I am a computer science student; I am studying the Algorithms course independently.
During the course, I saw this question:
Given an n-bit integer N, find a polynomial (in n) time algorithm that decides whether N is a power (that is, there are integers a and k > 1 so that a^k = N).
I thought of a first option that is exponential in n:
For all k , 1<k<N , try to divide N by k until I get result 1.
For example, if N = 27, I will start with k = 2 , because 2 doesn't divide 27, I will go to next k =3.
I will divide 27 / 3 to get 9, and divide it again until I will get 1. This is not a good solution because it is exponential in n.
My second option is using Modular arithmetic, using ak ≡ 1 mod (k+1) if gcd(a, k+1 ) = 1 (Euler's theorem). I don't know if a and k are relatively prime.
I am trying to write an algorithm, but I am struggling to do it:
function power(N)
Input: Positive integer N
Output: yes/no
Pick positive integers a_1, a_2, . . . , a_k < N at random
if (a_i)^N−1 ≡ 1 (mod N)
for all i = 1, 2, . . . , k:
return yes
else:
return no
I'm not sure if the algorithm is correct. How can I write this correctly?
Ignoring the cases when N is 0 or 1, you want to know if N is representable as a^b for a>1, b>1.
If you knew b, you could find a in O(log(N)) arithmetic operations (using binary search). Each arithmetic operation including exponentiation runs in polynomial time in log(N), so that would be polynomial too.
It's possible to bound b: it can be at most log_2(N)+1, otherwise a will be less than 2.
So simply try each b from 2 to floor(log_2(N)+1). Each try is polynomial in n (n ~= log_2(N)), and there's O(n) trials, so the resulting time is polynomial in n.
This looks like a simple math question. Suppose that we are given N = 96889010407 which is much less than Number.MAX_SAFE_INTEGER.
The question trys to figure out if N is a power where a**k === N for a > 1 and k > 1 . So we can also write it as
Math.log(a**k) === Math.log(N) yielding k*Math.log(a) === Math.log(N) yielding Math.log(a) === Math.log(N) / k where k is an Integer > 1.
Now remember the inverse logarithm. Math.log(y) = x yields y = Math.E**x.
This means we are looking for an Integer like a = Math.E**(Math.log(N) / k) for some k if exists. So start from k=2 and increment by 1.
k a = Math.E**(Math.log(N) / k)
___ _____________________________
2 311269.99599543784 -> NO
3 4592.947769836504 -> NO
4 557.9157606623403 -> NO
5 157.49069663608586 -> NO
6 67.77129015915592 -> NO
7 37.1080205641031 -> NO
8 23.62024048697092 -> NO
9 16.622531664172815 -> NO
10 12.54952973764698 -> NO
11 9.971310247420734 -> NO
12 8.232332000056601 -> NO
13 6.999999999999999 -> YES a is 7 and 96889010407 = 7^13
So for how long do we have to iterate? As long as Math.E**(Math.log(N) / k >= 2. In this case max 36 iterations since Math.E**(Math.log(96889010407) / 37 is 1.9811909632660634 and a must be an integer > 1.
This algorithm is probably the most efficient one for this job. It's time complexity is O(log2(N)) as we iterate k (the power). Had we chosen a to iterate then the time complexity would be O(sqrt(N)).
This is OK for Natural numbers but you can extend this to the Rationals as well.
Say, is 10.999671418529301 a perfect power?
All you have to do is to convert the decimal into a fraction the best way possible to get the rational form 4084101/371293 and apply both the numerator and the denominator to the mentioned algorithm above, to see if they both give the same power which in this case would be 5. 10.999671418529301 is 21^5/13^5.
Note: JS Math object is used in the example.
The number N cannot exceed 2^n. Hence you can initialize i=2, j=n and compute i^j with decreasing j until you arrive at N, then increase i and so on. A power is found in polynomial time.
