Dalam semangat Patch Gambar , inilah tantangan yang sama tetapi dengan teks.

Tantangan

Bit busuk telah mempengaruhi teks berharga Anda! Diberi paragraf yang terdiri dari karakter ASCII, dengan lubang persegi panjang di suatu tempat di dalamnya, program Anda harus mencoba mengisi lubang dengan teks yang sesuai, sehingga paragraf berpadu sebaik mungkin.

Definisi lebih lanjut

• Lubang akan selalu berbentuk persegi panjang, dan dapat menjangkau beberapa garis.
• Hanya akan ada satu lubang.
• Perhatikan bahwa lubang tidak harus jatuh pada batas kata (pada kenyataannya, biasanya tidak akan).
• Lubang tersebut paling banyak 25% dari paragraf input, tetapi mungkin tumpang tindih atau melampaui "akhir" dari teks "normal" (lihat contoh Euclid atau Badger di bawah).
• Karena menemukan lubang bukanlah titik utama dari tantangan ini, itu hanya akan terdiri dari tanda pagar #untuk memudahkan identifikasi.
• Tidak ada lokasi lain di paragraf masukan yang akan memiliki tanda pagar.
• Kode Anda tidak dapat menggunakan teks "normal" dalam contoh di bawah ini - ia hanya akan menerima dan memproses teks dengan lubang di dalamnya.
• Input dapat berupa string multi-baris tunggal, sebagai array string (satu elemen per baris), sebagai file, dll. - pilihan Anda tentang apa yang paling nyaman untuk bahasa Anda.
• Jika diinginkan, input tambahan opsional yang merinci koordinat lubang dapat diambil (misalnya, tupel koordinat atau sejenisnya).
• Tolong jelaskan algoritma Anda dalam kiriman Anda.

Pemungutan suara

Para pemilih diminta untuk menilai entri berdasarkan seberapa baik algoritma mengisi lubang teks. Beberapa saran termasuk yang berikut:

• Apakah area yang diisi sesuai dengan perkiraan ruang dan tanda baca sebagai sisa paragraf?
• Apakah area yang diisi memperkenalkan sintaksis yang salah? (misalnya, dua spasi berturut-turut, periode diikuti oleh tanda tanya, urutan yang keliru seperti , ,, dll.)
• Jika Anda menyipitkan mata (sehingga Anda tidak benar-benar membaca teks), dapatkah Anda melihat di mana lubang itu dulu?
• Jika tidak ada kata CamelCase di luar lubang, apakah lubang itu mengandung kata-kata? Jika tidak ada Huruf Kapital di luar lubang, apakah lubang itu berisi? Jika Ada Banyak Huruf Kapital Di Luar Lubang, apakah lubang itu mengandung jumlah yang proporsional?

Kriteria Validitas

Agar kiriman dianggap sah, ia tidak boleh mengubah teks paragraf di luar lubang (termasuk spasi tambahan). Satu baris tambahan di bagian paling akhir adalah opsional.

Uji Kasus

Format adalah paragraf asli dalam blok kode, diikuti oleh paragraf yang sama dengan sebuah lubang. Paragraf dengan lubang akan digunakan untuk input.

1 (Menambal Gambar)

In a popular image editing software there is a feature, that patches (The term
used in image processing is inpainting as @minxomat pointed out.) a selected
area of an image, based on the information outside of that patch. And it does a
quite good job, considering it is just a program. As a human, you can sometimes
see that something is wrong, but if you squeeze your eyes or just take a short
glance, the patch seems to fill in the gap quite well.

In a popular image editing software there is a feature, that patches (The term
used in image processing is inpainting as @minxomat pointed out.) a selected
area of an image, #############information outside of that patch. And it does a
quite good job, co#############is just a program. As a human, you can sometimes
see that something#############t if you squeeze your eyes or just take a short
glance, the patch seems to fill in the gap quite well.

2 (Alamat Gettysburg)

But, in a larger sense, we can not dedicate, we can not consecrate, we can not
hallow this ground. The brave men, living and dead, who struggled here, have
consecrated it, far above our poor power to add or detract. The world will
little note, nor long remember what we say here, but it can never forget what
they did here. It is for us the living, rather, to be dedicated here to the
unfinished work which they who fought here have thus far so nobly advanced. It
is rather for us to be here dedicated to the great task remaining before us-
that from these honored dead we take increased devotion to that cause for which
they gave the last full measure of devotion-that we here highly resolve that
these dead shall not have died in vain-that this nation, under God, shall have
a new birth of freedom-and that government of the people, by the people, for
the people, shall not perish from the earth.

But, in a larger sense, we can not dedicate, we can not consecrate, we can not
hallow this ground. The brave men, living and dead, who struggled here, have
consecrated it, far above our poor power to add or detract. The world will
little note, nor long remember what we say here, but it can never forget what
they did here. It is for us the living, rather, to be dedicated here to the
unfinished work which they who fought here h######################advanced. It
is rather for us to be here dedicated to the######################before us-
that from these honored dead we take increas######################use for which
they gave the last full measure of devotion-######################solve that
these dead shall not have died in vain-that ######################, shall have
a new birth of freedom-and that government of the people, by the people, for
the people, shall not perish from the earth.

3 (Lorem Ipsum)

Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do
eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim
ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut
aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit
in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur
sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt
mollit anim id est laborum.

Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do
eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim
ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut
aliquip ex ea commodo conse################irure dolor in reprehenderit
in voluptate velit esse cil################giat nulla pariatur. Excepteur
sint occaecat cupidatat non################in culpa qui officia deserunt
mollit anim id est laborum.

4 (Jabberwocky)

'Twas brillig, and the slithy toves
Did gyre and gimble in the wabe;
All mimsy were the borogoves,
And the mome raths outgrabe.

'Twas brillig, and the slithy toves
Did gyre a######### in the wabe;
All mimsy #########borogoves,
And the mome raths outgrabe.

