Programming Languages, Compilers & Formal Methods

Prompt:
You are given a linear scan register allocation problem with 3 physical registers: R0, R1, R2.

Live intervals for virtual registers:

V0: [0,7]
V1: [1,5]
V2: [2,9]
V3: [3,6]
V4: [4,11]
V5: [6,10]
V6: [8,14]
V7: [10,13]
V8: [12,15]
V9: [1,3]

Use the following deterministic rules:

1. Process intervals in ascending order of start point.
2. If two intervals have the same start point, process the one with the smaller end point first.
3. Before processing an interval with start point `s`, expire every active interval whose end point is less than or equal to `s`.
4. If one or more registers are free, assign the lowest-numbered free register.
5. If all registers are occupied:
   - Consider the active intervals plus the new interval.
   - Spill the interval with the largest end point.
   - If multiple intervals tie for the largest end point, spill the new interval.
   - If the new interval is spilled, it receives no register.
   - Otherwise, remove the spilled active interval and give its register to the new interval.
6. Record spill events in the exact order they occur.
7. For part C, evaluate the active-set size only at the distinct start indices that appear in the input: `0, 1, 2, 3, 4, 6, 8, 10, 12`.
   - At each such index, first finish processing all intervals that start there.
   - Then measure the number of active intervals.

Tasks:

A. List the intervals in the exact processing order.

B. List the spill events in order.

C. List all start indices from `0, 1, 2, 3, 4, 6, 8, 10, 12` where the number of active intervals is exactly 3 after all intervals starting at that index have been processed.

D. Provide the final register assignment for every interval that is not spilled.

Use this exact output format:

Processing Order: comma-separated interval names
Spill Events: comma-separated interval names, or `None` if there are no spills
Active Count 3 At: comma-separated start indices
Final Register Assignments:
one `interval -> register` entry per line, ordered by interval name

Lists the processing order exactly as V0, V9, V1, V2, V3, V4, V5, V6, V7, V8.

States that the total number of spill events is exactly 2.

Identifies the second spill event as V4.

Lists the part C indices exactly as 1, 2, 3, 4, 12 with no missing or extra indices.

Gives the R0 final assignments exactly as V0 -> R0 and V6 -> R0.

Gives the R1 final assignments exactly as V3 -> R1, V5 -> R1, V7 -> R1, and V9 -> R1.

Gives the R2 final assignments exactly as V1 -> R2 and V8 -> R2.

Processing Order: V0, V9, V1, V2, V3, V4, V5, V6, V7, V8
Spill Events: V2, V4
Active Count 3 At: 1, 2, 3, 4, 12
Final Register Assignments:
V0 -> R0
V1 -> R2
V3 -> R1
V5 -> R1
V6 -> R0
V7 -> R1
V8 -> R2
V9 -> R1

Explanation

Step 1: Determine processing order

Intervals are sorted by increasing start point, and ties are broken by smaller end point first.

- Start 0: V0
- Start 1: V9 before V1 because 3 < 5
- Start 2: V2
- Start 3: V3
- Start 4: V4
- Start 6: V5
- Start 8: V6
- Start 10: V7
- Start 12: V8

So the exact processing order is:
V0, V9, V1, V2, V3, V4, V5, V6, V7, V8

Step 2: Simulate the allocator

Start with no active intervals and all registers free.

1. Process V0 = [0,7]
- Expire intervals with end <= 0: none
- Lowest free register is R0
- Assign V0 -> R0
- Active set: V0(R0)

2. Process V9 = [1,3]
- Expire intervals with end <= 1: none
- Lowest free register is R1
- Assign V9 -> R1
- Active set: V0(R0), V9(R1)

3. Process V1 = [1,5]
- Expire intervals with end <= 1: none
- Lowest free register is R2
- Assign V1 -> R2
- Active set: V0(R0), V9(R1), V1(R2)

4. Process V2 = [2,9]
- Expire intervals with end <= 2: none
- No free register exists
- Compare end points of V0(7), V9(3), V1(5), V2(9)
- Largest end point is 9, so spill V2
- V2 gets no register
- Active set stays: V0(R0), V9(R1), V1(R2)

5. Process V3 = [3,6]
- Expire intervals with end <= 3: V9 expires, so R1 becomes free
- Lowest free register is R1
- Assign V3 -> R1
- Active set: V0(R0), V1(R2), V3(R1)

6. Process V4 = [4,11]
- Expire intervals with end <= 4: none
- No free register exists
- Compare end points of V0(7), V1(5), V3(6), V4(11)
- Largest end point is 11, so spill V4
- V4 gets no register
- Active set stays: V0(R0), V1(R2), V3(R1)

7. Process V5 = [6,10]
- Expire intervals with end <= 6: V1 expires and V3 expires
- R2 and R1 are now free
- Lowest free register is R1
- Assign V5 -> R1
- Active set: V0(R0), V5(R1)

8. Process V6 = [8,14]
- Expire intervals with end <= 8: V0 expires
- R0 becomes free
- Lowest free register is R0
- Assign V6 -> R0
- Active set: V5(R1), V6(R0)

9. Process V7 = [10,13]
- Expire intervals with end <= 10: V5 expires
- R1 becomes free
- Lowest free register is R1
- Assign V7 -> R1
- Active set: V6(R0), V7(R1)

10. Process V8 = [12,15]
- Expire intervals with end <= 12: none
- Lowest free register is R2
- Assign V8 -> R2
- Active set: V6(R0), V7(R1), V8(R2)

From the full trace, the spill events occur in this exact order:
V2, V4

Step 3: Compute part C exactly

Part C asks for the distinct start indices from the prompt, after all intervals beginning at that index have been processed.

- At start 0: active set is {V0} -> size 1
- At start 1: after processing both V9 and V1, active set is {V0, V9, V1} -> size 3
- At start 2: after processing V2 and spilling it, active set is still {V0, V9, V1} -> size 3
- At start 3: after expiring V9 and assigning V3, active set is {V0, V1, V3} -> size 3
- At start 4: after processing V4 and spilling it, active set is still {V0, V1, V3} -> size 3
- At start 6: active set is {V0, V5} -> size 2
- At start 8: active set is {V5, V6} -> size 2
- At start 10: active set is {V6, V7} -> size 2
- At start 12: active set is {V6, V7, V8} -> size 3

Therefore the exact indices for part C are:
1, 2, 3, 4, 12

Step 4: Final register assignments

Only non-spilled intervals appear in the final assignment list. Reading them from the simulation:

- V0 -> R0
- V1 -> R2
- V3 -> R1
- V5 -> R1
- V6 -> R0
- V7 -> R1
- V8 -> R2
- V9 -> R1

Spilled intervals:
V2, V4


Prompt:
Steve is trapped in a tunnel filled with hostile creepers. There are n creepers standing in a straight line, indexed from 1 to n.

For each position i, the creeper at that position has an explosive power e_i.
We use the term “creepers” consistently throughout; all entities referred to are the same.

Explosion Rule:
If Steve detonates the creeper at position i, it explodes and kills every creeper at position j such that:
|i − j| < e_i

This condition is strict. Positions where |i − j| = e_i are NOT affected.
Equivalently, detonating creeper i kills exactly the creepers in the interval:
[i − e_i + 1, i + e_i − 1], inclusive.

If e_i = 0, the interval is empty, meaning such a creeper cannot be detonated at all.
Dead creepers cannot be detonated again.

Objective:
Steve must kill all creepers using as few detonations as possible.

Input:
Each test contains multiple test cases.
The first line contains an integer t, the number of test cases.

For each test case:
- The first line contains an integer n (2 ≤ n ≤ 5·10^5), the number of creepers.
- The second line contains n integers e_1, e_2, …, e_n (0 ≤ e_i ≤ n), where e_i is the explosive power of the i-th creeper.

It is guaranteed that the sum of n over all test cases does not exceed 5·10^5.

Output:
For each test case:
- Output −1 if it is impossible to kill all creepers.
- Otherwise, output:
  1) An integer k, the minimum number of detonations.
  2) A sequence of k integers denoting the positions of the creepers to detonate.

If multiple valid solutions exist, any one may be printed.

Worked Example (Explosion Mechanics):

If n = 7 and e = [0, 2, 0, 3, 0, 0, 0]:

Detonating creeper at position 4 (e_4 = 3) kills creepers at positions:
|4 − j| < 3  →  j ∈ {2, 3, 4, 5, 6}

So the affected interval is [2, 6].

Does the answer explicitly define the detonation interval of creeper i as [i − e_i + 1, i + e_i − 1]?

Does the answer state that two creepers cannot both be detonated if one lies inside the other’s detonation interval ?

Does the answer define dp[i] as the minimum detonations needed to kill all creepers up to position (i + e_i − 1) with creeper i being the rightmost detonated creeper?

Does the answer list Case 1 transition conditions exactly as j + e_j − 1 < i and i − e_i + 1 ≤ j + e_j?

Does the answer list Case 2 transition conditions exactly as j < i − e_i + 1 and i − e_i + 1 ≤ j + e_j?

Does the answer state that dp[i] is computed using exactly two range minimum queries: [i − e_i, i − 1] and [i − e_i, i + e_i − 1]?

Does the answer state the dp formula dp[i] = 1 + min(query1, query2)?

Does the answer specify the exact point update rule segtree[i + e_i − 1] = min(current value, dp[i]) after computing dp[i]?

Does the answer state that the final result equals min(dp[i]) over all i satisfying i + e_i − 1 ≥ n, and outputs −1 if no finite dp[i] exists?

Does the answer state that reconstruction uses predecessor tracking and detonates the selected creepers in increasing order of explosive power e_i?

Notation:
We use e_i to denote the explosive power of the creeper at position i, exactly as defined in the task prompt.

Detonation Interval:
Detonating creeper i kills all creepers in the interval:
[i − e_i + 1, i + e_i − 1]

Two creepers cannot both be detonated if their detonation intervals overlap in a way that one would destroy the other before it is detonated.

Dynamic Programming Definition:

Define:
dp[i] = the minimum number of detonations required to kill all creepers up to position (i + e_i − 1),
with creeper i being the rightmost detonated creeper.

Exact Transition Conditions:

For a transition from a previously detonated creeper j < i to creeper i, exactly two valid cases exist.

Case 1: Creeper i is strictly to the right of j’s detonation interval
j + e_j − 1 < i
i − e_i + 1 ≤ j + e_j

Case 2: Creeper j is strictly to the left of i’s detonation interval
j < i − e_i + 1
i − e_i + 1 ≤ j + e_j

Segment Tree Optimization:

A naive implementation of these transitions is O(n²).
To optimize, process creepers from left to right and maintain a segment tree where:
segtree[x] stores the minimum detonations needed to kill all creepers up to position x.
All values are initialized to infinity.

Exact Computation of dp[i]:

When processing creeper i, compute dp[i] using exactly two range minimum queries:

1) Range [i − e_i, i − 1]
2) Range [i − e_i, i + e_i − 1]

Then:
dp[i] = 1 + min(result of both queries)

Exact Point Update:

After computing dp[i], update:
segtree[i + e_i − 1] = min(current value, dp[i])

Final Answer:

The minimum number of detonations required is:
min over all i such that i + e_i − 1 ≥ n of dp[i]

If no such i exists with a finite dp[i], output −1.

Reconstruction and Order:

Store predecessor information during DP to reconstruct the detonated creepers.
Detonate the chosen creepers in increasing order of explosive power e_i to ensure no detonation kills a later one.

Complexity:
Time Complexity: O(n log n)
Space Complexity: O(n)


Prompt:
# Odd-Quorum Rewrite Maximization — Extended Deterministic Output

You are given `t` test cases. In each test case:
- an array `a` of `n` **positive integers**
- a positive integer `k`

You may apply a "rewrite" operation that overwrites selected elements by a median.

---

## Rewrite game (per test case)

1. Choose one **odd** integer `x` with `1 ≤ x ≤ n`. This `x` is fixed for the entire test case.
2. You may perform the following operation **at most `k` times** (possibly zero), always using the same chosen `x`:
   - Choose any subsequence of the **current** array of length exactly `x` (indices strictly increasing).
   - Let `m` be the median of those `x` values (for odd `x`, median is the `((x+1)/2)`-th smallest).
   - Overwrite all `x` chosen positions with the value `m`.

Your objective is to maximize the final sum `sum(a[i])`.

---

## Worked example

Consider `a = [1, 5, 3, 2, 8]`, `n = 5`, `k = 1`.

**Step 1 — Sort and build prefix sums.**
Sorted: `b = [1, 2, 3, 5, 8]`. Prefix sums: `p = [0, 1, 3, 6, 11, 19]`. Base sum `S = 19`.

**Step 2 — Write the gain formula.**
Write `x = 2y + 1`. For median at sorted position `i` with half-window `y`:
- `L = min(k·y, i−1)` elements below `b[i]` get raised to `b[i]`.
- `y` elements `b[i+1..i+y]` are consumed (pulled down to `b[i]`).
- `leftGain = L·b[i] − p[L]`
- `rightLoss = (p[i+y] − p[i]) − y·b[i]`
- `F(i,y) = S + leftGain − rightLoss`

The formulation here is guaranteed to be correct for this problem and must be used.

**Step 3 — Enumerate all valid `(i, y)` pairs and pick the maximum.**

| i | b[i] | y | x | L | leftGain | rightLoss | F(i,y) |
|---|------|---|---|---|----------|-----------|--------|
| 4 | 5    | 1 | 3 | min(1,3)=1 | 1·5−1=4  | (11−6)−1·5=0 | 19+4−0=**23** |
| 5 | 8    | 0 | 1 | 0          | 0        | 0             | 19            |
| 3 | 3    | 1 | 3 | 1          | 3−1=2    | (11−6)−3=2    | 19            |
| 3 | 3    | 2 | 5 | 2          | 6−3=3    | (19−6)−6=7    | 15            |
| 2 | 2    | 1 | 3 | 1          | 2−1=1    | (6−3)−2=1     | 19            |

Wait — let me recompute `(i=4, y=1)`:
- `rightLoss = p[5]−p[4] − 1·5 = 19−11 − 5 = 3`
- `F = 19 + 4 − 3 = 20`

The maximum is `F(4,1) = 20` with `x=3`, median `b[4]=5`, `L=1`.

Interpretation on the sorted array: the `L=1` smallest value (`b[1]=1`) is raised to `5` (gain = 4), and the `y=1` value just above the median (`b[5]=8`) is pulled down to `5` (loss = 3). Net change = +1, new sum = 20.