E.g. with 7776 < 8192 = 2^13, you try 2^12 = 4096, then 3^12, 3^11, 3^10, 3^9, 3^8, then 4^8, 4^7, 4^6, 5^6, 5^5, 6^5 and you are done.
Generating prime numbers from 1 to n Python 3. How to improve efficiency and what is the complexity?
Input: A number, max (a large number)
Output: All the primes from 1 to max
Output is in the form of a list and will be [2,3,5,7,11,13,.......]
The code attempts to perform this task in an efficient way (least time complexity).
from math import sqrt
max = (10**6)*3
print("\nThis code prints all primes till: " , max , "\n")
list_primes=[2]
def am_i_prime(num):
"""
Input/Parameter the function takes: An integer number
Output: returns True, if the number is prime and False if not
"""
decision=True
i=0
while(list_primes[i] <= sqrt(num)): #Till sqrt(n) to save comparisons
if(num%list_primes[i]==0):
decision=False
break
#break is inserted so that we get out of comparisons faster
#Eg. for 1568, we should break from the loop as soon as we know that 1568%2==0
i+=1
return decision
for i in range(3,max,2): #starts from 3 as our list contains 2 from the beginning
if am_i_prime(i)==True:
list_primes.append(i) #if a number is found to be prime, we append it to our list of primes
print(list_primes)
How can I make this faster? Where can I improve?
What is the time complexity of this code? Which steps are inefficient?
In what ways is the Sieve of Eratosthenes more efficient than this?
Working for the first few iterations:-
We have a list_primes which contains prime numbers. It initially contains only 2.
We go to the next number, 3. Is 3 divisible by any of the numbers in list_primes? No! We append 3 to list_primes. Right now, list_primes=[2,3]
We go to the next number 4. Is 4 divisible by any of the numbers in list_primes? Yes (4 is divisible by 2). So, we don't do anything. Right now list_primes=[2,3]
We go to the next number, 5. Is 5 divisible by any of the numbers in list_primes? No! We append 5 to list_primes. Right now, list_primes=[2,3,5]
We go to the next number, 6. Is 6 divisible by any of the numbers in list_primes? Yes (6 is divisible by 2 and also divisible by 3). So, we don't do anything. Right now list_primes=[2,3,5]
And so on...
Interestingly, it takes a rather deep mathematical theorem to prove that your algorithm is correct at all. The theorem is: "For every n ≥ 2, there is a prime number between n and n^2". I know it has been proven, and much stricter bounds are proven, but I must admit I wouldn't know how to prove it myself. And if this theorem is not correct, then the loop in am_i_prime can go past the bounds of the array.
The number of primes ≤ k is O (k / log k) - this is again a very deep mathematical theorem. Again, beyond me to prove.
But anyway, there are about n / log n primes up to n, and for these primes the loop will iterate through all primes up to n^(1/2), and there are O (n^(1/2) / log n) of them.
So for the primes alone, the runtime is therefore O (n^1.5 / log^2 n), so that is a lower bound. With some effort it should be possible to prove that for all numbers, the runtime is asymptotically the same.
O (n^1.5 / log n) is obviously an upper bound, but experimentally the number of divisions to find all primes ≤ n seems to be ≤ 2 n^1.5 / log^2 n, where log is the natural logarithm.
The following rearrangement and optimization of your code will reach your maximum in nearly 1/2 the time of your original code. It combines your top level loop and predicate function into a single function to eliminate overhead and manages squares (square roots) more efficiently:
def get_primes(maximum):
primes = []
if maximum > 1:
primes.append(2)
squares = [4]
for number in range(3, maximum, 2):
i = 0
while squares[i] <= number:
if number % primes[i] == 0:
break
i += 1
else: # no break
primes.append(number)
squares.append(number * number)
return primes
maximum = 10 ** 6 * 3
print(get_primes(maximum))
However, a sieve-based algorithm will easily beat this (as it avoids division and/or multiplication). Your code has a bug: setting max = 1 will create the list [2] instead of the correct answer of an empty list. Always test both ends of your limits.
