We discussed true LRU implementations, but they are usually costly. In this post, we will explain how to implement pseudo LRU.

An example is shown below. The data structure is similar to a binary tree. For N ways in a cache set, we need to keep (N – 1) bits. If there are 8 sets in a way, then we will need 7 bits.

Upon a cache access, all tree nodes point to that cache way will be flipped. For example, if set 3 is accessed, then h0, h1 and h4 will be flipped in next cycle.

To find LRU, we can perform a depth-first-search starting from h0, and traverse nodes in lower levels. If the node is 0, then we traverse the left sub-tree; otherwise, we traverse the right sub-tree. In the diagram above, the LRU is set 3.

Once we understand the basic idea of pseudo LRU, it is easy to write code for pseudo LRU.


  1. for this assignment: assign lru_idx[0] = (~h0 & ~h1 & ~h3) | (~h0 & h1 & ~h4) | (h0 & ~h2 & ~h5) | (~h0 & h2 & ~h6);
    I think it should be assign lru_idx[0] = (~h0 & ~h1 & h3) | (~h0 & h1 & h4) | (h0 & ~h2 & h5) | (~h0 & h2 & h6);

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