5 (bukti Euclid tentang Teorema Pythagoras)

1.Let ACB be a right-angled triangle with right angle CAB.
2.On each of the sides BC, AB, and CA, squares are drawn,
CBDE, BAGF, and ACIH, in that order. The construction of
squares requires the immediately preceding theorems in Euclid,
and depends upon the parallel postulate. [footnote 14]
3.From A, draw a line parallel to BD and CE. It will
perpendicularly intersect BC and DE at K and L, respectively.
4.Join CF and AD, to form the triangles BCF and BDA.
5.Angles CAB and BAG are both right angles; therefore C, A,
and G are collinear. Similarly for B, A, and H.
6.Angles CBD and FBA are both right angles; therefore angle ABD
equals angle FBC, since both are the sum of a right angle and angle ABC.
7.Since AB is equal to FB and BD is equal to BC, triangle ABD
must be congruent to triangle FBC.
8.Since A-K-L is a straight line, parallel to BD, then rectangle
BDLK has twice the area of triangle ABD because they share the base
BD and have the same altitude BK, i.e., a line normal to their common
base, connecting the parallel lines BD and AL. (lemma 2)
9.Since C is collinear with A and G, square BAGF must be twice in area
to triangle FBC.
10.Therefore, rectangle BDLK must have the same area as square BAGF = AB^2.
11.Similarly, it can be shown that rectangle CKLE must have the same
area as square ACIH = AC^2.
12.Adding these two results, AB^2 + AC^2 = BD × BK + KL × KC
13.Since BD = KL, BD × BK + KL × KC = BD(BK + KC) = BD × BC
14.Therefore, AB^2 + AC^2 = BC^2, since CBDE is a square.

1.Let ACB be a right-angled triangle with right angle CAB.
2.On each of the sides BC, AB, and CA, squares are drawn,
CBDE, BAGF, and ACIH, in that order. The construction of
squares requires the immediately preceding theorems in Euclid,
and depends upon the parallel postulate. [footnote 14]
3.From A, draw a line parallel to BD and CE. It will
perpendicularly intersect BC and DE at K and L, respectively.
4.Join CF and AD, to form the triangles BCF and BDA.
5.Angles CAB and BAG are both right angles; therefore C, A,
and G are #############milarly for B, A, and H.
6.Angles C#############e both right angles; therefore angle ABD
equals ang############# both are the sum of a right angle and angle ABC.
7.Since AB#############FB and BD is equal to BC, triangle ABD
must be co#############iangle FBC.
8.Since A-#############ight line, parallel to BD, then rectangle
BDLK has t############# of triangle ABD because they share the base
BD and hav#############titude BK, i.e., a line normal to their common
base, conn#############rallel lines BD and AL. (lemma 2)
9.Since C #############with A and G, square BAGF must be twice in area
to triangl#############
10.Therefo############# BDLK must have the same area as square BAGF = AB^2.
11.Similar############# shown that rectangle CKLE must have the same
area as square ACIH = AC^2.
12.Adding these two results, AB^2 + AC^2 = BD × BK + KL × KC
13.Since BD = KL, BD × BK + KL × KC = BD(BK + KC) = BD × BC
14.Therefore, AB^2 + AC^2 = BC^2, since CBDE is a square.

6 (Badger, Badger, Badger oleh weebl)

Badger, badger, badger, badger, badger,
badger, badger, badger, badger, badger
Mushroom, mushroom, a-
Badger, badger, badger, badger, badger,
badger, badger, badger, badger, badger
Mushroom, mushroom, a-
Badger, badger, badger, badger, badger,
badger, badger, badger, badger, badger
Mush-mushroom, a
Badger, badger, badger, badger, badger,
badger, badger, badger, badger, badger
Argh! Snake, a snake!
Snaaake! A snaaaake, oooh its a snake!

Badger, badger, badger, badger, badger,
badger, badger, badger, badger, badger
Mushroom, mushroom, a-
Badger##################badger, badger,
badger##################badger, badger
Mushro##################
Badger##################badger, badger,
badger##################badger, badger
Mush-mushroom, a
Badger, badger, badger, badger, badger,
badger, badger, badger, badger, badger
Argh! Snake, a snake!
Snaaake! A snaaaake, oooh its a snake!

Python 2

Saya tahu bahwa @atlasologist sudah memposting solusi di Python 2, tetapi cara kerja saya agak berbeda. Ini bekerja dengan menelusuri semua lubang, dari atas ke bawah, kiri ke kanan, melihat 5 karakter ke belakang dan pada karakter di atas, dan menemukan karakter tempat yang cocok. Jika beberapa karakter ditemukan, itu memilih yang paling umum. Jika tidak ada karakter yang ditemukan, ini akan menghilangkan batasan karakter di atas. Jika masih tidak ada karakter yang ditemukan, ini mengurangi jumlah karakter yang terlihat kembali, dan berulang.

def fix(paragraph, holeChar = "#"):
    lines = paragraph.split("\n")
    maxLineWidth = max(map(len, lines))
    lines = [list(line + " " * (maxLineWidth - len(line))) for line in lines]
    holes = filter(lambda pos: lines[pos[0]][pos[1]] == holeChar, [[y, x] for x in range(maxLineWidth) for y in range(len(lines))])

    n = 0
    for hole in holes:
        for i in range(min(hole[1], 5), 0, -1):
            currCh = lines[hole[0]][hole[1]]
            over = lines[hole[0] - 1][hole[1]]
            left = lines[hole[0]][hole[1] - i : hole[1]]

            same = []
            almost = []
            for y, line in enumerate(lines):
                for x, ch in enumerate(line):
                    if ch == holeChar:
                        continue
                    if ch == left[-1] == " ":
                        continue
                    chOver = lines[y - 1][x]
                    chLeft = lines[y][x - i : x]
                    if chOver == over and chLeft == left:
                        same.append(ch)
                    if chLeft == left:
                        almost.append(ch)
            sortFunc = lambda x, lst: lst.count(x) / (paragraph.count(x) + 10) + lst.count(x)
            if same:
                newCh = sorted(same, key=lambda x: sortFunc(x, same))[-1]
            elif almost:
                newCh = sorted(almost, key=lambda x: sortFunc(x, almost))[-1]
            else:
                continue
            lines[hole[0]][hole[1]] = newCh
            break


    return "\n".join(map("".join, lines))

Inilah hasil Badger, Badger, Badger:

Badger, badger, badger, badger, badger,
badger, badger, badger, badger, badger 
Mushroom, mushroom, a-                 
Badger, badger, badger, badger, badger,
badger, badger, badger, badger, badger 
Mushroom, mushroom, a- b               
Badger, badger, badger, badger, badger,
badger, badger, badger, badger, badger 
Mush-mushroom, a                       
Badger, badger, badger, badger, badger,
badger, badger, badger, badger, badger 
Argh! Snake, a snake!                  
Snaaake! A snaaaake, oooh its a snake! 

