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Vitalik

Vitalik

4 years ago

An approximate introduction to how zk-SNARKs are possible (part 2)

If tasked with the problem of coming up with a zk-SNARK protocol, many people would make their way to this point and then get stuck and give up. How can a verifier possibly check every single piece of the computation, without looking at each piece of the computation individually? But it turns out that there is a clever solution.

Polynomials

Polynomials are a special class of algebraic expressions of the form:

  • x+5
  • x^4
  • x^3+3x^2+3x+1
  • 628x^{271}+318x^{270}+530x^{269}+…+69x+381

i.e. they are a sum of any (finite!) number of terms of the form cx^k

There are many things that are fascinating about polynomials. But here we are going to zoom in on a particular one: polynomials are a single mathematical object that can contain an unbounded amount of information (think of them as a list of integers and this is obvious). The fourth example above contained 816 digits of tau, and one can easily imagine a polynomial that contains far more.

Furthermore, a single equation between polynomials can represent an unbounded number of equations between numbers. For example, consider the equation A(x)+ B(x) = C(x). If this equation is true, then it's also true that:

  • A(0)+B(0)=C(0)
  • A(1)+B(1)=C(1)
  • A(2)+B(2)=C(2)
  • A(3)+B(3)=C(3)

And so on for every possible coordinate. You can even construct polynomials to deliberately represent sets of numbers so you can check many equations all at once. For example, suppose that you wanted to check:

  • 12+1=13
  • 10+8=18
  • 15+8=23
  • 15+13=28

You can use a procedure called Lagrange interpolation to construct polynomials A(x) that give (12,10,15,15) as outputs at some specific set of coordinates (eg. (0,1,2,3)), B(x) the outputs (1,8,8,13) on thos same coordinates, and so forth. In fact, here are the polynomials:

  • A(x)=-2x^3+\frac{19}{2}x^2-\frac{19}{2}x+12
  • B(x)=2x^3-\frac{19}{2}x^2+\frac{29}{2}x+1
  • C(x)=5x+13

Checking the equation A(x)+B(x)=C(x) with these polynomials checks all four above equations at the same time.

Comparing a polynomial to itself

You can even check relationships between a large number of adjacent evaluations of the same polynomial using a simple polynomial equation. This is slightly more advanced. Suppose that you want to check that, for a given polynomial F, F(x+2)=F(x)+F(x+1) with the integer range {0,1…89} (so if you also check F(0)=F(1)=1, then F(100) would be the 100th Fibonacci number)

As polynomials, F(x+2)-F(x+1)-F(x) would not be exactly zero, as it could give arbitrary answers outside the range x={0,1…98}. But we can do something clever. In general, there is a rule that if a polynomial P is zero across some set S=\{x_1,x_2…x_n\} then it can be expressed as P(x)=Z(x)*H(x), where Z(x)=(x-x_1)*(x-x_2)*…*(x-x_n) and H(x) is also a polynomial. In other words, any polynomial that equals zero across some set is a (polynomial) multiple of the simplest (lowest-degree) polynomial that equals zero across that same set.

Why is this the case? It is a nice corollary of polynomial long division: the factor theorem. We know that, when dividing P(x) by Z(x), we will get a quotient Q(x) and a remainder R(x) is strictly less than that of Z(x). Since we know that P is zero on all of S, it means that R has to be zero on all of S as well. So we can simply compute R(x) via polynomial interpolation, since it's a polynomial of degree at most n-1 and we know n values (the zeros at S). Interpolating a polynomial with all zeroes gives the zero polynomial, thus R(x)=0 and H(x)=Q(x).

Going back to our example, if we have a polynomial F that encodes Fibonacci numbers (so F(x+2)=F(x)+F(x+1) across x=\{0,1…98\}), then I can convince you that F actually satisfies this condition by proving that the polynomial P(x)=F(x+2)-F(x+1)-F(x) is zero over that range, by giving you the quotient:
H(x)=\frac{F(x+2)-F(x+1)-F(x)}{Z(x)}
Where Z(x) = (x-0)*(x-1)*…*(x-98).
You can calculate Z(x) yourself (ideally you would have it precomputed), check the equation, and if the check passes then F(x) satisfies the condition!

