So the task I have to solve is to calculate the binomial coefficient for 100>=n>k>=1 and then say how many solutions for n and k are over an under barrier of 123456789.
I have no problem in my formula of calculating the binomial coefficient but for high numbers n & k -> 100 the datatypes of c get to small to calculated this.
Do you have any suggestions how I can bypass this problem with overflowing the datatypes.
I thought about dividing by the under barrier straight away so the numbers don't get too big in the first place and I have to just check if the result is >=1 but i couldn't make it work.
Say your task is to determine how many binomial coefficients C(n, k) for 1 ≤ k < n ≤ 8 exceed a limit of m = 18. You can do this by using the recurrence C(n, k) = C(n − 1, k) + C(n − 1, k − 1) that can visualized in Pascal's triangle.
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 (20) 15 6 1
1 7 (21 35 35 21) 7 1
1 8 (28 56 70 56 28) 8 1
Start at the top and work your way down. Up to n = 5, everything is below the limit of 18. On the next line, the 20 exceeds the limit. From now on, more and more coefficients are beyond 18.
The triangle is symmetric and strictly increasing in the first half of each row. You only need to find the first element that exceeds the limit on each line in order to know how many items to count.
You don't have to store the whole triangle. It is enough to keey the last and current line. Alternatively, you can use the algorithm detailed [in this article][ot] to work your way from left to right on each row. Since you just want to count the coefficients that exceed a limit and don't care about their values, the regular integer types should be sufficient.
First, you'll need a type that can handle the result. The larget number you need to handle is C(100,50) = 100,891,344,545,564,193,334,812,497,256. This number requires 97 bits of precision, so your normal data types won't do the trick. A quad precision IEEE float would do the trick if your environment provides it. Otherwise, you'll need some form of high/arbitrary precision library.
Then, to keep the numbers within this size, you'll want cancel common terms in the numerator and the denominator. And you'll want to calculate the result using ( a / c ) * ( b / d ) * ... instead of ( a * b * ... ) / ( c * d * ... ).
I'm reading this editorial here: https://www.geeksforgeeks.org/given-an-array-arr-find-the-maximum-j-i-such-that-arrj-arri/ and I cannot understand the explanation for how the O(n) solution works. The paragraph describing it seems to contradict the code. I've gone through a sample array and manually make sure this seems to work but it does not seem intuitive to me at all.
Would someone more experienced with solving programming puzzles be willing to explain how and why this works, or explain what is wrong with it?
Thank you.
(Below is text from the link above:)
Problem:
Given an array arr[], find the maximum j – i such that arr[j] > arr[i]
Given an array arr[], find the maximum j – i such that arr[j] > arr[i].
Examples :
Input: {34, 8, 10, 3, 2, 80, 30, 33, 1}
Output: 6 (j = 7, i = 1)
Input: {9, 2, 3, 4, 5, 6, 7, 8, 18, 0}
Output: 8 ( j = 8, i = 0)
Method 2 (Efficient)
To solve this problem, we need to get two optimum indexes of arr[]: left index i and right index j. For an element arr[i], we do not need to consider arr[i] for left index if there is an element smaller than arr[i] on left side of arr[i]. Similarly, if there is a greater element on right side of arr[j] then we do not need to consider this j for right index. So we construct two auxiliary arrays LMin[] and RMax[] such that LMin[i] holds the smallest element on left side of arr[i] including arr[i], and RMax[j] holds the greatest element on right side of arr[j] including arr[j]. After constructing these two auxiliary arrays, we traverse both of these arrays from left to right. While traversing LMin[] and RMa[] if we see that LMin[i] is greater than RMax[j], then we must move ahead in LMin[] (or do i++) because all elements on left of LMin[i] are greater than or equal to LMin[i]. Otherwise we must move ahead in RMax[j] to look for a greater j – i value.
It works. The code author just took a confusing shortcut.
As the editorial points out, given indices i1 < i2 with arr[i1] ≤ arr[i2], there's no point in considering i = i2. We can always do better by setting i = i1 instead, since for all j,
j - i1 > j - i2, and
if arr[j] > arr[i2], then the fact that arr[i2] ≥ arr[i1] implies that arr[j] > arr[i1].