Inilah hasil dari buktinya:

1.Let ACB be a right-angled triangle with right angle CAB.                 
2.On each of the sides BC, AB, and CA, squares are drawn,                  
CBDE, BAGF, and ACIH, in that order. The construction of                   
squares requires the immediately preceding theorems in Euclid,             
and depends upon the parallel postulate. [footnote 14]                     
3.From A, draw a line parallel to BD and CE. It will                       
perpendicularly intersect BC and DE at K and L, respectively.              
4.Join CF and AD, to form the triangles BCF and BDA.                       
5.Angles CAB and BAG are both right angles; therefore C, A,                
and G are the same areamilarly for B, A, and H.                            
6.Angles CAB and CA, sqe both right angles; therefore angle ABD            
equals angle ABD becaus both are the sum of a right angle and angle ABC.   
7.Since ABD because theFB and BD is equal to BC, triangle ABD              
must be construction ofiangle FBC.                                         
8.Since A-angle ABD becight line, parallel to BD, then rectangle           
BDLK has the same area  of triangle ABD because they share the base        
BD and have the base thtitude BK, i.e., a line normal to their common      
base, conngle and G, sqrallel lines BD and AL. (lemma 2)                   
9.Since C = BD × BK + with A and G, square BAGF must be twice in area     
to triangle FBC. (lemma                                                    
10.Therefore angle and  BDLK must have the same area as square BAGF = AB^2.
11.Similarly for B, A,  shown that rectangle CKLE must have the same       
area as square ACIH = AC^2.                                                
12.Adding these two results, AB^2 + AC^2 = BD × BK + KL × KC             
13.Since BD = KL, BD × BK + KL × KC = BD(BK + KC) = BD × BC             
14.Therefore, AB^2 + AC^2 = BC^2, since CBDE is a square.

Dan hasil dari Jabberwocky:

'Twas brillig, and the slithy toves
Did gyre and the mo in the wabe;   
All mimsy toves, anborogoves,      
And the mome raths outgrabe.       

Python 2

Ini adalah solusi yang sangat mudah. Ini menciptakan string sampel yang terdiri dari kata-kata yang berada di antara panjang kata rata-rata A- ( A/ 2) dan A+ ( A/ 2), kemudian menerapkan potongan spasi terpimpin dan tertinggal dari sampel ke area tambalan. Itu tidak menangani kapitalisasi, dan saya yakin ada test case curveball di luar sana yang akan mematahkannya, tetapi tidak apa-apa pada contoh. Lihat tautan di bawah untuk menjalankan semua tes.

Saya juga memasukkan patch ke kode untuk mengukur baik.

def patch(paragraph):
    sample = [x.split() for x in paragraph if x.count('#') < 1]
    length = max([x.count('#') for x in paragraph if x.find('#')])
    s = sum(####################
    sample,[####################
    ])      ####################
    avg=sum(####################
    len(w)  ####################
    for w in####################
    s)//len(s)
    avg_range = range(avg-(avg//2),avg+(avg//2))
    sample = filter(lambda x:len(x) in avg_range, s)
    height=0
    for line in paragraph:
        if line.find('#'):height+=1
        print line.replace('#'*length,' '.join(sample)[(height-1)*length:height*length].strip())
    print '\n'

Lorem Ipsum, asli kemudian ditambal:

Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do
eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim
ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut
aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit
in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur
sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt
mollit anim id est laborum.

Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do
eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim
ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut
aliquip ex ea commodo conseore dolore magnairure dolor in reprehenderit
in voluptate velit esse cilenim minim quisgiat nulla pariatur. Excepteur
sint occaecat cupidatat nonnisi mollit aniin culpa qui officia deserunt
mollit anim id est laborum.

Cobalah

Java Shakespeare

Siapa yang butuh pemahaman tentang konvensi bahasa Inggris standar? Buat saja sendiri! Persis seperti Bard diizinkan untuk membuat kata-katanya sendiri. Bot ini tidak terlalu khawatir mengoreksi kata-kata yang terpotong, ia benar-benar hanya memasukkan kata-kata acak. Hasilnya adalah beberapa puisi yang indah. Sebagai fitur bonus, Bard memiliki kaliber lebih tinggi dan dapat menangani beberapa lubang asalkan ukurannya sama!

---

Input Sampel

 Dari makhluk paling adil yang kami inginkan meningkat,
  Bahwa dengan demikian mawar cantik tidak akan pernah mati,
  Tetapi sebagai riper harus pada saat meninggal,
  Tender-nya ############ menanggung ingatannya:
  Tapi kamu c ############ matamu sendiri,
  Pakan ############ ame dengan bahan bakar yang cukup besar,
  Membuat kelaparan di mana kelimpahan berada,
  Dirimu musuhmu, untuk diri manismu terlalu kejam:
  Engkau itu sekarang ornamen segar dunia,
  Dan hanya pemberita musim semi yang mencolok,
  Dalam budmu sendiri, isilah yang paling memberatkanmu,
  Dan tender churl mak'st adalah ############ ding:
    Kasihan dunia, kalau tidak t ############ menjadi,
    Untuk memakan makanan dunia, b ############# dan kamu.


                     2
  Ketika empat puluh musim dingin akan mengepung alismu,
  Dan gali parit yang dalam di ladang wanita cantikmu,
  Livery bangga pemuda Anda begitu menatap sekarang,
  Akan menjadi gulma compang-camping yang bernilai kecil dipegang:  
  Kemudian ditanya, di mana semua kecantikanmu berada,
  Di mana semua harta hari-hari sehatmu;
  Untuk mengatakan di dalam matamu sendiri tenggelam dalam,
  Malu memalukan, dan pujian yang hemat.
  Betapa lebih banyak pujian yang pantas digunakan oleh kecantikanmu,
  Jika kamu bisa menjawab 'Anakku yang adil ini
  Haruskah menghitung jumlah saya, dan membuat alasan lama saya '
  Membuktikan kecantikannya dengan suksesi Anda.
    Ini harus dibuat baru ketika kamu tua,
    Dan lihat darahmu hangat ketika kamu merasakannya dingin.


