Given a pyramid like:
0
1 2
3 4 5
6 7 8 9
...
and given the index of the pyramid i where i represents the ith number of the pyramid, is there a way to find the index of the row to which the ith element belongs? (e.g. if i = 6,7,8,9, it is in the 3rd row, starting from row 0)
There's a connection between the row numbers and the triangular numbers. The nth triangular number, denoted Tn, is given by Tn = n(n-1)/2. The first couple triangular numbers are 0, 1, 3, 6, 10, 15, etc., and if you'll notice, the starts of each row are given by the nth triangular number (the fact that they come from this triangle is where this name comes from.)
So really, the goal here is to determine the largest n such that Tn ≤ i. Without doing any clever math, you could solve this in time O(√n) by just computing T0, T1, T2, etc. until you find something bigger than i. Even better, you could binary search for it in time O(log n) by computing T1, T2, T4, T8, etc. until you overshoot, then binary searching on the range you found.
Alternatively, we could try to solve for this directly. Suppose we want to find the choice of n such that
n(n + 1) / 2 = i
Expanding, we get
n2 / 2 + n / 2 = i.
Equivalently,
n2 / 2 + n / 2 - i = 0,
or, more easily:
n2 + n - 2i = 0.
Now we use the quadratic formula:
n = (-1 ± √(1 + 8i)) / 2
The negative root we can ignore, so the value of n we want is
n = (-1 + √(1 + 8i)) / 2.
This number won't necessarily be an integer, so to find the row you want, we just round down:
row = ⌊(-1 + √(1 + 8i)) / 2⌋.
In code:
int row = int((-1 + sqrt(1 + 8 * i)) / 2);
Let's confirm that this works by testing it out a bit. Where does 9 go? Well, we have
(-1 + √(1 + 72)) / 2 = (-1 + √73) / 2 = 3.77
Rounding down, we see it goes in row 3 - which is correct!
Trying another one, where does 55 go? Well,
(-1 + √(1 + 440)) / 2 = (√441 - 1) / 2 = 10
So it should go in row 10. The tenth triangular number is T10 = 55, so in fact, 55 starts off that row. Looks like it works!
I get row = math.floor (√(2i + 0.25) - 0.5) where i is your number
Essentially the same as the guy above but I reduced n2 + n to (n + 0.5)2 - 0.25
I think ith element belongs nth row where n is number of n(n+1)/2 <= i < (n+1)(n+2)/2
For example, if i = 6, then n = 3 because n(n+1)/2 <= 6
and if i = 8, then n = 3 because n(n+1)/2 <= 8
Related
For n binary digits with base x, the maximum value will be:
x^(n-1) + x^(n-2) + ... + x^1 + x^0
By using geometric progression,
r=1/x
Using formula for sum of n finite numbers i get:
(x^n - 1) / (x - 1).
But my answer should have been x^n - 1.(formula: M= x^n - 1)
You are correct that the sum of the geometric series x0 + x1 + x2 + ... + xn-1 is indeed (xn - 1) / (x - 1). For example, if we pick x = 10 (base 10) and n = 3 (a three-digit base 10 number), we get back
1 + 10 + 100 = (1000 - 1) / 9 = 999 / 9 = 111.
However, the largest three-digit number is 999. And by looking at the above sum, you might get a sense of why we're off by a factor of 9. When writing out numbers in base 10, we'd maximize our number by having each digit be 9, not 1. And more generally, in base x, we'd maximize our value by having each digit be x - 1. That means that the maximum value is
(x - 1)(x0 + x1 + x2 + ... + xn-1) = (x - 1)(xn - 1)/(x - 1) = xn - 1.
Here's another, easier way to see this. What is the smallest number you can make with n+1 digits? That would be xn. Since that's the smallest (n+1)-digit number, the largest n-digit number must be that minus one, giving xn - 1 without needing to discuss geometric series.
Hi I was looking up solutions to a leetcode problem, I know how to solve the problem but someone else submitted this solution and I don't understand how this works.
