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更新距离与时间的计算关系（根据跳一跳源码计算） #841

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Feb 2, 2018
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更新距离与时间的计算关系（根据跳一跳源码计算） #841

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a9f5802
更新距离与时间的计算关系（根据源码总结）
baolw Jan 3, 2018
0341f96
调整格式、防止 dict 异常、更新版本号
binderclip Jan 3, 2018
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1 change: 1 addition & 0 deletions config/1920x1080/config.json
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Original file line number Diff line number Diff line change
Expand Up @@ -3,6 +3,7 @@
"press_coefficient": 1.392,
"piece_base_height_1_2": 20,
"piece_body_width": 70,
"head_diameter": 60,
"swipe": {
"x1": 500,
"y1": 1600,
Expand Down
35 changes: 20 additions & 15 deletions wechat_jump_auto.py
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Original file line number Diff line number Diff line change
Expand Up @@ -16,13 +16,15 @@
最后：根据两点的坐标算距离乘以系数来获取长按时间（似乎可以直接用 X 轴距离）
"""
from __future__ import print_function, division

import math
import os
import random
import sys
import time
import math
import random
from PIL import Image
from six.moves import input

try:
from common import debug, config, screenshot
except Exception as ex:
Expand All @@ -31,13 +33,11 @@
print('请检查项目根目录中的 common 文件夹是否存在')
exit(-1)


VERSION = "1.1.1"
VERSION = "1.2.1"

# DEBUG 开关，需要调试的时候请改为 True，不需要调试的时候为 False
DEBUG_SWITCH = False


# Magic Number，不设置可能无法正常执行，请根据具体截图从上到下按需
# 设置，设置保存在 config 文件夹中
config = config.open_accordant_config()
Expand All @@ -48,6 +48,8 @@
piece_base_height_1_2 = config['piece_base_height_1_2']
# 棋子的宽度，比截图中量到的稍微大一点比较安全，可能要调节
piece_body_width = config['piece_body_width']
# 图形中圆球的直径，可以利用系统自带画图工具，用直线测量像素，如果可以实现自动识别圆球直径，那么此处将可实现全自动。
head_diameter = config.get('head_diameter', 60)


def set_button_position(im):
Expand All @@ -58,17 +60,20 @@ def set_button_position(im):
w, h = im.size
left = int(w / 2)
top = int(1584 * (h / 1920.0))
left = int(random.uniform(left-50, left+50))
top = int(random.uniform(top-10, top+10)) # 随机防 ban
left = int(random.uniform(left - 50, left + 50))
top = int(random.uniform(top - 10, top + 10)) # 随机防 ban
swipe_x1, swipe_y1, swipe_x2, swipe_y2 = left, top, left, top


def jump(distance):
"""
跳跃一定的距离
"""
press_time = distance * press_coefficient
press_time = max(press_time, 200) # 设置 200ms 是最小的按压时间
# 计算程序长度与截图测得的距离的比例
scale = 0.945 * 2 / head_diameter
actual_distance = distance * scale * (math.sqrt(6) / 2)
press_time = (-945 + math.sqrt(945 ** 2 + 4 * 105 * 36 * actual_distance)) / (2 * 105) * 1000
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@chenjiandongx chenjiandongx Jan 3, 2018 •
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这是写死了吗，数字固定的？945 这个就是根据源码得来的？

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@chenjiandongx e.BOTTLE = {
headRadius: .945,
bodyWidth: 2.34,
bodyDepth: 2.34,
bodyHeight: 3.2,
reduction: .005,
minScale: .5,
velocityYIncrement: 15,
velocityY: 135,
velocityZIncrement: 70
})

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下面的计算公式其实是在解一个一元二次方程

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其实不是写死的，只是我把源码略掉了math.min的情况，源码对速度做了限制

