合肥生活安徽新聞合肥交通合肥房產生活服務合肥教育合肥招聘合肥旅游文化藝術合肥美食合肥地圖合肥社保合肥醫院企業服務合肥法律

        代做ECM2418、代寫 java,Python 程序設計

        時間:2023-12-10  來源:合肥網hfw.cc  作者:hfw.cc 我要糾錯



        ECM2418 Computer Languages and Representations Continuous Assessment
        Functional and Logic Programming
        Dr David Wakeling
        Handed out Handed in
        Thursday 26th October 2023 (T1:05) Thursday 14th December 2023 (T1:12)
        This Continuous Assessment is worth 40% of the module mark.
        Question 1: Light Show
        Every week, The Sunday Times newspaper publishes a Teaser. Teaser 3172, of Sunday 9th July 2023, was as follows.
        My bedside clock displays the time and date using eight digits; for example, at 9.43am on 28th June, the display would be
        Each digit in the electronic display lights up some (or all) of seven light seg- ments, the above display lighting up a total of 45 segments.
        On one occasion recently, the displayed digits were all different and the total number of lit segments was prime. The same was true exactly one day later. Then, just one minute after the second occasion, the number of lit segments was the average of the numbers of lit segments on those two previous occasions.
        What was that third display?
            1

        Question 1.1
        Show a Haskell function generator1 that returns a list of tuples (HR,MN,DY,MT) that may be solutions to the Teaser. That is, for which, HR, MN, DY and MT are valid hours, minutes, days (assume a non-leap year) and months.
        This function will be assessed by the number of tests that it passes, as counted by the function x_generator1 below. The expected answer is 10.
         x_generator1 :: Int x_generator1 =
        length [t | t <- ts, t ‘elem‘ g] where
         g = ts =
        generator1
        [ ( 2,15,14,11)
        , ( 4,31,27, 9)
        , ( 6,**,10, 8)
         , ( 9, 3,23, 6) , (11,19, 6, 5) , (13,35,19, 3) , (15,51, 2, 2) , (18, 6,16,12)
         , (20 ,22 ,29 ,10) , (22,38,11, 9) ]
        Question 1.2
        Show a Haskell function tester1 that returns true for tuples (HR,MN,DY,MT) that are solutions to the Teaser. That is, for which the tuple is “magic”, a second tuple exactly one day later is also “magic”, and just one minute on from this second tuple the number of lit segments on the display is the average of the number of lit segments of these two tuples. A tuple (HR,MN,DY,MT) is “magic” if the displayed digits of HR, MN, DY and MT are all be different, and the total number of lit segments in the display is prime.
        This function will be assessed by the number of tests that it passes, as counted by the func- tion x_tester1 below. Note that these test cases were NOT produced by generator1. The expected answer is 10.
         x_tester1 :: Int x_tester1 =
        length [t | t <- ts, tester1 t] where
         ts =
        [ ( 6,59,17,24)
        2

         , ( 6,59,17,34)
         , ( 6,59,27,14) , ( 6,59,27,41) , ( 8,59,12,46) , (16,59, 7,24) , (16,59, 7,42)
         , (16,59, 7,43) , (16 ,59 ,27 ,40) , (18,59, 2,46) ]
        Question 1.3
        On blue18.ex.ac.uk, my program
        computes [(16,59,27,4)], from which one can deduce the answer to Teaser 3172 is
        in 0.004 seconds. Tune your program so that on the same machine, it computes this answer within 1.000 seconds.
        (5 marks)
        Question 2: Digital Trio
        Teaser 3158, of Sunday 5th May 2023, was as follows.
        “I have a couple of subtraction problems for you”, George told Martha. Look: N**N2=N3andN3-N4=N5. CanyousolvethemifItellyouthatN1, N3 and N5 are all three-digit whole numbers whose sum is less than 2000, the same three non-zero digits appearing in all three numbers but no digit being repeated within any of those numbers? N2 and N4 are both two-digit whole numbers using two of the three digits mentioned above, and the first digit of N1 is not equal to the first digit of N2.
        What is N1?
         main :: IO () main =
         print (filter tester1 generator1)
         3