O(N**2)
Approximately speaking, the first call to am_I_prime does 1 comparison, the second does 2, ..., so the total count is 1 + 2 + ... + N, which is (N * (N-1)) / 2, which has order N-squared.
I have M intervals on the real line, each with a positive weight. I need to select N among them such that they don't overlap and give the maximum sum. How can I do that efficiently ?
If there is no subset of N non-overlapping intervals, there is no solution.
Without the non-overlap constraint, the question is trivial: pick the N largest weights. Because of the constraint, this doesn't work anymore. In my case, N and M are small (<20), but I hope that there is a more efficient solution than exhaustively trying all subsets.
You can solve it with dynamic programming. Let C(k, i) be the maximum sum of (up to) k weighted intervals, none of which has their left end less than i.
You can restrict i to be in the set of (real) start points for all the intervals, and k ranges from 0 to N.
Start by initializing C(k, max(start for start, end in interval)) to 0, and all other entries to -infinity.
Sort the intervals by start points, and iterate through them backwards.
For each interval (start, end) with weight w, and for each k:
C(start, k) = max(C(start, k), C(next(start), k), w + C(next'(end), k-1))
Here next(x) returns the smallest start point greater than x, and next'(x) returns the smallest start point greater than or equal to x. Both can be implemented by binary search (or linear scan if M is small).
Overall, this is going to take O(M*N*logM) time and O(M*N) space.
(This assumes the intervals aren't closed at both ends, so (0, 100) and (100, 200) don't overlap -- a small adjustment needs to be made if these are to be considered overlapping).
In progress:
Inspired by #PaulHankin's solution, here is a reformulation.
Sort all intervals by increasing right abscissa and iteratively find the largest possible sum up to the K-th right bound. Assume you have solved all optimization problems for the first K intervals and for all relevant N (from 0 to K).
When you take the next interval into consideration, you compute new candidate solutions as follows: for every N (up to the current nomber of intervals), lengthen all previous solutions that you can (including the empty one) without causing overlap, and keep the lenghtened solutions that are better.
Example:
The optimal sums up to the K-th right bound, by increasing N are
1: 2
2: 2 or 8 => 8 | -
3: 8 or 3 => 8 | 2 + 3 | -
4: 8 or 2 => 8 | 2 + 3 or 8 + 2 => 8 + 2 | 2 + 3 + 2 | -
5: 8 or 9 => 9 | 8 + 2 or 2 + 9 => 2 + 9 | 2 + 3 + 2 | - | -
Question spawned from this one. The problem can be formulated as follows:
Given two positive integers n and m, with m <= n, is there a way to find a suite of numbers, which cycles and covers all possible values from 0 to n?
As a basic example, if we take 3 as a number, for whatever number current between 0 and 3, we can compute the next value as:
next = (current+3) % 4
This will cycle. For instance: 1 -> 0 -> 3 -> 2 -> 1 etc. I found this solution by "chance" and it is even general ((i + n) % (n + 1) for any n), I cannot prove it mathematically. And it is a little too obvious.
Are there better ways to generate such a permutation?
I'm not sure what you intend m in the question to refer to, or how you're defining "a suite of numbers"). However, one way of getting a cycle of number is to use a recursion (or iteration) of the form:
next = f(current)
for some function f. For example, linear congruential RNGs use the iteration:
x = ( a · x + c ) mod m where 0 < a, c < m
They don't always produce all values from 0 to m-1, but under certain circumstances they do:
c and m are relatively prime
a - 1 is divisible by every prime factor of m (not including m)
if m is divisible by 4, a - 1 is divisible by 4.
(This is the Hull-Dobell theorem.)
