I've been having trouble with the following question
I have a given Binary Tree (not necessarily BST) and two pointers (x,y) and I need to find if X is Y's predecessor in a O(1) complexity, I can add as many fields as I want.
I was thinking about adding each predecessor as a field as I insert the next child to the tree but in that way how can I search if X is Y's predecessor in O(1) complexity.
If you use nodes, add an unsigned int field, call it L, starting at 1 with root.
When you recursively insert, take the value of the previous node and multiply by 2 then add 1 if you go right, or simply multiply by 2 if you go left.
You will get a tree of L values that looks like this:
1
/ \
/ \
/ \
/ \
10 11
/ \ / \
/ \ / \
100 101 110 111
\ / \
1001 1110 1111
/
10010
An ancestor P should have a value P.L such that P.L is a substring of C.L and the number of bits in P.L is strictly less than the bits in C.L.
The tree's L values in base-10 is:
1
/ \
/ \
/ \
/ \
2 3
/ \ / \
/ \ / \
4 5 6 7
\ / \
9 14 15
/
18
If you have both pointers, if you take log_2(L), You will get the # of bits in that number L, which if you notice, represents the level in the tree you are at.
So if:
// Parent (ancestor) has equal or more bits?
if (log(P.L) >= log(C.L)) {
// parent is not an ancestor because it
// is either lower in tree, or at same level
}
If that check passes, Subtract bits(P) from bits(C), this will tell you how many more bits C.L has than P.L. Or, how many levels C is lower than P.
int D = log(C.L) - log(P.L)
Since C is lower, and all we did to calculate C.L value was multiply parent' L values by two (shift left) some number of times, if we were to shift C back over to the right (divide by 2) D times, the first D bits should match.
// Divide by 2, D times
int c = C.L >> D
// Is P.L a substring of C.L?
if (c == P.L) {
// P.L is a substring of C.L
// means P is an ancestor of C
}
// If we get here, C is below P in the tree, but C
// is not in a subtree of P because the first `D bits don't match`
In essence, we use integers as strings to keep track of the path of insertion, and we use bit manipulation to check if C.L is a substring of P.L in constant time.
Note, if you used an array, then P.L and C.L are simply the indexes of the nodes you would like to check.
Related
I'm trying to implement Algorithm D from Knuth's "The Art of Computer Programming, Vol 2" in Rust although I'm having trouble understating how to implement the very last step of unnormalizing. My natural numbers are a class where each number is a vector of u64, in base u64::MAX. Addition, subtraction, and multiplication have been implemented.
Knuth's Algorithm D is a euclidean division algorithm which takes two natural numbers x and y and returns (q,r) where q = x / y (integer division) and r = x % y, the remainder. The algorithm depends on an approximation method which only works if the first digit of y is greater than b/2, where b is the base you're representing the numbers in. Since not all numbers are of this form, it uses a "normalizing trick", for example (if we were in base 10) instead of doing 200 / 23, we calculate a normalizer d and do (200 * d) / (23 * d) so that 23 * d has a first digit greater than b/2.
So when we use the approximation method, we end up with the desired q but the remainder is multiplied by a factor of d. So the last step is to divide r by d so that we can get the q and r we want. My problem is, I'm a bit confused at how we're suppose to do this last step as it requires division and the method it's part of is trying to implement division.
(Maybe helpful?):
The way that d is calculated is just by taking the integer floor of b-1 divided by the first digit of y. However, Knuth suggests that it's possible to make d a power of 2, as long as d * the first digit of y is greater than b / 2. I think he makes this suggestion so that instead of dividing, we can just do a binary shift for this last step. Although I don't think I can do that given that my numbers are represented as vectors of u64 values, instead of binary.
Any suggestions?
There are 4 items: A weights 2LB has profit $40, B weights 5LB has profit $30, C weights 10LB has profit $50, and D weights 5LB has profit $10. Compute the maximum total profit you can take from any of the 4 items with a knapsack weight 16LB. You cannot take any portions of an item but the whole.
