1 00:00:00,360 --> 00:00:06,450 Hello and welcome to this lecture, where we are going to generate a new population. 2 00:00:06,930 --> 00:00:16,170 Let's recap all the process we create is the initial population and then evaluate it's in this part 3 00:00:16,170 --> 00:00:17,670 of the implementation. 4 00:00:18,030 --> 00:00:27,960 We are not considering this stopping there yet, but now we have all these functions here, select various 5 00:00:27,960 --> 00:00:29,940 crossover mutation. 6 00:00:30,270 --> 00:00:39,090 Then, after generating a new population, which is the application of crossover and mutation, we need 7 00:00:39,100 --> 00:00:42,120 to evaluate the population again. 8 00:00:42,480 --> 00:00:53,460 But now we will evaluate the new population that will be generated using crossover and mutation, and 9 00:00:53,460 --> 00:00:57,240 that's what we are going to implement in this lecture. 10 00:00:57,570 --> 00:01:04,050 Moving back to our Google collab file, I will open a new set of codes. 11 00:01:04,440 --> 00:01:11,490 Let's create a new variable new population and am the least. 12 00:01:12,000 --> 00:01:23,350 Let's also define the mutational probability equals zero point zero one, which means one percent. 13 00:01:23,760 --> 00:01:38,280 And let's implement a far loop for new individuals in range from zero to genetic algorithm population 14 00:01:38,280 --> 00:01:39,060 size. 15 00:01:39,630 --> 00:01:44,490 Let's bring out the new individuals variable. 16 00:01:45,060 --> 00:01:55,500 We can run this code, see that we are going through position zero two position nine thin, which means 17 00:01:55,500 --> 00:02:05,440 this size of our population and to generate a new population, we are going to use the crossover function. 18 00:02:05,790 --> 00:02:09,210 We can take a look at the differential here. 19 00:02:09,540 --> 00:02:18,180 It belongs to the individual Class C that we are returning to Childs. 20 00:02:18,450 --> 00:02:26,580 So when we run crossover only once there is old is two new individuals. 21 00:02:27,060 --> 00:02:33,930 So if we run these codes that way, it is defined in the here. 22 00:02:34,290 --> 00:02:46,380 So after running this far loop, here we will and we've for the individuals because in each one of these 23 00:02:46,380 --> 00:02:51,480 runs here, the result will be two individuals. 24 00:02:51,840 --> 00:02:59,280 So we will increase the size of the population birds when we work with genetic algorithms. 25 00:02:59,610 --> 00:03:07,650 The most common is to keep this same size so we can put an additional parameter here. 26 00:03:08,160 --> 00:03:15,630 Number two, it means that we are not going to run these codes one by one. 27 00:03:15,990 --> 00:03:22,200 Now we will run two by two zero two four six eight. 28 00:03:22,200 --> 00:03:23,340 And so one. 29 00:03:23,790 --> 00:03:27,030 Now we need to select two parents. 30 00:03:28,050 --> 00:03:31,230 We will create the new variable variants. 31 00:03:31,230 --> 00:03:37,350 One equals genetic algorithm, select parents. 32 00:03:37,800 --> 00:03:48,740 Let's send the here this sum, and I will also create parents to genetic algorithm, select parents. 33 00:03:48,750 --> 00:03:50,910 And how will PWDs here year? 34 00:03:50,910 --> 00:03:57,780 Also this so we can just go up to the class, the genetic algorithm. 35 00:03:58,050 --> 00:04:05,460 And I will put this Branston comments just to be easier to visualize with the results. 36 00:04:05,880 --> 00:04:10,140 Let's create again the genetic algorithm glass. 37 00:04:10,930 --> 00:04:17,820 I will go back to these barred to here where we are creating the population. 38 00:04:18,060 --> 00:04:20,850 We can run all this code again. 39 00:04:21,570 --> 00:04:25,830 Here we are, set the the best in the video. 40 00:04:26,730 --> 00:04:33,540 We can see that the new some value is one hundred and seventy. 41 00:04:33,870 --> 00:04:37,440 And now let's go back to this part here. 42 00:04:37,740 --> 00:04:43,650 We can friends' parents one and parents. 43 00:04:43,650 --> 00:04:44,130 Sure. 44 00:04:44,550 --> 00:04:50,790 And I will put at the beginning this school months choose Keep the line. 45 00:04:51,480 --> 00:04:59,860 For example, in the first section, we will combine individual zero with individual. 46 00:04:59,980 --> 00:05:07,060 Almost zero as the result is the same, we will combine the same individuals. 47 00:05:07,300 --> 00:05:09,910 So the result will be the same. 48 00:05:10,270 --> 00:05:15,010 If you want, you can implement an additional all the here. 49 00:05:15,280 --> 00:05:22,150 If you don't want to return, this same parents in this section is a crucial. 50 00:05:22,510 --> 00:05:31,180 We will combine 11 with three five nine five two seven nine. 51 00:05:31,180 --> 00:05:41,770 And so one, as we saw in the theoretical lecture, the probability of selecting the best individuals 52 00:05:42,130 --> 00:05:49,870 are higher than the probability of selecting the worst individuals to hear at the end. 53 00:05:50,260 --> 00:06:01,300 Because of that, you can see that we are selecting the individuals that are at the first positions 54 00:06:01,300 --> 00:06:02,200 of the list. 55 00:06:02,620 --> 00:06:07,840 The worst individual that was selected was 11. 