---

## Deterministic reporting requirements

For each test case `c ∈ {1,2,3,4,5}`, you must output:
- `baseSum_c`: the initial sum of the given array
- `bestSum_c`: the maximum achievable final sum
- `gain_c`: `bestSum_c - baseSum_c`
- `bestX_c`: the odd `x` used by an optimal strategy
- `bestMedian_c`: the median value used by an optimal strategy
- `bestI_c`: the 1-indexed position `i` of `bestMedian_c` in the **sorted** array `b` (defined below), chosen by tie-break
- `bestL_c`: the value `L = min(k*y, i-1)` for the chosen optimum where `x = 2y+1`

### Sorting convention (for `bestI_c`)

Let `b` be `a` sorted in nondecreasing order (1-indexed: `b[1]..b[n]`).

### Tie-break rule (to make the answer unique)

Among all strategies achieving `bestSum_c`, pick the one with:
1. smallest `x`
2. then smallest `bestMedian`
3. then smallest `bestI` (in the sorted array)

---

## Required global outputs

Also output:
- `totalBaseSum`: sum of all `baseSum_c`
- `totalBestSum`: sum of all `bestSum_c`
- `totalGain`: `totalBestSum - totalBaseSum`

And output these complexity strings **exactly**:
- `perTestTime = "O(n log n)"`
- `totalTime = "O(S log S)"` where `S = Σn` over all test cases
- `extraSpace = "O(n)"` (extra space per test case, excluding input storage)

---

## Output format

Your final answer must include **one** JSON object containing **all** keys described above.
Additional explanations are allowed, but the JSON object must be clearly present and valid.

All `*_Sum_*`, `gain_*`, `bestX_*`, `bestMedian_*`, `bestI_*`, `bestL_*` must be integers.
All complexity values must be strings exactly as specified.

---

## Fixed input in this prompt

`t = 5`

### Test case 1
- `n = 12`, `k = 2`
- `a = [9, 1, 3, 50, 2, 10, 2, 11, 4, 10, 100, 3]`

### Test case 2
- `n = 9`, `k = 1`
- `a = [54, 1, 53, 1, 52, 100, 1, 51, 50]`

### Test case 3
- `n = 13`, `k = 3`
- `a = [2, 31, 150, 2, 33, 34, 30, 2, 151, 32, 100, 2, 2]`

### Test case 4
- `n = 14`, `k = 2`
- `a = [3, 21, 3, 20, 3, 22, 21, 20, 3, 101, 3, 100, 20, 3]`

### Test case 5
- `n = 15`, `k = 3`
- `a = [2, 45, 2, 91, 40, 2, 43, 44, 46, 1, 42, 90, 41, 2, 2]`

The reasoning contains an exhaustive enumeration of F(i, y) for every valid (i, y) pair in every one of the five test cases, i.e. for each test case and for every i in [1, n] every y in [0, min(i-1, n-i)] is listed together with its computed L, leftGain, rightLoss, and F(i, y) value. A partial enumeration that only shows 'top candidates' must fail this criterion.

The computation path in the reasoning matches the mandated one: for each test case the solver sorts a into b, builds the prefix-sum array p, computes S = p[n], uses x = 2y + 1 with L = min(k*y, i-1), and applies F(i, y) = S + L*b[i] - p[L] - (p[i+y] - p[i]) + y*b[i] (no alternative heuristic or greedy is used).

The tie-break rule is applied correctly and visibly: among all (i, y) pairs that achieve the maximum F in a test case, the solver selects the one with smallest x = 2y+1, then smallest median b[i], then smallest sorted index i, and justifies the choice against the enumeration table.

bestSum_c values are correct for all five test cases: bestSum_1 = 252, bestSum_2 = 507, bestSum_3 = 1030, bestSum_4 = 536, bestSum_5 = 757.

bestX_c values are correct for all five test cases: bestX_1 = 3, bestX_2 = 7, bestX_3 = 5, bestX_4 = 3, bestX_5 = 3.

bestI_c values are correct for all five test cases: bestI_1 = 11, bestI_2 = 5, bestI_3 = 11, bestI_4 = 13, bestI_5 = 14.

bestMedian_c values are correct for all five test cases: bestMedian_1 = 50, bestMedian_2 = 51, bestMedian_3 = 100, bestMedian_4 = 100, bestMedian_5 = 90.

bestL_c values are correct for all five test cases: bestL_1 = 2, bestL_2 = 3, bestL_3 = 6, bestL_4 = 2, bestL_5 = 3.

Global totals are correct and internally consistent: totalBaseSum = 1975, totalBestSum = 3082, totalGain = 1107, every per-test baseSum_c matches sum(a), and every gain_c equals bestSum_c - baseSum_c.

The final answer contains exactly one JSON object with all 38 required keys (baseSum_c, bestSum_c, gain_c, bestX_c, bestMedian_c, bestI_c, bestL_c for c=1..5, plus totalBaseSum, totalBestSum, totalGain, perTestTime, totalTime, extraSpace), numeric fields emitted as JSON integers and complexity fields emitted as the exact strings "O(n log n)", "O(S log S)", "O(n)".


Prompt:
# PROMPT

Simulate Tomasulo's Algorithm for out-of-order execution with the following configuration:

EXECUTION LATENCIES (by instruction type):
- ADD/SUB: 2 cycles
- MUL: 4 cycles
- DIV: 8 cycles
- LOAD: 3 cycles (includes memory access)

RESERVATION STATIONS:
- 3 Add/Sub stations: Add1, Add2, Add3
- 2 Mul/Div stations: Mul1, Mul2
- 2 Load stations: Load1, Load2

There are no structural hazards on functional units. Every reservation station that is ready to execute can begin execution immediately regardless of how many other stations of the same type are already executing.

REGISTER FILE: F0, F2, F4, F6, F8, F10 (all initially ready with value 0). Integer base registers R1 and R2 are always ready and are not tracked by the algorithm.

RULES:
1. Issue: One instruction issues per cycle in program order, provided a reservation station of the correct type is available. If no station is available, the instruction stalls and no later instruction can issue.
2. Execute: Begins the cycle AFTER all source operands are available. An operand broadcast on cycle T is considered available starting cycle T+1.
3. Write: Results broadcast on the CDB exactly one cycle after execution completes. If execution ends in cycle E, the write occurs in cycle E+1.
4. CDB Conflict: If multiple instructions are ready to write in the same cycle, the instruction in the station with the lowest name (alphabetical, then numeric) wins. Full priority order: Add1 < Add2 < Add3 < Load1 < Load2 < Mul1 < Mul2.
5. Stations freed: A reservation station is freed in the same cycle its instruction writes (broadcasts on CDB).
6. An instruction cannot begin execution in the same cycle it is issued.

INSTRUCTION SEQUENCE (I1-I8):
I1: LOAD  F2, 0(R1)
I2: LOAD  F4, 0(R2)
I3: MUL   F6, F2, F4
I4: ADD   F8, F2, F4
I5: DIV   F10, F6, F2
I6: SUB   F2, F8, F6
I7: ADD   F4, F2, F10
I8: MUL   F2, F4, F6

Track each instruction through Issue (I), Execute Start (X), Execute End (E), Write (W) cycles.

Compute these EXACT values:

1) In which cycle does I3 (MUL) begin execution?
2) In which cycle does I5 (DIV) write its result?
3) What is the SUM of write cycles for all 8 instructions?
4) In which cycle does I6 (SUB) begin execution?
5) What is the PRODUCT of issue cycles for I3, I5, and I7?
6) How many cycles have at least one instruction executing (between execute start and execute end, inclusive)?
7) In which cycle does I8 write its result?
8) What is the total number of cycles from I1 issue to I8 write (inclusive)?
9) What is the SUM of (Write cycle - Issue cycle) for instructions I1 through I4?
10) In cycle 10, how many reservation stations are occupied?

State each answer with its question number explicitly.

The response states that I3 (MUL) begins execution in exactly cycle 7.

The response states that I5 (DIV) writes its result in exactly cycle 20.

The response states that the sum of write cycles for all 8 instructions is exactly 116.

The response states that I6 (SUB) begins execution in exactly cycle 12.

The response states that the product of issue cycles for I3, I5, and I7 is exactly 105.

The response states that exactly 22 cycles have at least one instruction executing.

The response states that I8 (MUL) writes its result in exactly cycle 28.

The response states that the total number of cycles from I1 issue to I8 write (inclusive) is exactly 28.

The response states that the sum of (Write cycle - Issue cycle) for I1 through I4 is exactly 21.

The response states that exactly 4 reservation stations are occupied in cycle 10.

# GOLDEN ANSWER

## Step-by-step Trace

**I1: LOAD F2, 0(R1)** - Issues cycle 1 to Load1. R1 always ready, no data dependency. Cannot execute same cycle as issue (Rule 6), so execution starts cycle 2. Execution latency = 3 cycles: executes cycles 2-4. Writes cycle 5. F2 available from cycle 5 (usable for execution starting cycle 6).

**I2: LOAD F4, 0(R2)** - Issues cycle 2 to Load2. R2 always ready. Execution starts cycle 3 (cycle after issue). Executes cycles 3-5. Writes cycle 6. F4 available from cycle 6 (usable cycle 7).

**I3: MUL F6, F2, F4** - Issues cycle 3 to Mul1. Needs F2 (available cycle 5, usable cycle 6) and F4 (available cycle 6, usable cycle 7). Both operands usable cycle 7. Execution starts cycle 7. Latency = 4: executes cycles 7-10. Writes cycle 11. F6 available from cycle 11 (usable cycle 12).

**I4: ADD F8, F2, F4** - Issues cycle 4 to Add1. Needs F2 (usable cycle 6) and F4 (usable cycle 7). Both usable cycle 7. Execution starts cycle 7. Latency = 2: executes cycles 7-8. Writes cycle 9. (No CDB conflict - only I4 ready to write cycle 9.) F8 available from cycle 9 (usable cycle 10).

**I5: DIV F10, F6, F2** - Issues cycle 5 to Mul2. Needs F6 (usable cycle 12) and F2 (usable cycle 6). Both usable cycle 12. Execution starts cycle 12. Latency = 8: executes cycles 12-19. Writes cycle 20. F10 available from cycle 20 (usable cycle 21).

**I6: SUB F2, F8, F6** - Issues cycle 6 to Add2. Needs F8 (usable cycle 10) and F6 (usable cycle 12). Both usable cycle 12. Execution starts cycle 12. Latency = 2: executes cycles 12-13. Writes cycle 14. (No CDB conflict cycle 14.) F2 (new value) available cycle 14 (usable cycle 15).

**I7: ADD F4, F2, F10** - Issues cycle 7 to Add3. Needs F2 from I6 (usable cycle 15) and F10 from I5 (usable cycle 21). Both usable cycle 21. Execution starts cycle 21. Latency = 2: executes cycles 21-22. Writes cycle 23. F4 available cycle 23 (usable cycle 24).

**I8: MUL F2, F4, F6** - Needs a Mul station. At cycle 8: Mul1 has I3 (executing), Mul2 has I5 (waiting). Both occupied. Stalls cycles 8, 9, 10. Cycle 11: I3 writes and frees Mul1. I8 issues cycle 11 to Mul1. Needs F4 from I7 (usable cycle 24) and F6 (from I3, available cycle 11, usable cycle 12). Both usable cycle 24. Execution starts cycle 24. Latency = 4: executes cycles 24-27. Writes cycle 28.

## Summary Table

| Inst | Station | I  | X  | E  | W  |
|------|---------|----|----|----|----|
| I1   | Load1   | 1  | 2  | 4  | 5  |
| I2   | Load2   | 2  | 3  | 5  | 6  |
| I3   | Mul1    | 3  | 7  | 10 | 11 |
| I4   | Add1    | 4  | 7  | 8  | 9  |
| I5   | Mul2    | 5  | 12 | 19 | 20 |
| I6   | Add2    | 6  | 12 | 13 | 14 |
| I7   | Add3    | 7  | 21 | 22 | 23 |
| I8   | Mul1    | 11 | 24 | 27 | 28 |

## Answers

**1)** I3 begins execution in cycle **7**.

**2)** I5 writes its result in cycle **20**.

**3)** Sum of write cycles: 5 + 6 + 11 + 9 + 20 + 14 + 23 + 28 = **116**.

**4)** I6 begins execution in cycle **12**.

**5)** Product of issue cycles for I3, I5, I7: 3 x 5 x 7 = **105**.

**6)** Execution ranges (X to E inclusive): I1: 2-4, I2: 3-5, I3: 7-10, I4: 7-8, I5: 12-19, I6: 12-13, I7: 21-22, I8: 24-27. Union of cycles: {2,3,4,5, 7,8,9,10, 12,13,14,15,16,17,18,19, 21,22, 24,25,26,27} = **22** cycles.

**7)** I8 writes its result in cycle **28**.

**8)** I1 issues cycle 1, I8 writes cycle 28. Total inclusive: 28 - 1 + 1 = **28**.

**9)** (W-I) values: I1: 5-1=4, I2: 6-2=4, I3: 11-3=8, I4: 9-4=5. Sum = 4+4+8+5 = **21**.

**10)** Cycle 10 station occupancy: Mul1 has I3 (executing, writes cycle 11), Mul2 has I5 (waiting, issued cycle 5), Add2 has I6 (waiting, issued cycle 6), Add3 has I7 (waiting, issued cycle 7). Load1 freed cycle 5, Load2 freed cycle 6, Add1 freed cycle 9. I8 not yet issued. Total = **4** stations occupied.


Prompt:
Construct and analyze the FM-index of a cipher-transformed string.

Original: S = "abracadabra" (length 11)

Cipher at 0-indexed position i, where each character's alphabetic position is p (a=0, b=1, ..., z=25):

- i mod 3 = 0: shift forward by (i mod 5), i.e., new position = (p + (i mod 5)) mod 26
- i mod 3 = 1: shift backward by ((i * 2) mod 4), i.e., new position = (p - ((i * 2) mod 4)) mod 26
- i mod 3 = 2: mirror, i.e., new position = 25 - p

Apply the cipher to each character of S to obtain T' (length 11), then append '$' to form T = T' + '$' (length 12). All indices are 0-based. The sentinel '$' sorts before all letters.

Define Occ(c, k) = number of occurrences of character c in BWT[0..k-1] (i.e., the first k characters of BWT).