Now, step back and notice what we did here. We converted a 100-step-long computation into a single equation with polynomials. Of course, proving the N'th Fibonacci number is not an especially useful task, especially since Fibonacci numbers have a closed form. But you can use exactly the same basic technique, just with some extra polynomials and some more complicated equations, to encode arbitrary computations with an arbitrarily large number of steps.

see part 3

(Edited)

Hackernoon

Hackernoon

4 years ago


👏 Awesome post! When is part 3 coming?

Trent Lapinski

Trent Lapinski

4 years ago

Very complex topic, great explanation

More on Web3 & Crypto

Vitalik

Vitalik

4 years ago

An approximate introduction to how zk-SNARKs are possible (part 1)

You can make a proof for the statement "I know a secret number such that if you take the word ‘cow', add the number to the end, and SHA256 hash it 100 million times, the output starts with 0x57d00485aa". The verifier can verify the proof far more quickly than it would take for them to run 100 million hashes themselves, and the proof would also not reveal what the secret number is.

In the context of blockchains, this has 2 very powerful applications: Perhaps the most powerful cryptographic technology to come out of the last decade is general-purpose succinct zero knowledge proofs, usually called zk-SNARKs ("zero knowledge succinct arguments of knowledge"). A zk-SNARK allows you to generate a proof that some computation has some particular output, in such a way that the proof can be verified extremely quickly even if the underlying computation takes a very long time to run. The "ZK" part adds an additional feature: the proof can keep some of the inputs to the computation hidden.

You can make a proof for the statement "I know a secret number such that if you take the word ‘cow', add the number to the end, and SHA256 hash it 100 million times, the output starts with 0x57d00485aa". The verifier can verify the proof far more quickly than it would take for them to run 100 million hashes themselves, and the proof would also not reveal what the secret number is.

In the context of blockchains, this has two very powerful applications:

  1. Scalability: if a block takes a long time to verify, one person can verify it and generate a proof, and everyone else can just quickly verify the proof instead
  2. Privacy: you can prove that you have the right to transfer some asset (you received it, and you didn't already transfer it) without revealing the link to which asset you received. This ensures security without unduly leaking information about who is transacting with whom to the public.

But zk-SNARKs are quite complex; indeed, as recently as in 2014-17 they were still frequently called "moon math". The good news is that since then, the protocols have become simpler and our understanding of them has become much better. This post will try to explain how ZK-SNARKs work, in a way that should be understandable to someone with a medium level of understanding of mathematics.

Why ZK-SNARKs "should" be hard

Let us take the example that we started with: we have a number (we can encode "cow" followed by the secret input as an integer), we take the SHA256 hash of that number, then we do that again another 99,999,999 times, we get the output, and we check what its starting digits are. This is a huge computation.

A "succinct" proof is one where both the size of the proof and the time required to verify it grow much more slowly than the computation to be verified. If we want a "succinct" proof, we cannot require the verifier to do some work per round of hashing (because then the verification time would be proportional to the computation). Instead, the verifier must somehow check the whole computation without peeking into each individual piece of the computation.

One natural technique is random sampling: how about we just have the verifier peek into the computation in 500 different places, check that those parts are correct, and if all 500 checks pass then assume that the rest of the computation must with high probability be fine, too?

Such a procedure could even be turned into a non-interactive proof using the Fiat-Shamir heuristic: the prover computes a Merkle root of the computation, uses the Merkle root to pseudorandomly choose 500 indices, and provides the 500 corresponding Merkle branches of the data. The key idea is that the prover does not know which branches they will need to reveal until they have already "committed to" the data. If a malicious prover tries to fudge the data after learning which indices are going to be checked, that would change the Merkle root, which would result in a new set of random indices, which would require fudging the data again... trapping the malicious prover in an endless cycle.

But unfortunately there is a fatal flaw in naively applying random sampling to spot-check a computation in this way: computation is inherently fragile. If a malicious prover flips one bit somewhere in the middle of a computation, they can make it give a completely different result, and a random sampling verifier would almost never find out.


It only takes one deliberately inserted error, that a random check would almost never catch, to make a computation give a completely incorrect result.

If tasked with the problem of coming up with a zk-SNARK protocol, many people would make their way to this point and then get stuck and give up. How can a verifier possibly check every single piece of the computation, without looking at each piece of the computation individually? There is a clever solution.

see part 2

CyberPunkMetalHead

CyberPunkMetalHead

3 years ago

195 countries want Terra Luna founder Do Kwon

Interpol has issued a red alert on Terraform Labs' CEO, South Korean prosecutors said.