Note that a, c == 1 satisfies the above criteria for any m. Futhermore, if m is prime, any values of a and c satisify the criteria, and if m is a power of 2, then the criteria are satisfied by any a, c such that a == 1 mod 4 and c == 1 mod 2. However, for certain values of m (eg. 6), the only value of a which will work is 1.
This might not qualify as "stateless", but I don't think that there is any strictly stateless solution; for example, you might look for some function f such that:
f(0), f(1),... f(m-1)
is a permutation of
0, 1, ..., m-1
so that you could generate the cycle by calling f(i) for successive values of i. But that's still a state, since you have to remember the last value of i you used,
Incrementing each subsequent number by any number that does not share a common prime divisor with (n-m+1) would cover the sequence (e.g. for the sequence [2-11] (10 numbers) incrementing by 3, 7, or 9 would work but 2, 4, 5, 6, and 8 would not because they share a common divisor (2 and/or 5)
EDIT
I took out the shuffling idea since it seems that you want to increment by the same number each time. If you want a truly "random" sequence that has m at the first element just take m out and place it at the beginning. I'm not sure how that helps you, though.
My knowledge of big-O is limited, and when log terms show up in the equation it throws me off even more.
Can someone maybe explain to me in simple terms what a O(log n) algorithm is? Where does the logarithm come from?
This specifically came up when I was trying to solve this midterm practice question:
Let X(1..n) and Y(1..n) contain two lists of integers, each sorted in nondecreasing order. Give an O(log n)-time algorithm to find the median (or the nth smallest integer) of all 2n combined elements. For ex, X = (4, 5, 7, 8, 9) and Y = (3, 5, 8, 9, 10), then 7 is the median of the combined list (3, 4, 5, 5, 7, 8, 8, 9, 9, 10). [Hint: use concepts of binary search]
I have to agree that it's pretty weird the first time you see an O(log n) algorithm... where on earth does that logarithm come from? However, it turns out that there's several different ways that you can get a log term to show up in big-O notation. Here are a few:
Repeatedly dividing by a constant
Take any number n; say, 16. How many times can you divide n by two before you get a number less than or equal to one? For 16, we have that
16 / 2 = 8
8 / 2 = 4
4 / 2 = 2
2 / 2 = 1
Notice that this ends up taking four steps to complete. Interestingly, we also have that log2 16 = 4. Hmmm... what about 128?
128 / 2 = 64
64 / 2 = 32
32 / 2 = 16
16 / 2 = 8
8 / 2 = 4
4 / 2 = 2
2 / 2 = 1
This took seven steps, and log2 128 = 7. Is this a coincidence? Nope! There's a good reason for this. Suppose that we divide a number n by 2 i times. Then we get the number n / 2i. If we want to solve for the value of i where this value is at most 1, we get
n / 2i ≤ 1
n ≤ 2i
log2 n ≤ i
In other words, if we pick an integer i such that i ≥ log2 n, then after dividing n in half i times we'll have a value that is at most 1. The smallest i for which this is guaranteed is roughly log2 n, so if we have an algorithm that divides by 2 until the number gets sufficiently small, then we can say that it terminates in O(log n) steps.
An important detail is that it doesn't matter what constant you're dividing n by (as long as it's greater than one); if you divide by the constant k, it will take logk n steps to reach 1. Thus any algorithm that repeatedly divides the input size by some fraction will need O(log n) iterations to terminate. Those iterations might take a lot of time and so the net runtime needn't be O(log n), but the number of steps will be logarithmic.
So where does this come up? One classic example is binary search, a fast algorithm for searching a sorted array for a value. The algorithm works like this:
If the array is empty, return that the element isn't present in the array.
Otherwise:
Look at the middle element of the array.
If it's equal to the element we're looking for, return success.
If it's greater than the element we're looking for:
Throw away the second half of the array.
Repeat
If it's less than the the element we're looking for:
Throw away the first half of the array.
