From 4bbd576ec0f4c26a6710ecae6050e2522c84616c Mon Sep 17 00:00:00 2001
From: Wiktor Kuchta <wiktorkuchta@protonmail.com>
Date: Fri, 1 Feb 2019 21:50:12 +0100
Subject: [PATCH 1/8] Fix typo in Edmonds-Karp

---
 src/graph/edmonds_karp.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/src/graph/edmonds_karp.md b/src/graph/edmonds_karp.md
index 8c2c206be..07b45d8ba 100644
--- a/src/graph/edmonds_karp.md
+++ b/src/graph/edmonds_karp.md
@@ -36,7 +36,7 @@ The following image show a flow network.
 The first value of each edge represents the flow, which is initially 0, and the second value represents the capacity.
 <center>![Flow network](&imgroot&/Flow1.png)</center>
 
-We value of a flow of a network is the sum of all flows that gets produced in source $s$, or equivalent or the flows that is consumed in the sink $t$.
+The value of a flow of a network is the sum of all flows that gets produced in source $s$, or equivalently of the flows that are consumed in the sink $t$.
 A **maximal flow** is a flow with the maximal possible value.
 Finding this maximal flow of a flow network is the problem that we want to solve.
 

From 820c2b6fe8fd80a27362fc6eefe51a82684a1db8 Mon Sep 17 00:00:00 2001
From: Wiktor Kuchta <wiktorkuchta@protonmail.com>
Date: Fri, 1 Feb 2019 21:50:41 +0100
Subject: [PATCH 2/8] s/Lets/Let's/g

---
 src/algebra/fft.md                        | 2 +-
 src/data_structures/disjoint_set_union.md | 2 +-
 src/data_structures/segment_tree.md       | 2 +-
 src/graph/2SAT.md                         | 2 +-
 src/graph/fixed_length_paths.md           | 2 +-
 5 files changed, 5 insertions(+), 5 deletions(-)

diff --git a/src/algebra/fft.md b/src/algebra/fft.md
index 958ca11e5..91b4b028d 100644
--- a/src/algebra/fft.md
+++ b/src/algebra/fft.md
@@ -80,7 +80,7 @@ $$A(x) = A_0(x^2) + x A_1(x^2).$$
 The polynomials $A_0$ and $A_1$ are only half as much coefficients as the polynomial $A$.
 If we can compute the $\text{DFT}(A)$ in linear time using $\text{DFT}(A_0)$ and $\text{DFT}(A_1)$, then we get the recurrence $T_{\text{DFT}}(n) = 2 T_{\text{DFT}}\left(\frac{n}{2}\right) + O(n)$ for the time complexity, which results in $T_{\text{DFT}}(n) = O(n \log n)$ by the **master theorem**.
 
-Lets learn how we can accomplish that.
+Let's learn how we can accomplish that.
 
 Suppose we have computed the vectors $\left(y_k^0\right)\_{k=0}^{n/2-1} = \text{DFT}(A_0)$ and $\left(y_k^1\right)\_{k=0}^{n/2-1} = \text{DFT}(A_1)$.
 Let us find a expression for $\left(y_k\right)_{k=0}^{n-1} = \text{DFT}(A)$.
diff --git a/src/data_structures/disjoint_set_union.md b/src/data_structures/disjoint_set_union.md
index 3666b7bc3..56434f300 100644
--- a/src/data_structures/disjoint_set_union.md
+++ b/src/data_structures/disjoint_set_union.md
@@ -347,7 +347,7 @@ The only difficulty that we face is to compute the parity in the `union_find` me
 
 If we add an edge $(a, b)$ that connects two connected components into one, then when you attach one tree to another we need to adjust the parity.
 
-Lets derive a formula, which computes the parity issued to the leader of the set that will get attached to another set.
+Let's derive a formula, which computes the parity issued to the leader of the set that will get attached to another set.
 Let $x$ be the parity of the path length from vertex $a$ up to its leader $A$, and $y$ as the parity of the path length from vertex $b$ up to its leader $B$, and $t$ the desired parity that we have to assign to $B$ after the merge.
 The path contains the of the three parts:
 from $B$ to $b$, from $b$ to $a$, which is connected by one edge and therefore has parity $1$, and from $a$ to $A$.
diff --git a/src/data_structures/segment_tree.md b/src/data_structures/segment_tree.md
index 371c8b55d..77059896c 100644
--- a/src/data_structures/segment_tree.md
+++ b/src/data_structures/segment_tree.md
@@ -71,7 +71,7 @@ The time complexity of the construction is $O(n)$.
 For now we are going to answer sum queries. As an input we receive two integers $l$ and $r$, and we have to compute the sum of the segment $a[l \dots r]$ in $O(\log n)$ time. 
 