**Required:**

1. Transform each character of S. For any position where (p - shift) is negative, show the mod 26 computation explicitly. State the final transformed string T.

2. State which characters that appear in S do not appear in T' (the transformed string before appending '$').

3. Construct the suffix array (SA) and Burrows-Wheeler Transform (BWT) of T. List all 12 suffixes in sorted order with their starting positions.

4. Compute the C array (for each character c, the number of characters in T that are lexicographically smaller than c). Then compute: Occ('z', 5), Occ('z', 7), Occ('y', 8), Occ('y', 10).

5. Using LF(i) = C[BWT[i]] + Occ(BWT[i], i), compute LF(3), LF(5), LF(8), LF(11).

6. Perform backward search for the pattern "ze". Process characters right-to-left. Show the Occ values used at each step and the resulting SA range.

7. Compute the number of non-empty distinct substrings of T using the LCP array method: n(n+1)/2 - sum(LCP). Identify which consecutive suffix pairs in the sorted order contribute a non-zero LCP value, and state each pair's LCP.

The response correctly transforms S = 'abracadabra' to produce T' = 'azidczeyyva' (before appending '$'), yielding T = 'azidczeyyva$'.

The response identifies that the characters 'b' and 'r' from S do not appear in the transformed string T'.

The response produces SA = [11, 10, 0, 4, 3, 6, 2, 9, 8, 7, 5, 1] and BWT = 'av$dizzyyeca'.

The response computes Occ('z', 5) = 0 by counting zero 'z' characters in BWT[0..4] = 'av$di'.

The response computes Occ('z', 7) = 2 by counting two 'z' characters in BWT[0..6] = 'av$dizz'.

The response computes Occ('y', 8) = 1 by counting one 'y' character in BWT[0..7] = 'av$dizzy'.

The response computes Occ('y', 10) = 2 by counting two 'y' characters in BWT[0..9] = 'av$dizzyye'.

In backward search for 'ze', the response computes Occ('z', 6) = 1 and derives the final SA range [10, 10].

The response computes the non-empty distinct substring count as 75, using LCP sum = 3 subtracted from 78.

PHASE 1: CIPHER TRANSFORMATION

Original: S = "abracadabra" (length 11)

Rules:
- i mod 3 = 0: new_p = (p + (i mod 5)) mod 26
- i mod 3 = 1: new_p = (p - ((i*2) mod 4)) mod 26
- i mod 3 = 2: new_p = 25 - p

Character-by-character:

i=0:  'a' (p=0)   Rule 0: shift = 0 mod 5 = 0.  new_p = (0+0) mod 26 = 0 -> 'a'
i=1:  'b' (p=1)   Rule 1: shift = (2) mod 4 = 2.  new_p = (1-2) mod 26 = (-1) mod 26 = 25 -> 'z'
      [NEGATIVE: 1-2 = -1; -1 mod 26 = 25]
i=2:  'r' (p=17)  Rule 2: new_p = 25-17 = 8 -> 'i'
i=3:  'a' (p=0)   Rule 0: shift = 3 mod 5 = 3.  new_p = (0+3) mod 26 = 3 -> 'd'
i=4:  'c' (p=2)   Rule 1: shift = (8) mod 4 = 0.  new_p = (2-0) mod 26 = 2 -> 'c'
i=5:  'a' (p=0)   Rule 2: new_p = 25-0 = 25 -> 'z'
i=6:  'd' (p=3)   Rule 0: shift = 6 mod 5 = 1.  new_p = (3+1) mod 26 = 4 -> 'e'
i=7:  'a' (p=0)   Rule 1: shift = (14) mod 4 = 2.  new_p = (0-2) mod 26 = (-2) mod 26 = 24 -> 'y'
      [NEGATIVE: 0-2 = -2; -2 mod 26 = 24]
i=8:  'b' (p=1)   Rule 2: new_p = 25-1 = 24 -> 'y'
i=9:  'r' (p=17)  Rule 0: shift = 9 mod 5 = 4.  new_p = (17+4) mod 26 = 21 -> 'v'
i=10: 'a' (p=0)   Rule 1: shift = (20) mod 4 = 0.  new_p = (0-0) mod 26 = 0 -> 'a'

T' = "azidczeyyva"
T = "azidczeyyva$"

PHASE 2: CHARACTERS ELIMINATED FROM S

Characters in S: {a, b, c, d, r}
Characters in T': {a, c, d, e, i, v, y, z}
Missing: 'b' and 'r' do not appear in T'.
- 'b' at positions 1 and 8: transformed to 'z' (rule 1) and 'y' (rule 2)
- 'r' at positions 2 and 9: transformed to 'i' (rule 2) and 'v' (rule 0)

PHASE 3: SUFFIX ARRAY AND BWT

T = "azidczeyyva$"
     0123456789...

Suffixes sorted lexicographically:

SA[0]  = 11 | $
SA[1]  = 10 | a$
SA[2]  =  0 | azidczeyyva$
SA[3]  =  4 | czeyyva$
SA[4]  =  3 | dczeyyva$
SA[5]  =  6 | eyyva$
SA[6]  =  2 | idczeyyva$
SA[7]  =  9 | va$
SA[8]  =  8 | yva$
SA[9]  =  7 | yyva$
SA[10] =  5 | zeyyva$
SA[11] =  1 | zidczeyyva$

SA = [11, 10, 0, 4, 3, 6, 2, 9, 8, 7, 5, 1]

BWT[i] = T[(SA[i] - 1) mod 12]:
BWT[0]  = T[10] = 'a'
BWT[1]  = T[9]  = 'v'
BWT[2]  = T[11] = '$'
BWT[3]  = T[3]  = 'd'
BWT[4]  = T[2]  = 'i'
BWT[5]  = T[5]  = 'z'
BWT[6]  = T[1]  = 'z'
BWT[7]  = T[8]  = 'y'
BWT[8]  = T[7]  = 'y'
BWT[9]  = T[6]  = 'e' [TYPO-CHECK: T[6]='e' WRONG -> wait, T="azidczeyyva$", T[6]='e'. Yes correct.]
BWT[10] = T[4]  = 'c' [TYPO-CHECK: T[4]='c'. Yes.]
BWT[11] = T[0]  = 'a'

WAIT - let me recheck T indexing:
T[0]='a', T[1]='z', T[2]='i', T[3]='d', T[4]='c', T[5]='z', T[6]='e', T[7]='y', T[8]='y', T[9]='v', T[10]='a', T[11]='$'

BWT = "av$dizzyyeca"

PHASE 4: C ARRAY

Character frequencies in T: $ x1, a x2, c x1, d x1, e x1, i x1, v x1, y x2, z x2

C['$'] = 0
C['a'] = 1
C['c'] = 3
C['d'] = 4
C['e'] = 5
C['i'] = 6
C['v'] = 7
C['y'] = 8
C['z'] = 10

Occ COMPUTATIONS (Occ(c, k) = count of c in BWT[0..k-1]):

BWT = "av$dizzyyeca"
       positions: 0='a', 1='v', 2='$', 3='d', 4='i', 5='z', 6='z', 7='y', 8='y', 9='e', 10='c', 11='a'

Occ('z', 5) = count of 'z' in BWT[0..4] = "av$di" = 0
Occ('z', 7) = count of 'z' in BWT[0..6] = "av$dizz" = 2
Occ('y', 8) = count of 'y' in BWT[0..7] = "av$dizzy" = 1
Occ('y', 10) = count of 'y' in BWT[0..9] = "av$dizzyye" = 2

PHASE 5: LF-MAPPING

LF(i) = C[BWT[i]] + Occ(BWT[i], i)

LF(3): BWT[3]='d', C['d']=4, Occ('d',3)=0 (BWT[0..2]="av$" has no 'd') -> LF(3) = 4
LF(5): BWT[5]='z', C['z']=10, Occ('z',5)=0 -> LF(5) = 10
LF(8): BWT[8]='y', C['y']=8, Occ('y',8)=1 -> LF(8) = 9
LF(11): BWT[11]='a', C['a']=1, Occ('a',11)=1 (BWT[0..10]="av$dizzyyec" has one 'a') -> LF(11) = 2

PHASE 6: BACKWARD SEARCH FOR "ze"

Pattern "ze", processed right-to-left: first 'e', then 'z'.

Step 1 - Initialize with 'e':
  lo = C['e'] = 5
  hi = C['i'] - 1 = 6 - 1 = 5   (next character after 'e' in alphabet present is 'i')
  Range: [5, 5]

Step 2 - Extend with 'z':
  Occ('z', 5) = 0   (BWT[0..4] = "av$di")
  Occ('z', 6) = 1   (BWT[0..5] = "av$diz")
  new_lo = C['z'] + Occ('z', 5) = 10 + 0 = 10
  new_hi = C['z'] + Occ('z', 6) - 1 = 10 + 1 - 1 = 10
  Range: [10, 10]

Result: 1 occurrence. SA[10] = 5. T[5..] = "zeyyva$". Confirmed.

PHASE 7: LCP ARRAY AND DISTINCT SUBSTRINGS

LCP[i] = length of longest common prefix between suffix at SA[i-1] and SA[i].

LCP[0] = 0 (by definition)
LCP[1] = 0: "$" vs "a$" -> no common prefix
LCP[2] = 1: "a$" vs "azidczeyyva$" -> common prefix "a"
LCP[3] = 0: "azidczeyyva$" vs "czeyyva$" -> no common prefix
LCP[4] = 0: "czeyyva$" vs "dczeyyva$" -> no common prefix
LCP[5] = 0: "dczeyyva$" vs "eyyva$" -> no common prefix
LCP[6] = 0: "eyyva$" vs "idczeyyva$" -> no common prefix
LCP[7] = 0: "idczeyyva$" vs "va$" -> no common prefix
LCP[8] = 0: "va$" vs "yva$" -> no common prefix
LCP[9] = 1: "yva$" vs "yyva$" -> common prefix "y"
LCP[10] = 0: "yyva$" vs "zeyyva$" -> no common prefix
LCP[11] = 1: "zeyyva$" vs "zidczeyyva$" -> common prefix "z"

LCP = [0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1]

Contributing pairs (non-zero LCP):
- LCP[2] = 1: "a$" (pos 10) and "azidczeyyva$" (pos 0)
- LCP[9] = 1: "yva$" (pos 8) and "yyva$" (pos 7)
- LCP[11] = 1: "zeyyva$" (pos 5) and "zidczeyyva$" (pos 1)

LCP sum = 1 + 1 + 1 = 3

Non-empty distinct substrings = n(n+1)/2 - sum(LCP) = 12*13/2 - 3 = 78 - 3 = 75

FINAL ANSWERS:
1. T = "azidczeyyva$"
2. Characters from S absent in T': 'b' and 'r'
3. SA = [11, 10, 0, 4, 3, 6, 2, 9, 8, 7, 5, 1]; BWT = "av$dizzyyeca"
4. Occ('z',5) = 0, Occ('z',7) = 2, Occ('y',8) = 1, Occ('y',10) = 2
5. LF(3) = 4, LF(5) = 10, LF(8) = 9, LF(11) = 2
6. Backward search "ze": [5,5] -> [10,10], using Occ('z',5) = 0 and Occ('z',6) = 1
7. LCP sum = 3; contributing pairs at indices 2, 9, 11; non-empty distinct substrings = 75


Prompt:
You are given `k` simple undirected graphs `G1..Gk` on the *same* labeled vertex set `V = {1,2,...,n}`.

We say a graph `Q` is a **blow-up pattern** (called a *fun graph*) of a graph `H` if there exists a partition of `V(H)` into `|V(Q)|` nonempty parts, one part per vertex of `Q`, such that:

1. Each part induces either a **clique** or an **independent set** in `H`.
2. For two distinct vertices `x,y` of `Q` with corresponding parts `Px, Py`:
   - if `xy` is an edge of `Q`, then **every** vertex of `Px` is adjacent in `H` to **every** vertex of `Py`;
   - if `xy` is not an edge of `Q`, then **no** vertex of `Px` is adjacent in `H` to **any** vertex of `Py`.

A fun graph `Q` of `H` is called **super-minimal** (called *superb*) if among all fun graphs of `H`, it has the **fewest vertices**.

---

### The question

For each test case, decide whether there exist graphs `H1..Hk` that all share a common (unknown) vertex set `W`, such that:

- For every `i`, the given graph `Gi` is a **superb fun graph** of `Hi`.

Additionally, the following cross-graph restriction must hold for each label `v ∈ V`:

- If in some pair `(Gi,Hi)` the vertex `v` corresponds to an **independent-set part of size at least 2**, then there is **no** other pair `(Gj,Hj)` in which the same label `v` corresponds to a **clique part of size at least 2**.

Print `"Yes"` if such graphs `H1..Hk` exist, otherwise print `"No"`.

---

## Input
- The first line: integer `t` — number of test cases.
For each test case:
- One line: integers `n, k` (`1 ≤ n ≤ 300`, `1 ≤ k ≤ 10`)
- Then for each `i=1..k`:
  - One line: integer `m` — number of edges in `Gi`
  - Next `m` lines: edges `u v` (`1 ≤ u < v ≤ n`)

Constraints:
- Sum of `m` over all graphs over all test cases ≤ `2 * 10^5`
- Sum of `n` over all test cases ≤ `300`

## Output
For each test case output `Yes` or `No` (case-insensitive).

Does the solution define a Type-A twin pair as a nonedge pair (a,b) with identical open neighborhoods?

Does the solution define a Type-B twin pair as an edge pair (a,b) with identical closed neighborhoods?

Does the solution add the clause (x_a ∨ x_b) for each Type-A pair (a,b)?

Does the solution add the clause (¬x_a ∨ ¬x_b) for each Type-B pair (a,b)?

Does the solution state that satisfiability is decided using SCC computation on the implication graph?

Does the solution state that unsatisfiable occurs when x_v shares an SCC with ¬x_v for some v?

Does the solution describe detecting Type-A pairs using equality of adjacency bitsets?

Does the solution describe detecting Type-B pairs using equality of closed-neighborhood bitsets?

## Expert Solution (DSA — Graphs / 2-SAT)

### Key structural notion: twins inside `Gi`

Fix one graph `Gi` on vertex set `{1..n}`.

- **Type-A twin pair (open twins):** vertices `a ≠ b` such that
  `ab ∉ E(Gi)` and `N(a) = N(b)` (open neighborhoods identical).

- **Type-B twin pair (closed twins):** vertices `a ≠ b` such that
  `ab ∈ E(Gi)` and `N[a] = N[b]` (closed neighborhoods identical).

