Problem 814: Mezzo-forte
View on Project EulerProject Euler Problem 814 Solution
EulerSolve provides an optimized solution for Project Euler Problem 814, Mezzo-forte, with C++, Python, Java, and a step-by-step mathematical explanation.
Problem Summary The problem asks for the number \(S(n)\) of ways to assign to every vertex of a \(2\times 2n\) strip with a twisted wrap-around one of its three incident directions: left, right, or the vertical edge joining the two rows in the same column. An edge contributes \(1\) to the score when both of its endpoints choose each other, and we want the total score to be exactly \(n\). The answer is required modulo \(998244353\). A direct enumeration would inspect \(3^{4n}\) global assignments, because the strip has \(4n\) vertices and each vertex has three local options. The implementations avoid that exponential blow-up by processing the strip column by column and keeping only the boundary information that can still influence future mutual matches. Mathematical Approach The key observation is that once we cut the strip between two consecutive columns, the future only needs to know which horizontal choices are still waiting for a reciprocal answer from the next column. That turns the entire problem into a finite-state dynamic program with a score counter. Step 1: Interpret the local choices Each column contains two vertices, one in the top row and one in the bottom row....
Detailed mathematical approach
Problem Summary
The problem asks for the number \(S(n)\) of ways to assign to every vertex of a \(2\times 2n\) strip with a twisted wrap-around one of its three incident directions: left, right, or the vertical edge joining the two rows in the same column. An edge contributes \(1\) to the score when both of its endpoints choose each other, and we want the total score to be exactly \(n\). The answer is required modulo \(998244353\).
A direct enumeration would inspect \(3^{4n}\) global assignments, because the strip has \(4n\) vertices and each vertex has three local options. The implementations avoid that exponential blow-up by processing the strip column by column and keeping only the boundary information that can still influence future mutual matches.
Mathematical Approach
The key observation is that once we cut the strip between two consecutive columns, the future only needs to know which horizontal choices are still waiting for a reciprocal answer from the next column. That turns the entire problem into a finite-state dynamic program with a score counter.
Step 1: Interpret the local choices
Each column contains two vertices, one in the top row and one in the bottom row. Each vertex chooses exactly one of three directions:
$$L,\qquad R,\qquad V.$$
Here \(L\) means “choose the neighbor in the previous column”, \(R\) means “choose the neighbor in the next column”, and \(V\) means “choose the vertex in the other row of the same column”. Since the top and bottom vertices act independently, each column has
$$3^2=9$$
local configurations.
A score point appears only when an edge is chosen from both ends. A horizontal edge is counted when the previous column pointed right on some row and the current column answers left on the same row. A vertical edge is counted when both vertices of the current column choose \(V\).
Step 2: Encode the boundary by a 2-bit state
Cut the strip between two adjacent columns. At that boundary we only need to know whether the top row and the bottom row already have a pending right-pointing choice coming from the left. Represent that by a state
$$m=(a,b)\in\{0,1\}^2,$$
where \(a=1\) means “the top row is waiting for a left answer” and \(b=1\) means the same for the bottom row. So there are only four possible boundary states.
If the current column chooses directions \(x,y\in\{L,R,V\}\) for the top and bottom vertices, then the score increment is
$$\Delta(m;x,y)=[x=L]a+[y=L]b+[x=V\land y=V].$$
The outgoing state records which rows now point to the right:
$$m'=([x=R],[y=R]).$$
Step 3: Aggregate the 9 local configurations into transition multiplicities
For a fixed incoming state \(m\), many of the 9 local pairs \((x,y)\) lead to the same outgoing state and the same score increment. Therefore the implementations precompute only the multiplicity
$$T_{\mathrm{mid}}(m,m',d)=\#\{(x,y)\in\{L,R,V\}^2:\ m'=([x=R],[y=R]),\ \Delta(m;x,y)=d\}.$$
Because \(d\in\{0,1,2\}\), each incoming state has only a small number of distinct transition classes.