        Question 2.1
        Show a Haskell function generator2 that returns a list of tuples (N1,N2,N3,N4,N5) that may be solutions to the Teaser. That is, for which, N1, N3 and N5 are three-digit numbers, and N2 and N4 are two-digit numbers. The same three digits appear in N1, N3 and N5, two of these digits appear in N2, and two of them appear in N4. In each number, no digit is zero and none is repeated. The first digit of N1 is not equal to the first digit on N2.
        This function will be assessed by the number of tests that it passes, as counted by the function x_generator2 below. The expected answer is 10.
         x_generator2 :: Int x_generator2 =
         length [t | t <- ts, t ‘elem‘ g] where
        g = generator2
        ts =
        [ ("123","21","123","12","123")
         , ("**","26","261","12","621") , ("219","19","912","21","291") , ("**9","92","9**","**","239") , ("439","94","394","43","394") , ("549","95","945","95","945")
         , ("568","68","586","56","586")
        , ("769","67","679","97","796")
        , ("879","79","897","98","789")
        , ("987","79","789","79","789") ]
        (10 marks)
        Question 2.2
        Show a Haskell function tester2 that returns true for tuples (N1 , N2 , N3 , N4 , N5 ) that are solutions to the Teaser. That is, for which N1 − N2 = N3, N3 − N4 = N5 and N1 + N3 + N5 < 2000.
        This function will be assessed by the number of tests that it passes, as counted by the function x_tester2 below. The expected answer is 10.
         x_tester2 :: Int x_tester2 =
        length [t | t <- ts, tester2 t] where
         ts =
        [ ("138","01","137","50","87")
        4

         , ("143","01","142","52","**")
         , ("171","02","169","79","**") , ("152","03","149","54","95") , ("159","04","155","61","94") , ("161","05","156","63","93") , ("182","06","176","80","96")
         , ("151","07","144","57","87") , ("165","08","157","64","93") , ("174","09","165","71","94") ]
        (10 marks)
        Question 2.3
        On blue18.ex.ac.uk, my program
        computes [("594","45","549","54","495")] in 0.003 seconds. Tune your program so that on the same machine, it computes this answer within 1.000 seconds.
        (5 marks)
        Question 3: Easier to Ask the Audience
        Teaser 3145, of Sunday 1st January 2023, was as follows.
        “I have forgotten the phone number!” complained Martha, about to phone a friend. “So have I!” replied George, “but I have some vague memories of it. It is a perfect square with all the digits different, and the last digit is equal to the number of digits to be dialled. The last-but-one digit is odd and one of the digits is zero. Also the second and third and last-but-one digits are all exact multiples of the first digit. Maybe you can work it out.”
        Martha proceeded to dial the number correctly.
        What number did she dial?
         main :: IO () main =
         print (filter tester2 generator2)
        5

        Question 3.1
        Show a Prolog predicate generator3 that yields successive numbers N between 1,000 to 1,000,000 (inclusive) that may be solutions to the Teaser. That is, integers N that are perfect squares.
        This predicate will be assessed by the number of tests that it passes, as counted by the predicate x_generator3 below. The expected answer is 10.
         x_generator3( N ) :- x_generator3_loop(
        [ 1024 , 9409 , 23716 , 51529 , 123**4 , 185761 , 868624 , 962361
         , 982081, 1000000 ], 0, N ).
        x_generator3_loop( [], C, C ). x_generator3_loop( [T|TS], C, N ) :-
        generator3( T ),
         C1 is C + 1,
        x_generator3_loop( TS, C1, N ). x_generator3_loop( [_|TS], C, N ) :-
        x_generator3_loop( TS, C, N ).
        (10 marks)
        Question 3.2
        Show a Prolog predicate tester3 that is true for phone numbers N that are solutions to the Teaser. That is, for integers N where all of the digits are different, the last digit is equal to the number of digits, the last-but-one digit is odd and one of the digits is zero. In addition, the second and third and last-but-one digits are all exact multiples of the first digit.
        This predicate will be assessed by the number of tests that it passes, as counted by a predicate x_tester3 below. The expected answer is 10.
         x_tester3( N ) :- x_tester3_loop(
        [ 123056 , 128036 , 13**76 , 142076 , 148056 , 15**76 , 1**096 , 18**36
         , 193056, 198076 ], 0, N ).
        x_tester3_loop( [], C, C ). x_tester3_loop( [T|TS], C, N ) :-
        tester3( T ),
         C1 is C + 1,
        x_tester3_loop( TS, C1, N ).
        6