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@baolw 可以贴一下源码里时间到速度的计算公式吗？

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@baolw baolw Jan 3, 2018 •
edited by binderclip
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var i = (Date.now() - t.mouseDownTime) / 1e3;
t.bottle.velocity.vz = Math.min(i * d.BOTTLE.velocityZIncrement, 150), t.bottle.velocity.vz = +t.bottle.velocity.vz.toFixed(2), t.bottle.velocity.vy = Math.min(d.BOTTLE.velocityY + i * d.BOTTLE.velocityYIncrement, 180), t.bottle.velocity.vy = +t.bottle.velocity.vy.toFixed(2), t.direction = new o.Vector2(t.nextBlock.obj.position.x - t.bottle.obj.position.x, t.nextBlock.obj.position.z - t.bottle.obj.position.z), t.direction.x = +t.direction.x.toFixed(2), t.direction.y = +t.direction.y.toFixed(2);

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@baolw 能问下二次方程代表的什么吗？谢谢！

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这个公式应该也不是很精确，不知道有没有考虑台子压缩高度，我解出的解析解带虚数，不好用js解，但是可以用线性近似解 t=(1/28)d，js中测试过可以确保无限combo

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是的，确实没考虑

press_time = max(press_time, 200) # 设置 200ms 是最小的按压时间
press_time = int(press_time)
cmd = 'adb shell input swipe {x1} {y1} {x2} {y2} {duration}'.format(
x1=swipe_x1,
Expand Down Expand Up @@ -97,7 +102,7 @@ def find_piece_and_board(im):
scan_start_y = 0 # 扫描的起始 y 坐标
im_pixel = im.load()
# 以 50px 步长，尝试探测 scan_start_y
for i in range(int(h / 3), int(h*2 / 3), 50):
for i in range(int(h / 3), int(h * 2 / 3), 50):
last_pixel = im_pixel[0, i]
for j in range(1, w):
pixel = im_pixel[j, i]
Expand Down Expand Up @@ -129,7 +134,7 @@ def find_piece_and_board(im):
piece_y = piece_y_max - piece_base_height_1_2 # 上移棋子底盘高度的一半

# 限制棋盘扫描的横坐标，避免音符 bug
if piece_x < w/2:
if piece_x < w / 2:
board_x_start = piece_x
board_x_end = w
else:
Expand Down Expand Up @@ -162,19 +167,19 @@ def find_piece_and_board(im):
# 从上顶点往下 +274 的位置开始向上找颜色与上顶点一样的点，为下顶点
# 该方法对所有纯色平面和部分非纯色平面有效，对高尔夫草坪面、木纹桌面、
# 药瓶和非菱形的碟机（好像是）会判断错误
for k in range(i+274, i, -1): # 274 取开局时最大的方块的上下顶点距离
for k in range(i + 274, i, -1): # 274 取开局时最大的方块的上下顶点距离
pixel = im_pixel[board_x, k]
if abs(pixel[0] - last_pixel[0]) \
+ abs(pixel[1] - last_pixel[1]) \
+ abs(pixel[2] - last_pixel[2]) < 10:
break
board_y = int((i+k) / 2)
board_y = int((i + k) / 2)

# 如果上一跳命中中间，则下个目标中心会出现 r245 g245 b245 的点，利用这个
# 属性弥补上一段代码可能存在的判断错误
# 若上一跳由于某种原因没有跳到正中间，而下一跳恰好有无法正确识别花纹，则有
# 可能游戏失败，由于花纹面积通常比较大，失败概率较低
for j in range(i, i+200):
for j in range(i, i + 200):
pixel = im_pixel[board_x, j]
if abs(pixel[0] - 245) + abs(pixel[1] - 245) + abs(pixel[2] - 245) == 0:
board_y = j + 10
Expand All @@ -191,7 +196,7 @@ def yes_or_no(prompt, true_value='y', false_value='n', default=True):
"""
default_value = true_value if default else false_value
prompt = '{} {}/{} [{}]: '.format(prompt, true_value,
false_value, default_value)
false_value, default_value)
i = input(prompt)
if not i:
return default
Expand Down