These are exactly the pairs that can be merged in a fun-graph representation unless internal structure blocks the merge.

### Why “size ≥ 2” matters for superbness

Assume `Gi` is a superb fun graph of some `Hi` via parts `{P1..Pn}` (part `Pv` corresponds to label `v`).

- If `a,b` are **Type-A** in `Gi` and both `Pa` and `Pb` are **independent sets**, then `Pa ∪ Pb` is also independent, and it has the same external adjacency pattern (because `a` and `b` have identical neighborhoods in `Gi`).
  This produces a valid fun graph with fewer vertices than `Gi`, contradicting superbness.

  Therefore, for every Type-A pair `(a,b)` in `Gi`, at least one of `{Pa,Pb}` must be a **clique with size ≥ 2** (a singleton clique does not prevent `Pa ∪ Pb` from being independent).

- If `a,b` are **Type-B** in `Gi` and both `Pa` and `Pb` are **cliques**, then `Pa ∪ Pb` is a clique (edges inside parts are complete, and edges across parts are complete because `ab` is an edge of `Gi`).
  This again yields a smaller fun graph, contradicting superbness.

  Therefore, for every Type-B pair `(a,b)` in `Gi`, at least one of `{Pa,Pb}` must be an **independent set with size ≥ 2**.

### Collapsing the multi-graph constraint into one boolean per label

The extra global rule says a label `v` cannot be realized as:
- an **independent set of size ≥ 2** in some `Hi`
- and a **clique of size ≥ 2** in another `Hj`.

Introduce one boolean variable per label:
- `x_v = 1` means label `v` is allowed to be the “expanded clique label” (clique of size ≥ 2 whenever expansion is needed).
- `x_v = 0` means label `v` is allowed to be the “expanded independent label” (independent set of size ≥ 2 whenever expansion is needed).

Labels that end up size `1` in every `Hi` impose no restriction; they can be assigned either value.

Now translate each twin constraint into a 2-CNF clause on `{x_v}`:

- For every **Type-A** pair `(a,b)` in every `Gi`:
  at least one must be an expanded clique label
  ⇒ clause **`(x_a ∨ x_b)`**

- For every **Type-B** pair `(a,b)` in every `Gi`:
  at least one must be an expanded independent label
  ⇒ clause **`(¬x_a ∨ ¬x_b)`**

All constraints across all `k` graphs become a single **2-SAT instance with `n` variables**.

### Decision procedure (2-SAT)

Build the implication graph on `2n` literals.

- Clause `(p ∨ q)` adds implications `¬p → q` and `¬q → p`.

So:
- From `(x_a ∨ x_b)` add `¬x_a → x_b` and `¬x_b → x_a`.
- From `(¬x_a ∨ ¬x_b)` add `x_a → ¬x_b` and `x_b → ¬x_a`.

Run SCC (Kosaraju or Tarjan). The formula is satisfiable iff for every `v`,
`x_v` and `¬x_v` are in different SCCs.

If satisfiable print `Yes`, else print `No`.

### How to compute Type-A / Type-B pairs efficiently

For each `Gi`, store adjacency as bitsets `adj[v]` of length `n`.

For a pair `(a,b)`:
- `ab ∉ E(Gi)` and `adj[a] == adj[b]`  ⇒ Type-A
- `ab ∈ E(Gi)` and `(adj[a] ∪ {a}) == (adj[b] ∪ {b})` ⇒ Type-B

This is `O(k * n^2 * (n/word_size))`, which is fine for `n ≤ 300`.

### Complexity

Per test case:
- Building bitsets: `O(k * (n^2/word + m))`
- Checking all pairs: `O(k * n^2 * (n/word))`
- 2-SAT SCC: `O(n + number_of_clauses)`

Fits easily under the constraints.

### Output
For each test case output `Yes` iff the constructed 2-SAT instance is satisfiable.


Prompt:
You are given an undirected connected graph G with n vertices and exactly n edges. The graph has no self-loops and no multiple edges. All edges have unit weight. Each edge has a state: **on** or **off**. Initially all edges are **off**.

You are given m queries (v, u). For each query, you must **toggle** (change on to off, or off to on) the state of **all edges on the shortest path** from v to u in G (using minimum number of edges).

If there are multiple shortest paths, you must choose the **lexicographically minimal** one, where a path is represented as the vertex sequence v, v1, v2, ..., u and sequences are compared element-by-element from left to right.

After each query, output the number of connected components in the graph that contains all n vertices and **only the edges that are currently on**.

### Input
- First line: n m (3 <= n <= 10, 1 <= m <= 10)
- Next n lines: each line contains two integers a b (1 <= a, b <= n, a != b) describing an edge
- Next m lines: each line contains two integers v u (1 <= v, u <= n, v != u) describing a query

### Output
Print m lines. The i-th line is the number of connected components after the i-th query.

For the input "6 3\n1 2\n2 3\n3 4\n4 1\n2 5\n4 6\n2 4\n5 6\n3 1\n", does the answer output exactly "4\n4\n2"?

For the input "5 4\n1 2\n2 3\n3 1\n3 4\n4 5\n1 2\n2 3\n1 3\n5 1\n", does the answer output exactly "4\n3\n3\n1"?

For the input "4 2\n1 2\n2 3\n3 4\n4 1\n1 3\n2 4\n", does the answer output exactly "2\n2"?

For the input "5 2\n1 2\n2 3\n3 1\n1 4\n4 5\n5 2\n5 3\n", does the answer output exactly "2\n3"?

For the input "3 3\n1 2\n2 3\n3 1\n1 2\n2 3\n1 3\n", does the answer output exactly "2\n1\n1"?

Does the answer demonstrate that when two cycle arcs have equal length, the lexicographically smaller path is chosen by comparing the next vertex after the start?

Does the answer demonstrate state that a connected graph with n vertices and exactly n edges contains exactly one simple cycle?

## 1) Graph structure
A connected graph with n vertices and n edges contains exactly one simple cycle. All other edges are bridges forming trees attached to cycle vertices.

Let:
- C = set of cycle vertices, k = |C|
- onTreeEdges = number of ON edges not on the cycle
- onCycleEdges = number of ON edges on the cycle

## 2) Connected components formula
Tree edges alone give a forest:
  compsTrees = n - onTreeEdges

For the cycle on k vertices:
- If onCycleEdges = k (all cycle edges ON): compsCycle = 1
- If onCycleEdges < k: compsCycle = k - onCycleEdges

The k cycle vertices are k distinct components in the tree forest, merged by cycle edges:
  answer = compsTrees + compsCycle - k
  = (n - onTreeEdges) + compsCycle - k

## 3) Path structure
Let c(x) = the cycle vertex reached from x by following tree ancestors (or x itself if on cycle). Any shortest path v -> u consists of:
1. Unique tree path v -> c(v)
2. An arc on the cycle from c(v) to c(u)
3. Unique tree path c(u) -> u

Only the cycle arc direction is ambiguous. Choose the shorter arc. If both arcs have equal length, choose the direction whose next vertex after c(v) on the cycle has the smaller vertex id, which makes the full vertex sequence lexicographically minimal.

## 4) Implementation
Use HLD + segment trees with range flip/sum on tree edges, and a separate segment tree on the k cycle edges. Each query: flip tree paths via HLD, flip cycle arc, then compute answer using the formula above.

Complexity: O((n + m) log n) time, O(n) memory.

## 5) Verification

**Test A**
Input:
6 3
1 2
2 3
3 4
4 1
2 5
4 6
2 4
5 6
3 1

Cycle: 1-2-3-4-1 (k=4). Tree edges: 2-5, 4-6.

Query (2,4): cycle arcs both length 2. Next vertices: 3 vs 1. Choose 1 (path 2-1-4). Toggle 2-1, 1-4. onTree=0, onCycle=2. answer = 6-0+4-2-4 = 4.
Query (5,6): tree 5-2, tree 6-4, cycle 2-1-4 (same tie-break). Toggle tree 5-2, 4-6 ON; cycle 2-1, 1-4 OFF. onTree=2, onCycle=0. answer = 6-2+4-0-4 = 4.
Query (3,1): cycle arcs length 2. Next vertices: 2 vs 4. Choose 2 (path 3-2-1). Toggle 3-2, 2-1. onTree=2, onCycle=2. answer = 6-2+4-2-4 = 2.

Output:
4
4
2

**Test B**
Input:
5 4
1 2
2 3
3 1
3 4
4 5
1 2
2 3
1 3
5 1

Cycle: 1-2-3-1 (k=3). Tree edges: 3-4, 4-5.

Query (1,2): path 1-2 (length 1). Toggle 1-2. onTree=0, onCycle=1. answer = 5+3-1-3 = 4.
Query (2,3): path 2-3 (length 1). Toggle 2-3. onTree=0, onCycle=2. answer = 5+3-2-3 = 3.
Query (1,3): path 1-3 (length 1). Toggle 1-3. onTree=0, onCycle=3. answer = 5+1-3 = 3.
Query (5,1): tree 5-4-3, cycle 3-1 (length 1). Toggle tree 4-5, 3-4; cycle 3-1 OFF. onTree=2, onCycle=2. answer = 5-2+3-2-3 = 1.

Output:
4
3
3
1

**Test C**
Input:
4 2
1 2
2 3
3 4
4 1
1 3
2 4

Pure 4-cycle (k=4).

Query (1,3): arcs both length 2. Next vertex: 2 vs 4. Choose 2 (path 1-2-3). Toggle 1-2, 2-3. onCycle=2. answer = 4+4-2-4 = 2.
Query (2,4): arcs both length 2. Next vertex: 3 vs 1. Choose 1 (path 2-1-4). Toggle 2-1 (ON->OFF), 1-4 (OFF->ON). ON: {2-3, 1-4}. onCycle=2. answer = 4+4-2-4 = 2.

Output:
2
2

**Test D**
Input:
5 2
1 2
2 3
3 1
1 4
4 5
5 2
5 3

Cycle: 1-2-3 (k=3). Tree: 1-4, 4-5.

Query (5,2): tree 5-4-1, cycle 1-2 (length 1). Toggle tree 4-5, 1-4; cycle 1-2. onTree=2, onCycle=1. answer = 5-2+3-1-3 = 2.
Query (5,3): tree 5-4-1 (toggle back OFF), cycle 1-3 (length 1). Toggle 1-3. onTree=0, onCycle=2. answer = 5-0+3-2-3 = 3.

Output:
2
3

**Test E**
Input:
3 3
1 2
2 3
3 1
1 2
2 3
1 3

Pure triangle (k=3).

Query (1,2): toggle 1-2. onCycle=1. answer = 3+3-1-3 = 2.
Query (2,3): toggle 2-3. onCycle=2. answer = 3+3-2-3 = 1.
Query (1,3): path 1-3 (length 1). Toggle 3-1. onCycle=3. answer = 3+1-3 = 1.

Output:
2
1
1


Machine Learning & Artificial Intelligence

Prompt: A Deep Belief Network processes 784-dimensional inputs to 10 output classes. The architecture connects Input → L1 → L2 → L3 → Output, where each connection has a weight matrix.\n\n**Configurations:**\n\n| Config | L1 | L2 | L3 | η | k | epochs |\n|--------|-----|-----|-----|------|---|--------|\n| A | 500 | 500 | 500 | 0.10 | 1 | 15 |\n| B | 600 | 400 | 200 | 0.08 | 2 | 18 |\n| C | 400 | 400 | 400 | 0.05 | 3 | 20 |\n| D | 580 | 480 | 380 | 0.07 | 1 | 14 |\n| E | 550 | 450 | 350 | 0.06 | 1 | 22 |\n| F | 480 | 380 | 280 | 0.09 | 2 | 19 |\n| G | 520 | 420 | 320 | 0.07 | 1 | 19 |\n| H | 590 | 490 | 390 | 0.08 | 1 | 12 |\n| J | 540 | 440 | 340 | 0.06 | 1 | 17 |\n| K | 510 | 410 | 310 | 0.07 | 1 | 15 |\n\n**Observations (Ψ recorded before constraint validation):**\n\n| Config | M | T | Ψ |\n|--------|------|------|------|\n| A | 1794 | 22.5 | 68.4 |\n| B | 1585 | 21.6 | 69.0 |\n| D | 1839 | 20.2 | 69.9 |\n| G | 1527 | 23.9 | 67.5 |\n\n**Constraints:**\n\n| | |\n|---|---|\n| c1 | M ≤ 2000 |\n| c2 | T ≤ 28 |\n| c3 | η ≥ 0.10 → epochs ≤ 15 |\n| c4 | k ≥ 2 → T multiplied by (1 + 0.4×(k-1)) for c2 evaluation |\n| c5 | L1 ≥ 600 → M multiplied by 1.08 for c1 evaluation |\n| c6 | epochs > 20 → c2 threshold becomes 25 |\n| c7 | Ψ > 70 → M ≤ 1750 |\n\nDetermine the valid configuration that maximizes Ψ.

The response states that Config K is the optimal valid configuration.

The response states that Config K has Ψ equal to 71.0 or a value between 70.9 and 71.1.

The response derives that M equals 0.002 multiplied by the total parameters.

The response derives that T equals epochs multiplied by the sum of layer sizes divided by 1000.

The response derives that Ψ is a linear function of T with slope between -0.60 and -0.70.

The response states that Config H is invalid because it violates constraint c7.

The response calculates that Config H has M equal to 1893 or a value between 1890 and 1900.

The response identifies exactly five invalid configurations: B, C, E, F, and H.

The response applies the c4 time adjustment factor of 1.4 when k equals 2.


Prompt:
System overview
A single-core kernel manages five user processes P1 through P5 over discrete ticks.
At every tick, three subsystems interact: an MLFQ scheduler with three queues and dynamic Q0 quantum, demand-paged virtual memory with TLB and Enhanced Second Chance replacement, and a disk using C-LOOK with anti-starvation.
Physical memory has frames F1..F12. Disk has 64 sectors and arm starts at sector 0.
Page p of process Pk is stored at sector (k*13 + p*3 + 1) mod 64.

Initial conditions
At tick 0, P1, P2, and P4 are READY in Q0. P3 does not exist. P5 is NOT READY until tick 3.
Initial Q0 order is PID order among READY processes. No process starts blocked.
Pinned page-table frames are fixed and never evictable: F1=PT:P1, F2=PT:P2, F3=PT:P4.
When P5 becomes READY at tick 3 it receives F5=PT:P5. When P3 is created at tick 10 it receives F4=PT:P3.
Data frames are therefore exactly F6..F12 for the full run.

