I'm trying to write a function of the type:
pascal : int * int -> int
where the pair of ints represent the row and column, respectively, of Pascal's triangle.
Here's my attempt:
fun pascal(i : int, j : int) : int =
if (i = 0 andalso j = 0) orelse i = j orelse i = 0
then 1
else
pascal(i - 1, j - 1) + pascal(i - 1, j);
It works for my base cases but gives me strange output otherwise. For instance:
pascal(4, 2) gives me 11 and pascal(4, 1) gives me 15
It's a bit strange because, as long as the if clause fails and the else gets evaluated, I do want to return the sum of the element one row above and the element one row above and one element to the left.
What am I doing wrong?
Consider pascal 1 0. If you're using zero-based indexing for the table then this should be equal to 1. But:
pascal 1 0 = pascal 0 -1 + pascal 0 0 = 2
You should put some guards to deal with negative indices and indices where j is greater than i.
Related
I am running a loop equation that divides or subtracts. Every time it divides I want it to represent each division with the # 2 and every time it subtracts I want it to represent the subtractions with a 1. I then want that count string to be a value that I can manipulate with some basic math. Basically it'll look like this: 20/2 = 10 (2) 10/2 = 5 (2) 5/2 = 2.5 (2) 2.5-.5 = 2 (1) 2/2 = 1 (2)
22212 <=== that I want to make a new value but with the way I have it coded, it's not working. I think it may have something to do with the end='' in the code.
I've tried giving the value of the string = to a int value and tried joining the string but no luck so far.
num = 20
while num >= 1.5:
num /= 2
v = 1
print(v, end='')
if int(num) != num:
num -= .5
v = 2
print(v, end='') #trying to make the output here a value
nv = ''.join(str(int(v)))
nv = int(v) #trying to give the joined strs of nv a value
print(nv) #trying to get this to print the combined valued of v to something that math can be applied to.
print('')
The code doesn't give any errors I just can't figure out how to make the output and actual number that I can manipulate.
you are printing v = 1 after your division. In your post you said you want a 2 for division, I am assuming what you wrote in the post is the result you want.
a = ""
num = 20
while num >= 1.5:
num /= 2
a += "2"
if int(num) != num:
num -= .5
a += "1"
print(a)
now a is a string with your desired result. You can always convert that String to an int to do some math with it.
I'm coding the spigot algorithm for displaying digits of pi in ada, but my output is wrong and I can't figure out why
I've tried messing with the range of my loops and different ways to output my data but nothings worked properly
with ada.integer_text_io; use ada.integer_text_io;
with Ada.Text_IO; use Ada.Text_IO;
procedure Spigot is
n : constant Integer := 1000;
length : constant Integer := 10*n/3+1;
x,q,nines,predigit :Integer :=0;
a: array (0..length) of Integer;
begin
nines:=0;
predigit:=0;
for j in 0..length loop
a(j):=2;
end loop;
for j in 1..n loop
q:=0;
for i in reverse 1..length loop
x:=10*a(i) + q*i;
a(i):= x mod (2*i-1);
q:= x/(2*i-1);
end loop;
a(1):= q mod 10;
q:=q/10;
if q = 9 then
nines:=nines+1;
elsif q = 10 then
put(predigit+1);
for k in 0..nines loop
put("0");
end loop;
predigit:=0;
nines:=0;
else
put(predigit);
predigit:=q;
if nines/=0 then
for k in 0..nines loop
put("9");
end loop;
nines:=0;
end if;
end if;
end loop;
put(predigit);
end Spigot;
so it should just be displayed at 0 3 1 4 1 5 9 2 6 5 3 5 8 9... but the output i get is 0 3 1 4 1 599 2 6 5 3 5 89... it should only be 1 digit at a time and also the outputted values for pi aren't completely correct
I don't know the algorithm well enough to talk about why the digits are off, but I did notice some issues:
Your array is defined with bounds 0 .. Length, which would give you 1 extra element
In your loop that does the calculation, you loop from 1..length, which is ok, but you don't adjust the variable i consistently. The array indices need to be one less than the i's used in the actual calculations (keep in mind they still have to be correctly in bounds of your array). For example
x:=10*a(i) + q*i;
needs to either be
x:=10*a(i-1) + q*i;
or
x:=10*a(i) + q*(i+1);
depending on what you decide your array bounds to be. This applies to multiple lines in your code. See this Stackoverflow thread
You assign A(1) when your array starts at 0
Your loops to print out "0" and "9" should be either 1..length or 0 .. length-1
When you print the digits using Integer_Text_IO.Put, you need to specify a width of 1 to get rid of the spaces
There might be more, that's all I saw.
