Three hands of 17. Captured cards return to play. Can you predict a winner—or recognize a future that repeats?
Higher rank wins. A=1. Suits resolve ties: ♠ < ♥ < ♦ < ♣.
The account stays whole
Ⅰ stays outside the contest.Square sum of hand counts: 867.
Forgiveness puts one top card from each live hand into permanent escrow. At most 25 such interventions are possible. The eventual survivor can collect the escrow; if no hand remains, the result is a draw.
The observed continuation
Turn labels use the stored numerical heights; they do not decide card contests.
- The exact order within each hand is part of the state.
Rules you can use at a table
Set Ⅰ aside. Deal the other 51 cards round robin. Every nonempty hand reveals its first card. Larger rank wins, with A=1, J=11, Q=12, K=13. Equal ranks use the suit order above. Put the winning card on the bottom first, then other revealed cards in clockwise player order (1 → 2 → 3 → 1). Empty players are out.
The ordinary game has no choices after the shuffle. The holder of Ⅰ chooses whether an exact repeat remains a certified loop or triggers forgiveness. Recognizing a repeat requires the full ordered hands or a ledger. Hand counts alone are insufficient.
Put a number on the table. This exact arrangement decodes to the stored string 14.134725141735.
The order carries the record
S=♠, H=♥, D=♦, C=♣. Read left to right, then down. Card 1 is first; the null 2♣ must be last. This transport keeps its own fixed key even when the game tests an ace marker.
The recovered value
Permutation rank:
The agreed decimal scale is 10¹². The source identifies this record as a numerical zero ordinate. War sees the card order; it need not recognize that significance. Decoding does not calculate or certify a new zero.
What work was supplied?
At each of 51 positions, count the unused reference cards before that card. Multiply by the next factorial and add. The reference order is A,2,…,K, each in ♠,♥,♦,♣ order, with 2♣ omitted.
decoded value = R / 10¹²
This record's integer uses 44 binary digits. A whole active-card order has 51! possibilities, or about 219.88 bits of capacity. These measure different things.
Exact binary string and cost boundary
The direct decoder uses 51 position lookups. A simple linear-search implementation takes at most 1,326 card comparisons, plus factorial arithmetic. It does not enumerate 51! arrangements. This is one algorithm's cost, not a minimum running-time theorem. Computing the original zero is separate work; the card convention transports its recorded value.
A tail you can actually certify
An exact repeated state certifies every future move of this fixed card game. If the same three ordered hands occur twice and the rule remains unchanged, the continuation repeats. The full state supplies the certificate.
The published original example finishes after 144 turns. Swapping two cards before the same shuffle produces a 520-turn cycle: state 1015 equals state 1535. The inventory stays the same. The continuation changes.
Forgiveness is an explicit additional rule
When a state repeats, put one top card from every live hand under Ⅰ, permanently outside future contests. Then start a fresh recurrence ledger. Each intervention removes at least two circulating cards, so there can be at most 25. Every intervening finite deterministic phase finishes or repeats. Therefore this added rule guarantees eventual termination, as a survivor or a draw. It does not guarantee a short game.
| 128 tested starts | Finishes | Exact loops |
|---|---|---|
| Hold Ⅰ | 15 | 113 |
| Forgive every repeat | 128 | 0 |
The forgiveness runs took 55–28,834 turns and 0–24 interventions. These are results for the supplied starts and rule, not estimated probabilities for every shuffle. Withholding forgiveness preserves a detected loop; it does not make every starting deal loop.
The connection to RH
The paper studies what a measurement preserves about an underlying state. It gives ambient on-line and off-line points with exactly the same scalar weight. Here the same card inventory accommodates different integers and continuations. The reader can inspect the missing arrangement directly.
The card model's tail certificates rely on a complete finite state and its known transition. A corresponding claim about the zeta-zero tail would need a proved relation supplying those ingredients for the zeros. The game illustrates that requirement and proves the stated card results. It does not prove RH or an obstruction to proving RH.
What Ⅰ does repeatedly
A shared convention lets Ⅰ mark the boundary and preserve an account: cards in hands + escrow = 51; including Ⅰ gives 52 physical cards. Before any escrow, two surviving counts a+b=51 give 2(a²+b²)=51²+(a−b)². Assign Ⅰ physically to either hand and the difference changes by ±1. After escrow begins, include the bank in the account.
In the controlled ace/two test, each null gave 6 finishes and 58 cycles from 64 matched starts. Seven durations changed. Neither face had a termination advantage in that sample.
Game theory and the reader's prediction
Generous tit-for-tat concerns cooperative choices in the repeated Prisoner's Dilemma; its payoff setting differs from these forced card contests. Our termination argument comes from the explicit decreasing-circulation rule. The reader's choice and prediction are visible inputs, not an unmeasured claim about human cooperation. Related primary research.
Lay out the cards, move the pair, and check the recovered number. That verification is now in your hands. The statistics above come from the model; no reader-response statistic has been inferred from how the demonstration feels.
Read the v7.2 paper · Printable field guide · Printable card set · Zenodo 10.5281/zenodo.22851517 · Open the current Invariant shell