Process programs
P1: R(0), W(1), R(2), R(3), W(4), R(1), R(5), W(6), R(0), W(7), R(8), DONE.
P2: R(0), R(14), W(15), R(1), R(2), W(14), R(3), W(4), R(14), DONE.
P3: W(0), R(1), W(2), R(3), R(4), W(5), R(6), DONE.
P4: R(0), R(14), W(14), R(1), R(15), W(2), R(3), W(15), DONE.
P5: R(0), R(1), R(2), W(3), W(4), R(5), R(0), W(6), DONE.

Fork and shared mapping
At the start of tick 10, before scheduling, P3 is created from P1.
Fork blocks P1 for exactly tick 10.
During tick 10, P3 exists and is active for accounting, but P3 is not READY and is not schedulable.
At tick 11, both P1 and P3 are READY in Q0.
P3 inherits P1 mappings as read-only shared mappings. A write by P1 or P3 to an inherited shared page triggers COW.
P2 and P4 share VPN 14 and 15 in one shared frame pair.
The first of P2 or P4 to access VPN 14 establishes the shared mapping and allocates one physical frame for VPN 14 (loaded via the normal page-in path). VPN 15 is demand-paged independently into its own shared frame on first access. Whichever process did not trigger the original allocation attaches to the existing shared frame on its first access to VPN 14 or VPN 15; if that frame is still being loaded, the attaching process page-faults and waits for the same page-in to complete.
When one writes shared VPN 14 or 15, the other process's TLB entries for those VPNs are invalidated exactly at the next context-switch boundary away from the writer.

Per-tick execution order
At the start of each tick, execute the following order exactly.
1) Apply reference-bit decay on ticks 12, 24, 36, 48, 60.
2) Apply global boost on ticks 25 and 50.
3) Apply Q2 aging promotions.
4) Apply arrivals/unblocks that become READY at this tick.
5) If CPU is free, scheduler chooses next READY process.
6) If chosen PID differs from previously running PID or previous state was IDLE, consume one context-switch tick; exception: the first dispatch from initial IDLE at tick 0 consumes 0 ticks and is not counted.
7) Otherwise execute one unit of running work.
8) Disk progresses one unit of current seek/transfer if active; if idle, it selects next request immediately and also consumes first seek unit in the same tick.
Scheduler tie-break in any queue is lowest PID.

Scheduling rules
Q0 quantum is 2 + ((pf_non_cow + cow_faults_by_process) mod 3), sampled once at dispatch and fixed for that dispatch.
Q1 quantum is 5. Q2 quantum is 9. Quantum never changes mid-dispatch.
New READY and newly unblocked processes always enter Q0.
Blocking does not demote. On unblock, process enters Q0.
Quantum expiry without blocking demotes one level; Q2 remains Q2 on expiry.
Scheduler re-evaluates only when running process blocks, exhausts quantum, or terminates.
Aging rule: waiting 7 consecutive ticks in Q2 without dispatch promotes to Q0 at the start of the next tick.

Context-switch accounting
Count and charge 1 tick for every transition between distinct running states, including process-to-process, process-to-idle, and idle-to-process.
The only exception is the unique initial IDLE-to-first-process dispatch at tick 0, which has zero cost and is not counted.
Every charged context-switch tick flushes entire TLB and applies pending shared-region invalidations.

TLB rules
TLB capacity is 4 and fully associative.
Hit: memory operation costs 1 tick, increments entry hit_count by 1, and sets last_access_tick to current tick.
Miss with resident page: operation costs 2 ticks total (walk plus access), then insert/update entry with hit_count=0 and last_access_tick=current tick.
On insert when full, evict entry with minimum score (last_access_tick + 2*hit_count), tie by smallest VPN, then smallest PID.
Flush removes all entries and counters.
Any successful memory access (read or write, hit path or resident-miss path) sets the referenced frame's R bit immediately in that same tick before any replacement decision caused by that access.

Paging and replacement rules
Access to non-resident page causes immediate page fault and block; page-in request is enqueued in that same tick.
Enhanced Second Chance class order is fixed: C0=(R=0,D=0), C1=(R=0,D=1), C2=(R=1,D=0), C3=(R=1,D=1).
Victim selection is global over eligible data frames F6..F12.
Within the lowest non-empty class, choose oldest load_tick, tie lowest frame number.
This ordering is complete and does not depend on sweep start position.
Sweep does not modify reference bits. Reference bits change only on access and global decay ticks.
load_tick is defined as the tick at which the corresponding page-in transfer completes and the page becomes resident in its frame.

Minimum residency rule
Target floor is 1 data frame per active process, treated as a soft floor.
Active means process exists and is not terminated.
During victim eligibility, first consider only candidates whose owner currently holds more than 1 data frame (prefer to shrink larger holders). If no such candidate exists, allow eviction from owners holding exactly 1 data frame.
If still none (only possible when all data frames are PENDING), wait until any pending write-back completes.
This fallback order is mandatory and deterministic.

Dirty eviction and COW timing
If chosen victim is dirty, schedule write-back and mark frame PENDING immediately.
A PENDING frame cannot be reused until write-back completes.
The faulting page-in cannot claim that frame before completion.
COW fault blocks writer immediately, counts toward cow_faults and toward writer dynamic-quantum cumulative count, but not toward reported per-process page_faults.
Once destination frame is available, COW service consumes exactly 2 CPU ticks (copy then remap).
Those 2 ticks are kernel-work ticks executed while that process is RUNNING; they do not execute a user memory operation and do not decrement that dispatch quantum counter.
If replacement/write-back is needed, those disk delays occur before the 2 COW CPU ticks.
COW faults always forfeit remaining quantum at fault time and re-enter through Q0 when unblocked.
A write memory operation sets the frame D bit immediately in the tick of that write, regardless of whether translation came from a TLB hit or from a resident-page miss path.
For COW, the newly allocated destination frame starts clean after copy; the triggering write then sets D immediately when that write is applied.

Disk rules
Single arm and one active request at a time.
Request type is READ (page-in) or WRITE (write-back).
Same-tick enqueue order is deterministic by event order within tick; if event step ties, lower PID then lower VPN.
Priority rule: any request with wait_time > 6 is starved and has absolute priority; among starved requests use enqueue_tick then lower sector.
If no starved request exists, use C-LOOK in increasing-sector direction only: choose smallest pending sector >= current arm; if none exists, wrap to global smallest pending sector.
Seek time is ceil(abs(current-target)/2), minimum 1. Transfer time is 1.
At equal seek distance, READ beats WRITE; remaining ties use earlier enqueue_tick then lower sector.

Unblock timing
If page-in transfer completes at tick T, the process becomes READY at start of tick T+1 and enters Q0 at that time.
Requests generated during a context-switch tick are enqueued in that same tick and are eligible for disk selection in step 8 of that same tick.

Required output
Produce the unique deterministic execution trace and print exactly one key=value line per item, in this exact order:
page_faults=f1,f2,f3,f4,f5
tlb_hits=total
context_switches=count
cow_faults=count
dirty_writebacks=count
total_seek_ticks=sum
disk_arm_final=sector
completion_order=PID1,PID2,PID3,PID4,PID5
p3_first_cow_tick=tick
last_tick=tick
final_frame_map=F1:owner,F2:owner,F3:owner,F4:owner,F5:owner,F6:owner,F7:owner,F8:owner,F9:owner,F10:owner,F11:owner,F12:owner

Output value conventions
For final_frame_map owner, use Pk:n for process-page, PT:Pk for page-table frame, PENDING-WRITE:Pk:n if write-back is still in progress at simulation end, or FREE.
For p3_first_cow_tick use NA if P3 never triggers COW.

Does the answer report cow_faults exactly as 1?

Does the answer report tlb_hits exactly as 1?

Does the answer report p3_first_cow_tick exactly as 16?

Does the answer report context_switches exactly as 83?

Does the answer report dirty_writebacks exactly as 11?

Does the answer report disk_arm_final exactly as 39?

Does the answer report page_faults exactly as 11,10,6,7,8?

Does the answer report completion_order exactly as 3,4,1,5,2?

Does the answer report total_seek_ticks exactly as 564?

Does the answer report last_tick exactly as 623?

Does the answer report final_frame_map exactly as F1:PT:P1,F2:PT:P2,F3:PT:P4,F4:PT:P3,F5:PT:P5,F6:P1:7,F7:P2:14,F8:P2:4,F9:P4:15,F10:P5:6,F11:P1:8,F12:P2:3?

# Golden Solution: Single-Core Kernel Simulation (MLFQ + Demand-Paged VM + C-LOOK Disk)

## 1. Setup and Initialization

Physical memory holds 12 frames. Frames F1–F5 are pinned page tables:

| Frame | Role |
|-------|------|
| F1 | PT:P1 |
| F2 | PT:P2 |
| F3 | PT:P4 |
| F4 | PT:P3 (allocated at tick 10) |
| F5 | PT:P5 (allocated at tick 3) |

Data frames available throughout the run: F6–F12 (7 frames total).
Disk arm starts at sector 0. Sector formula: `(k*13 + p*3 + 1) mod 64`.

At tick 0: P1, P2, P4 are READY in Q0. P5 arrives at tick 3. P3 is forked from P1 at tick 10.

---

## 2. Detailed Tick-by-Tick Trace (Ticks 0–30)

### Tick 0

- Steps 1–4: Nothing applicable.
- Step 5: CPU idle. Pick P1 (lowest PID among {P1, P2, P4}).
- Step 6: First dispatch from initial IDLE — zero cost, not counted.
  Dispatch P1 with Q0 quantum = 2 + ((0+0) mod 3) = 2.
- Step 7: P1 executes R(0). TLB empty. Page table empty. **Page fault** (P1 pf=1). P1 blocks. Allocate F6 for P1:page0. Enqueue page-in sector (1·13 + 0·3 + 1) mod 64 = 14.
- Step 8: Disk idle. Select P1:page0 (only request). Seek from 0 to 14: ceil(14/2)=7. Begin, consume first seek unit (6 remaining).

### Tick 1

- Step 5: P1 blocked (prev_pid=P1). CPU free. Pick P2 (lowest PID among {P2, P4}).
- Step 6: **CS #1** P1→P2. TLB flush. Dispatch P2 Q0 quantum=2.
- Step 8: Disk seek (5 remaining).

### Tick 2

- Step 7: P2 executes R(0). Page fault (P2 pf=1). Allocate F7. Enqueue page-in sector 27.
- Step 8: Disk seek (4 remaining).

### Tick 3

- Step 4: P5 arrives, READY Q0. F5 assigned as PT:P5.
- Step 5: P2 blocked (prev_pid=P2). Pick P4 (pid 4 < pid 5).
- Step 6: **CS #2** P2→P4. Dispatch P4 Q0 quantum=2.
- Step 8: Disk seek (3 remaining).

### Tick 4

- Step 7: P4 executes R(0). Page fault (P4 pf=1). Allocate F8. Enqueue page-in sector 53.
- Step 8: Disk seek (2 remaining).

### Tick 5

- Step 5: P4 blocked. Pick P5.
- Step 6: **CS #3** P4→P5. Dispatch P5 Q0 quantum=2.
- Step 8: Disk seek (1 remaining).

### Tick 6

- Step 7: P5 executes R(0). Page fault (P5 pf=1). Allocate F9. Enqueue page-in sector 2.
- Step 8: Disk seek completes. Arm at 14. Transfer begins.

### Tick 7

- Step 5: P5 blocked. No READY processes.
- Step 6: **CS #4** P5→IDLE. TLB flush.
- Step 8: Disk transfer completes. P1:page0 loaded into F6 (load_tick=7). Unblock scheduled for tick 8.

### Tick 8

- Step 4: P1 unblocks → READY Q0.
- Step 5: Pick P1.
- Step 6: **CS #5** IDLE→P1. Dispatch P1 Q0 quantum = 2+((1+0) mod 3) = 3.
- Step 8: Disk idle. Select P2:page0 (sector 27). Starvation check: wait 8−2=6, not >6. C-LOOK: 27 ≥ 14. Seek ceil(13/2)=7. Begin (6 remaining).

### Tick 9

- Step 7: P1 R(0). TLB empty (flushed). Page table: page0→F6. F6 resident. TLB miss (walk tick). Set R=1 on F6. Quantum 3→2.
- Step 8: Disk seek (5 remaining).

### Tick 10

- Step 4: **Fork.** P3 created from P1. P1 blocked. P1's mid-TLB-miss state (op_rem) reset to 0. P3 inherits P1 page table: {page0→F6}. Both P1 and P3 mark page0 as COW. F4 allocated as PT:P3. P3 active from tick 10.
- Step 5: P1 blocked. P3 exists and is active for resource accounting, but not yet READY (unblocks at tick 11). No schedulable process.
- Step 6: **CS #6** P1→IDLE.
- Step 8: Disk seek (4 remaining).

### Tick 11

- Step 4: Fork unblock. P1 and P3 both READY in Q0.
- Step 5: Pick P1 (pid 1 < pid 3).
- Step 6: **CS #7** IDLE→P1. Dispatch P1 Q0 quantum=3.
- Step 8: Disk seek (3 remaining).

### Tick 12

- Step 1: **Ref-bit decay.** All R bits on F6–F12 cleared (F6.R becomes 0).
- Step 7: P1 R(0). TLB empty (flushed at CS). Page table: page0→F6. Resident. TLB miss walk. R set to 1. Quantum 3→2.
- Step 8: Disk seek (2 remaining).

### Tick 13

- Step 7: P1 R(0) TLB miss complete (access tick). R re-asserted, TLB insert for page0. PC advances to W(1). Quantum 2→1.
- Step 8: Disk seek completes. Arm at 27. Transfer begins.

### Tick 14

- Step 7: P1 W(1). Page1 not in page table. **Page fault** (P1 pf=2). Allocate F10. Enqueue page-in sector 17.
- Step 8: Disk transfer in progress for P2:page0.

### Tick 15

- Step 5: P1 blocked. Pick P3 (only READY Q0).
- Step 6: **CS #8** P1→P3. Dispatch P3 Q0 quantum = 2+((0+0) mod 3) = 2.
- Step 8: Disk transfer completes. P2:page0 → F7 (load_tick=15). Unblock at tick 16. Disk selects next: P4:page0 (sector 53, eq=4, wait=15−4=11>6, starved), P5:page0 (sector 2, eq=6, wait=9>6, starved), P1:page1 (sector 17, eq=14, wait=1, not starved). Starved requests: P4(eq=4), P5(eq=6). FCFS by enqueue_tick → P4(eq=4). Seek ceil((53−27)/2)=13. Begin (12 remaining).