         x_tester3_loop( [_|TS], C, N ) :-
         x_tester3_loop( TS, C, N ).
        (10 marks)
        Question 3.3
        On Swish Prolog, my program
        computes 1**056 in 0.40 seconds. Tune your program so that on the same system, it computes this answer within 2.00 seconds.
        (5 marks)
        Question 4: Cube Route
        Teaser 3149, of Sunday 29th January 2023, was as follows.
        I have a set of ten cards, each of which has a different digit written on it. All the cards have been used to make a set of prime numbers. After discarding the smallest prime, and without changing the order of any cards, I have placed the remaining primes in order of decreasing size to give a large number. It is possible, without changing the order of any cards, to break this number into a set composed entirely of cubes. Neither set contains a number with more than four digits.
        List, in order of decreasing size, my set of prime numbers.
        Question 4.1
        Show a Prolog predicate generator4 that yields arrangements of the digits 0 to 9 divided into runs of one, two, three or four digits that form prime numbers. Importantly (and somewhat surprisingly) leading zero digits do not count, so “251” is considered to be prime, but “0251” is not.
        This predicate will be assessed by the number of tests that it passes, as counted by a predicate x_generator4 below. The expected answer is 10.
         main :-
        generator3( N ), tester3( N ), write( N ).
        7

         x_generator4( N ) :-
         x_generator4_loop(
        [ [[9 ,6 ,7] ,[4 ,0 ,1] ,[2 ,8 ,3] ,[5]]
        , [[9 ,8 ,3] ,[6 ,0 ,1] ,[5] ,[4 ,7] ,[2]]
        , [[9 ,8 ,3] ,[6 ,7] ,[4 ,2 ,0 ,1] ,[5]]
        , [[9 ,8 ,5 ,1] ,[2] ,[4 ,3] ,[6 ,0 ,7]]
         , [[9 ,8 ,5 ,1] ,[2] ,[3] ,[6 ,0 ,4 ,7]]
        , [[9 ,8 ,5 ,1] ,[2] ,[7] ,[4 ,6 ,0 ,3]]
        , [[8 ,9] ,[7] ,[6 ,0 ,1] ,[2 ,5 ,4 ,3]]
        , [[8 ,9] ,[7] ,[5 ,6 ,3] ,[4 ,0 ,2 ,1]]
        , [[8 ,9] ,[5] ,[4 ,7] ,[6 ,0 ,1] ,[3] ,[2]]
         , [[3],[5],[6,0,7],[2],[4,1],[8,9]] ], 0, N ).
        x_generator4_loop( [], C, C ). x_generator4_loop( [T|TS], C, N ) :-
        generator4( T ),
         C1 is C + 1,
        x_generator4_loop( TS, C1, N ). x_generator4_loop( [_|TS], C, N ) :-
        x_generator4_loop( TS, C, N ).
        (10 marks)
        Question 4.2
        Show a Prolog predicate tester4 that is true for lists of lists of digits that form prime numbers and may be solutions to the Teaser. That is, for collections of prime numbers that after discarding the smallest prime, may be arranged in order of decreasing size to give a large number that may be divided into runs of one, two three or four digits that form cubes.
        This predicate will be assessed by the number of tests that it passes, as counted by the predicate x_tester4 below. The expected answer is 10.
         x_tester4( N ) :- x_tester4_loop(
        [ [[8 ,2 ,7] ,[6 ,1] ,[5 ,3] ,[4 ,0 ,9]] , [[8 ,2 ,7] ,[6 ,1] ,[4 ,0 ,9] ,[5 ,3]]
         , [[8 ,2 ,7] ,[5 ,3] ,[6 ,1] ,[4 ,0 ,9]]
        , [[8 ,2 ,7] ,[4 ,0 ,9] ,[6 ,1] ,[5 ,3]]
        , [[6 ,1] ,[8 ,2 ,7] ,[4 ,0 ,9] ,[5 ,3]]
        , [[6 ,1] ,[4 ,0 ,9] ,[5 ,3] ,[8 ,2 ,7]]
        , [[5 ,3] ,[6 ,1] ,[4 ,0 ,9] ,[8 ,2 ,7]]
         , [[5 ,3] ,[4 ,0 ,9] ,[6 ,1] ,[8 ,2 ,7]]
        , [[4 ,0 ,9] ,[5 ,3] ,[8 ,2 ,7] ,[6 ,1]]
        , [[4,0,9],[8,2,7],[6,1],[5,3]] ], 0, N ).
        8

