CN111185015B - Method for optimizing ten-player online competitive game matching mechanism - Google Patents

Method for optimizing ten-player online competitive game matching mechanism Download PDF

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CN111185015B
CN111185015B CN201911304653.8A CN201911304653A CN111185015B CN 111185015 B CN111185015 B CN 111185015B CN 201911304653 A CN201911304653 A CN 201911304653A CN 111185015 B CN111185015 B CN 111185015B
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CN111185015A (en
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万国春
米健
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Tongji University
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    • AHUMAN NECESSITIES
    • A63SPORTS; GAMES; AMUSEMENTS
    • A63FCARD, BOARD, OR ROULETTE GAMES; INDOOR GAMES USING SMALL MOVING PLAYING BODIES; VIDEO GAMES; GAMES NOT OTHERWISE PROVIDED FOR
    • A63F13/00Video games, i.e. games using an electronically generated display having two or more dimensions
    • A63F13/70Game security or game management aspects
    • A63F13/79Game security or game management aspects involving player-related data, e.g. identities, accounts, preferences or play histories
    • A63F13/795Game security or game management aspects involving player-related data, e.g. identities, accounts, preferences or play histories for finding other players; for building a team; for providing a buddy list
    • AHUMAN NECESSITIES
    • A63SPORTS; GAMES; AMUSEMENTS
    • A63FCARD, BOARD, OR ROULETTE GAMES; INDOOR GAMES USING SMALL MOVING PLAYING BODIES; VIDEO GAMES; GAMES NOT OTHERWISE PROVIDED FOR
    • A63F13/00Video games, i.e. games using an electronically generated display having two or more dimensions
    • A63F13/45Controlling the progress of the video game
    • AHUMAN NECESSITIES
    • A63SPORTS; GAMES; AMUSEMENTS
    • A63FCARD, BOARD, OR ROULETTE GAMES; INDOOR GAMES USING SMALL MOVING PLAYING BODIES; VIDEO GAMES; GAMES NOT OTHERWISE PROVIDED FOR
    • A63F13/00Video games, i.e. games using an electronically generated display having two or more dimensions
    • A63F13/70Game security or game management aspects
    • A63F13/79Game security or game management aspects involving player-related data, e.g. identities, accounts, preferences or play histories
    • A63F13/798Game security or game management aspects involving player-related data, e.g. identities, accounts, preferences or play histories for assessing skills or for ranking players, e.g. for generating a hall of fame
    • AHUMAN NECESSITIES
    • A63SPORTS; GAMES; AMUSEMENTS
    • A63FCARD, BOARD, OR ROULETTE GAMES; INDOOR GAMES USING SMALL MOVING PLAYING BODIES; VIDEO GAMES; GAMES NOT OTHERWISE PROVIDED FOR
    • A63F2300/00Features of games using an electronically generated display having two or more dimensions, e.g. on a television screen, showing representations related to the game
    • A63F2300/50Features of games using an electronically generated display having two or more dimensions, e.g. on a television screen, showing representations related to the game characterized by details of game servers
    • A63F2300/55Details of game data or player data management
    • A63F2300/5546Details of game data or player data management using player registration data, e.g. identification, account, preferences, game history
    • A63F2300/5566Details of game data or player data management using player registration data, e.g. identification, account, preferences, game history by matching opponents or finding partners to build a team, e.g. by skill level, geographical area, background, play style

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Abstract

本发明提供了一种用于优化十人在线竞技游戏匹配机制的方法,涉及游戏匹配机制优化技术。该方法采集5秒至60秒匹配时间之间进入的每名玩家最近30场的游戏信息,通过1‑D卷积、最大值函数、随机抽样、函数映射、算术匹配等算法,使得一方玩家的平均水平处于一个设定的区间中,并且最大限度的降低一方玩家的职业选择发生冲突的概率。使用本发明的方法,一方面增加了玩家的游戏体验,充分利用了玩家有限的游戏时间,实现了游戏服务于娱乐的目的。另一方面,对于本发明的运用,促进了十人在线竞技游戏在游戏市场的发展和壮大,对于游戏产业有一定促进意义,从更深一层来说,对于人类社会的和谐发展也有积极意义。

Figure 201911304653

The invention provides a method for optimizing the matching mechanism of a ten-player online competitive game, and relates to the optimization technology of the game matching mechanism. This method collects the last 30 game information of each player who has entered between 5 seconds and 60 seconds of matching time, and uses algorithms such as 1‑D convolution, maximum function, random sampling, function mapping, and arithmetic matching to make one player's The average level is in a set interval, and the probability of conflict between one player's career choice is minimized. Using the method of the present invention, on the one hand, the game experience of the player is increased, the limited game time of the player is fully utilized, and the purpose of the game serving entertainment is realized. On the other hand, the application of the present invention promotes the development and growth of the ten-player online competitive game in the game market, which has certain promotion significance for the game industry, and also has positive significance for the harmonious development of human society at a deeper level.

Figure 201911304653

Description

用于优化十人在线竞技游戏匹配机制的方法A method for optimizing the matching mechanism of a ten-player online competitive game

技术领域technical field

本发明涉及游戏匹配机制优化领域。The invention relates to the field of game matching mechanism optimization.