For example, when \(m=(0,0)\), the only way to score \(1\) inside the column is the pair \((V,V)\), while the other eight local assignments merely open or leave unmatched horizontal choices. The aggregated counts are
$$\begin{aligned} T_{\mathrm{mid}}((0,0),(0,0),0)&=3,\\ T_{\mathrm{mid}}((0,0),(0,0),1)&=1,\\ T_{\mathrm{mid}}((0,0),(1,0),0)&=2,\\ T_{\mathrm{mid}}((0,0),(0,1),0)&=2,\\ T_{\mathrm{mid}}((0,0),(1,1),0)&=1. \end{aligned}$$
Step 4: Run a score-tracking DP across the first \(2n-1\) columns
Choose a seam where the strip is cut open and fix its boundary state \(m_0\). Let \(D_j^{(m_0)}(m,k)\) denote the number of ways to process the first \(j\) columns so that the current boundary state is \(m\) and the accumulated score is \(k\).
The initial condition is
$$D_0^{(m_0)}(m,k)=\begin{cases}1,&m=m_0\text{ and }k=0,\\0,&\text{otherwise.}\end{cases}$$
For every ordinary column, each transition class contributes according to
$$D_{j+1}^{(m_0)}(m',k+d)\leftarrow D_{j+1}^{(m_0)}(m',k+d)+D_j^{(m_0)}(m,k)\,T_{\mathrm{mid}}(m,m',d).$$
This recurrence is the whole reason the method is fast: the exponentially many global configurations collapse to a quadratic-time dynamic program on four boundary states and one score axis.
Step 5: Handle the twisted wrap-around in the last column
The final column is special because the strip closes with a twist. A right-pointing choice from the top row returns to the bottom row at the seam, and a right-pointing choice from the bottom row returns to the top row. So the last-column transition table is
$$T_{\mathrm{last}}(m,m',d)=\#\{(x,y)\in\{L,R,V\}^2:\ m'=([y=R],[x=R]),\ \Delta(m;x,y)=d\}.$$
After applying that twisted update, a complete configuration is valid only if it returns to the same seam state and reaches total score \(n\). Summing over the four possible seam states yields
$$S(n)=\sum_{m_0\in\{0,1\}^2} D_{2n}^{(m_0)}(m_0,n)\pmod{998244353}.$$
Worked Example: \(n=1\)
When \(n=1\), the strip has \(2\) columns and the target score is \(1\). Start with seam state \(m_0=(0,0)\). For the first column, the five transition classes listed above occur with multiplicities \(3,1,2,2,1\).
To finish with total score \(1\) and return to \(m_0=(0,0)\) after the twisted last column, the valid combinations contribute
$$3\cdot 1+1\cdot 3+2\cdot 3+2\cdot 3+1\cdot 3=21.$$
Repeating the same calculation for the other seam states gives
$$m_0=(1,0)\Rightarrow 11,\qquad m_0=(0,1)\Rightarrow 11,\qquad m_0=(1,1)\Rightarrow 5.$$
Therefore
$$S(1)=21+11+11+5=48,$$
which matches the first checkpoint used by the implementation.
How the Code Works
The C++, Python, and Java implementations begin by enumerating the 9 local choices for each of the 4 incoming boundary states. They compress those raw possibilities into two tiny transition tables: one for ordinary columns and one for the twisted final column. This preprocessing takes constant time, but it removes repeated case analysis from the main loop.
Next, for each possible seam state, the implementation runs a layered dynamic program indexed by the current boundary state and the accumulated score. Only two layers are stored at any moment, because each column depends only on the previous one. After \(2n-1\) ordinary updates, one last twisted update is applied, and the entry that returns to the starting seam state with score exactly \(n\) is added to the final total.
All arithmetic is performed modulo \(998244353\). The three language versions implement the same recurrence and differ only in syntax and data-structure spelling.
Complexity Analysis
The number of boundary states is fixed at \(4\), the score axis has length \(n+1\), and there are \(2n\) columns. Each state has only a constant number of aggregated transitions, so the total running time is \(O(n^2)\). The rolling-array dynamic program stores only two layers of size \(4\times(n+1)\), so the memory usage is \(O(n)\).