 To do this, we will traverse the Segment Tree and use the precomputed sums of the segments.
-Lets assume that we are currently at the vertex that covers the segment $a[tl \dots tr]$.
+Let's assume that we are currently at the vertex that covers the segment $a[tl \dots tr]$.
 There are three possible cases. 
 
 The easiest case is when the segment $a[l \dots r]$ is equal to the corresponding segment of the current vertex (i.e. $a[l \dots r] = a[tl \dots tr]$), then we are finished and can return the precomputed sum that is stored in the vertex.
diff --git a/src/graph/2SAT.md b/src/graph/2SAT.md
index e2858a207..df2d670a1 100644
--- a/src/graph/2SAT.md
+++ b/src/graph/2SAT.md
@@ -22,7 +22,7 @@ We now construct a directed graph of these implications:
 for each variable $x$ there will be two vertices $v_x$ and $v_{\lnot x}$.
 The edges will correspond to the implications.
 
-Lets look at the example in 2-CNF form:
+Let's look at the example in 2-CNF form:
 
 $$(a \lor \lnot b) \land (\lnot a \lor b) \land (\lnot a \lor \lnot b) \land (a \lor \lnot c)$$
 
diff --git a/src/graph/fixed_length_paths.md b/src/graph/fixed_length_paths.md
index adee58ccd..d7ee69931 100644
--- a/src/graph/fixed_length_paths.md
+++ b/src/graph/fixed_length_paths.md
@@ -20,7 +20,7 @@ It is obvious that the constructed adjacency matrix if the answer to the problem
 It contains the number of paths of length $1$ between each pair of vertices.
 
 We will build the solution iteratively:
-Lets assume we know the answer for some $k$.
+Let's assume we know the answer for some $k$.
 Here we describe a method how we can construct the answer for $k + 1$.
 Denote by $C_k$ the matrix for the case $k$, and by $C_{k+1}$ the matrix we want to construct.
 With the following formula we can compute every entry of $C_{k+1}$:

From cddebce9a79c8c3dcc43c9460dc3b8cad307b37b Mon Sep 17 00:00:00 2001
From: Wiktor Kuchta <wiktorkuchta@protonmail.com>
Date: Fri, 1 Feb 2019 19:31:50 +0100
Subject: [PATCH 3/8] totient: Fix inconsistent use of phi and varphi

---
 src/algebra/phi-function.md | 4 ++--
 1 file changed, 2 insertions(+), 2 deletions(-)

diff --git a/src/algebra/phi-function.md b/src/algebra/phi-function.md
index 9087f2798..e1eef0cb7 100644
--- a/src/algebra/phi-function.md
+++ b/src/algebra/phi-function.md
@@ -75,7 +75,7 @@ This allows computing $x^n \bmod m$ for very big $n$, especially if $n$ is the r
 
 There is a less known version of the last equivalence, that allows computing $x^n \bmod m$ efficiently for not coprime $x$ and $m$.
 For arbitrary $x, m$ and $n \geq \log_2 m$:
-$$x^{n}\equiv x^{\varphi(m)+[n \bmod \varphi(m)]} \mod m$$
+$$x^{n}\equiv x^{\phi(m)+[n \bmod \phi(m)]} \mod m$$
 
 Proof:
 
@@ -100,7 +100,7 @@ $$x^n \bmod m = x^k\left(x^{n-k \bmod \phi(\frac{m}{a})} \bmod \frac{m}{a}\right
 This formula is difficult to apply, but we can use it to analyze the behavior of $x^n \bmod m$. We can see that the sequence of powers $(x^1 \bmod m, x^2 \bmod m, x^3 \bmod m, \dots)$ enters a cycle of length $\phi\left(\frac{m}{a}\right)$ after the first $k$ (or less) elements. 
 $\phi(m)$ divides $\phi\left(\frac{m}{a}\right)$ (because $a$ and $\frac{m}{a}$ are coprime we have $\phi(a) \cdot \phi\left(\frac{m}{a}\right) = \phi(m)$), therefore we can also say that the period has length $\phi(m)$.
 And since $\phi(m) \ge \log_2 m \ge k$, we can conclude the desired, much simpler, formula:
-$$ x^n \equiv x^{\varphi(m)} x^{(n - \varphi(m)) \bmod \varphi(m)} \bmod m \equiv x^{\varphi(m)+[n \bmod \varphi(m)]} \mod m.$$
+$$ x^n \equiv x^{\phi(m)} x^{(n - \phi(m)) \bmod \phi(m)} \bmod m \equiv x^{\phi(m)+[n \bmod \phi(m)]} \mod m.$$
 