技术背景technical background

随着生活节奏的加快,快节奏的在线竞技游戏已经成为人们特别是青少年日常生活的一部分,适当的游戏有益身心健康。然而,目前的匹配机制不完善,存在着双方玩家的游戏水平经常差距较大和玩家争抢角色职业的问题,造成很不好的对抗体验,背离了游戏服务于生活的目的,同时也阻碍着游戏产业前进的步伐。With the accelerated pace of life, fast-paced online competitive games have become a part of people's daily lives, especially teenagers. Appropriate games are beneficial to physical and mental health. However, the current matching mechanism is not perfect, and there is a problem that the game level of the players on both sides is often quite different and players compete for character occupations, resulting in a very bad confrontation experience, which deviates from the purpose of the game serving life, and also hinders the game. The pace of advancement of the industry.

当下,在线竞技游戏发展前景良好,人群基数必定会随着时间的推移而迅速增加。因此,如何优化匹配机制,解决双方玩家实力差距较大的现象,降低玩家角色职业选择冲突的概率,提高玩家的游戏体验成为游戏开发商急需解决的问题。At present, the development prospects of online competitive games are good, and the crowd base will definitely increase rapidly with the passage of time. Therefore, how to optimize the matching mechanism, solve the phenomenon of the large gap in the strength of the players on both sides, reduce the probability of conflict between the player's role and career choice, and improve the player's game experience has become an urgent problem for game developers to solve.

发明内容SUMMARY OF THE INVENTION

本发明的目的在于一种用于优化十人在线竞技游戏匹配机制的算法方法,可有效地提升玩家的对抗体验,增加竞技游戏的趣味性,促进游戏产业的发展。The purpose of the present invention is an algorithm method for optimizing the matching mechanism of a ten-player online competitive game, which can effectively improve the player's confrontation experience, increase the fun of the competitive game, and promote the development of the game industry.

为达到以上目的,本发明采用的技术方案是:In order to achieve the above purpose, the technical scheme adopted in the present invention is:

一种用于优化十人在线竞技游戏匹配机制的算法方法,其特征在于,包括以下步骤:An algorithm method for optimizing a ten-player online competitive game matching mechanism, characterized in that it comprises the following steps:

1)玩家游戏信息的采集1) Collection of player game information

采集某个时间进入的每名玩家最近N场的游戏信息,这里游戏信息指两个方面,一是最近N场游戏的评分score,二是最近N场游戏的角色职业选择情况;提供给步骤2);Collect the game information of each player who entered at a certain time in the last N games, where the game information refers to two aspects, one is the score of the last N games, and the other is the character occupation selection of the last N games; provide it to step 2 );

2)评分的1-D卷积2) 1-D convolution for scoring

用1行N列1-D模板分别对每名玩家的N场评分score进行卷积运算,所述模板的系数与每名玩家的N场评分score一一对应,得出与步骤1)中匹配人数相等的卷积运算结果,记卷积结果为G;所有玩家执行同样操作后,步骤1)中每名匹配玩家都有一个G值;提供给步骤3)、步骤6);Convolution operation is performed on each player's N-field rating score with a 1-row N-column 1-D template, the coefficients of the template are in one-to-one correspondence with each player's N-field rating score, and it is drawn that matches in step 1). The result of the convolution operation with the same number of people, denote the result of the convolution as G; after all players perform the same operation, each matching player in step 1) has a G value; provide it to step 3), step 6);

3)评分的区间设定3) The interval setting of the score

从步骤2)中的所有匹配玩家随机抽样得到十名玩家的G值,即抽取十个G值,利用图4模板对该十个G值进行1-D卷积,记卷积结果为A,设定A±20%为评分区间;提供给步骤6);The G values of ten players are randomly sampled from all matching players in step 2), that is, ten G values are extracted, and the ten G values are subjected to 1-D convolution using the template in Figure 4, and the convolution result is denoted as A, Set A±20% as the scoring interval; provide to step 6);

4)角色职业的函数映射4) Function mapping of role occupation

分别对步骤1)中匹配的所有玩家执行同样操作,对步骤1)中每名玩家N场游戏角色职业的选择进行函数映射,映射函数形式为t=E(角色职业),其中E为函数关系,t为映射结果,五类职业中每种职业的t具有唯一性,对相同的t进行统计,得到每个t对应的数量d,即图5t-d表,步骤1)中匹配的所有玩家都有一个各自的图5t-d表;提供给步骤5);Perform the same operation on all players matched in step 1) respectively, and perform function mapping on the choice of each player's role occupation in N games in step 1). The mapping function is in the form of t=E (role occupation), where E is the functional relationship , t is the mapping result, the t of each occupation in the five occupations is unique, and the same t is counted to obtain the number d corresponding to each t, that is, the table t-d in Figure 5, all players matched in step 1) have a respective Figure 5t-d table; provided to step 5);

5)玩家组合的生成5) Generation of player combinations

通过最大值函数得到步骤4)中每名玩家d(d1,d2,d3,d4,d5)值的最大值dmax,其对应的t为000、001、010、011、100中的一个,记为tdmax;算术匹配五个不同的tdmax为一个玩家组合,遍历步骤1)所有匹配玩家,生成五人一组的玩家组合,即得到初步玩家组合,提供给步骤6)筛选;Obtain the maximum value d max of each player's d (d1, d2, d3, d4, d5) value in step 4) through the maximum value function, and its corresponding t is one of 000, 001, 010, 011, and 100, denoted Be t dmax ; Arithmetic matching five different t dmax is a player combination, traverse step 1) all matching players, generate the player combination of a group of five, i.e. obtain preliminary player combination, provide step 6) screening;

6)最终玩家组合的得出6) The final player combination is obtained

将步骤5)得出的初步玩家组合中的每一组玩家的G值记为G1,G2,G3,G4,G5,计算Gave=(G1+G2+G3+G4+G5)/5,若Gave处在步骤3)中区间,那么该组玩家组合即为最终玩家组合。Denote the G value of each group of players in the preliminary player combination obtained in step 5) as G1, G2, G3, G4, G5, and calculate Gave = (G1+G2+G3+G4+G5)/5, if Gave is in the middle interval of step 3), then this group of players is the final player combination.

