Example 8 - Iterative Search

Summary

This example demonstrates the iterative process towards finding a suitable set of notes. Fees are calculated based on the structure of the transaction, but they also impact the transaction. Therefore, the algorithm needs to try out several combinations.

Let’s consider the following orders:

The number is the quantity for a given address type. If a row has more than one set of numbers, it’s the total amount.

• Order 1 is a t-addr for 10

Suppose we have the following notes:

• T: 5, 7
• S: 12
• O: 10

Note Selection

Let’s assume we choose (Sapling, Orchard, Transparent) in the Pool Usage Order and the change address is a UA with Transparent and Sapling receivers.

We begin with (12, 12, 10) in each pool respectively.

• Order 1: we use 10 from T-pool. Now we have (2, 12, 10)

Fee ZIP 327

In the T-pool, we have

• 2 inputs,
• 1 output for Order #1

In the S-pool, we have

• 0 input,
• 0 outputs

In the O-pool, we have

• 0 inputs,
• 0 outputs

1. T-pool contributes 2 logical actions = max(2, 1).
2. S-pool contributes 0 logical actions
3. O-pool contributes 0 logical actions

The number of logical actions = 2 and the fee is 2 * 5 = 10

However, we haven’t paid for the fee, and we haven’t considered the change outputs yet.

Paying for the fee and making change outputs

We pay for the fee using 10 from T&S-pool, following our pool usage preferences.

We also have change outputs:

• T: we use 12 (= 5+7), pay 10 to order #1 and 2 for fees,
• S: we use 12, pay 8 in fees, and we should get 4 back.

These outputs add up to the transaction and modify the fee.

1. T-pool contributes 2 logical actions = max(2, 1).
2. S-pool contributes 1 logical actions = max(1, 1).
3. O-pool contributes 0 logical actions

The number of logical actions = 3 and the fee is 3 * 5 = 15

Let’s adjust for the new fee.

But we can only pay 12 from the S-pool. We need to use the O-pool too now.

• T: pay 2 for fees. No T-change
• S: pay 12 for fees. No more S-change
• O: pay 1 for fees, get 9 in change

We have to add an orchard action!

1. T-pool contributes 2 logical actions = max(2, 2).
2. S-pool contributes 1 logical actions = max(1, 0).
3. O-pool contributes 1 logical actions

The number of logical actions = 4 and the fee is 4 * 5 = 20

We need to pay 5 more from the O-pool since the S-pool is empty.

• T: pay 2 for fees. No T-change
• S: pay 12 for fees. No S-change
• O: pay 6 for fees, get 4 in change

At last, we are not modifying the transaction structure anymore.

Final transaction

• Inputs: all notes are used
  • T: 5, 7
  • S: 12
  • O: 10
• Outputs:
  • Order 1: T/10
  • Change: O/4
• Fee: 20