From c132622a4c7c659759c69a999fb4ab4392b6bfc0 Mon Sep 17 00:00:00 2001 From: Tarun Khullar <36921843+roll-no-1@users.noreply.github.com> Date: Thu, 4 Oct 2018 19:58:05 +0530 Subject: [PATCH 1/5] Corrected a typo in the block sieve code Changed is_prime[j] = true to is_prime[j] = false --- src/algebra/sieve-of-eratosthenes.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/src/algebra/sieve-of-eratosthenes.md b/src/algebra/sieve-of-eratosthenes.md index b3430c789..7796764f3 100644 --- a/src/algebra/sieve-of-eratosthenes.md +++ b/src/algebra/sieve-of-eratosthenes.md @@ -145,7 +145,7 @@ int count_primes(int n) { if (is_prime[i]) { primes.push_back(i); for (int j = i * i; j <= nsqrt; j += i) - is_prime[j] = true; + is_prime[j] = false; } } From 534db0d0ab57a9c1c703f0f184c722cd9e134b79 Mon Sep 17 00:00:00 2001 From: Tarun Khullar <36921843+roll-no-1@users.noreply.github.com> Date: Fri, 5 Oct 2018 09:53:05 +0530 Subject: [PATCH 2/5] Reduced number of operations in block sieve code Changed max(start_idx, 2) * p to max(start_idx, p) * p. This is done so that we always start from the max. of the smallest multiple of p in the range [l, r] and p * p. It is more beneficial to start from p * p rather than 2 * p as lower multiples of p would have been covered by the smaller prime factors. --- src/algebra/sieve-of-eratosthenes.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/src/algebra/sieve-of-eratosthenes.md b/src/algebra/sieve-of-eratosthenes.md index 7796764f3..041a99750 100644 --- a/src/algebra/sieve-of-eratosthenes.md +++ b/src/algebra/sieve-of-eratosthenes.md @@ -156,7 +156,7 @@ int count_primes(int n) { int start = k * S; for (int p : primes) { int start_idx = (start + p - 1) / p; - int j = max(start_idx, 2) * p - start; + int j = max(start_idx, p) * p - start; for (; j < S; j += p) block[j] = false; } From bb9650e9eb2bb7e39308cf56deb580f2d9e84897 Mon Sep 17 00:00:00 2001 From: Tarun Khullar <36921843+roll-no-1@users.noreply.github.com> Date: Sun, 7 Oct 2018 16:26:44 +0530 Subject: [PATCH 3/5] Corrected a mistake in the text explaining block sieve code The text read: "Here we have an implementation that counts the number of primes smaller to n using block sieving" while actually the code counts the number of primes smaller than or equal to n. --- src/algebra/sieve-of-eratosthenes.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/src/algebra/sieve-of-eratosthenes.md b/src/algebra/sieve-of-eratosthenes.md index 041a99750..70e317eb2 100644 --- a/src/algebra/sieve-of-eratosthenes.md +++ b/src/algebra/sieve-of-eratosthenes.md @@ -132,7 +132,7 @@ We can work on blocks by turns, i.e. for every block $k$ we will go through all It is worth working on the first block differently: first, all the prime numbers from $[1; \sqrt n]$ shouldn't remove themselves; second, the numbers $0$ and $1$ should be marked as non-prime numbers. While working on the last block it should not be forgotten that the last needed number $n$ is not necessary located in the end of the block. -Here we have an implementation that counts the number of primes smaller than $n$ using block sieving. +Here we have an implementation that counts the number of primes smaller than or equal to $n$ using block sieving. ```cpp int count_primes(int n) { From 7a2c4eef4c1e209b92a9d0bbd4bb1c7cbd909200 Mon Sep 17 00:00:00 2001 From: Tarun Khullar <36921843+roll-no-1@users.noreply.github.com> Date: Mon, 15 Oct 2018 17:05:33 +0530 Subject: [PATCH 4/5] Corrected a typo in the application Two stripes Changed "arrays of with values values" to "arrays of values" in line 1 of this sub topic --- src/algebra/fft.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/src/algebra/fft.md b/src/algebra/fft.md index 0aef07f57..28fa86899 100644 --- a/src/algebra/fft.md +++ b/src/algebra/fft.md @@ -540,7 +540,7 @@ Thus these coefficients are the answer to the problem, and we were still able to ### Two stripes -We are given two Boolean stripes (cyclic arrays of with values values $0$ and $1$) $a$ and $b$. +We are given two Boolean stripes (cyclic arrays of values $0$ and $1$) $a$ and $b$. We want to find all ways to attach the first stripe to the second one, such that at no position we have a $1$ of the first stripe next to a $1$ of the second stripe. The problem doesn't actually differ much from the previous problem. From 9b788b47458492f01948dffc2d847af2b9d9f746 Mon Sep 17 00:00:00 2001 From: Tarun Khullar <36921843+roll-no-1@users.noreply.github.com> Date: Wed, 17 Oct 2018 15:44:10 +0530 Subject: [PATCH 5/5] Made some major changes to All possible scalar products subsection Here are the details and possible reasons for the change: * Changed (k-n-1) th shift to (k-(n-1))th shift. The former evaluates to -1th shift for k = n and is wrong. * Changed cyclic shift to cyclic left shift. I always got confused while reading this since I tend to forget whether it was the left shift or the right shift of b. So, I decided to add it explicitly since others may also face the same problem in the future without this info. * Changed index nos. of k from [n .. 2n-1] to [n-1 .. 2n-2]. Since c[n] == c[2n-1], it might make more sense to start from n-1 since after doing this we get shifts in the increasing order i.e., 0th shift, 1st shift, ... , (n-1)th shift. * Added a note in case anyone wonders about the value at c[2n-1]. --- src/algebra/fft.md | 5 +++-- 1 file changed, 3 insertions(+), 2 deletions(-) diff --git a/src/algebra/fft.md b/src/algebra/fft.md index 28fa86899..31c538b93 100644 --- a/src/algebra/fft.md +++ b/src/algebra/fft.md @@ -531,12 +531,13 @@ We have to compute the products of $a$ with every cyclic shift of $b$. We generate two new arrays of size $2n$: We reverse $a$ and append $n$ zeros to it. And we just append $b$ to itself. -When we multiply these two arrays as polynomials, and look at the coefficient $c[n],~ c[n+1],~ c[2n-1]$ of the product $c$, we get: +When we multiply these two arrays as polynomials, and look at the coefficient $c[n-1],~ c[n],~ c[2n-2]$ of the product $c$, we get: $$c[k] = \sum_{i+j=k} a[i] b[j]$$ And since all the elements $a[i] = 0$ for $i \ge n$: $$c[k] = \sum_{i=0}^{n-1} a[i] b[k-i]$$ -It is easy to see that this sum is just the scalar product of the vector $a$ with the $(k - n - 1)$-th cyclic shift. +It is easy to see that this sum is just the scalar product of the vector $a$ with the $(k - (n - 1))$-th cyclic left shift. Thus these coefficients are the answer to the problem, and we were still able to obtain it in $O(n \log n)$ time. +Note here that $c[2n-1]$ also gives us the $n$th cyclic shift but that is the same as the $0$th cyclic shift so we don't need to consider that seperately into our answer. ### Two stripes pFad - Phonifier reborn

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