> For the complete documentation index, see [llms.txt](https://alham-rizvi.gitbook.io/alhamrizvi/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://alham-rizvi.gitbook.io/alhamrizvi/cryptography/rsa/10.-partial-d-exposure.md).

# Partial d Exposure Attack

## 1. Problem Setup

We are given:

```
n = p · q
e = public exponent
```

But instead of full `d`, we know **part of it**:

```
d = known_part + unknown_part
```

Example:

```
Lower bits of d are known
Upper bits unknown
```

## 2. Core Idea

From RSA:

```
e · d ≡ 1 mod φ(n)
```

Rewrite:

```
e·d − 1 = k·φ(n)
```

Substitute:

```
d = d0 + x
```

Where:

```
d0 = known part
x  = unknown part
```

Then:

```
e(d0 + x) − 1 = k·φ(n)
```

Rearrange:

```
e·x = k·φ(n) − e·d0 + 1
```

## 3. Why this is useful

```
x is small (unknown bits are few)
```

So we can:

```
solve for x using lattice techniques
```

This is where **Coppersmith's Method** comes in.

## 4. Key Insight

```
We reduce RSA to a small root problem
```

Instead of factoring `n`, we solve:

```
f(x) ≡ 0 mod n
```

Where `x` is small.

***

## 5. When It Works

```
If enough bits of d are known
```

Typical condition:

```
Known bits ≥ ~50% of d
```

## 6. Attack Strategy

1. Express:

```
d = d0 + x
```

2. Substitute into:

```
e·d − 1 = k·φ(n)
```

3. Build polynomial in `x`
4. Use lattice (Coppersmith) to find small `x`
5. Recover full `d`
6. Decrypt normally

## 7. Python Reality

Important:

```
This is NOT practical in plain Python
```

You need:

* **SageMath**
* or advanced lattice libraries

## 8. SageMath Script (Typical)

```python
# SageMath required

n = ...
e = ...
d0 = ...  # known part

PR.<x> = PolynomialRing(Zmod(n))

f = e*(d0 + x) - 1

# Find small root
roots = f.small_roots(X=2^kbits, beta=0.5)

print(roots)
```

## 9. Simpler Case (Bruteforce version)

If unknown part is very small:

```python
from sympy import mod_inverse

n = ...
e = ...
d0 = ...
bits = 8  # unknown bits

for x in range(2**bits):
    d = d0 + x
    if (e*d - 1) % n != 0:
        continue

    print("Found d:", d)
```

This only works if:

```
Unknown bits are extremely small
```

## 10. When to Recognize in CTF

Look for:

```
"partial key leaked"
"some bits of d known"
"d starts with..."
"d ends with..."
```

## 11. Why This Breaks RSA

Because:

```
Even partial leakage reduces search space
```

And lattice methods exploit:

```
small unknown variables
```

## 12. Important Variants

### Known MSB of d

```
High bits known
```

### Known LSB of d

```
Low bits known
```

### Mixed leakage

```
Scattered bits known
```

All solvable with variations of Coppersmith.

## 13. Limitations

```
Too few known bits → attack fails
Requires lattice math
Not beginner-friendly in raw form
```

## 14. Summary

```
Partial d known → express d = d0 + x
→ build equation
→ solve small root (Coppersmith)
→ recover d
→ decrypt
```


---

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