Cuma, Ekim 05, 2007

100 Katlı Binadan Küreler Atılırsa?

Elimizde 100 katlı bir bina ve birbirinin aynı 2 tane küre var.
Tek bildiğimiz bu kürelerin binanın belli bir katına kadar yüksekten
atılınca kırılmayacağı, iki kürenin sağlamlık derecesi aynı (Örneğin, bir küre
50 kata kadar yükseklikten bırakılınca kırılmıyor ve 51. katta kırılıyorsa
diğer küre de aynı şekilde...)

Bizden istenen minimum sayıda deneme ile bu kürelerin en az kaçıncı kattan atılınca kırıldığını bulmak.

İpucu: Eğer bir küremiz olsaydı teker teker bütün katları denemek zorunda
kalacaktık (1,2,3,4,...,n) burada 100 deneme gerekiyor.

Ek Soru: Üç küre olsa deneme sayısı ne olurdu?

67 yorum:

tavuk dedi ki...

1 deneme yüzüncü kattan atılır bir küre sadece bir kez.

tavuk dedi ki...

özür tersten bakıyorum konsantrasyon sorunumda var şu ara, en az kaçıncı kattan diye olursa 1 den başlanır ve kırıldığı kat sayısı deneme sayısı olur. yani 1. katsa kırılırsa deneme 1 sayısı, 50 katsa kırılıyorsa deneme sayısı 50, 100. katta kırılıyorsa deneme sayısı 100 olur.

tavuk dedi ki...

küreler aynı olduğuna göre 1 kürede olsa, 2 kürede olsa, 3 kürede olsa, deneme sayısı her kattan bir kez olacağı için, sayı aynıdır...

tulin dedi ki...

Bence 2 şer 2 şer atılırsa daha az deneme sayısı olmuş olur. Misal 2 tane topumuz var 2. katta attık kırılmadı, 4. katta attık kırılmadı... fakat 78. katta kırıldı o zaman 77. kata kadar olduğunu anlarız ve diğer topu attığımızda kırılmayacaktır. DENEME SAYISI maksimum 50 olur. 3 topumuz oldaydı 3'er 3'er atardık. 33,333 defa maksimum deneme sayısı olurdu...

Loki dedi ki...

33.3333 diye deneme sayısı olmaz

3'er atlayarak deneme yapılır
3 farklı çözüm olabilir
çözüme 3n+c dersek (c mod3'te kırılan kat değeri)

kırılan cisim mod 3'te 0 ise
3.kat 6.kat ... 3n'inci katta kırıldığı bulunur
3n-1 denenip emin olunur n+1 deneme yapılır

mod 3'te 1 ise
3.kat 6.kat ... 3n+3'inci kat denenir kırılınca 2 deneme yapılır 3n+1'de kırılınca emin olunur (zaten 3n'inci katta kırılmamıştı n'inci denemede denemiştik)

mod 3'te 2 ise
3n+3te kırıldığı gözükür 2 deneme daha yapılır
3n+2de de kırılır
3n+1de kırılmayınca kırılan minimum katın 3n+2 olduğu gözükür

100'e kadar olan katlarda n+1,n+2,n+2 deneme yapılır
100 de ise 33 deneme yapılır ve 100'üncü kat da denenir
100'üncü katta 34
99'uncu katta 34
98'inci katta 35
97'inci katta 35 deneme yapılır

ozan dedi ki...

2 top varsa sıra su sekilde olmalıdır bence:
once 1. kattan atılır.
-kırılırsa problem degil hatta iyi bile olur :D
-kırılmazsa 4. kattan atılır.
--kırılırsa diger top ikinci kattan atılır
---diger top kırılırsa demekki en son 1. kata dayanıyomus deriz.
---kırılmazsa demekki 3. katta kırılıyo bu lanet.
--kırılmazsa busefer 8. kattan atılır vew bu boyle dewam eder
yani mantık su: bir top n=1,2,3,4,...25 için
4n inci kattan atılıyo
kırılmazsa ne ala ama kırılırsa bu sefer diger top 4n-2 inci kattan atılır.krılırsa bi altı kırılmazsa bir ustu sınırdır.
bu sekilde en fazla 26 deneme yapılır..

Loki_tr dedi ki...

Hodri meydan 3 topla maksimum 9 denemede bulacak yöntem:

İlk top 35. kata atılır
Eğer kırılırsa 2. atış 8. kata yapılır bunun kırılması durumunda 7 denemede bulmak kolay olur.
kırılmaması durumunda
bu sefer 15. kata (8+7) atış yapılır bu kırılırsa bulma icin 9. kattan başlayarak 14. kata kadar tek tek deneme yapılır.
Bir sonraki deneme 21 kata (15+6)
sonraki 26,30,33,34 şekilde bulunur..