After the May crash of Terra Luna revealed tax evasion issues, South Korean officials filed an arrest warrant for Do Kwon, but he is missing.

Do Kwon is now a fugitive in 195 countries after Seoul prosecutors placed him to Interpol's red list. Do Kwon hasn't commented since then. The red list allows any country's local authorities to apprehend Do Kwon.

Do Dwon and Terraform Labs were believed to have moved to Singapore days before the $40 billion wipeout, but Singapore authorities said he fled the country on September 17. Do Kwon tweeted that he wasn't on the run and cited privacy concerns.

Do Kwon was not on the red list at the time and said he wasn't "running," only to reply to his own tweet saying he hasn't jogged in a while and needed to trim calories.

Whether or not it makes sense to read too much into this, the reality is that Do Kwon is now on Interpol red list, despite the firmly asserts on twitter that he does absolutely nothing to hide.

UPDATE:

South Korean authorities are investigating alleged withdrawals of over $60 million U.S. and seeking to freeze these assets. Korean authorities believe a new wallet exchanged over 3000 BTC through OKX and Kucoin.

Do Kwon and the Luna Foundation Guard (of whom Do Kwon is a key member of) have declined all charges and dubbed this disinformation.

Singapore's Luna Foundation Guard (LFG) manages the Terra Ecosystem.

The Legal Situation

Multiple governments are searching for Do Kwon and five other Terraform Labs employees for financial markets legislation crimes.

South Korean authorities arrested a man suspected of tax fraud and Ponzi scheme.

The U.S. SEC is also examining Terraform Labs on how UST was advertised as a stablecoin. No legal precedent exists, so it's unclear what's illegal.

The future of Terraform Labs, Terra, and Terra 2 is unknown, and despite what Twitter shills say about LUNC, the company remains in limbo awaiting a decision that will determine its fate. This project isn't a wise investment.

OnChain Wizard

OnChain Wizard

3 years ago

How to make a >800 million dollars in crypto attacking the once 3rd largest stablecoin, Soros style

Everyone is talking about the $UST attack right now, including Janet Yellen. But no one is talking about how much money the attacker made (or how brilliant it was). Lets dig in.

Our story starts in late March, when the Luna Foundation Guard (or LFG) starts buying BTC to help back $UST. LFG started accumulating BTC on 3/22, and by March 26th had a $1bn+ BTC position. This is leg #1 that made this trade (or attack) brilliant.

The second leg comes in the form of the 4pool Frax announcement for $UST on April 1st. This added the second leg needed to help execute the strategy in a capital efficient way (liquidity will be lower and then the attack is on).

We don't know when the attacker borrowed 100k BTC to start the position, other than that it was sold into Kwon's buying (still speculation). LFG bought 15k BTC between March 27th and April 11th, so lets just take the average price between these dates ($42k).


So you have a ~$4.2bn short position built. Over the same time, the attacker builds a $1bn OTC position in $UST. The stage is now set to create a run on the bank and get paid on your BTC short. In anticipation of the 4pool, LFG initially removes $150mm from 3pool liquidity.

The liquidity was pulled on 5/8 and then the attacker uses $350mm of UST to drain curve liquidity (and LFG pulls another $100mm of liquidity).

But this only starts the de-pegging (down to 0.972 at the lows). LFG begins selling $BTC to defend the peg, causing downward pressure on BTC while the run on $UST was just getting started.

With the Curve liquidity drained, the attacker used the remainder of their $1b OTC $UST position ($650mm or so) to start offloading on Binance. As withdrawals from Anchor turned from concern into panic, this caused a real de-peg as people fled for the exits

So LFG is selling $BTC to restore the peg while the attacker is selling $UST on Binance. Eventually the chain gets congested and the CEXs suspend withdrawals of $UST, fueling the bank run panic. $UST de-pegs to 60c at the bottom, while $BTC bleeds out.


The crypto community panics as they wonder how much $BTC will be sold to keep the peg. There are liquidations across the board and LUNA pukes because of its redemption mechanism (the attacker very well could have shorted LUNA as well). BTC fell 25% from $42k on 4/11 to $31.3k

So how much did our attacker make? There aren't details on where they covered obviously, but if they are able to cover (or buy back) the entire position at ~$32k, that means they made $952mm on the short.

On the $350mm of $UST curve dumps I don't think they took much of a loss, lets assume 3% or just $11m. And lets assume that all the Binance dumps were done at 80c, thats another $125mm cost of doing business. For a grand total profit of $815mm (bf borrow cost).