Similarly, given indices j1 < j2 with arr[j1] ≤ arr[j2], there's no point in considering j = j1.
Let's examine the first sample input and eliminate all of these suboptimal candidates.
34 8 10 3 2 80 30 33 1
34 8 3 2 1 // candidate values of arr[i]
80 33 1 // candidate values of arr[j]
Observe that both subsequences decrease. This is a straightforward consequence of the conditions above for being a candidate.
Searching for the best i and j now involves a programming contest cliché. Let me organize the possibilities into a table. Note that the algorithm does not construct this table explicitly; it's just a visual aid.
80 33 1
-------------
34 √
8 √ √
3 √ √
2 √ √
1 √ √
The checkmark (√) means that arr[i] < arr[j]. Going down means increasing i and decreasing arr[i]. Going left means decreasing j and increasing arr[j]. Hence, wherever there is a checkmark, all squares below or to the left or some combination also have checkmarks. On the other hand, given a square with a checkmark that is below/to the left of another square with a checkmark, the latter square is necessarily a better solution, because i is less or j is greater or both. Therefore we are keenly interested in the boundary between checkmarks and no checkmarks, because that's where the optimal solution lies.
The algorithm for tracing the boundary is simple. Start in the upper left square. If the current square has a checkmark, go right. Otherwise, go down. I hope that you can get a sense of how this process visits the first checkmark in each column in time O(#rows + #columns). Here's another possible table.
33 8 7 3
---------------
34
8 √
3 √ √ √
2 √ √ √ √
1 √ √ √ √
To get to the implementation, observe that we're doing the same search but with some of the rows/columns duplicated to fill the empty space left by the non-candidates, saving us the trouble of minding the correspondence between indices. It's basically the same idea.
It's the magic of monotonic functions and the kind of insight one can get by visualising a problem's solution space and how those values align themselves. Let's plot one of the examples in the page you linked to; array indexes on the x-axis, LMin and RMax on the y-axis:
{34, 8,10, 3, 2,80,30,33, 1}
0 1 2 3 4 5 6 7 8
80 r r r r r r
.
34 l
33 r r
. ^
30
.
10
.
8 l l
. ^
3 l
2 l l l l
1 lr
0 1 2 3 4 5 6 7 8
An r value is not (necessarily) the value associated with the array index; rather, it's an indication of how far we can extend an interval to the right, guaranteeing no r on the right would be greater (meaning there might be a larger interval we'd miss). Similarly, an l is an indication that we are on the lowest left-ward value, and since we first explore moving along the rs, we are guaranteed not to have missed an earlier l for any interval in our search. Clearly, we iterate from left to right at most along all rs and all ls, so O(n).
The cycle leader iteration algorithm is an algorithm for shuffling an array by moving all even-numbered entries to the front and all odd-numbered entries to the back while preserving their relative order. For example, given this input:
a 1 b 2 c 3 d 4 e 5
the output would be
a b c d e 1 2 3 4 5
This algorithm runs in O(n) time and uses only O(1) space.
One unusual detail of the algorithm is that it works by splitting the array up into blocks of size 3k+1. Apparently this is critical for the algorithm to work correctly, but I have no idea why this is.
Why is the choice of 3k + 1 necessary in the algorithm?
Thanks!
This is going to be a long answer. The answer to your question isn't simple and requires some number theory to fully answer. I've spent about half a day working through the algorithm and I now have a good answer, but I'm not sure I can describe it succinctly.
The short version:
Breaking the input into blocks of size 3k + 1 essentially breaks the input apart into blocks of size 3k - 1 surrounded by two elements that do not end up moving.
The remaining 3k - 1 elements in the block move according to an interesting pattern: each element moves to the position given by dividing the index by two modulo 3k.
This particular motion pattern is connected to a concept from number theory and group theory called primitive roots.
Because the number two is a primitive root modulo 3k, beginning with the numbers 1, 3, 9, 27, etc. and running the pattern is guaranteed to cycle through all the elements of the array exactly once and put them into the proper place.
This pattern is highly dependent on the fact that 2 is a primitive root of 3k for any k ≥ 1. Changing the size of the array to another value will almost certainly break this because the wrong property is preserved.