                     3
  Lihat di gelasmu dan beri tahu wajahmu,
  Sekarang saatnya wajah itu harus membentuk yang lain,
  Perbaikan siapa yang baru jika sekarang kamu tidak memperbarui,
  Engkau memperdayai dunia, tanpa ibu, seorang ibu.
  Di mana dia begitu adil yang rahimnya belum pernah mendengar
  Meremehkan pengolahan tanah peternakan Anda?
  Atau siapa yang sangat disayanginya akan menjadi makam,
  Dari cintanya untuk menghentikan keturunan?  
  Engkau gelas ibumu dan dia di dalam kamu
  Memanggil kembali April yang indah dari masa jayanya,
  Jadi, kamu melalui jendela zamanmu akan melihat,
  Meskipun waktu emasmu ############.
    Tetapi jika ############# mbered tidak menjadi,
    Die singl ############ image mati bersamamu.

Output yang indah

 Dari makhluk paling adil yang kami inginkan meningkat,
  Bahwa dengan demikian mawar cantik tidak akan pernah mati,
  Tetapi sebagai riper harus pada saat meninggal,
  Tender-nya harus menanggung ingatannya:
  Tapi kamu c semua meratakan mata cerahmu sendiri,
  Beri makan Pembuktian Atau ame dengan bahan bakar substansial sendiri,
  Membuat kelaparan di mana kelimpahan berada,
  Dirimu musuhmu, untuk diri manismu terlalu kejam:
  Engkau itu sekarang ornamen segar dunia,
  Dan hanya pemberita musim semi yang mencolok,
  Dalam budmu sendiri, isilah yang paling memberatkanmu,
  Dan lembut churl mak'st adalah dia, ya Tuhan:
    Kasihan dunia, atau t Jadi,
    Untuk memakan hak dunia, bt begitu dan kamu.


                     2
  Ketika empat puluh musim dingin akan mengepung alismu,
  Dan gali parit yang dalam di ladang wanita cantikmu,
  Livery bangga pemuda Anda begitu menatap sekarang,
  Akan menjadi gulma compang-camping yang bernilai kecil dipegang:  
  Kemudian ditanya, di mana semua kecantikanmu berada,
  Di mana semua harta hari-hari sehatmu;
  Untuk mengatakan di dalam matamu sendiri tenggelam dalam,
  Malu memalukan, dan pujian yang hemat.
  Betapa lebih banyak pujian yang pantas digunakan oleh kecantikanmu,
  Jika kamu bisa menjawab 'Anakku yang adil ini
  Haruskah menghitung jumlah saya, dan membuat alasan lama saya '
  Membuktikan kecantikannya dengan suksesi Anda.
    Ini harus dibuat baru ketika kamu tua,
    Dan lihat darahmu hangat ketika kamu merasakannya dingin.


                     3
  Lihat di gelasmu dan beri tahu wajahmu,
  Sekarang saatnya wajah itu harus membentuk yang lain,
  Perbaikan siapa yang baru jika sekarang kamu tidak memperbarui,
  Engkau memperdayai dunia, tanpa ibu, seorang ibu.
  Di mana dia begitu adil yang rahimnya belum pernah mendengar
  Meremehkan pengolahan tanah peternakan Anda?
  Atau siapa yang sangat disayanginya akan menjadi makam,
  Dari cintanya untuk menghentikan keturunan?  
  Engkau gelas ibumu dan dia di dalam kamu
  Memanggil kembali April yang indah dari masa jayanya,
  Jadi, kamu melalui jendela zamanmu akan melihat,
  Meskipun Look menatap waktu emasmu.
    Tetapi jika itu terjadi, tidak menjadi,
    Die singl memperbaiki gambar mati bersamamu.

Baris pasangan terakhir sangat puitis jika saya mengatakannya sendiri. Ini bekerja dengan sangat baik pada alamat gettysburg juga.

But, in a larger sense, we can not dedicate, we can not consecrate, we can not
hallow this ground. The brave men, living and dead, who struggled here, have
consecrated it, far above our poor power to add or detract. The world will
little note, nor long remember what we say here, but it can never forget what
they did here. It is for us the living, rather, to be dedicated here to the
unfinished work which they who fought here h to of rather us of advanced. It
is rather for us to be here dedicated to the who be it, vain who before us 
that from these honored dead we take increas be dead the the what use for which
they gave the last full measure of devotion  dead government The solve that
these dead shall not have died in vain that  the take nor world , shall have
a new birth of freedom and that government of the people, by the people, for
the people, shall not perish from the earth.

---

Mari kita lihat apa yang membuat Shakespeare tergerak. Ini kodenya. Intinya dia berusaha membangun basis kosakata dari input. Dia kemudian menggunakan kata-kata ini dan secara acak menempatkannya di dalam lubang (memastikan bahwa itu cocok dengan baik). Dia deterministik karena dia menggunakan benih tetap untuk keacakan.

package stuff;

import java.io.File;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Iterator;
import java.util.Random;
import java.util.Scanner;
import java.util.Stack;

/**
 *
 * @author rohan
 */
public class PatchTheParagraph {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
Scanner in = new Scanner(System.in);
System.out.println("File Name :");
        String[] text = getWordsFromFile(in.nextLine());
System.out.println("==ORIGINAL==");
        for(String s:text){
    System.out.println(s);
}
                    int lengthOfHole= 0;
        int rows = 0;
            for(String s: text){
                s = s.replaceAll("[^#]", "");

//      System.out.println(s);
                if(s.length()>0){
                    lengthOfHole = s.length();
                rows++;
                }
            }
            ArrayList<String> words = new ArrayList<>();
            words.add("I");
            for(String s:text){
                String[] w = s.replaceAll("#", " ").split(" ");
for(String a :w){
    words.add(a);
            }

