What is Blockchain?

Go: open a terminal in a folder you can write to, e.g. cd ~/Desktop (Windows: cd %USERPROFILE%\Desktop).

Do: save this as chain.py. It prints nothing on its own. Notice the one idea in it: each block records the hash of the block before it.

import hashlib, json

def block_hash(block):
    return hashlib.sha256(json.dumps(block, sort_keys=True).encode()).hexdigest()

def make_chain(payments):
    chain, previous = [], "0" * 64
    for height, payment in enumerate(payments):
        block = {"height": height, "payment": payment, "previous": previous, "nonce": 0}
        block["hash"] = block_hash({k: v for k, v in block.items() if k != "hash"})
        previous = block["hash"]
        chain.append(block)
    return chain

def check(chain):
    previous = "0" * 64
    for block in chain:
        body = {k: v for k, v in block.items() if k != "hash"}
        if block_hash(body) != block["hash"]:
            return f"block {block['height']} has been altered"
        if block["previous"] != previous:
            return f"block {block['height']} does not follow block {block['height'] - 1}"
        previous = block["hash"]
    return "the whole chain verifies"

You should see: nothing. python3 chain.py should return you to the prompt silently, which means the file parses.

If not: IndentationError means the loop bodies lost their indent. If json is reported as undefined, the import line is missing its second name — it imports both hashlib and json.

Go: same folder.

Do: save this as build.py and run python3 build.py.

from chain import make_chain, check

chain = make_chain(["alice pays bob 5", "bob pays carol 2", "carol pays dan 1"])
for block in chain:
    print(f"block {block['height']}: {block['payment']:20s} hash {block['hash'][:16]}...")
print()
print(check(chain))

You should see: three linked blocks and a clean verdict:

block 0: alice pays bob 5     hash 4aab0bb5b82fb69f...
block 1: bob pays carol 2     hash 61355760bafc404b...
block 2: carol pays dan 1     hash e25a4b5a1f463847...

the whole chain verifies

Those hashes are not identifiers assigned by anybody — they are computed from the block’s contents. Change a single character of a payment and the hash becomes an entirely different number, which is the property the rest of this rests on.

If not: ModuleNotFoundError: No module named 'chain' means the two files are not in the same folder. Different hashes from those above mean a payment string differs; they must match character for character, including the spaces.

Go: same folder. Give yourself 500 coins instead of 5.

Do: save this as tamper.py and run it.

from chain import make_chain, check

chain = make_chain(["alice pays bob 5", "bob pays carol 2", "carol pays dan 1"])
chain[0]["payment"] = "alice pays bob 500"        # edit the history
print("after editing block 0:", check(chain))

You should see: the edit named immediately:

after editing block 0: block 0 has been altered

The block’s stored hash no longer matches its contents. Nobody needed a backup, a copy of the original, or a trusted record to notice — the block carries its own evidence.

If not: if it still says the chain verifies, the edit was applied to a copy — chain[0]["payment"] = ... must modify the list you then check.

Go: same folder. Obvious next move: recompute the altered block’s hash so it matches again.

Do: save this as cover.py and run it.

from chain import make_chain, check, block_hash

chain = make_chain(["alice pays bob 5", "bob pays carol 2", "carol pays dan 1"])
chain[0]["payment"] = "alice pays bob 500"
body = {k: v for k, v in chain[0].items() if k != "hash"}
chain[0]["hash"] = block_hash(body)               # patch up the block's own hash
print("after fixing block 0's hash:", check(chain))

You should see: the failure move one block along:

after fixing block 0's hash: block 1 does not follow block 0

This is the chain in blockchain. Block 1 recorded block 0’s original hash, so repairing block 0 breaks its link to block 1 — and repairing block 1 would break block 2, and so on to the end. One edit forces you to rebuild every block that came after it. With no cost attached to rebuilding, that would still be easy; the next step is where the cost comes from.

If not: if it now reports the whole chain verifies, the body dictionary still includes the old hash key — the comprehension must exclude it.

Go: same folder. Proof of work is a rule that a block only counts if its hash starts with a set number of zeros — which can only be found by guessing.

Do: save this as mine.py and run it. It gets slower with each line; that is the point.

import hashlib, time

def mine(data, zeros):
    start, nonce = time.time(), 0
    while True:
        h = hashlib.sha256(f"{data}{nonce}".encode()).hexdigest()
        if h.startswith("0" * zeros):
            return nonce, h, time.time() - start
        nonce += 1

for zeros in (1, 2, 3, 4, 5):
    nonce, h, secs = mine("block 0", zeros)
    print(f"{zeros} leading zero(s): {nonce:>9,} tries, {secs:>6.2f}s, hash {h[:20]}...")

You should see: each extra zero costing roughly sixteen times more work. The try counts are fixed; the times depend on your machine:

1 leading zero(s):        73 tries,   0.00s, hash 0bed7867503a98f7ef05...
2 leading zero(s):       148 tries,   0.00s, hash 0031abb64cd75b389a18...
3 leading zero(s):     3,209 tries,   0.00s, hash 0003497b0012eab4f7af...
4 leading zero(s):   113,511 tries,   0.08s, hash 0000258fb704df77bebb...
5 leading zero(s):   645,488 tries,   0.39s, hash 000005b0ed7319bf3102...

Now put steps 4 and 5 together. Changing one old payment means re-mining that block and every block after it, at this cost per block — while the honest network keeps adding new blocks you also have to catch up on. Bitcoin’s real difficulty is vastly beyond five zeros, which is what turns “you would have to redo the work” into “nobody can”. Immutability is not a promise; it is a bill.

If not: if the 5-zero line takes minutes, your machine is simply slower — the try count should still be 645,488, because the search is deterministic. If it never finishes, the nonce += 1 line is outside the loop.