The Long Version
To present this answer, I'm going to proceed in steps. First, I'm going to introduce cycle decompositions as a motivation for an algorithm that will efficiently shuffle the elements around in the right order, subject to an important caveat. Next, I'm going to point out an interesting property of how the elements happen to move around in the array when you apply this permutation. Then, I'll connect this to a number-theoretic concept called primitive roots to explain the challenges involved in implementing this algorithm correctly. Finally, I'll explain why this leads to the choice of 3k + 1 as the block size.
Cycle Decompositions
Let's suppose that you have an array A and a permutation of the elements of that array. Following the standard mathematical notation, we'll denote the permutation of that array as σ(A). We can line the initial array A up on top of the permuted array σ(A) to get a sense for where every element ended up. For example, here's an array and one of its permutations:
A 0 1 2 3 4
σ(A) 2 3 0 4 1
One way that we can describe a permutation is just to list off the new elements inside that permutation. However, from an algorithmic perspective, it's often more helpful to represent the permutation as a cycle decomposition, a way of writing out a permutation by showing how to form that permutation by beginning with the initial array and then cyclically permuting some of its elements.
Take a look at the above permutation. First, look at where the 0 ended up. In σ(A), the element 0 ended up taking the place of where the element 2 used to be. In turn, the element 2 ended up taking the place of where the element 0 used to be. We denote this by writing (0 2), indicating that 0 should go where 2 used to be, and 2 should go were 0 used to be.
Now, look at the element 1. The element 1 ended up where 4 used to be. The number 4 then ended up where 3 used to be, and the element 3 ended up where 1 used to be. We denote this by writing (1 4 3), that 1 should go where 4 used to be, that 4 should go where 3 used to be, and that 3 should go where 1 used to be.
Combining these together, we can represent the overall permutation of the above elements as (0 2)(1 4 3) - we should swap 0 and 2, then cyclically permute 1, 4, and 3. If we do that starting with the initial array, we'll end up at the permuted array that we want.
Cycle decompositions are extremely useful for permuting arrays in place because it's possible to permute any individual cycle in O(C) time and O(1) auxiliary space, where C is the number of elements in the cycle. For example, suppose that you have a cycle (1 6 8 4 2). You can permute the elements in the cycle with code like this:
int[] cycle = {1, 6, 8, 4, 2};
int temp = array[cycle[0]];
for (int i = 1; i < cycle.length; i++) {
swap(temp, array[cycle[i]]);
}
array[cycle[0]] = temp;
This works by just swapping everything around until everything comes to rest. Aside from the space usage required to store the cycle itself, it only needs O(1) auxiliary storage space.
In general, if you want to design an algorithm that applies a particular permutation to an array of elements, you can usually do so by using cycle decompositions. The general algorithm is the following:
for (each cycle in the cycle decomposition algorithm) {
apply the above algorithm to cycle those elements;
}
The overall time and space complexity for this algorithm depends on the following:
How quickly can we determine the cycle decomposition we want?
How efficiently can we store that cycle decomposition in memory?
To get an O(n)-time, O(1)-space algorithm for the problem at hand, we're going to show that there's a way to determine the cycle decomposition in O(1) time and space. Since everything will get moved exactly once, the overall runtime will be O(n) and the overall space complexity will be O(1). It's not easy to get there, as you'll see, but then again, it's not awful either.
The Permutation Structure
The overarching goal of this problem is to take an array of 2n elements and shuffle it so that even-positioned elements end up at the front of the array and odd-positioned elements end up at the end of the array. Let's suppose for now that we have 14 elements, like this:
0 1 2 3 4 5 6 7 8 9 10 11 12 13
We want to shuffle the elements so that they come out like this:
0 2 4 6 8 10 12 1 3 5 7 9 11 13
There are a couple of useful observations we can have about the way that this permutation arises. First, notice that the first element does not move in this permutation, because even-indexed elements are supposed to show up in the front of the array and it's the first even-indexed element. Next, notice that the last element does not move in this permutation, because odd-indexed elements are supposed to end up at the back of the array and it's the last odd-indexed element.
These two observations, put together, means that if we want to permute the elements of the array in the desired fashion, we actually only need to permute the subarray consisting of the overall array with the first and last elements dropped off. Therefore, going forward, we are purely going to focus on the problem of permuting the middle elements. If we can solve that problem, then we've solved the overall problem.
