Without further ado, we present to you forward-backward induction!

Forward-backward induction

Step 1 1 1: [The base case] We start by showing that our statement is true for an integer n 0 n_0 n0​. This isn't any different from standard induction.

Step 2 2 2: [The inductive step] This is made out of two parts. P ( k ) ⇒ P ( 2 k ) P(k)\Rightarrow P(2k) P(k)⇒P(2k): This is our "forward" part. This is where you show that if the statement is true for some integer k k k, it is also true for 2 k . 2k. 2k. P ( k ) ⇒ P ( k − 1 ) P(k)\Rightarrow P(k-1) P(k)⇒P(k−1): This is our "backward" part. Here you show that if the statement is true for some integer k k k, then it is also true for k − 1. k-1. k−1.



Completing steps 1 1 1 and 2 2 2 proves that the statement is true for all positive integers n n n.

Take some time to think about why this argument works. The first part of the inductive step shows that the statement is true for larger and larger values of n n n. But that leaves a lot of gaps in between. The second part ensures that all the gaps are taken care of.

Another way of viewing it is through the domino analogy. (Dominoes are a great way to think about induction!). Say, you have infinitely many dominoes arranged in a line. You have proved a very interesting property about these dominoes. If you knock over the k k k-th domino, for some strange reason, all the 2 n ⋅ k th 2^n\cdot k^{\text{th}} 2n⋅kth dominoes get knocked over. After you observe the dominoes for a little while longer, you prove that the dominoes are arranged in such a way that if one of them falls down, the one preceding it falls down as well. Now if you were to knock down the first domino, all of the dominoes (despite all their weirdness!) would eventually fall.

Convince yourself that the method of forward-backward induction does indeed work. Then continue reading.