I think you are translating this answer.
You need to be more careful of your indices and your loop ranges; for example, you’ve translated
for(int i = len; i > 0; --i) {
int x = 10 * A[i-1] + q*i;
A[i-1] = x % (2*i - 1);
q = x / (2*i - 1);
}
as
for i in reverse 1..length loop
x:=10*a(i) + q*i;
a(i):= x mod (2*i-1);
q:= x/(2*i-1);
end loop;
The loop ranges are the same. But in the seocnd line, the C code uses A[i-1], whereas yours uses a(i); similarly in the third line.
Later, for
for (int k = 0; k < nines; ++k) {
printf("%d", 0);
}
you have
for k in 0..nines loop
put("0");
end loop;
in which the C loop runs from 0 to nines - 1, but yours runs from 0 to nines. So you put out one more 0 than you should (and later on, likewise for 9s).
Also, you should use put (predigit, width=> 0).
It's about this dynamic programming challenge.
If you have a hard time to understand the Problem then see also on AbhishekVermaIIT's post
Basically, you get as input an array B and you construct array A. Fo this array A you need the maximum possible sum with absolute(A[i] - A[i-1]), for i = 1 to N. How to construct array A? --> You can choose for every element A[i] in array A either the values 1 or B[i]. (As you will deduce from the problem description any other value between these two values doesn't make any sense.)
And I came up with this recursive Java solution (without memoization):
static int costHelper(int[] arr, int i) {
if (i < 1) return 0;
int q = max(abs(1 - arr[i-1]) + costHelper(arr, i-1) , abs(arr[i] - arr[i-1]) + costHelper(arr, i-1));
int[] arr1 = new int[i];
for (int j = 0; j < arr1.length-1; j++) {
arr1[j] = arr[j];
}
arr1[i-1] = 1;
int r = max(abs(1 - 1) + costHelper(arr1, i-1) , abs(arr[i] - 1) + costHelper(arr1, i-1));
return max(q , r);
}
static int cost(int[] arr) {
return costHelper(arr, arr.length-1);
}
public static void main(String[] args) {
int[] arr = {55, 68, 31, 80, 57, 18, 34, 28, 76, 55};
int result = cost(arr);
System.out.println(result);
}
Basically, I start at the end of the array and check what is maximizing the sum of the last element minus last element - 1. But I have 4 cases:
(1 - arr[i-1])
(arr[i] - arr[i-1])
(1 - 1) // I know, it is not necessary.
(arr[i] -1)
For the 3rd or 4th case I construct a new array one element smaller in size than the input array and with a 1 as the last element.
Now, the result of arr = 55 68 31 80 57 18 34 28 76 55 according to Hackerrank should be 508. But I get 564.
Since it has to be 508 I guess the array should be 1 68 1 80 1 1 34 1 76 1.
For other arrays I get the right answer. For example:
79 6 40 68 68 16 40 63 93 49 91 --> 642 (OK)
100 2 100 2 100 --> 396 (OK)
I don't understand what is wrong with this algorithm.
I'm not sure exactly what's happening with your particular solution but I suspect it might be that the recursive function only has one dimension, i, since we need a way to identify the best previous solution, f(i-1), both if B_(i-1) was chosen and if 1 was chosen at that point, so we can choose the best among them vis-a-vis f(i). (It might help if you could add a description of your algorithm in words.)