Now, let's look at just the middle elements of the array. From our above example, that means that we're going to start with an array like this one:
Element 1 2 3 4 5 6 7 8 9 10 11 12
Index 1 2 3 4 5 6 7 8 9 10 11 12
We want to get the array to look like this:
Element 2 4 6 8 10 12 1 3 5 7 9 11
Index 1 2 3 4 5 6 7 8 9 10 11 12
Because this array was formed by taking a 0-indexed array and chopping off the very first and very last element, we can treat this as a one-indexed array. That's going to be critically important going forward, so be sure to keep that in mind.
So how exactly can we go about generating this permutation? Well, for starters, it doesn't hurt to take a look at each element and to try to figure out where it began and where it ended up. If we do so, we can write things out like this:
The element at position 1 ended up at position 7.
The element at position 2 ended up at position 1.
The element at position 3 ended up at position 8.
The element at position 4 ended up at position 2.
The element at position 5 ended up at position 9.
The element at position 6 ended up at position 3.
The element at position 7 ended up at position 10.
The element at position 8 ended up at position 4.
The element at position 9 ended up at position 11.
The element at position 10 ended up at position 5.
The element at position 11 ended up at position 12.
The element at position 12 ended up at position 6.
If you look at this list, you can spot a few patterns. First, notice that the final index of all the even-numbered elements is always half the position of that element. For example, the element at position 4 ended up at position 2, the element at position 12 ended up at position 6, etc. This makes sense - we pushed all the even elements to the front of the array, so half of the elements that came before them will have been displaced and moved out of the way.
Now, what about the odd-numbered elements? Well, there are 12 total elements. Each odd-numbered element gets pushed to the second half, so an odd-numbered element at position 2k+1 will get pushed to at least position 7. Its position within the second half is given by the value of k. Therefore, the elements at an odd position 2k+1 gets mapped to position 7 + k.
We can take a minute to generalize this idea. Suppose that the array we're permuting has length 2n. An element at position 2x will be mapped to position x (again, even numbers get halfed), and an element at position 2x+1 will be mapped to position n + 1 + x. Restating this:
The final position of an element at position p is determined as follows:
If p = 2x for some integer x, then 2x ↦ x
If p = 2x+1 for some integer x, then 2x+1 ↦ n + 1 + x
And now we're going to do something that's entirely crazy and unexpected. Right now, we have a piecewise rule for determining where each element ends up: we either divide by two, or we do something weird involving n + 1. However, from a number-theoretic perspective, there is a single, unified rule explaining where all elements are supposed to end up.
The insight we need is that in both cases, it seems like, in some way, we're dividing the index by two. For the even case, the new index really is formed by just dividing by two. For the odd case, the new index kinda looks like it's formed by dividing by two (notice that 2x+1 went to x + (n + 1)), but there's an extra term in there. In a number-theoretic sense, though, both of these really correspond to division by two. Here's why.
Rather than taking the source index and dividing by two to get the destination index, what if we take the destination index and multiply by two? If we do that, an interesting pattern emerges.
Suppose our original number was 2x. The destination is then x, and if we double the destination index to get back 2x, we end up with the source index.
Now suppose that our original number was 2x+1. The destination is then n + 1 + x. Now, what happens if we double the destination index? If we do that, we get back 2n + 2 + 2x. If we rearrange this, we can alternatively rewrite this as (2x+1) + (2n+1). In other words, we've gotten back the original index, plus an extra (2n+1) term.
Now for the kicker: what if all of our arithmetic is done modulo 2n + 1? In that case, if our original number was 2x + 1, then twice the destination index is (2x+1) + (2n+1) = 2x + 1 (modulo 2n+1). In other words, the destination index really is half of the source index, just done modulo 2n+1!
This leads us to a very, very interesting insight: the ultimate destination of each of the elements in a 2n-element array is given by dividing that number by two, modulo 2n+1. This means that there really is a nice, unified rule for determining where everything goes. We just need to be able to divide by two modulo 2n+1. It just happens to work out that in the even case, this is normal integer division, and in the odd case, it works out to taking the form n + 1 + x.
Consequently, we can reframe our problem in the following way: given a 1-indexed array of 2n elements, how do we permute the elements so that each element that was originally at index x ends up at position x/2 mod (2n+1)?