BTC was the perfect playground for the trade, as the liquidity was there to pull it off. While having LFG involved in BTC, and foreseeing they would sell to keep the peg (and prevent LUNA from dying) was the kicker.

Lastly, the liquidity being low on 3pool in advance of 4pool allowed the attacker to drain it with only $350mm, causing the broader panic in both BTC and $UST. Any shorts on LUNA would've added a lot of P&L here as well, with it falling -65% since 5/7.

And for the reply guys, yes I know a lot of this involves some speculation & assumptions. But a lot of money was made here either way, and I thought it would be cool to dive into how they did it.

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middlemarch.eth

middlemarch.eth

3 years ago

ERC721R: A new ERC721 contract for random minting so people don’t snipe all the rares!

That is, how to snipe all the rares without using ERC721R!

Introduction: Blessed and Lucky 

Mphers was the first mfers derivative, and as a Phunks derivative, I wanted one.

I wanted an alien. And there are only 8 in the 6,969 collection. I got one!

In case it wasn't clear from the tweet, I meant that I was lucky to have figured out how to 100% guarantee I'd get an alien without any extra luck.
Read on to find out how I did it, how you can too, and how developers can avoid it!
How to make rare NFTs without luck.

# How to mint rare NFTs without needing luck

The key to minting a rare NFT is knowing the token's id ahead of time.

For example, once I knew my alien was #4002, I simply refreshed the mint page until #3992 was minted, and then mint 10 mphers.

How did I know #4002 was extraterrestrial? Let's go back.

First, go to the mpher contract's Etherscan page and look up the tokenURI of a previously issued token, token #1:

As you can see, mphers creates metadata URIs by combining the token id and an IPFS hash.

This method gives you the collection's provenance in every URI, and while that URI can be changed, it affects everyone and is public.

Consider a token URI without a provenance hash, like https://mphers.art/api?tokenId=1.
As a collector, you couldn't be sure the devs weren't changing #1's metadata at will.
The API allows you to specify “if #4002 has not been minted, do not show any information about it”, whereas IPFS does not allow this.

It's possible to look up the metadata of any token, whether or not it's been minted.
Simply replace the trailing “1” with your desired id.


Mpher #4002

These files contain all the information about the mpher with the specified id. For my alien, we simply search all metadata files for the string “alien mpher.”

Take a look at the 6,969 meta-data files I'm using OpenSea's IPFS gateway, but you could use ipfs.io or something else.


Use curl to download ten files at once. Downloading thousands of files quickly can lead to duplicates or errors. But with a little tweaking, you should be able to get everything (and dupes are fine for our purposes).
Now that you have everything in one place, grep for aliens:


The numbers are the file names that contain “alien mpher” and thus the aliens' ids.
The entire process takes under ten minutes. This technique works on many NFTs currently minting.

In practice, manually minting at the right time to get the alien is difficult, especially when tokens mint quickly. Then write a bot to poll totalSupply() every second and submit the mint transaction at the exact right time.

You could even look for the token you need in the mempool before it is minted, and get your mint into the same block!

However, in my experience, the “big” approach wins 95% of the time—but not 100%.
“Am I being set up all along?”

Is a question you might ask yourself if you're new to this.
It's disheartening to think you had no chance of minting anything that someone else wanted.
But, did you have no opportunity? You had an equal chance as everyone else!
Take me, for instance: I figured this out using open-source tools and free public information. Anyone can do this, and not understanding how a contract works before minting will lead to much worse issues.

The mpher mint was fair.

While a fair game, “snipe the alien” may not have been everyone's cup of tea.
People may have had more fun playing the “mint lottery” where tokens were distributed at random and no one could gain an advantage over someone simply clicking the “mint” button.

How might we proceed?
Minting For Fashion Hats Punks, I wanted to create a random minting experience without sacrificing fairness. In my opinion, a predictable mint beats an unfair one. Above all, participants must be equal.

Sadly, the most common method of creating a random experience—the post-mint “reveal”—is deeply unfair. It works as follows:

  • During the mint, token metadata is unavailable. Instead, tokenURI() returns a blank JSON file for each id.
  • An IPFS hash is updated once all tokens are minted.
  • You can't tell how the contract owner chose which token ids got which metadata, so it appears random.

