随机采样生成凸函数的概率及网格采样构建1D图技术问询

Great question! Let's break this down step by step, since there's no universal closed-form formula for all values of M, N, and K, but we can unpack the problem structure and walk through how to compute the probability in different scenarios.

First, let's clarify a critical ambiguity in the problem statement to make sure we're on the same page:

• When you refer to "generating a convex function" from the sequentially connected points, we have to assume we're talking about a univariate convex function—meaning we first sort the K selected grid points by their x-coordinate (assuming matrix columns correspond to x, rows to y as function values) to get an ordered sequence $(x_1,y_1), (x_2,y_2), ..., (x_K,y_K)$ where $x_1 < x_2 < ... < x_K$. The piecewise linear function formed by connecting these sorted points is convex if its slopes are non-decreasing (for a standard convex, "upward-curving" function; if you meant concave, slopes are non-increasing, but we'll stick to standard convexity here).
• Note: If any two selected points share the same x-coordinate, they can't form a valid univariate function, so we'll exclude these cases from our convex function count (and account for their probability separately if needed).

1. Total Sample Space #

First, let's define our total possible outcomes:

• If we're selecting K distinct points (no replacement, matching the "set K elements to 1" description), the total number of unordered point sets is $\binom{M \times N}{K}$.
• If we're considering ordered sequences (the "connect points in selection order" description), each unordered set corresponds to $K!$ sequences, so the probability will be the same as the unordered case (since both numerator and denominator scale by $K!$).

We'll focus on unordered sets for simplicity, as convexity depends only on the point set itself, not the selection order.

2. Counting Convex Point Sets #

To find the number of convex point sets, we split the problem into two parts:

a. Choose distinct x-coordinates #

First, select K distinct x-values from the N available columns: there are $\binom{N}{K}$ ways to do this. Let's call a selected sequence of x-values $x_1 < x_2 < ... < x_K$, with gaps $d_i = x_{i+1} - x_i$ (each $d_i \geq 1$, and $\sum_{i=1}^{K-1} d_i \leq N-1$).

b. Count valid y-coordinate sequences #

For each fixed x-sequence, we need to count how many y-sequences $(y_1, y_2, ..., y_K)$ (each $y_i \in {1, 2, ..., M}$) satisfy the convexity condition:
For all $2 \leq i \leq K-1$, the slope between $(x_{i-1}, y_{i-1})$ and $(x_i, y_i)$ is less than or equal to the slope between $(x_i, y_i)$ and $(x_{i+1}, y_{i+1})$. Mathematically, this translates to:
$$\frac{y_i - y_{i-1}}{d_{i-1}} \leq \frac{y_{i+1} - y_i}{d_i}$$
Rearranged to avoid division (and handle integer values):
$$d_i(y_i - y_{i-1}) \leq d_{i-1}(y_{i+1} - y_i)$$

Dynamic Programming for Valid Y-Sequences

For general K, this count requires dynamic programming (DP), since each step's valid y-values depend on the previous two points' values and the gaps $d_i$. Here's a high-level DP setup:

• Define $dp[i][y][s]$ as the number of valid sequences of the first i points, where the i-th point has y-coordinate $y$, and the slope of the last segment (scaled by $d_{i-1}$ to avoid fractions) is $s = y - y_{i-1}$.
• Base cases:
  • For $i=1$: $dp[1][y][*] = 1$ for all $y \in {1, ..., M}$ (single points are trivially valid).
  • For $i=2$: $dp[2][y_2][y_2 - y_1] = 1$ for all pairs $(y_1, y_2)$ (two points form a line, which is both convex and concave).
• Recurrence: For $i \geq 3$, for each possible $y_{i-1}$ and slope $s_{i-1} = y_{i-1} - y_{i-2}$, we count all $y_i$ such that $d_{i-1} s_{i-1} \leq d_{i-2} (y_i - y_{i-1})$, then accumulate these counts into $dp[i][y_i][y_i - y_{i-1}]$.

Summing over all valid states in $dp[K][y][s]$ gives the number of valid y-sequences for the current x-sequence.

3. Special Cases (Simpler Probabilities) #

For small K, we can compute exact probabilities without full DP:

• K=1: Every single point trivially satisfies convexity (or is vacuously valid), so probability = 1.
• K=2: Any two distinct points form a linear function, which is convex, so probability = 1 (assuming K ≤ M×N, which ensures we can select two distinct points).
• K=3: For this case, we can compute the count manually for each x-sequence gap pair $(d_1, d_2)$. For example, if $d_1 = d_2 = 1$, the convexity condition simplifies to $y_3 \geq 2y_2 - y_1$. We can sum over all $(y_1, y_2)$ the number of valid $y_3$ values (from $\max(1, 2y_2 - y_1)$ to M), then divide by the total number of y-sequences ($M^3$) to get the proportion for that x-sequence.

4. Final Probability Calculation #

The total number of convex point sets is the sum of valid y-sequence counts over all possible K-length x-sequences. The probability is then:
$$\text{Probability} = \frac{\sum_{\text{all K-length x-sequences}} \text{Valid y-sequences count}}{\binom{M \times N}{K}}$$

For large M, N, or K, this requires computational implementation (using DP and combinatorial summation), but the structure above gives a clear path to compute the value.

内容的提问来源于stack exchange，提问作者dektorpan