Cycle Decompositions Revisited
At this point, we've made quite a lot of progress. Given any element, we know where that element should end up. If we can figure out a nice way to get a cycle decomposition of the overall permutation, we're done.
This is, unfortunately, where things get complicated. Suppose, for example, that our array has 10 elements. In that case, we want to transform the array like this:
Initial: 1 2 3 4 5 6 7 8 9 10
Final: 2 4 6 8 10 1 3 5 7 9
The cycle decomposition of this permutation is (1 6 3 7 9 10 5 8 4 2). If our array has 12 elements, we want to transform it like this:
Initial: 1 2 3 4 5 6 7 8 9 10 11 12
Final: 2 4 6 8 10 12 1 3 5 7 9 11
This has cycle decomposition (1 7 10 5 9 11 12 6 3 8 4 2 1). If our array has 14 elements, we want to transform it like this:
Initial: 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Final: 2 4 6 8 10 12 14 1 3 5 7 9 11 13
This has cycle decomposition (1 8 4 2)(3 9 12 6)(5 10)(7 11 13 14). If our array has 16 elements, we want to transform it like this:
Initial: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Final: 2 4 6 8 10 12 14 16 1 3 5 7 9 11 13 15
This has cycle decomposition (1 9 13 15 16 8 4 2)(3 10 5 11 14 7 12 6).
The problem here is that these cycles don't seem to follow any predictable patterns. This is a real problem if we're going to try to solve this problem in O(1) space and O(n) time. Even though given any individual element we can figure out what cycle contains it and we can efficiently shuffle that cycle, it's not clear how we figure out what elements belong to what cycles, how many different cycles there are, etc.
Primitive Roots
This is where number theory comes in. Remember that each element's new position is formed by dividing that number by two, modulo 2n+1. Thinking about this backwards, we can figure out which number will take the place of each number by multiplying by two modulo 2n+1. Therefore, we can think of this problem by finding the cycle decomposition in reverse: we pick a number, keep multiplying it by two and modding by 2n+1, and repeat until we're done with the cycle.
This gives rise to a well-studied problem. Suppose that we start with the number k and think about the sequence k, 2k, 22k, 23k, 24k, etc., all done modulo 2n+1. Doing this gives different patterns depending on what odd number 2n+1 you're modding by. This explains why the above cycle patterns seem somewhat arbitrary.
I have no idea how anyone figured this out, but it turns out that there's a beautiful result from number theory that talks about what happens if you take this pattern mod 3k for some number k:
Theorem: Consider the sequence 3s, 3s·2, 3s·22, 3s·23, 3s·24, etc. all modulo 3k for some k ≥ s. This sequence cycles through through every number between 1 and 3k, inclusive, that is divisible by 3s but not divisible by 3s+1.
We can try this out on a few examples. Let's work modulo 27 = 32. The theorem says that if we look at 3, 3 · 2, 3 · 4, etc. all modulo 27, then we should see all the numbers less than 27 that are divisible by 3 and not divisible by 9. Well, let'see what we get:
3 · 20 = 3 · 1 = 3 = 3 mod 27
3 · 21 = 3 · 2 = 6 = 6 mod 27
3 · 22 = 3 · 4 = 12 = 12 mod 27
3 · 23 = 3 · 8 = 24 = 24 mod 27
3 · 24 = 3 · 16 = 48 = 21 mod 27
3 · 25 = 3 · 32 = 96 = 15 mod 27
3 · 26 = 3 · 64 = 192 = 3 mod 27
We ended up seeing 3, 6, 12, 15, 21, and 24 (though not in that order), which are indeed all the numbers less than 27 that are divisible by 3 but not divisible by 9.