Let's look at the brute-force dynamic program: let m[i][j1] represent the best sum-of-abs-diff in A[0..i] when A_i is j1. Then, generally:
m[i][j1] = max(abs(j1 - j0) + m[i-1][j0])
for j0 in [1..B_(i-1)] and j1 in [1..B_i]
Python code:
def cost(arr):
if len(arr) == 1:
return 0
m = [[float('-inf')]*101 for i in xrange(len(arr))]
for i in xrange(1, len(arr)):
for j0 in xrange(1, arr[i-1] + 1):
for j1 in xrange(1, arr[i] + 1):
m[i][j1] = max(m[i][j1], abs(j1 - j0) + (m[i-1][j0] if i > 1 else 0))
return max(m[len(arr) - 1])
That works but times out since we are looping potentially 100*100*10^5 iterations.
I haven't thought through the proof for it, but, as you suggest, apparently we can choose only from either 1 or B_i for each A_i for an optimal solution. This allows us to choose between those directly in a significantly more efficient solution that won't time out:
def cost(arr):
if len(arr) == 1:
return 0
m = [[float('-inf')]*2 for i in xrange(len(arr))]
for i in xrange(1, len(arr)):
for j0 in [1, arr[i-1]]:
for j1 in [1, arr[i]]:
a_i = 0 if j1 == 1 else 1
b_i = 0 if j0 == 1 else 1
m[i][a_i] = max(m[i][a_i], abs(j1 - j0) + (m[i-1][b_i] if i > 1 else 0))
return max(m[len(arr) - 1])
This is a bottom-up tabulation but we could easily convert it to a recursive one using the same idea.
Here is the javascript code with memoization-
function cost(B,n,val) {
if(n==-1){
return 0;
}
let prev1=0,prev2=0;
if(n!=0){
if(dp[n-1][0]==-1)
dp[n-1][0] = cost(B,n-1,1);
if(dp[n-1][1]==-1)
dp[n-1][1] = cost(B,n-1,B[n]);
prev1=dp[n-1][0];
prev2=dp[n-1][1];
}
prev1 = prev1 + Math.abs(val-1);
prev2 = prev2+ Math.abs(val-B[n]);
return Math.max(prev1,prev2);
}
where B->given array,n->total length,val-> 1 or B[n], value considered by the calling function.
Initial call -> Math.max(cost(B,n-2,1),cost(B,n-2,B[n-1]));
BTW, this took me around 3hrs, rather could have easily done with iteration method. :p
//dp[][0] is when a[i]=b[i]
dp[i][0]=max((dp[i-1][0]+abs(b[i]-b[i-1])),(dp[i-1][1]+abs(b[i]-1)));
dp[i][1]=max((dp[i-1][1]+abs(1-1)),(dp[i-1][0]+abs(b[i-1]-1)));
Initially all the elements in dp have the value of 0.
We know that we will get the answer if at any i the value is b[i] or 1. So the final answer is :
max(dp[n-1][0],dp[n-1][1])
dp[i][0] signifies a[i]=b[i] and dp[i][1] signifies a[i]=1.
So at every i we want the maximum of [i-1][0] (previous element is b[i-1]) or [i-1][1] (previous element is 1)
I have a 2-dimensional array that looks like this:
1 1 0 0 1
1 0 1 1 0
0 0 1 1 0
1 1 0 1 1
0 0 1 1 1
I'm trying to figure out a way to identify the longest contiguous chain of 1's going either across or down. In this case, it starts at column 4, row 2, and its length is 4, going down.
I was thinking of using recursion, but I'm running into some issues keeping track of position, especially when encountering a 0.
So far, I have something along the lines of this (for checking across only):
main() {
...
for(i = 0; i < n; i++)
for(j = 0; j < n; j++)
if (G[i][j] == 1) {
CheckAcross(i, j, n);
}
...
}
void CheckAcross (int i, int j, int n) {
if (i < 0 || i >= n || j < 0 || j >= n) return; // outside of grid
if (G[i][j] == 0 ) return; //0 encountered
G[i][j] = WordCount + 1;
CheckAcross(i, j + 1, n);
}
where G[][] is the 2-dimensional array containing the 1's and 0's, n is the number of rows/columns, i is the row number and j is the column number.
Thanks for any assistance in advance!
Your current answer will take O(n3) time; to evaluate a single line, you check every possible start and end position (O(n) possibilities for each), and there are n lines.
Your algorithm is correct, but let's see if we can improve on the running time.
