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In all articles which I read, neighbor to process first is "closest" neighbor. But finally it's needed to visit all nodes to figure out all possible paths. So, the question is - why do we do this? I believe the same result can be achieved if we simply traverse Graph in BFS way and will perform calculation of costs. For example:
first step- 0, costs table:
2 - 6 |
3 - 2 |
second step- 2, costs table:
2 - 6 |
3 - 2 |
1 - 9 |
third step- 3, costs table:
2 - 6 |
3 - 2 |
1 - 9 |
4 - 12 |
forth step- 1, costs table:
2 - 6 |
3 - 2 |
1 - 9 |
4 - 12 |
5 - 12 |
fifth step- 4, costs table:
2 - 6 |
3 - 2 |
1 - 9 |
4 - 12 |
5 - 12 |
With simple BFS traversing the cheapest way was find out. What do I missing?
Suppose the path from A to B and B to C are both cost 1, and the direct route from A to C is cost 3. (In the real world, the first two are highways that bypass a mountain while the third is a tiny trail over a mountain pass.)
Dijkstra will route you A -> B -> C for a total cost of 2 while BFS will route you A -> C for a total cost of 3.
Therefore you have to process lowest cost first to get the right answer.
At each step, Dijkstra's algorithm extends the lowest-cost path known so far. Thus, when you finally encounter the goal state, you know that all other, unfinished paths have a greater cost. Therefore, the one you just found is the shortest path.
I cannot get my head around this:
Say I got a number 9. I want to know the minimum steps needed to split it so that no number is greater than 3.
I always thought that the most efficient way is to halve it every loop.
So, 9 -> 4,5 -> 2,2,5 -> 2,2,2,3 so 3 steps in total. However, I just realised a smarter way: 9 -> 3,6 -> 3,3,3 which is 2 steps only...
After some research, the number of steps is in fact (n-1)/target, where target=3 in my example.
Can someone please explain this behaviour to me?
If we want to cut a stick of length L into pieces of size no greater than S, we need ceiling(L/S) pieces. Each time we make a new cut, we increase the number of pieces by 1. It doesn't matter what order we make the cuts in, only where. For example, if we want to break a stick of length 10 into pieces of size 2 or less:
-------------------
0 1 2 3 4 5 6 7 8 9 10
we should cut it in the following places:
---|---|---|---|---
0 1 2 3 4 5 6 7 8 9 10
and any order of cuts is fine, as long as these are the cuts that are made. On the other hand, if we start by breaking it in half:
---------|---------
0 1 2 3 4 5 6 7 8 9 10
we have made a cut that isn't part of the optimal solution, and we have wasted our time.
I really like #user2357112's explanation of why cutting in half is not the right first step, but I also like algebra, and you can prove that ceil(n / target) - 1 is optimal using induction.
Let's prove first that you can always do it in ceil(n / target) - 1 steps.
If n <= target, obviously no step are required, so the formula works. Suppose n > target. Split n into target and n - target (1 step). By induction, n - target can be split in ceil((n - target)/target) - 1 steps. Therefore the total number of steps is
1 + ceil((n - target) / target) - 1
= 1 + ceil(n / target) - target/target - 1
= ceil(n / target) - 1.
Now let's prove that you can't do it in fewer than ceil(n / target) - 1 steps. This is obvious if n <= target. Suppose n > target and the first step is n -> a + b. By induction, a requires at least ceil(a / target) - 1 steps and b requires at least ceil(b / target) - 1 steps. The minimum number of steps required is therefore at least
1 + ceil(a / target) - 1 + ceil(b / target) - 1
>= ceil((a + b) / target) - 1 using ceil(x) + ceil(y) >= ceil(x + y)
= ceil(n / target) - 1 using a + b = n
Every n can be thought of as a priority queue of \lfloor n/target \rfloor target elements placed first on the queue and one element whose value is n%target. Every time you remove an element from the queue, you place it back on the queue. Remove all but the last element: you have clearly removed \lfloor (n-1)/target \rfloor elements. If the last element is less than or equal to the target, we are done. If it is greater than the target, we have a contradiction. So, after \lfloor (n-1)/target \rfloor steps we have a queue consisting only of elements less than or equal to target.
I am trying to understand the n log n solution. It seems that you dont have to store the entire temp lists at any given point, just the last element would do.
How does that work?
I didnt want to post on the existing posts, so created a new one.
Spent a lot of time understanding it. . :(
you don't have to store the entire temp lists at any given point, just the last element would do.
