1]
# 6 [0, 1, 1] [1, 1, 1]
# 7 [1, 1, 1] [0, 0, 0, 1]
# 8 [0, 0, 0, 1] [1, 0, 0, 1]
# 9 [1, 0, 0, 1] [0, 1, 0, 1]
Aside: converting the little endian representation back to decimal is a little different. I provide this to show that use cases for recursion exist everywhere
def decimal (bin, power = 0):
if not bin:
return 0
else:
return (bin[0] << power) + decimal (bin[1:], power + 1)
This part of the answer gives you cake and allows you to eat it too. You get big endian bit order and a recursive increment that steps through the bits in left-to-right order – You should use either implementation above for a number of reasons, but this aims to show you that even though your problem is complex, it's still possible to think about it recursively. No reverse or arr[::-1] was misused in the making of this function.
def binary (dec): # big endian bit order
if dec < 2:
return [ dec ]
else:
return binary (dec >> 1) + [ dec & 1 ]
def increment (bin, cont = lambda b, carry: [1] + b if carry else b):
if bin == [0]:
return cont ([1], 0)
elif bin == [1]:
return cont ([0], 1)
else:
n, *rest = bin
return increment (rest, lambda b, carry:
cont ([n ^ carry] + b, n & carry))
for x in range (10):
print (x, binary (x), increment (binary (x)))
# 0 [0] [1]
# 1 [1] [1, 0]
# 2 [1, 0] [1, 1]
# 3 [1, 1] [1, 0, 0]
# 4 [1, 0, 0] [1, 0, 1]
# 5 [1, 0, 1] [1, 1, 0]
# 6 [1, 1, 0] [1, 1, 1]
# 7 [1, 1, 1] [1, 0, 0, 0]
# 8 [1, 0, 0, 0] [1, 0, 0, 1]
# 9 [1, 0, 0, 1] [1, 0, 1, 0]
We start by breaking the problem up into smaller parts; n is the first problem, and rest is the rest of the problems. But the key to thinking with continuations (like cont above) is to think big.
In this particular problem, n gets updated based on whether rest gets updated. So we immediately recur on rest and pass a continuation that will receive the result of the subproblem. Our continuation receives the answer to the subproblem b, and whether or not that subproblem results in a carry.
...
else:
n, *rest = bin
return increment (rest, lambda b, carry:
cont ([n ^ carry] + b, n & carry))
The n ^ carry and n & carry expressions determine what the answer to this subproblem is and what the next carry will be. The following truth table shows that ^ and & encodes our answer and next_carry respectively. For example, if n is 0 and carry is 1, the carry can be consumed. The answer will be [1] + the answer to the subproblem and the next carry will be 0.
n carry (answer, next_carry) n ^ carry n & carry
0 0 ([0] + b, 0) 0 0
0 1 ([1] + b, 0) 1 0
1 0 ([1] + b, 0) 1 0
1 1 ([0] + b, 1) 0 1
The base cases are simple. If the subproblem is [0], the answer is [1] and no carry of 0. If the subproblem is [1], then the answer is [0]with a carry of 1
...
if bin == [0]:
return cont ([1], 0)
elif bin == [1]:
return cont ([0], 1)
Lastly, design the default continuation – if the answer to the problem b results in a carry, simply prepend [1] to the answer, otherwise just return the answer.
cont = lambda b, carry: [1] + b if carry else b
You are asking for the increment/successor/next function that will generate a sequence of sequences. Since other have given code, I will give a general method for developing such functions.
First, develop a multiple recursion (2 or more recursive calls) for calculating, say, all sequences of the type of length N. For binary sequences (bs) in big-endian order, from N 0s to N 1s, the base case bs(0) expression is [[]], the sequence that contains the one sequence with no binary digits. The double recursion for bs(n) in terms of bs(n-1) is ([0] concatenated to all members of bs(n-1) (in order)) plus ([1] contanenated to all members of bs(n-1)).
Next, focus on the transition between the subsequences returned by adjacent recursive calls. Here there is just one: 0, 1, ..., 1 to 1, 0, ..., 0. To increment across this boundary, we need to replace 0 followed by 0 or more 1s by 1 followed by the same number of 0s. We find such breaks by scanning from the right for the first 0, as others have shown.
It turns out the every increment crosses the boundary between adjacent bs(k) calls for some value of k, which is to say, at some level of the tree of calls resulting from double recursion.
So far, that I know of, the same idea works for designing the increment function for sequences of grey codes, sequences of conbinations (n things taken k at a time), and sequences of permutations.