Please show how can the above problem be solved using knapsack problem approach.
A simple solution is to consider all subsets of items and calculate the total weight and value of all subsets. Then you should consider taking only the subsets that weighs less or equal to the capacity of your knapsack. From all such subsets, pick the maximum value subset.
To consider all subsets of items, there can be two cases for every item:
the item is included in the optimal subset,
the item is not included in the optimal subset.
Let's say the capacity of your knapsack is W. Therefore, the maximum value that can be obtained from n items is max of following two values.
Maximum value obtained by n-1 items and W weight (excluding the nth item)
Value of nth item plus maximum value obtained by n-1 items and W minus weight of the nth item (including nth item).
If weight of the nth item is greater than W, then the nth item can't be included and case 1 is the only option. This would be a naive approach and the solution would take 2n time.
Now for the overlapping subproblem:
weight = {2, 5, 10, 5}
Capacity = 16
The recursion tree would look like:
// Here n,k -> items remaining, capacity remaining
// Going to left child -> including the item at hand
// Going to right child -> excluding the item at hand
_______4,16______
/ \
/ \
3,14 3,16
/ \ / \
/ \ / \
2,9 2,14 2,5 2,16
\ / \ \ / \
\ / \ \ / \__
1,9 1,4 1,14 1,5 1,6 1,16
/\ /\ /\ /\ / \
/ \ / \ / \ / \ / \
0,4 0,9 0,9 0,14 0,0 0,0 0,1 0,6 0,11 0,16
Since there are overlapping subproblems in the leaf, we can solve it using dynamic programming. If you store the values, it will be efficient to use them later. Here the match occurs in leaf nodes, if you take other examples, you'll see a match can occur far before the leaf nodes.
The pseudo-code would look like:
Procedure Knapsack(n, W): //here, n = number of items, W = capacity of Knapsack
for i from 0 up to n
for j from 0 up to W
if i == 0 or j == 0
table[i][j] :=0
else if weight[i-1] <= j
table[i][j] := max(profit[i-1] + table[i-1][w-weight[i-1]], table[i-1][j])
else
table[i][j] := table[i-1][j]
end if
end for
end for
Return table[n][W]
In Verilog code
case ({Q[0], Q_1})
2'b0_1 :begin
A<=sum[7]; Q<=sum; Q_1<=Q;
end
2'b1_0 : begin
A<=difference[7]; Q<=difference; Q_1<=Q;
end
default: begin
A<=A[7]; Q<=A; Q_1<=Q;
end
endcase
is above code is same as below code
case ({Q[0], Q_1})
2'b0_1 : {A, Q, Q_1} <= {sum[7], sum, Q};
2'b1_0 : {A, Q, Q_1} <= {difference[7], difference, Q};
default: {A, Q, Q_1} <= {A[7], A, Q};
endcase
If yes then why i am getting different result?
Edit:-A, Q, sum and difference are all 8-bit values and Q_1 is a 1-bit value.
No, these are not the same. The concatenation operator ({ ... }) allows you to create vectors from several different signals, allowing you to both use these vectors and assign to these vectors, resulting in the assignment of the component signals will the appropriate bits from the result. From your previous question (Please Explain these verilog code?), I see that A, Q, sum and difference are all 8-bit values and Q_1 is a 1-bit value. Lets examine the first assignment (noting that the other three work the same way):
{A, Q, Q_1} <= {sum[7], sum, Q};
If we look at the right-hand side, we can see that the result of the concatenation is a 17-bit vector, as sum[7] is 1 bit (the MSb of sum), sum is 8 bits, and Q is 8 bits (1 + 8 + 8 = 17). Lets say sum = 8'b10100101 and Q = 8'b00110110, what would {sum[7], sum, Q} look like? Well, its the concatenation of the values from sum and Q so it would be 17'b1_10100101_00110110, the first bit coming from sum[7], the next 8 bits from sum and the final 8 bits from Q.