            }
                        Iterator<String> j = words.iterator();
            while(j.hasNext()){
                String o;
                if((o = j.next()).equals("")){
                    j.remove();
                }
            }
            System.out.println(words);
            Stack<String> out = new Stack<>();
            String hashRow = "";
            for(int i = 0;i<lengthOfHole;i++){
                hashRow+="#";
            }

        for(int i = 0;i<rows;i++){
            int length = lengthOfHole-1; 
            String outPut = " ";
            while(length>2){
String wordAttempt = words.get(getRandom(words.size()-1));
while(wordAttempt.length()>length-1){
 wordAttempt = words.get(getRandom(words.size()-1));
}           
length -= wordAttempt.length()+1;
            outPut+=wordAttempt;
                outPut+=" ";
            }
        out.push(outPut);
    }
System.out.println("==PATCHED==");
        for(String s : text){
            if(s.contains(hashRow)){
                System.out.println(s.replaceAll(hashRow,out.pop()));
            }else{
                System.out.println(s);
            }
        }
                                    }
public static final Random r = new Random(42);
    public static int getRandom(int max){
    return (int) (max*r.nextDouble());
}
    /**
     *
     * @param fileName is the path to the file or just the name if it is local
     * @return the number of lines in fileName
     */
    public static int getLengthOfFile(String fileName) {
        int length = 0;
        try {
            File textFile = new File(fileName);
            Scanner sc = new Scanner(textFile);
            while (sc.hasNextLine()) {
                sc.nextLine();
                length++;
            }
        } catch (Exception e) {
System.err.println(e);
        }
        return length;
    }

    /**
     *
     * @param fileName is the path to the file or just the name if it is local
     * @return an array of Strings where each string is one line from the file
     * fileName.
     */
    public static String[] getWordsFromFile(String fileName) {
        int lengthOfFile = getLengthOfFile(fileName);
        String[] wordBank = new String[lengthOfFile];
        int i = 0;
        try {
            File textFile = new File(fileName);
            Scanner sc = new Scanner(textFile);
            for (i = 0; i < lengthOfFile; i++) {
                wordBank[i] = sc.nextLine();
            }
            return wordBank;
        } catch (Exception e) {
            System.err.println(e);
            System.exit(55);
        }
        return null;
    }
}

---

Sebagian besar puisi Shakespeare adalah ranah publik.

Python 2.7

Solusi Python lain dengan pendekatan berbeda. Program saya melihat teks sebagai rantai Markov , di mana setiap huruf diikuti oleh huruf lain dengan probabilitas tertentu. Jadi langkah pertama adalah membangun tabel probabilitas. Langkah selanjutnya adalah menerapkan probabilitas itu ke tambalan.

Kode lengkap, termasuk satu contoh teks di bawah ini. Karena satu contoh menggunakan karakter unicode, saya menyertakan codepage eksplisit (utf-8) untuk kompatibilitas dengan contoh itu.

#!/usr/bin/env python
# -*- coding: utf-8 -*-

from collections import defaultdict
import numpy

texts = [
"""'Twas brillig, and the slithy toves
Did gyre a######### in the wabe;
All mimsy #########borogoves,
And the mome raths outgrabe."""
]

class Patcher:
    def __init__(self):
        self.mapper = defaultdict(lambda: defaultdict(int))

    def add_mapping(self, from_value, to_value):
        self.mapper[from_value][to_value] += 1

    def get_patch(self, from_value):
        if from_value in self.mapper:
            sum_freq = sum(self.mapper[from_value].values())
            return numpy.random.choice(
                self.mapper[from_value].keys(),
                p = numpy.array(
                    self.mapper[from_value].values(),dtype=numpy.float64) / sum_freq)
        else:
            return None

def add_text_mappings(text_string, patcher = Patcher(), ignore_characters = ''):
    previous_letter = text_string[0]
    for letter in text_string[1:]:
        if not letter in ignore_characters:
            patcher.add_mapping(previous_letter, letter)
            previous_letter = letter
    patcher.add_mapping(text_string[-1], '\n')

def patch_text(text_string, patcher, patch_characters = '#'):
    result = previous_letter = text_string[0]
    for letter in text_string[1:]:
        if letter in patch_characters:
            result += patcher.get_patch(previous_letter)
        else:
            result += letter
        previous_letter = result[-1]
    return result

def main():
    for text in texts:
        patcher = Patcher()
        add_text_mappings(text, patcher, '#')
        print patch_text(text, patcher, '#')
        print "\n"

if __name__ == '__main__':
    main()

Contoh keluaran untuk Lorem Ipsum:

Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do
eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim
ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut
aliquip ex ea commodo conse Exe eut ccadamairure dolor in reprehenderit
in voluptate velit esse cilore indipserexepgiat nulla pariatur. Excepteur
sint occaecat cupidatat non upir alostat adin culpa qui officia deserunt
mollit anim id est laborum.

Garis puitis ekstra di Jabberwocky:

'Twas brillig, and the slithy toves
Did gyre and me the in the wabe;
All mimsy was
An inborogoves,
And the mome raths outgrabe.

C # 5 masif seperti biasa

Saya melemparkan ini bersama-sama, ini agak berantakan, tetapi beberapa kali menghasilkan hasil yang OK. Ini sebagian besar algoritma deterministik, tetapi dengan beberapa (tetap-seed) keacakan ditambahkan untuk menghindarinya menghasilkan string yang sama untuk kesenjangan yang sama. Perlu beberapa upaya untuk mencoba menghindari hanya memiliki kolom ruang di kedua sisi celah.

Ini bekerja dengan tokenizing input ke dalam kata-kata dan tanda baca (tanda baca berasal dari daftar yang dimasukkan secara manual, karena saya tidak dapat diganggu untuk bekerja jika Unicode dapat melakukan ini untuk saya), sehingga dapat menempatkan spasi sebelum kata-kata, dan tidak sebelum tanda baca, karena ini cukup khas. Itu terbagi pada ruang putih khas. Dalam jalur rantai markov (saya pikir), ini menghitung seberapa sering setiap token mengikuti satu sama lain, dan kemudian tidak menghitung probabilitas untuk ini (saya pikir karena dokumen sangat kecil, kami akan lebih baik untuk bias terhadap hal-hal kita melihat banyak di mana kita bisa). Kemudian kami melakukan pencarian luas pertama, mengisi ruang yang ditinggalkan oleh hash dan kata-kata 'parsial' di kedua sisi, dengan biaya yang dihitung sebagai -fabness(last, cur) * len(cur_with_space), di mana fabnesspengembalian berapa kali curdiikutilastuntuk setiap token yang ditambahkan dalam string yang dihasilkan. Secara alami, kami berusaha meminimalkan biaya. Karena kita tidak selalu dapat mengisi celah dengan kata-kata dan tanda baca yang ditemukan dalam dokumen, itu juga mempertimbangkan sejumlah token 'khusus' dari negara tertentu, termasuk string parsial di kedua sisi, yang kami bias terhadap dengan kenaikan biaya sewenang-wenang.

