Argent Blockchain Engineer Assessment
In this repository there are solutions for the Argent Blockchain Engineer Technical interview.
Assessment A: Algorithm & Data Structure problem
The solution is coded in Python. Find it in assessment_a/solution.py.
Assessment B: Smart-contract problem
Point 1
function draw():
# check if time limit is done
if actual block > block number limit:
# check everyone has submitted the secret
if submitted_secrets equal tickets_bought:
winner_address = XOR tickets
winner_address.send(balance)
# revert transactions for fairness
else:
for ticket in tickets_bought:
ticket.send(balance / tickets_bought.length)
# reset all
del tickets_bought
del balance
del submitted_secrets
return winner_address
The strategy is secure because it is needed that participants send a hash computed locally with the chosen number. The sender and the linked hash is stored so we can assure only one number per participant to make the process fair. There is a time limit of confirmed block to be able to start the lottery to avoid exploits from miners. Also, if all the ticket buyers don't commit their submission when this function is called the lottery reverts as the last committer may have advantage to previous commiters.
Point 2
Array secrets for mapping addresses to secrets
Array balances for mapping addresses to eth sent
Array accounts for participant index to address
Array tickets for commited participants
The random number would be generated from: rand = n1 XOR n2 XOR n3 .. XOR nn where n is the commit number by the participant. Then we would apply a module operation to the commiters length. tickets[rand%tickets.length] to choose the winner.
Point 3
Having a constructor in the contract where owner = msg.sender() and require that msg.sender() == owner when calling the draw method.
Assessment 3: DeFi problem
Point a
Following the Aave documentation of Health Risk definition:
((100 + (100,000 / 2,000)) * 0.8 / (150,000 / 2,000) = 1.6
Point b
When ETH price is < 880$ the Health Factor would be below 1 which is not recommended. When ETH price < 500$ the position would be liquidated.
HF below 1:
((100 + (100,000 / 880)) * 0.8 / (150,000 / 880) = 1
Liquidation HF:
((100 + (100,000 / 500)) * 0.8 / (150,000 / 500) = 0.8
Point c
For keeping the HF above 1.2 the borrowed USDC must be below 200k USDC. So, 50k more USDC could be borrowed while keeping HF above 1.2.
((100 + (100,000 / 500)) * 0.8 / (150,000 / 500) = 0.8
Point d
If ~280,000 USDC are borrowed the HF would be 0.85 and the position would be liquidated
Point e
Coded solution in assessment_c/point_e.py
- Number of years to be in risk of liquidation: 6 with HF: 0.93
- Number of years to be liquidated: 8 with HF: 0.78