### Tick 16

- Step 4: P2 unblocks → READY Q0.
- Step 7: P3 executes W(0). Page0 is in cow_pages (inherited from P1). **COW fault** (cow_faults=1). p3_first_cow_tick = 16. Allocate F11 for P3:page0. Frame immediately available (F11 free). P3:page0→F11. load_tick=16. P3 → READY Q0 with cow_work_remaining=2.
- Step 8: Disk seek (11 remaining).

### Ticks 17–20

- **Tick 17:** CS #9 P3→P2. Dispatch P2 Q0 quantum=3 (pf=1).
- **Tick 18:** P2 R(0) TLB miss walk on F7. Quantum 3→2.
- **Tick 19:** P2 R(0) miss complete. TLB insert. Quantum 2→1.
- **Tick 20:** P2 R(14). VPN 14 not in page table. Not in mmap yet (first access). **Page fault** (P2 pf=2). Allocate F12. Enqueue page-in sector (2·13+14·3+1) mod 64 = 5. Establishes shared mmap: mmap[14]=F12.

### Ticks 21–23

- **Tick 21:** CS #10 P2→P3. Dispatch P3 Q0 quantum=3 (pf=0, cow=1). cow_work_remaining=2.
- **Tick 22:** P3 COW kernel tick 1 (no quantum decrement). cow_work_remaining=1.
- **Tick 23:** P3 COW kernel tick 2 complete. Write applied to F11: R=1, D=1. PC advances past W(0).

### Tick 24

- Step 1: **Ref-bit decay.** All R cleared.
- Step 7: P3 R(1). Page1 not in page table. **Page fault** (P3 pf=1). Need frame: all data frames occupied or reserved. Non-reserved frames: F6(P1:page0, C0, load=7), F7(P2:page0, C0, load=15), F11(P3:page0, C1, load=16). Soft-floor: each active owner has 1 frame. No owner >1. Fallback to exactly 1. Lowest class C0: F6(load=7) vs F7(load=15). Oldest → **evict F6** (P1:page0, clean). Enqueue page-in for P3:page1, sector 43.

### Tick 25

- Step 2: **Global boost** (all READY→Q0).
- Step 5: P3 blocked. No READY. CS #11 P3→IDLE.

### Ticks 26–29: IDLE

Disk servicing P4:page0 (sector 53, seek=13).

### Tick 29

- Step 8: Disk transfer completes. P4:page0 → F8 (load_tick=29). Unblock at tick 30.

### Tick 30

- Step 4: P4 unblocks → READY Q0.
- CS #12 IDLE→P4. Dispatch P4 Q0 quantum = 2+((1+0) mod 3) = 3.
- Step 8: Disk selects P5:page0 (sector 2, starved since eq=6, wait=24). Seek ceil((53−2)/2)=ceil(51/2)=26. Begin.

---

## 3. Mid-Simulation Highlights (Ticks 30–523)

From this point the simulation enters a pattern of: process dispatches, attempts one or two memory operations (each requiring TLB miss = 2 ticks), page-faults, blocks, and waits for long disk seeks. The 7 data frames are under constant pressure from 5 active processes.

**Shared mmap resolution (tick 33):** P4 attempts R(14). The shared frame F12 (VPN 14) is still reserved (P2's page-in in progress). P4 page-faults (P4 pf=2) and waits for the same page-in — no separate disk request. This is the mmap piggyback: 42 total page faults but only 41 disk READs.

**Tick 73:** Both P2 and P4 unblock when F12's page-in completes at tick 72. P2 gets dispatched first (lower PID).

**Tick 80:** P4 W(14) — TLB hit (the only TLB hit in the entire simulation). The TLB entry for VPN 14 was inserted at tick 79 after the resident-miss completed. The write sets F12.D=1 and schedules invalidation of P2's TLB entries for VPN 14/15 at next CS.

**VPN 15 handling:** P2 first accesses VPN 15 at tick 76 (W(15) page fault). F9 allocated, page-in sector 8. P4 later accesses VPN 15 at tick 219, finding it mmap-resident in F9 and attaching without a new page-in.

**Thrashing examples:** P2 suffers two double-faults. R(1) first faults at tick 132, page-in completes at tick 252, but F12 is evicted at tick 253 by P1's R(1) page fault — P2 must re-fault at tick 255. R(2) first faults at tick 396, loaded at tick 434, but evicted at tick 436 — re-fault at tick 438. P4's W(15) experiences similar thrashing: faults at tick 411, loaded at tick 436, evicted at tick 438, then re-faults at tick 440.

**No process ever reaches Q2.** Every quantum expiry demotes Q0→Q1, but the process always page-faults before exhausting Q1's quantum of 5, which sends it back to Q0 on unblock. Aging never triggers.

### Complete quantum expiry list

| Tick | Process | Transition |
|------|---------|------------|
| 125 | P1 | Q0→Q1 |
| 132 | P2 | Q0→Q1 |
| 153 | P3 | Q0→Q1 |
| 173 | P5 | Q0→Q1 |
| 219 | P4 | Q0→Q1 |
| 306 | P1 | Q0→Q1 |
| 416 | P3 | Q0→Q1 |
| 476 | P5 | Q0→Q1 |
| 479 | P1 | Q0→Q1 |
| 616 | P2 | Q0→Q1 |

Of these 10 expiries, 9 result in re-dispatch of the same process (it is the only READY candidate). The exception is tick 476 where P5 expires and P1 (newly unblocked) is selected instead (CS #64 P5→P1).

---

## 4. Complete Disk I/O Schedule

52 disk operations total: 41 READs (page-ins) + 11 WRITEs (dirty writebacks). One additional page fault (P4's mmap piggyback at tick 33) produces no disk request.

| # | Begin | Done | Kind | Sector | Process:VPN | Frame | Seek | Arm After |
|---|-------|------|------|--------|-------------|-------|------|-----------|
| 1 | 0 | 7 | READ | 14 | P1:0 | F6 | 7 | 14 |
| 2 | 8 | 15 | READ | 27 | P2:0 | F7 | 7 | 27 |
| 3 | 16 | 29 | READ | 53 | P4:0 | F8 | 13 | 53 |
| 4 | 30 | 56 | READ | 2 | P5:0 | F9 | 26 | 2 |
| 5 | 57 | 65 | READ | 17 | P1:1 | F10 | 8 | 17 |
| 6 | 66 | 72 | READ | 5 | P2:14 | F12 | 6 | 5 |
| 7 | 73 | 92 | READ | 43 | P3:1 | F6 | 19 | 43 |
| 8 | 93 | 112 | READ | 5 | P5:1 | F7 | 19 | 5 |
| 9 | 113 | 121 | READ | 20 | P1:2 | F8 | 8 | 20 |
| 10 | 122 | 128 | READ | 8 | P2:15 | F9 | 6 | 8 |
| 11 | 129 | 145 | WRITE | 40 | P3:0 | F11 | 16 | 40 |
| 12 | 146 | 149 | READ | 46 | P3:2 | F6 | 3 | 46 |
| 13 | 150 | 169 | READ | 8 | P5:2 | F7 | 19 | 8 |
| 14 | 170 | 178 | READ | 23 | P1:3 | F8 | 8 | 23 |
| 15 | 179 | 188 | WRITE | 5 | P2:14 | F12 | 9 | 5 |
| 16 | 189 | 215 | READ | 56 | P4:1 | F11 | 26 | 56 |
| 17 | 216 | 236 | WRITE | 17 | P1:1 | F10 | 20 | 17 |
| 18 | 237 | 240 | READ | 11 | P5:3 | F7 | 3 | 11 |
| 19 | 241 | 249 | READ | 26 | P1:4 | F8 | 8 | 26 |
| 20 | 250 | 252 | READ | 30 | P2:1 | F12 | 2 | 30 |
| 21 | 253 | 268 | READ | 59 | P4:2 | F11 | 15 | 59 |
| 22 | 269 | 274 | READ | 49 | P3:3 | F10 | 5 | 49 |
| 23 | 275 | 296 | WRITE | 8 | P2:15 | F9 | 21 | 8 |
| 24 | 297 | 302 | READ | 17 | P1:1 | F12 | 5 | 17 |
| 25 | 303 | 318 | WRITE | 46 | P3:2 | F6 | 15 | 46 |
| 26 | 319 | 337 | WRITE | 11 | P5:3 | F7 | 18 | 11 |
| 27 | 338 | 359 | READ | 52 | P3:4 | F10 | 21 | 52 |
| 28 | 360 | 379 | READ | 14 | P5:4 | F9 | 19 | 14 |
| 29 | 380 | 388 | READ | 29 | P1:5 | F12 | 8 | 29 |
| 30 | 389 | 390 | READ | 30 | P2:1 | F6 | 1 | 30 |
| 31 | 391 | 407 | READ | 62 | P4:3 | F7 | 16 | 62 |
| 32 | 408 | 412 | READ | 55 | P3:5 | F10 | 4 | 55 |
| 33 | 413 | 428 | WRITE | 26 | P1:4 | F8 | 15 | 26 |
| 34 | 429 | 432 | READ | 32 | P1:6 | F12 | 3 | 32 |
| 35 | 433 | 434 | READ | 33 | P2:2 | F6 | 1 | 33 |
| 36 | 435 | 436 | READ | 34 | P4:15 | F7 | 1 | 34 |
| 37 | 437 | 450 | WRITE | 59 | P4:2 | F11 | 13 | 59 |
| 38 | 451 | 472 | READ | 17 | P5:5 | F8 | 21 | 17 |
| 39 | 473 | 475 | READ | 14 | P1:0 | F6 | 2 | 14 |
| 40 | 476 | 486 | READ | 33 | P2:2 | F7 | 10 | 33 |
| 41 | 487 | 497 | WRITE | 14 | P5:4 | F9 | 10 | 14 |
| 42 | 498 | 520 | READ | 58 | P3:6 | F11 | 22 | 58 |
| 43 | 521 | 533 | READ | 35 | P1:7 | F6 | 12 | 35 |
| 44 | 534 | 551 | READ | 2 | P5:0 | F8 | 17 | 2 |
| 45 | 552 | 554 | READ | 5 | P2:14 | F7 | 2 | 5 |
| 46 | 555 | 570 | READ | 34 | P4:15 | F9 | 15 | 34 |
| 47 | 571 | 573 | READ | 38 | P1:8 | F11 | 2 | 38 |
| 48 | 574 | 583 | WRITE | 55 | P3:5 | F10 | 9 | 55 |
| 49 | 584 | 596 | WRITE | 32 | P1:6 | F12 | 12 | 32 |
| 50 | 597 | 603 | READ | 20 | P5:6 | F10 | 6 | 20 |
| 51 | 604 | 612 | READ | 36 | P2:3 | F12 | 8 | 36 |
| 52 | 616 | 618 | READ | 39 | P2:4 | F8 | 2 | 39 |

Total seek ticks: 564. Final arm position: sector 39.

---

## 5. Complete Eviction Log

35 evictions over the simulation. Each eviction uses Enhanced Second Chance (lowest class, oldest load_tick, lowest frame number) with minimum-residency soft floor. After P3 terminates at tick 523, P3's frames become freely evictable (soft floor does not protect terminated processes).

| Tick | Frame | Owner:VPN | R | D | Class | Reason |
|------|-------|-----------|---|---|-------|--------|
| 24 | F6 | P1:0 | 0 | 0 | C0 | P3:page1 needs frame |
| 60 | F7 | P2:0 | 0 | 0 | C0 | P5:page1 needs frame |
| 69 | F8 | P4:0 | 0 | 0 | C0 | P1:page2 needs frame |
| 76 | F9 | P5:0 | 0 | 0 | C0 | P2:page15 needs frame |
| 81 | F11 | P3:0 | 0 | 1 | C1 | P4:page1; dirty writeback |
| 96 | F6 | P3:1 | 1 | 0 | C2 | P3:page2 needs frame |
| 116 | F7 | P5:1 | 1 | 0 | C2 | P5:page2 needs frame |
| 125 | F8 | P1:2 | 1 | 0 | C2 | P1:page3 needs frame |
| 132 | F12 | P2:14 | 1 | 1 | C3 | P2:page1; dirty writeback |
| 153 | F10 | P1:1 | 1 | 1 | C3 | P3:page3; dirty writeback |
| 173 | F7 | P5:2 | 1 | 0 | C2 | P5:page3 needs frame |
| 182 | F8 | P1:3 | 1 | 0 | C2 | P1:page4 needs frame |
| 221 | F11 | P4:1 | 1 | 0 | C2 | P4:page2 needs frame |
| 244 | F9 | P2:15 | 1 | 1 | C3 | P5:page4; dirty writeback |
| 253 | F12 | P2:1 | 1 | 0 | C2 | P1:page1 reload |
| 255 | F6 | P3:2 | 1 | 1 | C3 | P2:page1; dirty writeback |
| 272 | F7 | P5:3 | 1 | 1 | C3 | P4:page3; dirty writeback |
| 278 | F10 | P3:3 | 1 | 0 | C2 | P3:page4 needs frame |
| 306 | F12 | P1:1 | 1 | 0 | C2 | P1:page5 needs frame |
| 363 | F10 | P3:4 | 1 | 0 | C2 | P3:page5 needs frame |
| 383 | F8 | P1:4 | 1 | 1 | C3 | P5:page5; dirty writeback |
| 392 | F12 | P1:5 | 1 | 0 | C2 | P1:page6 needs frame |
| 396 | F6 | P2:1 | 1 | 0 | C2 | P2:page2 needs frame |
| 411 | F7 | P4:3 | 1 | 0 | C2 | P4:page15 needs frame |
| 416 | F11 | P4:2 | 1 | 1 | C3 | P3:page6; dirty writeback |
| 436 | F6 | P2:2 | 1 | 0 | C2 | P1:page0 reload |
| 438 | F7 | P4:15 | 1 | 0 | C2 | P2:page2 reload |
| 440 | F9 | P5:4 | 1 | 1 | C3 | P4:page15 reload; dirty writeback |
| 479 | F6 | P1:0 | 1 | 0 | C2 | P1:page7 needs frame |
| 481 | F8 | P5:5 | 1 | 0 | C2 | P5:page0 reload |
| 490 | F7 | P2:2 | 1 | 0 | C2 | P2:page14 reload |
| 537 | F11 | P3:6 | 1 | 0 | C2 | P1:page8 needs frame; P3 terminated |
| 555 | F10 | P3:5 | 1 | 1 | C3 | P5:page6 needs frame; P3 terminated; dirty writeback |
| 559 | F12 | P1:6 | 1 | 1 | C3 | P2:page3; dirty writeback |
| 616 | F8 | P5:0 | 1 | 0 | C2 | P2:page4 needs frame |