We can also try this working mod 27 and considering 1, 2, 22, 23, 24 mod 27, and we should see all the numbers less than 27 that are divisible by 1 and not divisible by 3. In other words, this should give back all the numbers less than 27 that aren't divisible by 3. Let's see if that's true:
20 = 1 = 1 mod 27
21 = 2 = 2 mod 27
22 = 4 = 4 mod 27
23 = 8 = 8 mod 27
24 = 16 = 16 mod 27
25 = 32 = 5 mod 27
26 = 64 = 10 mod 27
27 = 128 = 20 mod 27
28 = 256 = 13 mod 27
29 = 512 = 26 mod 27
210 = 1024 = 25 mod 27
211 = 2048 = 23 mod 27
212 = 4096 = 19 mod 27
213 = 8192 = 11 mod 27
214 = 16384 = 22 mod 27
215 = 32768 = 17 mod 27
216 = 65536 = 7 mod 27
217 = 131072 = 14 mod 27
218 = 262144 = 1 mod 27
Sorting these, we got back the numbers 1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20, 22, 23, 25, 26 (though not in that order). These are exactly the numbers between 1 and 26 that aren't multiples of three!
This theorem is crucial to the algorithm for the following reason: if 2n+1 = 3k for some number k, then if we process the cycle containing 1, it will properly shuffle all numbers that aren't multiples of three. If we then start the cycle at 3, it will properly shuffle all numbers that are divisible by 3 but not by 9. If we then start the cycle at 9, it will properly shuffle all numbers that are divisible by 9 but not by 27. More generally, if we use the cycle shuffle algorithm on the numbers 1, 3, 9, 27, 81, etc., then we will properly reposition all the elements in the array exactly once and will not have to worry that we missed anything.
So how does this connect to 3k + 1? Well, we need to have that 2n + 1 = 3k, so we need to have that 2n = 3k - 1. But remember - we dropped the very first and very last element of the array when we did this! Adding those back in tells us that we need blocks of size 3k + 1 for this procedure to work correctly. If the blocks are this size, then we know for certain that the cycle decomposition will consist of a cycle containing 1, a nonoverlapping cycle containing 3, a nonoverlapping cycle containing 9, etc. and that these cycles will contain all the elements of the array. Consequently, we can just start cycling 1, 3, 9, 27, etc. and be absolutely guaranteed that everything gets shuffled around correctly. That's amazing!
And why is this theorem true? It turns out that a number k for which 1, k, k2, k3, etc. mod pn that cycles through all the numbers that aren't multiples of p (assuming p is prime) is called a primitive root of the number pn. There's a theorem that says that 2 is a primitive root of 3k for all numbers k, which is why this trick works. If I have time, I'd like to come back and edit this answer to include a proof of this result, though unfortunately my number theory isn't at a level where I know how to do this.
Summary
This problem was tons of fun to work on. It involves cute tricks with dividing by two modulo an odd numbers, cycle decompositions, primitive roots, and powers of three. I'm indebted to this arXiv paper which described a similar (though quite different) algorithm and gave me a sense for the key trick behind the technique, which then let me work out the details for the algorithm you described.
Hope this helps!
Here is most of the mathematical argument missing from templatetypedef’s
answer. (The rest is comparatively boring.)
Lemma: for all integers k >= 1, we have
2^(2*3^(k-1)) = 1 + 3^k mod 3^(k+1).
Proof: by induction on k.
Base case (k = 1): we have 2^(2*3^(1-1)) = 4 = 1 + 3^1 mod 3^(1+1).
Inductive case (k >= 2): if 2^(2*3^(k-2)) = 1 + 3^(k-1) mod 3^k,
then q = (2^(2*3^(k-2)) - (1 + 3^(k-1)))/3^k.
2^(2*3^(k-1)) = (2^(2*3^(k-2)))^3
= (1 + 3^(k-1) + 3^k*q)^3
= 1 + 3*(3^(k-1)) + 3*(3^(k-1))^2 + (3^(k-1))^3
+ 3*(1+3^(k-1))^2*(3^k*q) + 3*(1+3^(k-1))*(3^k*q)^2 + (3^k*q)^3
= 1 + 3^k mod 3^(k+1).
Theorem: for all integers i >= 0 and k >= 1, we have
2^i = 1 mod 3^k if and only if i = 0 mod 2*3^(k-1).
Proof: the “if” direction follows from the Lemma. If
i = 0 mod 2*3^(k-1), then
2^i = (2^(2*3^(k-1)))^(i/(2*3^(k-1)))
= (1+3^k)^(i/(2*3^(k-1))) mod 3^(k+1)
= 1 mod 3^k.
The “only if” direction is by induction on k.
Base case (k = 1): if i != 0 mod 2, then i = 1 mod 2, and
2^i = (2^2)^((i-1)/2)*2
= 4^((i-1)/2)*2
= 2 mod 3
!= 1 mod 3.
