🎰 Issues figuring out how to fix C++ Blackjack program - Stack Overflow

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I have tried using a do-while loop so that it asks for a new card whenever the total < 21 but it ends up the same way. I'm at a loss at how to solve.


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An assignment is discussed in which students must develop a blackjack application in both Excel and Mathematica. Common solutions are considered in relation to general problem-solving strategies, and the The blackjack assignme​nt.


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*C E] In poker, a straight flush (see Exercise C35) beats four of a kind, C Q E In a game of Blackjack (Twenty One), using a single ordinary deck of fifty.


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C programming tutorial 10 - Intro to Problem Solving

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Mark C. Lewis, Lisa Lacher For this whole problem remember that you can treat Strings as sequences. cards, without the dealer getting blackjack (note.


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There is no known algorithm for weird problems like this. It's really an exercise in problem solving, for you. So, how to proceed? 1) Instructors.


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Mark C. Lewis, Lisa Lacher For this whole problem remember that you can treat Strings as sequences. cards, without the dealer getting blackjack (note.


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500 years of NOT teaching THE CUBIC FORMULA. What is it they think you can't handle?

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val = true sets the content of val to true. val==true checks the content of value. They are two very different things! In your while loop and in the.


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An assignment is discussed in which students must develop a blackjack application in both Excel and Mathematica. Common solutions are considered in relation to general problem-solving strategies, and the The blackjack assignme​nt.


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The Martingale Problem

He's given us three integers with , there MAY or may not be a 1 which can be treated as a 1 or And yes, it is quite funny how he says three cards when blackjack is a two card game. Similarly for dealing with aces, take the first card value and div by Then do the sum and check again, storing results in an array of 4. You may need to "read between the lines" of the chalkboard, a bit - they might just be a subtle verbal reference of some kind. What about switch statements? The fact that everybody else does it some other way only means that they are wrong". I hate these "let's have a sack race, but you can't use your legs", kind of problems. Replies: 9 Last Post: , AM. Talk to your fellow students about them. What about using a trigraph, instead of an if statement, is that allowed? Considering aces as 1 or 11, there are at worst eight possible sums, when all three are aces. I can do that, however, i do not know how to do so without crossing the threshold. The problem? However, may i ask how to proceed? You can use two indices, one for results array, and one for cards array, then you can use the result of div'ing the card by 11 to advance results index or not to avoid repeating values. Hey Adak, Oh yes! I really appreciate it Oh btw, branchless aint allowed , i think. These sums which for aces may be 1 or 11 can be calculated in a "branchless" fashion by using an implicit conversion from bool to int and a multiplication. Anyways thanks for the response. Each holds a number from 2 - Or a number say 1 , which can be treated as a 11 or a 1. There is no known algorithm for weird problems like this. Last edited by HeirofRome; at PM. It's really an exercise in problem solving, for you. Hey iMalc, That actually makes a lot of sense. I get what you are saying, make three 'sum' variables considering all the sums and select the one that is largest. How do i know if an integer is a 1 or an 11 without selection. Hey everyone, So i'll come out and say it. After calculating all three sums, many of which will typically be the same unless you did have aces, and then use implicit conversion from bool to int combined with multiplication again, to mathematically "select" the largest one that is not above the threshold. All up this would take 7 lines of code. My homepage Advice: Take only as directed - If symptoms persist, please see your debugger Linus Torvalds: "But it clearly is the only right way. All times are GMT The time now is AM. I did spend some time seething about what the heck the instructor wanted from us but im afraid this is the problem definition, no hins whatsoever. ATTEMPT: Quite simple i tried using math by first adding the three integers and trying various combinations like mod 21, divide by 21 etc. It's difficult to get help with these kinds of problems, because it's like asking someone to play soccer while holding their right foot, with their right hand, all the time - just seems a bit silly. Well, not so much. All rights reserved.{/INSERTKEYS}{/PARAGRAPH} So i do have a program written that can do this perfectly. Thanks HeirofRome. Thanks for the reply guys! Trying to solve both of these at the same time, is a hefty load to carry, sometimes. So yeah, totally do-able. Sounds simple enough right? Look for those. It uses atleast one or two if statements. This is a homework assignment I and didnt understand at all. I am completely bowled over by this problem I would greatly appreciate any help in understanding this algorithm. {PARAGRAPH}{INSERTKEYS}Remember Me? Move to the second card, and so on. So, how to proceed? Replies: 1 Last Post: , PM. BlackJack algorithm - Homework kinda. As in, how do i put the threshold condition in? If the sum is less than 22 it won't get shifted, otherwise you'll be left with zero. Replies: 38 Last Post: , PM. I used chars and just chose 6 as N. And i cant get around it mathematically. Just my 2 cents. When you have the algorithm for that well in hand whether the program is finished or not , then try to solve the part about the restrictions. Thank You. Given those three integers, we have to print out the best hand i. If there is no way we can get 21 or less than 21 with our hand , we print a 0. The catch our brilliant instructor put in , is to do this problem without selection so no ifs, whiles, for, etc , no recursion, no external functions. Reason: Additional Info. Realising that when you have multiple aces that at most one of them can be an 11 or else you exceed 21, and with three aces a total of 3 is pointless, reduces that down to just three sums that are worth considering. Similar Threads Anyone else think it's kinda weird By Babkockdood in forum General Discussions. I think branchless qualifies as a selection statement. Welcome to the forum, however! The deadline has passed, and i could not figure this out, but the problem still bugs me. Thread: BlackJack algorithm - Homework kinda.