Now we have to assign this 17-bit value to the left hand side. On the left, we have {A, Q, Q_1}, which is also 17 bits (A is 8 bits, Q is 8 bits and Q_1 is 1 bit). However, we have to assign the bits from our 17-bit value we got above to the proper signals that make up this new 17-bit vector, that means the 8 most significant bits go into A, the next 8 bits go into Q and the least significant bit go into Q_1. So, if we take our value from above (17'b1_10100101_00110110), and split it up this way (17'b11010010_10011011_0), we see A = 8'b11010010, Q = 8'b10011011 and Q_1 = 1'b0. Thus, this is not the same as assigning A = sum[7], Q = sum and Q_1 = Q (this would result in A = 8'b00000001, Q = 8'b10100101, Q_1 = 1'b0, with many bits of Q being lost and A having 7 extra bits).
However, this doesnt mean we cant split up the left-hand side concatenation, it would just look like this:
A <= {sum[7], sum[7:1]};
Q <= {sum[0], Q[7:1]};
Q_1 <= Q[0];
Yes, they are same. For example try this small code and check, the output is same :
module test;
wire A,B,C;
reg p,q,r;
initial
begin
p=1; q=1; r=0;
end
assign {A,B,C} = {p,q,r};
initial #1 $display("%b %b %b",A,B,C);
endmodule
In general if you want to understand concatenation operator, you can refer here
Edit : I have assumed A and p , B and q, C and r of same length.
I would like to know where I can read about algorithms for solving this problem efficiently:
Four directions allowed: up, down, left, right
Cells containing zero can't be visited.
Visiting the same cell twice is illegal.
Moves wraps around the edges:
(first row is connected with last row)
(first col is connected with last col)
Example, 5x5 and 5 steps:
9 1 3 1 9
6 3 2 4 1
0 7 * 7 7
5 4 9 4 9
7 9 1 5 5
Starting point: *
Solution: down,left,down,left,down. That is 9 + 4 + 9 + 7 + 9 = 38
[9] 1 3 1 9
6 3 2 4 1
0 7 * 7 7
5 [4][9] 4 9
[7][9] 1 5 5
This problem is probably not related to:
Finding the maximum sub matrix
Dynamic programming
You specified in comments that you wanted a sub-second way of finding the best value 20-step path out of a 5x5 matrix. I've implemented a basic recursive search tree that does this. Ultimately, the difficulty of this is still O(3^k), but highly saturated cases like yours (21 out of 24 allowed nodes visited) will solve much faster because the problem simplifies to "skip the n*n-z-k-1 lowest value cells" (in this case, n=5, z=1 and k+1 = 21; the winning path skips three 1's).
The problem instance in your question solves in 0.231seconds on a 3 year old i5 laptop and about half a second on ideone.com. I've provided code here http://ideone.com/5kOyxq (note that 'up' and 'down' are reversed because of the way I input the data).
For less saturated problems you may need to add a Bound/Cut method. You can generate a Bound as follows:
First, run over the NxN matrix and collect the K highest value elements (can be done in N² log K) and sort them by max-first. Then, cumulatively calculate the value UB[t] = SUM[i::0->t] SortedElements[i]. So, any t-length path has a UB of UB[t] (max t elements).
At step T, the current Branch's UB is UB[t]. If ValueSoFar[T] + UB[K-T] <= BestPathValue, you can stop that branch.
There may be better ways, but this should be sufficient for reasonably sized matrices and path lengths.
Game or puzzle. Given a matrix, number of steps and a sum, find the path.
Would be nice if there is a real world application for this, but i haven't found it.
Games tend to "burn in" knowledge in young brains, so why not burn in something useful, like addition?