步骤5)中,所述算数匹配算法实现五个不同tdmax的组合,由以下步骤实现:In step 5), described arithmetic matching algorithm realizes the combination of five different t dmax , is realized by the following steps:

步骤5.2.1,将1)中匹配的所有玩家的tdmax组成一个队列,队列的长度与玩家数量一致,设定一个标志p指向队列的第一名玩家的tdmaxStep 5.2.1, form a queue with the t dmax of all players matched in 1), the length of the queue is the same as the number of players, and set a flag p to point to the t dmax of the first player in the queue.

步骤5.2.2,利用p取出第一名玩家的tdmax,记为t1,保留t1,第一名玩家的tdmax从队列中消失,随后p向后移动一位,指向第二名玩家的tdmaxStep 5.2.2, use p to take out the t dmax of the first player, record it as t1, keep t1, the t dmax of the first player disappears from the queue, and then p moves backward one place, pointing to the t of the second player dmax .

步骤5.2.3,再次利用p取出第二名玩家的tdmax,若t1与第二名玩家的tdmax不同,则保留第二名玩家的tdmax,记为t2,t2与t1形成组合,第二名玩家的tdmax也从队列中消失;否则,将第二名玩家的tdmax退回队列的原位置。将p后移一位。Step 5.2.3, use p again to take out the t dmax of the second player. If t1 is different from the t dmax of the second player, keep the t dmax of the second player and record it as t2. t2 and t1 form a combination. The second player's t dmax is also removed from the queue; otherwise, the second player's t dmax is returned to its original position in the queue. Shift p back one place.

步骤5.2.4,重复步骤5.2.3,得到t3,t4,t5,那么t1,t2,t3,t4,t5对应的5名玩家即为一组玩家组合。Step 5.2.4, repeat step 5.2.3 to get t3, t4, t5, then 5 players corresponding to t1, t2, t3, t4, t5 are a group of players.

步骤5.2.4-1如果p指向队列最后一名玩家的tdmax,即第一次遍历结束,就发现无法组成一组玩家组合,那么将已经取出的玩家、该队列剩余的玩家和下一批进入匹配的玩家重新执行本发明所有流程,不执行步骤5.2.5、步骤5.2.6以及步骤6);如果第一次遍历之后至少生成了一组玩家组合,则继续执行步骤5.2.5、步骤5.2.6以及步骤6)。Step 5.2.4-1 If p points to the t dmax of the last player in the queue, that is, at the end of the first traversal, it is found that a group of players cannot be formed, then the players that have been taken out, the remaining players in the queue and the next batch The players who have entered the match re-execute all the processes of the present invention, and do not execute steps 5.2.5, 5.2.6 and 6); if at least one group of player combinations is generated after the first traversal, continue to execute steps 5.2.5 and 5.2.5. 5.2.6 and step 6).

步骤5.2.5,此时再次将p移动指向队列剩余玩家中第一名玩家的tdmax,重复以上步骤5.2.1至步骤5.2.4步骤即可实现遍历所有玩家这一目标。Step 5.2.5, move p again to point to the t dmax of the first player among the remaining players in the queue, and repeat the above steps 5.2.1 to 5.2.4 to achieve the goal of traversing all players.

步骤5.2.6,当p指向队列最后一名玩家的tdmax时,发现已经无法再得到一组t1,t2,t3,t4,t5组合,那么将已经取出的玩家、该队列剩余的玩家、步骤6)中不满足步骤3)中评分区间的玩家组合和下一批进入匹配的玩家重新执行本发明所有流程。Step 5.2.6, when p points to the t dmax of the last player in the queue, it is found that a group of t1, t2, t3, t4, t5 combinations can no longer be obtained, then the players that have been taken out, the remaining players in the queue, and the steps In 6), the combination of players that do not meet the scoring interval in step 3) and the next batch of players who enter the matching process re-execute all the processes of the present invention.

采用本发明可以有效地减小双方玩家平均游戏水平差距,并且降低了一方玩家角色职业选择发生冲突的概率,提高了双方玩家的游戏体验。本发明意义:一方面,对于玩家个人来说,在有限的时间里更加充分地放松身心,增加了玩家的游戏体验,符合游戏服务于生活的产业宗旨,同时也一定程度上加快了和谐社会的建设步伐。另一方面,多人在线竞技游戏随着时间的推移,其人群基数必定不断扩大。而本发明的使用,促进了该类游戏在游戏市场发展和壮大,间接地推动了整个游戏产业的发展。The invention can effectively reduce the difference between the average game levels of the two players, and reduce the probability of conflict between one player's role choice and occupation, and improve the game experience of the two players. Significance of the present invention: On the one hand, for individual players, they can relax their mind and body more fully in a limited time, increase the player's game experience, conform to the industrial purpose that games serve life, and also speed up the development of a harmonious society to a certain extent. Construction pace. On the other hand, the crowd base of multiplayer online competitive games must continue to expand over time. The use of the present invention promotes the development and expansion of such games in the game market, and indirectly promotes the development of the entire game industry.