First, recall the O(n2) solution: you set up an array L that has at each element i the length of the longest non-decreasing subsequence of A ending at element i. Here is an example:
A: 2 5 7 3 8 2 9 6 9
L: 1 2 3 2 4 2 5 3 6
Now imagine setting up an array M, which at each index k stores the element of A that ends the longest non-decreasing subsequence of length k. Here is how M would look at each step (dash - shows places that are not filled in)
M (step 0) - - - - - - - - -
M (step 1) 2 - - - - - - - -
M (step 2) 2 5 - - - - - - -
M (step 3) 2 5 7 - - - - - -
M (step 4) 2 3 7 - - - - - -
M (step 5) 2 3 7 8 - - - - -
M (step 6) 2 2 7 8 - - - - -
M (step 7) 2 2 7 8 9 - - - -
M (step 8) 2 2 6 8 9 - - - -
M (step 9) 2 2 6 8 9 9 - - -
Work through the example manually to understand the mechanics of filling out the array M.
Now comes the key observation: at each step, M is ordered in non-decreasing order. Intuitively, this is clear, because otherwise the larger number could be attached to a longer sequence, and be moved up the array M.
This lets you construct your algorithm:
Store the last position maxPos of M that has been filled in
When you arrive at A[i], run a binary search in M[0..maxPos] for a place where A[i] should be placed
If you end up in the middle of the array, replace the old value with min(A[i], oldValue)
If the element is larger than or equal to all elements added to M so far, add A[i] to the end, and increment maxPos
It is easy to see that the above algorithm is O(n*log(n)), because each of its n steps uses a binary search, which is log(n).
I have already read What is an "external node" of a "magic" 3-gon ring? and I have solved problems up until 90 but this n-gon thing totally baffles me as I don't understand the question at all.
So I take this ring and I understand that the external circles are 4, 5, 6 as they are outside the inner circle. Now he says there are eight solutions. And the eight solutions are without much explanation listed below. Let me take
9 4,2,3; 5,3,1; 6,1,2
9 4,3,2; 6,2,1; 5,1,3
So how do we arrive at the 2 solutions? I understand 4, 3, 2, is in straight line and 6,2,1 is in straight line and 5, 1, 3 are in a straight line and they are in clockwise so the second solution makes sense.
Questions
Why does the first solution 4,2,3; 5,3,1; 6,1,2 go anti clock wise? Should it not be 423 612 and then 531?
How do we arrive at 8 solutions. Is it just randomly picking three numbers? What exactly does it mean to solve a "N-gon"?
The first doesn't go anti-clockwise. It's what you get from the configuration
4
\
2
/ \
1---3---5
/
6
when you go clockwise, starting with the smallest number in the outer ring.
How do we arrive at 8 solutions. Is it just randomly picking three numbers? What exactly does it mean to solve a "N-gon"?
For an N-gon, you have an inner N-gon, and for each side of the N-gon one spike, like
X
|
X---X---X
| |
X---X---X
|
X
so that the spike together with the side of the inner N-gon connects a group of three places. A "solution" of the N-gon is a configuration where you placed the numbers from 1 to 2*N so that each of the N groups sums to the same value.
The places at the end of the spikes appear in only one group each, the places on the vertices of the inner N-gon in two. So the sum of the sums of all groups is
N
∑ k + ∑{ numbers on vertices }
k=1
The sum of the numbers on the vertices of the inner N-gon is at least 1 + 2 + ... + N = N*(N+1)/2 and at most (N+1) + (N+2) + ... + 2*N = N² + N*(N+1)/2 = N*(3*N+1)/2.
Hence the sum of the sums of all groups is between
N*(2*N+1) + N*(N+1)/2 = N*(5*N+3)/2
and
N*(2*N+1) + N*(3*N+1)/2 = N*(7*N+3)/2
inclusive, and the sum per group must be between
(5*N+3)/2
and
(7*N+3)/2
again inclusive.
For the triangle - N = 3 - the bounds are (5*3+3)/2 = 9 and (7*3+3)/2 = 12. For a square - N = 4 - the bounds are (5*4+3)/2 = 11.5 and (7*4+3)/2 = 15.5 - since the sum must be an integer, the possible sums are 12, 13, 14, 15.
Going back to the triangle, if the sum of each group is 9, the sum of the sums is 27, and the sum of the numbers on the vertices must be 27 - (1+2+3+4+5+6) = 27 - 21 = 6 = 1+2+3, so the numbers on the vertices are 1, 2 and 3.
For the sum to be 9, the value at the end of the spike for the side connecting 1 and 2 must be 6, for the side connecting 1 and 3, the spike value must be 5, and 4 for the side connecting 2 and 3.