56 00:06:08,110 --> 00:06:16,030 We can take a look at the value here, individual 11 and here is the solution. 57 00:06:16,330 --> 00:06:23,140 But in none of these equations we select, it's these bad solutions. 58 00:06:23,440 --> 00:06:29,320 For example, 14 15 where this score equals one. 59 00:06:29,620 --> 00:06:36,460 So these solutions are bad because we cannot load the products on the truck. 60 00:06:36,790 --> 00:06:41,560 This is the first step to generate a new population. 61 00:06:41,860 --> 00:06:44,620 We need to select the parents. 62 00:06:45,010 --> 00:06:53,860 Now let's create the children equals genetic algorithm population. 63 00:06:54,310 --> 00:07:06,050 Let's select parents one and we are going to call the cross over functional and we'll sans population. 64 00:07:06,850 --> 00:07:07,480 Parents. 65 00:07:08,860 --> 00:07:10,690 In this saccone's equation. 66 00:07:11,020 --> 00:07:15,700 We are combining 11 with three in the third. 67 00:07:15,910 --> 00:07:17,860 Five with four nine. 68 00:07:17,860 --> 00:07:21,580 And so one, we can bring the results. 69 00:07:22,880 --> 00:07:36,680 Brent's genetic algorithm, but population parents, one but chromosome we can all brands, population, 70 00:07:37,070 --> 00:07:43,280 parents, true, but chromosome we can run. 71 00:07:43,280 --> 00:07:47,360 This code suggests you see the partial results. 72 00:07:47,660 --> 00:07:50,030 Now we have different values. 73 00:07:50,450 --> 00:07:59,300 We selected Position Zero, which is the best solution because it is in the first position of the list 74 00:07:59,660 --> 00:08:04,580 and we are combining with solution number five. 75 00:08:05,210 --> 00:08:16,460 Then we select then, which is the these chromosome here and we are combining with nine these chromosome 76 00:08:16,460 --> 00:08:16,910 here. 77 00:08:17,180 --> 00:08:23,840 And we keep combining the chromosomes using crossover after dads. 78 00:08:24,050 --> 00:08:28,030 We can see the children that were generated. 79 00:08:28,580 --> 00:08:45,380 Brian's children was these zero dots chromosome and also children in position one dots chromosome. 80 00:08:45,860 --> 00:08:48,170 Let's run this code again. 81 00:08:48,440 --> 00:08:51,110 We will get different values. 82 00:08:51,590 --> 00:08:58,460 We are combining individual number three with individual number one. 83 00:08:59,330 --> 00:09:07,010 These two lines represent the chromosome of the regional parents. 84 00:09:07,340 --> 00:09:10,340 And these are their two lines. 85 00:09:10,340 --> 00:09:13,610 Each represents the combination. 86 00:09:13,820 --> 00:09:17,390 We are applying the cross over function. 87 00:09:17,900 --> 00:09:23,360 And now we need to access the new population list. 88 00:09:23,870 --> 00:09:25,190 Let's set bands. 89 00:09:25,190 --> 00:09:30,320 The new individuals, children position zero. 90 00:09:30,590 --> 00:09:37,430 And then we can apply the mutation in order to change some of the genes. 91 00:09:37,910 --> 00:09:53,060 Let's send the here mutation probability new population dots, bands, children position one again mutation, 92 00:09:53,060 --> 00:09:57,050 and let's send the here mutation probability. 93 00:09:57,410 --> 00:09:59,660 Let's run this code again. 94 00:10:00,110 --> 00:10:05,150 We can see the massagers related to the mutation. 95 00:10:05,690 --> 00:10:10,670 For example, let's review this first execution. 96 00:10:10,970 --> 00:10:16,760 We are selecting solution number three of the original population. 97 00:10:17,240 --> 00:10:27,830 It is this chromosome here and solution our individual number nine, which is the first chromosome here. 98 00:10:28,290 --> 00:10:31,190 Then we apply crossover. 99 00:10:31,760 --> 00:10:37,460 And as a result, we have these two combinations here. 100 00:10:37,700 --> 00:10:41,540 We are combining three with nine. 101 00:10:41,960 --> 00:10:47,660 Then we are applying mutation with one percent probability. 102 00:10:47,930 --> 00:10:50,600 We have a before and after. 103 00:10:51,080 --> 00:10:55,370 We can see that it was a curated for our children. 104 00:10:55,370 --> 00:11:00,350 Number one, the original value was zero. 105 00:11:00,590 --> 00:11:10,520 And then we change its H1 and then you can see all the results here, according to the individuals that 106 00:11:10,520 --> 00:11:11,810 were selected. 107 00:11:12,230 --> 00:11:16,520 And this is how you can create a new generation. 108 00:11:17,120 --> 00:11:27,500 However, that codes is creating only one new generation, and we need to create several new generations. 109 00:11:27,890 --> 00:11:38,600 So in the next lecture, we are going to start working with this stop being reburial so we can define 110 00:11:38,600 --> 00:11:48,170 the number of generations that we may use in order to have the final result of the genetic algorithm. 111 00:11:48,620 --> 00:11:49,340 So there?