Note 1: the 1->0 transitions can be done 1 at a time or all at once.
Note 2: the binary bit testing and flipping is the turing machine algorithm for count + 1. Stephen Wolfram, A New Kind of Science, presents, as I remember, 3 different implementations in the TM chapter.
Note 3 (added in edit): If one switches from prepending 0 first to prepending 1 first, or from prepending to appending, or both, one gets 3 other sequence of sequences, with 3 other increment functions.
You do not need recursion in this case. Let us start with the simplest implementation:
implement a full adder which will operate on bits.
implement a ripple adder using the full adder which will operate on 2 lists of bits.
The full adder implementation is straight forrward.
def full_adder(a,b,cin):
sum_ = a^(b^cin)
cout = (a&b) | (a|b)&cin
return sum_, cout
Tests to make sure the full adder conforms to the specs:
>>> zero_one = (0,1)
>>> [full_adder(*x) for x in [(a,b,c) for a in zero_one for b in zero_one for c in zero_one]]
[(0, 0), (1, 0), (1, 0), (0, 1), (1, 0), (0, 1), (0, 1), (1, 1)]
Since the parameters of the ripple adder are lists, we need to ensure the list lengths match before the addition. This is done by padding the shorter list with leading zeros.
def ripple_adder(xs,ys):
x, y = map(len, (xs, ys))
alen = max(x, y)
ax, by = map(lambda f: f if len(f) == alen else [0]*(alen-len(f)) + f, (xs, ys))
cout = 0
res = [0]*(alen)
for i in range(alen-1, -1, -1):
a, b, cin = ax[i], by[i], cout
s, cout = full_adder(a, b, cin)
res[i] = s
if cout:
res = [1] + res
return res
Finally, we define bin_inc the binary increment function in terms of the ripple adder
def bin_inc(bin_lst):
return ripple_adder(bin_lst, [1])
>>> bin_inc([1,1,1])
[1, 0, 0, 0]
>>> bin_inc([1,0,0,1])
[1, 0, 1, 0]
Now for a solution that is simpler but requires a little insight. Consider the following
obervations for an input xs with length L and output ys
if xs is all ones, ys will have length L+1, the first element of ys will be 1 and the rest will be zeros.
if xs has a zero element, let p be the index of the last zero element in xs.
Then ys will be the list consisting of the first p elements of xs followed by a 1 followed by zeros of length L - p. That is
ys = xs[:p] + [1] + [0]*(L-p).
We can translate this to python code easily. We use python's list.index method to find the last occurence of zero by working on the reverse of the list and adjust the algorithm appropriately:
def bin_inc_2(xs):
if all(xs):
return [1] + [0]*len(xs)
p = xs[::-1].index(0)
return xs[:-p-1] + [1] + [0]*p
This works and is simpler than the ripple_adder based implementation. One minor drawback you might notice is when xs has a zero element, we traverse it to check if it is all ones, then traverse it again to find the first occurence of zero.
We can simplify the implementation and make it more pythonic atthe same time by using a try except block:
def bin_inc_3(bin_lst):
try:
p = bin_lst[::-1].index(0)
return bin_lst[:-p-1] + [1] + [0]*p
except ValueError:
return [1] + [0]*len(bin_lst)
This implementation is simple in terms of source text, and idiomatic python. Now we test it against the ripple_adder based adder to make sure it works well.
>>> z_n = (0,1)
>>> xs = [[a,b,c,d,e,f,g,h] for a in z_n for b in z_n for c in z_n for d in z_n
for e in z_n for f in z_n for g in z_n for h in z_n ]
>>> print(all(ripple_adder(x, [1]) == bin_inc_3(x) for x in xs))
True
Fantastic, it works as intended and correctly handles leading zeros as evidenced by the tests (which increments every number from 0 to 255).
There's only a need to recurse when there is a carry:
def f(n):
# Base cases
if not n:
return [1]
if n == [1]:
return [1, 0]
if n[-1] == 0:
return n[:-1] + [1]
if n[-2:] == [0, 1]:
return n[:-2] + [1, 0]
# Recurse
return f(n[:-2]) + [0, 0]
numbers = [[1, 0, 1], [1,1,1], [1,0,0,1], [1, 0, 1, 1]]
for n in numbers:
print n, f(n)
I have a nested list of 0s and ones such as :
L = [[1, 0, 1],
[0, 0, 0],
[0, 0, 1]]
I want to check if any ones are diagonal to each other. The nested list can come in any size, as long as the length of the sublists is equal to the length of the whole list (it's a square). So if I ran it on the above list, it would return False, because of two ones being diagonal.