If you start with the smallest value on the spikes - 4 - you know you have to place 2 and 3 on the vertices of the side that spike protrudes from. There are two ways to arrange the two numbers there, leading to the two solutions for sum 9.
If the sum of each group is 10, the sum of the sums is 30, and the sum of the numbers on the vertices must be 9. To represent 9 as the sum of three distinct numbers from 1 to 6, you have the possibilities
1 + 2 + 6
1 + 3 + 5
2 + 3 + 4
For the first group, you have one side connecting 1 and 2, so you'd need a 7 on the end of the spike to make 10 - no solution.
For the third group, the minimal sum of two of the numbers is 5, but 5+6 = 11 > 10, so there's no place for the 6 - no solution.
For the second group, the sums of the sides are
1 + 3 = 4 -- 6 on the spike
1 + 5 = 6 -- 4 on the spike
3 + 5 = 8 -- 2 on the spike
and you have two ways to arrange 3 and 5, so that the group is either 2-3-5 or 2-5-3, the rest follows again.
The solutions for the sums 11 and 12 can be obtained similarly, or by replacing k with 7-k in the solutions for the sums 9 resp. 10.
To solve the problem, you must now find out
what it means to obtain a 16-digit string or a 17-digit string
which sum for the groups gives rise to the largest value when the numbers are concatenated in the prescribed way.
(And use pencil and paper for the fastest solution.)
given a sorted array of distinct integers, what is the minimum number of steps required to make the integers contiguous? Here the condition is that: in a step , only one element can be changed and can be either increased or decreased by 1 . For example, if we have 2,4,5,6 then '2' can be made '3' thus making the elements contiguous(3,4,5,6) .Hence the minimum steps here is 1 . Similarly for the array: 2,4,5,8:
Step 1: '2' can be made '3'
Step 2: '8' can be made '7'
Step 3: '7' can be made '6'
Thus the sequence now is 3,4,5,6 and the number of steps is 3.
I tried as follows but am not sure if its correct?
//n is the number of elements in array a
int count=a[n-1]-a[0]-1;
for(i=1;i<=n-2;i++)
{
count--;
}
printf("%d\n",count);
Thanks.
The intuitive guess is that the "center" of the optimal sequence will be the arithmetic average, but this is not the case. Let's find the correct solution with some vector math:
Part 1: Assuming the first number is to be left alone (we'll deal with this assumption later), calculate the differences, so 1 12 3 14 5 16-1 2 3 4 5 6 would yield 0 -10 0 -10 0 -10.
sidenote: Notice that a "contiguous" array by your implied definition would be an increasing arithmetic sequence with difference 1. (Note that there are other reasonable interpretations of your question: some people may consider 5 4 3 2 1 to be contiguous, or 5 3 1 to be contiguous, or 1 2 3 2 3 to be contiguous. You also did not specify if negative numbers should be treated any differently.)
theorem: The contiguous numbers must lie between the minimum and maximum number. [proof left to reader]
Part 2: Now returning to our example, assuming we took the 30 steps (sum(abs(0 -10 0 -10 0 -10))=30) required to turn 1 12 3 14 5 16 into 1 2 3 4 5 6. This is one correct answer. But 0 -10 0 -10 0 -10+c is also an answer which yields an arithmetic sequence of difference 1, for any constant c. In order to minimize the number of "steps", we must pick an appropriate c. In this case, each time we increase or decrease c, we increase the number of steps by N=6 (the length of the vector). So for example if we wanted to turn our original sequence 1 12 3 14 5 16 into 3 4 5 6 7 8 (c=2), then the differences would have been 2 -8 2 -8 2 -8, and sum(abs(2 -8 2 -8 2 -8))=30.
Now this is very clear if you could picture it visually, but it's sort of hard to type out in text. First we took our difference vector. Imagine you drew it like so:
4|
3| *
2| * |
1| | | *
0+--+--+--+--+--*
-1| |
-2| *
We are free to "shift" this vector up and down by adding or subtracting 1 from everything. (This is equivalent to finding c.) We wish to find the shift which minimizes the number of | you see (the area between the curve and the x-axis). This is NOT the average (that would be minimizing the standard deviation or RMS error, not the absolute error). To find the minimizing c, let's think of this as a function and consider its derivative. If the differences are all far away from the x-axis (we're trying to make 101 112 103 114 105 116), it makes sense to just not add this extra stuff, so we shift the function down towards the x-axis. Each time we decrease c, we improve the solution by 6. Now suppose that one of the *s passes the x axis. Each time we decrease c, we improve the solution by 5-1=4 (we save 5 steps of work, but have to do 1 extra step of work for the * below the x-axis). Eventually when HALF the *s are past the x-axis, we can NO LONGER IMPROVE THE SOLUTION (derivative: 3-3=0). (In fact soon we begin to make the solution worse, and can never make it better again. Not only have we found the minimum of this function, but we can see it is a global minimum.)