#include<iostream>
#include<climits>
#define R 3
#define C 3
int MAX(int x, int y, int z);
int Max_Cost(int cost[R][C], int m, int n)
{
if (n < 0 || m < 0)
return INT_MIN;
else if (m == 0 && n == 0)
return cost[m][n];
else
return cost[m][n] + MIN( Max_Cost(cost, m-1, n-1),
Max_Cost(cost, m-1, n),
Max_Cost(cost, m, n-1)
);
}
int MAX(int x, int y, int z)
{
return max(max(x, y), z);
}
int main()
{
int cost[R][C] = { {3, 2, 5},
{5, 8, 2},
{9, 7, 1}
};
cout<<Max_Cost(cost, 2, 1);
return 0;
}
I know the modulus (%) operator calculates the remainder of a division. How can I identify a situation where I would need to use the modulus operator?
I know I can use the modulus operator to see whether a number is even or odd and prime or composite, but that's about it. I don't often think in terms of remainders. I'm sure the modulus operator is useful, and I would like to learn to take advantage of it.
I just have problems identifying where the modulus operator is applicable. In various programming situations, it is difficult for me to see a problem and realize "Hey! The remainder of division would work here!".
Imagine that you have an elapsed time in seconds and you want to convert this to hours, minutes, and seconds:
h = s / 3600;
m = (s / 60) % 60;
s = s % 60;
0 % 3 = 0;
1 % 3 = 1;
2 % 3 = 2;
3 % 3 = 0;
Did you see what it did? At the last step it went back to zero. This could be used in situations like:
To check if N is divisible by M (for example, odd or even)
or
N is a multiple of M.
To put a cap of a particular value. In this case 3.
To get the last M digits of a number -> N % (10^M).
I use it for progress bars and the like that mark progress through a big loop. The progress is only reported every nth time through the loop, or when count%n == 0.
I've used it when restricting a number to a certain multiple:
temp = x - (x % 10); //Restrict x to being a multiple of 10
Wrapping values (like a clock).
Provide finite fields to symmetric key algorithms.
Bitwise operations.
And so on.
One use case I saw recently was when you need to reverse a number. So that 123456 becomes 654321 for example.
int number = 123456;
int reversed = 0;
while ( number > 0 ) {
# The modulus here retrieves the last digit in the specified number
# In the first iteration of this loop it's going to be 6, then 5, ...
# We are multiplying reversed by 10 first, to move the number one decimal place to the left.
# For example, if we are at the second iteration of this loop,
# reversed gonna be 6, so 6 * 10 + 12345 % 10 => 60 + 5
reversed = reversed * 10 + number % 10;
number = number / 10;
}
Example. You have message of X bytes, but in your protocol maximum size is Y and Y < X. Try to write small app that splits message into packets and you will run into mod :)
There are many instances where it is useful.
If you need to restrict a number to be within a certain range you can use mod. For example, to generate a random number between 0 and 99 you might say:
num = MyRandFunction() % 100;
Any time you have division and want to express the remainder other than in decimal, the mod operator is appropriate. Things that come to mind are generally when you want to do something human-readable with the remainder. Listing how many items you could put into buckets and saying "5 left over" is good.
Also, if you're ever in a situation where you may be accruing rounding errors, modulo division is good. If you're dividing by 3 quite often, for example, you don't want to be passing .33333 around as the remainder. Passing the remainder and divisor (i.e. the fraction) is appropriate.
As #jweyrich says, wrapping values. I've found mod very handy when I have a finite list and I want to iterate over it in a loop - like a fixed list of colors for some UI elements, like chart series, where I want all the series to be different, to the extent possible, but when I've run out of colors, just to start over at the beginning. This can also be used with, say, patterns, so that the second time red comes around, it's dashed; the third time, dotted, etc. - but mod is just used to get red, green, blue, red, green, blue, forever.