Eğer 35. kat olmazsa bu sefer 64. katdan atış yapılır kırılması durumunda:
Önce 42. kat sonra sonuç alınana kadar sırasıyla 42+6,48+5,53+4,57+3,60+2,62+1 katlarına atış yapılır

64. katta da kırılmazsa bu sefer 85. bu da olmazsa 100. katdan atış yapılır.

Bu yöntemle 120 katlı binaya kadar 9 atışta en az hangi kattan kırılacağı bulunabilinir..

Çağrı dedi ki...

İlk küreyi 50. kattan yüksekten denerim kırılırsa ikinci küreyi 25.
kattan denerim eğer o da kırılırsa binanın %25 inde kırılıyor derim.

Çağrı dedi ki...

ama eğer 50 . katta kırılmassa 75 de denerim 75 te kırılırsa binanın %50 si ile % 75 i arasındaki herangi bir kattan atılır

Çağrı dedi ki...

Bunun daha bir çok cevabı var aklımda ama ben bu secenekleri gösterdim.

sercan dedi ki...

Minimum deneme yapmak için kürenin kırılmaması gerekir. 2 küre olduğu için ilk hakkı 50de kullanırız. kırılırsa 49 seçenek var demek ve tek küre kaldığından 1 den başlamak lazım. Fakat kırılmaz ise 75 den atarız. (Kırılırsa 26 seçenek kalır) 88, 94, 97 ve 99 a kadar kırılmaz ise minimum 6 kez atma ile 100 de kırıldığı sonucunu buluruz.

armağan dedi ki...

x ve y kürelerimiz olsun
* x küresini ilk baş 10. kattan atarız
Kırılırsa; y küresini 1.kat 2.kat,,,,, deneriz ve en fazla 9. denemede buluruz
Kırılmazsa; 20. katı deneriz
*2. aşamada x küresini 20 kattan atarız
Kırılırsa;1. denemede 10 katta kırılmadığını bildiğimiz için 2 şamaya 11. kattan başlarız 11.kat,12.kat,13,14,,,19 deneriz v bunuda en fazla 10. denemede buluruz
Kırılmazsa 30. katı deneriz

***yani sonuç olarak 0-10 arasında 9 denemede buluruz. 10-20 arasında ise 10 denemde,20-30 arasında 11,30-40 12.denemde,100-90 arasında ise 18. denemede buluruz.yalnız top 100.katta ise 2 küre bulunduğu için elimizde har zamanki gibi riske atamadığımız için en az 10 denemde bulabiliriz.
Şimdi yöntemi deniyelim.Mesela 67. katta olsun
10 dan başlarız
(her deneme için yıldız koyacağım)
70 * 61 e geç
.
.
67 * yıldızların top=14 dur
66 *
65 *
64 *
63 *
62 *
61 *
60 *buradaki denemden sonra 70 geç
50 *
40 *
30 *
20 *
10 *
0
sonuç olarak deneme sayımız
max 18 ,min 9 dur bu rakam kürelerin kırılma noktasına göre değişir.:)

GÜRSEL dedi ki...

yine 100 deneme topların hepsi aynı kattan atınca kırıldığı için 100 denemede buluruz. A topu x. katta kırılırsa B topu da , C topu da x. katta kırılır.

freeozgur dedi ki...

ilk denemenin birini 100. diğerini 50.kattan atarız. eğer ikiside kırıldıysa diğer atışta 49 ve 25. katları deneriz.100. kattaki kırılıp 50. kattaki kırılmadıysa ikinci atışı 99 ve 75. kattan deneriz. ve bu yöntemle sonuca en az denemeyle ulaşırız. bir anlamda HEDEF DARALTMA yöntemide denebilir buna.

freeozgur dedi ki...

üç küremiz olsaydı bir önceki yorumumun mantığıyla birinci küreyi 100. ikinci küreyi 66. üçüncü küreyi 33. kattan atarız. kürelerin kırılıp kırılmamasına göre hedefi yine üçe bölerek yani kırılan ve kırılmayan katlar aras ımesafeyi üçe bölerek devam ederiz

bthn35-1907@hotmail.com = batuhan dedi ki...