Footnotes and References
- Problem page: https://projecteuler.net/problem=814
- Dynamic programming: Wikipedia — Dynamic programming
- Finite-state machine: Wikipedia — Finite-state machine
- Mobius strip: Wikipedia — Mobius strip
Problem 814 source code
C++
#include <array>
#include <cassert>
#include <cstdint>
#include <iostream>
#include <vector>
using i64 = std::int64_t;
static constexpr i64 kMod = 998244353;
enum Dir : int { L = 0, R = 1, V = 2 };
static int swap_low2(int mask) {
return ((mask & 1) << 1) | ((mask >> 1) & 1);
}
static std::array<std::vector<std::array<int, 3>>, 4> build_transitions(bool last_column) {
std::array<std::vector<std::array<int, 3>>, 4> grouped;
for (int in = 0; in < 4; ++in) {
std::array<std::array<int, 3>, 4> cnt{};
const int in_top = in & 1;
const int in_bottom = (in >> 1) & 1;
for (int top = 0; top < 3; ++top) {
for (int bot = 0; bot < 3; ++bot) {
int delta = 0;
if (top == L && in_top) {
++delta;
}
if (bot == L && in_bottom) {
++delta;
}
if (top == V && bot == V) {
++delta;
}
int out = 0;
if (top == R) {
out |= 1;
}
if (bot == R) {
out |= 2;
}
if (last_column) {
out = swap_low2(out);
}
++cnt[out][delta];
}
}
for (int out = 0; out < 4; ++out) {
for (int d = 0; d <= 2; ++d) {
if (cnt[out][d] > 0) {
grouped[in].push_back({out, d, cnt[out][d]});
}
}
}
}
return grouped;
}
static i64 S_mod(int n) {
const int cols = 2 * n;
const int target = n;
const auto trans_mid = build_transitions(false);
const auto trans_last = build_transitions(true);
i64 total = 0;
for (int init = 0; init < 4; ++init) {
std::array<std::vector<i64>, 4> dp;
std::array<std::vector<i64>, 4> next;
for (int s = 0; s < 4; ++s) {
dp[s].assign(target + 1, 0);
next[s].assign(target + 1, 0);
}
dp[init][0] = 1;
for (int col = 0; col < cols - 1; ++col) {
for (int s = 0; s < 4; ++s) {
std::fill(next[s].begin(), next[s].end(), 0);
}
for (int in = 0; in < 4; ++in) {
const auto& cur = dp[in];
for (const auto& tr : trans_mid[in]) {
const int out = tr[0];
const int delta = tr[1];
const int ways = tr[2];
for (int k = 0; k + delta <= target; ++k) {
if (cur[k] == 0) {
continue;
}
i64 add = (cur[k] * ways) % kMod;
i64& dst = next[out][k + delta];
dst += add;
if (dst >= kMod) {
dst -= kMod;
}
}
}
}
dp.swap(next);
}
for (int s = 0; s < 4; ++s) {
std::fill(next[s].begin(), next[s].end(), 0);
}
for (int in = 0; in < 4; ++in) {
const auto& cur = dp[in];
for (const auto& tr : trans_last[in]) {
const int out = tr[0];
const int delta = tr[1];
const int ways = tr[2];
for (int k = 0; k + delta <= target; ++k) {
if (cur[k] == 0) {
continue;
}
i64 add = (cur[k] * ways) % kMod;
i64& dst = next[out][k + delta];
dst += add;
if (dst >= kMod) {
dst -= kMod;
}
}
}
}
total += next[init][target];
total %= kMod;
}
return total;
}
int main() {
assert(S_mod(1) == 48);
assert(S_mod(10) == 420121075);
std::cout << S_mod(1000) << '\n';
return 0;
}
Python
def solve():
MOD = 998244353
n = 1000; cols = 2*n; target = n
L, R, V = 0, 1, 2
def swap_low2(mask): return ((mask&1)<<1)|((mask>>1)&1)
def build_trans(last_col):
grouped = [[] for _ in range(4)]
for inp in range(4):
cnt = [[0]*3 for _ in range(4)]
it, ib = inp&1, (inp>>1)&1
for top in range(3):
for bot in range(3):
d = 0
if top == L and it: d += 1
if bot == L and ib: d += 1
if top == V and bot == V: d += 1
out = 0
if top == R: out |= 1
if bot == R: out |= 2
if last_col: out = swap_low2(out)
cnt[out][d] += 1
for out in range(4):
for d in range(3):