My current code is:
for num, i in enumerate(List):
for count, m in enumerate(i):
key1 = True
key2 = True
key3 = True
key4 = True
num1 = num
num2 = num
num3 = num
num4 = num
count1 = count
count2 = count
count3 = count
count4 = count
if m == 1:
while key1 or key2 or key3 or key4:
#print(key1, key2, key3, key4)
try:
if List[num1 + 1][count1 + 1] == 1:
print(List[num1 + 1][count1 + 1])
return False
num1 += 1
count1 += 1
except IndexError:
key1 = False
try:
if List[num2 - 1][count2 + 1] == 1:
if num2 > 0:
print(List[num2 - 1][count2 + 1])
return False
num2 -= 1
count2 += 1
except IndexError:
key2 = False
try:
if List[num3 + 1][count3 - 1] == 1:
if count3 > 0:
print(List[num3 + 1][count3 - 1])
print(num3 + 1, count3 - 1)
return False
num3 += 1
count3 -= 1
except IndexError:
key3 = False
try:
if List[num4 - 1][count4 - 1] == 1:
if count4 > 0 and num4 > 0:
print(List[num4 - 1][count4 - 1])
return False
num4 -= 1
count4 -=1
except IndexError:
key4 = False
return True
The code scans the list for 1s, and when one is found, it looks at its four corners and searches them. It continues to search in the direction of the corner, moving one grid at a time. If another 1 is found, it returns false. Once all possible diagonal grids are searched (up to returning index error), the code moves to the next 1 that is found. Once every single 1 is searched and none of them find any diagonal is found, the code returns true.
It feels clunky and inefficient, but I'm not sure how to compact it. Is there a shorter way to do this?
You can access the diagonals like so:
L = [[1, 0, 1],
[0, 0, 0],
[0, 0, 1]]
diag1 = [L[i][i] for i in range(len(L))]
diag2 = [L[i][len(L[0])-i-1] for i in range(len(L))]
if diag1 == diag2:
pass #do something here
You can start by thinking of a means of naming the diagonals. For example, in a 3x3 matrix the indices (x, y) go like this:
(0,0) (1,0) (2,0)
(0,1) (1,1) (2,1)
(0,2) (1,2) (2,2)
If we follow the diagonals, their indices have a simple pattern. The diagonals that run lower left to upper right are:
Diag #1 (0,0)
Diag #2 (0,1) (1,0)
Diag #3 (0,2) (1,1) (2,0)
Diag #4 (1,2) (2,1)
Diag #5 (2,2)
From upper left to lower right they are:
#1 (0,2)
#2 (0,1) (1,2)
#3 (0,0) (1,1) (2,2)
#4 (1,0) (2,1)
#5 (2,0)
Notice that in the lower left to upper right case, every cell (x, y) on the same diagonal has the same value of x + y. In the other case, every cell on the same diagonal has the same value of x - y. Obviously this will be true for any size matrix. We can use x+y and x-y to name the individual diagonals. As we step through the matrix, each time we encounter a "1" we can immediately calculate the names of the two diagonals that intersect there.
This suggests an efficient algorithm for deciding if any two "1"s are on the same diagonal. In Python we can use two sets to keep track of the "occupied" diagonals. If we encounter a "1" on an already-occupied diagonal, we return True, otherwise false. We can step through the matrix in any order provided we visit all the elements.
def has_diagonal_ones(a):
# a is a 2-dimensional square array of 0's and 1's
lower_left_upper_right = set()
upper_left_lower_right = set()
for x in range(len(a)):
for y in range(len(a)):
if a[x][y] == 0:
continue
i_lower_to_upper = x + y
i_upper_to_lower = x - y
if i_lower_to_upper in lower_left_upper_right:
return True
if i_upper_to_lower in upper_left_lower_right:
return True
lower_left_upper_right.add(i_lower_to_upper)
upper_left_lower_right.add(i_upper_to_lower)
return False
L = [[1, 0, 1],
[0, 0, 0],
[0, 0, 1]]
print(has_diagonal_ones(L))
>>> True
you can use numpy to solve this problem.
import numpy as np
def diagonal(arr):
d = np.diag(a)
set_len = len(set(d))
d_len = len(d)
if set_len < d_len:
return False
else:
return True
a = np.array([[1, 0, 1], [0, 0, 0], [0, 0, 1]])
print(diagonal(a))
I have a lib for this. Checkout diag and adiag functions! https://github.com/dhilst/funcyou/blob/master/funcyou/indexers.py