Jika BFS gagal menemukan solusi, maka kami secara naif mencoba untuk memilih kata keterangan acak, atau hanya memasukkan spasi untuk mengisi ruang.

Hasil

Semua 6 dapat ditemukan di sini: https://gist.github.com/anonymous/5277db726d3f9bdd950b173b19fec82a

Kasus uji Euclid tidak berjalan dengan baik ...

Menambal Gambar

In a popular image editing software there is a feature, that patches (The term
used in image processing is inpainting as @minxomat pointed out.) a selected
area of an image, that patches information outside of that patch. And it does a
quite good job, co the patch a is just a program. As a human, you can sometimes
see that something In a short it if you squeeze your eyes or just take a short
glance, the patch seems to fill in the gap quite well.

Jabberwocky

'Twas brillig, and the slithy toves
Did gyre and the in in the wabe;
All mimsy the mome borogoves,
And the mome raths outgrabe.

Luak

Badger, badger, badger, badger, badger,
badger, badger, badger, badger, badger
Mushroom, mushroom, a-
Badger, badger, badger, badger, badger,
badger, badger, badger, badger, badger
Mushroom, badger, badger
Badger, badger, badger, badger, badger,
badger, badger, badger, badger, badger
Mush-mushroom, a
Badger, badger, badger, badger, badger,
badger, badger, badger, badger, badger
Argh! Snake, a snake!
Snaaake! A snaaaake, oooh its a snake!

_Saya senang dengan bagaimana yang satu ini ternyata ... itu meramalkan bahwa "luak, luak," cocok, atau yang ini tidak akan melakukannya dengan baik

Kode

Jalankan dengan

csc ParaPatch.cs
ParaPatch.exe infile outfile

Ada cukup banyak. Satu-satunya yang menarik adalah Fillmetode. Saya menyertakan implementasi heap, karena .NET tidak memilikinya (MENGAPA MS MENGAPA ?!).

using System;
using System.Collections.Generic;
using System.Linq;

namespace ParaPatch
{
    class Program
    {
        private static string[] Filler = new string[] { "may", "will", "maybe", "rather", "perhaps", "reliably", "nineword?", "definitely", "elevenword?", "inexplicably" }; // adverbs
        private static char[] Breaking = new char[] { ' ', '\n', '\r', '\t' };
        private static char[] Punctuation = new char[] { ',', '.', '{', '}', '(', ')', '/', '?', ':', ';', '\'', '\\', '"', ',', '!', '-', '+', '[', ']', '£', '$', '%', '^', '—' };

        private static IEnumerable<string> TokenizeStream(System.IO.StreamReader reader)
        {
            System.Text.StringBuilder sb = new System.Text.StringBuilder();

            HashSet<char> breaking = new HashSet<char>(Breaking);
            HashSet<char> punctuation = new HashSet<char>(Punctuation);

            while (!reader.EndOfStream)
            {
                int ci = reader.Read();
                if (ci == -1) // sanity
                    break;

                char c = (char)ci;

                if (breaking.Contains(c))
                {
                    if (sb.Length > 0)
                        yield return sb.ToString();
                    sb.Clear();
                }
                else if (punctuation.Contains(c))
                {
                    if (sb.Length > 0)
                        yield return sb.ToString();
                    yield return ""+c;
                    sb.Clear();
                }
                else
                {

                    sb.Append(c);
                }
            }

            if (sb.Length > 0)
                yield return sb.ToString();
        }

        private enum DocTokenTypes
        {
            Known,
            LeftPartial,
            RightPartial,
            Unknown,
        }

        private class DocToken
        {
            public DocTokenTypes TokenType { get; private set; }
            public string StringPart { get; private set; }
            public int Length { get; private set; }

            public DocToken(DocTokenTypes tokenType, string stringPart, int length)
            {
                TokenType = tokenType;
                StringPart = stringPart;
                Length = length;
            }
        }

        private static IEnumerable<DocToken> DocumentTokens(IEnumerable<string> tokens)
        {
            foreach (string token in tokens)
            {
                if (token.Contains("#"))
                {
                    int l = token.IndexOf("#");
                    int r = token.LastIndexOf("#");

                    if (l > 0)
                        yield return new DocToken(DocTokenTypes.LeftPartial, token.Substring(0, l), l);

                    yield return new DocToken(DocTokenTypes.Unknown, null, r - l + 1);

                    if (r < token.Length - 1)
                        yield return new DocToken(DocTokenTypes.RightPartial, token.Substring(r + 1), token.Length - r - 1);
                }
                else
                    yield return new DocToken(DocTokenTypes.Known, token, token.Length);
            }
        }

        private class State : IComparable<State>
        {
            // missing readonly params already... maybe C#6 isn't so bad
            public int Remaining { get; private set; }
            public int Position { get; private set; }
            public State Prev { get; private set; }
            public string Token { get; private set; }
            public double H { get; private set; }
            public double Fabness { get; private set; }
            public string FullFilling { get; private set; }

            public State(int remaining, int position, Program.State prev, double fabness, double h, string token, string toAdd)
            {
                Remaining = remaining;
                Position = position;
                Prev = prev;
                H = h;
                Fabness = fabness;
                Token = token;

                FullFilling = prev != null ? prev.FullFilling + toAdd : toAdd;
            }

            public int CompareTo(State other)
            {
                return H.CompareTo(other.H);
            }
        }

        public static void Main(string[] args)
        {
            if (args.Length < 2)
                args = new string[] { "test.txt", "testout.txt" };

            List<DocToken> document;
            using (System.IO.StreamReader reader = new System.IO.StreamReader(args[0], System.Text.Encoding.UTF8))
            {
                document = DocumentTokens(TokenizeStream(reader)).ToList();
            }

            foreach (DocToken cur in document)
            {
                Console.WriteLine(cur.StringPart + " " + cur.TokenType);
            }

            // these are small docs, don't bother with more than 1 ply
            Dictionary<string, Dictionary<string, int>> FollowCounts = new Dictionary<string, Dictionary<string, int>>();
            Dictionary<string, Dictionary<string, int>> PreceedCounts = new Dictionary<string, Dictionary<string, int>>(); // mirror (might be useful)