Repeat
For example, to search for 5 in the array
1 3 5 7 9 11 13
We'd first look at the middle element:
1 3 5 7 9 11 13
^
Since 7 > 5, and since the array is sorted, we know for a fact that the number 5 can't be in the back half of the array, so we can just discard it. This leaves
1 3 5
So now we look at the middle element here:
1 3 5
^
Since 3 < 5, we know that 5 can't appear in the first half of the array, so we can throw the first half array to leave
5
Again we look at the middle of this array:
5
^
Since this is exactly the number we're looking for, we can report that 5 is indeed in the array.
So how efficient is this? Well, on each iteration we're throwing away at least half of the remaining array elements. The algorithm stops as soon as the array is empty or we find the value we want. In the worst case, the element isn't there, so we keep halving the size of the array until we run out of elements. How long does this take? Well, since we keep cutting the array in half over and over again, we will be done in at most O(log n) iterations, since we can't cut the array in half more than O(log n) times before we run out of array elements.
Algorithms following the general technique of divide-and-conquer (cutting the problem into pieces, solving those pieces, then putting the problem back together) tend to have logarithmic terms in them for this same reason - you can't keep cutting some object in half more than O(log n) times. You might want to look at merge sort as a great example of this.
Processing values one digit at a time
How many digits are in the base-10 number n? Well, if there are k digits in the number, then we'd have that the biggest digit is some multiple of 10k. The largest k-digit number is 999...9, k times, and this is equal to 10k + 1 - 1. Consequently, if we know that n has k digits in it, then we know that the value of n is at most 10k + 1 - 1. If we want to solve for k in terms of n, we get
n ≤ 10k+1 - 1
n + 1 ≤ 10k+1
log10 (n + 1) ≤ k + 1
(log10 (n + 1)) - 1 ≤ k
From which we get that k is approximately the base-10 logarithm of n. In other words, the number of digits in n is O(log n).
For example, let's think about the complexity of adding two large numbers that are too big to fit into a machine word. Suppose that we have those numbers represented in base 10, and we'll call the numbers m and n. One way to add them is through the grade-school method - write the numbers out one digit at a time, then work from the right to the left. For example, to add 1337 and 2065, we'd start by writing the numbers out as
1 3 3 7
+ 2 0 6 5
==============
We add the last digit and carry the 1:
1
1 3 3 7
+ 2 0 6 5
==============
2
Then we add the second-to-last ("penultimate") digit and carry the 1:
1 1
1 3 3 7
+ 2 0 6 5
==============
0 2
Next, we add the third-to-last ("antepenultimate") digit:
1 1
1 3 3 7
+ 2 0 6 5
==============
4 0 2
Finally, we add the fourth-to-last ("preantepenultimate"... I love English) digit:
1 1
1 3 3 7
+ 2 0 6 5
==============
3 4 0 2
Now, how much work did we do? We do a total of O(1) work per digit (that is, a constant amount of work), and there are O(max{log n, log m}) total digits that need to be processed. This gives a total of O(max{log n, log m}) complexity, because we need to visit each digit in the two numbers.
Many algorithms get an O(log n) term in them from working one digit at a time in some base. A classic example is radix sort, which sorts integers one digit at a time. There are many flavors of radix sort, but they usually run in time O(n log U), where U is the largest possible integer that's being sorted. The reason for this is that each pass of the sort takes O(n) time, and there are a total of O(log U) iterations required to process each of the O(log U) digits of the largest number being sorted. Many advanced algorithms, such as Gabow's shortest-paths algorithm or the scaling version of the Ford-Fulkerson max-flow algorithm, have a log term in their complexity because they work one digit at a time.
As to your second question about how you solve that problem, you may want to look at this related question which explores a more advanced application. Given the general structure of problems that are described here, you now can have a better sense of how to think about problems when you know there's a log term in the result, so I would advise against looking at the answer until you've given it some thought.
