Puppy Raffle

AI First Flight #1
Beginner FriendlyFoundrySolidityNFT
EXP
View results
Submission Details
Severity: medium
Valid

O(n^2) duplicate check in enterRaffle makes entry progressively more expensive, down to a block-gas-limit DoS

Summary

Every enterRaffle call re-checks every pair of addresses in the whole players array, so gas grows with the square of the number of entrants. Late entrants pay far more than early ones, and an attacker can fill the round early to price out or block everyone after them.

Description

src/PuppyRaffle.sol:86-90 loops over the entire array, not just the new players, costing about n(n−1)/2 storage-read pairs per call:

for (uint256 i = 0; i < players.length - 1; i++) {
for (uint256 j = i + 1; j < players.length; j++) {
require(players[i] != players[j], "PuppyRaffle: Duplicate player");

Measured: the first 100 entrants cost 6.57M gas; the next 100 cost 19.25M; one single entrant after 200 players costs 17.29M, which is more than half a 30M block.

Risk

Likelihood: Medium — grows with normal participation, and an attacker can accelerate it by entering many addresses.

Impact: Medium — unfair costs, then denial of service of the core entry function. No funds are stolen directly.

Proof of Concept

function test_PoC_EnterRaffleGasGrowsQuadratically() public {
uint256 g0 = gasleft();
_enter(1, 100);
uint256 firstBatch = g0 - gasleft();
g0 = gasleft();
_enter(101, 100);
assertGt(g0 - gasleft(), firstBatch * 2);
g0 = gasleft();
_enter(201, 1);
assertGt(g0 - gasleft(), firstBatch); // 1 late ticket > 100 early tickets
}

Run with forge test --match-test test_PoC_EnterRaffleGasGrowsQuadratically -vvv. It passes.

Recommended Mitigation

Use a mapping keyed by raffle round for an O(1) duplicate check:

+ uint256 public raffleId;
+ mapping(address => uint256) public enteredInRaffle; // raffleId + 1
...
for (uint256 i = 0; i < newPlayers.length; i++) {
+ require(enteredInRaffle[newPlayers[i]] != raffleId + 1, "PuppyRaffle: Duplicate player");
+ enteredInRaffle[newPlayers[i]] = raffleId + 1;
players.push(newPlayers[i]);
}
- for (uint256 i = 0; i < players.length - 1; i++) { ... }

Increment raffleId in selectWinner.

Updates

Lead Judging Commences

ai-first-flight-judge Lead Judge 40 minutes ago
Submission Judgement Published
Validated
Assigned finding tags:

[M-01] `PuppyRaffle: enterRaffle` Use of gas extensive duplicate check leads to Denial of Service, making subsequent participants to spend much more gas than prev ones to enter

## Description `enterRaffle` function uses gas inefficient duplicate check that causes leads to Denial of Service, making subsequent participants to spend much more gas than previous users to enter. ## Vulnerability Details In the `enterRaffle` function, to check duplicates, it loops through the `players` array. As the `player` array grows, it will make more checks, which leads the later user to pay more gas than the earlier one. More users in the Raffle, more checks a user have to make leads to pay more gas. ## Impact As the arrays grows significantly over time, it will make the function unusable due to block gas limit. This is not a fair approach and lead to bad user experience. ## POC In existing test suit, add this test to see the difference b/w gas for users. once added run `forge test --match-test testEnterRaffleIsGasInefficient -vvvvv` in terminal. you will be able to see logs in terminal. ```solidity function testEnterRaffleIsGasInefficient() public { vm.startPrank(owner); vm.txGasPrice(1); /// First we enter 100 participants uint256 firstBatch = 100; address[] memory firstBatchPlayers = new address[](firstBatch); for(uint256 i = 0; i < firstBatchPlayers; i++) { firstBatch[i] = address(i); } uint256 gasStart = gasleft(); puppyRaffle.enterRaffle{value: entranceFee * firstBatch}(firstBatchPlayers); uint256 gasEnd = gasleft(); uint256 gasUsedForFirstBatch = (gasStart - gasEnd) * txPrice; console.log("Gas cost of the first 100 partipants is:", gasUsedForFirstBatch); /// Now we enter 100 more participants uint256 secondBatch = 200; address[] memory secondBatchPlayers = new address[](secondBatch); for(uint256 i = 100; i < secondBatchPlayers; i++) { secondBatch[i] = address(i); } gasStart = gasleft(); puppyRaffle.enterRaffle{value: entranceFee * secondBatch}(secondBatchPlayers); gasEnd = gasleft(); uint256 gasUsedForSecondBatch = (gasStart - gasEnd) * txPrice; console.log("Gas cost of the next 100 participant is:", gasUsedForSecondBatch); vm.stopPrank(owner); } ``` ## Recommendations Here are some of recommendations, any one of that can be used to mitigate this risk. 1. User a mapping to check duplicates. For this approach you to declare a variable `uint256 raffleID`, that way each raffle will have unique id. Add a mapping from player address to raffle id to keep of users for particular round. ```diff + uint256 public raffleID; + mapping (address => uint256) public usersToRaffleId; . . function enterRaffle(address[] memory newPlayers) public payable { require(msg.value == entranceFee * newPlayers.length, "PuppyRaffle: Must send enough to enter raffle"); for (uint256 i = 0; i < newPlayers.length; i++) { players.push(newPlayers[i]); + usersToRaffleId[newPlayers[i]] = true; } // Check for duplicates + for (uint256 i = 0; i < newPlayers.length; i++){ + require(usersToRaffleId[i] != raffleID, "PuppyRaffle: Already a participant"); - for (uint256 i = 0; i < players.length - 1; i++) { - for (uint256 j = i + 1; j < players.length; j++) { - require(players[i] != players[j], "PuppyRaffle: Duplicate player"); - } } emit RaffleEnter(newPlayers); } . . . function selectWinner() external { //Existing code + raffleID = raffleID + 1; } ``` 2. Allow duplicates participants, As technically you can't stop people participants more than once. As players can use new address to enter. ```solidity function enterRaffle(address[] memory newPlayers) public payable { require(msg.value == entranceFee * newPlayers.length, "PuppyRaffle: Must send enough to enter raffle"); for (uint256 i = 0; i < newPlayers.length; i++) { players.push(newPlayers[i]); } emit RaffleEnter(newPlayers); } ```

Support

FAQs

Can't find an answer? Chat with us on Discord, Twitter or Linkedin.

Give us feedback!