            HashSet<string> knowns = new HashSet<string>(); // useful to have lying around

            // build counts
            DocToken last = null;
            foreach (DocToken cur in document)
            {
                if (cur.TokenType == DocTokenTypes.Known)
                {
                    knowns.Add(cur.StringPart);
                }

                if (last != null && last.TokenType == DocTokenTypes.Known && cur.TokenType == DocTokenTypes.Known)
                {
                    {
                        Dictionary<string, int> ltable;
                        if (!FollowCounts.TryGetValue(last.StringPart, out ltable))
                        {
                            FollowCounts.Add(last.StringPart, ltable = new Dictionary<string, int>());
                        }

                        int count;
                        if (!ltable.TryGetValue(cur.StringPart, out count))
                        {
                            count = 0;
                        }
                        ltable[cur.StringPart] = count + 1;
                    }


                    {
                        Dictionary<string, int> ctable;
                        if (!PreceedCounts.TryGetValue(cur.StringPart, out ctable))
                        {
                            PreceedCounts.Add(cur.StringPart, ctable = new Dictionary<string, int>());
                        }

                        int count;
                        if (!ctable.TryGetValue(last.StringPart, out count))
                        {
                            count = 0;
                        }
                        ctable[last.StringPart] = count + 1;
                    }
                }

                last = cur;
            }

            // build probability grid (none of this efficient table filling dynamic programming nonsense, A* all the way!)
            // hmm... can't be bothered
            Dictionary<string, Dictionary<string, double>> fabTable = new Dictionary<string, Dictionary<string, double>>();
            foreach (var k in FollowCounts)
            {
                Dictionary<string, double> t = new Dictionary<string, double>();

                // very naive
                foreach (var k2 in k.Value)
                {
                    t.Add(k2.Key, (double)k2.Value);
                }

                fabTable.Add(k.Key, t);
            }

            string[] knarr = knowns.ToArray();
            Random rnd = new Random("ParaPatch".GetHashCode());

            List<string> fillings = new List<string>();
            for (int i = 0; i < document.Count; i++)
            {
                if (document[i].TokenType == DocTokenTypes.Unknown)
                {
                    // shuffle knarr
                    for (int j = 0; j < knarr.Length; j++)
                    {
                        string t = knarr[j];
                        int o = rnd.Next(knarr.Length);
                        knarr[j] = knarr[o];
                        knarr[o] = t;
                    }

                    fillings.Add(Fill(document, fabTable, knarr, i));
                    Console.WriteLine(fillings.Last());
                }
            }

            string filling = string.Join("", fillings);

            int fi = 0;

            using (System.IO.StreamWriter writer = new System.IO.StreamWriter(args[1]))
            using (System.IO.StreamReader reader = new System.IO.StreamReader(args[0]))
            {
                while (!reader.EndOfStream)
                {
                    int ci = reader.Read();
                    if (ci == -1)
                        break;

                    char c = (char)ci;
                    c = c == '#' ? filling[fi++] : c;

                    writer.Write(c);
                    Console.Write(c);
                }
            }

//            using (System.IO.StreamWriter writer = new System.IO.StreamWriter(args[1], false, System.Text.Encoding.UTF8))
//            using (System.IO.StreamReader reader = new System.IO.StreamReader(args[0]))
//            {
//                foreach (char cc in reader.ReadToEnd())
//                {
//                    char c = cc;
//                    c = c == '#' ? filling[fi++] : c;
//                    
//                    writer.Write(c);
//                    Console.Write(c);
//                }
//            }

            if (args[0] == "test.txt")
                Console.ReadKey(true);
        }

        private static string Fill(List<DocToken> document, Dictionary<string, Dictionary<string, double>> fabTable, string[] knowns, int unknownIndex)
        {
            HashSet<char> breaking = new HashSet<char>(Breaking);
            HashSet<char> punctuation = new HashSet<char>(Punctuation);

            Heap<State> due = new Heap<Program.State>(knowns.Length);

            Func<string, string, double> fabness = (prev, next) =>
            {
                Dictionary<string, double> table;
                if (!fabTable.TryGetValue(prev, out table))
                    return 0; // not fab
                double fab;
                if (!table.TryGetValue(next, out fab))
                    return 0; // not fab
                return fab; // yes fab
            };

            DocToken mostLeft = unknownIndex > 2 ? document[unknownIndex - 2] : null;
            DocToken left = unknownIndex > 1 ? document[unknownIndex - 1] : null;
            DocToken unknown = document[unknownIndex];
            DocToken right = unknownIndex < document.Count - 2 ? document[unknownIndex + 1] : null;
            DocToken mostRight = unknownIndex < document.Count - 3 ? document[unknownIndex + 2] : null;

            // sum of empty space and partials' lengths
            int spaceSize = document[unknownIndex].Length
                + (left != null && left.TokenType == DocTokenTypes.LeftPartial ? left.Length : 0)
                + (right != null && right.TokenType == DocTokenTypes.RightPartial ? right.Length : 0);

            int l = left != null && left.TokenType == DocTokenTypes.LeftPartial ? left.Length : 0;
            int r = l + unknown.Length;

            string defaultPrev =
                left != null && left.TokenType == DocTokenTypes.Known ? left.StringPart :
                mostLeft != null && mostLeft.TokenType == DocTokenTypes.Known ? mostLeft.StringPart :
                "";

            string defaultLast =
                right != null && right.TokenType == DocTokenTypes.Known ? right.StringPart :
                mostRight != null && mostRight.TokenType == DocTokenTypes.Known ? mostRight.StringPart :
                "";

            Func<string, string> topAndTail = str =>
            {
                return str.Substring(l, r - l);
            };

            Func<State, string, double, bool> tryMove = (State prev, string token, double specialFabness) => 
            {
                bool isPunctionuation = token.Length == 1 && punctuation.Contains(token[0]);
                string addStr = isPunctionuation || prev == null ? token : " " + token;
                int addLen = addStr.Length;

                int newRemaining = prev != null ? prev.Remaining - addLen : spaceSize - addLen;
                int oldPosition = prev != null ? prev.Position : 0;
                int newPosition = oldPosition + addLen;

                // check length
                if (newRemaining < 0)
                    return false;