Because they alone decide who gets what, the person setting the metadata clearly has a huge unfair advantage over the people minting. Unlike the mpher mint, you have no chance of winning here.
But what if it's a well-known, trusted, doxxed dev team? Are reveals okay here?
No! No one should be trusted with such power. Even if someone isn't consciously trying to cheat, they have unconscious biases. They might also make a mistake and not realize it until it's too late, for example.

You should also not trust yourself. Imagine doing a reveal, thinking you did it correctly (nothing is 100%! ), and getting the rarest NFT. Isn't that a tad odd Do you think you deserve it? An NFT developer like myself would hate to be in this situation.

Reveals are bad*

UNLESS they are done without trust, meaning everyone can verify their fairness without relying on the developers (which you should never do).
An on-chain reveal powered by randomness that is verifiably outside of anyone's control is the most common way to achieve a trustless reveal (e.g., through Chainlink).

Tubby Cats did an excellent job on this reveal, and I highly recommend their contract and launch reflections. Their reveal was also cool because it was progressive—you didn't have to wait until the end of the mint to find out.

In his post-launch reflections, @DefiLlama stated that he made the contract as trustless as possible, removing as much trust as possible from the team.

In my opinion, everyone should know the rules of the game and trust that they will not be changed mid-stream, while trust minimization is critical because smart contracts were designed to reduce trust (and it makes it impossible to hack even if the team is compromised). This was a huge mistake because it limited our flexibility and our ability to correct mistakes.

And @DefiLlama is a superstar developer. Imagine how much stress maximizing trustlessness will cause you!

That leaves me with a bad solution that works in 99 percent of cases and is much easier to implement: random token assignments.

Introducing ERC721R: A fully compliant IERC721 implementation that picks token ids at random.

ERC721R implements the opposite of a reveal: we mint token ids randomly and assign metadata deterministically.
This allows us to reveal all metadata prior to minting while reducing snipe chances.
Then import the contract and use this code:

What is ERC721R and how does it work

First, a disclaimer: ERC721R isn't truly random. In this sense, it creates the same “game” as the mpher situation, where minters compete to exploit the mint. However, ERC721R is a much more difficult game.
To game ERC721R, you need to be able to predict a hash value using these inputs:

This is impossible for a normal person because it requires knowledge of the block timestamp of your mint, which you do not have.

To do this, a miner must set the timestamp to a value in the future, and whatever they do is dependent on the previous block's hash, which expires in about ten seconds when the next block is mined.

This pseudo-randomness is “good enough,” but if big money is involved, it will be gamed. Of course, the system it replaces—predictable minting—can be manipulated.
The token id is chosen in a clever implementation of the Fisher–Yates shuffle algorithm that I copied from CryptoPhunksV2.

Consider first the naive solution: (a 10,000 item collection is assumed):

  1. Make an array with 0–9999.
  2. To create a token, pick a random item from the array and use that as the token's id.
  3. Remove that value from the array and shorten it by one so that every index corresponds to an available token id.

This works, but it uses too much gas because changing an array's length and storing a large array of non-zero values is expensive.

How do we avoid them both? What if we started with a cheap 10,000-zero array? Let's assign an id to each index in that array.

Assume we pick index #6500 at random—#6500 is our token id, and we replace the 0 with a 1.

But what if we chose #6500 again? A 1 would indicate #6500 was taken, but then what? We can't just "roll again" because gas will be unpredictable and high, especially later mints.

This allows us to pick a token id 100% of the time without having to keep a separate list. Here's how it works:

  1. Make a 10,000 0 array.
  2. Create a 10,000 uint numAvailableTokens.
  3. Pick a number between 0 and numAvailableTokens. -1
  4. Think of #6500—look at index #6500. If it's 0, the next token id is #6500. If not, the value at index #6500 is your next token id (weird!)
  5. Examine the array's last value, numAvailableTokens — 1. If it's 0, move the value at #6500 to the end of the array (#9999 if it's the first token). If the array's last value is not zero, update index #6500 to store it.
  6. numAvailableTokens is decreased by 1.
  7. Repeat 3–6 for the next token id.

So there you go! The array stays the same size, but we can choose an available id reliably. The Solidity code is as follows:


GitHub url

Unfortunately, this algorithm uses more gas than the leading sequential mint solution, ERC721A.

This is most noticeable when minting multiple tokens in one transaction—a 10 token mint on ERC721R costs 5x more than on ERC721A. That said, ERC721A has been optimized much further than ERC721R so there is probably room for improvement.

