EVM Puzzle 10 solution

This is Part 10 of the “Let’s play EVM Puzzles” series, where I will explain how to solve each puzzle challenge.

EVM Puzzles is a project developed by Franco Victorio (@fvictorio_nan) that is a perfect fit if you are in the process of learning how the Ethereum EVM works, and you want to apply some of the knowledge you have just acquired.

EVM Puzzle 10

00      38          CODESIZE
01 34 CALLVALUE
02 90 SWAP1
03 11 GT
04 6008 PUSH1 08
06 57 JUMPI
07 FD REVERT
08 5B JUMPDEST
09 36 CALLDATASIZE
0A 610003 PUSH2 0003
0D 90 SWAP1
0E 06 MOD
0F 15 ISZERO
10 34 CALLVALUE
11 600A PUSH1 0A
13 01 ADD
14 57 JUMPI
15 FD REVERT
16 FD REVERT
17 FD REVERT
18 FD REVERT
19 5B JUMPDEST
1A 00 STOP

This puzzle is similar to the Puzzle 9 we have just completed. It’s mostly about understanding what opcodes do and solve a system of equations.

Let’s see what new opcodes have been introduced:

- GT: pop 2 values from the stack and push the result of 1 to the stack. If the result is true it push 1 otherwise 0
- MOD: pop 2 values from the stack and push back to the stack the result of value0 % value1. Note that the denominator (value1) is 0 the result will be 0
- ISZERO: pop a value from the stack and push the result of value0 === 0 to the stack

Block 1: check calldata size and call value

00 38 CODESIZE
01 34 CALLVALUE
02 90 SWAP1
03 11 GT
04 6008 PUSH1 08
06 57 JUMPI
07 FD REVERT
08 5B JUMPDEST

The block adds the size of the code to the stack, add the value sent with the transaction to the stack, swap them in position (you could have achieved the same result with less gas) and then perform GT(CALLVALUE, CODESIZE).

If the result of that is 0 it will not follow the JUMPI jump and revert.
`CODESIZE` push to the stack the amount of bytes of the contract’s code. In this case, it will push to the stack the value 0x1b (27 in decimal).

Note: The number of code’s instructions are 24 (so 24 bytes) but you must add to those also the bytes pushed by the PUSH* opcodes. In this case, we have 2 PUSH1 and 1 PUSH2 so in total we need to add 3 bytes. That’s why the CODESIZE return 27 → 24 bytes for the number of instructions + 3 bytes from the values of the `PUSH` in the code.

We have found our first equation to not revert: GT(27, CALLVALUE) = 1 so we must have CALLVALUE <= 27 to not revert.

Block 2: check the calldata size

08 5B JUMPDEST
09 36 CALLDATASIZE
0A 610003 PUSH2 0003
0D 90 SWAP1
0E 06 MOD
0F 15 ISZERO

The opcodes push to the stack the CALLDATASIZE, push 0x0003, swap them, perform a MOD(0x0003, CALLDATASIZE) and perform ISZERO on the value0 present in the stack. Because we have just performed the MOD operation, it will be ISZERO(MOD(0x0003, CALLDATASIZE))

This value will be used by the JUMPI from the instruction in position 14. If the result of the ISZERO is not 1 the contract will revert because it will not perform the jump.

The size of our calldata must be a multiple of 3 to make MODE(3, CALLDATASIZE) be equal to 0.

This is the second part of the system of equations.

Block 3: find the correct call value to jump to a valid JUMPDEST

10 34 CALLVALUE
11 600A PUSH1 0A
13 01 ADD
14 57 JUMPI

Currently, in our stack we have the result of ISZERO(MOD(0x0003, CALLDATASIZE)) and we know that it will be 1 otherwise we are going to revert.

Performing the other operation will make the stack be like

PUSH 0A
CALLVALUE
ISZERO(MOD(0x0003, CALLDATASIZE))

At this point, we perform the ADD so we have the stack that will be

ADD(0A, CALLVALUE)
ISZERO(MOD(0x0003, CALLDATASIZE))

JUMPI will perform a jump to the position with value ADD(0x0A, CALLVALUE). The JUMPDEST that we want to reach is the one in position 19 (25 in decimal).

This mean that ADD(0x0A, CALLVALUE) === 19. The only possible value for that is that our CALLVALUE is 10 (in hex, is 0x0F)

Solution

The system of equations we have to solve is this:

  • CODESIZE = 27 (1b in hex) is always
  • CALLVALUE must be <= 27 to make GT(CALLVALUE, CODESIZE) return 1
  • CALLVALUE = 15 (0F in hex) to make ADD(0A, CALLVALUE) return 19
  • CALLDATASIZE = multiple of 3 to make ISZERO(MOD(0x0003, CALLDATASIZE)) return 1

A possible solution could be:

  • CALLVALUE = 15
  • CALLDATA = 0xFFFFFF

Here’s the link to the solution of Puzzle 10 on EVM Codes website to simulate it.

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