Example 10 - Multi Payments

Summary

This example combines multiple orders of different address types. It also demonstrates how the algorithm handles multiple receivers in a unified address.

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
• Order 4 is a ua with transparent and sapling receivers for 40
• Order 7 is a ua with transparent, sapling and orchard receivers for 70 in total across T, S and O receivers

Suppose we have the following notes:

• T: 100
• S: 160
• O: 70 & 50

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 (100, 160, 120) in each pool respectively.

• Order 1: we use 10 from T-pool. Now we have (90, 160, [70, 50])
• Order 2: we use 20 from S-pool. Now we have (90, 140, [70, 50])
• Order 3: we use 30 from O-pool. Now we have (90, 140, [40, 50])
• Order 4: we use 30 from S-pool. S is preferred over T. Now we have (90, 100, [40, 50]).
• Order 5: we use 50 from O-pool. O is preferred over T. Now we have (90, 100, 40).
• Order 6: we use 60 from O-pool. S is preferred over O. Now we have (90, 40, 40).
• Order 7: we use 40 from S-pool and 30 from the O-pool. S is preferred over O, but we don’t have enough and we need the O-pool too. Now we have (90, 0, 10).

Fee ZIP 327

In the T-pool, we have

• 1 input,
• 1 output for Order #1

In the S-pool, we have

• 1 input,
• 4 outputs for Order #2, #4, #6, #7

In the O-pool, we have

• 2 inputs,
• 3 outputs for Order #3, #5, #7

1. T-pool contributes 1 logical actions = max(1, 1).
2. S-pool contributes 4 logical actions = max(1, 4)
3. O-pool contributes 3 logical actions = max(2, 3)

The number of logical actions = 8 and the fee is 8 * 5 = 40

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 O-pool and 30 from T-pool, following our pool usage preferences.

So now, we should have (60, 0, 0). But we spent all our inputs therefore, we need a change output of 60 to our t-addr.

This T-output increases the number of T-logical actions to 2 = max(1, 2). Then the total number of logical actions is now 9. The fee becomes 45.

The change is also adjusted to 55.

Final transaction

• Inputs: all notes are used
  • T: 100
  • S: 160
  • O: 70 & 50
• Outputs:
  • Order 1: T/10
  • Order 2: S/20
  • Order 3: O/30
  • Order 4: S/40
  • Order 5: O/50
  • Order 6: S/60
  • Order 7: S/40, O/30
  • Change: T/55
• Fee: 45

Test Code

fn test_example10() {
  env_logger::init();
  let mut config = NoteSelectConfig::new(CHANGE_ADDRESS);
  config.use_transparent = true;
  config.privacy_policy = PrivacyPolicy::AnyPool;
  config.precedence = [ Pool::Sapling, Pool::Orchard, Pool::Transparent ];

  let utxos = [utxo!(1, 100), sapling!(2, 160), orchard!(3, 70), orchard!(4, 50)];
  let mut orders = [t!(1, 10), s!(2, 20), o!(3, 30), ts!(4, 40), to!(5, 50), so!(6, 60), tso!(7, 70)];

  let tx_plan = note_select_with_fee::<FeeZIP327>(&utxos, &mut orders, &config).unwrap();
  println!("{}", serde_json::to_string(&tx_plan).unwrap());
}