The problem might become simpler if we break it into simpler problems, i.e. "What is the longest contiguous chain of 1s in this 1-dimensional array?". If we solve it 2n times, then we have our answer, so we just need to get this one down to smaller than O(n2) for an improvement.
Well, we can simply go through the line, remembering the position (start and end) and length of the longest sequence of 1s. This takes O(n) time, and is optimal (if the sequence is all 1s or 0s, we would have to read every element to know where the start/end of the longest sequence is).
Then we can simply solve this for every row and every column, in O(n2) time.
Create a new n-by-n matrix called V. This will store, for each cell, the number of 1s at that cell and immediately above it. This will be O(n^2).
checkAllVertical(int n) {
V = malloc(....) // create V, an n-by-n matrix initialized to zero
for(int r=0; r<n; r++) {
for(int c=0; c<n; c++) {
if(G[r][c]=1) {
if(r==0)
V[r][c] = 1;
else
V[r][c] = 1 + V[r][c];
}
}
}
}
You don't really need to allocate all of V. One row at a time would suffice.
I have a small question about vector and matrix.
Suppose a vector V = {v1, v2, ..., vn}. I generate a n-by-n distance matrix M defined as:
M_ij = | v_i - v_j | such that i,j belong to [1, n].
That is, each element M_ij in the square matrix is the absolute distance of two elements in V.
For example, I have a vector V = {1, 3, 3, 5}, the distance matrix will be
M=[
0 2 2 4;
2 0 0 2;
2 0 0 2;
4 2 2 0; ]
It seems pretty simple. Now comes to the question. Given such a matrix M, how to obtain the initial V?
Thank you.
Based on some answer for this question, it seems that the answer is not unique. So, now suppose that all the initial vector has been normalized to 0 mean and 1 variance. The question is: Given such a symmetric distance matrix M, how to decide the initial normalized vector?
You can't. To give you an idea of why, consider these two cases:
V1 = {1,2,3}
M1 = [ 0 1 2 ; 1 0 1 ; 2 1 0 ]
V2 = {3,4,5}
M2 = [ 0 1 2 ; 1 0 1 ; 2 1 0 ]
As you can see, a single M could be the result of more than one V. Therefore, you can't map backwards.
There is no way to determine the answer uniquely, since the distance matrix is invariant to adding a constant to all elements and to multiplying all the values by -1. Assuming that element 1 is equal to 0, and that the first nonzero element is positive, however, you can find an answer. Here is the pseudocode:
# Assume v[1] is 0
v[1] = 0
# e is value of first non-zero vector element
e = 0
# ei is index of first non-zero vector element
ei = 0
for i = 2...n:
# if all vector elements have been 0 so far
if e == 0:
# get the current distance from element 1 and its index
# this new element may still be 0
e = d[1,i]
ei = i
v[i] = e
elseif d[1,i] == d[ei,i] + v[ei]: # v[i] <= v[1]
# v[i] is to the left of v[1] (assuming v[ei] > v[1])
v[i] = -d[1,i]
else:
# some other case; v[i] is to the right of v[1]
v[i] = d[1,i]
I don't think it is possible to find the original vector, but you can find a translation of the vector by taking the first row of the matrix.
If you let M_ij = | v_i - v_j | and you translate all v_k for k\in [1,n] you will get
M_ij = | v-i + 1 - v_j + 1 |
= | v_i - v_j |
Hence, just take the first row as the vector and find one initial point to translate the vector to.
Correction:
Let v_1 = 0, and let l_k = | v_k | for k\in [2,n] and p_k the parity of v_k
Let p_1 = 1
for(int i = 2; i < n; i++)
if( | l_i - l_(i+1) | != M_i(i+1) )
p_(i+1) = - p_i
else
p_(i+1) = p_i
doing this for all v_k for k\in [2,n] in order will show the parity of each v_k in respect to the others
Then you can find a translation of the original vector with the same or opposite direction
Update (For Normalized vector):
Let d = Sqrt(v_1^2 + v_2^2 + ... + v_n^2)
Vector = {0, v_1 / d, v_2 / d, ... , v_n / d}
or
{0, -v_1 / d, -v_2 / d, ... , -v_n / d}