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#103348: "Better algorithm in order to get less repeats"
#103348: "Better algorithm in order to get less repeats"
這是關於哪方面的案件?
發生什麼事? 請從下方選擇
建議:依我所見,有些調整將大幅增進遊戲完成度
發生什麼事? 請從下方選擇
建議:依我所見,有些調整將大幅增進遊戲完成度
請檢查是否已有同課題案件
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# | Status | Votes | Game | Type | Title | Last update |
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細節描述
• 如果可以的話,請轉貼螢幕顯示的錯誤訊息。
As discussed here: boardgamearena.com/forum/viewtopic.php?t=27740&start=30
Possible solutions:
a) Algorithm 1 (13 distinct cards out of 110 are chosen randomly and for each chosen card 1 word out of 5 is chosen randomly)
or
b) Algorithm 2 (13 distinct words out of 550 are chosen randomly)
or
c) Give players the opportunity to choose 1 out of 5 words (as in the physical game)• 請說明你當時想做什麼,你做了什麼,然後發生了什麼事
• 你的瀏覽器是什麼?
Mozilla v5
• 請轉貼未翻譯的英文字句。 建議將此錯誤的螢幕截圖上傳到 Imgur.com 並轉貼連結。
As discussed here: boardgamearena.com/forum/viewtopic.php?t=27740&start=30
Possible solutions:
a) Algorithm 1 (13 distinct cards out of 110 are chosen randomly and for each chosen card 1 word out of 5 is chosen randomly)
or
b) Algorithm 2 (13 distinct words out of 550 are chosen randomly)
or
c) Give players the opportunity to choose 1 out of 5 words (as in the physical game)• 這些文字存在於 翻譯系統 中嗎?若為真,其是否已被翻譯超過 24 小時?
• 你的瀏覽器是什麼?
Mozilla v5
• 請簡明精確地解釋你的建議,以便讓人一目了然。
As discussed here: boardgamearena.com/forum/viewtopic.php?t=27740&start=30
Possible solutions:
a) Algorithm 1 (13 distinct cards out of 110 are chosen randomly and for each chosen card 1 word out of 5 is chosen randomly)
or
b) Algorithm 2 (13 distinct words out of 550 are chosen randomly)
or
c) Give players the opportunity to choose 1 out of 5 words (as in the physical game)• 你的瀏覽器是什麼?
Mozilla v5
• 當你不能動作時,螢幕上顯示什麼?(螢幕全黑?部份遊戲介面?錯誤訊息?)
As discussed here: boardgamearena.com/forum/viewtopic.php?t=27740&start=30
Possible solutions:
a) Algorithm 1 (13 distinct cards out of 110 are chosen randomly and for each chosen card 1 word out of 5 is chosen randomly)
or
b) Algorithm 2 (13 distinct words out of 550 are chosen randomly)
or
c) Give players the opportunity to choose 1 out of 5 words (as in the physical game)• 你的瀏覽器是什麼?
Mozilla v5
• 遊戲規則的哪部分在 BGA 版本有所錯漏?
As discussed here: boardgamearena.com/forum/viewtopic.php?t=27740&start=30
Possible solutions:
a) Algorithm 1 (13 distinct cards out of 110 are chosen randomly and for each chosen card 1 word out of 5 is chosen randomly)
or
b) Algorithm 2 (13 distinct words out of 550 are chosen randomly)
or
c) Give players the opportunity to choose 1 out of 5 words (as in the physical game)• 這項違反規則之處可否在遊戲重播中看到?若可以是在哪步?(重播時左上角資訊)
• 你的瀏覽器是什麼?
Mozilla v5
• 你當時想採取哪個遊戲行動?
As discussed here: boardgamearena.com/forum/viewtopic.php?t=27740&start=30
Possible solutions:
a) Algorithm 1 (13 distinct cards out of 110 are chosen randomly and for each chosen card 1 word out of 5 is chosen randomly)
or
b) Algorithm 2 (13 distinct words out of 550 are chosen randomly)
or
c) Give players the opportunity to choose 1 out of 5 words (as in the physical game)• 你想觸發這個遊戲行動時做了什麼?
• 當你試著這麼做時發生了什麼(錯誤訊息、遊戲狀態條訊息...)?
• 你的瀏覽器是什麼?
Mozilla v5
• 問題發生在遊戲的哪一步?(當前遊戲指示是什麼)
As discussed here: boardgamearena.com/forum/viewtopic.php?t=27740&start=30
Possible solutions:
a) Algorithm 1 (13 distinct cards out of 110 are chosen randomly and for each chosen card 1 word out of 5 is chosen randomly)
or
b) Algorithm 2 (13 distinct words out of 550 are chosen randomly)
or
c) Give players the opportunity to choose 1 out of 5 words (as in the physical game)• 當你試著進行遊戲動作時發生了什麼(錯誤訊息、遊戲狀態條訊息...)?
• 你的瀏覽器是什麼?
Mozilla v5
• 請描述顯示問題。 建議將此錯誤的螢幕截圖上傳到 Imgur.com 並轉貼連結。
As discussed here: boardgamearena.com/forum/viewtopic.php?t=27740&start=30
Possible solutions:
a) Algorithm 1 (13 distinct cards out of 110 are chosen randomly and for each chosen card 1 word out of 5 is chosen randomly)
or
b) Algorithm 2 (13 distinct words out of 550 are chosen randomly)
or
c) Give players the opportunity to choose 1 out of 5 words (as in the physical game)• 你的瀏覽器是什麼?