eger soruyu yanlış anlamadıysam 2 küreyi atıgımızda 1 deneme oluyor galıba ona göre bakarsak,2 kureyle 4 demede bulurum.
1. deneme
1. topu 100 kattan atarım mesala kırıldı ,2. topu 50 den atarım bu sefer kırılmadı mesala boylece 1. denemeyı yaptıktan sonra 50. katla 100. kat arasını arıyacagım.
2. deneme
75. kattan atarın 1. topu saglam mı cıktı dıyelım ozaman kırılıcak yer 75le 100 arasında olur. 2. topumu 87. kattan atarım dıyelım kı gene saglam cıktı ozaman kırılıcak yer 87 ile 100 arasında
3.denmeme
1. topu 93 ten atarım(100le 87 nın ortası) bu sefer kırıldı dıyelım, ozaman kırılma yerı 87 ile 93 arası 2. topu 90. katta denerım gene kırıldıgını varsayalım ozaman topun kırılıcagı 87 ile 90 arasında
4. deneme
1. topu 89 da denerım burdada kırıldıysa 2. topu 88 de denerım we kacıncı katta kırılıcagını bulabılırım...

yukardakı hesaplamalarda hep yarısını alamaya calıştım ama bazen mesala 75 ile 100 arasını bulurken 2 sayının toplamı tek oldugunu için tam ortasını bulamadım yanı ya 87 yada 88 arasında karar vermek zorunda kaldım eger 88 i seçmiş olsaydım deneme sayısı 5 olabılırdı yanı burda seçıme baglı bana göre tam sonuç yok...

OZGUR dedi ki...

cevap 18 deneme sercan arkadaşım mantığın doğru ancak bir sorunun kesin bir cevabı olması gerekli
1. topu 10. kattan atarsın kırılır ise 2. topla 1,2,3....9'a kadar 9 hakkın kalır 1. deneme ile toplam 10 eder kırılmaz ise 20. kattan atarız kırılr ise 2. topla 11,12,13....19'a kadar 9 hakkın kalır toplam 2+9 11 deneme yapmış oluruz kırılmaz ise 30. kattan atarız kırılırsa 3+9 12 deneme kırılamaz ise 40. kattan atarız kırılırsa 9+4 toplam 13 denem bu şekilde 50.katada 14 60.katta 15 70. katta 16 80. katta 17 90. katta kırılır ise 18 deneme yapmak zorundayız kırılmaz ise 95. kattan atarız kırılır ise yaptığımız 10 deneme + 91,92,93,94 katlarda 4 denem toplam 14 kırılmaz ise 98. kattan atarız kırılırsa 11 deneme + 96,97. katlarda 2 deneme toplam 13 deneme kırılmaz ise 99. kattan atarız kırılırsa 12 dene kırılmaz ise 100. kattan atarız kırılır toplam 13 deneme yapmış oluruz ancak 90. katta kırılma ihtimali olduğundan en az 18 deneme yapmak kesin sonuç verir

kensei dedi ki...
Bu yorum yazar tarafından silindi.
barış dağdaş dedi ki...

minumum 99 adet gerekiyor. 100 denemeye gerek yok �nk� 99 da olmassa 100 kesin olsu�undan 99 yeter.

peace_34@hotmail.com dedi ki...

elinizde 1 tanede olsa 99 tanede olsa her katta ayr� ayr� deneme yapmak zorundas�n�z bu da demektirki en az 99 adet deneme gerekiyor. 99'a kadar olmam�sa ki bu olas�l�k var 100. deneme zaten kesin k�r�lma yeri olaca�ndan gerek yok. yani minumum 99 dur.

ilhan hoca dedi ki...

1. deneme 50. kat
kırılırsa 2. kat 25. kat
kırılmassa 75. kat.

25 te kırılırsa bulunamaz.kırılmassa 37. kattan denenir.
bu işleme devam edelirse 6. denemede en az bulunabilir. yani her seferinde önceki denemelerin tam ortasındaki kat alınmaya çalısılır,matiloz_4@hotmail.com

kensei dedi ki...

2 top için cevap 19 dur.

max deneme sayısı = (100/n)+(n-1) dir

(burada "n" topları ne kadar kat arayla atacağımızdır.)

diyelim ki 5 kat arayla atıyoruz o zaman max deneme sayısı = (100/5)+(5-1) = 24 dür.

bu ifadenin "n" ye göre türevini "0" yapan değer bize minumum deneme sayısını verecek olan atış şeklini verecektir.

ifadenin türevi alınırsa;

((-100)/(n^2))+1 = 0 olur
bu ifadeyi "0" yapan değer n=10 da gerçekleşir.

Sonuç olarak 1.küreyi 10 kat da bir atmamız gerekir.