---

## 6. Per-Process Execution Summary

### P1 (11 operations, 11 page faults, 0 COW)

| Op# | Instruction | Tick Start | Event | Frame |
|-----|------------|------------|-------|-------|
| 0 | R(0) | 0 | Page fault → F6 (loaded tick 7) | F6 |
| 0 | R(0) | 9 | TLB miss walk; interrupted by fork at tick 10 | F6 |
| 0 | R(0) | 12–13 | TLB miss walk+access (restart after fork) | F6 |
| 1 | W(1) | 14 | Page fault → F10 (loaded tick 65) | F10 |
| 2 | R(2) | 69 | Page fault → F8 (loaded tick 121); evicts P4:0 | F8 |
| 3 | R(3) | 125 | Page fault → F8 (loaded tick 178); evicts P1:2 | F8 |
| 4 | W(4) | 182 | Page fault → F8 (loaded tick 249); evicts P1:3 | F8 |
| 5 | R(1) | 253 | Page fault → F12 (loaded tick 302); reload | F12 |
| 6 | R(5) | 306 | Page fault → F12 (loaded tick 388); evicts P1:1 | F12 |
| 7 | W(6) | 392 | Page fault → F12 (loaded tick 432); evicts P1:5 | F12 |
| 8 | R(0) | 436 | Page fault → F6 (loaded tick 475); reload | F6 |
| 9 | W(7) | 479 | Page fault → F6 (loaded tick 533); evicts P1:0 | F6 |
| 10 | R(8) | 537 | Page fault → F11 (loaded tick 573); evicts P3:6 | F11 |

P1 completes at **tick 576** after R(8) TLB miss access completes.

### P2 (9 operations, 10 page faults, 0 COW)

| Op# | Instruction | Key Event |
|-----|------------|-----------|
| 0 | R(0) | PF tick 2 → F7 (loaded tick 15) |
| 1 | R(14) | PF tick 20 → F12 (loaded tick 72). Shared mmap established |
| 2 | W(15) | PF tick 76 → F9 (loaded tick 128). Shared VPN 15 |
| 3 | R(1) | PF tick 132. Dirty evict F12, WB wait. Loaded tick 252 into F12. Evicted at tick 253 by P1! PF again tick 255. Dirty evict F6, WB wait. Loaded tick 390 into F6 |
| 4 | R(2) | PF tick 396 → F6 (loaded tick 434). Evicted at tick 436! PF again tick 438 → F7 (loaded tick 486) |
| 5 | W(14) | PF tick 490 → F7 (loaded tick 554). Reload shared VPN 14 |
| 6 | R(3) | PF tick 559. Dirty evict F12, WB wait. Loaded tick 612 into F12 |
| 7 | W(4) | PF tick 616 → F8 (loaded tick 618); evicts P5:0 |
| 8 | R(14) | TLB miss ticks 622–623. Resident in F7. No page fault |

P2 has 10 page faults because R(1) and R(2) each fault twice (pages evicted between first load and access). P2 completes at **tick 623**.

### P3 (7 operations, 6 page faults, 1 COW)

| Op# | Instruction | Key Event |
|-----|------------|-----------|
| 0 | W(0) | COW fault tick 16. F11 allocated immediately. 2 kernel ticks at ticks 22–23 |
| 1 | R(1) | PF tick 24 → F6 (loaded tick 92) |
| 2 | W(2) | PF tick 96 → F6 (loaded tick 149) |
| 3 | R(3) | PF tick 153 → F10 (loaded tick 274) |
| 4 | R(4) | PF tick 278 → F10 (loaded tick 359) |
| 5 | W(5) | PF tick 363 → F10 (loaded tick 412) |
| 6 | R(6) | PF tick 416 → F11 (loaded tick 520) |

P3 completes at **tick 523**. First to finish — shortest program and COW resolved quickly.

### P4 (8 operations, 7 page faults, 0 COW)

| Op# | Instruction | Key Event |
|-----|------------|-----------|
| 0 | R(0) | PF tick 4 → F8 (loaded tick 29) |
| 1 | R(14) | PF tick 33 (mmap piggyback on P2's F12 page-in, loaded tick 72) |
| 2 | W(14) | TLB hit tick 80. The only TLB hit in the simulation |
| 3 | R(1) | PF tick 81 → F11 (loaded tick 215); dirty evict P3:0 |
| 4 | R(15) | TLB miss ticks 219–220. Attaches to shared F9 (loaded by P2) |
| 5 | W(2) | PF tick 221 → F11 (loaded tick 268) |
| 6 | R(3) | PF tick 272 → F7 (loaded tick 407) |
| 7 | W(15) | PF tick 411 → F7 (loaded tick 436). Evicted at tick 438! PF again tick 440. Dirty evict F9, WB wait. Loaded tick 570 into F9 |

P4 completes at **tick 573**.

### P5 (8 operations, 8 page faults, 0 COW)

| Op# | Instruction | Key Event |
|-----|------------|-----------|
| 0 | R(0) | PF tick 6 → F9 (loaded tick 56) |
| 1 | R(1) | PF tick 60 → F7 (loaded tick 112) |
| 2 | R(2) | PF tick 116 → F7 (loaded tick 169) |
| 3 | W(3) | PF tick 173 → F7 (loaded tick 240) |
| 4 | W(4) | PF tick 244 → F9. Dirty evict P2:15, WB wait. Loaded tick 379 |
| 5 | R(5) | PF tick 383 → F8. Dirty evict P1:4, WB wait. Loaded tick 472 |
| 6 | R(0) | PF tick 481 → F8 (loaded tick 551). Reload page 0 |
| 7 | W(6) | PF tick 555 → F10. Dirty evict P3:5, WB wait. Loaded tick 603 |

P5 completes at **tick 606**.

---

## 7. Context-Switch Breakdown

83 total context switches. The tick-0 first dispatch is free and not counted.

Breakdown by type:

- Process → Process: 15 transitions (e.g., CS #1 P1→P2, CS #9 P3→P2, CS #64 P5→P1, ...)
- Process → IDLE: 34 transitions (every page fault that leaves no READY process)
- IDLE → Process: 34 transitions (each unblock from an idle state)

10 quantum expiries trigger scheduling re-evaluation. In 9 of these, the expired process is the only candidate and is re-dispatched without a context switch. At tick 476, P5 expires but P1 (newly unblocked) has a lower PID and is selected, producing CS #64.

---

## 8. Dirty Writeback Summary

11 dirty writebacks occurred when evicting frames with D=1:

| Tick Evicted | Frame | Content | Sector | WB Begin | WB Done |
|-------------|-------|---------|--------|----------|---------|
| 81 | F11 | P3:0 | 40 | 129 | 145 |
| 132 | F12 | P2:14 | 5 | 179 | 188 |
| 153 | F10 | P1:1 | 17 | 216 | 236 |
| 244 | F9 | P2:15 | 8 | 275 | 296 |
| 255 | F6 | P3:2 | 46 | 303 | 318 |
| 272 | F7 | P5:3 | 11 | 319 | 337 |
| 383 | F8 | P1:4 | 26 | 413 | 428 |
| 416 | F11 | P4:2 | 59 | 437 | 450 |
| 440 | F9 | P5:4 | 14 | 487 | 497 |
| 555 | F10 | P3:5 | 55 | 574 | 583 |
| 559 | F12 | P1:6 | 32 | 584 | 596 |

---

## 9. Final State at Tick 623

### Frame Map

| Frame | Contents |
|-------|----------|
| F1 | PT:P1 |
| F2 | PT:P2 |
| F3 | PT:P4 |
| F4 | PT:P3 |
| F5 | PT:P5 |
| F6 | P1:7 |
| F7 | P2:14 |
| F8 | P2:4 |
| F9 | P4:15 |
| F10 | P5:6 |
| F11 | P1:8 |
| F12 | P2:3 |

### Completion Order

| Rank | Process | Tick |
|------|---------|------|
| 1st | P3 | 523 |
| 2nd | P4 | 573 |
| 3rd | P1 | 576 |
| 4th | P5 | 606 |
| 5th | P2 | 623 |

---

## 10. Required Output

```text
page_faults=11,10,6,7,8
tlb_hits=1
context_switches=83
cow_faults=1
dirty_writebacks=11
total_seek_ticks=564
disk_arm_final=39
completion_order=3,4,1,5,2
p3_first_cow_tick=16
last_tick=623
final_frame_map=F1:PT:P1,F2:PT:P2,F3:PT:P4,F4:PT:P3,F5:PT:P5,F6:P1:7,F7:P2:14,F8:P2:4,F9:P4:15,F10:P5:6,F11:P1:8,F12:P2:3
```


Prompt:
You are given a permutation `p` of `{1, 2, ..., n}`. It is **not** sorted in ascending order and **not** sorted in descending order.

Initially, all elements are colored **red**. Players alternate turns. On each turn, a player may do one of:

1. **Rearrange** the permutation so that all **red** elements keep their positions (blue elements may be permuted arbitrarily).
2. Change the color of **one** red element to **blue**.
3. Skip the turn.

The **first** player wins if the permutation becomes `[1, 2, ..., n]`.
The **second** player wins if the permutation becomes `[n, n-1, ..., 1]`.
If the game lasts `100500` turns without a winner, it is a draw.

Determine the result under optimal play:
- output **1** if the first player wins,
- output **2** if the second player wins,
- output **0** if it is a draw.

### Input
- `n` (`3 ≤ n ≤ 5·10^5`)
- `p_1, ..., p_n`: a permutation of `1..n`, not fully ascending/descending

### Output
One integer: `1`, `2`, or `0`.

Does the answer state that using the “rearrange while keeping red positions fixed” operation is only useful if it produces an immediate win on that same move (because the opponent can undo any non-winning rearrangement next move)?

Does the answer define the set of indices needed for the first player (ascending) as S_asc = { i | p[i] ≠ i}?

Does the answer define the set of indices needed for the second player (descending) as S_desc = { i | p[i] ≠ n−i+1}?

Does the answer define a as the count of indices only the first player needs to make blue, i.e., a = |S_asc \\ S_desc|?

Does the answer define b as the count of indices only the second player needs to make blue, i.e., b = |S_desc \\ S_asc|?

Does the answer define c as the count of indices both players need to make blue, i.e., c = |S_asc ∩ S_desc|?

Does the answer state the first player winning condition as a + c ≤ b?

Does the answer state the second player winning condition as b + c < a?

Does the answer state the final decision rule: output 1 if (a+c ≤ b), else output 2 if (b+c < a), otherwise output 0?

### Expert Solution

A “rearrange (keeping red positions fixed)” move is only useful if it produces an **immediate win** on that move; otherwise the opponent can undo any non-winning rearrangement on their next move because the set of fixed (red) positions hasn’t changed.

So the game reduces to a race to recolor enough positions to **blue** so that a winning permutation can be formed by permuting only blue positions.

---

## Required-blue index sets

- To reach ascending `[1, 2, ..., n]`, any index `i` with `p_i ≠ i` must become blue (a red mismatch can never be fixed):
  `S_asc = { i | p_i ≠ i }`

- To reach descending `[n, n-1, ..., 1]`, any index `i` with `p_i ≠ (n - i + 1)` must become blue:
  `S_desc = { i | p_i ≠ (n - i + 1) }`

Partition into counts:
- `a = |S_asc \ S_desc|`
- `b = |S_desc \ S_asc|`
- `c = |S_asc ∩ S_desc|`

---

## Optimal-play outcome

- First player wins iff:
  `a + c ≤ b`

- Second player wins iff:
  `b + c < a`

- Otherwise neither can force a win, so the result is a draw.

**Output rule:**
- if `a + c ≤ b` output **1**
- else if `b + c < a` output **2**
- else output **0**

All counts are computed in one pass over `i = 1..n` in `O(n)`.


Prompt:
Let `a` and `b` be two arrays of lengths `n` and `m`, with no elements in common. Define `merge(a, b)` recursively:

- `merge(∅, b) = b` and `merge(a, ∅) = a`.
- If both are non-empty and `a_1 < b_1`, then
  `merge(a, b) = [a_1] + merge([a_2, ..., a_n], b)`.
- If both are non-empty and `a_1 > b_1`, then
  `merge(a, b) = [b_1] + merge(a, [b_2, ..., b_m])`.

You are given a permutation `p` of length `2n` (it contains each integer `1` to `2n` exactly once). Determine if there exist two arrays `a` and `b`, each of length `n` and with no common elements, such that:
`p = merge(a, b)`

### Input
- `n` (`1 ≤ n ≤ 2000`)
- `p_1, ..., p_{2n}`: a permutation of `1, ..., 2n`

### Output
Print **YES** if such arrays `a, b` exist, otherwise print **NO**.

Does the answer state that merge outputs the smaller of the two current first elements (a1 vs b1) when both arrays are non-empty?

Does the answer define the block starts as indices i where p[i] is a new prefix maximum (p[i] > all previous p[1..i-1])?

Does the answer state that these prefix-maximum indices partition p into contiguous blocks (segments)?

Does the answer describe how to compute each block length from consecutive block-start indices (and the last block to the end)?

Does the answer explicitly state that the block lengths add up to 2n?

Does the answer state the check to perform as: whether some subset of block lengths sums to n?

Does the answer define dp[s] as a boolean meaning “some subset of processed blocks has total length s”?

Does the answer include the DP initialization dp[0] = true?

Does the answer give the DP update for each block length L as setting dp[s+L] = true when dp[s] is true (i.e., subset-sum transition)?

Does the answer state the final decision rule: print YES iff dp[n] is true, else NO, and state time complexity O(n^2)?

### Expert Solution

In `merge(a, b)`, whenever both arrays are non-empty, the next output element is `min(a_1, b_1)`, and that element is removed from its array.

---

## 1) Key lemma (using the maximum)

Let `M` be the **maximum** value among the elements that have not yet been output.
Assume `M` is currently the first element of one array (say it belongs to `a`).

