Hybrid Argument proof

Question

I am trying to understand what the Hybrid Argument is in cryptography and why is it useful.

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By the definition of the Hybrid Argument we know that to prove that if two distributions $D = D_1, D_2, ..., D_k = D'$ can be distinguished, then $\exists D_i, D_{i+1} $ that can be distinguished.

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Let:

1. $\alpha, \beta, \gamma \in \mathbf{N}$
2. $0 < \alpha < \beta < \gamma$
3. $g_1$ be a $(\alpha, \beta)$-pseudo-random generator
4. $g_2$ be a $(\beta, \gamma)$-pseudo-random generator

Consider $g(x) = g_2(g_1(x))$, then we want to prove by using the Hybrid Argument that $g(x)$ is $(\alpha-\gamma)$-pseudo-random generator.

Let's also define the following distributions on $ \{ 0,1 \}^\gamma$:

• $D_1: y$ uniform in $ \{ 0,1 \}^\gamma$
• $D_2: y = g_1(x)$ for $x$ uniform in $ \{ 0,1 \}^\alpha$
• $D_3: y = g_2(z)$ for $z$ uniform in $ \{ 0,1 \}^\beta$

Here we can prove that g is a pseudo-random generator by using the triangle inequality:

By reduction to absurd, there's a distinguisher $A$ between $D_1$ and $D_2 \rightarrow A$ must either:

• Distinguish between $D_1$ and $D_3$
• Distinguish between $D_2$ and $D_3$

It's in this point that the triangle inequality is used to show the contradiction. Wouldn't it be easy for the distinguisher A to differentiate the two distributions as they have different length?

Also, what's the triangle inequality in this context and why does it help us deduce that there's a contradiction with this statement?