The question is how many number of stacks can you form with n number of coins where k-th row has k number of coins. https://leetcode.com/explore/challenge/card/july-leetcoding-challenge/544/week-1-july-1st-july-7th/3377/
Returning the above formula works, can anyone explain it to me?
The sum of the first N natural numbers (1 + 2 + 3 + ... + N) is known to be equal to N(N+1)2
The game says that if you have 6 coins you have to stack them in this way:
x (1)
x x (2)
x x x (3)
and 6 is equal to 1 + 2 + 3.
If you are given K coins and you know that N(N+1)/2 = K then you know that you can have N rows. Now the question is, given K, how can you find N?
Let's do the math:
N(N+1)/2 = K
N^2 + N = 2*K
N^2 + N -2*K = 0
N = (-1 + sqrt(1 + 8K))/2
N = -1/2 + sqrt(1/4 + 2K)
N = sqrt(2*K + 0.25) - 0.5
I have this recurrence formula:
P(n) = ( P(n-1) + 2^(n/2) ) % (X)
s.t. P(1) = 2;
where n/2 is computer integer division i.e. floor of x/2
Since i am taking mod X, this relation should repeat at least with in X outputs.
but it can start repeating before that.
How to find this value?
It needn't repeat within x terms, consider x = 3:
P(1) = 2
P(2) = (P(1) + 2^(2/2)) % 3 = 4 % 3 = 1
P(3) = (P(2) + 2^(3/2)) % 3 = (1 + 2) % 3 = 0
P(4) = (P(3) + 2^(4/2)) % 3 = 4 % 3 = 1
P(5) = (P(4) + 2^(5/2)) % 3 = (1 + 4) % 3 = 2
P(6) = (P(5) + 2^(6/2)) % 3 = (2 + 8) % 3 = 1
P(7) = (P(6) + 2^(7/2)) % 3 = (1 + 8) % 3 = 0
P(8) = (P(7) + 2^(8/2)) % 3 = 16 % 3 = 1
P(9) = (P(8) + 2^(9/2)) % 3 = (1 + 16) % 3 = 2
P(10) = (P(9) + 2^(10/2)) % 3 = (2 + 32) % 3 = 1
P(11) = (P(10) + 2^(11/2)) % 3 = (1 + 32) % 3 = 0
P(12) = (P(11) + 2^(12/2)) % 3 = (0 + 64) % 3 = 1
and you see that the period is 4.
Generally (suppose X is odd, it's a bit more involved for even X), let k be the period of 2 modulo X, i.e. k > 0, 2^k % X = 1, and k is minimal with these properties (see below).
Consider all arithmetic modulo X. Then
n
P(n) = 2 + ∑ 2^(j/2)
j=2
It is easier to see when we separately consider odd and even n:
m m
P(2*m+1) = 2 + 2 * ∑ 2^i = 2 * ∑ 2^i = 2*(2^(m+1) - 1) = 2^((n+2)/2) + 2^((n+1)/2) - 2
i=1 i=0
since each 2^j appears twice, for j = 2*i and j = 2*i+1. For even n = 2*m, there's one summand 2^m missing, so
P(2*m) = 2^(m+1) + 2^m - 2 = 2^((n+2)/2) + 2^((n+1)/2) - 2
and we see that the length of the period is 2*k, since the changing parts 2^((n+1)/2) and 2^((n+2)/2) have that period. The period immediately begins, there is no pre-period part (there can be a pre-period for even X).
Now k <= φ(X) by Euler's generalisation of Fermat's theorem, so the period is at most 2 * φ(X).
(φ is Euler's totient function, i.e. φ(n) is the number of integers 1 <= k <= n with gcd(n,k) = 1.)
What makes it possible that the period is longer than X is that P(n+1) is not completely determined by P(n), the value of n also plays a role in determining P(n+1), in this case the dependence is simple, each power of 2 being used twice in succession doubles the period of the pure powers of 2.