Calculation of prime numbers
The modulo can be useful to convert and split total minutes to "hours and minutes":
hours = minutes / 60
minutes_left = minutes % 60
In the hours bit we need to strip the decimal portion and that will depend on the language you are using.
We can then rearrange the output accordingly.
Converting linear data structure to matrix structure:
where a is index of linear data, and b is number of items per row:
row = a/b
column = a mod b
Note above is simplified logic: a must be offset -1 before dividing & the result must be normalized +1.
Example: (3 rows of 4)
1 2 3 4
5 6 7 8
9 10 11 12
(7 - 1)/4 + 1 = 2
7 is in row 2
(7 - 1) mod 4 + 1 = 3
7 is in column 3
Another common use of modulus: hashing a number by place. Suppose you wanted to store year & month in a six digit number 195810. month = 195810 mod 100 all digits 3rd from right are divisible by 100 so the remainder is the 2 rightmost digits in this case the month is 10. To extract the year 195810 / 100 yields 1958.
Modulus is also very useful if for some crazy reason you need to do integer division and get a decimal out, and you can't convert the integer into a number that supports decimal division, or if you need to return a fraction instead of a decimal.
I'll be using % as the modulus operator
For example
2/4 = 0
where doing this
2/4 = 0 and 2 % 4 = 2
So you can be really crazy and let's say that you want to allow the user to input a numerator and a divisor, and then show them the result as a whole number, and then a fractional number.
whole Number = numerator/divisor
fractionNumerator = numerator % divisor
fractionDenominator = divisor
Another case where modulus division is useful is if you are increasing or decreasing a number and you want to contain the number to a certain range of number, but when you get to the top or bottom you don't want to just stop. You want to loop up to the bottom or top of the list respectively.
Imagine a function where you are looping through an array.
Function increase Or Decrease(variable As Integer) As Void
n = (n + variable) % (listString.maxIndex + 1)
Print listString[n]
End Function
The reason that it is n = (n + variable) % (listString.maxIndex + 1) is to allow for the max index to be accounted.
Those are just a few of the things that I have had to use modulus for in my programming of not just desktop applications, but in robotics and simulation environments.
Computing the greatest common divisor
Determining if a number is a palindrome
Determining if a number consists of only ...
Determining how many ... a number consists of...
My favorite use is for iteration.
Say you have a counter you are incrementing and want to then grab from a known list a corresponding items, but you only have n items to choose from and you want to repeat a cycle.
var indexFromB = (counter-1)%n+1;
Results (counter=indexFromB) given n=3:
`1=1`
`2=2`
`3=3`
`4=1`
`5=2`
`6=3`
...
Best use of modulus operator I have seen so for is to check if the Array we have is a rotated version of original array.
A = [1,2,3,4,5,6]
B = [5,6,1,2,3,4]
Now how to check if B is rotated version of A ?
Step 1: If A's length is not same as B's length then for sure its not a rotated version.
Step 2: Check the index of first element of A in B. Here first element of A is 1. And its index in B is 2(assuming your programming language has zero based index).
lets store that index in variable "Key"
Step 3: Now how to check that if B is rotated version of A how ??
This is where modulus function rocks :
for (int i = 0; i< A.length; i++)
{
// here modulus function would check the proper order. Key here is 2 which we recieved from Step 2
int j = [Key+i]%A.length;
if (A[i] != B[j])
{
return false;
}
}
return true;
It's an easy way to tell if a number is even or odd. Just do # mod 2, if it is 0 it is even, 1 it is odd.
Often, in a loop, you want to do something every k'th iteration, where k is 0 < k < n, assuming 0 is the start index and n is the length of the loop.
So, you'd do something like:
int k = 5;
int n = 50;
for(int i = 0;i < n;++i)
{
if(i % k == 0) // true at 0, 5, 10, 15..
{
// do something
}
}
Or, you want to keep something whitin a certain bound. Remember, when you take an arbitrary number mod something, it must produce a value between 0 and that number - 1.