                // check start
                if (oldPosition < l) // implies left is LeftPartial
                {
                    int s = oldPosition;
                    int e = newPosition > l ? l : newPosition;
                    int len = e - s;
                    if (addStr.Substring(0, len) != left.StringPart.Substring(s, len))
                        return false; // doesn't match LeftPartial
                }

                // check end
                if (newPosition > r) // implies right is RightPartial
                {
                    int s = oldPosition > r ? oldPosition : r;
                    int e = newPosition;
                    int len = e - s;
                    if (addStr.Substring(s - oldPosition, len) != right.StringPart.Substring(s - r, len))
                        return false; // doesn't match RightPartial
                }

                if (newRemaining == 0)
                {
                    // could try to do something here (need to change H)
                }

                string prevToken = prev != null ? prev.Token : defaultPrev;
                bool isLastunctionuation = prevToken.Length == 1 && punctuation.Contains(prevToken[0]);

                if (isLastunctionuation && isPunctionuation) // I hate this check, it's too aggresive to be realistic
                    specialFabness -= 50;

                double fab = fabness(prevToken, token);

                if (fab < 1 && (token == prevToken))
                    fab = -1; // bias against unrecognised repeats

                double newFabness = (prev != null ? prev.Fabness : 0.0)
                    - specialFabness // ... whatever this is
                    - fab * addLen; // how probabilistic

                double h = newFabness; // no h for now

                State newState = new Program.State(newRemaining, newPosition, prev, newFabness, h, token, addStr);

//                Console.WriteLine((prev != null ? prev.Fabness : 0) + "\t" + specialFabness);
//                Console.WriteLine(newFabness + "\t" + h + "\t" + due.Count + "\t" + fab + "*" + addLen + "\t" + newState.FullFilling);

                due.Add(newState);
                return true;
            };

            // just try everything everything
            foreach (string t in knowns)
                tryMove(null, t, 0);

            if (left != null && left.TokenType == DocTokenTypes.LeftPartial)
                tryMove(null, left.StringPart, -1);

            while (!due.Empty)
            {
                State next = due.RemoveMin();

                if (next.Remaining == 0)
                {
                    // we have a winner!!
                    return topAndTail(next.FullFilling);
                }

                // just try everything
                foreach (string t in knowns)
                    tryMove(next, t, 0);
                if (right != null && right.TokenType == DocTokenTypes.RightPartial)
                    tryMove(next, right.StringPart, -5); // big bias
            }

            // make this a tad less stupid, non?
            return Filler.FirstOrDefault(f => f.Length == unknown.Length) ?? new String(' ', unknown.Length); // oh dear...
        }
    }

    //
    // Ultilities
    //

    public class Heap<T> : System.Collections.IEnumerable where T : IComparable<T>
    {
        // arr is treated as offset by 1, all idxes stored need to be -1'd to get index in arr
        private T[] arr;
        private int end = 0;

        private void s(int idx, T val)
        {
            arr[idx - 1] = val;
        }

        private T g(int idx)
        {
            return arr[idx - 1];
        }

        public Heap(int isize)
        {
            if (isize < 1)
                throw new ArgumentException("Cannot be less than 1", "isize");

            arr = new T[isize];
        }

        private int up(int idx)
        {
            return idx / 2;
        }

        private int downLeft(int idx)
        {
            return idx * 2;
        }

        private int downRight(int idx)
        {
            return idx * 2 + 1;
        }

        private void swap(int a, int b)
        {
            T t = g(a);
            s(a, g(b));
            s(b, t);
        }

        private void moveUp(int idx, T t)
        {
        again:
            if (idx == 1)
            {
                s(1, t);
                return; // at end
            }

            int nextUp = up(idx);
            T n = g(nextUp);
            if (n.CompareTo(t) > 0)
            {
                s(idx, n);
                idx = nextUp;
                goto again;
            }
            else
            {
                s(idx, t);
            }
        }

        private void moveDown(int idx, T t)
        {
        again:
            int nextLeft = downLeft(idx);
            int nextRight = downRight(idx);

            if (nextLeft > end)
            {
                s(idx, t);
                return; // at end
            }
            else if (nextLeft == end)
            { // only need to check left
                T l = g(nextLeft);

                if (l.CompareTo(t) < 0)
                {
                    s(idx, l);
                    idx = nextLeft;
                    goto again;
                }
                else
                {
                    s(idx, t);
                }
            }
            else
            { // check both
                T l = g(nextLeft);
                T r = g(nextRight);

                if (l.CompareTo(r) < 0)
                { // left smaller (favour going right if we can)
                    if (l.CompareTo(t) < 0)
                    {
                        s(idx, l);
                        idx = nextLeft;
                        goto again;
                    }
                    else
                    {
                        s(idx, t);
                    }
                }
                else
                { // right smaller or same
                    if (r.CompareTo(t) < 0)
                    {
                        s(idx, r);
                        idx = nextRight;
                        goto again;
                    }
                    else
                    {
                        s(idx, t);
                    }
                }
            }
        }

        public void Clear()
        {
            end = 0;
        }

        public void Trim()
        {
            if (end == 0)
                arr = new T[1]; // don't /ever/ make arr len 0
            else
            {
                T[] narr = new T[end];
                for (int i = 0; i < end; i++)
                    narr[i] = arr[i];
                arr = narr;
            }
        }

        private void doubleSize()
        {
            T[] narr = new T[arr.Length * 2];
            for (int i = 0; i < end; i++)
                narr[i] = arr[i];
            arr = narr;
        }

        public void Add(T item)
        {
            if (end == arr.Length)
            {
                // resize
                doubleSize();
            }

            end++;
            moveUp(end, item);
        }

        public T RemoveMin()
        {
            if (end < 1)
                throw new Exception("No items, mate.");

            T min = g(1);

            end--;
            if (end > 0)
                moveDown(1, g(end + 1));

            return min;
        }

        public bool Empty
        {
            get
            {
                return end == 0;
            }
        }

        public int Count
        {
            get
            {
                return end;
            }
        }

        System.Collections.IEnumerator System.Collections.IEnumerable.GetEnumerator()
        {
            return GetEnumerator();
        }

        public IEnumerator<T> GetEnumerator()
        {
            return (IEnumerator<T>)arr.GetEnumerator();
        }
    }
}