附图说明Description of drawings

图1是本发明的操作流程示意图Fig. 1 is the operation flow schematic diagram of the present invention

图2是实施例的1行30列的1-D模板系数图Fig. 2 is a 1-D template coefficient diagram of 1 row and 30 columns according to the embodiment

图3是实施例的1行30列1-D模板示意图3 is a schematic diagram of a 1-D template with 1 row and 30 columns of an embodiment

图4是实施例的1行10列滤波模板FIG. 4 is a filter template of 1 row and 10 columns of an embodiment

图5是实施例的t-d示意图Fig. 5 is the t-d schematic diagram of the embodiment

图6是本发明步骤5)流程示意图Fig. 6 is the present invention step 5) schematic flow chart

具体实施方式Detailed ways

实施例Example

一种用于优化十人在线竞技游戏匹配机制的算法方法,其特征在于,包括以下步骤:An algorithm method for optimizing a ten-player online competitive game matching mechanism, characterized in that it comprises the following steps:

1)玩家游戏信息的采集1) Collection of player game information

采集在某匹配时间之间(本实施例5S-60S)进入的每名玩家最近若干N场(本实施例N设定30场)的游戏信息,这里游戏信息指两个方面,一是最近N场(实施例30场)游戏的评分score,二是最近N场(实施例30场)游戏的角色职业选择情况;提供给步骤2);Collect the game information of each player who entered between a certain matching time (5S-60S in this embodiment) in the most recent N games (30 games in this embodiment), where game information refers to two aspects, one is the most recent N games. The score of the game (30 games in the embodiment), and the second is the role career choice of the most recent N games (30 games in the embodiment); provided to step 2);

2)评分的1-D卷积2) 1-D convolution for scoring

用图3中的1行30列1-D模板分别对每名玩家的30场评分score进行卷积运算,所述模板的系数与每名玩家的30场评分score一一对应,得出与步骤1)中匹配人数相等的卷积运算结果,记卷积结果为G;所有玩家执行同样操作后,步骤1)中每名匹配玩家都有一个G值;提供给步骤3)、步骤6);Use the 1-D template of 1 row and 30 columns in Fig. 3 to carry out convolution operation on each player's 30 field scoring scores respectively, and the coefficients of the template correspond to each player's 30 field scoring scores one-to-one, and the steps are as follows: 1) The result of the convolution operation that matches the number of people is equal, and the convolution result is denoted as G; after all players perform the same operation, each matching player in step 1) has a G value; provide to step 3), step 6);

3)评分的区间设定3) The interval setting of the score

从步骤2)中的所有匹配玩家随机抽样得到十名玩家的G值,即抽取十个G值,利用图4模板对该十个G值进行1-D卷积,记卷积结果为A,设定A±20%为评分区间;提供给步骤6);The G values of ten players are randomly sampled from all matching players in step 2), that is, ten G values are extracted, and the ten G values are subjected to 1-D convolution using the template in Figure 4, and the convolution result is denoted as A, Set A±20% as the scoring interval; provide to step 6);

4)角色职业的函数映射4) Function mapping of role occupation

分别对步骤1)中匹配的所有玩家执行同样操作,对步骤1)中每名玩家三十场游戏角色职业的选择进行函数映射,映射函数形式为t=E(角色职业),其中E为函数关系,t为映射结果,五类职业中每种职业的t具有唯一性,对相同的t进行统计,得到每个t对应的数量d,即图5t-d表,步骤1)中匹配的所有玩家都有一个各自的图5t-d表;提供给步骤5);Perform the same operation on all players matched in step 1) respectively, and perform function mapping on the choice of each player's 30-game character occupation in step 1). The mapping function is in the form of t=E (character occupation), where E is a function relationship, t is the mapping result, the t of each occupation in the five types of occupations is unique, and the same t is counted to obtain the number d corresponding to each t, that is, the table t-d in Figure 5, all matching in step 1) Players have a respective figure 5t-d table; provided to step 5);

5)玩家组合的生成5) Generation of player combinations

通过最大值函数得到步骤4)中每名玩家d(d1,d2,d3,d4,d5)值的最大值dmax,其对应的t为000、001、010、011、100中的一个,记为tdmax;算术匹配五个不同的tdmax为一个玩家组合,遍历步骤1)所有匹配玩家,生成五人一组的玩家组合,即得到初步玩家组合,提供给步骤6)筛选;Obtain the maximum value d max of each player's d (d1, d2, d3, d4, d5) value in step 4) through the maximum value function, and its corresponding t is one of 000, 001, 010, 011, and 100, denoted Be t dmax ; Arithmetic matching five different t dmax is a player combination, traverse step 1) all matching players, generate the player combination of a group of five, i.e. obtain preliminary player combination, provide step 6) screening;

6)最终玩家组合的得出6) The final player combination is obtained

将步骤5)得出的初步玩家组合中的每一组玩家的G值记为G1,G2,G3,G4,G5,计算Gave=(G1+G2+G3+G4+G5)/5,若Gave处在步骤3)中区间,那么该组玩家组合即为最终玩家组合。Denote the G value of each group of players in the preliminary player combination obtained in step 5) as G1, G2, G3, G4, G5, and calculate Gave = (G1+G2+G3+G4+G5)/5, if Gave is in the middle interval of step 3), then this group of players is the final player combination.

各步骤以下详述之Each step is detailed below

步骤2)中,每名玩家评分的卷积运算:图3模板的尺寸应和游戏场次一致,即为30。另外,根据时间上越近的游戏场次与当前游戏水平相关度越大的特点,图3模板系数和步骤1)中的每名玩家的30场评分一一对应指的是图3模板由大到小的系数和时间上由近及远的每名玩家的30场评分一一对应。In step 2), the convolution operation of each player's score: the size of the template in Figure 3 should be consistent with the game session, that is, 30. In addition, according to the feature that the closer the game times in time are more relevant to the current game level, the one-to-one correspondence between the template coefficients in Figure 3 and the 30-game scores of each player in step 1) means that the template in Figure 3 is from large to The small coefficient corresponds to the 30-game score of each player from near to far in time.