Deneyelim diyelim ki küre 99.kata kadar dayanabiliyor. (99. katta kırılması en kötü olasılıktır)
O zaman;

1.deneme --> 10. kat (kırılmadı)
2.deneme --> 20. kat (kırılmadı)
3.deneme --> 30. kat (kırılmadı)
4.deneme --> 40. kat (kırılmadı)
5.deneme --> 50. kat (kırılmadı)
6.deneme --> 60. kat (kırılmadı)
7.deneme --> 70. kat (kırılmadı)
8.deneme --> 80. kat (kırılmadı)
9.deneme --> 90. kat (kırılmadı)
10.deneme --> 100. kat KIRILDI
11.deneme --> 91. kat (kırılmadı)
12.deneme --> 92. kat (kırılmadı)
13.deneme --> 93. kat (kırılmadı)
14.deneme --> 94. kat (kırılmadı)
15.deneme --> 95. kat (kırılmadı)
16.deneme --> 96. kat (kırılmadı)
17.deneme --> 97. kat (kırılmadı)
18.deneme --> 98. kat (kırılmadı)
19.deneme --> 99. kat KIRILDI



3 top için cevap 12 dir.

2 top için yaptığımız denklemi burada da kuracak olursak;

Max deneme sayısı (y) :
y =(100/n)+((n-1)/x)+(x-1)
burada "n" 1.küreyi kaçar kat ara ile attacağımız
x ise 2. küreyi kaçar kat ara ile atacağımız olsun.

diyelim ki 1.yi 25 2.yi 4 kat arayla atıyoruz o zaman max deneme sayısı =
(100/25)+(25-1)/4+(4-1)=13 dür

bu ifadenin "x" ve "n" göre türevini alıp "0" eşitlersek, bulacağımız "n" ve "x" değerleri bize minumum deneme sayısını verecek değerlerdir.

ifadenin türevi alınırsa;

((-100)/(n^2))+((x-n+1)/(x^2))+1=0
olur
Fakat bu değeri sıfır yapan değer tam sayı olmadığı için buna en yakın değerleri aldım

n=34 x=5
n=26 x=5
n=21 x=5
n=17 x=4
n=14 x=4
gibi değerlerin hepsi 12 denemeyi vermektedir

n=26 x=5 ile deneyelim diyelim ki küre 76.kata kadar dayanabiliyor. (76. katta kırılması en kötü olasılıktır)
O zaman;

1.deneme --> 26. kat (kırılmadı)
2.deneme --> 52. kat (kırılmadı)
3.deneme --> 78. kat KIRILDI
şimdi 52'den 5'er 5'er deneyelim
4.deneme --> 57. kat (kırılmadı)
5.deneme --> 62. kat (kırılmadı)
6.deneme --> 65. kat (kırılmadı)
7.deneme --> 72. kat (kırılmadı)
8.deneme --> 77. kat KIRILDI
Şimdi 72'den 1'er 1'er deneyelim
9.deneme --> 73. kat (kırılmadı)
10.deneme -->74. kat (kırılmadı)
11.deneme -->75. kat (kırılmadı)
12.deneme -->76. kat KIRILDI

M586 dedi ki...
Bu yorum yazar tarafından silindi.
Adsız dedi ki...

BAKIN ÖNCE MANTIKLI DÜŞÜNECEK OLURSAK EĞER ARDAŞIK OLARAK ATARIZ BU TOPLARI YANİ 1,3,5,7 DİE TEK RAKAMLI ARDAŞIK GÖSTERDİĞİM GİBİ EĞER TOP 1 DE KIRILMIYOR AMA 3 DE KIRILMIYOSA KAT 3 AMA MESELA 3 KATTDA KIRILMIYOR AMA CATLIYOSA 4 KATTA KIRILIR DEMEKTİR BU KÜRELER

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Adsız dedi ki...

bence loki nin cevabı doğru ve gayet mantıklı

Adsız dedi ki...

küreler eşit olduğundan küre sayısı birde olsa ikide olsa üçte olsa deneme sayısı aynıdır yani yine 100 dür

fasafiso dedi ki...

dikkatsevrhortladi@hotmail.com

fasafiso dedi ki...

Bir kere en fazla 51 deneme olacağı kesin. Şöyleki 50.kattan enin birini atarız eğer birşey olmuyorsa denemeye 50.kattan yukarıdan başlarız. Eğer kırılıyorsa denemeye 50.kattan aşağıdan başlarız ve 10'un katları olarak devam ederiz. böylelikle kürenin kırıldığı aralığı en az denemede bulabiliriz.

Adsız dedi ki...

1. küreyi 1. kattan atarız eğer küre kırılmazsa 10. kattan atarız 10. katta kırılırsa diğer küreyi 2,3,4....9 kata kadar atarız kırılırsa zaten kırılır bizde sonucu buluruz ama eğer 1. küre 10. kattan atıldığında kırılmıyorsa bu işlemi 10-20,20-30.....90-100 kadar yaparız minimum ihtimal 10 defa maximum ihtimal 20 olur.(eğer cevap doğruysa çok sevicem çünkü daha 12 yaşındayım)

Adsız dedi ki...

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Adsız dedi ki...

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Adsız dedi ki...

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Greetings

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