 ## Practice Problems  
 

From 2731dc23b68e46c8982254d21151aa7035b5e6a6 Mon Sep 17 00:00:00 2001
From: Wiktor Kuchta <wiktorkuchta@protonmail.com>
Date: Fri, 1 Feb 2019 21:27:19 +0100
Subject: [PATCH 4/8] Fix typos in discrete log

---
 src/algebra/discrete-log.md | 7 +++++--
 1 file changed, 5 insertions(+), 2 deletions(-)

diff --git a/src/algebra/discrete-log.md b/src/algebra/discrete-log.md
index 19045d9ec..9d323a1a2 100644
--- a/src/algebra/discrete-log.md
+++ b/src/algebra/discrete-log.md
@@ -85,7 +85,7 @@ int solve(int a, int b, int m) {
 	for (int p = n; p >= 1; --p)
 		vals[powmod(a, p * n, m)] = p;
 	for (int q = 0; q <= n; ++q) {
-		int cur = (powmod (a, q, m) * b) % m;
+		int cur = (powmod(a, q, m) * b) % m;
 		if (vals.count(cur)) {
 			int ans = vals[cur] * n - q;
 			return ans;
@@ -106,7 +106,10 @@ We also need to change the second step accordingly.
 
 ## Improved implementation
 
-A possible improvement is to get rid of binary exponentiation. This can be done by keeping a variable that is multiplied by $a$ each time we increase $q$ and a variable that is multiplied by $a^n$ each time we increase $p$. With this change, the complexity of the algorithm is still the same, but now the $\log$ factor is only for `map`. Instead of a `map`, we can also use a hash table (`unordered_map` in C++) which has the average time complexity $O(1)$ for inserting and searching.
+A possible improvement is to get rid of binary exponentiation.
+This can be done by keeping a variable that is multiplied by $a$ each time we increase $q$ and a variable that is multiplied by $a^n$ each time we increase $p$.
+With this change, the complexity of the algorithm is still the same, but now the $\log$ factor is only for the `map`.
+Instead of a `map`, we can also use a hash table (`unordered_map` in C++) which has the average time complexity $O(1)$ for inserting and searching.
 
 ```cpp
 int solve(int a, int b, int m) {

From 210d4fac0e8a93eb6664f7ffca6ba4476cfbc1b8 Mon Sep 17 00:00:00 2001
From: Wiktor Kuchta <wiktorkuchta@protonmail.com>
Date: Fri, 1 Feb 2019 21:37:22 +0100
Subject: [PATCH 5/8] Mention reducing the result in discrete log

Probably better than the `ans < m` check, reasoning of which was lost in
translation.
---
 src/algebra/discrete-log.md | 4 +++-
 1 file changed, 3 insertions(+), 1 deletion(-)

diff --git a/src/algebra/discrete-log.md b/src/algebra/discrete-log.md
index 9d323a1a2..f77bf2f1b 100644
--- a/src/algebra/discrete-log.md
+++ b/src/algebra/discrete-log.md
@@ -64,7 +64,9 @@ $$O(\sqrt {m} \log m).$$
 
 ### The simplest implementation
 
-In the following code, the function `powmod` calculates $a^b \pmod m$ and the function `solve` produces a proper solution to the problem. It returns $-1$ if there is no solution and returns one of the possible solutions otherwise.
+In the following code, the function `powmod` calculates $a^b \pmod m$ and the function `solve` produces a proper solution to the problem.
+It returns $-1$ if there is no solution and returns one of the possible solutions otherwise.
+The resulting discrete logarithm can be big, but you can make it smaller using [Euler's theorem](./algebra/phi-function.html#toc-tgt-2).
 