Inductive case (k >= 2): if 2^i = 1 mod 3^k, then
2^i = 1 mod 3^(k-1), and the inductive hypothesis implies that
i = 0 mod 2*3^(k-2). Let j = i/(2*3^(k-2)). By the Lemma,
1 = 2^i mod 3^k
= (1+3^(k-1))^j mod 3^k
= 1 + j*3^(k-1) mod 3^k,
where the dropped terms are divisible by (3^(k-1))^2, so
j = 0 mod 3, and i = 0 mod 2*3^(k-1).
I have a Sorted array .Lets assume
{4,7,9,12,23,34,56,78} Given min and max I want to find elements in array between min and max in efficient way.
Cases:min=23 and max is 78 op:{23,34,56,78}
min =10 max is 65 op:{12,23,34,56}
min 0 and max is 100 op:{4,7,9,12,23,34,56,78}
Min 30 max= 300:{34,56,78}
Min =100 max=300 :{} //empty
I want to find efficient way to do this?I am not asking code any algorithm which i can use here like DP exponential search?
Since it's sorted, you can easily find the lowest element greater than or equal to the minimum desired, by using a binary search over the entire array.
A binary search basically reduces the serch space by half with each iteration. Given your first example of 10, you start as follows with the midpoint on the 12:
0 1 2 3 4 5 6 7 <- index
4 7 9 12 23 34 56 78
^^
Since the element you're looking at is higher than 10 and the next lowest is lesser, you've found it.
Then, you can use a similar binary search but only over that section from the element you just found to the end. This time you're looking for the highest element less than or equal to the maximum desired.
On the same example as previously mentioned, you start with:
3 4 5 6 7 <- index
12 23 34 56 78
^^
Since that's less than 65 and the following one is also, you need to increase the pointer to the halfway point of 34..78:
3 4 5 6 7 <- index
12 23 34 56 78
^^
And there you have it, because that number is less and the following number is more (than 65)
Then you have the start at stop indexes (3 and 6) for extracting the values.
0 1 2 3 4 5 6 7 <- index
4 7 9 ((12 23 34 56)) 78
-----------
The time complexity of the algorithm is O(log N). Though keep in mind that this really only becomes important when dealing with larger data sets. If your data sets do consist of only about eight elements, you may as well use a linear search since (1) it'll be easier to write; and (2) the time differential will be irrelevant.
I tend not to worry about time complexity unless the operations are really expensive, the data set size gets into the thousands, or I'm having to do it thousands of times a second.
Since it is sorted, this should do:
List<Integer> subarray = new ArrayList<Integer>();
for (int n : numbers) {
if (n >= MIN && n <= MAX) subarray.add(n);
}
It's O(n) as you only look at every number once.
I am new to MatLab and programming in general. I have been set the following problem:
4 players take part in a competition each person starts with 100 points.
Each person randomly plays another player.
If a player wins they get 1 point and the losing player loses 1 point from their total.
When a player has zero points they are elimated.
The game is over when there is only 1 player left.
I assume I need to create some sort of matrix array with 4 players and their 4 scores
A = [1 2 3 4; 100 100 100 100]
Is this correct and how would i remove a player and their score once it reaches zero.
Any help would be greatly recieved. Thanks
I think you are on the right track, assuming you define A as such, you can remove the players with nonpositive score as such:
A = [1 2 3 4; 100 100 0 100]
idx = A(2,:)<0.01 %To prevent suffering from rounding errors
A = A(:,~idx)
I just came across this page while looking for this answer myself (I'm also new to Matlab).
However, the above answers seem too complicated; Matlab's matrix addressing and assignment operators solve this problem very simply, as I found out after a bit of experimentation:
A = [1 2 3 4; 100 0 100 100]
A =
1 2 3 4
100 0 100 100
A(:,A(2,:)==0)=[]
A =
1 3 4
100 100 100
I hope this is useful to someone.
The 1st answer is absolutely correct. However, in order to approach this problem, there is a straight-forward syntax built for this purpose, using sparse you will also be able to extract the player with 0-points directly:
A = [1 2 3 4; 100 100 0 100]
ans= sparse (A(2,:))
Good luck ;)