if cnt[out][d] > 0:
grouped[inp].append((out, d, cnt[out][d]))
return grouped
tm = build_trans(False); tl = build_trans(True)
total = 0
for init in range(4):
dp = [[0]*(target+1) for _ in range(4)]
dp[init][0] = 1
for col in range(cols - 1):
nxt = [[0]*(target+1) for _ in range(4)]
for inp in range(4):
cur = dp[inp]
for out, delta, ways in tm[inp]:
for k in range(target - delta + 1):
if cur[k] == 0: continue
nxt[out][k+delta] = (nxt[out][k+delta] + cur[k]*ways) % MOD
dp = nxt
# last column
nxt = [[0]*(target+1) for _ in range(4)]
for inp in range(4):
cur = dp[inp]
for out, delta, ways in tl[inp]:
for k in range(target - delta + 1):
if cur[k] == 0: continue
nxt[out][k+delta] = (nxt[out][k+delta] + cur[k]*ways) % MOD
total = (total + nxt[init][target]) % MOD
return str(total)
if __name__ == '__main__':
print(solve())
Java
import java.util.ArrayList;
import java.util.List;
public class Euler814 {
static final long kMod = 998244353;
static final int L = 0;
static final int R = 1;
static final int V = 2;
static int swapLow2(int mask) {
return ((mask & 1) << 1) | ((mask >> 1) & 1);
}
static class Transition {
int out, delta, ways;
Transition(int out, int delta, int ways) {
this.out = out;
this.delta = delta;
this.ways = ways;
}
}
static List<Transition>[] buildTransitions(boolean lastColumn) {
List<Transition>[] grouped = new ArrayList[4];
for (int i = 0; i < 4; i++) {
grouped[i] = new ArrayList<>();
}
for (int in = 0; in < 4; ++in) {
int[][] cnt = new int[4][3];
int inTop = in & 1;
int inBottom = (in >> 1) & 1;
for (int top = 0; top < 3; ++top) {
for (int bot = 0; bot < 3; ++bot) {
int delta = 0;
if (top == L && inTop == 1)
++delta;
if (bot == L && inBottom == 1)
++delta;
if (top == V && bot == V)
++delta;
int out = 0;
if (top == R)
out |= 1;
if (bot == R)
out |= 2;
if (lastColumn)
out = swapLow2(out);
cnt[out][delta]++;
}
}
for (int out = 0; out < 4; ++out) {
for (int d = 0; d <= 2; ++d) {
if (cnt[out][d] > 0) {
grouped[in].add(new Transition(out, d, cnt[out][d]));
}
}
}
}
return grouped;
}
static long SMod(int n) {
int cols = 2 * n;
int target = n;
List<Transition>[] transMid = buildTransitions(false);
List<Transition>[] transLast = buildTransitions(true);
long total = 0;
for (int init = 0; init < 4; ++init) {
long[][] dp = new long[4][target + 1];
dp[init][0] = 1;
for (int col = 0; col < cols - 1; ++col) {
long[][] next = new long[4][target + 1];
for (int in = 0; in < 4; ++in) {
long[] cur = dp[in];
for (Transition tr : transMid[in]) {
int out = tr.out;
int delta = tr.delta;
long ways = tr.ways;
long[] nextDst = next[out];
for (int k = 0; k + delta <= target; ++k) {
if (cur[k] == 0)
continue;
long add = (cur[k] * ways) % kMod;
nextDst[k + delta] += add;
if (nextDst[k + delta] >= kMod) {
nextDst[k + delta] -= kMod;
}
}
}
}
dp = next;
}
long[][] next = new long[4][target + 1];
for (int in = 0; in < 4; ++in) {
long[] cur = dp[in];
for (Transition tr : transLast[in]) {
int out = tr.out;
int delta = tr.delta;
long ways = tr.ways;
long[] nextDst = next[out];
for (int k = 0; k + delta <= target; ++k) {
if (cur[k] == 0)
continue;
long add = (cur[k] * ways) % kMod;
nextDst[k + delta] += add;
if (nextDst[k + delta] >= kMod) {
nextDst[k + delta] -= kMod;
}
}
}
}
total += next[init][target];
total %= kMod;
}
return total;
}
public static String solve() {
return Long.toString(SMod(1000));
}
public static void main(String[] args) {
System.out.println(solve());
}
}