Since all values are distinct, `M` is larger than every remaining element of `b`.
So while `b` is non-empty, `min(a_1, b_1)` cannot be `M`; merge must keep outputting elements from `b` until `b` becomes empty, and **only then** can it output `M`.

Therefore, in the final permutation `p`, the position where the current maximum `M` appears begins a **contiguous suffix** that must come entirely from the array containing `M`.

---

## 2) Blocks from prefix maxima

Repeatedly removing such suffixes partitions `p` into contiguous blocks.

Equivalently, the **block starts** are exactly the indices `i` where `p_i` is a new prefix maximum:
`p_i > max(p_1, ..., p_{i-1})`

Let these indices be `i_1 < i_2 < ... < i_k` (with `i_1 = 1`).
Then the block lengths are:
- `L_1 = i_2 - i_1`
- `L_2 = i_3 - i_2`
- ...
- `L_{k-1} = i_k - i_{k-1}`
- `L_k = 2n - i_k + 1`

They sum to `2n`, and each block must be assigned wholly to either `a` or `b`.

---

## 3) Reduction to subset sum

Since `|a| = |b| = n`, we need to choose a subset of block lengths whose sum is exactly `n`
(the remaining blocks then sum to `n` automatically).

- This is **necessary** because blocks cannot be split between arrays.
- This is **sufficient** because assigning whole blocks to `a` and `b` makes merge reproduce `p` block-by-block by the lemma above.

---

## 4) DP

Subset-sum DP:

- `dp[s]` = whether we can reach total length `s`
- Initialize `dp[0] = true`
- For each block length `L`, for `s` from `n - L` down to `0`:
  `dp[s + L] |= dp[s]`

Output **YES** iff `dp[n]` is true, else **NO**.

---

## 5) Complexity

At most `2n` blocks and `n ≤ 2000`, so the DP is:
`O(n^2)`
time and `O(n)` memory.


Prompt:
Given integers n, m, and v, consider all n x m matrices whose entries are integers in [1, v].

A matrix a is called **friendly** if:

  min_{1 <= i <= n} ( max_{1 <= j <= m} a_{i,j} )  <=  max_{1 <= j <= m} ( min_{1 <= i <= n} a_{i,j} )

In words:
- For each row, compute its maximum element. Take the minimum of these row-maximums.
- For each column, compute its minimum element. Take the maximum of these column-minimums.
- The matrix is friendly if the first quantity is less than or equal to the second.

**Task:** For each test case, compute the number of friendly n x m matrices with entries in [1, v], modulo 998244353.

Additionally, state the closed-form formula you use (in terms of n, m, v, and a summation variable k), and explain the key mathematical insight that enables an efficient O(n * v * log m) algorithm. Specifically:
1. Why the friendly condition reduces to equality of the two quantities.
2. What geometric structure the set of "extremal" positions forms.
3. How the binomial theorem eliminates the inner summation over column subsets.

**Input:**
- First line: integer t (number of test cases).
- Each test case: one line with three integers n, m, v.

**Constraints:**
- 1 <= t <= 5
- 1 <= n <= 250
- 1 <= m <= 10^9
- 1 <= v <= 4000
- n * v <= 10^6

**Output:**
For each test case, output one integer: the count of friendly matrices modulo 998244353.

**Test Input:**
5
2 2 2
3 3 3
2 1000000000 2
3 4 5
4 100 3

The response outputs 14 as the answer for the test case n=2, m=2, v=2.

The response outputs 11463 as the answer for the test case n=3, m=3, v=3.

The response outputs 123732870 as the answer for the test case n=2, m=1000000000, v=2.

The response outputs 92839345 as the answer for the test case n=3, m=4, v=5.

The response outputs 410518279 as the answer for the test case n=4, m=100, v=3.

The response correctly explains that the friendly condition (min of row-maxes <= max of col-mins) holds if and only if the two quantities are equal, using the minimax inequality.

The response identifies that the set of positions simultaneously achieving the row-maximum and column-minimum value k forms a complete subrectangle (i.e., a Cartesian product of a subset of rows and a subset of columns).

The response applies the binomial theorem to collapse the summation over column subsets into a closed-form expression, eliminating the explicit loop over j.

The response states that the overall time complexity is O(n * v * log m) per test case.

We count n x m matrices with entries in [1, v] satisfying:
  min_i(max_j a[i][j]) <= max_j(min_i a[i][j])

1. Reduction to Equality

Let L = min_i(max_j a[i][j]) achieved at row r1, and
    R = max_j(min_i a[i][j]) achieved at column c1.

Consider cell (r1, c1):
- a[r1][c1] <= max_j a[r1][j] = L  (since L is the max of row r1)
- a[r1][c1] >= min_i a[i][c1] >= R  (wait, min_i a[i][c1] is the column min, and R is the max of column mins)

Actually more carefully:
- L = max_j a[r1][j], so a[r1][c1] <= L.
- min_i a[i][c1] <= a[r1][c1], and R = max_j(min_i a[i][j]) >= min_i a[i][c1].

This gives us: L >= a[r1][c1] >= min_i a[i][c1], and
independently R >= min_i a[i][c1].

The key fact is the minimax inequality: for any matrix,
  min_i(max_j a[i][j]) >= max_j(min_i a[i][j])

Proof: For any row i and column j, max_j' a[i][j'] >= a[i][j] >= min_i' a[i'][j].
Taking min over i on the left and max over j on the right gives L >= R.

Therefore the friendly condition L <= R combined with L >= R gives L = R.
The matrix is friendly if and only if the two quantities are equal.

2. Structure: Extremal Positions Form a Subrectangle

Fix a value k in [1, v]. Consider positions (i, j) where:
- a[i][j] = k
- k is the maximum of row i (so all entries in row i are <= k)
- k is the minimum of column j (so all entries in column j are >= k)

Claim: these positions form a complete (combinatorial) subrectangle.

Proof: Take (r1, c1) and (r2, c2) both in this set with r1 != r2, c1 != c2.
Consider (r1, c2):
- Row r1 has max = k, so a[r1][c2] <= k.
- Column c2 has min = k (from (r2, c2)), so a[r1][c2] >= k.
Thus a[r1][c2] = k, and (r1, c2) is also in the set.

By induction, if the set involves rows R and columns C, then every
(r, c) with r in R, c in C is in the set. So the set is an
|R| x |C| complete subrectangle.

3. Counting via Inclusion-Exclusion

Fix k, and choose i rows and j columns to form the subrectangle.

Cell constraints:
- Inside the subrectangle (i*j cells): fixed to k.
- Chosen rows, outside chosen columns (i*(m-j) cells): must be <= k, so k choices each.
- Chosen columns, outside chosen rows ((n-i)*j cells): must be >= k, so (v-k+1) choices each.
- Remaining cells ((n-i)*(m-j) cells): unrestricted, v choices each.

To ensure at least 1 row and at least 1 column, apply inclusion-exclusion:

  sum_{k=1}^{v} sum_{i=1}^{n} sum_{j=1}^{m}
    (-1)^{i+j} * C(n,i) * C(m,j)
    * k^{i*(m-j)}
    * (v-k+1)^{(n-i)*j}
    * v^{(n-i)*(m-j)}

4. Binomial Theorem Eliminates the j-Loop

For fixed i, rewrite the sum over j (extending to j=0 then subtracting):

  sum_{j=0}^{m} C(m,j) * [-(v-k+1)^{n-i}]^j * [k^i * v^{n-i}]^{m-j}

By the binomial theorem this equals:

  (k^i * v^{n-i} - (v-k+1)^{n-i})^m

Subtracting the j=0 term [k^i * v^{n-i}]^m, the final formula is:

  answer = sum_{k=1}^{v} sum_{i=1}^{n}
    (-1)^i * C(n,i) * [ (k^i * v^{n-i} - (v-k+1)^{n-i})^m  -  (k^i * v^{n-i})^m ]

All exponentiations use fast modular exponentiation (mod 998244353).

5. Complexity

- k ranges over v values, i ranges over n values.
- Each term requires O(log m) for modular exponentiation.
- Total: O(n * v * log m).
- Since n*v <= 10^6 and log m <= 30, this is efficient.

6. Verification Code (Python)

```python
import sys
from math import comb

MOD = 998244353

def solve(n, m, v):
    ans = 0
    for k in range(1, v + 1):
        for i in range(1, n + 1):
            sign = (-1) ** i
            c = comb(n, i) % MOD
            A = pow(k, i, MOD) * pow(v, n - i, MOD) % MOD   # k^i * v^{n-i}
            B = pow(v - k + 1, n - i, MOD)                   # (v-k+1)^{n-i}
            base = (A - B) % MOD
            term = sign * c % MOD * ((pow(base, m, MOD) - pow(A, m, MOD)) % MOD) % MOD
            ans = (ans + term) % MOD
    return ans % MOD

t = int(input())
for _ in range(t):
    n, m, v = map(int, input().split())
    print(solve(n, m, v))
```

7. Test Case Verification

Input:
5
2 2 2
3 3 3
2 1000000000 2
3 4 5
4 100 3

Output:
14          (brute-force verified: 14 out of 16 matrices are friendly)
11463       (brute-force verified: 11463 out of 19683 matrices)
123732870   (requires modular exponentiation)
92839345
410518279

The first two outputs were independently verified by brute-force enumeration.


Prompt:
## Reach a Regular Bracket Sequence by Flipping Equal Adjacent Pairs

You are given a bracket string `s` of length `n` consisting only of characters `(` and `)`.

You may apply the following operations any number of times (possibly zero), in any order:

1. Choose an index `i` (`1 ≤ i < n`) such that `s[i] = s[i+1] = '('`, and replace both characters with `')'`.
2. Choose an index `i` (`1 ≤ i < n`) such that `s[i] = s[i+1] = ')'`, and replace both characters with `'('`.

For each test case, output any **regular bracket sequence** `t` that can be obtained from `s` by these operations, or output `-1` if it is impossible.

A bracket sequence is **regular** iff it is a correct bracket sequence (balanced and every prefix has at least as many `(` as `)`).

### Input
Multiple test cases:
- First line: `t` — number of test cases (`1 ≤ t ≤ 10^4`)
For each test case:
- Line 1: integer `n` (`1 ≤ n ≤ 2·10^5`)
- Line 2: string `s` of length `n`

It is guaranteed that the sum of `n` over all test cases does not exceed `2·10^5`.

### Output
For each test case, print a regular bracket sequence `t` obtainable from `s`, or `-1`.

The answer defines a transformation that flips the bracket character on every even position of the string.

The answer states that in the transformed string the operation can rewrite the substring \"()\" into \")(\".

The answer states that in the transformed string the operation can rewrite the substring \")(\" into \"()\".

The answer states that if n is odd then the output is −1.

The answer states that a necessary feasibility condition is that the number of '(' characters in the transformed string is even.

The answer states that the transformed operations allow reaching any permutation of the transformed string’s characters.

The answer states that if the transformed string contains zero '(' characters then it is impossible to obtain a regular bracket sequence.

The answer defines x as open/2 where open is the number of '(' characters in the transformed string.

The answer gives an explicit construction of t as \"()\" repeated (x−1) times, followed by \"(\", followed by \"()\" repeated (n/2−x) times, followed by \")\".

The answer states a per-test-case time complexity of O(n)

## Expert Solution

### 1) Key transformation: flip even positions

Define a transformed string `s'` (1-indexed) by flipping the bracket on every even position:
- if `s[i] = '('` then `s'[i] = ')'` for even `i`
- if `s[i] = ')'` then `s'[i] = '('` for even `i`
- odd positions stay unchanged

This transformation is invertible (apply it again to return to the original domain).

---

### 2) Operations become adjacent swaps in the transformed string

Consider an operation on equal adjacent brackets in `s`:

- If `s[i]=s[i+1]='('`, after flipping even positions the pair becomes either `"()"` or `")("` (depending on parity of `i`).
  After the operation the pair becomes `"))"`, which flips to the opposite mixed pair.
- Similarly, `"))"` operations also flip between `"()"` and `")("`.

Therefore, in `s'` the operation is exactly:
\[
"()" \leftrightarrow ")("
\]
which is an adjacent swap of two different characters.

Hence, using these operations we can perform arbitrary adjacent swaps in `s'`, so we can reach **any permutation of `s'`**.
Equivalently, the only invariant is the multiset of characters, i.e. the number of `'('` in `s'`.

Let:
- `open = count('(' in s')`.

Then `s` can reach a target `t` iff `s'` can be permuted into `t'` where `t'` is `t` with even positions flipped, which holds iff `t'` has exactly `open` opening brackets.

---

### 3) Characterize when a regular bracket sequence is possible

A regular bracket sequence must have even length, so if `n` is odd the answer is `-1`.

Also, for any regular bracket sequence `t`, after flipping even positions the number of `'('` in `t'` is always even (it equals `2 * (# of '(' on odd positions in t)`).
So we must have `open` even.

If `open = 0`, then `t'` has no `'('`, which is impossible for any regular bracket sequence `t` (the first character of `t` must be `'('` and position 1 is not flipped).
So we also need `open >= 2`.

Thus feasibility conditions are:
- `n` is even
- `open` is even
- `open > 0` (equivalently `open >= 2`)

If any fails, print `-1`.

---

### 4) Construct a regular bracket sequence with a prescribed `open`

Let `open = count('(' in s')` and define:
\[
x = open/2.
\]
We construct a regular bracket sequence `t` of length `n` such that after flipping even positions, `t'` has exactly `open` opening brackets.

Construction:
- Output `"()"` repeated `(x-1)` times
- Then output `"("`
- Then output `"()"` repeated `(n/2 - x)` times
- Then output `")"`

Formally:
\[
t = \underbrace{()() \dots ()}_{x-1\ \text{times}} \; + \; ( \; + \; \underbrace{()() \dots ()}_{(n/2-x)\ \text{times}} \; + \; )
\]

This is a concatenation of regular bracket sequences, so `t` is regular.

By construction, the number of `'('` on odd positions in `t` is exactly `x`, hence `t'` has `2x = open` opening brackets.

Since `s'` can be permuted into any string with the same number of `'('`, we can permute `s'` into `t'`, and applying the inverse flip yields `t` obtainable from `s`.

Output `t`.

---

### 5) Complexity

For each test case:
- building `s'` and counting `open`: `O(n)`
- constructing `t`: `O(n)`

Total over all test cases: `O(sum n)`.