Conclusion

Listed below are your options:

  • ERC721A: Minters pay lower gas but must spend time and energy devising and executing a competitive minting strategy or be comfortable with worse minting results.
  • ERC721R: Higher gas, but the easy minting strategy of just clicking the button is optimal in all but the most extreme cases. If miners game ERC721R it’s the worst of both worlds: higher gas and a ton of work to compete.
  • ERC721A + standard reveal: Low gas, but not verifiably fair. Please do not do this!
  • ERC721A + trustless reveal: The best solution if done correctly, highly-challenging for dev, potential for difficult-to-correct errors.

Did I miss something? Comment or tweet me @dumbnamenumbers.
Check out the code on GitHub to learn more! Pull requests are welcome—I'm sure I've missed many gas-saving opportunities.

Thanks!

Read the original post here

Ash Parrish

Ash Parrish

3 years ago

Sonic Prime and indie games on Netflix

Netflix will stream Spiritfarer, Raji: An Ancient Epic, and Lucky Luna.

Netflix's Geeked Week brought a slew of announcements. The flurry of reveals for The Sandman, The Umbrella Academy season 3, One Piece, and more also included game and game-adjacent announcements.

Netflix released a teaser for Cuphead season 2 ahead of its August premiere, featuring more of Grey DeLisle's Ms. Chalice. DOTA: Dragon's Blood season 3 hits Netflix in August. Tekken, the fighting game that throws kids off cliffs, gets an anime, Tekken: Bloodline.

Netflix debuted a clip of Sonic Prime before Sonic Origins in June and Sonic Frontiers in 2022.

Castlevania: Nocturne will follow Richter Belmont.

Netflix is reviving licensed games with titles based on its shows. There's a Queen's Gambit chess game, a Shadow and Bone RPG, a La Casa de Papel heist adventure, and a Too Hot to Handle game where a pregnant woman must choose between stabbing her cheating ex or forgiving him.

Riot's rhythm platformer Hextech Mayhem debuted on Netflix last year, and now Netflix is adding games from Devolver Digital. Reigns: Three Kingdoms is a card game that lets players choose the fate of Three Kingdoms-era China by swiping left or right on cards. Spiritfarer, the "cozy game about death" from 2020, and Raji: An Ancient Epic are coming to Netflix. Poinpy, a vertical climber from the creator of Downwell, is now on Netflix.

Desta: The Memories Between is a turn-based strategy game set in dreams and memories.

Snowman's Lucky Luna will also be added soon.

With these games, Netflix is expanding beyond dinky mobile games — it plans to have 50 by the end of the year — and could be a serious platform for indies that want to expand into mobile. It takes gaming seriously.

Jenn Leach

Jenn Leach

3 years ago

In November, I made an effort to pitch 10 brands per day. Here's what I discovered.

Photo by Nubelson Fernandes on Unsplash

I pitched 10 brands per workday for a total of 200.

How did I do?

It was difficult.

I've never pitched so much.

What did this challenge teach me?

  • the superiority of quality over quantity

  • When you need help, outsource

  • Don't disregard burnout in order to complete a challenge because it exists.

First, pitching brands for brand deals requires quality. Find firms that align with your brand to expose to your audience.

If you associate with any company, you'll lose audience loyalty. I didn't lose sight of that, but I couldn't resist finishing the task.

Outsourcing.

Delegating work to teammates is effective.

I wish I'd done it.

Three people can pitch 200 companies a month significantly faster than one.

One person does research, one to two do outreach, and one to two do follow-up and negotiating.

Simple.

In 2022, I'll outsource everything.

Burnout.

I felt this, so I slowed down at the end of the month.

Thanksgiving week in November was slow.

I was buying and decorating for Christmas. First time putting up outdoor holiday lights was fun.

Much was happening.

I'm not perfect.

I'm being honest.

The Outcomes

Less than 50 brands pitched.

Result: A deal with 3 brands.

I hoped for 4 brands with reaching out to 200 companies, so three with under 50 is wonderful.

That’s a 6% conversion rate!

Whoo-hoo!

I needed 2%.

Here's a screenshot from one of the deals I booked.

These companies fit my company well. Each campaign is different, but I've booked $2,450 in brand work with a couple of pending transactions for December and January.

$2,450 in brand work booked!

How did I do? You tell me.

Is this something you’d try yourself?