Mozilla v5
• 請轉貼未翻譯的英文字句。 建議將此錯誤的螢幕截圖上傳到 Imgur.com 並轉貼連結。
As discussed here: boardgamearena.com/forum/viewtopic.php?t=27740&start=30
Possible solutions:
a) Algorithm 1 (13 distinct cards out of 110 are chosen randomly and for each chosen card 1 word out of 5 is chosen randomly)
or
b) Algorithm 2 (13 distinct words out of 550 are chosen randomly)
or
c) Give players the opportunity to choose 1 out of 5 words (as in the physical game)• 這些文字存在於 翻譯系統 中嗎?若為真,其是否已被翻譯超過 24 小時?
• 你的瀏覽器是什麼?
Mozilla v5
• 請簡明精確地解釋你的建議,以便讓人一目了然。
As discussed here: boardgamearena.com/forum/viewtopic.php?t=27740&start=30
Possible solutions:
a) Algorithm 1 (13 distinct cards out of 110 are chosen randomly and for each chosen card 1 word out of 5 is chosen randomly)
or
b) Algorithm 2 (13 distinct words out of 550 are chosen randomly)
or
c) Give players the opportunity to choose 1 out of 5 words (as in the physical game)• 你的瀏覽器是什麼?
Mozilla v5
案件沿革
2023年11月 7日 11:28 • KuWizard • 開發者需要更詳細的建議內容:
As I already mentioned in the thread mentioned, if there's any developer who would like to invest their time - I'm ready to share the code and lead to the correct direction.
Sorry for that but I'm not ready to invest my time in Just One anymore
Sorry for that but I'm not ready to invest my time in Just One anymore
2023年11月 7日 13:28 • Mr Karotte • 開發者需要更詳細的建議內容:
I'd be game to include changes. Still wondering whether it might not be easier to just add a couple hundred extra words.
2023年11月 7日 14:23 • KuWizard • 開發者需要更詳細的建議內容:
@Mr Karotte, It will be easier technically of course. But the publisher is against it.
2023年11月 7日 14:31 • Dorian Gray2 • 開發者需要更詳細的建議內容:
KuWizard: Like I Poster in the forum:
So for example, if $wordID is 1, it would take the first word for every 13 card?
That would be like 5 options for the card index combinations.
I guess the fix would be to use a different word index for every of the 13 cards. that would be like 5^13 = 1.220.703.125 options.
So, you would have to recalculate the wordID for each card, instead of computing it once on game start. Seems to be a pretty trivial change, isn't it?
If you would share code I could take a look at that.
So for example, if $wordID is 1, it would take the first word for every 13 card?
That would be like 5 options for the card index combinations.
I guess the fix would be to use a different word index for every of the 13 cards. that would be like 5^13 = 1.220.703.125 options.
So, you would have to recalculate the wordID for each card, instead of computing it once on game start. Seems to be a pretty trivial change, isn't it?
If you would share code I could take a look at that.
2023年11月 7日 15:26 • KuWizard • 開發者需要更詳細的建議內容:
@Dorian Gray2, this is a trivial change indeed however I just don't get how double shuffling gets a better chance of getting a new word. I already shuffle "cards" so it's randomised. Randomising a number to get a word (choosing a random number from 0-4) looks like a unnecessary complication. But if you have a proof that this does increase randomness - I'll add this, no problem.
2023年11月 7日 15:46 • Dorian Gray2 • 開發者需要更詳細的建議內容:
Well, each wod should have the same likelihood. Im just talking about the number of game variation. It would still be good to make that change, as that would be closer to a real game behaviour.
I don't have an idea why this algo would be biased towards specific words. I guess that either your card shuffling or the index selection is flawed. Are you using random_int(4) for that wordID? And how do you shuffle cards?
I don't have an idea why this algo would be biased towards specific words. I guess that either your card shuffling or the index selection is flawed. Are you using random_int(4) for that wordID? And how do you shuffle cards?
2023年11月 7日 15:54 • KuWizard • 開發者需要更詳細的建議內容:
en.doc.boardgamearena.com/Main_game_logic:_yourgamename.game.php#Dice_and_bga_rand
We are forced to use bga_rand() and I'm using it.
However for cardIds I create an array from 0 to 549, use PHP shuffle() method and take first 13 or first 550 depending on a game option.
We are forced to use bga_rand() and I'm using it.
However for cardIds I create an array from 0 to 549, use PHP shuffle() method and take first 13 or first 550 depending on a game option.
2023年11月 8日 13:00 • Dorian Gray2 • 開發者需要更詳細的建議內容:
KuWizard: Proof for less repetitions is as follows:
Consider two different games.
Assume you drew [b]two[/b] equal [b]cards[/b] in both games.
The propability that [b]both[/b] cards have the same word (two repetitions) is:
- 20 % in your variant (you have to select the same index in the second game as in the first, and both words will repeat)
- 4 % in the variant with indices per card (as you select a different index for each card).
If you have 3 same cards, it's always 20% in your variant, and 0.8% in the index-per-card-variant.
I guess that would just feel more random to players. If you have one repetition, you don't notice it. If you have [b]two[/b] repetitions, you might notice it.
So, could you pretty please change it? :)
Consider two different games.
Assume you drew [b]two[/b] equal [b]cards[/b] in both games.
The propability that [b]both[/b] cards have the same word (two repetitions) is:
- 20 % in your variant (you have to select the same index in the second game as in the first, and both words will repeat)
- 4 % in the variant with indices per card (as you select a different index for each card).
If you have 3 same cards, it's always 20% in your variant, and 0.8% in the index-per-card-variant.
I guess that would just feel more random to players. If you have one repetition, you don't notice it. If you have [b]two[/b] repetitions, you might notice it.
So, could you pretty please change it? :)
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- 問題是否發生了好幾次?每次都發生?時好時壞?
- 建議將此錯誤的螢幕截圖上傳到 Imgur.com 並轉貼連結。
更改案件狀態為:
bug?id=103348