具体方法为:The specific method is:

通过图2模板卷积得出的G值应该反映出步骤1)每名匹配玩家当前的游戏水平。而当前游戏水平和时间点越近的评分相关度必然越高,和时间点越远的评分相关度必然越低,这是符合逻辑的。为了实现以上逻辑,将步骤1)中每名玩家的30场评分按照时间点由远及近的顺序记为第1场、第2场......第n场......第30场,与图3模板系数f(1)、f(2)......f(n)......f(30)相对应,f(n)的值由下式确定:The G value obtained by convolution of the template in Figure 2 should reflect the current game level of each matched player in step 1). The closer the current game level and the time point are, the higher the correlation is, and the farther the time point is, the lower the correlation is, which is logical. In order to realize the above logic, the 30 game scores of each player in step 1) are recorded as the 1st game, the 2nd game...the nth game... The 30th field corresponds to the template coefficients f(1), f(2)...f(n)...f(30) in Fig. 3, and the value of f(n) is given by the following formula Sure:

f(n)=(1/2)f(n-1)+(1/4)f(n-2)+(1/8)f(n-3)+......+(1/2n-1)f(1)+1(f(1)=1,n=2,3,4......30),最终确定图2也就是图3中模板系数的值。f(n)=(1/2)f(n-1)+(1/4)f(n-2)+(1/8)f(n-3)+......+(1 /2 n-1 )f(1)+1(f(1)=1,n=2,3,4...30), and finally determine the value of the template coefficient in Fig. 2, that is, Fig. 3.

score(n)表示第n场的评分,则对于步骤1)每名匹配玩家G=(1/(f(1)+f(2)+......+f(30)))*(f(1)*score(1)+f(2)*score(2)+......+f(30)*score(30))。score(n) represents the score of the nth game, then for each matching player in step 1) G=(1/(f(1)+f(2)+...+f(30)))* (f(1)*score(1)+f(2)*score(2)+...+f(30)*score(30)).

步骤3)意义:Step 3) Meaning:

步骤3)中的卷积结果A代表了步骤1)中所有匹配玩家的平均水平,为实现这一目标,可进行如下步骤:The convolution result A in step 3) represents the average level of all matched players in step 1). To achieve this, the following steps can be performed:

步骤3.1,记步骤3)该十名玩家的G值为G0,G1,......G9,图4模板的系数分别为A0,A1,A2......A9;Step 3.1, record step 3) The G values of the ten players are G0, G1,...G9, and the coefficients of the template in Figure 4 are A0, A1, A2...A9;

步骤3.2,G0和A0,G1和A1......G9和A9一一对应;Step 3.2, G0 and A0, G1 and A1...G9 and A9 correspond one by one;

步骤3.3,将图4模板系数归一化:记T=G0+G1+......+G9,Step 3.3, normalize the template coefficients in Figure 4: mark T=G0+G1+...+G9,

通过计算A0=G0/T,A1=G1/T,......A9=G9/T确定图4模板系数;Determine the template coefficient of Figure 4 by calculating A0=G0/T, A1=G1/T, ...... A9=G9/T;

步骤3.4,之后A由下式确定:A=G0*A0+G1*A1+......+G9*A9。Step 3.4, after which A is determined by the following formula: A=G0*A0+G1*A1+...+G9*A9.

步骤4):Step 4):

所述角色职业,可总体分为五类,这与十人在线竞技游戏一方的玩家数量相一致。本实施例是针对十人在线竞技游戏而提出的,玩家分为两个阵营,每个阵营为五人,每个人都有其游戏任务,对应五种不同职业。The character occupations can be generally divided into five categories, which is consistent with the number of players on one side of the ten-player online competitive game. This embodiment is proposed for a ten-player online competitive game. Players are divided into two camps, and each camp has five players, each of whom has a game task corresponding to five different occupations.

步骤4)中,t采取三位二进制编码,五个不同的t分别对应于000、001、010、011、100,对应于d1、d2、d3、d4、d5。In step 4), t adopts a three-bit binary code, and five different t correspond to 000, 001, 010, 011, and 100, respectively, and correspond to d1, d2, d3, d4, and d5.

t采取000 001 010 011 100这五个三位二进制数进行编码,分别和五种角色职业一一对应,因此函数自变量和因变量的对应关系E为单对单。t uses five three-digit binary numbers of 000 001 010 011 100 for coding, which correspond to the five roles and occupations one-to-one, so the corresponding relationship E between the independent variable of the function and the dependent variable is one-to-one.

步骤1)中匹配的所有玩家都各有一张图5所示的匹配自身的t-d表,确切反应最近三十场次玩家对各职业的使用频率,d值越大,代表玩家更青睐于对应职业,d值越小,代表玩家更抵触对应职业。All players matched in step 1) each have a t-d table for matching themselves as shown in Figure 5, which exactly reflects the frequency of players using each occupation in the last 30 games. The larger the d value, the more the player prefers the corresponding occupation. The smaller the d value, the more resistance the player is to the corresponding profession.