 ```cpp
 int powmod(int a, int b, int m) {

From 7cb31bfeaed527d60b08a26fad121fdc37bb37e1 Mon Sep 17 00:00:00 2001
From: Wiktor Kuchta <wiktorkuchta@protonmail.com>
Date: Fri, 1 Feb 2019 22:21:58 +0100
Subject: [PATCH 6/8] Improve code style in discrete root

More consistent and modern, without type casting noise
---
 src/algebra/discrete-root.md | 84 ++++++++++++++++++++----------------
 1 file changed, 46 insertions(+), 38 deletions(-)

diff --git a/src/algebra/discrete-root.md b/src/algebra/discrete-root.md
index 21d916bbb..14c8fb51d 100644
--- a/src/algebra/discrete-root.md
+++ b/src/algebra/discrete-root.md
@@ -53,81 +53,89 @@ This is the final formula for all solutions of discrete root problem.
 Here is a full implementation, including routines for finding the primitive root, discrete log and finding and printing all solutions.
 
 ```cpp
-int gcd (int a, int b) {
-	return a ? gcd (b%a, a) : b;
+int gcd(int a, int b) {
+	return a ? gcd(b % a, a) : b;
 }
  
-int powmod (int a, int b, int p) {
+int powmod(int a, int b, int p) {
 	int res = 1;
-	while (b)
-		if (b & 1)
-			res = int (res * 1ll * a % p),  --b;
-		else
-			a = int (a * 1ll * a % p),  b >>= 1;
+	while (b > 0) {
+		if (b & 1) {
+			res = res * a % p;
+		}
+		a = a * a % p;
+		b >>= 1;
+	}
 	return res;
 }
  
-int generator (int p) {
+// Finds the primitive root modulo p
+int generator(int p) {
 	vector<int> fact;
-	int phi = p-1,  n = phi;
-	for (int i=2; i*i<=n; ++i)
+	int phi = p-1, n = phi;
+	for (int i = 2; i * i <= n; ++i) {
 		if (n % i == 0) {
-			fact.push_back (i);
+			fact.push_back(i);
 			while (n % i == 0)
 				n /= i;
 		}
+	}
 	if (n > 1)
-		fact.push_back (n);
+		fact.push_back(n);
  
-	for (int res=2; res<=p; ++res) {
+	for (int res = 2; res <= p; ++res) {
 		bool ok = true;
-		for (size_t i=0; i<fact.size() && ok; ++i)
-			ok &= powmod (res, phi / fact[i], p) != 1;
-		if (ok)  return res;
+		for (int factor : fact) {
+			if (powmod(res, phi / factor, p) != 1) {
+				ok = false;
+				break;
+			}
+		}
+		if (ok) return res;
 	}
 	return -1;
 }
  
+// This program finds all numbers x such that x^k = a (mod n)
 int main() {
- 
 	int n, k, a;
-	cin >> n >> k >> a;
+	scanf("%d %d %d", &n, &k, &a);
 	if (a == 0) {
-		puts ("1\n0");
+		puts("1\n0");
 		return 0;
 	}
  
-	int g = generator (n);
+	int g = generator(n);
  
+	// Baby-step giant-step discrete logarithm algorithm
 	int sq = (int) sqrt (n + .0) + 1;
-	vector < pair<int,int> > dec (sq);
-	for (int i=1; i<=sq; ++i)
-		dec[i-1] = make_pair (powmod (g, int (i * sq * 1ll * k % (n - 1)), n), i);
-	sort (dec.begin(), dec.end());
+	vector<pair<int, int>> dec(sq);
+	for (int i = 1; i <= sq; ++i)
+		dec[i-1] = {powmod(g, i * sq * k % (n - 1), n), i};
+	sort(dec.begin(), dec.end());
 	int any_ans = -1;
-	for (int i=0; i<sq; ++i) {
-		int my = int (powmod (g, int (i * 1ll * k % (n - 1)), n) * 1ll * a % n);
-		vector < pair<int,int> >::iterator it =
-			lower_bound (dec.begin(), dec.end(), make_pair (my, 0));
+	for (int i = 0; i < sq; ++i) {
+		int my = powmod(g, i * k % (n - 1), n) * a % n;
+		auto it = lower_bound(dec.begin(), dec.end(), make_pair(my, 0));
 		if (it != dec.end() && it->first == my) {
 			any_ans = it->second * sq - i;
 			break;
 		}
 	}
 	if (any_ans == -1) {
-		puts ("0");
+		puts("0");
 		return 0;
 	}
  
-	int delta = (n-1) / gcd (k, n-1);
+	// Print all possible answers
+	int delta = (n-1) / gcd(k, n-1);
 	vector<int> ans;
-	for (int cur=any_ans%delta; cur<n-1; cur+=delta)
-		ans.push_back (powmod (g, cur, n));
-	sort (ans.begin(), ans.end());
-	printf ("%d\n", ans.size());
-	for (size_t i=0; i<ans.size(); ++i)
-		printf ("%d ", ans[i]);
- 
+	for (int cur = any_ans % delta; cur < n-1; cur += delta)
+		ans.push_back(powmod(g, cur, n));
+	sort(ans.begin(), ans.end());
+	printf("%d\n", ans.size());
+	for (int answer : ans)
+		printf("%d ", answer);
 }
 ```
 

