TexSAW CTF 2026 Excellent Neurons
Find the flag by reverse engineering this neural network. Oh, and its in Excel.
Flag format:
texsaw{flag}(e.g.,texsaw{orthogonal})
前置知识
本题涉及神经网络逆向工程，且模型实现在 Excel 电子表格中。以下概念有助于理解：
| 概念 | 说明 |
|---|---|
| 前馈神经网络 | 输入层→隐藏层→输出层，数据逐层前向传递 |
| ReLU | \(f(x)=\max(0,x)\)，负输入→零输出（关键逻辑门控！） |
| Sigmoid | \(\sigma(x)=\frac{1}{1+e^{-x}}\)，输出映射到 \((0,1)\)，\(>0.5\) 表示正输入 |
| 权重与偏置 | 网络参数，本例中 W1 是极端稀疏的 +1/-1 矩阵 |
| Excel 逆向 | .xlsx 本质是 ZIP，内部 XML 含公式与数据 |
| 前向传播公式 | 归一化字符与权重相乘，再加偏置；公式见表后。 |
前向传播约束：
\[ \frac{\mathrm{ASCII}}{127} \times \mathrm{weight} + \mathrm{bias} = 0 \]
当 \(\mathrm{weight} \neq 0\) 时，还原字符值：
\[ \mathrm{ASCII} = \operatorname{round}\!\left(\frac{-\mathrm{bias} \times 127}{\mathrm{weight}}\right) \]
Solution
1. Understanding the Success Condition
Examining the spreadsheet, specifically Row 112, we find the conditions for a valid flag: - \(F > 0.5\), \(L < 0.5\), \(A > 0.5\), \(G < 0.5\)
Since the output layer uses a Sigmoid activation
function: - To get output \(> 0.5\),
the input to the sigmoid (z3) must be positive (\(> 0\)). - To get output \(< 0.5\), the input to the sigmoid
(z3) must be negative (\(<
0\)).
Analyzing the final weights (W3) and biases
(b3), we find that all weights in W3 are large
(300 or -300) and the biases are small (between 0.49 and 0.59). For
these specific constraints to be met, the output of Layer 2
(a2) must be an extremely small positive number, very close
to zero.
2. Reverse-Engineering Layer 2
Layer 2 uses the ReLU activation function: \(a_2 = \operatorname{ReLU}(z_2)\), where \(z_2 = \operatorname{sum}(a_1 \cdot W_2) + b_2\).
Key observations: - All weights in W2 are
negative (e.g., -0.954667). - The bias
b2 is a very small positive number (approx. \(\frac{1}{254}\)). - a1 is the
output of Layer 1's ReLU, so \(a_1 \ge
0\).
Negative W2 weights reduce z2 whenever an
a1 component is positive. Setting every a1
component to zero is a sufficient way to preserve the positive bias
contribution. Proving that this is necessary requires the exact weights,
biases, and discrete input step; the retained values alone do not
establish that bound.
3. Analyzing Layer 1
To force \(a_1 = 0\), we must ensure that \(z_1 = x \cdot W_1 + b_1 \le 0\) for all neurons in the first layer.
Looking closely at the biases (b1), they appear to be
"random" decimals: - h1[0] = -0.795276 -
h1[3] = -0.90551 - h1[12] = -0.968504 -
h1[23] = 0.91339
If we multiply these values by 127 (the maximum
standard ASCII value), they resolve into integers. This reveals that
W1 and b1 are essentially implementing
boundary checks for each character of the input string.
Converting these non-zero biases back to ASCII: -
-0.795276 * 127 \(\approx\) -101 \(\rightarrow\)
e - -0.90551 * 127 \(\approx\) -115 \(\rightarrow\)
s - -0.968504 * 127 \(\approx\) -123 \(\rightarrow\)
{ - 0.91339 * 127 \(\approx\) 116 \(\rightarrow\)
t
4. Reconstructing the Flag
Extracting all valid b1 values and converting them
yields a multiset of 22 characters:
t, e, x, s, a, w, {, }, s, v, r, r, r, 3, 3, 3, 3, _, n, l, 4, u
We know the flag format is texsaw{flag}. Removing the
wrapper characters (t, e, x, s, a, w, {, }), we are left
with: s, v, r, r, r, 3, 3, 3, 3, _, n, l, 4, u
The character multiset suggests a Leetspeak phrase, but does not
determine character order. A reproducible reconstruction must associate
each nonzero W1 entry with its input column and intersect
that column’s bounds, then evaluate the candidate through the complete
network and the Row 112 predicate. The spreadsheet, extraction script,
and forward-check output are not retained here; this is a method note
rather than a complete solver. The flag below is retained from the
existing record, without a new acceptance or model verification
claim.