Consider the sequence a[k] = (2^k) % X for odd X > 1. It has the simple recurrence
a[0] = 1
a[k+1] = (2 * a[k]) % X
so each value completely determines the next, thus the entire following part of the sequence. (Since X is assumed odd, it also determines the previous value [if k > 0] and thus the entire previous part of the sequence. With H = (X+1)/2, we have a[k-1] = (H * a[k]) % X.)
Hence if the sequence assumes one value twice (and since there are only X possible values, that must happen within the first X+1 values), at indices i and j = i+p > i, say, the sequence repeats and we have a[k+p] = a[k] for all k >= i. For odd X, we can go back in the sequence, therefore a[k+p] = a[k] also holds for 0 <= k < i. Thus the first value that occurs twice in the sequence is a[0] = 1.
Let p be the smallest positive integer with a[p] = 1. Then p is the length of the smallest period of the sequence a, and a[k] = 1 if and only if k is a multiple of p, thus the set of periods of a is the set of multiples of p. Euler's theorem says that a[φ(X)] = 1, from that we can conclude that p is a divisor of φ(X), in particular p <= φ(X) < X.
Now back to the original sequence.
P(n) = 2 + a[1] + a[1] + a[2] + a[2] + ... + a[n/2]
= a[0] + a[0] + a[1] + a[1] + a[2] + a[2] + ... + a[n/2]
Since each a[k] is used twice in succession, it is natural to examine the subsequences for even and odd indices separately,
E[m] = P(2*m)
O[m] = P(2*m+1)
then the transition from one value to the next is more regular. For the even indices we find
E[m+1] = E[m] + a[m] + a[m+1] = E[m] + 3*a[m]
and for the odd indices
O[m+1] = O[m] + a[m+1] + a[m+1] = O[m] + 2*a[m+1]
Now if we ignore the modulus for the moment, both E and O are geometric sums, so there's an easy closed formula for the terms. They have been given above (in slightly different form),
E[m] = 3 * 2^m - 2 = 3 * a[m] - 2
O[m] = 2 * 2^(m+1) - 2 = 2 * a[m+1] - 2 = a[m+2] - 2
So we see that O has the same (minimal) period as a, namely p, and E also has that period. Unless maybe if X is divisible by 3, that is also the minimal (positive) period of E (if X is divisible by 3, the minimal positive period of E could be a proper divisor of p, for X = 3 e.g., E is constant).
Thus we see that 2*p is a period of the sequence P obtained by interlacing E and O.
It remains to be seen that 2*p is the minimal positive period of P. Let m be the minimal positive period. Then m is a divisor of 2*p.
Suppose m were odd, m = 2*j+1. Then
P(1) = P(m+1) = P(2*m+1)
P(2) = P(m+2) = P(2*m+2)
and consequently
P(2) - P(1) = P(m+2) - P(m+1) = P(2*m+2) - P(2*m+1)
But P(2) - P(1) = a[1] and
P(m+2) - P(m+1) = a[(m+2)/2] = a[j+1]
P(2*m+2) - P(2*m+1) = a[(2*m+2)/2] = a[m+1] = a[2*j+2]
So we must have a[1] = a[j+1], hence j is a period of a, and a[j+1] = a[2*j+2], hence j+1 is a period of a too. But that means that 1 is a period of a, which implies X = 1, a contradiction.
Therefore m is even, m = 2*j. But then j is a period of O (and of E), thus a multiple of p. On the other hand, m <= 2*p implies j <= p, and the only (positive) multiple of p satisfying that inequality is p itself, hence j = p, m = 2*p.
I would like to compute the average of three numbers, like:
d = int(round((a + b + c) / 3.0))
Where a, b, c, d are integers.
Is it possible to get the same result using just integers?
I'm interested in this because of performance reasons, I assume doing the math using integers should be faster than using floats.
The example above converts the integers to floats, calculates the result, rounds it and converts back to integer. Is it possible to avoid the int <-> float conversions?