When we talk about big-Oh descriptions, we are usually talking about the time it takes to solve problems of a given size. And usually, for simple problems, that size is just characterized by the number of input elements, and that's usually called n, or N. (Obviously that's not always true-- problems with graphs are often characterized in numbers of vertices, V, and number of edges, E; but for now, we'll talk about lists of objects, with N objects in the lists.)
We say that a problem "is big-Oh of (some function of N)" if and only if:
For all N > some arbitrary N_0, there is some constant c, such that the runtime of the algorithm is less than that constant c times (some function of N.)
In other words, don't think about small problems where the "constant overhead" of setting up the problem matters, think about big problems. And when thinking about big problems, big-Oh of (some function of N) means that the run-time is still always less than some constant times that function. Always.
In short, that function is an upper bound, up to a constant factor.
So, "big-Oh of log(n)" means the same thing that I said above, except "some function of N" is replaced with "log(n)."
So, your problem tells you to think about binary search, so let's think about that. Let's assume you have, say, a list of N elements that are sorted in increasing order. You want to find out if some given number exists in that list. One way to do that which is not a binary search is to just scan each element of the list and see if it's your target number. You might get lucky and find it on the first try. But in the worst case, you'll check N different times. This is not binary search, and it is not big-Oh of log(N) because there's no way to force it into the criteria we sketched out above.
You can pick that arbitrary constant to be c=10, and if your list has N=32 elements, you're fine: 10*log(32) = 50, which is greater than the runtime of 32. But if N=64, 10*log(64) = 60, which is less than the runtime of 64. You can pick c=100, or 1000, or a gazillion, and you'll still be able to find some N that violates that requirement. In other words, there is no N_0.
If we do a binary search, though, we pick the middle element, and make a comparison. Then we throw out half the numbers, and do it again, and again, and so on. If your N=32, you can only do that about 5 times, which is log(32). If your N=64, you can only do this about 6 times, etc. Now you can pick that arbitrary constant c, in such a way that the requirement is always met for large values of N.
With all that background, what O(log(N)) usually means is that you have some way to do a simple thing, which cuts your problem size in half. Just like the binary search is doing above. Once you cut the problem in half, you can cut it in half again, and again, and again. But, critically, what you can't do is some preprocessing step that would take longer than that O(log(N)) time. So for instance, you can't shuffle your two lists into one big list, unless you can find a way to do that in O(log(N)) time, too.
(NOTE: Nearly always, Log(N) means log-base-two, which is what I assume above.)
In the following solution, all the lines with a recursive call are done on
half of the given sizes of the sub-arrays of X and Y.
Other lines are done in a constant time.
The recursive function is T(2n)=T(2n/2)+c=T(n)+c=O(lg(2n))=O(lgn).
You start with MEDIAN(X, 1, n, Y, 1, n).
MEDIAN(X, p, r, Y, i, k)
if X[r]<Y[i]
return X[r]
if Y[k]<X[p]
return Y[k]
q=floor((p+r)/2)
j=floor((i+k)/2)
if r-p+1 is even
if X[q+1]>Y[j] and Y[j+1]>X[q]
if X[q]>Y[j]
return X[q]
else
return Y[j]
if X[q+1]<Y[j-1]
return MEDIAN(X, q+1, r, Y, i, j)
else
return MEDIAN(X, p, q, Y, j+1, k)
else
if X[q]>Y[j] and Y[j+1]>X[q-1]
return Y[j]
if Y[j]>X[q] and X[q+1]>Y[j-1]
return X[q]
if X[q+1]<Y[j-1]
return MEDIAN(X, q, r, Y, i, j)
else
return MEDIAN(X, p, q, Y, j, k)
The Log term pops up very often in algorithm complexity analysis. Here are some explanations:
1. How do you represent a number?
Lets take the number X = 245436. This notation of “245436” has implicit information in it. Making that information explicit:
X = 2 * 10 ^ 5 + 4 * 10 ^ 4 + 5 * 10 ^ 3 + 4 * 10 ^ 2 + 3 * 10 ^ 1 + 6 * 10 ^ 0
Which is the decimal expansion of the number. So, the minimum amount of information we need to represent this number is 6 digits. This is no coincidence, as any number less than 10^d can be represented in d digits.