步骤5):Step 5):

所述的算术匹配,其特点是,可以从与步骤1)中匹配玩家个数相一致的所有tdmax中匹配五个不同的tdmax作为一组玩家组合且直到剩余玩家无法满足五个不同的tdmax这一条件为止。The arithmetic matching is characterized in that five different t dmax can be matched as a group of player combinations from all t dmax that are consistent with the number of players matched in step 1) until the remaining players cannot satisfy the five different t dmax. t dmax until this condition.

假设现在步骤1)匹配了101名玩家,他们各自dmax对应了各自的tdmax(dmax对应的t记为tdmax),也就是说,共有101个tdmaxSuppose now that step 1) matches 101 players, and their respective d max corresponds to their respective t dmax (the t corresponding to d max is recorded as t dmax ), that is to say, there are 101 t dmax in total,

此时在这101个tdmax中,可能甚至没办法找到五个不同的tdmax组成一组,也有可能找到一组两组三组......但是最多找到二十组,因为必然受进场总人数101限制的。无论如何都会剩余至少一名玩家无法参与组合,之后只需把这些没法组合的玩家和下一轮匹配的玩家再次进行所有流程即可。At this time, in these 101 t dmax , it may not even be possible to find five different t dmax to form a group, or it is possible to find a group of two groups of three groups... But at most twenty groups can be found, because it is bound to be affected by The total number of people entering the venue is limited to 101. In any case, there will be at least one player left who cannot participate in the combination, and then it is only necessary to perform all the processes again with the players who cannot be combined and the players who are matched in the next round.

步骤1)中匹配的玩家数量和步骤5)生成的tdmax的数量一致,该算数匹配算法实现五个不同tdmax的组合,由以下步骤实现:The number of players matched in step 1) is the same as the number of t dmax generated in step 5). This arithmetic matching algorithm realizes the combination of five different t dmax , which is realized by the following steps:

步骤5.2.1,将1)中匹配的所有玩家的tdmax组成一个队列,队列的长度与玩家数量一致,设定一个标志p指向队列的第一名玩家的tdmaxStep 5.2.1, form a queue with the t dmax of all players matched in 1), the length of the queue is the same as the number of players, and set a flag p to point to the t dmax of the first player in the queue.

步骤5.2.2,利用p取出第一名玩家的tdmax,记为t1,保留t1,第一名玩家的tdmax从队列中消失,随后p向后移动一位,指向第二名玩家的tdmaxStep 5.2.2, use p to take out the t dmax of the first player, record it as t1, keep t1, the t dmax of the first player disappears from the queue, and then p moves backward one place, pointing to the t of the second player dmax .

步骤5.2.3,再次利用p取出第二名玩家的tdmax,若t1与第二名玩家的tdmax不同,则保留第二名玩家的tdmax,记为t2,t2与t1形成组合,第二名玩家的tdmax也从队列中消失;否则,将第二名玩家的tdmax退回队列的原位置。将p后移一位。Step 5.2.3, use p again to take out the t dmax of the second player. If t1 is different from the t dmax of the second player, keep the t dmax of the second player and record it as t2. t2 and t1 form a combination. The second player's t dmax is also removed from the queue; otherwise, the second player's t dmax is returned to its original position in the queue. Shift p back one place.

步骤5.2.4,重复步骤5.2.3,得到t3,t4,t5,那么t1,t2,t3,t4,t5对应的5名玩家即为一组玩家组合。Step 5.2.4, repeat step 5.2.3 to get t3, t4, t5, then 5 players corresponding to t1, t2, t3, t4, t5 are a group of players.

步骤5.2.4-1如果p指向队列最后一名玩家的tdmax,即第一次遍历结束,就发现无法组成一组玩家组合,那么将已经取出的玩家、该队列剩余的玩家和下一批进入匹配的玩家重新执行本发明所有流程,不执行步骤5.2.5、步骤5.2.6以及步骤6);如果第一次遍历之后至少生成了一组玩家组合,则继续执行步骤5.2.5、步骤5.2.6以及步骤6)。Step 5.2.4-1 If p points to the t dmax of the last player in the queue, that is, at the end of the first traversal, it is found that a group of players cannot be formed, then the players that have been taken out, the remaining players in the queue and the next batch The players who have entered the match re-execute all the processes of the present invention, and do not execute steps 5.2.5, 5.2.6 and 6); if at least one group of player combinations is generated after the first traversal, continue to execute steps 5.2.5 and 5.2.5. 5.2.6 and step 6).

步骤5.2.5,此时再次将p移动指向队列剩余玩家中第一名玩家的tdmax,重复以上步骤5.2.1至步骤5.2.4步骤即可实现遍历所有玩家这一目标。Step 5.2.5, move p again to point to the t dmax of the first player among the remaining players in the queue, and repeat the above steps 5.2.1 to 5.2.4 to achieve the goal of traversing all players.

步骤5.2.6,当p指向队列最后一名玩家的tdmax时,发现已经无法再得到一组t1,t2,t3,t4,t5组合,那么将已经取出的玩家、该队列剩余的玩家、步骤6)中不满足步骤3)中评分区间的玩家组合和下一批进入匹配的玩家重新执行本发明所有流程。Step 5.2.6, when p points to the t dmax of the last player in the queue, it is found that a group of t1, t2, t3, t4, t5 combinations can no longer be obtained, then the players that have been taken out, the remaining players in the queue, and the steps In 6), the combination of players that do not meet the scoring interval in step 3) and the next batch of players who enter the matching process re-execute all the processes of the present invention.