Given the requirements for 1, 1, 2 -> 1; 1, 2, 2 -> 2 then this can be done using integer division.
Using // for the integer division and n for number of elements.
average = ( a+ b + c + .... + n//2 ) // n
ie sum up all the values and then add a number to deal with rounding.
As noted in #Henrik's answer this assumes that all numbers are positive.
(a + b + c + 1) / 3
Explanation: if (a + b + c) % 3 == 1, it's rounded down; if (a + b + c) % 3 == 2, it's rounded up.
At least this should work for a + b + c >= 0. You may need to treat negative values separately.
A variant of Mahmoud's answer using only one division:
d = (((a+b+c) * 10) + 15) / 30
It should be as simple as that:
d = (((a+b+c) * 10) / 3 + 5) / 10;
Suppose there is number s=12 , now i want to make sequence with the element a1+a2+.....+an=12.
The criteria is as follows-
n must be minimum.
a1 and an must be 1;
ai can differs a(i-1) by only 1,0 and -1.
for s=12 the result is 6.
So how to find the minimum value of n.
Algorithm for finding n from given s:
1.Find q = FLOOR( SQRT(s-1) )
2.Find r = q^2 + q
3.If s <= r then n = 2q, else n = 2q + 1
Example: s = 12
q = FLOOR( SQRT(12-1) ) = FLOOR(SQRT(11) = 3
r = 3^2 + 3 = 12
12 <= 12, therefore n = 2*3 = 6
Example: s = 160
q = FLOOR( SQRT(160-1) ) = FLOOR(SQRT(159) = 12
r = 12^2 + 12 = 156
159 > 156, therefore n = 2*12 + 1 = 25
and the 25-numbers sequence for
159: 1,2,3,4,5,6,7,8,9,10,10,10,9,10,10,10,9,8,7,6,5,4,3,2,1
Here's a way to visualize the solution.
First, draw the smallest triangle (rows containing successful odd numbers of stars) that has a greater or equal number of stars to n. In this case, we draw a 16-star triangle.
*
***
*****
*******
Then we have to remove 16 - 12 = 4 more stars. We do this diagonally starting from the top.
1
**2
****3
******4
The result is:
**
****
******
Finally, add up the column heights to get the final answer:
1, 2, 3, 3, 2, 1.
There are two cases: s odd and s even. When s is odd, you have the sequence:
1, 2, 3, ..., (s-1)/2, (s-1)/2, (s-1)/2-1, (s-1)/2-2, ..., 1
when n is even you have:
1, 2, 3, ..., s/2, s/2-1, s/2-2, ..., 1
The maximum possible for any given series of length n is:
n is even => (n^2+2n)/4
n is odd => (n+1)^2/4
These two results are arrived at easily enough by looking at the simple arithmetic sum of series where in the case of n even it is twice the sum of the series 1...n/2. In the case of n odd it is twice the sum of the series 1...(n-1)/2 and add on n+1/2 (the middle element).
Clearly you can generate any positive number that is less than this max as long as n>3.
So the problem then becomes finding the smallest n with a max greater than your target.
Algorithmically I'd go for:
Find (sqrt(4*s)-1) and round up to the next odd number. Call this M. This is an easy to work out value and will represent the lowest odd n that will work.
Check M-1 to see if its max sum is greater than s. If so then that your n is M-1. Otherwise your n is M.
Thank all you answer me. I derived a simpler solution. The algorithm looks like-
First find what is the maximum sum that can be made using n element-
if n=1 -> 1 sum=1;
if n=2 -> 1,1 sum=2;
if n=3 -> 1,2,1 sum=4;
if n=4 -> 1,2,2,1 sum=6;
if n=5 -> 1,2,3,2,1 sum=9;
if n=6 -> 1,2,3,3,2,1 sum=12;
So from observation it is clear that form any number,n 9<n<=12 can be
made using 6 element, similarly number
6<n<=9 can be made at using 5 element.
So it require only a binary search to find the number of
element that make a particular number.