So how many digits are required to represent X? Thats equal to the largest exponent of 10 in X plus 1.
==> 10 ^ d > X
==> log (10 ^ d) > log(X)
==> d* log(10) > log(X)
==> d > log(X) // And log appears again...
==> d = floor(log(x)) + 1
Also note that this is the most concise way to denote the number in this range. Any reduction will lead to information loss, as a missing digit can be mapped to 10 other numbers. For example: 12* can be mapped to 120, 121, 122, …, 129.
2. How do you search for a number in (0, N - 1)?
Taking N = 10^d, we use our most important observation:
The minimum amount of information to uniquely identify a value in a range between 0 to N - 1 = log(N) digits.
This implies that, when asked to search for a number on the integer line, ranging from 0 to N - 1, we need at least log(N) tries to find it. Why? Any search algorithm will need to choose one digit after another in its search for the number.
The minimum number of digits it needs to choose is log(N). Hence the minimum number of operations taken to search for a number in a space of size N is log(N).
Can you guess the order complexities of binary search, ternary search or deca search? Its O(log(N))!
3. How do you sort a set of numbers?
When asked to sort a set of numbers A into an array B, here’s what it looks like ->
Permute Elements
Every element in the original array has to be mapped to it’s corresponding index in the sorted array. So, for the first element, we have n positions. To correctly find the corresponding index in this range from 0 to n - 1, we need…log(n) operations.
The next element needs log(n-1) operations, the next log(n-2) and so on. The total comes to be:
==> log(n) + log(n - 1) + log(n - 2) + … + log(1)Using log(a) + log(b) = log(a * b), ==> log(n!)
This can be approximated to nlog(n) - n. Which is O(n*log(n))!
Hence we conclude that there can be no sorting algorithm that can do better than O(n*log(n)). And some algorithms having this complexity are the popular Merge Sort and Heap Sort!
These are some of the reasons why we see log(n) pop up so often in the complexity analysis of algorithms. The same can be extended to binary numbers. I made a video on that here.
Why does log(n) appear so often during algorithm complexity analysis?
Cheers!
We call the time complexity O(log n), when the solution is based on iterations over n, where the work done in each iteration is a fraction of the previous iteration, as the algorithm works towards the solution.
Can't comment yet... necro it is!
Avi Cohen's answer is incorrect, try:
X = 1 3 4 5 8
Y = 2 5 6 7 9
None of the conditions are true, so MEDIAN(X, p, q, Y, j, k) will cut both the fives. These are nondecreasing sequences, not all values are distinct.
Also try this even-length example with distinct values:
X = 1 3 4 7
Y = 2 5 6 8
Now MEDIAN(X, p, q, Y, j+1, k) will cut the four.
Instead I offer this algorithm, call it with MEDIAN(1,n,1,n):
MEDIAN(startx, endx, starty, endy){
if (startx == endx)
return min(X[startx], y[starty])
odd = (startx + endx) % 2 //0 if even, 1 if odd
m = (startx+endx - odd)/2
n = (starty+endy - odd)/2
x = X[m]
y = Y[n]
if x == y
//then there are n-2{+1} total elements smaller than or equal to both x and y
//so this value is the nth smallest
//we have found the median.
return x
if (x < y)
//if we remove some numbers smaller then the median,
//and remove the same amount of numbers bigger than the median,
//the median will not change
//we know the elements before x are smaller than the median,
//and the elements after y are bigger than the median,
//so we discard these and continue the search:
return MEDIAN(m, endx, starty, n + 1 - odd)
else (x > y)
return MEDIAN(startx, m + 1 - odd, n, endy)
}