所述最大值函数形式为dmax=F(d1,d2,d3,d4,d5),作用是得到每名玩家d1,d2,d3,d4,d5中的最大值或者最大值之一。F是函数对应关系,其运算法则由以下步骤组成:The maximum value function is in the form of d max =F(d1, d2, d3, d4, d5), and the function is to obtain the maximum value or one of the maximum values among d1, d2, d3, d4, and d5 of each player. F is the function correspondence, and its algorithm consists of the following steps:

步骤5.1.1,首先dmax=d1,即假定d1是最大值或者最大值之一。Step 5.1.1, first d max = d1, ie d1 is assumed to be the maximum value or one of the maximum values.

步骤5.1.2,将dmax与d2相比较,若d2大于dmax,则令dmax=d2,否则dmax值不变。Step 5.1.2, compare d max with d2 , if d2 is greater than d max , set d max =d2, otherwise the value of d max remains unchanged.

重复步骤步骤5.1.2,dmax依次和d3,d4,d5比较,最终得到d1,d2,d3,d4,d5中的最大值或者最大值之一dmaxStep 5.1.2 is repeated, and d max is compared with d3, d4, and d5 in turn, and finally the maximum value or one of the maximum values d max among d1, d2, d3, d4, and d5 is obtained.

举例某名玩家,他有一份图5所示的t-d表,这份表中有5个d值,记为d1,d2,d3,d4,d5,5个d值中的最大值或者最大值之一记为dmax For example, a player has a td table as shown in Figure 5. There are 5 d values in this table, which are recorded as d1, d2, d3, d4, d5, the maximum or the maximum value among the 5 d values. One is denoted as d max

那应该怎么确定dmax呢?So how should we determine d max ?

假设此玩家的d值分别是4 5 6 6 5,Suppose the player's d values are 4 5 6 6 5, respectively,

假定dmax=4,4相当于d1,dmax和d2比较,即4和5比较,因为5>4,所以更新dmax,让dmax=5;dmax和d3比较,即5和6比较(因为dmax已经更新了),因为6>5,继续更新dmax,dmax=6;dmax和d4比较,即6和6比较,因为6=6,不更新dmax,dmax仍然是6;dmax和d5比较,即6和5比较,因为5<6,不更新dmax,dmax仍然是6;Assuming d max = 4, 4 is equivalent to d1, d max is compared with d2, that is, 4 is compared with 5, because 5>4, so update d max , let d max =5; d max is compared with d3, that is, 5 and 6 are compared (because d max has been updated), because 6>5, continue to update d max , d max =6; compare d max with d4, that is, compare 6 with 6, because 6=6, do not update d max , d max is still 6; d max is compared with d5, that is, 6 is compared with 5, because 5<6, d max is not updated, and d max is still 6;

因此dmax=6,即d3,对应的t为010。Therefore, d max =6, that is, d3, and the corresponding t is 010.

Claims (2)