Thus the solution is as follows: Pretend the first number is in place. Calculate the vector of differences. Minimize the sum of the absolute value of this vector; do this by finding the median OF THE DIFFERENCES and subtracting that off from the differences to obtain an improved differences-vector. The sum of the absolute value of the "improved" vector is your answer. This is O(N) The solutions of equal optimality will (as per the above) always be "adjacent". A unique solution exists only if there are an odd number of numbers; otherwise if there are an even number of numbers, AND the median-of-differences is not an integer, the equally-optimal solutions will have difference-vectors with corrective factors of any number between the two medians.
So I guess this wouldn't be complete without a final example.
input: 2 3 4 10 14 14 15 100
difference vector: 2 3 4 5 6 7 8 9-2 3 4 10 14 14 15 100 = 0 0 0 -5 -8 -7 -7 -91
note that the medians of the difference-vector are not in the middle anymore, we need to perform an O(N) median-finding algorithm to extract them...
medians of difference-vector are -5 and -7
let us take -5 to be our correction factor (any number between the medians, such as -6 or -7, would also be a valid choice)
thus our new goal is 2 3 4 5 6 7 8 9+5=7 8 9 10 11 12 13 14, and the new differences are 5 5 5 0 -3 -2 -2 -86*
this means we will need to do 5+5+5+0+3+2+2+86=108 steps
*(we obtain this by repeating step 2 with our new target, or by adding 5 to each number of the previous difference... but since you only care about the sum, we'd just add 8*5 (vector length times correct factor) to the previously calculated sum)
Alternatively, we could have also taken -6 or -7 to be our correction factor. Let's say we took -7...
then the new goal would have been 2 3 4 5 6 7 8 9+7=9 10 11 12 13 14 15 16, and the new differences would have been 7 7 7 2 1 0 0 -84
this would have meant we'd need to do 7+7+7+2+1+0+0+84=108 steps, the same as above
If you simulate this yourself, can see the number of steps becomes >108 as we take offsets further away from the range [-5,-7].
Pseudocode:
def minSteps(array A of size N):
A' = [0,1,...,N-1]
diffs = A'-A
medianOfDiffs = leftMedian(diffs)
return sum(abs(diffs-medianOfDiffs))
Python:
leftMedian = lambda x:sorted(x)[len(x)//2]
def minSteps(array):
target = range(len(array))
diffs = [t-a for t,a in zip(target,array)]
medianOfDiffs = leftMedian(diffs)
return sum(abs(d-medianOfDiffs) for d in diffs)
edit:
It turns out that for arrays of distinct integers, this is equivalent to a simpler solution: picking one of the (up to 2) medians, assuming it doesn't move, and moving other numbers accordingly. This simpler method often gives incorrect answers if you have any duplicates, but the OP didn't ask that, so that would be a simpler and more elegant solution. Additionally we can use the proof I've given in this solution to justify the "assume the median doesn't move" solution as follows: the corrective factor will always be in the center of the array (i.e. the median of the differences will be from the median of the numbers). Thus any restriction which also guarantees this can be used to create variations of this brainteaser.
Get one of the medians of all the numbers. As the numbers are already sorted, this shouldn't be a big deal. Assume that median does not move. Then compute the total cost of moving all the numbers accordingly. This should give the answer.
community edit:
def minSteps(a):
"""INPUT: list of sorted unique integers"""
oneMedian = a[floor(n/2)]
aTarget = [oneMedian + (i-floor(n/2)) for i in range(len(a))]
# aTargets looks roughly like [m-n/2?, ..., m-1, m, m+1, ..., m+n/2]
return sum(abs(aTarget[i]-a[i]) for i in range(len(a)))
This is probably not an ideal solution, but a first idea.
Given a sorted sequence [x1, x2, …, xn]:
Write a function that returns the differences of an element to the previous and to the next element, i.e. (xn – xn–1, xn+1 – xn).
If the difference to the previous element is > 1, you would have to increase all previous elements by xn – xn–1 – 1. That is, the number of necessary steps would increase by the number of previous elements × (xn – xn–1 – 1). Let's call this number a.
If the difference to the next element is >1, you would have to decrease all subsequent elements by xn+1 – xn – 1. That is, the number of necessary steps would increase by the number of subsequent elements × (xn+1 – xn – 1). Let's call this number b.