          x_tester4_loop( [], C, C ). x_tester4_loop( [T|TS], C, N ) :-
        tester4( T ),
        C1 is C + 1,
        x_tester4_loop( TS, C1, N ).
         x_tester4_loop( [_|TS], C, N ) :- x_tester4_loop( TS, C, N ).
        (10 marks)
        Question 4.3
        On Swish Prolog, my program
        computes a first result in in 33.0 seconds, from which one can deduce the answer to the Teaser is 827, 409, 61, 53. Tune your program so that on the same system, it computes this answer within 120.0 seconds.
        (5 marks)
        Submission
        You should submit a single “.zip” file to the ELE system. Other compression formats, such as “.rar”, “.7z”, “.gz” and “.bz2” are unacceptable, and will receive a mark of zero. The “.zip” file should contain four completed text files “Light.hs” (containing the answer to Question 1), “Trio.hs” (containing the answer to Question 2), “Audience.pl” (con- taining the answer to Question 3) and “Cube.pl” (containing the answer to Question 4).
        If there is any question as to whether your functional programs compute the correct result, these questions will be answered on the implementation at
           https://www.tutorialspoint.com/compile_haskell_online.php
        If there is any question as to whether your logic programs compute the correct result, these questions will be answered on the implementation at
           https://swish.swi-prolog
         main :-
        generator4( XS ), tester4( XS ), write( XS ).
        9

        All students are reminded of the University regulations on academic honesty and plagiarism.
        In particular, functions an predicates clearly intended ONLY to pass the given tests will be treated as malpractice (“an attempt to deceive the examiners”).
        請加QQ:99515681 或郵箱:99515681@qq.com   WX:codehelp

        掃一掃在手機打開當前頁
      1. 上一篇:CSC8204 代寫、Secure Software程序
      2. 下一篇:INT305 代做、代寫 Python 語言編程
      3. 無相關信息
        合肥生活資訊

        合肥圖文信息
        挖掘機濾芯提升發動機性能
        挖掘機濾芯提升發動機性能
        戴納斯帝壁掛爐全國售后服務電話24小時官網400(全國服務熱線)
        戴納斯帝壁掛爐全國售后服務電話24小時官網
        菲斯曼壁掛爐全國統一400售后維修服務電話24小時服務熱線
        菲斯曼壁掛爐全國統一400售后維修服務電話2
        美的熱水器售后服務技術咨詢電話全國24小時客服熱線
        美的熱水器售后服務技術咨詢電話全國24小時
        海信羅馬假日洗衣機亮相AWE  復古美學與現代科技完美結合
        海信羅馬假日洗衣機亮相AWE 復古美學與現代
        合肥機場巴士4號線
        合肥機場巴士4號線
        合肥機場巴士3號線
        合肥機場巴士3號線
        合肥機場巴士2號線
        合肥機場巴士2號線
      4. 幣安app官網下載 短信驗證碼

        關于我們 | 打賞支持 | 廣告服務 | 聯系我們 | 網站地圖 | 免責聲明 | 幫助中心 | 友情鏈接 |

        Copyright © 2024 hfw.cc Inc. All Rights Reserved. 合肥網 版權所有
        ICP備06013414號-3 公安備 42010502001045

        主站蜘蛛池模板: 国产福利电影一区二区三区久久老子无码午夜伦不 | 一区二区三区精密机械| 国产精品va无码一区二区| 国精品无码一区二区三区在线 | 亚洲国产一区国产亚洲| 末成年女A∨片一区二区| 亚洲av综合av一区二区三区| 精品久久久久久中文字幕一区| 免费人人潮人人爽一区二区| 在线视频一区二区三区三区不卡 | 亚洲视频一区网站| 成人精品一区久久久久| 精品国产一区二区22| 无码人妻一区二区三区免费看| 精品少妇人妻AV一区二区三区| 一区二区三区免费视频观看| 国产a∨精品一区二区三区不卡 | 国产伦精品一区二区三区免费迷| 亚洲Aⅴ无码一区二区二三区软件| 国产成人无码精品一区二区三区| 无码少妇A片一区二区三区| 国产情侣一区二区三区| 在线成人一区二区| 曰韩精品无码一区二区三区| 亚洲国产AV无码一区二区三区| 久久久久国产一区二区三区| 国产精品福利一区二区久久| 国产无码一区二区在线| 人妻互换精品一区二区| 国产乱码精品一区二区三区中文 | 一区二区手机视频| 久久久久人妻一区二区三区 | 国产亚洲一区二区精品| 韩国精品一区二区三区无码视频 | 手机福利视频一区二区| 久久99热狠狠色精品一区| 美女视频一区二区| 糖心vlog精品一区二区三区| 一本大道东京热无码一区| 美女福利视频一区二区| 最新欧美精品一区二区三区|