1.一种用于优化十人在线竞技游戏匹配机制的方法,其特征在于,包括以下步骤:1. a method for optimizing ten online competitive game matching mechanism, is characterized in that, comprises the following steps: 1)玩家游戏信息的采集1) Collection of player game information 采集某个时间进入的每名玩家最近N场的游戏信息,这里游戏信息指两个方面,一是最近N场游戏的评分score,二是最近N场游戏的角色职业选择情况;提供给步骤2);Collect the game information of each player who entered at a certain time in the last N games, where the game information refers to two aspects, one is the score of the last N games, and the other is the character occupation selection of the last N games; provide it to step 2 ); 2)评分的1-D卷积2) 1-D convolution for scoring 用1行N列1-D模板分别对每名玩家的N场评分score进行卷积运算,所述模板的系数与每名玩家的N场评分score一一对应,得出与步骤1)中匹配人数相等的卷积运算结果,记卷积结果为G;所有玩家执行同样操作后,步骤1)中每名匹配玩家都有一个G值;提供给步骤3)、步骤6);Convolution operation is performed on each player's N-field rating score with a 1-row N-column 1-D template, the coefficients of the template are in one-to-one correspondence with each player's N-field rating score, and it is drawn that matches in step 1). The result of the convolution operation with the same number of people, denote the result of the convolution as G; after all players perform the same operation, each matching player in step 1) has a G value; provide it to step 3), step 6); 3)评分的区间设定3) The interval setting of the score 从步骤2)中的所有匹配玩家随机抽样得到十名玩家的G值,即抽取十个G值,利用如下模板The G values of ten players are randomly sampled from all matching players in step 2), that is, ten G values are drawn, and the following template is used A0A0 A1A1 A2A2 A3A3 A4A4 A5A5 A6A6 A7A7 A8A8 A9 A9
对该十个G值进行1-D卷积,记卷积结果为A,设定A+20%为评分区间;提供给步骤6);Perform 1-D convolution on the ten G values, denote the convolution result as A, and set A + 20% as the scoring interval; provide it to step 6); 4)角色职业的函数映射4) Function mapping of role occupation 分别对步骤1)中匹配的所有玩家执行同样操作,对步骤1)中每名玩家N场游戏角色职业的选择进行函数映射,映射函数形式为t=E,E为角色职业,其中E为函数关系,t为映射结果,五类职业中每种职业的t具有唯一性,对相同的t进行统计,得到每个t对应的数量d,即下t-d表,Perform the same operation on all players matched in step 1) respectively, and perform function mapping on the choice of each player's role occupation in N games in step 1), the mapping function is in the form of t=E, E is the role occupation, and E is the function relationship, t is the mapping result, the t of each occupation in the five types of occupations is unique, and the same t is counted to obtain the quantity d corresponding to each t, that is, the following t-d table, tt 000000 001001 010010 011011 100100 dd d1d1 d2d2 d3d3 d4d4 d5 d5
步骤1)中匹配的所有玩家都有一个各自的如下t-d表;All players matched in step 1) have a respective t-d table as follows; tt 000000 001001 010010 011011 100100 dd d1d1 d2d2 d3d3 d4d4 d5 d5
提供给步骤5);provided to step 5); 5)玩家组合的生成5) Generation of player combinations 通过最大值函数得到步骤4)中每名玩家d(d1,d2,d3,d4,d5)值的最大值dmax,其对应的t为000、001、010、011、100中的一个,记为tdmax;算术匹配五个不同的tdmax为一个玩家组合,遍历步骤1)所有匹配玩家,生成五人一组的玩家组合,即得到初步玩家组合,提供给步骤6)筛选;Obtain the maximum value d max of each player's d (d1, d2, d3, d4, d5) value in step 4) through the maximum value function, and its corresponding t is one of 000, 001, 010, 011, and 100, denoted Be t dmax ; Arithmetic matching five different t dmax is a player combination, traverse step 1) all matching players, generate the player combination of a group of five, i.e. obtain preliminary player combination, provide step 6) screening; 6)最终玩家组合的得出6) The final player combination is obtained 将步骤5)得出的初步玩家组合中的每一组玩家的G值记为G1,G2,G3,G4,G5,计算Gave=(G1+G2+G3+G4+G5)/5,若Gave处在步骤3)中区间,那么该玩家组合即为最终玩家组合。Denote the G value of each group of players in the preliminary player combination obtained in step 5) as G1, G2, G3, G4, G5, and calculate Gave = (G1+G2+G3+G4+G5)/5, if Gave is in the interval in step 3), then the player combination is the final player combination.
2.如权利要求1所述的方法,其特征在于,步骤5)中,2. method as claimed in claim 1, is characterized in that, in step 5), 所述算术匹配算法实现五个不同tdmax的组合,由以下步骤实现:The arithmetic matching algorithm realizes the combination of five different t dmax by the following steps: 步骤5.2.1,将1)中匹配的所有玩家的tdmax组成一个队列,队列的长度与玩家数量一致,设定一个标志p指向队列的第一名玩家的tdmaxStep 5.2.1, form a queue with the t dmax of all players matched in 1), the length of the queue is the same as the number of players, and set a flag p to point to the t dmax of the first player in the queue; 步骤5.2.2,利用p取出第一名玩家的tdmax,记为t1,保留t1,第一名玩家的tdmax从队列中消失,随后p向后移动一位,指向第二名玩家的tdmaxStep 5.2.2, use p to take out the t dmax of the first player, record it as t1, keep t1, the t dmax of the first player disappears from the queue, and then p moves backward one place, pointing to the t of the second player dmax ; 步骤5.2.3,再次利用p取出第二名玩家的tdmax,若t1与第二名玩家的tdmax不同,则保留第二名玩家的tdmax,记为t2,t2与t1形成组合,第二名玩家的tdmax也从队列中消失;否则,将第二名玩家的tdmax退回队列的原位置;将p后移一位;Step 5.2.3, use p again to take out the t dmax of the second player. If t1 is different from the t dmax of the second player, keep the t dmax of the second player and record it as t2. t2 and t1 form a combination. The t dmax of the second player also disappears from the queue; otherwise, the t dmax of the second player is returned to the original position of the queue; p is moved back one place; 步骤5.2.4,重复步骤5.2.3,得到t3,t4,t5,那么t1,t2,t3,t4,t5对应的5名玩家即为一组玩家组合;Step 5.2.4, repeat step 5.2.3 to get t3, t4, t5, then 5 players corresponding to t1, t2, t3, t4, t5 are a group of players; 步骤5.2.4-1如果p指向队列最后一名玩家的tdmax,即第一次遍历结束,就发现无法组成一组玩家组合,那么将已经取出的玩家、该队列剩余的玩家和下一批进入匹配的玩家重新执行上述所有步骤,不执行步骤5.2.5、步骤5.2.6以及步骤6);如果第一次遍历之后至少生成了一组玩家组合,则继续执行步骤5.2.5、步骤5.2.6以及步骤6);Step 5.2.4-1 If p points to the t dmax of the last player in the queue, that is, at the end of the first traversal, it is found that a group of players cannot be formed, then the players that have been taken out, the remaining players in the queue and the next batch Players who have entered the match re-execute all the above steps, but do not execute steps 5.2.5, 5.2.6 and 6); if at least one set of player combinations is generated after the first traversal, continue to execute steps 5.2.5 and 5.2 .6 and step 6); 步骤5.2.5,此时再次将p移动指向队列剩余玩家中第一名玩家的tdmax,重复以上步骤5.2.1至步骤5.2.4步骤即可实现遍历所有玩家这一目标;Step 5.2.5, move p again to the t dmax of the first player among the remaining players in the queue, and repeat the above steps 5.2.1 to 5.2.4 to achieve the goal of traversing all players; 步骤5.2.6,当p指向队列最后一名玩家的tdmax时,发现已经无法再得到一组t1,t2,t3,t4,t5组合,那么将已经取出的玩家、该队列剩余的玩家、步骤6)中不满足步骤3)中评分区间的玩家组合和下一批进入匹配的玩家重新执行上述所有步骤。Step 5.2.6, when p points to the t dmax of the last player in the queue, it is found that a group of t1, t2, t3, t4, t5 combinations can no longer be obtained, then the players that have been taken out, the remaining players in the queue, and the steps In 6), the combination of players who do not meet the scoring interval in step 3) and the next batch of players who enter the match will perform all the above steps again.
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