If a < b, then increase all previous elements until they are contiguous to the current element. If a > b, then decrease all subsequent elements until they are contiguous to the current element. If a = b, it doesn't matter which of these two actions is chosen.
Add up the number of steps taken in the previous step (by increasing the total number of necessary steps by either a or b), and repeat until all elements are contiguous.
First of all, imagine that we pick an arbitrary target of contiguous increasing values and then calculate the cost (number of steps required) for modifying the array the array to match.
Original: 3 5 7 8 10 16
Target: 4 5 6 7 8 9
Difference: +1 0 -1 -1 -2 -7 -> Cost = 12
Sign: + 0 - - - -
Because the input array is already ordered and distinct, it is strictly increasing. Because of this, it can be shown that the differences will always be non-increasing.
If we change the target by increasing it by 1, the cost will change. Each position in which the difference is currently positive or zero will incur an increase in cost by 1. Each position in which the difference is currently negative will yield a decrease in cost by 1:
Original: 3 5 7 8 10 16
New target: 5 6 7 8 9 10
New Difference: +2 +1 0 0 -1 -6 -> Cost = 10 (decrease by 2)
Conversely, if we decrease the target by 1, each position in which the difference is currently positive will yield a decrease in cost by 1, while each position in which the difference is zero or negative will incur an increase in cost by 1:
Original: 3 5 7 8 10 16
New target: 3 4 5 6 7 8
New Difference: 0 -1 -2 -2 -3 -8 -> Cost = 16 (increase by 4)
In order to find the optimal values for the target array, we must find a target such that any change (increment or decrement) will not decrease the cost. Note that an increment of the target can only decrease the cost when there are more positions with negative difference than there are with zero or positive difference. A decrement can only decrease the cost when there are more positions with a positive difference than with a zero or negative difference.
Here are some example distributions of difference signs. Remember that the differences array is non-increasing, so positives always have to be first and negatives last:
C C
+ + + - - - optimal
+ + 0 - - - optimal
0 0 0 - - - optimal
+ 0 - - - - can increment (negatives exceed positives & zeroes)
+ + + 0 0 0 optimal
+ + + + - - can decrement (positives exceed negatives & zeroes)
+ + 0 0 - - optimal
+ 0 0 0 0 0 optimal
C C
Observe that if one of the central elements (marked C) is zero, the target must be optimal. In such a circumstance, at best any increment or decrement will not change the cost, but it may increase it. This result is important, because it gives us a trivial solution. We pick a target such that a[n/2] remains unchanged. There may be other possible targets that yield the same cost, but there are definitely none that are better. Here's the original code modified to calculate this cost:
//n is the number of elements in array a
int targetValue;
int cost = 0;
int middle = n / 2;
int startValue = a[middle] - middle;
for (i = 0; i < n; i++)
{
targetValue = startValue + i;
cost += abs(targetValue - a[i]);
}
printf("%d\n",cost);
You can not do it by iterating once on the array, that's for sure.
You need first to check the difference between each two numbers, for example:
2,7,8,9 can be 2,3,4,5 with 18 steps or 6,7,8,9 with 4 steps.
Create a new array with the difference like so: for 2,7,8,9 it wiil be 4,1,1. Now you can decide whether to increase or decrease the first number.
Lets assume that the contiguous array looks something like this -
c c+1 c+2 c+3 .. and so on
Now lets take an example -
5 7 8 10
The contiguous array in this case will be -
c c+1 c+2 c+3
In order to get the minimum steps, the sum of the modulus of the difference of the integers(before and after) w.r.t the ith index should be the minimum. In which case,
(c-5)^2 + (c-6)^2 + (c-6)^2 + (c-7)^2 should be minimum
Let f(c) = (c-5)^2 + (c-6)^2 + (c-6)^2 + (c-7)^2
= 4c^2 - 48c + 146
Applying differential calculus to get the minima,
f'(c) = 8c - 48 = 0
=> c = 6
So our contiguous array is 6 7 8 9 and the minimum cost here is 2.
To sum it up, just generate f(c), get the first differential and find out c.
This should take O(n).
Brute force approach O(N*M)
If one draws a line through each point in the array a then y0 is a value where each line starts at index 0. Then the answer is the minimum among number of steps reqired to get from a to every line that starts at y0, in Python:
y0s = set((y - i) for i, y in enumerate(a))
nsteps = min(sum(abs(y-(y0+i)) for i, y in enumerate(a))
for y0 in xrange(min(y0s), max(y0s)+1)))
Input
2,4,5,6
2,4,5,8
Output
1
3