From 0f073326724d9ee6ec43b01fb1f775ceeb80b5ae Mon Sep 17 00:00:00 2001
From: Wiktor Kuchta <wiktorkuchta@protonmail.com>
Date: Fri, 1 Feb 2019 22:23:07 +0100
Subject: [PATCH 7/8] Fix typos in discrete root

---
 src/algebra/discrete-root.md | 10 +++++-----
 1 file changed, 5 insertions(+), 5 deletions(-)

diff --git a/src/algebra/discrete-root.md b/src/algebra/discrete-root.md
index 14c8fb51d..8494737ef 100644
--- a/src/algebra/discrete-root.md
+++ b/src/algebra/discrete-root.md
@@ -2,7 +2,7 @@
 
 # Discrete Root
 
-The problem of finding discrete root is defined as follows. Given a prime $n$ and two integers $a$ and $k$, find all $x$ for which:
+The problem of finding a discrete root is defined as follows. Given a prime $n$ and two integers $a$ and $k$, find all $x$ for which:
 
 $x^k \equiv a \pmod n$
 
@@ -14,7 +14,7 @@ Let's apply the concept of a [primitive root](./algebra/primitive-root.html) mod
 
 We can easily discard the case where $a = 0$. In this case, obviously there is only one answer: $x = 0$.
 
-Since we jnow that $n$ is a prime, any number between 1 and $n-1$ can be represented as a power of the primitive root, and we can represent the discrete root problem as follows:
+Since we know that $n$ is a prime and any number between 1 and $n-1$ can be represented as a power of the primitive root, we can represent the discrete root problem as follows:
 
 $(g^y)^k \equiv a \pmod n$
 
@@ -26,7 +26,7 @@ This, in turn, can be rewritten as
 
 $(g^k)^y \equiv a \pmod n$
 
-Now we have one unknown $y$, which is a discrete logarithm problem. The solution can be found using Shanks' baby-step-giant-step algorithm in $O(\sqrt {n} \log n)$ (or we can verify that there are no solutions).
+Now we have one unknown $y$, which is a discrete logarithm problem. The solution can be found using Shanks' baby-step giant-step algorithm in $O(\sqrt {n} \log n)$ (or we can verify that there are no solutions).
 
 Having found one solution $y_0$, one of solutions of discrete root problem will be $x_0 = g^{y_0} \pmod n$.
 
@@ -46,11 +46,11 @@ where $l$ is chosen such that the fraction must be an integer. For this to be tr
 
 $x = g^{y_0 + i \frac {\phi (n)}{gcd(k, \phi (n))}} \pmod n \forall i \in Z$.
 
-This is the final formula for all solutions of discrete root problem.
+This is the final formula for all solutions of the discrete root problem.
 
 ## Implementation
 
-Here is a full implementation, including routines for finding the primitive root, discrete log and finding and printing all solutions.
+Here is a full implementation, including procedures for finding the primitive root, discrete log and finding and printing all solutions.
 
 ```cpp
 int gcd(int a, int b) {

From 40f9c37d4b677aaffe24b0d0e3e500a87c0f2349 Mon Sep 17 00:00:00 2001
From: Wiktor Kuchta <wiktorkuchta@protonmail.com>
Date: Fri, 1 Feb 2019 22:26:40 +0100
Subject: [PATCH 8/8] Add practice problem to discrete root

---
 src/algebra/discrete-root.md | 3 +++
 1 file changed, 3 insertions(+)

diff --git a/src/algebra/discrete-root.md b/src/algebra/discrete-root.md
index 8494737ef..daab06030 100644
--- a/src/algebra/discrete-root.md
+++ b/src/algebra/discrete-root.md
@@ -139,3 +139,6 @@ int main() {
 }
 ```
 
+## Practice problems
+
+* [Codeforces - Lunar New Year and a Recursive Sequence](https://codeforces.com/contest/1106/problem/F)

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